HISTORY OF THE SCIENCE.

The properties of bodies may be resolved into two classes; one comprehending those which belong to all bodies whatsoever, and another, such as belong only to particular bodies. Amongst the properties of the first class we may consider extension, magnitude, figure, divisibility, impenetrability, inertia, weight, mobility, &c. Some of those of the second are solidity, liquidity, transparency, and the like. Of these properties, extension, magnitude, figure, and divisibility, are the subject of Geometry. The discussion of other properties of body belongs to Physics, called also Natural Philosophy.

The objects of geometry having been continually presented to the human mind, and being of such a nature as could at all times be perfectly comprehended, it may be supposed that the science would in some form or other exist from the very beginning of society. Indeed there is a natural geometry which all possess, and which must have been employed in the divisions of property and the erection of dwellings in every age. The name of the science indeed indicates its origin, for it is derived from γεωμετρία, the science of land-measuring.

According to the testimony of Herodotus, the science was first cultivated in Egypt. He had been informed at Memphis and Thebes, that Sesostris the king had divided the lands amongst his subjects, giving to each an allotment, for which an annual tribute was to be paid. The overflowing of the Nile, however, disturbed the landmarks, and rendered it necessary to re-adjust them by measurement; and hence the origin of geometry, which in the course of time passed from Egypt into Greece. There are two things to be here noticed; the assertion of a verification depending on geometry, and the particular opinion of Herodotus as to the origin of the science. If, as some chronologers have supposed, Sesostris be the same as the King Sesac, or Shishak, who made war on Rehoboam, the son of Solomon, it would follow from Herodotus, that geometry had its origin not more than about a thousand years before the Christian era. It might, however, be earlier; for the operation of measuring the land indicated a science which had made some progress. Aristotle has attributed the invention of geometry to the Egyptian priests, who, living secluded from the world, had leisure for study. But however this might be, all ancient writers are agreed in giving the Egyptians the credit of having been the earliest cultivators of geometry.

The philosopher Thales, who lived 640 years before Christ, brought the sciences, and particularly mathematics, into Greece from Egypt, whither he had gone in quest of knowledge at an advanced period of life. Diogenes Laertius relates that he there measured the height of the obelisks by means of their shadows; and Plutarch says, that this display of skill in geometry astonished the king Amasis. This shows that the Egyptians had not then advanced far in geometrical knowledge. Proclus also has recorded that Thales, by geometrical principles, determined the distance of vessels remote from the shore. On his return to Greece he founded the Ionian school, so called from Ionia, his native country. His celebrity for learning excited the attention of his countrymen, and drew to him disciples.

There appears to have been some slight traces of geometry in Greece at a still earlier period. We allude to the geometrical construction of a triangle by Euphorbus of Phrygia, and the discovery of some properties of figures. These, however, were probably just the natural geometry common to all ages. The rule and compasses, most important instruments, have been referred to the ages of fable, the honour of the invention of the latter having been ascribed to the nephew of Dadalus, and that of the square and level to Theodorus of Samos, one of the architects of the temple of Ephesus. But there are no certain indications of geometry as a science before Thales. He laid its foundation in Greece, and infused into his countrymen a taste for the science. He is said to have applied a circle to the measurement of angles, and to have discovered various properties of triangles, by comparing them one with another. In particular, he discovered the important proposition that all angles in a semicircle are right angles, a discovery which greatly delighted him, and for which he expressed his gratitude to the Muses by a sacrifice. Proclus has recorded that he made many discoveries in the science, which, however, have gone to the common stock of geometrical knowledge that had accumulated before Euclid collected it into a system, and the authors of which are unknown. It is probable that the greater part of the disciples of Thales were geometers, but few of their names have descended to us. Ameristus and Anaximander are the only ones now known. Of the former we only know that he was a skilful geometer. The latter wrote a kind of elementary treatise or introduction to geometry, the earliest work of the kind that is known to have existed.

Thales was succeeded in his school by Anaximander, who is said to have invented the sphere, the gnomon, and geographical charts. These required a considerable knowledge of geometrical science. It has been said that he attempted the grand practical geometrical problem of measuring the magnitude of the earth. This probably has originated in his being the inventor of representations of its surface. The honour of being the first to measure the earth seems to belong to another. Anaximander was followed by Anaximenes, born at Miletus 540 years B.C. He is said to have invented sun-dials, and thus must have known both astronomy and geometry; indeed a desire to understand the former would prove a strong stimulus to the study of the latter. He had as his disciple and successor Anaxagoras, who being persecuted for his opinions, employed himself in prison in trying to square the circle. This is the first notice on record of an attempt to resolve what appears now to be an insoluble problem, on which no modern geometer will spend his time. It might, however, have a very different aspect to the ancient mathematicians.

Pythagoras, born about 580 years B.C., was one of the greatest men of antiquity. According to some he was a Tuscan, whilst others say he was a Tyrian; but his name is better known than his origin. At the age of eighteen he became the disciple of Thales, and imbibed deeply his opinion of the importance of temperance and economy of time as necessary to success in the study of philosophy. In the infancy of science knowledge was only to be acquired by travel, the men of different countries standing in the place of our modern books as the depositories of learning. Pythagoras visited Phenicia, Chaldea, and India. In the annals of this last country the remembrance of the philosophical traveller is still preserved. He afterwards went into Egypt, where he is said to have remained twenty-two years, holding intercourse with the priests. The long duration of his abode is not probable, as the knowledge he seems to have acquired did not correspond to so protracted a residence. Whilst History, in Egypt, he is said to have consulted the columns of Sothis, on which that celebrated person had inscribed the principles of geometry. He returned to Samos, his adopted country, and there opened his school; but not finding auditors, he removed to Italy, and took up his abode in Cortona, a town in the territory of Tarentum, where he found many disciples, and acquired a great reputation. He is said to have discovered the important theorem in geometry, that in a right-angled triangle the squares on the sides containing the right angle are together equal to the square on the side opposite to it; and on this account he is said to have sacrificed an hundred oxen in gratitude to the Muses. The sacrifice is probably a fable. The shedding of blood was contrary to his moral principles, and so many oxen were not likely to be at the disposal of a philosopher. In his school, geometry made great progress, and was augmented by several new theories, such as that of the incommensurability of certain lines, in particular that of the side of a square to its diagonal. The theory of the regular solids had its origin also in the Pythagorean school, a doctrine which requires an extensive knowledge of geometry. This subject, now neglected amidst the riches of modern mathematics, because of its small utility, was in the beginning of great importance, on account of other discoveries which must have been incidentally made in its prosecution. Diogenes Laertius has attributed to Pythagoras the discovery that of all figures having the same boundary, the circle among plane figures, and the sphere among solids, are the most capacious. If this was so (a thing doubtful), he is the first on record who considered isoperimetrical problems.

The Pythagorean school sent forth many philosophers and mathematicians: amongst the latter, Philolaus and Archytas held a distinguished rank. Of Philolaus we know but little. He held to its full extent the Pythagorean doctrine of the motion of the earth round the sun as a centre, and was the first to unveil it. He composed a work on Mechanics, and thus has a claim to be associated with Eudoxus and Archytas as inventors of that part of mathematics. He had a tragic end, having been massacred by the people of the small republic of which he had been a legislator. We know more of Archytas. He was the author of a solution of the problem of two mean proportions. He was also one of the first who made use of the geometrical analysis, which, according to Proclus, he had learned from Plato, and by its assistance he made many discoveries in geometry. We pass over his mechanical inventions, one of which, the artificial pigeon, which winged its way through the air, certainly savours of the marvellous. Archytas is said to have been blamed by Plato for applying geometry to mechanics; but if this was so, it might be for the manner of the application. His solution of the problem of finding two mean proportions gives support to this conjecture; for although ingenious, it has the defect of requiring a motion which cannot be executed.

Democritus of Abdera cultivated geometry extensively. It has been conjectured that he was one of the first who treated of the contact of circles and spheres, and of irrational lines and solids.

Hippocrates, who lived about the year 380 B.C., was originally a merchant; but having no turn for commerce, he did not prosper. To renovate his affairs, he went to Athens, and there for the first time became acquainted with geometry, which seemed to suit his particular turn of mind. In the prosecution of this study he discovered that the curvilinear space, called from its form a lune, comprehended by half the circumference of one circle and the fourth of the circumference of another (their concavities being turned the same way), was equal to a rectilineal space, viz. the area of a right-angled triangle, whose hypotenuse was the diameter of the circle, and its sides each equal to the radius of the quadrantial arc. This could not be considered as a true quadrature of a curvilinear space; it was merely a geometrical juggle, by which a common space being taken from two equal curvilinear figures, one of the remainders was by a sort of chance a rectilineal figure. The first true quadrature of a curvilinear space is due to Archimedes. Hippocrates attempted the quadrature of the circle; but, if his mode of reasoning has been truly handed down to us, he committed a great oversight. This is the first recorded paralogism in geometry. He was more successful in treating of the duplication of the cube, which he showed to depend on the finding of two mean proportionals between two given lines. He was also the first who composed elements of geometry, but these have been lost. He appears to have retained the mercantile spirit, insomuch that he accepted money for teaching geometry. On this account he was expelled the Pythagorean school; a measure somewhat hard, considering his reduced circumstances.

Two geometers, Bryson and Antiphon, appear to have lived about the time of Hippocrates, and a little before Aristotle. They are only known by some animadversions of the latter on their attempts to square the circle. Before this time geometers knew that the area of a circle was equal to a triangle whose base is equal to its circumference, and altitude equal to its radius. This truth could not escape the observation of the early geometers. Bryson erroneously believed that, by a geometrical construction, he could find the circumference, in which case the quadrature would have been easy. Antiphon, it appears, proceeded by increasing continually the number of sides of an inscribed regular polygon, and considered the circle to be equal to the ultimate result; a process perfectly correct, and the same in effect as that by which Archimedes found the area of the parabola.

From this brief view of the progress of geometry during the first two centuries after its introduction into Greece, we pass to the school of Plato, in which geometry completely changed its character, and advanced with increased vigour and more rapid strides. Hitherto its subjects had been of the most elementary nature; now, however, the views of geometers became more enlarged, and a new era in the science commenced. Although the disciple and successor of Socrates, who had no taste either for mathematical or physical science, Plato held them in the highest estimation; and, imitating the earlier sages of Greece, he undertook voyages and journeys to improve his mathematical knowledge. He visited Egypt to converse with the priests, and Italy to consult with the celebrated Pythagoreans Philolaus, Timaeus of Locri, and Archytas. With the last of these he contracted a particular friendship. He went also to Cyrene to hear the discourses of the mathematician Theodorus; and on his return to Greece, when he had founded his school, he made the mathematics, particularly geometry, the basis of his instructions; and his disciples, encouraged by his example and his exhortations, applied with ardour to the study of that science. He never allowed a day to pass without making his disciples acquainted with some new truth. He placed over his school this inscription, *Let no one ignorant of geometry enter here.* He held that the Divinity continually geometrizes; meaning thereby, no doubt, that the laws by which the universe are governed are in accordance with the doctrines of mathematics. It does not appear that Plato composed any work expressly on the mathematics; but his invention of the geometrical analysis may be considered as one of the greatest improvements the science has received. The theory of Conic Sections originated in the Platonic school; and a third discovery was that of Geometrical Loci, a theory not only beautiful and interesting on account of the abstract truths which it unfolds, but also by reason of its great importance in the resolution of geometrical problems. The celebrated pro- Geometry.

History. blem concerning the duplication of the cube was agitated about his time; but its origin was earlier; for Hippocrates, as has been already stated, had reduced it to the finding of two continued mean proportionals. Plato himself gave a solution of the problem; and it was also resolved by Archytas, Eudoxus, Eratosthenes, and Menachmus. The solutions of eleven of the ancient geometers have been preserved in the Commentary of Eutocius on Archimedes' Treatise concerning the Sphere and Cylinder. It is probable that the trisection of an angle, a problem of the same degree of difficulty as the duplication of the cube, excited the efforts of the Platonic school. We have indeed no positive testimony in favour of this conjecture; but the progress of the human mind does not allow us to doubt that the problem was of great antiquity. After the solution of the very simple problem, the bisection of an angle, it was natural to think of dividing it into three equal angles; and those who first thought of the latter problem would no doubt be astonished to find, that two problems so like in appearance should yet differ so much in the difficulty of their solution. The trisection of an angle, and the duplication of a cube, are in fact problems of such a nature that they cannot be resolved by straight lines and circles alone, the only lines admitted by the ancients into their elements of geometry. The reason of this impossibility would not be apparent to the ancient geometers, although it is made evident by the modern doctrines of algebra.

The names of the individuals to whom we owe the particular discoveries which have been specified have not come down to us; they must therefore be regarded as belonging to the Platonic school generally. Proclus has recorded the names of a number of the geometers of that school. Some of advanced years attended it as friends of its celebrated head, or from respect to his doctrines; others, chiefly young persons, as disciples. Of the first class were Laodamus, Archytas, and Theetetus; Laodamus was one of the earliest to whom Plato communicated his method of analysis before he made it public; and he is said to have profited much by the possession of this instrument of discovery. Archytas was a Pythagorean, having extensive knowledge in geometry and mechanics. He frequently visited Plato at Athens; but in one of his voyages he perished by shipwreck. Theetetus was a rich citizen of Athens, who had studied along with Plato under Socrates. He appears to have cultivated the theory of the regular solids, called now the Platonic bodies.

The progress of geometry now required that its elements should be new modelled. This was done by Leon, a scholar of Neocles, or Neoclides, a philosopher who had studied under Plato. To Leon has been ascribed that part of the solution of a problem called its determination, which treats of the cases in which the problem is possible, and of those in which it cannot be resolved. Eudoxus of Cnidus was one of the friends and contemporaries of Plato; he generalized many theorems, and thereby greatly advanced geometry. It is believed that he cultivated the conic sections; and even the invention of these has been ascribed to him. He resolved the problem of the duplication of the cube; and it is to be regretted that Eutocius, who speaks with contempt of his solution, has not recorded it with others in his Commentary on Archimedes. Diogenes Laertius has attributed to him the invention of curve lines in general; from which we may infer, that other curves besides the conic sections were known in the school of Plato. Archimedes, in his Treatise on the Sphere and Cylinder, says that Eudoxus found the measure of the pyramid and cone, and that he was especially occupied with the contemplation of solids. Some have given him the credit of being the writer of the fifth book of Euclid, which treats of proportion. Amongst the numerous mathematicians of the Platonic school, there was one, Cratistus, who is particularly mentioned by Proclus. He was a remarkable person; for his knowledge of his science was in a manner innate. There was no problem of that time, however difficult, which he could not resolve by his natural geometry. He was the Pascal of antiquity. There were two brothers, Menachmus and Dinostratus, who also greatly distinguished themselves. The former, a particular disciple of Plato, extended the theory of the conic sections. Eratosthenes seems to attribute to him the merit of having invented them; at any rate, he was the first who solved the problem of doubling a cube by means of these curves. Dinostratus followed in the path of his brother, and advanced geometry by his discoveries. He is particularly known as the reputed inventor of a curve called the quadratrix, by which he seems to have intended to divide an angle or an arc of a circle in any given ratio. The curve takes its name from a property by which, could it be constructed geometrically, the quadrature of the circle would be obtained. According to Pappus, Dinostratus discovered this property of the curve; and probably this is the reason why it has been called the quadratrix of Dinostratus, just as we say the spiral of Archimedes. Indeed there is some ground for believing that the curve was invented by Hippas, a philosopher and skilful geometer, contemporary with Socrates.

It was chiefly in geometry that the followers of Plato excelled. In this they imitated their leader, who directed his attention to abstract speculations, rather than to the study of nature. The conic sections, one of their most refined and profound geometrical theories, remained merely an intellectual speculation from its first invention till the days of Kepler and Newton, when it acquired a high importance on account of the discovery that the orbits of the planets are conic sections.

The school of Plato, after the death of its founder, was divided into two others, which, although opposed to each other on various points, yet agreed in holding the mathematics in esteem, and in regarding them as an indispensable preparation for the study of philosophy. It is recorded, that a person ignorant of geometry and arithmetic having presented himself for admission into the school of Xenocrates, the successor of Plato after Speusippus, he was repelled by its chief, with an intimation that he was not properly prepared for the study of philosophy: Anas philosophorum non habes, said the philosopher.

The mathematics, thus encouraged and protected, continued to be cultivated and improved. The celebrated geometer Euclid was of this school; and it may be conjectured, from the age in which he lived, that he had acquired his knowledge under the tuition of the first successors of Plato. So also, we may presume, was Aristaeus, another celebrated geometer of antiquity, though little known in our time, because his works have perished. We learn, however, from Pappus, that he was one of the ancients who contributed the most to the progress of the sublime geometry. He was the author of two excellent works; one a treatise on The Conic Sections, in five books, which contained a great part of what Apollonius has given in the first four books of his work; the other, also in five books, treated of Solid Loci. Pappus, in prescribing a course of study to his son, places this last work next in order after the Conics of Apollonius. Hence we may infer that the propositions which it contained were of a higher order of difficulty, and required a previous knowledge of the conic sections. Euclid entertained great esteem for Aristaeus. We may hence conclude that he was either his disciple or his intimate friend.

The pure mathematics were less cultivated in the school of Aristotle than they had been in that of Plato; they were not, however, neglected; for, unlike the modern peripa- Although the mathematics received but little addition in the Aristotelian school, still they had some cultivators. Of these, Theophrastus was the principal. He wrote some treatises relating to them, particularly a complete history of these sciences down to his time, a work which has unfortunately been lost. It consisted of four books on the history of geometry, six on that of astronomy, and one on that of arithmetic. We have great reason to regret the want of the sure light which this work, had it come down to our times, would have thrown on the origin and progress of these sciences, instead of which, we have only a few glimmerings, that serve to make the darkness sensible, but not to dissipate the obscurity. Another disciple of Aristotle, Eudemus, composed a history of the mathematics. This consisted of six books on the history of geometry, and as many on that of astronomy. We owe to these all we now know concerning the origin of those sciences; for it is from this source that Proclus, Theon, and Diogenes Laertius have drawn the few notices which they have transmitted to us.

We come next to a new epoch in the history of the ancient mathematics. This was the institution of the school of Alexandria. However memorable this event may have been in the history of general literature, it is more especially remarkable in that of the mathematics; for in this celebrated school all the mathematical sciences were cultivated with a degree of care not inferior to that which had been bestowed on pure geometry in the school of Plato, and with corresponding success.

The period which immediately followed the death of Alexander proved one of trouble and confusion. The vast empire founded by that conqueror was dismembered by his principal captains; and Egypt was the portion which fell to one of them, Ptolemy Lagus. As soon as he had established tranquillity in his dominions, Lagus turned his attention to the sciences; and by his encouragement of their cultivators, he drew to him many of the learned of Greece; and his capital soon rivalled Athens as the resort of men of knowledge and talent. These constituted the Alexandrian school. The perfection of this celebrated establishment is mainly due to his son and successor Ptolemy Philadelphus, who bestowed on the learned men he had induced to settle in his capital, protection and liberal encouragement. Amongst those who resorted to this institution, none has been more celebrated than the Greek geometer Euclid. His country is unknown; and of the events of his life no notice has reached us. It appears that he had resided in Greece, and studied geometry under the disciples of Plato. Thence he went to Alexandria, induced probably by the liberality of the first Ptolemy. Pappus has described him as gentle and modest, and entertaining a particular regard for those who could contribute to the progress of the mathematics. His character formed a contrast to that of Apollonius, another geometer, who was vain, and delighted in depreciating his contemporaries. We may suppose that Euclid was not much of a courtier, from his reply to Ptolemy, who inquired of him whether there was not an easier way to a knowledge of geometry than the study of his Elements. "There is no royal road to geometry," was his answer. We have elsewhere (see Euclid) given a full account of this distinguished geometer and his writings.

The great excellence of the Elements of Euclid must increase our regret for the loss of a treatise which he composed on Porisms. Of this work, all that has come down to modern times is an abstract, inserted by Pappus Alexandrinus in his Mathematical Collections. This, however, has suffered so much from time, that all we can immediately learn from it is, that the ancients put a high value on porisms, and regarded them as an important part of their analysis. The efforts of modern geometers had been exerted in endeavouring to discover the nature of porisms, but without much success. At length Dr Robert Simson, with a zeal and perseverance not to be surpassed, succeeded in divining their nature, and even in restoring a great number of Euclid's propositions. These form a part of the posthumous works of this distinguished geometer.

Leaving the Alexandrian school, our attention is drawn to Sicily, which gave birth to Archimedes, a geometer whose genius has been the admiration of all who have come after him. His life was written by Heracles; but this precious piece of biography, so well calculated to interest our curiosity, has unfortunately perished. Archimedes was born 287 years before our era. He was the relation and friend of Hiero, king of Syracuse. His skill as a mechanician is universally known; but it is only as a geometer that we shall speak of him here. Endowed with a mind of the highest order, he exerted it in the extension of mathematical science. The measure of curvilineal magnitude was a subject new in his time, concerning which but little was known. Archimedes attached himself to this branch by predilection, and opened new views, which have engaged the attention and exercised the genius of such men in modern times as Kepler, Cavalieri, and Dr Wallis. The last of these has called him a man of prodigious sagacity, who first laid the foundations of almost all that had given celebrity to the seventeenth century of our era.

The writings of Archimedes were numerous. We have two books by him on the sphere and cylinder, which terminate in the fine geometrical discoveries, that the solidity and surface of the sphere are respectively two thirds of the solidity and surface of the circumscribing cylinder, with which he was so much delighted, that he desired that these figures should be inscribed on his tomb. His book on the measure of the circle is a kind of supplement to his treatise on the sphere, which supposed a knowledge of that measure. The determination of the exact ratio of the diameter to the circumference was, as at present, a problem not to be resolved; but he found limits to that ratio, and an approximation to it sufficiently accurate for the ordinary wants of the arts, which was all that he attempted. He might, however, have carried his approximation farther, as was afterwards done by Apollonius, and by Philo, another geometer less known.

Archimedes having succeeded in measuring the sphere and cylinder, bodies long the study of geometers, he opened a new field of inquiry in his Treatise on Conoids and Spheroids. All his determinations are now familiar to geometers; his processes of reasoning are highly ingenious, but withal somewhat intricate. Indeed we question whether many modern mathematicians have had patience to go through his investigations, considering that the same conclusions can now be reached by much shorter methods.

The properties of the spiral, invented by his friend Conon, and the quadrature of the parabola, are among the additions made to the ancient geometry by Archimedes. This last was a grand step in the progress of geometry; because it was the first curvilinear space legitimately squared, and the only one, until the application of the modern methods of analysis. In latter times the notion of infinity has been introduced into geometry, and has contributed greatly to its extension. The ancients, however, carefully avoided this term, which might have injured the stability of the science. Their theory of exhaustions supplied its place, and was a safe, although a laborious and indirect way of establishing truth. The writings of Archi- medes give the finest examples of its application. We pass over the mechanical and optical inventions of this great man, and only observe that they indicate a state of considerable perfection in the geometrical science of that period.

The writings of Archimedes are the most important of the few classic remains of ancient geometry which we now possess. The subjects of which they treat, being less elementary, are now less studied than those in the Elements of Euclid. Every geometer, however, should have read them at least once in his life. The earliest edition of his writings was published at Basle in 1644. It was in Greek, with a Latin translation by Venatorius, the two versions being printed as separate volumes, and accompanied by the Commentary of Eutocius. There have been many editions of his works. Of these we shall only notice two; that of Torelli, in Greek and Latin, published at Oxford in 1792, and a French translation of his writings by Peyrard, the learned editor and translator of Euclid's works.

Returning to the school of Alexandria, we have to mention Eratosthenes, a geometer as well as an astronomer. Indeed he was one of those uncommon men, whose genius embraces all subjects; for he was also an orator, a poet, an antiquary, and a philosopher. His extensive knowledge induced the third Ptolemy to make him his librarian. He chiefly cultivated geometry and astronomy, and has been classed with the three great geometers of antiquity, Aristotle, Euclid, and Apollonius, who had improved geometrical analysis. Pappus mentions a work by Eratosthenes on this subject, in two books. He has told us its title, but has not described its exact object, which can therefore only be conjectured. He gave a solution of the problem of the duplication of the cube, which Eutocius has preserved in his Commentaries on Archimedes; and he applied his geometrical and astronomical science to the solution of the grand problem, the measurement of the magnitude of the earth, which, however, had been attempted before his time.

Apollonius of Perga appeared as a geometer in the Alexandrian school, about the time when Archimedes had finished his career. It might be a question which of these two great men was endowed with the higher genius; for such was the estimation in which Apollonius was held at the period in which he lived, that he obtained the appellation of the great geometer. He was the most profound and fertile writer that has ever treated of geometry. His works which have survived the ravages of time, or have been restored by modern geometers, with the addition of the Data and Porisms of Euclid, this last also a restoration, constitute nearly all which we now know of that beautiful subject, the ancient geometrical analysis. Whoever would understand it, and acquire skill and facility in its application, would do well to study with care the various works of Apollonius which have been restored, if not exactly according to the letter, at least in the true spirit of the originals. A full account of the writings of this distinguished geometer has been given as an article of biography (see Apollonius); it is therefore unnecessary to enumerate them here.

The ancient mathematicians addressed their writings to others, their particular friends. It is in this way that the names of some of their contemporaries have descended to us, although their writings have been lost. Apollonius dedicated the first three books of his Conics to Eu德mus, a geometer, and in such terms as show that he addressed himself to one well acquainted with the subject. He says that he had been encouraged to study the subject by Naucrates; and he requests Endemus to communicate what he had written to Philonides, also a geometer. Endemus having died, he addressed the fourth book to Atalus, and speaks of Thrasydeus, with whom Conon of Samos, a little anterior to his time, had held a correspondence on the conics; also of Nicoteles the Cyrenean, who it seems had a controversy with Conon. The regret which Archimedes expresses for the loss of Conon gives us reason to believe that he was a profound geometer. We know also that he was an astronomer. Archimedes dedicated several of his works to Dositheus, another geometer, whose name has thus become known to us.

It was about this period that the geometer Nicomedes lived. He was the inventor of the Conchoid, a curve which served to resolve two remarkable problems, the duplication of the cube, and the trisection of an angle.

The period which comprehended the ages of Euclid, Archimedes, and Apollonius, was that in which the science of geometry shone forth with the greatest splendour. It had been in a state of gradual improvement from the time of Thales. Now, however, it seems to have arrived at a degree of perfection beyond which its cultivators did not carry it. We may suppose that the science of astronomy took its place, and presented a new and more inviting field of discovery. Hipparchus, the father of astronomy, was also a geometer; for we find him applying its principles, combined with arithmetic, to his science.

There were geometers who were also mechanicians. These, as well as the astronomers, probably cultivated geometry alone as connected with their own science. Gemimus of Rhodes lived between the time of Hipparchus and the beginning of our era. He was the author of a work on geometry, and another on astronomy. The former, which is lost, was probably a historical commentary, a sort of development of geometrical discoveries. Proclus seems to have drawn from it all that he has said on the history and metaphysics of geometry. Its loss is matter of regret. Ctesibius, and Hero his disciple, were celebrated as mechanicians. The former lived in the middle of the second century before Christ. Philo of Byzantium was also a celebrated mechanician. Posidonium was at once geometer, astronomer, mechanician, and geographer. He deserved well of geometry, for having repelled the attack of Zeno the Epicurean, who had attempted to invalidate its principles and certainty. Dionysodorus was a skilful geometer, as appears by his solution of a difficult problem of Archimedes. Theodosius was the author of a work on the geometry of the sphere, which has descended to our times. It appears to have been studied as a classic in the English universities. Dr Barrow gave an edition of it, along with the writings of Archimedes, and the four books of the Conics of Apollonius, in 1615. The best edition is that of Hunt, printed at Oxford 1707.

The study of the mathematical sciences, which languished in the first century of the Christian era, revived a little in the beginning of the second. Menelaus, an astronomer, wrote on trigonometry and spherical geometry. This work has reached our times by a translation from the Arabic. Dr Halley, holding in high esteem every vestige of the ancient geometry, had prepared a new edition of this geometer, corrected from a Hebrew manuscript. It did not appear, however, until 1758, by the care of Costard, the author of a history of astronomy. Menelaus, as we learn from Pappus, cultivated the theory of curve lines; a subject which may be regarded as belonging to the higher geometry among the ancients.

Ptolemy, if not an inventive geometer, must at least have been perfectly acquainted with its theories, for he had the merit of applying them to astronomy. His Almagest has handed down from antiquity a beautiful geometrical theorem, the foundation of the Greek trigonometry. It is this; the rectangle contained in the diagonals of a quadrilateral inscribed in a circle, is equal to both the rectangles contained by the opposite sides. Several mathematicians of no great note lived in the third and fourth centuries of the Christian era. Such were Serenus, who wrote two books on cones and cylinders; Hypsicles of Alexandria, the supposed author of two books on regular solids, sometimes ascribed to Euclid; and Perseus Criticus, who wrote on certain lines called Spiriques, formed from a solid generated by an arc of a circle revolving about a fixed axis. Philo of Thyaneus, and Demetrius of Alexandria, both wrote on curve lines.

We are now come to a period in which the sciences had begun to decline; when, instead of original treatises, writers were content to compose commentaries and annotations on the works of their more illustrious predecessors. It was in this way that Pappus and Theon of Alexandria, who lived towards the end of the fourth century, rendered service to the mathematical sciences. The first, however, must not be classed with the ordinary scholastics; for in his Mathematical Collections he has shown that he was an expert geometer. His object in composing this work was to collect into one body many scattered discoveries, and to illustrate and supply deficiencies in the writings of the celebrated mathematicians who had preceded him. He has done this in regard to Apollonius, Archimedes, Euclid, and Theodosius, by a number of lemmas and important propositions which they had assumed as known. His writings have made us acquainted with the various attempts made by the ancients to double the cube, and to divide an angle into three or more equal parts. We are indebted to him for almost all we know of these and other important matters in the ancient geometry; and it is through his writings that the names of many geometers have come down to our days, which, but for him, would have been lost in the obscurity of time. The preface to his seventh book is of inestimable value, inasmuch as it has preserved from oblivion many of the analytic works of the ancients. His brief analysis of these works, for that is all we have, has given continuity to the history of geometry; and his indications have served as vestiges of the steps by which the ancients proceeded in their discoveries. Guided by these brief notices, the ingenuity of the moderns has succeeded in restoring the writings of Apollonius and Euclid, which had disappeared for ages. It is not improbable, that in this renovated form, the works of these geometers surpass in excellence the originals. The restorers have had the advantage of an improved state of the science, and the works of later geometers, to assist them in their researches.

It is probable that at this time there is not an entire copy of the eight books of Pappus in existence; certainly there is none in Britain. In a catalogue of the manuscripts of England and Ireland, published at Oxford in 1698, there is mention made of two copies in the Savilian library. These have been more than two centuries in Oxford. One, marked No. 3, is in folio, and is well written on paper; it contains the Greek text of the third, fourth, fifth, sixth, seventh, and eighth books, and in addition some works of Theodosius and Autolycus. There is no memorial to indicate when it was written, or by whom. The other copy, with a mark No. 9, is not so well written; it has been copied by various hands, on separate sheets of paper. This copy, besides the books preserved in the former, contains also a part of the second book. Dr Wallis has spoken of three manuscripts of Pappus at Oxford, one in the Bodleian and two in the Savilian library. No traces of a third copy can now be found in the Bodleian library. The late Dr Trail, in his life of Dr Robert Simson, has described a manuscript now in the Advocates' Library, Edinburgh. It contains five books of Pappus, viz. the third, fourth, fifth, sixth, and eighth; but unfortunately the seventh, the most valuable portion of the work, is wanting. This was purchased at Paris by Dr James Moore in 1748.

The British Museum possesses a manuscript copy of Pappus, which had been the property of the Saliente family at Verona. None of the manuscripts of which we have any details are of a very early date. There are some points of resemblance in all that have been examined, which seem to indicate a common origin. It is well known that early manuscripts are few in number; sometimes only one is known. This may have been the case in regard to the text of Pappus, and will account for the agreement of all the manuscripts in particular places. We owe the edition which we possess of this ancient work to the labour of Commandine, who translated and enriched it with notes, but died when he was about to give it to the world. It afterwards appeared in 1588, under the patronage and by the pecuniary aid of Francisco Maria, duke of Urbano. That part of the second book which has been preserved was published in the third volume of the works of Dr Wallis: it treats of the ancient arithmetic.

We had formerly occasion to mention Hypatia, the daughter of Theon, as a commentator on Diophantus, in our account of the progress of Algebra. She has a claim to our notice also in the history of geometry, by her having composed a commentary on the writings of Apollonius.

The philosopher Proclus, the chief of the Platonic school at Athens, made it in some measure the seat of the mathematical sciences about the middle of the fifth century. In imitation of the chief of his sect, he held the sciences in great estimation. He was not, however, distinguished for original discoveries; but his commentary on the first book of Euclid, setting aside the fault of its prolixity, is valuable, because of the information it contains on the history and metaphysics of geometry at a remote period.

We pass over several geometers of this period, of whom hardly any thing but their names is known. We mention, however, Diocles as the inventor of a curve called the cissoid, contrived for the solution of the celebrated problem, to find two mean proportional.

Geometrical science had by this time been long on the decline. The Alexandrian school, however, still existed; and a repetition of such a period as that in which Apollonius and Euclid flourished might have been hoped for, had it not been for the troubles which agitated the East. The taking of Alexandria by the Saracens in the year 640 gave a deathblow to the sciences, not only in that celebrated city, but throughout the Greek empire. The library was delivered over to destruction by the command of the ignorant fanatic Omar; and this, the finest monument of human genius, the accumulated store of knowledge that had been collecting for ages, was employed during six months in warming the four thousand baths of Alexandria. Such was the end of the Alexandrian school, which had for nearly ten centuries contributed to the advancement of the human mind in knowledge.

About the eighth century we find Hero, called the younger, an engineer and mathematician. He wrote a treatise on Geodesia (so practical geometry was called), which had no great merit; it is, however, remarkable for containing, without demonstration, a rule for finding the area of a triangle when its three sides are known. There is no notice of this important theorem in the writings of the ancients which have come down to us. It was, however, probably found by some geometer earlier than Hero. It appears to have been afterwards lost, and rediscovered by Tartalea.

We have an unfavourable opinion of the modern Arabians, but there was a period when they had a very different character. When the sciences were neglected by the Greeks, and existed only in manuscripts buried in the dust of libraries, the Arabians brought them forth from obscurity, and gave them an honourable asylum. They were the depositaries of knowledge during the ages of barbarism and ignorance which overspread Europe; and we owe to our intercourse with them the first glimpses of light which penetrated the obscurity of the eleventh, twelfth, and thirteenth centuries. The Arabians diligently cultivated the science of astronomy; and as a corresponding knowledge of other branches of science was necessary to success, they became geometers, opticians, and even algebraists. The greater part of the Greek geometers, particularly those necessary to the study of astronomy, as Euclid, Theodosius, Hypsicles, Menelaus, were translated into the Arabic language in the time of Almamon, or soon after him. They even began to cultivate the sublime geometry of the ancients, for the four first books of the Conics of Apollonius were translated by order of that prince. They added to their literature the writings of Archimedes, and at least three of the last books of the Conics of Apollonius; and all these works are to be found in libraries rich in oriental manuscripts. It is from the last-mentioned work, corrected and augmented by the notes of Nassreddin, a Persian geometer, that Dr Halley has enriched our geometry with the fifth, sixth, and seventh books of the ancient Greek author; the remaining books, viz. the eighth, appear however to have been entirely lost. The Arabian writers quote the writings of several Greek geometers of which we know nothing, such as a treatise on parallel lines, another on triangles, and a third on the division of the circle. We must allow to the Arabians the merit of having given to our trigonometry its present form. They also simplified the practice of its operations, by employing the sines of the arcs instead of the chords of the double arcs employed by the Greeks. This was even one of their earliest inventions, for we find it in the writings of their astronomer Albatenius.

The Arabian historians have recorded the names of several of their cultivators of geometry; but these may be passed over, ignorant as we are of their contributions to the science. The optician Alhazen however deserves notice, because of the knowledge of geometry shown in his writings, and also on account of a geometrical problem applicable to optics which bears his name. The solution he gives was probably derived from the Greek geometers. It must be confessed the Arabians do not appear to have been an inventive people; almost always commentators or compilers, they seldom rose higher than the labours which such functions impose.

The Persians have also had their geometers. The most celebrated was Nassir-Eddin Al-Tussi. The science is indebted to him for some good works, particularly a learned commentary on Euclid, which was written in Arabic, and printed in 1590 in Italy. Another geometrical work is a revision of the Conics of Apollonius, with a commentary. Dr Halley found this useful in remodelling the fifth, sixth, and seventh books of that precious treatise. The Persian geometers have long known the principal Greek writers; and they even profess to have some of their writings which we have not known. It appears that their geometers had a peculiar taste in treating the doctrines of abstract science, for they have given every proposition of the elements a name expressive of one of its uses, or of some fanciful analogy it may have to things foreign to geometry. The forty-seventh of the first book of Euclid, for example, they call the figure of the bride, and the forty-eighth the bride's sister; and, with some reason, they call the mathematics the difficult science. Mathematical knowledge among the Turks has been nearly in the same state as among the Persians. The libraries of Constantinople contain Arabic and some Turkish translations of the Greek writers. Geometry is taught in their schools, but it does not appear that they go beyond the Elements of Euclid. There are hardly any traces of geometry amongst the Hebrews. We know that when Solomon's temple was built, Hiram king of Tyre furnished architects and navigators. From this we may infer that the Jews had no geometry at that period. It was not before their second dispersion that they took an interest in the sciences. However, when mixed with other nations, some of their learned men cultivated geometry, but never proceeded beyond the elements. About the time of Almamon, in imitation of the Arabians, they translated the writings of some of the Greek mathematicians into their language, in particular those of Euclid and Archimedes. They however made no addition themselves to geometrical science.

The researches of Europeans concerning the mathematical and astronomical science of India have discovered among the Hindus some works on geometry of great antiquity. We have already given some account of these interesting writings in our sketch of the history of Algebra; and we have there described the Lilavati, a work on arithmetic and geometry by Bhascara Acharya, a Hindu mathematician, who lived about the year 1150 of the Christian era. In another work, reputed more ancient, viz. the Surya Siddhanta, we find a rational system of trigonometry combined with fable and absurdity. There has been great difference of opinion as to the time when this last treatise, the most ancient on the Indian astronomy, was written. The French astronomer Bailly believed it to have been composed more than 3000 years before the Christian era; but more sober inquirers have judged it to be about 750 years old. The trigonometry it contains is probably much older. The Hindu books on geometry are remarkable for not containing demonstrations. The writers knew the celebrated proposition concerning the squares on the sides of a right-angled triangle, the discovery of which is ascribed to Pythagoras; and they also knew the rule for finding the area of a triangle from its sides, which was not known to the early Greeks. It appears from the Institutes of Akbar, that the Indians supposed the diameter of a circle to be to its circumference as 1250 to 3927; which is the proportion of 1 to 3:1416, and is more accurate than the determination of Archimedes, or indeed than any known before the sixteenth century.

The Chinese have cultivated astronomy from a remote antiquity, yet it does not appear that they have made any progress in geometry. They knew the proposition of Pythagoras, it would seem, earlier than it was known to the Greeks, yet it remained sterile in their hands. They had no spherical trigonometry before the thirteenth century, and then it was not of native growth; it was probably derived from the Arabians or Persians.

The Romans, unlike the Greeks, gave little attention to the sciences. The mathematics in particular were disregarded at Rome; and geometry, scarcely known there, did not extend beyond the art of measuring land and of fixing boundaries. The mathematics were, however, not unknown to Cicero. The respect with which he speaks of them, and the veneration he manifested for the memory of Archimedes in searching for his tomb, on which he knew were inscribed a sphere and cylinder, indicate that he had given some attention to the subject of geometry.

The Roman geographer Strabo must be numbered amongst the geometers of his day; and Cicero mentions another, Didymus, who was blind; but he was a Greek by birth. Vitruvius has given much information, in his architecture, on matters connected with geometry.

We pass over those ages in which the sciences had nearly disappeared in Europe, and hail with satisfaction their revival in the eighth century. Mathematical science was at first dimly recognised, like the dawn of a new day after a gloomy night, in the writings of Boethius, Bede, his disciple Alcuin, and Gerbert, who travelled into Arabia, the only country where a knowledge of the mathematical sciences could be acquired. In this he was imitated by the English monk Adelard, or Athelard, in the twelfth century, who went into Spain and Egypt; also by other History. Englishmen, such as Daniel Morlay, Robert of Reading, William Shell or William de Corchis, Clement Langton, &c. Adelard on his return translated Euclid, and he appears to have been the first who made this writer known in the West. His work was never printed. Plato of Tivoli translated the Spherics of Theodosius into Latin about 1120; but this translation was not printed before 1518. It is worthy of remark, that almost all the restorers of the sciences were ecclesiastics, who in the seclusion of the cloisters studied the works of the ancients which had been preserved in the monasteries. This at least is one benefit the world has received from the leisure of the monks of that period, which all of them did not spend in idleness.

The sciences found many cultivators, and much encouragement from sovereigns, in the thirteenth century; Jordanus Nemorarius, who lived about 1230, was well versed in geometry and arithmetic. John of Halifax, known also by the name of Sacro-Bosco, his contemporary, was a mathematician. Campanus of Novara, the celebrated translator and commentator of Euclid's Elements, was of this age. Almost all the early editions of this work were made from his version and his manuscript commentary. We have a treatise by him on the Quadrature of the Circle. But he deviates from geometrical accuracy, by mistaking the approximate ratio found by Archimedes for the true and exact ratio. The paradoxism which he gives for a demonstration may, however, be excused in the geometers of his time. They were few, and their writings must not be too nicely scanned. The name of Roger Bacon is so identified with the progress of science, that we must reckon him among the geometers of his day. In support of this claim we record his treatise on Perspective, a geometrical theory.

Leonardus of Pisa, who in the fifteenth century brought algebra into Italy, was also known as a cultivator of geometry. He composed a treatise on the subject, which Commandine thought worthy of publication, and had prepared it for the press; but his death prevented its appearance.

In the beginning of this century, Cardinal Cusa acquired reputation in geometry, by a pretended quadrature of the circle, and other writings. These, however, were only a tissue of paradoxisms. It appears from the introduction to a tract of this writer, that Pope Nicolas V. who occupied the pontifical throne from 1447 to 1455, was a geometer, and had translated the works of Archimedes from Greek into Latin. The true restorers of mathematical science in the fifteenth century were Purbach and Regiomontanus. Purbach banished the use of sexagenary calculation from trigonometry, which he enriched with several new propositions. He supposed the radius divided into 600,000 parts, instead of the divisions used by the ancients; and in place of the chords of the double arcs, expressed in sexagenary parts of the radius, he calculated the sines in 600,000ths of the radius. He was the inventor of the geometrical square, an instrument used in practical geometry; and he appears to have been the first who applied the plumb-line to mark the divisions of an instrument. Regiomontanus, whose true name was John Muller (called also John of Mont-Royal), was born in 1436. At the age of fourteen he was captivated with the mathematical sciences, and put himself under the guidance of Purbach. The preceptor and his pupil made a journey into Italy to study Greek, in order that they might draw their knowledge from the pure sources of antiquity. Purbach died there, but his disciple followed out his purpose. He afterwards translated the Spherics of Menelaus and Theodosius into Latin; he corrected from the Greek text the ancient version of Archimedes made by Gerard of Cremona; he translated the Conics of Apollonius, and the Cylindrics of Serenus, besides other works on mixed mathematics; he commented on the books of Archimedes, which Eutocius had not touched. He defended Euclid against the imputations of Campanus, and he confuted the pretended quadrature of Cardinal Cusa. We pass over for the present his improvements in trigonometry.

Lucas Pacioli, or Lucas de Burgo, the author of the earliest printed book on algebra, contributed by the same work to the establishment of geometry. It was first printed in 1494, and was therefore one of the early printed books on the science, but not the earliest, for the first edition of Euclid was in 1482.

The decline of the Greek empire, and the taking of Constantinople, which happened in 1453, scattered the learned men of that country, many of whom took shelter in Italy, and brought with them their language, and the precious remains of ancient learning. The art of printing soon spread abroad these treasures, and in the sixteenth century geometry was very generally cultivated. Amongst the writers on the science in this century we may, in particular, reckon Zamberti, John Baptiste Memmains, Commandine, Guido Ubaldo, Maurolycus, and Tartalea. These and others laboured usefully on the ancients, by translations and commentaries. The Jesuit Clavius deserves particular mention. Besides his labour on Euclid, we owe to him a good work for the time on practical geometry in eight books.

We pass a multitude of names, and come to the French geometer Vieta, who was deeply versed in the ancient as well as in the modern geometry, and the restorer of the tangencies, one of the lost works of Apollonius. Adrian Metius, a Dutch geometer, deserves to be mentioned for his convenient and accurate approximate ratio (113 to 355) of the diameter to the circumference of a circle, given in a work on practical geometry. At this period the geometer Nonius, a professor at Coimbra, laboured with zeal in spreading a knowledge of the science in Portugal.

The seventeenth century opened propitiously for the progress of geometry. Lucas Valerius, an Italian, took up a subject which Archimedes had neglected, namely, the centre of gravity of solids; and Marinus Ghetaldus, another Italian, made an attempt to restore the lost book of Apollonius, entitled De Inclinationibus; but he left the labour incomplete. He also composed a work De Resolutione et Compositione Mathematica, which proves him to have imbibed the true spirit of the ancient geometry. A Scottish geometer, Alexander Anderson, lived at this time. He had a decided taste for the ancient analysis, of which he gave an essay in his Supplementum Apollonii Redivivi, in which he supplied what Ghetaldus had left imperfect. The Netherlands could boast of some geometers who did credit to their time. Ludolph Van Ceulen has been celebrated for his approximation to the ratio of the diameter of a circle to its circumference (ALGEBRA, art. 272), a prodigious effort of labour, but requiring little genius. Willebrord Snellius was a more distinguished geometer. He undertook the restoration of the lost book De Sectione Determinata of Apollonius, and accomplished his task creditably. He improved greatly the way in which Ceulen had approximated to the ratio of the diameter to the circumference, and verified the labour of that model of patience in calculation. Albert Girard has acquired great reputation in geometry by his supposed divination of the Porisms of Euclid; for in his edition of Stevinus he positively avers that he had restored the three books, which, however, never appeared; and indeed it may be doubted whether he understood the true nature of porisms. John Neper might be reckoned amongst geometers, but he has still higher claims to the notice of posterity for his invention of logarithms.

The science was sedulously cultivated in England at this period by Robert Record, John Dee, Leonard and Thomas Digges, and Henry Billingsley. Record was the author of the Pathway of Knowledge, the earliest book in the English language on geometry. It was first printed in Geometry.

History. 1551, and is dedicated to Edward VI. Edward Wright, the inventor of what is called Mercator's Chart, was an English geometer whose knowledge of the science greatly exceeded what was common in his time.

Germany had many geometers at this period, but none of a high order. John Werner of Nuremberg cultivated the ancient geometry, and excelled in a knowledge of its analysis. Every writer on geometry at this period gave a system of trigonometry nearly the same as we now have it in its most elementary form. To George Joachim Rheticus mathematical science is particularly indebted for a trigonometrical table, more extensive than any that had been given before him. It was printed in 1594, with this title, Opus Palatium de Triangulis; but the author was then dead. The work was published by Valentine Otho.

Pitiscus, another German geometer, extended and reprinted this work in 1613, with the title of Thesaurus Mathematicus, sive canon sinuum ad radium 1,000,000,000, &c. This is indeed a treasure, and one of the most remarkable monuments of human patience extant.

Want of space compels us to pass in silence many names not less celebrated than some we have given, but less remarkable, because of their abundance in respect of the age to which they gave splendour. Indeed we may reckon that every astronomer and algebraist was then a geometer.

Kepler opened an unexplored way into the field of geometry, by boldly introducing the notion of infinity into that science. We may suppose that Archimedes must have entertained in effect the same idea in his speculations on the sphere and cylinder, but was prevented from following it to its full extent by the strict laws according to which the early mathematicians had constructed their demonstrations. The views of Kepler were eagerly seized by Roberval in France, and by Cavalieri in Italy. They were the fertile germs which, under the culture of such men, led, in the course of the seventeenth century, to a rich harvest of discoveries.

We have now arrived at a period at which the discoveries of the ancients formed but a small part of the whole body of geometrical science, augmented as it had been by the moderns, and which was continually increasing. The new views which were opened soon produced the methods of indivisibles of Cavalieri and Roberval, the theory of tangents of Fermat, the arithmetic of infinites of Wallis, the geometry of curve lines of Descartes, and, finally, the method of fluxions of Newton, and the differential calculus of Leibnitz. The origin and progress of these have been fully explained in our introductions to Algebra and Fluxions, and require not to be here repeated. The ancient geometry, however, still had great value. Huygens and Newton delivered their sublime discoveries in its language, and represented the relations of time, space, velocity, force, the objects of their physical inquiries, by geometrical diagrams. The rigorous mode of ancient demonstrations served likewise as a model on which to form the processes of reasoning by which the new truths in geometry and physics were to be established.

Not to pass altogether in silence the names of geometers whose discoveries have made the seventeenth century ever memorable, we observe, that to Guidin, a Jesuit, we owe the discovery of a property of the centre of gravity applicable to the measurement of solids formed by revolution; and to Descartes, Fermat, Roberval, and Barrow, a method of determining the tangents of curves. Galileo first suggested the cycloid, the nature of which was afterwards fully disclosed by the investigations of Roberval, Fermat, Pascal, Huygens, Wren, and Wallis. James Gregory first suggested the logarithmic curve; and John Bernoulli and Leibnitz showed the true nature of the catenary, which Galileo could not discover; afterwards, David Gregory demonstrated its properties by the new geometry.

The ancient geometry came in for a share of that general attention which was bestowed on the modern theories. Newton, as has been already stated, held it in high esteem; and David Gregory and Dr Halley employed their genius and learning in restoring to their pristine excellence the precious remains of Euclid and Apollonius.

Previously to the time of Newton, pure geometry and algebra were the only subjects which mathematicians had for the exercise of their genius; but his sublime discoveries presented a new and an immense field for investigation. The fact that the orbits of the planets are ellipses, naturally connected the doctrines of astronomy with the conic sections; and the two were wrought up into the beautiful physico-mathematical theory of central forces delivered in the Principia. When this came to be fully understood, it served as a model of reasoning in all speculations on physics, and the geometrical method of Newton was adopted and imitated by his followers. Thus, we have the Elements of Physical and Geometrical Astronomy of David Gregory, and the Phoronomia of Herman, and many other like treatises, composed in the geometrical style. It was, however, soon discovered, and first by foreign mathematicians, that the geometrical method, which, when carried a certain length, is perspicuous and easy, is yet unsuitable to the more difficult speculations of mechanical philosophy, by reason of its clumsiness. It was therefore abandoned by Leibnitz, the Bernoullis, Euler, and their followers, and the greatly more manageable modern methods of the fluxional or differential calculus adopted in its stead. The mathematicians in Britain, who had imbibed deeply and acquired a relish for the spirit of the geometrical method, however, still adhered wholly or in part to that method. Macaulain, the expositor of Newton's calculus of fluxions, defended its principles, and delivered many of his own fine speculations in the pure and mixed mathematics, in the prolix style of the ancient geometry. But he also gave a theory of the fluxional calculus in the more concise language of the modern analysis, which served as a practical example of the advantage which the latter has over the former as an instrument of invention. Dr Robert Simson, the editor of Euclid, greatly preferred the geometrical to the modern analysis; and his disciple, Dr Mathew Stewart, the author of General Theorems, Tracts relating to Physical Astronomy, and Propositiones Geometricae more Veteran Demonstratae, fully agreed with him in this sentiment. At a later period another Scottish professor, the late Mr Playfair, with a profound knowledge of both the ancient and modern analysis, and with no prejudice in favour of the former, yet gave an elegant example of its beauty, in his paper on the origin and investigation of porisms.

These distinguished Scottish mathematicians, by their writings, contributed greatly to keep up the taste for geometry in Britain during the last century. It is proper to acknowledge also, that much is due to Dr A. Robertson, the late professor of astronomy at Oxford, to Dr Horsley, bishop of Rochester, and others, for the estimation in which the geometrical analysis is now held in Britain.

The nature and objects of the ancient geometry have been fully understood for a century past, and the only question now is, how it ought to be studied. There have been a great variety of treatises within that time sent into the world. We do not, however, suppose that the science has gained by this diversity of guides. The settled public opinion seems to be in favour of some standard book; and Euclid's Elements have been, by almost universal consent, chosen as that standard in this country. There is another work of great excellence in the French language, Éléments de Géométrie, by the late Legendre, a mathematician of the highest celebrity. Were we to choose a book different from Euclid for purposes of instruction, History. It would be Legendre's Geometry, of which there is an English translation. It may be useful to read more than one book on a subject which we wish to understand thoroughly, and those who hold this opinion can find no better book than that of the French geometer. We have kept him constantly in view in composing this treatise.

In tracing the progress of a science, it is always interesting to know by what steps it has advanced in the career of improvement. These may be well ascertained by consulting the authors who have written on the science, from its infancy downwards. With this view we have given a catalogue of writers on algebra and fluxions, and we now do the same in regard to geometry. The early writers are not so much mentioned for their excellence, as from a wish to give a faithful portraiture of the state and progress of the science at different periods. The works on geometry in the last century are too numerous to be all specified. Many have fallen into obscurity. Almost all are mere transcripts from Euclid; indeed, generally speaking, their excellence is in proportion to the fidelity with which they have copied the ancient.

Ancient Writers on Geometry.

Euclid, Elements of Geometry, flourished b.c. 272 First edition in Latin, that of Ratdolt, 1482 First edition of Greek text, by Hervage, 1533 Edit. in Greek, Latin, and French, by Peyrard, 1814 (For other editions, see Euclid.) Porisms, restored by Dr Simson, Opera quaedam Reliqua, 1776 Apollonius, Conics, and various works, fl. b.c. 244 For an account of his writings, see Apollonius. Archimedes on the Sphere and Cylinder, &c., born b.c. 287 First edition of Greek text by Venatorius, 1544 Oxford edition, Gr. and Lat., by Torelli, 1792 Edition in French, by Peyrard, 1808 Eratosthenes, Geometria cum Annot. 1672, fl. b.c. 194 Theon, Commentator on Euclid, fl. b.c. 117 Theodosius, Sphericorum libri tres, printed at Oxford, 1707 Menelaus, Spherics, edited by Maurolycus, fl. a.c. 100 Serenus, De Sectione Cylindri et Coni, fl. a.c. 200 His treatise is in Oxford edition of Apollonius. Pappus, Mathematicae Collectiones, 1588, 1660, fl. a.c. 380 Proclus, Commentator on Euclid, fl. a.c. 450 Taylor gave a translation in English, 1788. Veterum Mathematicorum Athenai, Bitonis, Apollodori, Heronis, Philonis, et aliorum, Opera, Gr. et Lat., 1693

Writers in Modern Times.

Lucas de Burgo, Summa de Arithmetica, &c., 1494 Brauardinus, Geometria Speculativa, 1495 Bovile, L'Art et Science de Géométrie, 1514 N. de Cusa, De Geometricis Transmutationibus, 1514 Albert Durer, Institutionum Geometricarum lib. iv., 1532 Orontius Fineus, Liber de Geometrica Practica, 1544 Record, The Pathway to Knowledge, 1551 Buteo, Opuscula quaedam Geometrica, 1554 Ramus, Arithmeticæ lib. ii. Geometricæ lib. xxvi., 1580 Finckius, Geometricæ Rotunde lib. iii., 1583 Stevinus, Problematum Geometricorum lib. v., 1583 Barotius, Geom. Problema Tredecem Modis Dem., 1586 Vieta, Opera Mathematica, 1589 Digges (Tho.), Pantometria, a geometrical treatise, 1591 Baptista Porta, Elementorum Curvilineorum lib. ii., 1601 Ramus, Geometria, 1604 Clavius, Geometria Practica, 1606 Ghetaldus, Apollonius Redivivus, 1607 Anderson, Supplementum Apollonii Redivivi, 1612 Kepler, Nova Stereometria, &c., 1618 Van Ceulen, De Circulo et Adscriptis, 1619 Snellius, Cyclometricus, 1621 Metius, Arith. libri ii. et Geom. libri vi., 1626 Guildin, De Centro Gravitatis, &c., 1635 Cavalieri, Geometria Indivisibilibus promota, 1635 — Exercitationes Geometricæ sex, 1647 Descartes, Geometria, 1637 Toricielli, De Sphaera et Solidis Sphaeralibus, &c., 1644 Gregory St Vincent, Opus Geom. quadrat. Circuli, 1647 Rudd, Practical Geometry, 1650 Wallis, Arithmetica Infinitorum, 1656 — De Cycloide et Cissoidi, 1659 Pascal (A. Dettonville), Lettres (on the Cycloid), 1658 Ricci, Exerc. Geom. de Maximis et Minimis, 1666 J. Gregory, Vera Circ. et Hyp. Quadratura, 1667 — Geometricæ Pars Universalis, 1668 Slusius, Mesolabium, &c., 1668 Huygens, De Linearum Curvarum evolutione et dimensione (in Horol. Oscil.), 1673 — Opera (collected by S'Gravesande), 1751 Barrow, Lectiones Geometricæ, 1674 Viviani, Enodatio Problematum Gallicorum, 1677 De Omerique, Analysis Geometrica, 1698 Sharpe, Geometry Improved, 1718 Clairaut, Éléments de Géométrie, 1746 Math. Stewart, General Theorems, 1746 — Propositiones Geometricæ, 1763 Montucla, Histoire des Recherches sur la quadrature du Cercle (in Histoire de Mathématiques), 1754 T. Simpson, Elements of Geometry, 1752 Emerson, Elements of Geometry, 1763 Hutton, A Treatise on Mensuration, 1770 Lawson on the Geometrical Analysis of the Ancients, 1774 R. Simson, Opera quaedam Reliqua, 1776 West, Elements of Mathematics, 1784 Playfair, Origin and Investigation of Porisms (Edin. Phil. Trans. vol. iii.), 1794 — Elements of Geometry, 1795 Wallace, Geometrical Porisms (in Edin. Phil. Tr. vol. iv.), 1796 La Croix, Éléments de Géométrie, 1795 Mascheroni, Géométrie du Compas, 1798 — Problèmes pour les Arpenteurs, 1803 Carnot's Géométrie de Position, 1803 Kampf, Éléments de Géométrie, 1805 Leslie, Elements of Geometry, Geometrical Analysis, &c., 1809 L'Huillier, Éléments d'Analyse Géométrique, &c., 1809 Diesterweg, Geometrische Aufgaben nach der Methode der Griechen, 1825 Vincent, Cours de Géométrie Élémentaire, 1827 Duncan, Elements of Plane and Solid Geometry, 1833 Legendre, Éléments de Géométrie (15th edition), 1852

General Notions of Geometrical Magnitude.

Our notions of geometrical magnitudes are obtained by the contemplation of a body, or solid. We readily understand that this is extended in three directions; that is, it has length, breadth, and thickness. The outside or boundary of a solid, that which separates the particular space it occupies from space in general, is called a surface or superficies. A surface, then, is not conceived as having any thickness; it has only length and breadth.

A surface has a boundary; something which encloses it, or separates any portion of it from the remainder. This boundary, which has no thickness, nor breadth, but length only, is a line.

Again, there is something which terminates a line, indicating where it begins and where it ends, or which may PART I.—OF LINES AND FIGURES UPON A PLANE.

SECT. I.—PRINCIPLES OF GEOMETRY.

Definitions.

I. Geometry is a science which treats of the properties and relations of quantities having extension, and which are called magnitudes. Extension is distinguished into length, breadth, and thickness.

II. A Point is that which has position, but not magnitude.

III. A Line is that which has only length. Hence the extremities of a line are points, and the intersections of one line with another are also points.

IV. A Straight or Right Line is the shortest way from one point to another.

V. Every line which is neither straight nor composed of straight lines is a Curve Line. Thus AB is a straight line, ACDB is a line made up of straight lines, and AEB is a curve line.

VI. A Superficies or Surface is that which has only length and breadth. Hence the extremities of a superficies are lines, and the intersections of one superficies with another are also lines.

VII. A Plane Superficies is that in which any two points being taken, the straight line between them lies wholly in that superficies.

VIII. Every superficies which is neither plane nor composed of plane superficies, is a Curve Superficies.

IX. A Solid is that which has length, breadth, and thickness. Hence the boundaries of a solid are superficies; and the boundary which is common to two solids which are contiguous, is a superficies.

X. A Plane Rectilineal Angle is the inclination of two straight lines to one another, which meet together, but are not in the same straight line. The point in which the lines meet one another is called the Vertex of the angle.

When there is only one angle at a point, it may be expressed by the letter placed at that point; thus the angle contained by the lines EF and EG may be called the angle E. If, however, there be several angles, as at B, then each is expressed by three letters, one of which is the letter that stands at the vertex of the angle, and the others are the letters that stand somewhere upon the lines containing the angle, the letter at the vertex being placed between the other two. Thus the angle contained by the lines BA and BD is called the angle ABD or DBA.

Angles, in common with other quantities, admit of addition, subtraction, multiplication, and division. Thus the sum of the angles ABD and ABC is the angle DBC; the difference of the angles DBC and ABD is the angle ABC.

XI. When a straight line standing on another straight line makes the adjacent angles equal to one another, each of which is called a Right Angle, and the straight line which stands on the other is called a Perpendicular to it. Thus, if DC meet AB, and make the angles ACD, DCB equal to one another, each of them is a right angle, and DC is a perpendicular to AB.

XII. An Obtuse Angle is that which is greater than a right angle, and an Acute Angle is that which is less than a right angle. Thus ABC being supposed a right angle, DBC is an obtuse angle, and EBC an acute angle.

XIII. Parallel Straight Lines are such as are in the same plane, and which being produced ever so far both ways, do not meet.

XIV. A Plane Figure is a plane terminated everywhere by lines.

If the lines be straight, the space which they enclose is called a Rectilineal figure, or a Polygon, and the lines themselves constitute the Perimeter of the polygon.

XV. When a polygon has three sides (which is the smallest number it can have), it is called a Triangle; when it has four, it is called a Quadrilateral; when it has five, a Pentagon; when six, a Hexagon, &c.

XVI. An Equilateral triangle is that which has three equal sides (fig. 7); an Isosceles triangle is that which has only two equal sides (fig. 8); and a Scalene triangle is that which has all its sides unequal (fig. 10). XVII. A Right-angled triangle is that which has a right angle; the side opposite to the right angle is called the Hypotenuse. Thus in the triangle ABC, having the angle at B a right angle, the side AC is the hypotenuse.

XVIII. An Obtuse-angled triangle is that which has an obtuse angle (fig. 10); and an acute-angled triangle is that which has three acute angles (fig. 11).

XIX. Of quadrilateral figures, a square is that which has all its sides equal, and all its angles right angles (fig. 12). A Rectangle is that which has all its angles right angles, but not all its sides equal (fig. 13). A Rhombus is that which has all its sides equal, but its angles are not right angles (fig. 14). A Parallelogram, or Rhomboid, is that which has its opposite sides parallel (fig. 15). A Trapezoid is that which has only two of its opposite sides parallel (fig. 16).

XX. A Diagonal is a straight line which joins the vertices of two angles, which are not adjacent to each other, such as AC in fig. 42.

XXI. An Equilateral Polygon is that which has all its sides equal; and an Equiangular Polygon is that which has all its angles equal. If a polygon be both equilateral and equiangular, it is called a Regular Polygon.

XXII. Two polygons are equilateral between themselves when the sides of the one are equal to the sides of the other, each to each, and in the same order; that is, when in going about each of the figures in the same direction, the first side of the one is equal to the first side of the other; the second side of the one is equal to the second side of the other; the third to the third, and so on. The same is to be understood of two polygons which are equiangular between themselves.

Explanation of Terms.

An Axiom is a proposition the truth of which is evident at first sight.

A Theorem is a truth which becomes evident by a process of reasoning called Demonstration.

A Problem is a question proposed, which requires a solution.

A Lemma is a subsidiary truth employed in the demonstration of a theorem, or the solution of a problem.

The common name, Proposition, is given indifferently to theorems, problems, and lemmas.

A Corollary is a consequence which follows from one or several propositions.

A Scholium is a remark upon one or more propositions that have gone before, tending to show their connection, their restriction, their extension, or the manner of their application.

A Hypothesis is a supposition made either in the enunciation of a proposition, or in the course of a demonstration.

Explanation of Signs.

That the demonstrations may be more concise, we shall make use of the following signs borrowed from algebra; and in employing them we shall take for granted that the reader is acquainted with at least the notation and first principles of that branch of mathematics.

To express that two quantities are equal, the sign = is put between them; thus \( A = B \) signifies that the quantity denoted by \( A \) is equal to the quantity denoted by \( B \).

To express that \( A \) is less than \( B \) they are written thus, \( A < B \).

To express that \( A \) is greater than \( B \) they are written thus, \( A > B \).

The sign + (read plus) written between the letters which denote two quantities, indicates that the quantities are to be added together; thus \( A + B \) means the sum of the quantities \( A \) and \( B \).

The sign — (read minus) written between two letters means the excess of the one quantity above the other; thus \( A - B \) means the excess of the quantity denoted by \( A \) above the quantity denoted by \( B \). The signs + and — will sometimes occur in the same expression; thus \( A + C - D \) means that \( D \) is to be subtracted from the sum of \( A \) and \( C \); also \( A - D + C \) means the same thing.

The sign \( \times \) put between two quantities means their product if they be considered as numbers; but if they be considered as lines, it signifies a rectangle having these lines for its length and breadth; thus \( A \times B \) means the product of two numbers \( A \) and \( B \); or else a rectangle having \( A \) and \( B \) for the sides about one of its right angles.

We shall likewise indicate the product of two quantities, in some cases, by writing the letters close together; thus \( mA \) will be used to express the product of \( m \) and \( A \), and Lines and so on with other expressions, agreeably to the common notation in algebra.

The expression $A^2$ means the square of the quantity $A$, and $A^3$ means the cube of $A$; also $PQ^2$ and $PQ^3$ mean the one the square, and the other the cube, of a line whose extremities are the points $P$ and $Q$.

On the other hand, the sign $\sqrt{}$ indicates a root to be extracted; thus $\sqrt{A \times B}$ means the square root of the product of $A$ and $B$.

**Axioms.**

1. Two quantities, each of which is equal to a third, are equal to one another. 2. The whole is greater than its part. 3. The whole is equal to the sum of all its parts. 4. Only one straight line can be drawn between two points. 5. Two magnitudes, whether they be lines, surfaces, or solids, are equal, when, being applied the one to the other, they coincide with one another entirely, that is, when they exactly fill the same space. 6. All right angles are equal to one another.

**Note.**—The references are to be understood thus: (7) refers to the seventh proposition of the section in which it occurs; (4, 2) means the fourth proposition of the second section; (2 Cor. 28, 4) means the second corollary to the twenty-eighth proposition of the fourth section.

**Theorem I.**—A straight line $CD$, which meets with another $AB$, makes with it two adjacent angles, which, taken together, are equal to two right angles.

At the point $C$ let $CE$ be perpendicular to $AB$. The angle $ACD$ is the sum of the angles $ACE$, $ECD$; therefore $ACD + BCD$ is the sum of the three angles $ACE$, $ECD$, $BCD$. The first of these is a right angle, and the two others are together equal to a right angle; therefore the sum of the two angles $ACD$, $BCD$, is equal to two right angles.

**Cor. 1.** If one of the angles is a right angle, the other is also a right angle.

**Cor. 2.** All the angles $ACE$, $ECD$, $DCF$, $FCB$, at the same point $C$, on the same side of the line $AB$, are, taken together, equal to two right angles. For their sum is equal to the two angles $ACD$, $DCB$.

**Theorem II.**—Two straight lines which coincide with each other in two points, also coincide in all their extent, and form but one and the same straight line.

Let the points which are common to the two lines be $A$ and $B$; in the first place it is evident that they must coincide entirely between $A$ and $B$; otherwise two straight lines could be drawn from $A$ to $B$, which is impossible (Axiom 4). Now let us suppose, if possible, that the lines when produced separate from each other at a point $C$, the one becoming $ACD$, and the other $ACE$. At the point $C$ let $CF$ be drawn, so as to make the angle $ACF$ a right angle; then $ACE$ being a straight line, the angle $FCE$ is a right angle (1 Cor. 1); and because $ACD$ is a straight line, the angle $FCD$ is also a right angle, therefore the angle $FCE$ is equal to $FCD$, a part to the whole, which is impossible; therefore the straight lines which have the common points $A$, $B$ cannot separate when produced, therefore they must form one and the same straight line.

**Theorem III.**—If two adjacent angles $ACD$, $DCB$ make together two right angles; the two exterior lines $AC$, $CB$, which form these angles, are in the same straight line.

For if $CB$ is not the line $AC$ produced, let $CE$ be that line produced, then, $ACE$ being a straight line, the angles $ACD$, $DCE$ are together equal to two right angles (1); but, by hypothesis, the angles $ACD$, $DCB$ are together equal to two right angles; therefore $ACD + DCB = ACD + DCE$. From these equals take away the common angle $ACD$, and the remaining angles $DCB$, $DCE$ are equal, that is, a part equal to the whole, which is impossible; therefore $CB$ is the line $AC$ produced.

**Theorem IV.**—If two straight lines $AB$, $DE$ cut each other; the vertical or opposite angles are equal.

For since $DE$ is a straight line, the sum of the angles $ACD$, $ACE$ is equal to two right angles (1); and since $AB$ is a straight line, the sum of the angles $ACE$, $BCE$ is equal to two right angles; therefore the sum $ACD + ACE$ is equal to the sum $ACE + BCE$. From each of these take away the same angle $ACE$, and there remains the angle $ACD$ equal to its opposite angle $BCE$.

In like manner it may be demonstrated, that the angle $ACE$ is equal to its opposite angle $BCD$.

**Cor. 1.** From this it appears, that if two straight lines cut one another, the angles they make at the point of their intersection are together equal to four right angles.

**Cor. 2.** And hence all the angles made by any number of lines meeting in one point are together equal to four right angles.

**Theorem V.**—Two triangles are equal when they have an angle, and the two sides containing it of the one, equal to an angle, and the two sides containing it of the other, each to each.

Let the triangles $ABC$, $DEF$ have the angle $A$ equal to the angle $D$, the side $AB$ equal to $DE$, and the side $AC$ equal to $DF$; the triangles shall be equal. For if the triangle $ABC$ be applied to the triangle $DEF$, so that the point $A$ may be on $D$, and the line $AB$ upon $DE$, then the point $B$ shall coincide with $E$, because $AB = DE$; and the line $AC$ shall coincide with $DF$, because the angle $BAC$ is equal to $EDF$; and the point $C$ shall coincide with $F$, because $AC = DF$; and since $B$ coincides with $E$, and $C$ with $F$, the line $BC$ shall coincide with $EF$, and the two triangles shall coincide exactly, the one with the other; therefore they are equal (Ax. 5).

Cor. Hence it follows that the bases, or third sides BC, EF of the triangles are equal; and the remaining angles B, C of the one are equal to the remaining angles E, F of the other, each to each, namely, those to which the equal sides are opposite.

**Theorem VI.**—Two triangles are equal, when they have a side, and the two adjacent angles of the one equal to a side, and the two adjacent angles of the other, each to each. (See fig. 22.)

Let the side BC be equal to the side EF, the angle B to the angle E, and the angle C to the angle F; the triangle ABC shall be equal to the triangle DEF. For if the triangle ABC be applied to the triangle DEF, so that the equal sides BC, EF may coincide, then, because the angle B is equal to E, the side BA shall coincide with ED, and therefore the point A shall be somewhere in ED; and because the angle C is equal to F, the side CA shall coincide with FD, and therefore the point A shall be somewhere in FD; now the point A being somewhere in the lines ED and FD, it can only be at D their intersection; therefore the two triangles ABC, DEF must entirely coincide, and be equal to one another.

Cor. Hence it appears that the remaining angles A, D of the triangles are equal; and the remaining sides AB, AC of the one are equal to the remaining sides DE, DF of the other, each to each, viz. those to which the equal angles are opposite.

**Theorem VII.**—Any two sides of a triangle are together greater than the third. (Fig. 22.)

For the side BC, for example, being the shortest way between the points B, C (Def. 4), it must be less than BA + AC.

**Theorem VIII.**—If from a point O, within a triangle ABC, there be drawn straight lines OB, OC to the extremities of BC, one of its sides; the sum of these lines shall be less than that of AB, AC, the two other sides.

Let BO be produced to meet CA in D; because the straight line OC is less than OD + DC, to each of these add BO, and BO + OC < BO + OD + DC; that is, BO + OC < BD + DC.

Again, since BD < BA + AD, to each of these add DC, and we have BD + DC < BA + AC; but it has been shown that BO + OC < BD + DC, much more then is BO + OC < BA + AC.

**Theorem IX.**—If two sides AB, AC of a triangle ABC be equal to two sides DE, DF of another triangle DEF, each to each, but the angle BAC contained by the former greater than the angle EDF contained by the latter; the third side BC of the first triangle shall be greater than the third side EF of the second.

Suppose AG drawn so that the angle CAG = D, take AG = DE and join CG; then the triangle GAC is equal to the triangle EDF (5), and therefore GC = EF. Now there may be three cases, Lines and according as the point G falls without the triangle BAC, fig. 24, or on the side BC, fig. 25, or within the same triangle, fig. 26.

**Case I.** Because GC < GI + IC, and AB < AI + IB (7), therefore GC + AB < GI + AI + IC + IB, that is, GC + AB < AG + BC; from each of these unequal quantities take away the equal quantities AB, AG, and there remains GC < BC, therefore EF < BC.

**Case II.** If the point G fall upon the side BC, then it is evident that GC, or its equal EF, is less than BC.

**Case III.** Lastly, if the point G fall within the triangle BAC, then AG + GC < AB + BC(8), therefore, taking away the equal quantities AG, AB, there remains GC < BC or EF < BC.

Cor. Hence, conversely, if EF be less than BC, the angle EDF is less than BAC; for the angle EDF cannot be equal to BAC, because then (5), EF would be equal to BC; neither can the angle EDF be greater than BAC, for then (by the Theor.) EF would be greater than BC.

**Theorem X.**—Two triangles are equal, when the three sides of the one are equal to the three sides of the other, each to each. (Fig. 22.)

Let the side AB = DE, AC = DF, and BC = EF; then shall the angle A = D, B = E, C = F.

For if the angle A were greater than D; since the sides AB, AC are equal to DE, DF, each to each, it would follow (9), that BC would be greater than EF; and if the angle A were less than the angle D, then BC would be less than EF; but BC is equal to EF, therefore the angle A can neither be greater nor less than the angle D, therefore it must be equal to it. In the same manner it may be proved that the angle B = E, and that the angle C = F.

**Scholium.** It may be remarked, as in Theorem V. and Theorem VI. that the equal angles are opposite to the equal sides.

**Theorem XI.**—In an isosceles triangle, the angles opposite to the equal sides are equal to one another.

Let the side AB = AC, then shall the angle C = B

Suppose a straight line drawn from A the vertex of the triangle, to D the middle of its base; the two triangles ABD, ACD have the three sides of the one equal to the three sides of the other, each to each, namely, AD common to both, AB = AC by hypothesis, and BD = DC by construction, therefore (preced. Theor.) the angle B is equal to the angle C.

Cor. Hence, every equilateral triangle is also equiangular. Scholium. From the equality of the triangles ABD and ACD, it follows that the angle BAD = DAC, and that the angle BDA = ADC; therefore these two last are right angles. Hence it appears that a straight line drawn from the vertex of an isosceles triangle to the middle of its base is perpendicular to that base, and divides the vertical angle into two equal parts.

In a triangle that is not isosceles, any one of its three sides may be taken indifferently for a base; and then its vertex is that of the opposite angle. In an isosceles triangle, the base is that side which is not equal to one of the others.

**Theorem XII.** If two angles of a triangle be equal; the opposite sides are equal, and the triangle is isosceles.

Let the angle ABC = ACB; the side AC shall be equal to the side AB. For if the sides are not equal, let AB be the greater of the two; take BD = AC, and join CD; the angle DBC is by hypothesis equal to ACB, and the two sides DB, BC are equal to the two sides AC, BC each to each; therefore the triangle DBC is equal to the triangle ACB (5); but a part cannot be equal to the whole; therefore the sides AB, AC cannot be unequal, that is, they are equal, and the triangle is isosceles.

**Theorem XIII.** Of the two sides of a triangle, that is the greater which is opposite to the greater angle; and conversely, of the two angles of a triangle, that is the greater which is opposite to the greater side.

First, let the angle C > B; then shall the side AB opposite to C be greater than the side AC opposite to B. Suppose CD drawn, so that the angle BCD = B; in the triangle BDC, BD is equal to DC (12); but AD + DC = AC, and AD + DC = AD + DB = AB; therefore AB > AC.

Next, let the side AB > AC; then shall the angle C opposite to AB, be greater than the angle B, opposite to AC. For if C were less than B, then, by what has been demonstrated, AB < AC, which is contrary to the hypothesis of the proposition, therefore C is not less than B; and if C were equal to B, then it would follow that AC = AB (12), which is also contrary to the hypothesis; therefore C is greater than B.

**Theorem XIV.** From a point A without a straight line DE, no more than one perpendicular can be drawn to that line.

For suppose it possible to draw two, AB and AC; produce one of them AB, so that BF = AB, and join CF. The triangle CBF is equal to the triangle ABC, for the angle CBF is a right angle, as well as CBA, and the side BF = BA; therefore the triangles are equal (5), and hence the angle BCF = BCA; but the angle BCA is by hypothesis a right angle; therefore the angle BCF is also a right angle; hence AC and CF lie in a straight line (3), and consequently two straight lines ACF, ABF may be drawn between two points A, F, which is impossible (Axiom 4); therefore it is equally impossible that two perpendiculars can be drawn from the same point to the same straight line.

**Theorem XV.** If from a point A, without a straight line DE, a perpendicular AB be drawn upon that line, and also different oblique lines AE, AC, AD, &c. to different points of the same line.

First, The perpendicular AB shall be shorter than any one of the oblique lines.

Secondly, The two oblique lines AC, AE, which meet the line DE on opposite sides of the perpendicular, and at equal distances BC, BE from it, are equal to one another.

Lastly, Of any two oblique lines AC, AD, or AE, AD, that which is more remote from the perpendicular is the greater.

Produce the perpendicular AB, so that BF = BA, and join FC, FD.

1. The triangle BCF is equal to the triangle BCA; for the right angle CBF = CBA, the side CB is common, and the side BF = BA, therefore the third side CF = AC (5); but AF < AC + CF (7), that is, 2AB < 2AC; therefore AB < AC; that is, the perpendicular is shorter than any one of the oblique lines.

2. If BE = BC, then, as AB is common to the two triangles ABE, ABC, and the right angle ABE = ABC, the triangles ABE, ABC shall be equal (5), and AE = AC.

3. In the triangle DFA, the sum of the lines AD, DF is greater than the sum of AC, CF (8), that is, 2AD > 2AC; therefore AD > AC, that is, the oblique line, which is more remote from the perpendicular, is greater than that which is nearer.

Cor. 1. The perpendicular measures the distance of any point from a straight line.

Cor. 2. From the same point three equal straight lines cannot be drawn to terminate in the same straight line; for if they could be drawn, then two of them would be on the same side of the perpendicular, and equal to each other, which is impossible.

**Theorem XVI.** If from C, the middle of a straight line AB, a perpendicular CD be drawn to that line; first, every point in the perpendicular is equally distant from the extremities of the line AB; secondly, every point without the perpendicular is at unequal distances from the same extremities A, B.

1. Let D be any point in CD; then, because the two oblique lines DA, DB are equally distant from the perpendicular, they are equal to one another (15); therefore every point in CD is equally distant from the extremities of AB.

2. Let E be a point out of the perpendicular; join EA, EB; one of these lines must cut the perpendicular in F; join BF, then AF = BF, and AE = BF + FE; but BF + FE > BE (7), therefore AE > BE, that is, E any point out of the perpendicular is at unequal distances from the extremities of AB.

**Theorem XVII.** Two right-angled triangles are equal, when the hypotenuse and a side of the one are equal to the hypotenuse and a side of the other, each to each. Let the hypotenuse AC = DF, and the side AB = DE; the triangle ABC shall be equal to DEF. The proposition will evidently be true (10), if the remaining sides BC, EF are equal. Now, if it be possible to suppose that they are unequal, let BC be the greater, take BG = EF, and join AG; then the triangles ABG, DEF, having the side AB = DE, BG = EF, and the angle B = E, will be equal to one another (5), and will have the remaining side AG = DF; but by hypothesis DF = AC; therefore AG = AC; but AG cannot be equal to AC (15), therefore it is impossible that BC can be unequal to EF, and therefore the triangles ABC, DEF, are equal to one another.

**Theorem XVIII.**—Two straight lines AC, ED, which are perpendicular to a third straight line AE, are parallel to each other.

For if they could meet at a point O, then two perpendiculars OA, OE, might be drawn from the same point O, to the straight line AE, which is impossible (14).

In the next theorem, it is necessary to assume another axiom, in addition to those already laid down in the beginning of this section.

**Axiom 7.**

If two points E, G in a straight line AB are situated at unequal distances EH, GP from another straight line CD in the same plane; these two lines, when indefinitely produced, on the side of the least distance GH, will meet each other.

**Theorem XIX.**—If two straight lines AB, CD be parallel; perpendiculars EF, GH to one of the lines, which are terminated by the other line, are equal, and are perpendicular to both the parallels.

For if EF and GH, which are perpendicular to CD, were unequal, the lines AB, CD would meet each other (by the above axiom), which is contrary to the supposition that they are parallel. And if EF, GH be not perpendicular to AB, let EK be perpendicular to EF, meeting GH in K; then, because EK and FH are perpendicular to EF, they are parallel (18); and therefore, by what has been just shown, the perpendiculars EF, KH must be equal; but by hypothesis EF = GH, therefore KH = GH, which is impossible; therefore EF is perpendicular to AB; and in the same way it may be shown that GH is perpendicular to AB.

Cor. Hence it appears, that through the same point E, no more than one parallel can be drawn to the same straight line CD.

**Theorem XX.**—Straight lines AB, EF, which are parallel to the same straight line CD, are parallel to each other.

For let HKG be perpendicular to CD, it will also be perpendicular to both AB and EF (19); therefore these last lines are parallel to each other.

**Theorem XXI.**—If a straight line EF meet two parallel lines and straight lines AB, CD; it makes the alternate angles AEF, EFD equal.

Let EH and GF be perpendicular to CD, then these lines will be parallel (18), and also at right angles to AB (19), and therefore FH and GE are equal to one another (19); therefore the triangles FGE, FHE, having the side FG = HE, and GE = FH, and FE common to both, will be equal; and hence the angle FEG will be equal to EFH, that is, FEA, will be equal to EFD.

Cor. 1. Hence if a straight line KL intersect two parallel straight lines AB, CD; it makes the exterior angle KEB equal to the interior and opposite angle EFD on the same side of the line. For the angle AEF = KEB, and it has been shown that AEF = EFD; therefore KEB = EFD.

Cor. 2. Hence also, if a straight line EF meet two parallel straight lines AB, CD; it makes the two interior angles BEF, EFD on the same side together, equal to two right angles. For the angle AEF has been shown to be equal to EFD; therefore, adding the angle FEB to both, AEF + FEB = EFD + FEB; but AEF + FEB is equal to two right angles, therefore the sum EFD + FEB is also equal to two right angles.

**Theorem XXII.**—If a straight line EF, meeting two other straight lines AB, CD, makes the alternate angles AEF, EFD equal; those lines shall be parallel.

For if AE is not parallel to CD, suppose, if possible, that some other line KE can be drawn through E, parallel to CD; then the angle KEF must be equal to EFD (21), that is (by hypothesis), to AEF; which is impossible; therefore, neither KE, nor any other line drawn through E, except AB, can be parallel to CD.

Cor. If a straight line EF intersecting two other straight lines AB, CD, makes the exterior angle GEB equal to the interior and opposite angle EFD on the same side; or the two interior angles BEF, EFD on the same side equal to two right angles; in either case the lines are parallel. For if the angle GEB = EFD, then also AEF = EFD (4). And if BEF + EFD = two right angles, then, because BEF + AEF = two right angles (1), BEF + EFD = BEF + AEF, and taking BEF from both, EFD = AEF, therefore (by the theorem) in each case the lines are parallel.

**Theorem XXIII.**—If a side AC of a triangle ABC be produced towards D; the exterior angle BCD is equal to both the interior and opposite angles BAC, ABC.

Let CE be parallel to AB, then the angle B = BCE (21), and the angle A = ECD (1 Cor. 21), therefore B + A = BCE + ECD = BCD.

Cor. The exterior angle of a triangle is greater than either of the interior opposite angles.

**Theorem XXIV.**—The three interior angles of a triangle ABC taken together are equal to two right angles. For if AC be produced to D, then A + B = BCD (23); to each of these equal quantities add ACB, then shall A + B + ACB = BCD + BCA; but BCD + BCA = two right angles (1), therefore A + B + ACB = two right angles.

Cor. 1. If two angles of one triangle be equal to two angles of another triangle, each to each; the third angle of the one shall be equal to the third angle of the other, and the triangles shall be equiangular.

Cor. 2. If two angles of a triangle, or their sum, be given, the third angle may be found, by subtracting their sum from two right angles.

Cor. 3. In a right-angled triangle, the sum of the two acute angles is equal to a right angle.

Cor. 4. In an equilateral triangle, each of the angles is equal to the third part of two right angles, or to two thirds of one right angle.

Theorem XXV.—The sum of all the interior angles of a polygon is equal to twice as many right angles, wanting four, as the figure has sides.

Let ABCDE be a polygon; from a point F within it draw straight lines to all its angles, then the polygon shall be divided into as many triangles as it has sides; but the sum of the angles of each triangle is equal to two right angles (24); therefore, the sum of all the angles of the triangles is equal to twice as many right angles as there are triangles, that is, as the figure has sides; but the sum of all the angles of the triangles is equal to the sum of all the angles of the polygon, together with the sum of the angles at the point F, which last sum is equal to four right angles (2 Cor. 4); therefore the sum of all the angles of the polygon, together with four right angles, is equal to twice as many right angles as the figure has sides, and consequently the sum of the angles of the polygon is equal to twice as many right angles, wanting four, as the figure has sides.

Cor. The four interior angles of a quadrilateral are, taken together, equal to four right angles.

Theorem XXVI.—The opposite sides of a parallelogram are equal, and the opposite angles are also equal.

Draw the diagonal BD; the two triangles ADB, DBC have the side BD common to both, and AB, DC being parallel, the angle ABD = BDC (21), also, AD, BC being parallel, the angle ADB = DBC; therefore the two triangles are equal (6), and the side AB, opposite to the angle ADB, is equal to DC, opposite to the equal angle DBC. In like manner the third side AD is equal to the third side BC, therefore the opposite sides of a parallelogram are equal.

In the next place, because of the equality of the same triangles, the angle A is equal to the angle C, and also the angle ADC composed of the two angles ADB, BDC is equal to the angle ABC composed of the angles CBD, DBA; therefore the opposite angles of a parallelogram are also equal.

Theorem XXVII.—If the opposite sides of a quadrilateral ABCD be equal, so that AB = DC, and AD = BC; Lines and Figures upon a Plane.

Draw the diagonal BD. The two triangles ABD, CDB have the three sides of the one equal to the three sides of the other, each to each, therefore the triangles are equal (10); and the angle ADB, opposite to AB, is equal to DBC opposite to DC; therefore the side AD is parallel to BC (22). For a similar reason AB is parallel to DC; therefore the quadrilateral ABCD is a parallelogram.

Theorem XXVIII.—If two opposite sides AB, DC of a quadrilateral be equal and parallel; the two other sides are in like manner equal and parallel; and the figure is a parallelogram. (Fig. 42.)

Draw the diagonal BD. Because AB is parallel to CD, the alternate angles ABD, BDC are equal (21); now the side AB = DC, and DB is common to the triangles ABD, BDC, therefore these triangles are equal (5), and hence the side AD = BC, and the angle ADB = DBC, consequently AD is parallel to BC (22), therefore the figure ABCD is a parallelogram.

Sect. II.—Of the Circle.

Definitions.

I. A Circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal to one another.

And this point is called the centre of the circle.

II. Every straight line CA, CE, CD, &c., drawn from the centre to the circumference, is called a Radius or Semidiameter; and every straight line, such as AB, which passes through the centre, and is terminated both ways by the circumference, is called a Diameter.

Hence it follows that all the radii of a circle are equal, and all the diameters are also equal, each being the double of the radius.

III. An Arch of a circle is any portion of its circumference, as FHG.

The chord or subtense of an arch is the straight line FG which joins its extremities.

IV. A Segment of a circle is the figure contained by an arch and its chord. If the figure be the half of the circle it is called a Semicircle.

Note. Every chord corresponds to two arches, and consequently to two segments; but in speaking of these, it is always the smallest that is meant, unless the contrary be expressed.

V. A Sector of a circle is the figure contained by an arch DE and the two radii CD, CE, drawn to the extremities of the arch. If the radii be at right angles to each other, it is called a Quadrant.

VI. A straight line is said to be placed, or applied, in a circle, when its extremities are in the circumference of the circle, as FG.

VII. A rectilineal figure is said to be inscribed in a circle, when the vertices of all its angles are upon the circumference of the circle; in this case, the circle is said to be circumscribed about the figure.

VIII. A straight line is said to touch a circle, or to be a Lines and tangent to a circle, when it meets the circumference in one point only; such, for example, is BD, fig. 49. The point A, which is common to the straight line and circle, is called the Point of Contact.

IX. A polygon is said to be described or circumscribed about a circle, when all its sides are tangents to the circle; and in this case, the circle is said to be inscribed in the polygon.

Theorem I.—Any diameter AB divides the circle, and its circumference, into two equal parts. (Fig. 43.)

For if the figure AEB be applied to AFB, so that the base AB may be common to both, the curve line AEB must fall exactly upon the curve line AEB; otherwise there would be points in the one, or the other, unequally distant from the centre, which is contrary to the definition of a circle.

Theorem II.—Every chord is less than the diameter.

Let the radii CA, CD be drawn from the centre to the extremities of the chord AD; then the straight line AD is less than AC + CD, that is, AD < AB.

Theorem III.—A straight line cannot meet the circumference of a circle in more than two points.

For if it could meet it in three, these three points would be equally distant from the centre, and therefore three equal straight lines might be drawn from the same point to the same straight line, which is impossible (2 Cor. 15, 1).

Theorem IV.—In the same circle, or in equal circles, equal arches are subtended by equal chords; and, conversely, equal chords subtend equal arches.

If the radius AC be equal to the radius EO, and the arch AMD equal to the arch ENG; the chord AD shall be equal to the chord EG.

For the diameter AB being equal to the diameter EF, the semicircle AMDB may be applied exactly upon the semicircle ENGF, and then the curve line AMDB shall coincide entirely with the curve line ENGF; but the arch AMD being supposed equal to ENG, the point D must fall upon G, therefore the chord AD is equal to the chord EG.

Conversely, if the chord AD = EG, the arch AMD is equal to the arch ENG.

For if the radii CD, OG be drawn, the two triangles ACD, EOG have three sides of the one equal to the three sides of the other, each to each, viz. AC = EO, CD = OG, and AD = EG; therefore, these triangles are equal (10, 1), and hence the angle ACD = EOG. Now, if the semicircle ADB be placed upon EGF, because the angle ACD = EOG, it is evident that the radius CD will fall upon the radius OG, and the point D upon G; therefore the Lines and arch AMD is equal to the arch ENG.

Theorem V.—In the same circle, or in equal circles, the greater arch is subtended by the greater chord; and, conversely (if the arch be less than half the circumference), the greater chord subtends the greater arch. (Fig. 45.)

For let the arch AH be greater than AD, and let the chords AD, AH, and the radii CD, CH, be drawn. The two sides AC, CH, of the triangle ACH, are equal to the two sides AC, CD of the triangle ACD; and the angle ACH is greater than ACD; therefore the third side AH is greater than the third side AD (9, 1); therefore the chord which subtends the greater arch is the greater. Conversely, if the chord AH be greater than AD, it may be inferred (Cor. 9, 1) from the same triangles, that the angle ACH is greater than ACD, and that thus the arch AH is greater than AD.

Note. Each of the arches is here supposed less than half the circumference; if they were greater, the contrary property would have place, the arch increasing as the chord diminishes.

Theorem VI.—The radius CG, perpendicular to a chord AB, bisects the chord (or divides it into two equal parts); it also bisects the arch AGB subtended by the chord.

Draw the radii CA, CB; these radii are two equal oblique lines in respect of the perpendicular CD, therefore they are equally distant from the perpendicular (15, 1), that is, AD = DB.

In the next place, because CG is perpendicular to the middle of AB, every point in CG is at equal distances from A and B (16, 1), therefore, if GA, GB be drawn, these lines are equal, and as they are the chords of the arches AG, BG, the arches are also equal (4).

Scholium. Since C the centre, D the middle of the chord AB, and G the middle of the arch subtended by that chord, are three points situated in the same straight line perpendicular to that chord; and that two points in a straight line are sufficient to determine its position; it follows that a straight line which passes through any two of these points must necessarily pass through the third, and must be perpendicular to the chord. It also follows, that a perpendicular to the middle of a chord passes through the centre, and the middle of the arch subtended by that chord.

Theorem VII.—If three points A, B, C be taken in the circumference of a circle; no other circumference which does not coincide with the former can be made to pass through the same three points.

Let the chords AB, BC be drawn, and let OD, OF be drawn from the centre, perpendicular to, and consequently bisecting these chords. The centre of every circle passing through A and B, must necessarily be somewhere in the perpendicular DO (last theor.), and in like manner the centre of every circle passing through B and C must be somewhere in the perpendicular OF; therefore the centre of a circle passing through A, B, and C, must be in the intersection of the perpendiculars DO, FO, and consequently can only be at one and the same point O; Lines and therefore, only one circle can be made to pass through the same three points A, B, C.

Cor. One circumference of a circle cannot intersect another in more than two points; for if they could have three common points, they would have the same centre, and consequently would coincide with each other.

**Theorem VIII.**—Two equal chords are equally distant from the centre; and of unequal chords, the less is farther from the centre.

Let the chord \( AB = DE \); suppose the chords bisected by the perpendiculars \( CF, CG \) from the centre, and draw the radii \( CA, CD \). The right angled triangles \( CAF, CDG \) have equal hypotenuses \( CA, CD \); the side \( AF (= \frac{1}{2} AB) \) of the one is also equal to the side \( DG (= \frac{1}{2} DE) \) of the other, therefore, their remaining sides \( CF, CG \) (which are the distances of the chords from the centre) are equal (17, 1).

Next, let the chord \( AH \) be greater than \( DE \); the arch \( AKH \) shall be greater than \( DME \); upon the arch \( AKH \) take \( ANB \) equal to \( DME \), draw the chord \( AB \), and suppose \( COF \) drawn from the centre perpendicular to \( AB \), and \( CI \) perpendicular to \( AH \). It is evident that \( CF > CO \), and (15, 1) \( CO > CI \); much more then is \( CF > CI \); but \( CF = CG \), because the chords \( AB, DE \) are equal; therefore \( CG > CI \); that is, of the unequal chords, the less is farther from the centre.

**Theorem IX.**—The perpendicular \( BD \), drawn at the extremity of a radius \( CA \), is a tangent to the circle.

For any oblique line \( CE \) is greater than the perpendicular \( CA \) (15, 1), therefore the point \( E \) is without the circle; therefore the line \( BD \) has but one point \( A \) common with the circumference, and consequently it is a tangent to the circle. (Def. 8.)

**Scholium.** Through the same point \( A \), only one tangent, \( AD \), can be drawn to the circle. For if it be possible to draw another, let \( AG \) be that other tangent; draw \( CF \) perpendicular to \( AG \); then \( CF \) shall be less than \( CA \) (15, 1), therefore \( F \) must be within the circle; and consequently \( AF \) when produced must necessarily meet the circle in another point besides \( A \); therefore it cannot be a tangent.

**Theorem X.**—If \( BC \), the distance of the centres of two circles, be less than the sum of their radii, and also the greater radius less than the sum of the distance of their centres and the lesser radius, the two circles intersect each other.

For, that the circles may intersect each other in a point \( A \), it is necessary that the triangle \( ABC \) be possible; therefore, not only must \( CB \) be less than \( CA + AB \), but also the greater radius \( AB \) must be less than \( AC + CB \) (7, 1); and it is evident, that as often as the triangle \( ABC \) can be constructed, the circumferences described on the centres \( B, C \) shall intersect each other in two points \( A, D \).

**Theorem XI.**—If the distance \( CB \) of the centres of two circles be equal to the sum of the radii \( CA, BA \); the circles shall touch each other externally.

It is evident that they have a common point \( A \); but they cannot have more; for if they had two, then the distance of the centres must necessarily be less than the sum of the radii.

**Theorem XII.**—If the distance \( CB \) of the centres of two circles be equal to the difference of the radii; the two circles shall touch each other internally.

In the first place, it is evident that the point \( A \) is common to them both; they cannot, however, have another; for that this may happen, it is necessary that the greater radius \( AB \) be smaller than the sum of the radius \( AC \) and the distance \( CB \) of the centre (10), which is not the case.

Cor. Therefore, if two circles touch each other, either internally or externally, their centres and the point of contact are in the same straight line.

**Theorem XIII.**—In the same circle, or in equal circles, equal angles \( ACB, DCE \), at the centres, intercept upon the circumference equal arches \( AB, DE \). And, conversely, if the arches \( AB, DE \) be equal, the angles \( ACB, DCE \) are equal.

First, if the angle \( ACB \) be equal to \( DCE \), the one angle may be applied upon the other; and as the lines containing them are equal, it is manifest that the point \( A \) will fall upon \( D \), and the point \( B \) upon \( E \); thus the arch \( AB \) will coincide with and be equal to the arch \( DE \).

Next, if the arch \( AB \) be equal to \( DE \); the angle \( ACB \) is equal to \( DCE \): for if the angles are not equal, let \( ACB \) be the greater, and let \( ACI \) be taken equal to \( DCE \); then, by what has been already demonstrated, the arch \( AI = DE \); but by hypothesis \( AB = DE \); therefore, \( AI = AB \), which is impossible; therefore the angle \( ACB = DCE \).

**Theorem XIV.**—The angle \( BCD \) at the centre of a circle is double the angle \( BAD \) at the circumference, when both stand on the same arch \( BD \).

First, let the centre of the circle be within the angle... Lines and BAD; draw the diameter AE. The exterior angle BCE of the triangle BCA is equal to both the inward and opposite angles BAC, CBA (23, 1); but the triangle BCA being isosceles, the angle BAC = CBA; therefore the angle BCE is double of the angle BAC. For the same reason, the angle DCE is double of the angle DAE; therefore the whole angle BCD is double of the whole angle BAD.

**Fig. 55.**


**Fig. 56.**


Suppose, in the next place, that the centre is without the angle BAD; then, drawing the diameter AE, it may be demonstrated, as in the first case, that the angle ECD is double of the angle EAD, and that the angle ECB, a part of the first, is double the angle EAB, a part of the other; therefore the remaining angle BCD is double the remaining angle BAD.

**Theorem XV.**—All angles BAD, BFD in the same segment BAFD of a circle are equal to one another.

**Fig. 57.**


**Fig. 58.**


If the segment be greater than a semicircle, from the centre C draw CB and CD; then the angles BAD and BFD being (by last theorem) each equal to half BCD, they must be equal to one another.

But if the segment BAFD be less than a semicircle, let H be the intersection of BF and AD; then, the triangles ABH and FDH having the angle AHB of the one equal to FHD of the other (4, 1), and AB = FD (by case I) will have the remaining angle of the one equal to the remaining angle of the other; that is, the angle BAH = HFD, or BAD = BFD.

**Theorem XVI.**—The opposite angles of any quadrilateral figure ABCD described in a circle are together equal to two right angles.

Draw the diagonals AC, BD; because the angle ABD = ACD, and CBD = CAD (last theor.), the sum ABD + CBD = ACD + CAD; or ABC = ACD + CAD; to each of these equals add ADC, and ABC + ADC = ACD + CAD + ADC; but the last three angles being the angles of the triangle ADC, are taken together equal to two right angles (24, 1), therefore ABC + ADC = two right angles.

In the same manner the angles BAD, BCD may be shown to be together equal to two right angles.

**Theorem XVII.**—In a circle, the angle BAD in a semi-circle is a right angle, but the angle ABD in a segment greater than a semicircle is less than a right angle; and the angle AED in a segment less than a semicircle is greater than a right angle.

Let C be the centre, join CA, and produce BA to F. Because CB = CA, the angle CAB = CBA (11, 1); and because CD = CA, the angle CAD = CDA, therefore, the whole angle BAD = CBA + CDA; but these two last angles are together equal to DAF (23, 1), therefore the angle BAD = DAF; and hence each of them is a right angle.

And because ABD + ADB is a right angle, therefore ABD, an angle in a segment greater than a semicircle, is less than a right angle.

And because ADBE is a quadrilateral in a circle, the opposite angles B and E are equal to two right angles (last theor.), but B is less than a right angle; therefore the angle E, which is in a segment less than a semicircle, is greater than a right angle.

**Theorem XVIII.**—The angle BAC contained by AC, a tangent, and AB, a chord drawn from the point of contact, is equal to any angle ADB in the alternate segment of the circle.

Draw the diameter AE, and join DE. The angles EAC, EDA, being right angles (last theor.), are equal to one another; and of these, EAB, a part of the one, is equal to EDB, a part of the other (15), therefore the remainder, BAC, of the former is equal to the remainder, BDA, of the latter.

**Sect. III.—Of Proportion.**

In comparing two quantities of the same kind, in respect of magnitude, we are led to consider how often the one quantity contains either the whole or some part of the other quantity. To resolve this question, we must find a common measure of the quantities; that is, a quantity that is contained in each a certain number of times exactly.

Let the quantities be two given straight lines AB, CD; their common measure (if they have one) may be found by proceeding as follows:

Take the less line CD out of the greater AB, as often as possible, and let BE be the remainder. Next take the remainder BE out of the line CD, as often as it can be had, and let DF be the second remainder.

Again, take the second remainder DF out of the first BE, as often as it can be had, and let BG be the third remainder. Take the third remainder BG out of the second DF as often as possible, and proceed in this way, until a remainder be found, which is contained an exact number of times in the preceding. This last remainder will be a common measure of the two lines.

For example, let CD be contained twice in AB, with the remainder BE; and BE once in CD, with a remain- Lines and der DF; and DF once in BE, with a remainder BG, and let BG be contained twice exactly in DF; then

\[ DF = 2 \cdot BG \] \[ EB = DF + BG = 3 \cdot BG \] \[ CD = EB + DF = 5 \cdot BG \] \[ AB = 2 \cdot CD + EB = 13 \cdot BG. \]

Thus it appears, that BG is contained thirteen times in AB, and five times in CD; therefore BG is a common measure of the two lines. Hence also we see, that the relation which AB bears to CD, in regard to magnitude, consists in the former containing thirteen such parts as the latter contains five.

In this example it has been supposed that a remainder is at last found, which is contained an exact number of times in the remainder immediately preceding. But this will not always happen. If the given lines be the diagonal of a square, and its side, and in innumerable other cases, it may be demonstrated, that no one of the succeeding remainders can be contained an exact number of times in that before it: therefore such quantities cannot have a common measure. Hence quantities are distinguished into two kinds, viz. such as admit of a common measure, and such as do not admit of a common measure. The former are said to be commensurable, and the latter incommensurable.

The doctrine of proportion, as delivered in the 5th book of Euclid's Elements, applies alike to commensurable and incommensurable quantities; the theory may, however, be rendered considerably easier, by confining it to commensurable quantities. It is this limited view of the subject that is here given. Experience proves that Euclid's fifth book is difficult to be understood by the generality of readers; and this fact may serve as an excuse for deviating, unwillingly, from the illustrious ancient.

Definitions.

I. When one quantity contains another a certain number of times, without a remainder, the former is said to be a multiple of the latter, and the latter a part of the former.

II. When several quantities are multiples of as many others, and each contains its part the same number of times, the former are called equimultiples of the latter, and the latter like parts of the former.

III. Between any two finite magnitudes of the same kind there subsists a certain relation, in respect of quantity, which is called their ratio. The two magnitudes compared are called the terms of the ratio. The first the antecedent, and the last the consequent.

IV. If there be four quantities, and if the first contain a part of the second (without a remainder) just as often as the third contains a like part of the fourth, the ratio of the first to the second is said to be equal to the ratio of the third to the fourth; or the first is said to have to the second the same ratio which the third has to the fourth.

V. The terms of any number of equal ratios are called proportionals.

Note. When four quantities A, B, C, D, are proportionals, it is usual to say, that A is to B, as C to D, and to write them thus, \( A : B :: C : D \); or thus, \( A : B = C : D \).

VI. Quantities are said to be continual proportionals when the ratio of the first to the second, of the second to the third, of the third to the fourth, and so on, are all equal.

VII. When three quantities are continual proportionals, the second is said to be a mean proportional between the other two.

VIII. In proportionals, the antecedent terms are called homologous to one another; as also the consequents to one another.

IX. When there is any number of quantities of the same kind, the first is said to have to the last the ratio compounded of the ratio which the first has to the second, and of the ratio which the second has to the third, and of the ratio which the third has to the fourth, and so on, to the last magnitude.

X. If three magnitudes are continual proportionals, the ratio of the first to the third is said to be duplicate of the ratio of the first to the second.

XI. If four magnitudes are continual proportionals, the ratio of the first to the fourth is said to be triplicate of the ratio of the first to the second, or of the ratio of the second to the third, &c.

The following technical words are employed to signify certain ways of changing either the order or magnitude of proportionals, so as that they still continue proportionals.

XII. If four quantities are proportionals, they are said to be proportionals by inversion, when it is inferred that the second is to the first, as the fourth to the third.

XIII. They are proportionals by alternation, when the first is to the third, as the second to the fourth.

XIV. By composition, when the sum of the first and second is to the second, as the sum of the third and fourth to the fourth.

XV. By division, when the difference of the first and second is to the second, as the difference of the third and fourth to the fourth.

XVI. By conversion, when the first is to the difference of the first and second, as the third to the difference of the third and fourth.

XVII. By mixing, when the sum of the first and second is to their difference, as the sum of the third and fourth to their difference.

Axioms.

1. Equal quantities have the same ratio to the same quantity, and the same quantity has the same ratio to equal quantities.

2. Quantities having the same ratio to the same quantity, or to equal quantities, are equal among themselves; and those to which the same quantity has the same ratio are equal to one another.

3. Ratios equal to one and the same ratio are also equal one to the other.

Theorem I.—Quantities have to one another the same ratio which their equimultiples have.

Let A and B be two quantities, and supposing m to denote any number, let mA and mB (that is, m times A and m times B) be any equimultiples of these quantities; the ratio of A to B is equal to the ratio of mA to mB; or \( A : B = mA : mB \).

For, let A contain three such parts, each equal to X, as B contains four, so that

\[ A = X + X + X; \quad B = X + X + X + X. \]

Then \( mA = mX + mX + mX, \) \[ mB = mX + mX + mX + mX, \]

because a whole taken any number of times, is equal to the sum of its parts taken as often. Now, because A contains the one fourth of B three times, and mA evidently contains one fourth of mB also three times; A contains a part of B exactly as often as mA contains a like part of mB. Therefore (Def. 4), the ratio of A to B is equal to the ratio of mA to mB.

In the same manner, if A and B be supposed any other multiples of X, it may be proved that \( A : B = mA : mB \).

Cor. Like parts of magnitudes have to each other the same ratio as the whole; for A and B are like parts of mA and mB. Lines and Theorem II.—If four quantities be proportional, they are proportional also when taken inversely.

Let A, B, C, D be four quantities, such that \( A : B = C : D \); then \( B : A = D : C \).

For, suppose that C contains two such equal parts as D contains three, then, because the ratio of A to B is equal to the ratio of C to D, A will contain two such parts as B contains three. (Def. 4.) Therefore B will contain three such parts as A contains two; and D will contain three such parts as C contains two. Thus B will contain a part of A as often as D contains a like part of C; therefore the ratio of B to A is equal to the ratio of D to C. (Def. 4.)

Theorem III.—If four quantities of the same kind be proportional, they will also be proportional when taken alternately.

Let \( A : B = C : D \), then, if the quantities be all of the same kind, \( A : C = B : D \).

For, let C contain three such parts as D contains four; then (Def. 4) A will contain three such parts as B contains four. Let each of the equal parts contained in A and B be X, and let each of the equal parts contained in C and D be Y; then

\[ A = 3X, \quad B = 4X \\ C = 3Y, \quad D = 4Y. \]

And because A and C contain X and Y the same number of times, A and C are equimultiples of X and Y. (Def. 2.) Also, because B and D contain X and Y the same number of times, B and D are equimultiples of X and Y.

Therefore \( A : C = X : Y \) Prop. I.

And \( B : D = X : Y \)

Hence \( A : C = B : D \) (Axion 3.)

Cor. If the first of four proportionals be greater than the third, the second shall be greater than the fourth; and if the first be equal to the third, the second shall be equal to the fourth; and if the first be less than the third, the second shall be less than the fourth.

Theorem IV.—If four quantities be proportional, they shall also be proportional by composition.

Let \( A : B = C : D \); then, by composition, \( A + B : B = C + D : D \).

Let C contain five such parts as D contains three, then (Def. 4) A will contain five such parts as B contains three. Let each of the equal parts contained in A and B be X, and let each of the equal parts contained in C and D be Y. Then, because

\[ A = 5X, \quad B = 3X, \quad C = 5Y, \quad D = 3Y, \]

by adding, \( A + B = 8X \), and \( C + D = 8Y \).

Now, \( B = 3X \), and \( D = 3Y \);

Therefore \( A + B \) contains one third of B eight times, and \( C + D \) contains one third of D also eight times, and, in general, \( A + B \) will contain a part of B as often as \( C + D \) contains a like part of D; therefore (Def. 4) \( A + B : B = C + D : D \).

Theorem V.—If four quantities be proportional, they will also be proportional by division.

Let \( A : B = C : D \), and let A be greater than B, and C greater than D; then \( A - B : B = C - D : D \).

For, making the same supposition as in last proposition, so that \( A = 5X, \quad B = 3X, \quad C = 5Y, \quad D = 3Y \),

by subtracting, \( A - B = 2X \), and \( C - D = 2Y \).

Now, \( B = 3X \), \( D = 3Y \).

Thus it appears, that in this case \( A - B \) contains one third of B twice, and that \( C - D \) contains one third of D also twice; and, in general, that \( A - B \) will always contain some part of B as often as \( C - D \) contains a like part of D; therefore (Def. 4) \( A - B : B = C - D : D \).

Theorem VI.—If four quantities be proportional, they shall also be proportional by conversion.

Let \( A : B = C : D \); then \( A : A - B = C : C - D \).

For, making the same supposition as in the two last propositions, because

\[ A = 5X, \quad B = 3X, \quad C = 5Y, \quad D = 3Y, \]

therefore \( A - B = 2X, \quad C - D = 2Y \).

Hence it appears, that in this case \( A \) contains the half of \( A - B \) five times, and that \( C \) contains the half of \( C - D \) also five times; and, in general, that \( A \) will contain a part of \( A - B \) as often as \( C \) contains a like part of \( C - D \); therefore (Def. 4) \( A : A - B = C : C - D \).

Theorem VII.—If the first of four magnitudes has the same ratio to the second which the third has to the fourth, and if any equimultiples whatever be taken of the first and third, and any whatever of the second and fourth; the multiple of the first shall have the same ratio to the multiple of the second that the multiple of the third has to the multiple of the fourth.

Let \( A : B = C : D \), and supposing \( m \) and \( n \) to denote any two numbers, let the antecedents A and C be taken each \( m \) times, and let the consequents B and D be taken each \( n \) times; \( mA : nB = mC : nD \).

Suppose that C contains two such parts as D contains three, then (Def. 4) A will contain two such parts as B contains three. Let each of the equal parts contained in A and B be X, and let each of the equal parts contained in C and D be Y, so that \( A = 2X, \quad B = 3X, \quad C = 2Y, \quad D = 3Y \).

Then \( mA = 2mX, \quad nB = 3nX \)

\( mC = 2mY, \quad nD = 3nY \).

From these expressions, it appears that X and Y are like parts of \( nB \) and \( nD \); for each is contained in its multiple three \( n \) times; also, that \( mA \) and \( mC \) are equimultiples of X and Y; for each contains its part two \( m \) times.

Therefore \( mA \) contains a part of \( nB \), exactly as often as \( mC \) contains a like part of \( nD \); and hence (Def. 4) \( mA : nB = mC : nD \).

Cor. If \( A : B = C : D \), then \( mA : nB = mC : nD \).

Theorem VIII.—If there be any number of quantities, and as many others, which, taken two and two in order, have the same ratio; the first will have to the last of the first quantities the same ratio which the first of the others has to the last.

First, let there be three quantities, A, B, C, and other three, H, K, L, such that \( A : B = H : K \), and \( B : C = K : L \); then \( A : C = H : L \).

For, suppose that H contains two such parts as K contains three and L seven, then (Def. 4) A will contain two such parts as B contains three, and C seven.

Let the equal parts contained in A, B, C be X, and let the equal parts contained in H, K, L be Y, so that

\[ A = 2X, \quad B = 3X, \quad C = 7X, \quad H = 2Y, \quad K = 3Y, \quad L = 7Y. \]

In this case, A contains one seventh of C twice, and H contains one seventh of L also twice; and, in general, A will contain a part of C as often as H contains the same part of L; therefore \( A : C = H : L \) (Def. 4).

Next, let there be four quantities A, B, C, D, and other four H, K, L, M, such that \( A : B = H : K \), and \( B : C = K : L \), and \( C : D = L : M \); then \( A : D = H : M \). For, since A, B, C are three quantities, and H, K, L other three quantities, which, taken two and two, have the same ratio; by the foregoing case, A : C = H : L. And because A : C = H : L; and C : D = L : M; therefore, by that same case, A : D = H : M. In the same way, the demonstration may be extended to any number of quantities.

Note. This proposition is usually cited by the words *ex aequali* or *ex aequo*.

**Theorem IX.**—If there be any number of quantities, and as many others, which, taken two and two, in a cross order, have the same ratio; the first will have to the last of the first quantities, the same ratio which the first of the others has to the last.

First, let there be three quantities A, B, C, and other three H, K, L, such that A : B = K : L, and B : C = H : K; then A : C = H : L.

For, suppose K to contain two such equal parts as L contains three, then A will contain two such parts as B contains three. (Def. 4.) Let each of the equal parts contained in A and B be X, and let each of the equal parts contained in K and L be Y, so that

\[ A = 2X, \quad B = 3X, \quad K = 2Y, \quad L = 3Y. \]

Also, let Z be the same part of C that Y is of L, and let V be the same part of H that X is of A; so that, in this case,

\[ C = 3Z, \quad H = 2V. \]

Then, because B : C = H : K; that is, 3X : 3Z = 2V : 2Y, and (by Prop. I.) 3X : 3Z = X : Z, and 2V : 2Y = V : Y; therefore X : Z = V : Y (Ax. 3). Hence (by Prop. VII.) 2X : 3Z = 2V : 3Y; but 2X = A, 3Z = C, 2V = H, 3Y = L; therefore A : C = H : L.

Next, let there be four quantities, A, B, C, D, and other four H, K, L, M, such that A : B : L : M, and B : C = K : L; and C : D = H : K; then A : D = H : M.

For since A, B, C are three quantities, and K, L, M other three, which, taken two and two in a cross order, have the same ratio; by the first case, A : C = K : M; but C : D = H : K; therefore, again, by the first case, A : D = H : M.

In the same manner, the demonstration may be extended to any number of quantities.

Note. This proposition is usually cited by the words *ex aequali in proportionem perturbata*, or *ex aequo inversely*.

**Theorem X.**—If the first has to the second the same ratio which the third has to the fourth, and the fifth to the second the same ratio which the sixth has to the fourth; the first and fifth together shall have to the second the same ratio which the third and sixth have to the fourth.

Let \( A : B = C : D \), and \( E : B = F : D \); then \( A + E : B = C + F : D \).

Because \( E : B = F : D \), by inversion, \( B : E = D : F \) (2); but, by hypothesis, \( A : B = C : D \); therefore *ex aequali* (Prop. VIII.), \( A : E = C : F \); and, by composition, \( A + E : E = C + F : F \). Again, by hypothesis, \( E : B = F : D \); therefore *ex aequo*. \( A + E : B = C + D : F \).

**Theorem XI.**—If four quantities be proportional, the sum of the first two is to their difference as the sum of the other two to their difference.

Let \( A : B = C : D \); then, if A be greater than B, \( A + B : A - B = C + D : C - D \).

For, since \( A : B = C : D \), by division (5) \( A - B : B = C - D : D \), and inversion (2) \( B : A - B = D : C - D \), and by composition (4) \( A + B : B = C + D : D \); therefore, *ex aequo*. \( A + B : A - B = C + D : C - D \).

In like manner, if B be greater than A, it may be proved that \( A + B : B - A = C + D : D - C \).

**Theorem XII.**—If any number of quantities be proportional; as one of the antecedents is to its consequent, so is the sum of all the antecedents to the sum of all the consequents.

Let \( A : B = C : D = E : F \); then \( A : B = A + C + E : B + D + F \). For, suppose that A contains two such parts, each = X, as B contains three, and that C contains two such parts, each = Y, as D contains three, and that E contains two such parts, each = Z, as F contains three, and so on; then

\[ \begin{align*} A &= 2X \\ B &= 3X \\ C &= 2Y \\ D &= 3Y \\ E &= 2Z \\ F &= 3Z. \end{align*} \]

Hence, by addition,

\[ A + C + E = 2X + 2Y + 2Z = 2(X + Y + Z); \quad B + D + F = 3X + 3Y + 3Z = 3(X + Y + Z). \]

Thus it appears that A contains one third of B twice, and that A + C + E contains one third of B + D + F also twice; and, in general, that A will contain a part of B just as often as A + C + E contains a like part of B + D + F; therefore \( A : B = A + C + E : B + D + F \).

In treating of proportion, we have supposed that the antecedent contains some part of the consequent a certain number of times exactly, which part is therefore a common measure of the antecedent and consequent. But there are quantities which cannot have a common measure, and which are therefore said to be *incommensurable*; such, for example, are the sides of two squares one of which has its surface double that of the other.

Although the ratio of two incommensurable quantities cannot be expressed in numbers, yet we can always assign a ratio in numbers which shall be as near to that ratio as we please. For let A and B be any two quantities whatever, and suppose that x is such a part of A that \( A = px \); then if q denote the number of times that x can be taken from B, and d the remainder, we have \( B = qx + d \), and \( qx = B - d \); and because \( p : q = px : qx \), therefore \( p : q = A : B - d \). Now, as d is less than x, by taking x sufficiently small, d may be less than any proposed quantity; so that \( B - d \) may differ from B by less than any given quantity; therefore two numbers p and q may always be assigned, such, that the ratio of p to q shall be the same as the ratio of A to a quantity that differs less from B than by any given quantity, however small that quantity may be.

Hence we may conclude, that whatever has been delivered in this section relating to commensurable quantities, may be considered as applying equally to such as are incommensurable.

**Sect. IV.—The Proportions of Figures.**

**Definitions.**

I. *Equivalent Figures* are such as have equal surfaces. Two figures may be equivalent, although very dissimilar; thus a circle may be equivalent to a square, a triangle to a rectangle, and so of other figures.

We shall give the denomination of *equal figures* to those which, being applied the one upon the other, coincide entirely; thus, two circles having the same radius are equal; and two triangles having three sides of the one equal to three sides of the other, each to each, are also equal. II. Two figures are similar, when the angles of the one are equal to the angles of the other, each to each; and the homologous sides proportional. The homologous sides are those which have the same position in the two figures; or which are adjacent to the equal angles. The angles themselves are called homologous angles.

Two equal figures are always similar, but similar figures may be very unequal.

III. In two different circles, similar sectors, similar arches, similar segments, are such as correspond to equal angles at the centre. Thus the angle A being equal to the angle O, the arch BC is similar to the arch DE, and the sector ABC to the sector ODE, &c.

IV. The Altitude of a parallelogram is the perpendicular which measures the distance between the opposite sides or bases AB, CD.

V. The Altitude of a triangle is the perpendicular AD drawn from the vertical angle A upon the base BC.

VI. The Altitude of a trapezoid is the perpendicular EF drawn between its two parallel bases AB, CD.

VII. The Area and the surface of a figure are terms of nearly the same signification. The term area, however, is more particularly used to denote the superficial quantity of the figure in respect of its being measured, or compared with other surfaces.

**Theorem I.**—Parallelograms which have equal bases and equal altitudes are equivalent.

Let AB be the common base of the parallelograms ABCD, EBAF, which being supposed to have the same altitude, the sides DC, FE opposite to the bases will lie in DE a line parallel to AB. Now, from the nature of a parallelogram, AD = BC, and AE = BE; for the same reason DC = AB, and FE = AB; therefore DC = FE, and taking away DC and FE from the same line DE, the remainders CE and DF are equal; hence the triangles DAF, CBE have three sides of the one equal to three sides of the other, each to each; and consequently are equal (10, 1). Now, if from the quadrilateral ABED the triangle ADF be taken away, there will remain the parallelogram ABEF; and if from the same quadrilateral ABED, the triangle CBE, equal to the former, be taken away, there will remain the parallelogram ABCD; therefore the two parallelograms ABCD, ABEF, which have the same base and the same altitude, are equivalent.

Cor. Every parallelogram is equivalent to a rectangle of the same base and altitude.

**Theorem II.**—Every triangle ABC is the half of a parallelogram ABCD, having the same base and altitude.

For the triangles ABC, ACD are equal (28, 1).

Cor. 1. Therefore a triangle ABC is the half of a rectangle BCEF of the same base and altitude.

Cor. 2. All triangles having equal bases and equal altitudes are equivalent.

**Theorem III.**—Two rectangles of the same altitude are to each other as their bases.

Let ABCD, AEFD be two rectangles, which have a common altitude AD; the rectangle ABCD shall have to the rectangle AEFD the same ratio that the base AB has to the base AE.

Let the base AB have to the base AE the ratio of the number p (which we shall suppose 7) to the number q (which may be 4), that is, let AB contain p (7), such equal parts as AE contains q (4); then, if perpendiculars be drawn to AB and AE at the points of division, the rectangles ABCD and AEFD will be divided, the former into p and the latter into q rectangles, which will be all equal (1); for they have equal bases and the same altitude. Thus the rectangle ABCD will also contain p such equal parts as the rectangle AEFD contains q; therefore the rectangle ABCD is to AEFD as the number p to the number q (Ax. 3, 3), that is, as the base AB to AE.

**Theorem IV.**—Any two rectangles are to each other as the products of any numbers proportional to their sides.

Let the numbers m, n, p, q have among themselves the same ratios that the sides of the rectangles ABCD, AEFG have to each other; that is, let AB contain m such equal parts, whereof AD contains n; and AE contains p, and AG contains q, then shall

\[ \text{ABCD : AEFG} = mn : pq. \]

Let the rectangles be so placed that the sides AB, AE may be in a straight line, then AD and AG will also lie in a straight line (3, 1). Now (3)

\[ \text{ABCD : AEHD} = \frac{AB}{AE} = m : p, \] but \( m : p = mn : np \) (1, 3),

therefore \( \text{ABCD : AEHD} = mn : np. \)

Again, \( \text{AEHD : AEFG} = \frac{AD}{AG} = n : q; \) but \( n : q = pn : pq; \) therefore \( \text{AEHD : AEFG} = pn : pq; \)

and it was shown that

\[ \text{ABCD : AEHD} = mn : np \text{ or } pn, \] therefore (8, 3) \( \text{ABCD : AEFG} = mn : pq. \)

**Scholium.**

Hence it appears, that the product of the base by the altitude of a rectangle may be taken for its measure, observing that by such product is meant that of the number of linear units in the base by the number of linear units in the altitude. This measure is however not absolute, but relative; for it must be supposed, that in comparing Lines and one rectangle with another, the sides of both are measured by the same linear unit. For example, if the base of a rectangle A be three units, and its altitude 10, the rectangle is represented by $3 \times 10$, or 30; this number considered by itself has no meaning; but if we have a second rectangle B, the base of which is twelve units and the altitude seven, this second rectangle shall be represented by the number $12 \times 7$, or 84; and hence it may be concluded that the two rectangles are to each other as 30 to 84; therefore, if in estimating any superficies the rectangle A be taken for the measuring unit, the rectangle B shall have for its absolute measure $\frac{84}{30}$, that is, it shall be $\frac{84}{30}$ superficial units.

It is more common, as well as more simple, to take for a superficial unit a square, the side of which is an unit in length, and then the measure which we have regarded only as relative becomes absolute: for example, the number 30, which is the measure of the rectangle A, represents thirty superficial units, or thirty squares, each having its side equal to an unit. To illustrate this, see fig. 70.

**Theorem V.**—The area of any parallelogram is equal to the product of its base by its altitude. (Fig. 67.)

For the parallelogram ABCD is equivalent to the rectangle FBCE, which has the same base BC, and the same altitude AO (Cor. 1); but the measure of the rectangle is $BC \times AO$ (Schol. 4), therefore the area of the parallelogram is $BC \times AO$.

**Cor.** Parallelograms having the same base, or equal bases, are to each other as their altitudes; and parallelograms having the same altitude are to each other as their bases; for in the former case put B for the common base, and A and A' for the altitudes, then the areas of the figures are $B \times A$ and $B \times A'$; and it is manifest that $B \times A : B \times A' = A : A'$; and in the latter case, putting A for the common altitude, and B and B' for the bases, it is evident that $B \times A : B' \times A = B : B'$.

**Theorem VI.**—The area of a triangle is equal to the product of its base by the half of its altitude. (Fig. 67.)

For the triangle ABC is half of the parallelogram ABCD, which has the same base BC, and the same altitude AO (2); but the area of the parallelogram is $BC \times AO$ (5), therefore that of the triangle is $\frac{1}{2} BC \times AO$, or $BC \times \frac{1}{2} AO$.

**Cor.** Two triangles of the same altitude are to each other as their bases; and two triangles having the same base are to each other as their altitudes.

**Theorem VII.**—The area of a trapezoid ABCD is equal to the product of its altitude EF by half the sum of its parallel sides AB, CD.

Through the point I, the middle of BC, draw KL parallel to the opposite side AD, and produce DC to meet KL. In the triangles IBL, ICK, IB is equal to IC by construction, and the angle ICK = IBL (21, 1); therefore these triangles are equal; and hence the trapezoid ABCD is equivalent to the parallelogram ALKD, and has for its measure $AL \times EP$. But $AL = DK$, and because the triangle IBL is equal to the triangle KCI, the side BL = CK; therefore $AB + CD = AL + DK = 2AL$; hence $AL$ is half the sum of the parallel sides AB, CD; and as the area of the trapezoid is equal to $FE \times AL$, it is also equal to $FE \times \left(\frac{AB + CD}{2}\right)$.

**Theorem VIII.**—If four straight lines AB, AC, AD, AE be proportionals; the rectangle ABFE, contained by the two extremes, is equivalent to the rectangle ACGD contained by the means. And conversely, if the rectangle contained by AB, AE, the extremes, be equivalent to the rectangle contained by AC, AD, the means, the four lines are proportionals.

Let the rectangles be so placed as to have the common angle A, and let BF, DG intersect each other in H. Because the rectangles ABHD, ACGD have the same altitude AD,

$$ABHD : ACGD = AB : AC \quad (3);$$

and because the rectangles ABHD, ABFE have the same altitude AB, for the same reason,

$$ABHD : ABFE = AD : AE;$$

but by hypothesis $AB : AC = AD : AE$, therefore (Ax. 3, 3) $ABHD : ACGD = ABHD : ABFE$; therefore (Ax. 2, 3) the rectangle ACGD = ABFE.

Next suppose that the rectangle ACGD = ABFE; then

$$ABHD : ACGD = ABHD : ABFE \quad (\text{Ax.1, 3});$$

but $ABHD : ACGD = AB : AC \quad (3)$, and $ABHD : ABFE = AD : AE$; therefore $AB : AC = AD : AE$.

**Cor.** If three straight lines be proportionals, the rectangle contained by the extremes is equal to the square of the mean; and if the rectangle contained by the extremes be equal to the square of the mean, the three straight lines are proportionals.

**Theorem IX.**—If four straight lines be proportionals, and also other four; the rectangles contained by the corresponding terms shall be proportionals; that is, if $AB : BC = CD : DE$, and $BF : BG = DH : DI$, then shall rectangle AF : rect. BM = rect. CH : rect. DQ.

For in BG and DI produced, if necessary, take BF = BF, and DH = DH, and let FP be parallel to BC, and HN to DE; then (3),

$$\text{rect. AF : rect. BP = AB : BC},$$

and rect. CH : rect. DN = CD : DE;

but $AB : BC = CD : DE$ (by hypothesis), therefore,

$$\text{rect. AF : rect. BP = rect. CH : rect. DN};$$

now (3) rect. BP : rect. BM = BF : BG, and rect. DN : rect. DQ = DH : DI; but BF : BG = DH : DI (by hypoth.) therefore,

$$\text{rect. BP : rect. BM = rect. DN : rect. DQ};$$

but it has been shown that

$$\text{rect. AF : rect. BP = rect. CH : rect. DN},$$

therefore (8, 3),

$$\text{rect. AF : rect. BM = rect. CH : rect. DQ}.$$

**Cor.** Hence the squares of four proportional straight lines are themselves proportionals.

**Theorem X.**—If a straight line AC be divided into any two parts at B, the square made upon the whole line... AC shall be equal to the squares made upon the two parts AB, BC, together with twice the rectangle contained by these two parts; which may be expressed thus,

\[ AC^2 = AB^2 + BC^2 + 2AB \times BC. \]

Suppose the square ACDE to be constructed; take AF = AB, draw FG parallel to AC, and BH parallel to CD.

The square ACDE is made up of four parts; the first ABIF is the square upon AB, because AF = AB; the second IGDH is the square upon BC, for AC = AE, and AB = AF; therefore AC — AB = AE — AF, that is, BC = EF; but BC = IG, and EF = DG (26, 1); therefore IGDH is the square upon BC; and the remaining two parts are the two rectangles BCGI, FEHI, which have each for their measure \( AB \times BC \); therefore the square upon AC is equal to the squares upon AB and BC, and twice the rectangle \( AB \times BC \).

**THEOREM XI.**—If a straight line AC be the difference of two straight lines AB, BC, the square made upon AC shall be equal to the excess of the two squares upon AB and BC above twice the rectangle contained by AB and BC; that is,

\[ AC^2 = AB^2 + BC^2 - 2AB \times BC. \]

Construct the square ABIE, take AE = AC, and draw CG parallel to BI, and HK parallel to AB; and complete the square EFLK. The two rectangles CBIG, GLKD have each \( AB \times BC \) for their measure; and if these be taken from the whole figure ABILKEA, that is, from \( AB^2 + BC^2 \), there will remain the square ACDE, that is, the square upon AC.

**THEOREM XII.**—The rectangle contained by the sum and the difference of two straight lines is equal to the difference of the squares upon those lines; that is,

\[ (AB + BC)(AB - BC) = AB^2 - BC^2. \]

Construct upon AB and AC the squares ABIF, ACDE; produce AB, so that BK = BC, and complete the rectangle AKLE. The base AK of the rectangle is the sum of the two lines AB, BC; and its altitude AE is the difference of the same lines; therefore the rectangle AKLE = \((AB + BC)(AB - BC)\); but the same rectangle is composed of two parts ABHE + BHLK, of which BHLK is equal to the rectangle EDGF; for BH = DE, and BK = FE; therefore AKLE = ABHE + EDGF; but these two parts constitute the excess of the square ABIF above the square DHIG, the former of which is the square upon AB, and the latter the square upon BC; therefore \((AB + BC)(AB - BC) = AB^2 - BC^2\).

**THEOREM XIII.**—The square upon the hypotenuse of a right-angled triangle is equal to the sum of the squares upon the two other sides.

Let ABC be a right-angled triangle; having formed the squares upon its three sides, draw a perpendicular AD from the right angle upon the hypotenuse, and produce it to E, and draw the diagonals AF, CH. The angle ABF is evidently the sum of ABC and a right angle, and the angle HBC is also the sum of ABC and a right angle; therefore the angle ABF = HBC. Now AB = BH, for they are sides of the same square, and BC = BF for the same reason; therefore the triangles ABF, HBC have two sides, and the included angle of the one equal to two sides, and the included angle of the other, each to each, therefore the triangles are equal (5, 1); but the triangle ABF is the half of the rectangle BDEF (which for brevity's sake we shall call BE), because it has the same base BF, and the same altitude BD (2); and the triangle HBC is in like manner half of the square AH; for the angles BAC, BAL being both right angles, CA and AL constitute a straight line parallel to BH (3, 1); and thus the triangle HBC and the square AH have the same base HB, and the same altitude AB; from which it follows that the triangle is half of the square (2). It has now been proved that the triangle ABF is equal to the triangle HBC; and that the rectangle BE is double of the former, and the square AH double of the latter; therefore the rectangle BE is equal to the square AH. It may be demonstrated in like manner that the rectangle CDEG, or CE, is equal to the square AI; but the rectangles BE, CE make up the square BCGF; therefore the square BCGF upon the hypotenuse is equal to the squares ALHB, AKIC upon the other two sides.

**THEOREM XIV.**—In a triangle ABC, if the angle C be acute; the square of the opposite side AB is less than the squares of the sides which contain the angle C; and if AD a perpendicular be drawn to BC from the opposite angle, the difference shall be equal to twice the rectangle \( BC \times CD \); that is,

\[ AB^2 = AC^2 + CB^2 - 2BC \times CD. \]

First, suppose AD to fall within the triangle, then BD = BC — CD, and consequently (11) \( BD^2 = BC^2 + CD^2 - 2BC \times CD \); to each of these equals add \( AD^2 \); then, observing that \( BD^2 + DA^2 = BA^2 \), and \( CD^2 + DA^2 = CA^2 \),

\[ AB^2 = BC^2 + CA^2 - 2BC \times CD. \]

Next, suppose AD to fall without the triangle, so that BD = CD — BC, and therefore \( BD^2 = CD^2 + BC^2 - 2BC \times CD \) (11), to each of these add \( AD^2 \) as before, and we get

\[ AB^2 = BC^2 + CA^2 - 2BC \times CD. \]

**THEOREM XV.**—In a triangle ABC, if the angle C be ob- tuse; the square of the opposite side \( AB \) is greater than the sum of the squares of the sides which contain the angle \( C \); and if \( AD \) a perpendicular be drawn to \( BC \) from the opposite angle; the difference shall be equal to twice the rectangle \( BC \times CD \); that is,

\[ AB^2 = AC^2 + BC^2 + 2BC \times CD. \]

For \( BD = BC + CD \), and therefore (10), \( BD^2 = BC^2 + CD^2 + 2BC \times CD \); to each of these equals add \( AD^2 \), then, observing that \( AD^2 + DB^2 = AB^2 \), and \( AD^2 + DC^2 = AC^2 \),

\[ AB^2 = BC^2 + CA^2 + 2BC \times CD. \]

**Scholium.** It is only when a triangle has one of its angles a right angle, that the sum of the squares of two of its sides can be equal to the square of the third side; for if the angle contained by those sides be acute, the sum of their squares is greater than the square of the opposite side, and if the angle be obtuse, that sum is less than the square of the opposite side.

**Theorem XVI.**—If a straight line \( AE \) be drawn from the vertex of any triangle \( ABC \) to the middle of its base \( BC \); the sum of the squares of the sides is equal to twice the square of half the base, and twice the square of the line drawn from the vertex to the middle of the base; that is,

\[ AB^2 + AC^2 = 2BE^2 + 2AE^2. \]

Draw \( AD \) perpendicular to \( BC \), then

\[ AB^2 = BE^2 + EA^2 = 2BE \times ED \quad (14), \] and \( AC^2 = CE^2 + EA^2 + 2CE \times ED \quad (15)\);

therefore, by adding equals to equals, and observing that \( BE = CE \), and therefore \( BE^2 = CE^2 \), and \( 2BE \times ED = 2CE \times ED \),

\[ AB^2 + AC^2 = 2BE^2 + 2AE^2. \]

**Theorem XVII.**—A straight line \( DE \) drawn parallel to one of the sides of a triangle \( ABC \), divides the other two sides \( AB, AC \) proportionally, so that \( AD : DB = AE : EC \).

Join \( BE \) and \( CD \). The triangles \( BDE, CDE \), having the same base \( DE \), and the same altitude, are equivalent (2), and the triangles \( ADE, BDE \), having the same altitude, are to one another as their bases (6), that is, \( ADE : BDE = AD : DB \); the triangles \( ADE, CDE \), having also the same altitude, are to one another as their bases; that is, \( ADE : CDE = AE : EC \), but the triangle \( BDE \) has been proved equal to \( CDE \); therefore, because of the common ratio in the two proportions, we have (Ax. 3, 3)

\[ AD : DB = AE : EC. \]

Cor. Hence, by composition, \( AB : AD = AC : AE \); and \( AB : BD = AC : CE \).

**Theorem XVIII.**—Conversely, if two of the sides \( AB, AC \) of a triangle be divided proportionally by the straight line \( DE \), so that \( AD : DB = AE : EC \); then shall the line \( DE \) be parallel to the remaining side \( BC \).

For if \( DE \) is not parallel to \( BC \), suppose some other line \( DO \) to be parallel to \( BC \); then, \( AB : BD = AC : CO \) (17), and since by hypothesis \( AD : DB = AE : EC \); and consequently, by composition, \( AB : BD = AC : CE \); therefore, \( AC : CO = AC : CE \); therefore, \( CO = CE \) (2 Ax. 3), which is impossible; therefore \( DO \) is not parallel to \( BC \).

Cor. If it be supposed that \( BA : AD = CA : AE \); still \( DE \) will be parallel to \( BC \); for by division \( BD : AD = CE : AE \), this proportion being the same as in the theorem, the conclusion must be the same.

**Theorem XIX.**—A straight line \( AD \), which bisects the angle \( BAC \) of a triangle, divides the base \( BC \) into two segments proportional to the adjacent sides \( BA, AC \); that is, \( BD : DC = BA : AC \).

Through the point \( C \) draw \( CE \) parallel to \( AD \), so as to meet \( BA \) produced. In the triangle \( BCE \), the line \( AD \) is parallel to one of its sides \( CE \), therefore \( BD : DC = BA : AE \); now the triangle \( ACE \) is isosceles, for, because of the parallels \( AD, CE \), the angle \( ACE = DAC \), and the angle \( AEC = BAD \), (21, 1); but by hypothesis \( DAC = BAD \), therefore \( ACE = AEC \); and consequently, \( AE = AC \) (12, 1), therefore, substituting \( AC \) instead of \( AE \) in the above proportion, it becomes \( BD : DC = BA : AC \).

**Theorem XX.**—If two triangles be equiangular, their homologous sides are proportional, and the triangles are similar.

Let \( ABC, CDE \) be two equiangular triangles, which have the angle \( BAC = CDE \), \( ABC = DEC \), and \( ACB = DCE \); the homologous sides, or the sides adjacent to the equal angles, shall be proportional; that is, \( BC : CE = AB : CD = AC : DE \).

Place the homologous sides \( BC, CE \) in the same direction, and produce the sides \( BA, ED \), till they meet in \( F \). Because \( BCE \) is a straight line, and the angle \( BCA \) is equal to \( CED \), the lines \( CA, EF \) are parallel (22, 1); and in like manner, because the angle \( ABC = DCE \), the lines \( BF, CD \) are parallel; therefore the figure \( ACDF \) is a parallelogram, and hence \( AF = CD \), and \( CA = DF \) (26, 1). In the triangle \( BFE \) the line \( AC \) is parallel to the side \( FE \), therefore \( BC : CE = BA : AF \); or since \( AF = CD, BC : CE = BA : CD \). Again, in the same triangle, because \( CD \) is parallel to the side \( BF, BC : CE = FD : DE \), or, since \( FD = AC, BC : CE = AC : DE \); having now shown that \( BC : CE = BA : CD \), and that \( BC : CE = AC : DE \), it follows that \( BA : CD = AC : DE \); therefore the equiangular triangles \( BAC, CDE \) have their homologous sides proportional, and hence (Def. 2) the triangles are similar.

**Scholium.** It is manifest that the homologous sides are opposite to the equal angles.

**Theorem XXI.**—If two triangles have their homologous sides proportional, they are equiangular and similar.

Suppose that \( BC : EF = AB : DE = AC : DF \); then shall \( A = D, B = E, C = F \). At the point \( E \) make the Lines and angle FEG = B, and at the point F make EFG = C; then the third angle G shall be equal to the third angle A, and the two triangles ABC, GEF shall be equiangular; therefore, by the last theorem, BC : EF = AB : GE; but by hy- pothesis BC : EF = AB : DE; therefore GE = DE (Ax. 2, 3). In like manner, because by the same the- orem BC : EF = CA : FG; and by hypothesis BC : EF = CA : FD; therefore FG = FD; but it was shown that EG = ED, therefore, the triangles GEF, DEF, having the sides of the one equal to those of the other, each to each, are equal; but, by construction, the triangle GEF is equi- angular to ABC, therefore also the triangles DEF, ABC are equiangular and similar.

**Theorem XXII.**—Two triangles which have an angle of the one equal to an angle of the other, and the sides about these angles proportional, are similar.

Let the angle A = D, and let AB : DE = AC : DF,

Fig. 85.

the triangle ABC is similar to DEF. Take AG = DE, and draw GH parallel to BC, then the angle AGH = ABC (21, I), therefore the triangle AGH is equiangular to ABC, and consequently (20) AB : AG = AC : AH; but by hypothesis AB : DE = AC : DF, and by construction AG = DE, therefore AH = DF; the two triangles AGH, DEF are therefore equal (5, 1), but the triangle AGH is similar to ABC, therefore DEF is similar to ABC.

**Theorem XXIII.**—In a right-angled triangle, if a perpen- dicular AD be drawn from the right angle upon the hypothe- nuse, then,

1. The triangles ABD, CAD on each side of the perpendicular are similar to the whole tri- angle BAC, and to one an- other.

2. Each side AB or AC is a mean proportional between the hypotenuse BC, and the adjacent segment BD or DC.

3. The perpendicular AD is a mean proportional between the two segments BD, DC.

1. The triangles BAD, BAC have the common angle B; besides, the right angle BAC is equal to the right angle BDA, therefore the third angle BAD of the one is equal to the third angle BCA of the other; therefore, these triangles are equiangular and similar; and in the same manner it may be shown that the triangle DAC is equiangular and similar to BAC; therefore the three tri- angles are equiangular and similar to each other.

2. Because the triangle BAD is similar to the triangle BAC, their homologous sides are proportional. Now the side BD of the lesser triangle is homologous to the side BA of the greater, because they are opposite to the equal angles BAD, BCA; in like manner BA, considered as a side of the lesser triangle, is homologous to the side BC of the greater, each being opposite to a right angle; there- fore, BD : BA = BA : BC. In the same manner it may be shown that CD : CA = CA : CB, therefore each side is a mean proportional between the hypotenuse and the segment adjacent to that side.

3. By comparing the homologous sides of the two simi- lar triangles ABD, ACD, it appears that BD : DA = DA : DC; therefore the perpendicular is a mean proportional between the segments of the hypotenuse.

**Theorem XXIV.**—Two triangles which have an angle of the one equal to an angle of the other, are to each other as the rectangles of the sides which con- tain the equal angles; that is, the triangle ABC is to the triangle ADE as the rectangle AB × AC to the rectangle AD × AE.

Join BE; because the triangles ABE, ADE, have a common vertex E, they have the same altitude, there- fore ABE : ADE = AB : AD (Cor. to 6); but AB : AD = AB × AE : AD × AE (3), there- fore,

ABE : ADE = AB × AE : AD × AE.

In the same manner, it may be demonstrated that

ABC : ABE = AB × AC : AB × AE;

therefore (8, 3) ABC : ADE = AB × AC : AD × AE.

Con. Therefore the two triangles are equivalent, if the rectangle AB × AC = AD × AE, or (8), if AB : AD = AE : AC, in which case the sides about the equal angles are said to be reciprocally proportional.

**Scholium.** What has been proved of triangles is also true of parallelograms, they being the doubles of such tri- angles.

**Theorem XXV.**—Two similar triangles are to each other as the squares of their homologous sides. (See fig. 85.)

Let the angle A = D, the angle B = E, and therefore the angle C = F,

then (20) AB : DE = AC : DF; now AB : DE = AB : DE,

for the two ratios are identical, therefore (9),

AB² : DE² = AB × AC : DE × DF;

but ABC : DEF = AB × AC : DE × DF (24), therefore ABC : DEF = AB² : DE² (Ax. 3, 3), therefore the two similar triangles ABC, DEF are to each other as the squares of the homologous sides AB, DE, or as the squares of any of the other homologous sides.

**Theorem XXVI.**—Similar polygons are composed of the same number of triangles which are similar and similar- ly situated.

In the polygon ABCDE, draw from one of the angles A the diagonals AC, AD to all the other angles. In the po- lygon FGHIK, draw in like manner from the angle F, homologous to A, the diagonals FH, FI to the other angles.

Because the polygons are similar, the angle ABC is equal to its homologous angle FGH (Def. 2), also the sides AB, BC are proportional to FG, GH; so that AB : FG = BC : GH, therefore the triangles ABC, FGH are similar (22); Lines and therefore the angle BCA = GHF, and these being taken from the equal angles BCD, GHI, the remainders ACD, FHI are equal; but the triangles ABC, FGH being similar, AC : FH = BC : GH, besides, because of the similarity of the polygons, BC : GH = CD : HI; therefore AC : FH = CD : HI; now it has been already shown that the angle ACD = FHI, therefore the triangles ACD, FHI are similar (22). It may be demonstrated in the same manner that the remaining triangles are similar, whatever be the number of sides of the polygon; therefore two similar polygons are composed of the same number of triangles similar to each other, and similarly situated.

**Theorem XXVII.**—The perimeters of similar polygons are as the homologous sides; and the polygons themselves are as the squares of the homologous sides. (See fig. 88.)

For, since by the nature of similar figures AB : FG = BC : GH = CD : HI, &c. therefore (12, 3) \(AB + BC + CD\), &c. the perimeter of the first figure, is to \(FG + GH + HI\), &c. the perimeter of the second, as the side AB to its homologous side FG.

Again, because the triangles ABC, FGH are similar, \(ABC : FGH = AC^2 : FH^2\) (25), in like manner \(ACD : FHI = AC^2 : FH^2\); therefore,

\[ABC : FGH = ACD : FHI.\]

By the same manner of reasoning,

\[ACD : FHI = ADE : FIK,\]

and so on if there be more triangles; hence, from this series of equal ratios, it follows (12, 3) that \(ABC + ACD + ADE\), or the polygon ABCDE, is to \(FGH + FHI + FIK\), or the polygon FGHIK, as one of the antecedents ABC, is to its consequent FGH, or as \(AB^2\) to \(FG^2\); therefore similar polygons are to each other as the squares of their homologous sides.

Cor. 1. If three similar figures have their homologous sides equal to the three sides of a right-angled triangle, the figure having the greatest side shall be equal to the two others; for these three figures are proportional to the squares of their homologous sides, and the square of the hypotenuse is equal to the squares of the other two sides.

Cor. 2. Similar polygons have to each other the duplicate ratio of their homologous sides. For let L be a third proportional to the homologous sides AB, FG, then (Def. 10, 3) AB has to L the duplicate ratio of AB to FG; but \(AB : L = AB^2 : AB \times L\) (3), or since \(AB \times L = FG^2\), (Cor. to 8) \(AB : L = AB^2 : FG^2 = ABCDE : FGHIK\); therefore the figure ABCDE has to the figure FGHIK, the duplicate ratio of \(AB\) to \(FG\).

**Theorem XXVIII.**—The segments of two chords AB, CD, which cut each other within a circle, are reciprocally proportional, that is, \(AO : DO = CO : OB\).

Join AC and BD; and because the triangles AOC, BOD have the angles at O equal (4, 1), and the angle A = D and the angle C = B (15, 2), the triangles are similar; therefore the homologous sides are proportional (20), that is, \(AO : DO = CO : BO\).

Con. Hence \(AO \times BO = CO \times DO\) (8), that is, the rectangle contained by the segments of the one chord is equal to the rectangle contained by the segments of the other.

**Theorem XXIX.**—If from a point O without a circle two straight lines be drawn, terminating in the concave arch Lines and BC; the whole lines shall be reciprocally proportional to the parts of them without the circle, that is, \(OB : OC = OD : OA\).

Join AC, BD; then the triangles OAC, OBD have the common angle O, also the angle B = C (15, 2), therefore the triangles are similar, and the homologous sides are proportional, that is, \(OB : OC = OD : OA\).

Con. Therefore (8) \(OA \times OB = OC \times OD\), that is, the rectangles contained by the whole lines, and the parts of them without the circle, are equal to one another.

**Theorem XXX.**—If from a point O without a circle a straight line OA be drawn touching the circle, and also a straight line OC cutting it; the tangent shall be a mean proportional between the whole line which cuts the circle, and the part of it without the circle; that is, \(OC : OA = OA : OD\).

For if AC, AD be joined, the triangles OAD, OCA, have the angle at O common to both, also the angle ACD or ACO equal to DAO (18, 2), therefore the triangles are similar (20), and consequently \(CO : OA = OA : OD\).

Con. Therefore (Cor. to 8) \(CO \times OD = OA^2\); that is, the square of the tangent is equal to the rectangle contained by the whole line which cuts the circle, and the part of it without the circle.

**Theorem XXXI.**—In the same circle, or in equal circles, any angles ACB, DEF are to each other as the arches AB, DF of the circles intercepted between the lines which contain the angles.

Suppose the arch AB to have to the arch DF the ratio of the number \(p\) to the number \(q\); then the arch AB being supposed divided into equal parts \(Ag, gh, hB\), the number of which is \(p\), the arch DF shall contain \(q\) equal parts \(Dk, kl, lm, mn, nF\), each of which is equal to any one of the equal parts into which AB is divided. Draw straight lines from the centres of the circles to the points of division; these lines will divide ACB into \(p\) angles and DEF into \(q\) angles, which are all equal (13, 2); therefore the angle ACB has to the angle DEF the ratio of the number \(p\) to the number \(q\), which ratio is the same as that of the arch AB to the arch DF.

Con. Hence it appears that angles may be measured and compared with each other by means of arches of circles described on the vertices of the angles as centres; observing, however, that the radii of the circles must be equal. **SECT. V.—PROBLEMS.**

**Problem I.**—To bisect a given straight line \( AB \); that is, to divide it into two equal parts.

From the points \( A \) and \( B \) as centres, with any radius greater than the half of \( AB \), describe arches, cutting each other in \( D \) and \( D \) on each side of the line \( AB \). Draw a straight line through the points \( D, D \), cutting \( AB \) in \( C \); the line \( AB \) is bisected in \( C \).

For the points \( D, D \), being equally distant from the extremities of the line \( AB \), are each in a straight line perpendicular to the middle of \( AB \) (16, 1), therefore the line \( DCD \) is that perpendicular, and consequently \( C \) is the middle of \( AB \).

**Problem II.**—To draw a perpendicular to a given straight line \( BC \), from a given point \( A \) in that line.

Take the points \( B \) and \( C \) at equal distances from \( A \); and on \( B \) and \( C \) as centres, with any radius greater than \( BA \), describe arches, cutting each other in \( D \); draw a straight line from \( A \) through \( D \), which will be the perpendicular required. For the point \( D \), being at equal distances from the extremities of the line \( BC \), must be in a perpendicular to the middle of \( BC \) (16, 1), therefore \( AD \) is the perpendicular required.

**Problem III.**—To draw a perpendicular to a given line \( BD \), from a given point \( A \) without that line.

On \( A \) as a centre, with a radius sufficiently great, describe an arch, cutting the given line in two points \( B, D \); and on \( B \) and \( D \) as centres, with a radius greater than the half of \( BD \), describe two arches, cutting each other in \( F \); draw a straight line through the points \( A \) and \( F \), meeting \( BD \) in \( C \); the line \( AC \) is the perpendicular required.

For the two points \( A \) and \( F \) are each at equal distances from \( B \) and \( D \); therefore a line passing through \( A \) and \( F \) is perpendicular to the middle of \( BD \) (16, 1).

**Problem IV.**—At a given point \( A \), in a given line \( AB \), to make an angle equal to a given angle \( K \).

On \( K \) as a centre, with any radius, describe an arch to meet the lines containing the angle \( K \) in \( L \) and \( I \); and on \( A \) as a centre, with the same radius, describe an indefinite arch \( BO \); on \( B \) as a centre, with a radius equal to the chord \( LI \), describe an arch, cutting the arch \( BO \) in \( D \); draw \( AD \), and the angle \( DAB \) shall be equal to \( K \).

For the arches \( BD, LI \) having equal radii and equal chords, the arches themselves are equal (4, 2), therefore the angles \( A \) and \( K \) are also equal (13, 2).

**Problem V.**—To bisect a given arch \( AB \), or a given angle \( C \).

First, to bisect the arch \( AB \), on \( A \) and \( B \) as centres, with one and the same radius, describe arches to intersect in \( D \); join \( CD \), cutting the arch in \( E \), and the arch \( AE \) shall be equal to \( EB \).

For, since the points \( C \) and \( D \) are at equal distances from \( A \), and also from \( B \), the line which joins them is perpendicular to the middle of the chord \( AB \) (16, 1), therefore the arch \( AB \) is bisected at \( E \) (6, 2).

Secondly, to bisect the angle \( C \); on \( C \) as a centre, with any distance, describe an arch, meeting the lines containing the angle in \( A \) and \( B \); then find the point \( D \) as before, and the line \( CD \) will manifestly bisect the angle \( C \) as required.

**Scholium.** By the same construction we may bisect each of the arches \( AE, EB \); and again we may bisect each of the halves of these arches, and so on; thus, by successive subdivisions, an arch may be divided into four, eight, sixteen parts, &c.

**Problem VI.**—Through a given point \( A \), to draw a straight line parallel to a given straight line \( BC \).

On \( A \) as a centre, with a radius sufficiently large, describe the indefinite arch \( EO \); on \( E \) for a centre, with the same radius, describe the arch \( AF \); in \( EO \) take \( ED \) equal to \( AF \), draw a line from \( A \) through \( D \), and \( AD \) will be parallel to \( BC \).

For if \( AE \) be joined, the angle \( EAD \) is equal to \( AEB \) (13, 2), and they are alternate angles, therefore \( AD \) is parallel to \( BC \) (22, 1).

**Problem VII.**—To construct a triangle, the sides of which may be equal to three given lines \( A, B, C \).

Take a straight line \( DE \), equal to one of the given lines \( A \); on \( D \) as a centre, with a radius equal to another of the lines \( B \), describe an arch; on \( E \) as a centre, with a radius equal to the remaining line \( C \), describe another arch, cutting the former in \( F \); join \( DF \) and \( EF \), and \( DEF \) will be the triangle required, as is sufficiently evident.

**Scholium.** It is necessary that the sum of any two of the lines be greater than the third line (7, 1).

**Problem VIII.**—To construct a parallelogram, the adjacent sides of which may be equal to two given lines \( A, B \), and the angle they contain equal to a given angle \( C \). Draw the straight line DE = A; make the angle GDE = C, and take DG = B; describe two arches, one on G as a centre, with a radius GF = DE, and the other on E, with a radius EF = DG; then DEFG shall be the parallelogram required.

For by construction the opposite sides are equal; therefore the figure is a parallelogram (27, 1), and it is so constructed, that the adjacent sides and the angle they contain have the magnitudes given in the problem.

Cor. If the given angle be a right angle, the figure will be a rectangle; and if the adjacent sides be also equal, the figure will be a square.

**Problem IX.—To find the centre of a given circle, or of a circle of which an arch is given.**

Take any three points A, B, D, in the circumference of the circle, or in the given arch, and having drawn the straight lines AB, BD, bisect them by the perpendiculars EG, FH; the point C where the perpendiculars intersect each other is the centre of the circle, as is evident from Theorem VI. Sect. II.

**Scholium.** By the very same construction, a circle may be found that shall pass through three given points A, B, C; or that shall be described about a given triangle ADC.

**Problem X.—To draw a tangent to a given circle through a given point A.**

If the given point A be in the circumference (fig. 102), draw the radius AC; and through A, draw AD perpendicular to AC, and AD will be a tangent to the circle (9, 2). But if the given point A be without the circle (fig. 103), draw AC to the centre, and bisect AC in O, and on O as a centre, with OA or OC as a radius, describe a circle which will cut the given circle in two points D and D'; join AD and AD', and each of the lines AD, AD' will be a tangent to the circle.

For, draw the radii CD, CD', then each of the angles ADC, ADC' is a right angle (17, 2); therefore AD and AD' are both tangents to the circle (9, 2).

Cor. The two tangents AD, AD' are equal to one another (17, 1).

**Problem XI.—To inscribe a circle in a given triangle ABC.**

Bisect A and B any two angles of the triangle by the straight lines AO, BO, which meet each other in O; from O draw OD, OE, OF, perpendiculars to its sides; these lines shall be equal to one another.

For in the triangles ODB, OEB, the angle ODB = OEB, and the angle OBD = OBE; therefore, the remaining angles BOD, BOE, are equal; and as the side OB is common to both triangles, they are equal to one another (6, 1), therefore the side OD = OE; in the same manner it may be demonstrated, that OD = OF; therefore the lines OD, OE, OF, are equal to one another; and consequently a circle described on O as a centre, with OD as a radius, will pass through E and F; and as the sides of the triangle are tangents to the circle (9, 2), it will be inscribed in the triangle.

**Problem XII.—Upon a given straight line AB, to describe a segment of a circle that may contain an angle equal to a given angle C.**

Produce AB towards D, and at the point B make the angle DBE equal to the given angle C; draw BO perpendicular to BE, and GO perpendicular to the middle of AB, meeting BO in O; on O as a centre, with OB as a radius, describe a circle, which will pass through A, and AMB shall be the segment required.

For since FE is perpendicular to BO, FE is a tangent to the circle; therefore the angle EBD (which is equal to C by construction) is equal to any angle AMB in the alternate segment (18, 2).

**Problem XIII.—To divide a straight line AB into any proposed number of equal parts; or into parts having to each other the same ratios that given lines have.**

First, let it be proposed to divide the line AB (fig. 106) into five equal parts. Through the extremity A draw an indefinite line AG, take AC of any magnitude, and take CD, DE, EF, and FG, each equal to AC, that is, take AG equal to five times AC; join GB, and draw CI parallel to GB, the line AI shall be one fifth part of AB, and AI being taken five times in AB, the line AB shall be divided into five equal parts.

For since CI is parallel to GB, the sides AG and AB are cut proportionally in C and I; but AC is the fifth part of AG; therefore AI is the fifth part of AB.

Next, let it be proposed to divide AB (fig. 107) into parts, having to each other the ratios that the lines P, Q, Problem XIV.—To find a fourth proportional to three given lines A, B, C.

Draw two straight lines DE, DF, containing any angle; on DE take DA = A, and DB = B, and on DF take DC = C; join AC and draw BX parallel to AC; then DX shall be the fourth proportional required.

For because BX is parallel to AC, DA : DB = DC : DX (17, 4), that is, A : B = C : DX; therefore DX is a fourth proportional to A, B, and C.

Cor. The same construction serves to find a third proportional to two lines A and B; for it is the same as a fourth proportional to the lines A, B, and B.

Problem XV.—To find a mean proportional between two straight lines A, B.

Upon any straight line DF take DE = A, and EF = B; and on DF as a diameter describe a semicircle DGF; draw EG perpendicular to DF, meeting the circle in G; the line EG shall be the mean proportional required.

For, if DG, FG be joined, the angle DGF is a right angle (17, 2), therefore, in the right-angled triangle DGF, GE is a mean proportional between DE and EF (23, 4).

Problem XVI.—To divide a given straight line AB into two parts, so that the greater may be a mean proportional between the whole line and the other part.

At B, one of the extremities of the line, draw BC perpendicular to AB, and equal to the half of AB; on C as a centre, with CB as a radius, describe a circle; join AC, Lines and Figures upon a Plane.

Fig. 107.

R have. Through A draw AG, and in AG take AC = P, CD = Q, DE = R; join EB, and draw CI and DK parallel to EB; the line AB shall be divided as required.

For, because of the parallels CI, DK, EB; the parts AI, IK, KB have to each other the same ratios that the parts AC, CD, DE have (17, 4), which parts are by construction equal to the given lines P, Q, R.

Problem XVII.—To make a square equivalent to a given parallelogram, or to a given triangle.

First, let ABCD be a given parallelogram, the base of which is AB, and altitude DE; find XY a mean proportional between AB and DE (by Problem 15), and XY shall be the side of the square required.

For since by construction AB : XY = XY : DE, therefore XY² = AB × DE (8, 4) = parallelogram ABCD (5, 4).

Next, let ABC be a given triangle (fig. 113), BC its base, and AD its altitude; find XY a mean proportional between half the base and the altitude, and XY shall be the side of the square required.

For since \( \frac{1}{2} BC : XY = XY : AD \); therefore (8, 4) XY² = \( \frac{1}{2} BC \times AD = \triangle ABC \) (6, 4).

Problem XVIII.—Upon a given line EF, to construct a rectangle EFGX equivalent to a given rectangle ABCD.

Find a fourth proportional to the three lines EF, AB, and AD (by Problem 14); draw EX perpendicular to EF, and equal to that fourth proportional, and complete the rectangle EFGX, which will have the magnitude required.

For since EF : AB = AD : EX, therefore (8, 4) EF × EX = AB × AD; that is, the rectangle EFGX is equal to the rectangle ABCD.

Problem XIX.—To make a triangle equivalent to a given polygon ABCDE.

First, draw the diagonal CE, so as to cut off the triangle CDE; draw DG parallel to CE, to meet AE produced in G; join CG, and the given polygon ABCDE shall be equivalent to another polygon ABCG, which has one side fewer.

For since DG is parallel to CE, the triangle CGE is equivalent to the triangle CDE (2 Cor. 2, 4); to each add the polygon ABCE, and the polygon ABCDE shall be equivalent to the polygon ABCG.

In like manner, if the diagonal CA be drawn, also BF parallel to CA, meeting EA produced, and CF be joined, the triangle CFA is equivalent to the triangle CBA, and thus the polygon ABCDE is transformed to the triangle CFG.

In this way a triangle may be found equivalent to any other polygon; for by transforming the figure into another equivalent figure that has one side fewer, and repeating the operation, a figure will at last be found which has only three sides.

Scholium. As a square may be found equivalent to a triangle, by combining the problem with Prob. XVII, a square may be found equivalent to any rectilineal figure whatever.

Problem XX.—Upon a given line FG to construct a polygon similar to a given polygon ABCDE. (Fig. 88.)

Draw the diagonals AC, AD; at the point F make the angle GFH = BAC, and at the point G make the angle FGH = ABC; thus a triangle FGH will be constructed similar to ABC. Again, on FH construct in like manner a triangle FIH, similar to ADC and similarly situated; and on FI construct a triangle FKI similar to AED, and similarly situated; and these triangles FGHI, FHI, FIK shall form a polygon FGHIK similar to ABCDE (26, 4).

Problem XXI.—To inscribe a square in a given circle.

Draw two diameters AC, BD, so as to intersect each other at right angles; join the extremities of the diameters A, B, C, D, and the figure ABCD shall be a square inscribed in the circle.

For the angles AOB, BOC, &c. being all equal, the chords AB, BC, CD, DA are equal; and as each of the angles of the figure ABCD is in a semicircle, it is a right angle (17, 2), therefore the figure is a square.

Problem XXII.—To inscribe a regular hexagon, and also an equilateral triangle, in a given circle.

From any point A in the circumference, apply AB and BC each equal to AO the radius; draw the three diameters AD, BE, CF, and join their adjacent extremities by the lines AB, BC, &c. and the figure ABCDEF thus formed is the hexagon required.

For the triangles AOB, BOC being by construction equilateral, each of the angles AOB, BOC is one third of two right angles (4, Cor. 24, 1); and since AOB + BOC + COD = two right angles, therefore COD = one third of two right angles; therefore the three angles AOB, BOC, COD are equal, and as these are equal to the angles AOF, FOE, EOD, the six angles at the centre are all equal; therefore the chords AB, BC, CD, DE, EF, FA are all equal; thus the figure is equilateral. It is also equiangular, for the angles FAB, ABC, &c. are in equal segments, each having for its base the chords of two sixths of the circumference; therefore the angles A, B, &c. are equal (13, 2).

If straight lines be drawn joining A, C, E, the vertices of the alternate angles of the hexagon, there will be formed an equilateral triangle inscribed in a circle, as is sufficiently evident.

Scholium. As the same manner of reasoning will apply alike to any equilateral polygon; it may be inferred that every equilateral polygon inscribed in a circle is also equiangular.

Problem XXIII.—To inscribe a regular pentagon and decagon in a given circle.

Draw any radius AO, and divide it into two parts AF, FO, such that AO : OF = OF : AF (16); from A place AG in the circumference equal to OF; join OG, and draw the chord AHB perpendicular to OG, the chord AB shall be a side of the pentagon required.

Join GF, and because AO : OF = OF : AF, and that AG = OF, therefore AO : AG = AG : AF; now the angle A is common to the two triangles OAG, GAF, and it has been shown that the sides about that angle in the two triangles are proportional; therefore (22, 4), the triangles are similar, and the triangle AOG being isosceles, the triangle AGF is also isosceles, so that AG = GF; but AG = FO (by construction), therefore, GF = FO, and the angle FOG = FGO, and FOG + FGO = 2FOG; but AFG = FOG + FGO (23, 1), and AFG = FAG, therefore FAG = 2FOG; hence in the isosceles triangle AOG, each of the angles at the base is double the vertical angle AOG; therefore the sum of all the angles is equal to five times the vertical angle AOG; but the sum of all the angles is equal to two right angles (24, 1), therefore the angle AOG is one fifth of two right angles, and consequently AOB = 2AOG = two fifths of two right angles, equal one fifth of four right angles, therefore the arch AB is one fifth of the whole circumference. If we now suppose straight lines BC, CD, DE, to be applied in the circle each equal to AB, the chord of one fifth of the circumference, and AE to be joined, the figure thus formed will be an equilateral pentagon, and it is also equiangular (Schol. 22). It is evident that AG is the side of a decagon inscribed in the circle.

Problem XXIV.—Having given ABCD, &c. a regular polygon inscribed in a circle; to describe a regular polygon of the same number of sides about the circle.

Draw GH a tangent to the circle at T the middle of the arch AB; do the same at the middle of each of the other arches BC, CD, &c.; these tangents shall form a regular polygon GHIK, &c. described about the circle.

Join OG, OH, &c. also OT and ON. In the triangles OTH, ONH, the side OT = ON, and OH is common to both, and OTH, ONH, are right angles, therefore the triangles are equal (17, 1) and the angle TOH = NOH; now B is the middle of the arch TN, therefore OH passes through B; and in the same manner it appears that I is in the line OC produced, &c. Now Lines and because OT bisects the arch AB, it is perpendicular to the chord AB (6, 2); therefore GH is parallel to AB (9, 2, and 18, 1), and HI to BC, therefore the angle GHO = ABO, and IHO = CBO, and hence GHI = ABC; and in like manner it appears that HIK = BCD, &c.; therefore the angles of the circumscribed polygon are equal to those of the inscribed polygon. And because of the parallels, GH : AB = OH : OB, and HI : BC = OH : OB, therefore GH : AB = HI : BC; but AB = BC; therefore GH = HI. For the same reason HI = IK, &c., therefore the polygon is regular, and similar to the inscribed polygon.

SECT. VI.—OF THE QUADRATURE OF THE CIRCLE.

AXIOM.

If ABC be an arch of a circle, and AD, CD be two tangents at its extremities, intersecting each other in D; the sum of the tangents AD, DC is greater than the arch ABC.

Cor. Hence the perimeter of any polygon described about a circle, is greater than the circumference of the circle.

PROP. I. THEOREM.—Equilateral polygons, ABCDEF, GHIKLM, of the same number of sides inscribed in circles are similar, and are to one another as the squares of the radii of the circles.

As each of the polygons is by hypothesis equilateral, it will also be equiangular (Schol. 22, 5). Let us suppose, for example, that the polygons are hexagons; then, as the sum of the angles is the same in both, viz. eight right angles (25, 1), the angle A will be one sixth part of eight right angles, and the angle G will be the same; therefore A = G; in like manner B = H, C = I, &c., and as the figures are equilateral, AB : GH = BC : HI = CD : IK, &c., therefore (2, Def. 4) the figures are similar. Draw AO, BO, GP, HP to the centres of the circles; then, because the angle AOB is the same part of four right angles that the arch AB is of the whole circumference, and the angle GPH the same part of four right angles that GH is of the whole circumference (13, 2), the angles AOB, GPH are each the same part of four right angles; therefore they are equal; the isosceles triangles AOB, GPH are therefore similar (22, 4), and consequently AB : GH = AO : GP; therefore (9 and 27, 4) polygon ABCDEF : polygon GHIKLM = AO² : GP².

PROP. II. THEOREM.—A circle being given, two similar polygons may be found, the one inscribed in the circle, and the other described about it, which shall differ from each other by a space less than any given space.

Let AG be the side of a square equal to the given space; and let ABG be such an arch of the given circle, Lines and that AG is its chord. Bisect the fourth part of the circumference (5, 5), then bisect one of its halves, and proceed in this manner till, by repeated bisections, there will at length be found an arch AB less than AG. As the arch thus found will be contained in the circumference.

Fig. 121.

Let ABC, &c. be any equilateral polygon inscribed in the circle, and DEF, &c. a similar polygon described about it; draw lines from the extremities of AB and DE, a side of each polygon, to O the centre; and let OKH be perpendicular to these sides. Put P for the perimeter of Lines and the polygon DEF, &c., and \( p \) for the perimeter of the polygon ABC, &c., and \( n \) for the number of the sides of each. Then, because \( n \times \frac{1}{2}DE = \frac{1}{2}P \), \( n \times \frac{1}{2}DE \times OH = \frac{1}{2}P \times OH \), but \( n \times \frac{1}{2}DE \times OH = n \times \text{triangle } DOE = \text{polygon } DEF, &c.; \) therefore, \( \frac{1}{2}P \times OH = \text{polygon } DEF, &c.; \) and in like manner it appears that \( \frac{1}{2}P \times OK = \text{polygon } ABC, &c. \)

Now let Q denote the circumference of the circle, then, because \( \frac{1}{2}Q > \frac{1}{2}p \), and \( OH > OK \), therefore \( \frac{1}{2}Q \times OH > \frac{1}{2}p \times OK \), that is, \( \frac{1}{2}Q \times OH \) is greater than the inscribed polygon. Again, because \( \frac{1}{2}Q < \frac{1}{2}P \) (Axiom), therefore \( \frac{1}{2}Q \times OH < \frac{1}{2}P \times OH \), that is, \( \frac{1}{2}Q \times OH \) is less than the circumscribed polygon.

Thus it appears that \( \frac{1}{2}Q \times OH \) is greater than any polygon inscribed in the circle, but less than any polygon described about it; therefore, \( \frac{1}{2}Q \times OH \) is equal to the circle (2 Cor. to 2).

**Prop. IV. Theorem.**—The areas of circles are to one another as the squares of their radii. (Fig. 120, 122.)

Let ABCDEF and GHIKLM be equilateral polygons of the same number of sides inscribed in the circles, and OA, PG their radii; and let Q be such a space, that \( AO^2 : GP^2 = \text{circle } ABE : Q \); then, because \( AO^2 : GP^2 = \text{polygon } ABCDEF : \text{polygon } GHIKLM \), and \( AO^2 : GP^2 = \text{circle } ABE : Q \), therefore \( \text{polygon } ABCDEF : \text{polygon } GHIKLM = \text{circle } ABE : Q \); but circle ABE \( \supseteq \) polygon GHIKLM; therefore \( Q \supseteq \text{polygon } GHIKLM \); that is, Q is greater than any polygon inscribed in the circle GHL. In the same manner it is demonstrated that Q is less than any polygon described about the circle GHL; therefore Q is equal to the circle GHL (2). And because \( AO^2 : GP^2 = \text{circle } ABE : Q \), therefore \( AO^2 : GP^2 = \text{circle } ABE : \text{circle } GHL \).

**Cor. 1.** The circumferences of circles are to one another as their radii. Put M for the circumference of the circle ABE and N for the circumference of GKL; then, circle ABE : circle GHL = \( AO^2 : GP^2 \); but \( \frac{1}{2}M \times AO = \text{circle } ABE \), also \( \frac{1}{2}N \times GP = \text{circle } GHL \) (3), therefore \( \frac{1}{2}M \times AO : \frac{1}{2}N \times GP = AO^2 : GP^2 \); and by alternation \( \frac{1}{2}M \times AO : \frac{1}{2}N \times GP = AO^2 : GP^2 \); therefore \( \frac{1}{2}M : AO = \frac{1}{2}N : GP \); and again, by alternation, \( \frac{1}{2}M : \frac{1}{2}N = AO : GP \), therefore \( M : N = AO : GP \).

**Cor. 2.** A circle described with the hypotenuse of a right-angled triangle as a radius, is equal to two circles described with the other two sides as radii. Let the sides of the triangle be \( a, b \) and the hypotenuse \( h \), and let the circles described with these lines as radii be A, B, and H.

Because \( A : H = a^2 : h^2 \) and \( B : H = b^2 : h^2 \),

therefore \( A + B : H = a^2 + b^2 : h^2 \) (10, 3);

but \( a^2 + b^2 = h^2 \) (13, 4), therefore \( A + B = H \).

**Prop. V. Problem.**—Having given the area of a regular polygon inscribed in a circle, and also the area of a similar polygon described about it; to find the areas of regular inscribed and circumscribed polygons, each of double the number of sides.

Let AB be the side of the given inscribed polygon, and EF parallel to AB that of the similar circumscribed polygon, and C the centre of the circle; if the chord AM, and the tangents AP, BQ be drawn, the chord AM shall be the side of the inscribed polygon of double the number of sides; and PQ or 2PM that of the similar circumscribed polygon. Put A for the area of the polygon, of which AB is a side, and B for the area of the circumscribed polygon; also \( a \) for the area of the polygon of which AM is a side, and \( b \) for the area of the similar circumscribed polygon; then A and B Figures upon a Plane.

**Fig. 123.**

I. The triangles ACD, ACM, which have a common vertex A, are to one another as their bases CD, CM; besides, these triangles are to one another as the polygons, of which they form like parts, therefore \( A : a = CD : CM \).

The triangles CAM, CME, which have a common vertex M, are to each other as their bases CA, CE; they are also to one another as the polygons \( a \) and B, of which they are like parts; therefore, \( a : B = CA : CE \); but because of the parallels DA, ME, CD : CM = CA : CE; therefore, \( A : a = a : B \); therefore the polygon \( a \), which is one of the two required, is a mean proportional between the two known polygons A and B, so that \( a = \sqrt{A \times B} \).

II. The triangles CPM, CPE, having the same altitude CM, are to one another as PM to PE. But as CP bisects the angle MCE, \( PM : PE = CM : CE \) (19, 4) = \( CD : CA = A : a \); therefore CPM : CPE = \( A : a \); and consequently CPM + CPE, or CME : CPM = \( A + a : A \), and CME : 2CPM = \( A + a : 2A \); but CME and 2CPM, or CMPA, are to one another as the polygons B and \( b \), of which they are like parts; therefore \( A + a : 2A = B : b \).

Now the polygon \( a \) has been already found, therefore by this last proportion the polygon \( b \) is determined; that is,

\[ b = \frac{2A \times B}{A + a} \]

**Prop. VI. Problem.**—To find nearly the ratio of the circumference of a circle to its diameter.

Let the radius of the circle = 1, then the sides of the inscribed square being the hypotenuse of a right-angled triangle of which the radii are the sides (see fig. 115), the area of the inscribed square will be 2 (13, 4), and the circumscribed square, being the square of the diameter, will be 4. Now, retaining the notation of last problem, if we make \( A = 2 \), and \( B = 4 \), the formula \( a = \sqrt{A \times B} \),

\[ b = \frac{2A \times B}{A + a} \]

gives us \( a = 2.8284271 \), &c. the area of the inscribed octagon, and \( b = 3.13137085 \), &c. the area of the circumscribed octagon. By substituting these numbers in the formula, instead of A and B, we shall obtain the areas of the inscribed and circumscribing polygons of 16 sides; and thence we may find those of 32 sides, and so on, as in the following table:

| No. of Sides | Ins. Polygons | Circ. Polygons | |--------------|---------------|---------------| | 4 | 2.000000 | 4.000000 | | 8 | 2.8284271 | 3.3137085 | | 16 | 3.0614674 | 3.1825979 | | 32 | 3.1214451 | 3.1517249 | | 64 | 3.1365485 | 3.1441184 | | 128 | 3.1403311 | 3.1422236 | | 256 | 3.1412772 | 3.1417504 | | 512 | 3.1415138 | 3.1416321 | | 1024 | 3.1415729 | 3.1416025 | | 2048 | 3.1415877 | 3.1415951 | | 4096 | 3.1415914 | 3.1415913 | | 8192 | 3.1415923 | 3.1415928 | | 16384 | 3.1415925 | 3.1415927 | | 32768 | 3.1415926 | 3.1415926 |

Hence it appears that the area of a regular polygon of 32768 sides inscribed in the circle, and of a similar poly- Geometry, gon described about it, differ so little from each other that the numbers which express them are the same as far as the eighth decimal place. And as the circle is greater than the one polygon, and less than the other, its area will be nearly $3\cdot1415926$. But the area is the product of the radius and the half of the circumference; therefore the radius being unity, half the circumference is $3\cdot1415926$ nearly; and the radius is to half the circumference, or the diameter is to the circumference, nearly as 1 to $3\cdot1415926$.

Archimedes, by means of inscribed and circumscribed polygons of 96 sides, found that the diameter is to the circumference as 7 to 22 nearly, which ratio is nearer to the truth than can be expressed by any smaller numbers; and Metius found it to be more nearly as 113 to 355. Both of these expressions are convenient on account of the smallness of the numbers, but later mathematicians Geometry, have carried the approximation to a much greater degree of accuracy. Dr Rutherford has lately calculated the ratio of the circumference of a circle to its diameter to the extent of 208 places of decimals, which is 58 places farther than it had been previously carried. The result is contained in a paper which was published in the Transactions of the Royal Society, and is as follows: $\pi = 3\cdot14159265358979$ $3238462643383279502884197169399375105809749459$ $2307816406286280899662803482524211706798214808651$ $328230664709384460955058223172535940812848473781$ $392038638380215747399600825931269129401832806517$ $44$, &c. The formula by which the above was calculated is $\frac{\pi}{4} = 4 \tan^{-1} \frac{1}{70} - \tan^{-1} \frac{1}{99}$.

PART II.—GEOMETRY OF SOLIDS.

SECT. I.—OF PLANES AND SOLID ANGLES.

Definitions.

I. A straight line is perpendicular, or at right angles, to a plane, when it is perpendicular to every straight line which it meets in that plane. The plane is also perpen- dicular to the line.

II. A line is parallel to a plane when they cannot meet each other, although both be produced. The plane is also parallel to the line.

III. Parallel planes are such as cannot meet each other, though produced.

IV. It will be demonstrated (Theor. 3) that the com- mon section of two planes is a straight line; this being premised, the inclination of two planes is the angle con- tained by two straight lines drawn perpendicular to the line which is their common section, from any point in it, the one perpendicular being drawn in the one plane, and the other in the other plane.

This angle may be either acute or obtuse.

V. If it be a right angle, the two planes are perpendicu- lar to each other.

VI. A solid angle is that which is made by the meet- ing of more than two plane angles, which are not in the same plane, in one point. Thus the solid angle V is formed by the plane angles AVB, BVC, CVD, DVA. (Fig. 133.)

Theorem I.—One part of a straight line cannot be in a plane and another part above it.

For from the definition of a plane (Sect. 1, Part 1), it is manifest, that if a straight line coincide with a plane in two points, it must be wholly in the plane.

Theorem II.—Two straight lines which cut each other in a plane determine its position; that is, the plane can coincide with these lines only in one position.

Let the straight lines AB, AC cut each other in A; conceive a plane to pass through AB, and to be turned about that line, till it pass through the point C; and this it can manifestly do only in one position; then, as the points A and C are in the plane, the whole line AC must be in the plane; there- fore there is only one position in which the plane can coincide with the same two lines AB, AC.

Cor. Therefore a triangle ABC, or three points A, B C, not in a straight line, determine the position of plane.

Theorem III.—If two planes AB, CD intersect each other; their intersection is a straight line.

Let E and F be two points in the line of common section, and let a straight line EF be drawn between them; then the line EF must be in the plane AB (Def. 1), and the same line must also be in the plane CD; therefore it must be the common section of them both.

Theorem IV.—If a straight line AP be perpendicular to two straight lines PB, PC at P, the point of their in- tersection; it will also be perpendicular to the plane MN, in which these lines are.

Draw any other line PQ in the plane MN, and from Q, any point in that line, draw QD parallel to PB; make DC = DP; join CQ, meet- ing PB in B; and join AB, AQ, AC. Because DQ is parallel to PB, and PD = DC; therefore BQ = QC, and BC is bisected in Q. Hence in the triangle BAC, $$AB^2 + AC^2 = 2AQ^2 + 2BQ^2$$ (16, 4), and in the like manner, in the triangle PBC, $$PB^2 + PC^2 = 2PQ^2 + 2CQ^2$$; therefore, taking equal quantities from equal quantities, that is, subtracting the two last quantities, which are put equal to each other, from the two first, and observing, that APB, APC are by hypothesis right-angled triangles, $$AB^2 - BP^2 = AP^2$$, and $$AC^2 - CP^2 = AP^2$$, we have $$AP^2 + AP^2 = 2AQ^2 - 2PQ^2$$, and therefore $$AP^2 = AQ^2 - PQ^2$$, or $$AP^2 + PQ^2 = AQ^2$$; therefore the triangle APQ is right-angled at P (Schol. 15, 4, Part 1), and consequently AP is perpendicular to the plane MN (Def. 1). Geometry Cor. 1. The perpendicular AP is shorter than any oblique line AQ; therefore it measures the distance of the point A from the plane.

Cor. 2. From the same point P in a plane no more than one perpendicular can be drawn. For if it be possible that there can be two perpendiculars, conceive a plane to pass through them, and to intersect the plane MN in the straight line PQ; then these perpendiculars will be in the same plane, and both perpendicular to the same line PQ, at the same point P in that line, which is impossible.

It is also impossible that from a point without a plane two perpendiculars can be drawn to the plane; for if the straight lines AP, AQ could be two such perpendiculars, then the triangle APQ would have two right angles, which is impossible.

Theorem V.—If a straight line AP be perpendicular to a plane MN, every straight line DE parallel to AP is perpendicular to the same plane.

Let a plane pass through the parallel lines AP, DE, and intersect the plane MN in the line PD; through D draw BC at right angles to PD; take DC = DB, and join PB, PC, AB, AC, AD. Because DB = DC, therefore PB = PC (Cor. 5, I, Part L); and because AP is perpendicular to the plane MN, so that APB, APC are right angles, AB = AC, therefore ABC is an isosceles triangle; and since its base BC is bisected at D, BC is perpendicular to AD (Schol. II, I, P. L); but by construction BC is perpendicular to PD; therefore (4) BC or BD is perpendicular to the plane passing through the lines AD and PD, or AP and DE; hence BD is perpendicular to DE, but PD is also perpendicular to DE (19, I, P. L), therefore DE is perpendicular to the two lines DP, DB; and therefore it is perpendicular to the plane MN passing through them.

Cor. 1. Conversely, if the straight lines AP, DE be perpendicular to the same plane MN, they are parallel; for if not, through D draw a parallel to AP; this parallel will be perpendicular to the plane MN (by the theorem), therefore, from the same point D two perpendiculars may be drawn to a plane, which is impossible (4).

Cor. 2. Two straight lines A and B which are parallel to a third line C, though not in the same plane, are parallel to each other. For suppose a plane to be perpendicular to the line C, the lines A and B parallel to this perpendicular are perpendicular to the same plane; therefore, by the preceding corollary, they are parallel between themselves.

Theorem VI.—Two planes MN, PQ, perpendicular to the same straight line AB, are parallel to each other.

For, if they can meet each other, let O be a point common to both, and join OA, OB; then the line AB, which is perpendicular to the plane MN, must be perpendicular to AO, a line drawn in the plane MN from the point in which AB meets that plane. For the same reason AB is perpendicular to BO; therefore OA, OB are two perpendiculars drawn from the same point O, AB, which is impossible.

Theorem VII.—The intersections EF, GH of two parallel planes MN, PQ, with a third plane FG, are parallel.

For if the lines EF, GH, situated in the same plane, are not parallel, they must meet if produced; therefore the planes MN, PQ, in which they are, must also meet, which is contrary to the hypothesis of their being parallel.

Theorem VIII.—Any straight line AB, perpendicular to MN one of two parallel planes MN, PQ, is also perpendicular to PQ the other plane. (Fig. 128.)

From B draw any straight line BC in the plane PQ, and let a plane pass through the lines AB, BC, and meet the plane MN in the line AD, then AD will be parallel to BC (7); and since AB is perpendicular to the plane MN, it must be perpendicular to the line AD, therefore it is also perpendicular to BC (19, I, P. L); hence (Def. I) the line AB is perpendicular to the plane PQ.

Theorem IX.—Parallel straight lines EG, FH, comprehended between two parallel planes MN, PQ, are equal. (Fig. 129.)

Let a plane pass through the lines EG, FH, and meet the parallel planes in EF and GH; then EF and GH are parallel (7) as well as EG and FH; therefore EGHF is a parallelogram, and EG = HF.

Cor. Hence two parallel planes are everywhere at the same distance from each other. For if EG and FH be perpendicular to the two planes, they are parallel (I Cor. 5); therefore they are equal.

Theorem X.—If two straight lines CA, EA, meeting one another, be parallel to two other lines DB, FB, that meet one another, though not in the same plane with the first two; the first two and the other two shall contain equal angles, and the plane passing through the first two shall be parallel to the plane passing through the other two.

Take AC = BD, AE = BF, and join CE, DF, AB, CD, EF. Because AC is equal and parallel to BD, the figure ABDG is a parallelogram; therefore CD is equal and parallel to AB. For a similar reason EF is equal and parallel to AB; therefore also CE is equal and parallel to DF (Cor. 5, and 28, I, P. L); therefore the triangles CAE, DBF are equal (10, I, P. L), hence the angle CAE = DBF.

In the next place, the plane ACE is parallel to the plane BDF; for suppose that the plane parallel to BDF, passing through the point A, meets the lines CD, EF in any other points than C and E (for example in G and H), then (9) the three lines AB, GD, FH are equal; but the three lines AB, CD, EF have been shown to be equal; therefore CD = GD, and FH = EF, which is absurd, therefore the plane ACE is parallel to BDF.

Theorem XI.—If three straight lines AB, CD, EF, not situated in the same plane, are equal and parallel; the triangles ACE, BDF formed by joining the extremities of these lines are equal, and their planes parallel. (See fig. 130.) Geometry of Solids.

For since \(AB\) is equal and parallel to \(CD\), the figure \(ABDC\) is a parallelogram; therefore the side \(AC\) is equal and parallel to \(BD\). In like manner it may be shown that the sides \(AE, BF\) are equal and parallel; also \(CE, DF\); therefore the triangles \(CAE, BDF\) are equal. It may be demonstrated, as in last proposition, that their planes are parallel.

**Theorem XII.**—If a straight line \(AP\) be perpendicular to a plane \(MN\); any plane \(APB\), passing through \(AP\), shall be perpendicular to the plane \(MN\).

Let \(BC\) be the intersection of the planes \(AB, MN\): if in the plane \(MN\) the line \(DE\) be drawn perpendicular to \(BP\), the line \(AP\), being perpendicular to the plane \(MN\), shall be perpendicular to each of the straight lines \(BC, DE\); therefore the angle \(APD\) is a right angle; now \(PA\) and \(PD\) are drawn in the planes \(AB, MN\) perpendicular to their common section, therefore (5 Def.) the planes \(AB, MN\) are perpendicular to each other.

**Scholium.**—When three straight lines, such as \(AP, BP, DP\), are perpendicular to each other, each is perpendicular to the plane of the two other lines.

**Theorem XIII.**—If the plane \(AB\) be perpendicular to the plane \(MN\), and if in the plane \(AB\) a straight line \(PA\) be drawn perpendicular to \(BP\), the common intersection of the planes; then shall \(PA\) be perpendicular to the plane \(MN\). (Fig. 131.)

For, if in the plane \(MN\), a line \(PD\) be drawn perpendicular to \(PB\), the angle \(APD\) shall be a right angle, because the planes are perpendicular to each other; therefore the line \(AP\) is perpendicular to the two lines \(PB, PD\), and therefore it is perpendicular to their plane \(MN\).

Con. If the plane \(AB\) be perpendicular to the plane \(MN\), and from any point \(P\), in their common intersection, a perpendicular be drawn to the plane \(MN\); this perpendicular shall be in the plane \(AB\); for if it is not, a perpendicular \(AP\) may be drawn in the plane \(AB\) to the common intersection \(BP\), which will be at the same time perpendicular to the plane \(MN\); therefore, at the same point \(P\), there may be two perpendiculars to a plane \(NM\), which is impossible (4).

**Theorem XIV.**—If two planes \(AB, AD\) be perpendicular to a third; their common intersection \(AP\) is perpendicular to the third plane. (Fig. 131.)

For, if through the point \(P\) a perpendicular be drawn to the plane \(MN\), this perpendicular shall be in the plane \(AB\), and also in the plane \(AD\) (Cor. 12); therefore it is at their common intersection \(AP\).

**Theorem XV.**—If two straight lines be cut by parallel planes; they shall be cut in the same ratio.

Let the line \(AB\) meet the planes \(MN, PQ, RS\), in \(A, E, B\); and let \(CD\) meet them in \(C, F, D\); then shall \(AE : EB = CF : FD\). For draw \(AD\) meeting the plane \(PQ\) in \(G\), and join \(AC, EG, GF, BD\); the lines \(EG, BD\) being the common sections of the plane of the triangle \(ABD\) and the parallel planes \(PQ, RS\), they are parallel (7); and in Geometry like manner it appears, that \(AC, GF\) are parallel; therefore \(AE : EB (= AG : GD) = CF : FD\).

**Theorem XVI.**—If a solid angle be contained by three plane angles; the sum of any two of these is greater than the third.

It is evidently only necessary to demonstrate the theorem, when the plane angle which is compared with the sum of the other two is greater than either of them; for, if it were equal to or less than one of them, the theorem would be manifest; therefore let \(S\) be a solid angle formed by three plane angles \(ASB, ASC, BSC\), of which \(ASB\) is the greatest. In the plane \(ASB\) make the angle \(BSD = BSC\); draw any straight line \(ADB\), and having taken \(SC = SD\), join \(AC, BC\); the triangles \(BSC, BSD\), having two sides, and the included angle of the one equal to two sides, and the included angle of the other, each to each, are equal (5, 1, P.I.), therefore \(BD = BC\); now \(AB \perp AC + BC\), therefore, taking \(BD\) from the first of these unequal quantities, and \(BC\) from the second, we get \(AD \perp AC\); and as the triangles \(ASD, ASC\) have \(SD = SC\), and \(SA\) common to both, and \(AD \perp AC\), therefore (9, 1, P. I.) the angle \(ASD \perp ASC\); and, adding \(DSB\) to the one, and \(CSB\) to the other, \(ASB \perp ASC + BSC\).

**Theorem XVII.**—The sum of all the plane angles which form any solid angle is less than four right angles.

Let the solid angle \(V\) be cut by any plane \(ABCDE\); from a point \(O\) taken in this plane, draw to all its angles the lines \(OA, OB, OC, OD, OE\). The sum of all the angles of the triangles \(AVB, BVC, \ldots\) formed about the vertex \(V\) is equivalent to the sum of the angles of a like number of triangles \(AOB, BOC, \ldots\) formed about the point \(O\). Now at \(B\) the angle \(ABC\), which is the sum of \(OBA\) and \(OBC\), is less than the sum of the angles \(VBA, VBC\) (16); in like manner at \(C, OCB + OCD < VCB + VCD\), and so on with all the angles of the polygon \(ABCDE\). Hence, in the triangles of which the common vertex is \(O\), the sum of the angles at their bases is less than the sum of the angles at the bases of the triangles which have their vertex at \(V\); therefore the sum of the angles about the point \(O\) is greater than the sum of the angles about the point \(V\). But the sum of the angles about \(O\) is equal to four right angles, therefore the sum of the plane angles which form the solid angle about the vertex \(V\) is less than four right angles.

**Scholium.** This demonstration supposes that the solid angle is convex, or that it lies all on one side of the plane of any one of its faces. If it were otherwise, the sum of the solid angles would be unlimited.

**Theorem XVIII.**—If each of two solid angles be contained by three plane angles equal to one another, each to each; the planes in which the equal angles are, have the same inclination to one another. Let the angle ASB = DTE, the angle ASC = DTF, and the angle BSC = ETF; the two planes ASB, ASC shall have to each other the same inclination as the two planes DTE, DTF.

Take A any point in SA, and in the two planes ASB, ASC draw AB and AC perpendiculars to AS, then (Def. 4) the angle BAC is the inclination of these planes; again, take TD = SA, and in the planes TDE, TDF draw DE and DF perpendiculars to TD, and the angle EDF shall be the inclination of these other planes; join BC, EF. The triangles ASB, DTE have the side AS = DT, the angle SAB = TDE and ASB = DTE, therefore the triangles are equal, and thus AB = DE, and SB = TE. In like manner it appears that the triangles ASC, DTF are equal, and therefore that AC = DF, and SC = TF. Now the triangles BSC, ETF, having BS = TE, SC = TF, and the angle BSC = ETF, are also equal, and therefore BC = EF; but it has been shown that AB = DE, and that AC = DF; therefore the triangles BAC, EDF are equal, and consequently the angle BAC = EDF; that is, the inclination of the planes ASB and ASC is equal to the inclination of the planes DTE and DTF. In the same manner it may be proved that the other planes have the same inclination to one another.

Scholium. If the three plane angles which contain the solid angles are equal, each to each; and if the angles are also disposed in the same order in the two solid angles, then these angles when applied to one another will coincide and be equal. But if the plane angles be disposed in a contrary order, the solid angles will not coincide, although the theorem is equally true in both cases. In this last case, the solid angles are called Symmetrical angles, and they must be accounted equal, because they are equal in every thing that determines their magnitude.

SECT. II.—OF SOLIDS BOUNDED BY PLANES.

Definitions.

I. A Solid is that which has length, breadth, and thickness.

II. A Prism is a solid contained by plane figures, of which two that are opposite are equal, similar, and parallel; and the others are parallelograms.

To construct this solid, let ABCDE be any polygon, fig. 136; if in a plane parallel to ABC there be drawn straight lines FG, GH, HI, &c. equal and parallel to the sides AB, BC, CD, &c. so as to form a polygon FGHIK equal to ABCDE, and straight lines AF, BG, CH, &c. be drawn, joining the vertices of the homologous angles in the two planes; the planes or faces ABGF, BCHG, &c. thus formed will be parallelograms; and the solid ABCDEFGHIK contained by these parallelograms and the two polygons is the prism itself.

III. The equal and parallel polygons ABCDE, FGHIK are called the Bases of the prism; and the distance between the bases is its Altitude.

IV. When the base of a prism is a parallelogram, and consequently the figure has all its faces parallelograms; it is called a parallelopiped. A parallelopiped is rectangular, when all its faces are rectangles.

V. A Cube is a rectangular parallelopiped contained by six equal squares.

VI. A Pyramid is a solid contained by several planes, which meet in the same point V, and terminate in a polygonal plane ABCDE. Fig. 134.

VII. The polygon ABCDE is called the Base of the pyramid; the point V is its Vertex; and a perpendicular let fall from the vertex upon the base is called its Altitude.

VIII. Two solids are similar when they are contained by the same number of similar planes, similarly situated, and having like inclinations to one another.

Theorem I.—Two prisms are equal when the three planes which contain a solid angle of the one are equal to the three planes which contain a solid angle of the other, each to each, and are similarly situated.

Let the base ABCDE be equal to the base abcde, the parallelogram ABGF equal to the parallelogram abgf, and the parallelogram BCHG equal to the parallelogram bchg; the prism ABCI shall be equal to the prism abci.

For let the base ABCDE be applied to its equal the base abcde, so that they may coincide with each other; then, because the three plane angles which form the solid angle B are equal to the three plane angles which form the angle b, each to each, viz. ABC = abc, ABG = abg, and GBC = gbc, and because these angles are similarly situated, the solid angles B and b are equal (15, 1), therefore the side BG shall fall upon the side bg; and because the parallelograms ABGF, abgf are equal, the side FG shall fall upon its equal fg; in like manner it may be shown that GH falls upon gh, therefore the upper base FGHIK coincides entirely with its equal fghik, and the two solids coincide with each other, or occupy the same space, therefore the prisms are equal.

Cor. Two right prisms which have equal bases and equal altitudes are equal to one another. If the equal angles of the lower bases follow each other in the same order, then the three planes which contain each solid angle of the one prism will be respectively equal to the three planes which contain a solid angle of the other prism, and will be similarly situated; and when the one solid angle is applied to the other, these planes will coincide, and the prisms will exactly coincide. If the equal angles of the lower base follow each other in a contrary order, then, by inverting one of the prisms, so that its upper may become its lower base, the angles of the two bases will follow each other in the same order; so that in either case the prisms coincide and are equal.

Theorem II.—In any parallelopiped, the opposite planes are equal and parallel.

From the nature of the solid (4 Def.), the bases ABCD, EFGH are equal parallelograms, and their sides are parallel; therefore the planes AC, EG are parallel; and because AD is equal and parallel to BC, and AE is equal and parallel to BF, the Geometry angle DAE = CBF, and the plane DAE is parallel to the plane CBF (10, 1); therefore also the parallelogram DAHEH is equal to the parallelogram CBFG. It may in like manner be demonstrated that the opposite parallelograms ABFE, DCGH are equal and parallel.

Cor. Any two opposite faces of a parallelopiped may be taken for its bases.

Theorem III.—In every prism ABCDE-FGHIK, the sections NOPQR, STVXY, made by parallel planes, are equal polygons.

Fig. 138.

Because the parallel planes NOPQR, STVXY are cut by a third plane NOST; the sections NO, ST are parallel; and these lines are included between the parallels NS, OT, which are sides of the prism; therefore NO is equal to ST.

For a like reason the sides OP, PQ, QR, &c. of the section NOPQR, are respectively equal to the sides TV, VX, XY, &c. of the section STVXY. And besides, these equal sides being also parallel, the angles NOP, OPQ, &c. of the first section are respectively equal to the angles STV, TXV, &c. of the second section. Therefore the two sections NOPQR, STVXY are equal polygons.

Cor. Every section of an upright prism, by a plane parallel to the base, is equal to that base.

Theorem IV.—If a parallelopiped AG be cut by a plane passing through BD, FH, the diagonals of two of the opposite planes; it will be cut into two equivalent prisms ABDHEF, GHFBCD equal to one another.

Through B and F, the extremities of one of the sides, draw the planes Bade, Fehg, perpendicular to BF, to meet the three other sides of the solid in a, d, c, and in e, h, g. These sections are equal (3), because the planes are perpendicular to FB, and therefore parallel. They are also parallelograms (7, 1), because aB, dc, the opposite sides of the same section, are the intersections of two parallel planes ABFE, DCGH by the same plane. For a like reason, the figure BaeF is a parallelogram, as also the other lateral faces BFGC, cdhg, adhe of the solid Bade-Fehg. Therefore this solid is a prism (Def. 2), and it is a right prism, because BF is perpendicular to the plane of the base.

From the constitution of the figure, the right prism Bh is divided into two right triangular prisms aBd-eFh, cBd-gFh, and it is now to be proved that the oblique triangular prism ABCD-EFGH is equal to the right triangular prism aBd-eFh. Since these two prisms have a common part ABD-eFh, it is only necessary to prove that the remainders, viz. the solids BaAdd, FeHh are equivalent to each other.

Because BAEF, BaeF are parallelograms, therefore AE = BF = ae; hence Aa = Ex. In like manner it may be proved that Dd = Hh. Conceive now that Feh, the base of the solid Feh, is placed on Bad, the base of the solid BaAdd; then, the point e falling on a, and h on d; the lines eE, hH will coincide with their equals aA, dD, because they are perpendicular to the same plane. Therefore, the two solids will coincide entirely, the one with the other; and hence it follows that the oblique prism BAD-FEH is equivalent to the right prism Bad-Feh.

In the same manner it may be demonstrated that the Geometry oblique prism BCD-FGH is equal to the right prism Bod-Fgh. But the two prisms are equal (Cor. to 1), since they have the same altitude, BF and their bases are equal, they being halves of the same parallelogram; therefore the two triangular prisms BAD-FEH, BCD-FGH, which are equivalent to them, are also equivalent to each other.

Cor. Every triangular prism ABD-EFH is half a parallelopiped AG, having the same solid angle A with the same edges AB, AD, AE.

Scholium. Although the triangular prisms, into which the oblique parallelopiped is divided, are contained by equal planes, and have their solid angles equal, yet they cannot be made to coincide. The reason is, the plane angles about the corresponding solid angles are not placed in the same order. These solid angles are therefore symmetrical, and cannot be brought to coincide (18, 1). Two prisms, or two solids of any kind so constituted, are called symmetrical solids. An exact notion of their relation to each other may be acquired, by considering that any object, and its image reflected from a plane mirror, are symmetrical figures. They resemble each other exactly; but every part is placed in a reverse order; thus, the reflected image of a right hand is a left hand.

In symmetrical solids every circumstance on which the magnitude of each depends is the very same in both; hence it might be assumed, as an axiom in the geometry of solids, that they are equivalent.

Theorem V.—If two parallelopipeds AG, AL have a common base ABCD, and have their upper bases EFGH, IKLM in the same plane, and between the same parallels EK, HL; the two parallelopipeds are equivalent to each other.

There may be three cases, according as EI is greater or smaller than EF, or equal to it; but the demonstration is the same for them all. In the first place we shall prove that the triangular prism AEI-DHM is equal to the triangular prism BFK-CGL.

Since AE is parallel to BF, and HE to GF, the angle AEI = BFK, HEI = GFK, and HEA = GFB. Of these six angles, the three first form the solid angle E, and the three others form the solid angle F; therefore, since the solid angles E and F are equal. Now, if the prism AEI-DHM be applied to the prism BFK-CGL, so that their bases AEI, BFK, which are equal, may coincide with each other; then, because the solid angle E is equal to the solid angle F, the side EH shall fall upon FG, and this is all that is necessary to prove that the two prisms coincide entirely; for the base AEI and the edge EH determine the prism AEM, and the base BFK and the edge FG determine the prism BFL (1); therefore the prisms are equal. Now, if from the solid AEL, the prism AEM be taken away, there will remain the parallelopiped AIL; and if from the same solid AEL, the prism BFL be taken away, there will remain the parallelopiped AEG; therefore the parallelopipeds AIL, AEG are equivalent to each other.

Theorem VI.—Two parallelopipeds upon the same base, and having the same altitude, are equivalent to one another. Let \(ABCD\) be the common base of the two parallelopipeds \(AG, AL\), which, because they have the same altitude, will have their upper bases in the same plane. Then, because \(EF\) and \(AB\) are equal and parallel, as also \(IK\) and \(AB\), \(EF\) is equal and parallel to \(IK\) (Cor. 2, 5, 1); for a similar reason \(GF\) is equal and parallel to \(LK\). Let the sides \(EF, HG\), as also the sides \(LK, IM\), be produced, so as to form by their intersections the parallelogram \(NOPQ\); it is manifest that this parallelogram is equal to each of the bases \(EFGH, IKLM\). Now, if we suppose a third parallelopiped, which, with the same lower base \(ABCD\), has for its upper base \(NOPQ\), this third parallelopiped will be equivalent to the parallelopiped \(AG\) (4); for the same reason the third parallelopiped will be equivalent to the parallelopiped \(AL\); therefore the two parallelopipeds \(AG, AL\), which have the same base and the same altitude, are equivalent to one another.

**Theorem VII.**—Any parallelopiped may be changed into a rectangular parallelopiped having the same altitude, and an equivalent base. (Fig. 141 and 142.)

At the points \(A, B, C, D\) (fig. 141) let \(AI, BK, CL, DM\) be drawn perpendicular to the plane \(ABCD\), and terminating in the plane of the upper base; then, \(IK, KL, LM, MI\), being joined, a parallelopiped \(AL\) will thus be formed, which will manifestly have its lateral faces \(AK, BL, CM, DI\) rectangles; and if the base \(AC\) be a rectangle, the solid \(AL\) will be a rectangular parallelopiped equivalent to the parallelopiped \(AG\).

But if \(ABCD\) be not a rectangle (fig. 142), draw \(AO\) and \(BN\) perpendicular to \(CD\), and \(OQ\) and \(NP\) perpendicular to \(DC\), meeting \(ML\) in \(Q\) and \(P\); the solid \(ABNOIKPQ\) will manifestly be a rectangular parallelopiped, which will be equal to the parallelopiped \(AL\); for they have the same base \(ABKI\), and the same altitude, viz. \(AO\); therefore the rectangular parallelopiped \(AP\) is equivalent to the parallelopiped \(AG\) (fig. 141), and they have the same altitude, and the base \(ABNO\) of the former is equivalent to the base \(ABCD\) of the latter.

**Theorem VIII.**—Two rectangular parallelopipeds \(AG, AL\), which have the same base \(ABCD\), are to each other as their altitudes \(AE, AI\).

Suppose that the altitudes \(AE, AI\) are to each other as the numbers \(m\) and \(n\), so that \(AE\) contains \(m\) such equal parts \(pq\), &c. as \(AI\) contains \(n\). Let \(AE\) and \(AI\) be divided into \(m\) and \(n\) equal parts respectively, and let planes pass through the points of division parallel to the base \(ABCD\); thus the parallelopiped \(AG\) will be divided into \(m\) solids, which will also be parallelopipeds having equal bases (3) and equal altitudes, therefore they will be equal among themselves; and in like manner the parallelopiped \(AL\) will be divided into \(n\) equal solids; and as each of the solids in \(AG\) is equal to each of the solids in \(AL\), the parallelopiped \(AG\) will contain \(m\) such equal parts as the parallelopiped \(AL\) contains \(n\); therefore the parallelopiped \(AG\) will be to the parallelopiped \(AL\) as the number \(m\) to the number \(n\), that is, as \(AE\) the altitude of the former to \(AI\) the altitude of the latter.

**Theorem IX.**—Two rectangular parallelopipeds \(AG, AK\), which have the same altitude \(AE\), are to each other as their bases \(ABCD, AMNO\).

Let the two solids be placed, the one by the side of the other, as represented in the figure, and let the plane \(ONKI\) be produced, so as to meet the plane \(DCGH\) in \(PQ\), thus forming a third parallelopiped \(AQ\), which may be compared with each of the parallelopipeds \(AG, AK\). The two solids \(AG, AQ\), having the same base \(ADHE\), are to each other as their altitudes \(AB, AO\) (8); and, in like manner, the two solids \(AQ, AK\) having the same base \(AOLE\), are to each other as their altitudes \(AD, AM\); that is,

\[ \text{sol. } AG : \text{ sol. } AQ = AB : AO, \] \[ \text{sol. } AQ : \text{ sol. } AK = AD : AM; \]

but \(AB : AO = \text{base } AC : \text{base } AP\) (3, 4, Part I.) and \(AD : AM = \text{base } AP : \text{base } AN\),

therefore,

\[ \text{sol. } AG : \text{ sol. } AQ = \text{base } AC : \text{base } AP, \] \[ \text{sol. } AQ : \text{ sol. } AK = \text{base } AP : \text{base } AN, \]

therefore, ex aequo,

\[ \text{sol. } AG : \text{ sol. } AK = \text{base } AC : \text{base } AN. \]

**Theorem X.**—Two rectangular parallelopipeds are to each other as the products of numbers proportional to their bases and altitudes, or as the products of the numbers which express their three dimensions. (Fig. 144.)

Let \(AG\) be a parallelopiped, the three dimensions of which are expressed by the lines \(AB, AD, AE\), and \(AZ\) another parallelopiped the dimensions of which are expressed by the lines \(AO, AM, AX\). Let the two solids \(AG, AZ\) be so placed, that their surfaces may have a common angle \(BAE\); produce such of the planes as are necessary, so as to form a third parallelopiped \(AK\), having the same altitude as the parallelopiped \(AG\). By the last proposition,

\[ \text{sol. } AG : \text{ sol. } AK = \text{base } AC : \text{base } AN, \]

and by the last theorem but one,

\[ \text{sol. } AK : \text{ sol. } AZ = AE : AX; \]

but, considering the bases \(AC, AN\) as measured by numbers, as also the altitudes \(AE, AX\), by 1 of Sect. III. Part I.

\[ \text{base } AC : \text{base } AN = AE \times \text{base } AC : AE \times \text{base } AN, \] \[ \text{and } AE : AX = AE \times \text{base } AN : AX \times \text{base } AN, \]

therefore,

\[ \text{sol. } AG : \text{ sol. } AK = AE \times \text{base } AC : AE \times \text{base } AN, \] \[ \text{sol. } AK : \text{ sol. } AZ = AE \times \text{base } AN : AX \times \text{base } AN, \]

therefore, ex aequo,

\[ \text{sol. } AG : \text{ sol. } AZ = AE \times \text{base } AC : AX \times \text{base } AN; \]

which proportion, by substituting for the bases \(AC, AN\) Geometry their numerical values \( AB \times AD \) and \( AO \times AM \), becomes

\[ \text{Scholium. Hence it appears that the product of the base of a rectangular parallelepiped by its altitude, or the product of its three dimensions, may be taken as its numerical measure; and it is upon this principle that all other solids are estimated. When two parallelepipeds are compared by means of their bases and altitudes, their bases must be considered as measured by the same superficial unit, and their altitudes by the same linear unit; thus, if \( P \) and \( Q \) denote two parallelepipeds, and the base of \( P \) contain three such equal spaces as that of \( Q \) contains four; and the altitude of \( P \) contains two such equal lines as that of \( Q \) contains five, then } P : Q = 3 \times 2 : 4 \times 5 = 6 : 20.

If all the dimensions of each solid be used in comparing them, then, the same linear unit must be employed in estimating all the dimensions of both solids; thus, if the length, breadth, and height of the solid \( P \) be 4, 3, and 6 linear units, respectively; and those of \( Q \), 7, 2, and 5, of the same units; then \( P : Q = 4 \times 3 \times 6 : 7 \times 2 \times 5 = 72 : 70 \).

As lines are compared by considering how often each contains some other line taken as a measuring unit, and surfaces, by considering how often each contains a square whose side is that unit; so solids may be compared, by considering how often each contains a cube, the side or edge of which is the same linear unit. Accordingly, the dimensions of the parallelepipeds \( P \) and \( Q \) being as above, the proportion \( P : Q = 72 : 70 \) may be considered as indicating that \( P \) contains 72 such equal cubes as \( Q \) contains 70.

The magnitude of a solid, its bulk, or its extension, constitutes its solidity, or its content; thus, we say, that the solidity or the content of a rectangular parallelepiped is equal to the product of its base by its altitude, or to the product of its three dimensions.

**Theorem XI.**—The solidity of any parallelopiped, or in general of any prism, is equal to the product of its base by its altitude.

1. Any parallelopiped is equivalent to a rectangular parallelopiped of the same altitude, and an equivalent base (7); and it has been shown that the solidity of such a parallelopiped is equal to the product of its base by its altitude.

2. Every triangular prism is the half of a parallelopiped of the same altitude, but having its base double that of the prism (3); therefore the solidity of the prism is half that of the parallelopiped, or it is half the product of the base of the parallelopiped by its altitude, that is, it is equal to the product of the base of the prism by its altitude.

3. Any other prism may be divided into as many triangular prisms as there can be triangles in the polygon which forms its base; but the solidity of each prism is equal to the product of its base by their common altitude; therefore the solidity of the whole prism is equal to the product of the sum of all their bases by the common altitude, or it is equal to the product of the base of the prism, which is the sum of them all, by its altitude.

Con. Two prisms having the same altitude are to each other as their bases; and two prisms having the same bases are to each other as their altitudes.

Note. The cube of a line \( AB \) is expressed thus; \( AB \times AB \times AB \), but more commonly by \( AB^3 \).

**Theorem XII.**—Similar prisms are to one another as the cubes of their homologous sides.

Let \( P \) and \( p \) be two prisms, of which \( BC, bc \) are the homologous sides; the Geometry prism \( P \) is to the prism \( p \) of Solids, as a cube on the line \( BC \) to a cube on \( bc \).

From \( A \) and \( a \), homologous angles of the two prisms, draw \( AH, ah \) perpendicular to their bases \( BCD, bcd \). Join \( BH \); take \( Ba = ba \), and in the plane \( BHA \) draw \( ah \) perpendicular to \( BH \); then \( ah \) shall be perpendicular to the plane \( CBD \) (13, 1), and equal to \( ah \), the altitude of the other prism; for if the solid angles \( B \) and \( b \) were applied the one to the other, the planes which contain them, and consequently the perpendiculars \( ah, ah' \) would coincide (Schol. 18, I).

Now, because of the similar triangles \( ABH, abh \), and the similar figures \( AC, ac \),

\[ AH : ah = AB : ab = BC : bc, \]

and because of the similar bases,

\[ \text{base } BCD : \text{base } bcd = BC^2 : bc^2 \quad (25, 4, \text{Part I}). \]

From these two proportions, by considering all the quantities, as expressed by numbers (by 11, 3, Part I),

\[ AH \times \text{base } BCD : ah \times \text{base } BCD = BC^2 : bc \times BC^2, \]

therefore, ex aequo,

\[ AH \times \text{base } BCD : ah \times \text{base } bcd = BC^3 : bc^3. \]

But \( AH \times \text{base } BCD \) expresses the content of the prism \( P \); and \( ah \times \text{base } bcd \) expresses that of the other prism \( p \); therefore

\[ \text{prism } P : \text{prism } p = BC^3 : bc^3. \]

Con. Similar prisms are to one another in the triplicate ratio of their homologous sides. Let \( S \) and \( s \) denote the homologous sides of two similar prisms \( P \) and \( p \). It is manifest that

\[ S^3, S^2s, Ss^2, s^3, \]

are four continual proportionals; therefore (Def. 11, Sect. III, Part I.), the ratio of \( S^3 \), the first, to \( s^3 \), the last, is the triplicate of the ratio of \( S^2 \), the first, to \( Ss \), the second; now, by the proposition, \( P : p = S^3 : s^3 \), and the ratio of \( S^3 \) to \( Ss^2 \) is the same as the ratio of \( S \) to \( s \); therefore, the ratio of \( P \) to \( p \) is triplicate of the ratio of \( S \) to \( s \).

**Prop. XIII. Theor.**—If a triangular pyramid \( A-BCD \) be cut by a plane parallel to its base, the section \( FGH \) is similar to the base.

For because the parallel planes \( BCD, FGH \) are cut by a third plane \( ABC \), the sections \( FG, BC \) are parallel (7, 1). In like manner it appears that \( FH \) is parallel to \( BD \); therefore the angle \( HFG \) is equal to the angle \( DBC \) (10, 1.) And because the triangle \( ABC \) is similar to the triangle \( AFH \), we have

\[ BC : BA = FG : FA, \text{ and } BA : BD = FA : FH. \]

Therefore, ex aequo, \( BC : BD = FG : FH \); now the angle \( DBC \) has been shown to be equal to the angle \( HFG \); therefore the triangles \( DBC, HFG \) are equiangular (20, 4, Part I.).

**Prop. XIV. Theor.**—If two triangular pyramids \( A-BCD, a-bcd \), which have equivalent bases and equal altitudes, be cut by planes that are parallel to the bases, and at equal distances from them; the sections \( FGH, fgh \) will be equal. (Fig. 146.) Geometry Draw AKE, \(a\)k perpendicular to the bases BCD, \(b\)cd, meeting the cutting planes in K and \(k\); then, because of the parallel planes, we have AE: AK = AB: AF, and ae: ak = ab: af (15, 1); but, by hypothesis, AE = ae, and AK = ak; therefore, AB: AF = ab: af. Again, because of similar triangles, AB: AF = BC: FG, and ab: af = be: fg; therefore BC: FG = be: fg; and hence BC²: FG² = be²: fg² (23, 4, Part I); but because of the similar triangles BDC, FHG, BC²: FG² = trian. BDC : trian. FHG, and in like manner be²: fg² = trian. bcd : trian. fgh (25, 4, Part I), therefore trian. BCD : trian. FHG = trian. bcd : trian. fgh.

Now, trian. BCD = trian. bcd (by hypothesis), therefore the triangle FHG is equal to the triangle fgh.

Theorem XV.—A series of prisms of the same altitude may be circumscribed about any pyramid ABCD, such that the sum of the prisms shall exceed the pyramid by a solid less than any given solid Z.

Let Z denote a prism standing on the same base with the pyramid, viz. the triangle BCD, and having for its altitude the perpendicular drawn from a certain point E in the line AC upon the plane BCD. It is evident that CE multiplied by a certain number \(m\) will be greater than AC; divide CA into as many equal parts as there are units in \(m\), and let these be CF, FG, GH, HA, each of which will be less than CE. Through each of the points F, G, H, let planes be made to pass parallel to the plane BCD, making with the sides of the pyramid the sections FPQ, GR, HTU, which will be all similar to one another, and to the base BCD (13). From the point B draw in the plane of the triangle ABC the straight line BK parallel to CF, meeting FP produced in K. In like manner, from D draw DL parallel to CF, meeting FQ in L; join KL, and it is plain that the solid KBKDLF is a prism. By the same construction let the prisms PM, RO, TV be described. Also let the straight line IP, which is in the plane of the triangle ABC, be produced till it meet BC in h; and let the line MQ be produced till it meet DC in g. Join hg, then hCg QFP is a prism; and is equal to the prism PM (Cor. II). In the same manner is described the prism mS equal to the prism RO, and the prism qU equal to the prism TV. The sum, therefore, of all the inscribed prisms hQ, mS, and qU is equal to the sum of the prisms PM, RO, and TV, that is, to the sum of all the circumscribed prisms except the prism BL; therefore, BL is the excess of the prisms circumscribed about the pyramid above the prisms inscribed within it. But the prism BL is less than the prism which has the triangle BCD for its base, and for its altitude the perpendicular from E upon the plane BCD, which prism is, by hypothesis, equal to the given solid Z; therefore the excess of the circumscribed above the inscribed prisms is less than the solid Z. But the excess of the circumscribed prisms above the inscribed is greater than their excess above the pyramid ABCD, because ABCD is greater than the sum of the inscribed prisms; much more therefore is the excess of the circumscribed prisms above the pyramid less than the solid Z. A series of prisms of the same altitude has therefore been circumscribed about the pyramid ABCD exceeding it by a solid less than the given solid Z.

Prop. XVI. Theor.—Pyramids that have equal bases and altitudes are equal to one another.

Let A-BCD, \(a\)bcd be two pyramids that have equal bases BCD, \(b\)cd, and equal altitudes; viz. the perpendiculars drawn from the vertices A and \(a\) upon the planes BCD, \(b\)cd, the pyramid A-BCD is equal to the pyramid \(a\)bcd.

For if they are not equal, let Z represent the solid which is equal to the excess of one of them, \(a\)bcd, above the other A-BCD; and let a series of prisms CE, FG, HK, LM, of the same altitude be circumscribed about the pyramid A-BCD, so as to exceed it by a solid less than Z, which is always possible (15); also let a series of prisms, ce, fg, hk, lm, equal in number to the other, and of the same altitude, be circumscribed about the pyramid \(a\)bcd. And because the pyramids have equal altitudes, and the number of prisms described about each is the same, the altitudes of the prisms will be all equal, and the bases of the corresponding prisms in the two pyramids, as EF, ef, will be sections of the pyramids at equal distances from their bases, therefore they are equal (14), and the prisms themselves are equal (1), and the sum of all the prisms described about the one pyramid is equal to the sum of all the prisms described about the other pyramid. To abridge, put P and p to denote the pyramids A-BCD, and \(a\)bcd respectively, and Q and q to express the sums of the prisms described about them. Then, because by hypothesis Z = p — P, and by construction Z = Q — P, therefore p — P = Q — P, hence p must be greater than Q; but Q is equal to q, therefore p must be greater than q, that is, the pyramid p is greater than q, the sum of the prisms described about it, which is impossible; therefore the pyramids P, p are not unequal, that is, they are equal.

Theorem XVII.—Every prism having a triangular base may be divided into three pyramids that have triangular bases, and that are equal to one another.

Let ABC, DEF be the opposite bases of a triangular prism. Join AE, EC, CD; and because ABED is a parallelogram, of which AE is the diameter, the triangle ADE is equal to the triangle ABE; therefore the pyramid of which the base is the triangle ADE and vertex the point C, is equal to the pyramid of which the base is the triangle ABE, and vertex the point C. But the pyramid of which the base is the triangle ABE and vertex the point C, that is, the pyramid ABCE, is equal to the pyramid DEFC (16), for they have equal bases, viz. the triangles ABC, DFE, and the same altitude, viz. the altitude of the prism ABCDEF. Therefore, the three pyramids ADEC, ABEC, DFEC are equal to one Geometry another; but these pyramids make up the whole prism of Solids. ABCDEF; therefore, the prism ABCDEF is divided into three equal pyramids.

Cor. 1. From this it is manifest that every pyramid is the third part of a prism which has the same base and the same altitude with it; for if the base of the prism be any other figure than a triangle, it may be divided into prisms having triangular bases.

Cor. 2. Pyramids having equal altitudes are to one another as their bases; because the prisms upon the same bases, and of the same altitude, are to one another as their bases.

SECT. III.—OF CYLINDERS, CONES, AND THE SPHERE.

Definitions.

I. A Cylinder is a solid figure described by the revolution of a right-angled parallelogram about one of its sides, which remains fixed.

The Axis of the cylinder is the fixed straight line about which the parallelogram revolves.

The Bases of the cylinder are the circles described by the two revolving opposite sides of the parallelogram.

II. A Cone is a solid figure described by the revolution of a right-angled triangle about one of the sides containing the right angle, which side remains fixed.

The Axis of the cone is the fixed line about which the triangle revolves.

The Base of the cone is the circle described by that side containing the right angle which revolves.

III. A Sphere is a solid figure described by the revolution of a semicircle about a diameter.

The Axis of a sphere is the fixed line about which the semicircle revolves.

The Centre of a sphere is the same with that of the semicircle.

The Diameter of a sphere is any straight line which passes through the centre, and is terminated both ways by the superficies of the sphere.

IV. Similar cones and cylinders are those which have their axes and diameters of their bases proportional.

Theorem I. If from any point E in the circumference of the base of a cylinder ABCD, a perpendicular EF be drawn to the plane of the base AEB; the straight line EF is wholly in the cylindric superficies.

Let GH be the axis, and AGHD the rectangle, which by its revolution describes the cylinder. Because HG is perpendicular to AG in every position of the revolving rectangle, it is perpendicular to the plane of the circle described by AG; and because AD, the line which describes the cylindric superficies, is parallel to GH, it is also perpendicular to the plane of that circle (5, 1).

Now when by the revolution of the rectangle AGHD the point A coincides with the point E, the line EF will coincide with AD, and thus will be wholly in the cylindric superficies; for otherwise two perpendiculars might be drawn to the same plane, from the same point, which is impossible (2 Cor. 4, 1).

Theorem II.—A cylinder and a parallelopiped having equivalent bases, and the same altitude, are equal to one another.

Let ABCD be a cylinder and EF a parallelopiped having equivalent bases, viz. the circle AGB and the parallelogram EH, and having also equal altitudes; the cylinder ABCD is equal to the parallelopiped EF. If not, let them be unequal; and first let the cylinder be less than the parallelopiped EF; and from the parallelopiped EF let there be cut off a part EQ by a plane PQ parallel to NF, equal to the cylinder ABCD. In the circle AGB inscribe the polygon AGKBLM that shall differ from the circle by a space less than the parallelogram PH (1 Cor. 2, 6, P. L), and cut off from the parallelogram EH a part OR equal to the polygon AGKBLM, then it is manifest that the parallelogram OR is greater than the parallelogram OP, therefore the point R will fall between P and N. On the polygon AGKBLM let an upright prism be constituted of the same altitude with the cylinder, which will therefore be less than the cylinder, because it is within it (1); and if through the point R a plane RS parallel to NF be made to pass, it will cut off the parallelopiped ES equal to the prism AGBC, because its base is equal to that of the prism, and its altitude is the same. But the prism AGBC is less than the cylinder ABCD, and the cylinder ABCD is equal to the parallelopiped EQ, by hypothesis; therefore, ES is less than EQ, and it is also greater, which is impossible. The cylinder ABCD therefore is not less than the parallelopiped EF; and in the same manner it may be shown not to be greater than EF, therefore they are equal.

Theorem III.—If a cone and cylinder have the same base and the same altitude; the cone is the third part of the cylinder.

Let the cone ABCD and the cylinder BFKG have the same base, viz. the circle BCD, and the same altitude, viz. the perpendicular from the point A upon the plane BCD; the cone ABCD is the third part of the cylinder BFKG. If not, let the cone ABCD be the third part of another cylinder LMNO having the same altitude with the cylinder BFKG; but let the bases BCD, LIM be unequal, and first let BCD be greater than LIM. Then, because the circle BCD is greater than the circle LIM, a polygon may be inscribed in BCD that shall differ from it less than LIM does (1 Cor. 2, 6, P. L), and which therefore will be greater than Geometry LIM. Let this be the polygon BECFD, and upon BECFD let there be constituted the pyramid ABECFD, and the prism BCFKHG. Because the polygon BECFD is greater than the circle LIM, the prism BCFKHG is greater than the cylinder LMNO, for they have the same altitude, but the prism has the greater base. But the pyramid ABECFD is the third part of the prism BCFHG (17, 1); therefore it is greater than the third part of the cylinder LMNO.

Now the cone ABECFD is by hypothesis the third part of the cylinder LMNO; therefore, the pyramid ABECFD is greater than the cone ABCD, and it is also less, because it is inscribed to the cone, which is impossible. Therefore the cone ABCD is not less than the third part of the cylinder BFKG. And in the same manner, by circumscribing a polygon about the circle BCD, it may be shown, that the cone ABCD is not greater than the third part of the cylinder BFKG; therefore, it is equal to the third part of the cylinder.

Theorem IV.—Let ABDC be a plane figure, bounded by a straight line CD, a line of any kind AB, which is terminated by perpendiculars at the extremities of CD, and by these perpendiculars AC, BC. Let AB be a solid generated by the revolution of this figure about CD as an axis; a series of cylinders may be described about the solid, and another series may be inscribed in it, having all the same altitude, and such that the sum of the circumscribed cylinders shall exceed the sum of the inscribed cylinders by less than any given solid S.

Let S denote a solid which is a cylinder, having Bb for the diameter of its base, and DP for its height. Suppose the fixed axis CD to be divided into a number of equal parts DK, KG, GE, EC, each less than DP. In the plane of the figure ABDC, draw perpendiculars EF, GH, KL to meet the line AB in F, H, L. Construct the inscribed rectangles AE, FG, HK, LD, also the circumscribed rectangles CF, EH, GL, KB. By the rotation of the plane figure about the axis CD, these rectangles will evidently generate a series of cylinders inscribed in the solid, and another series described about it. Let the circumscribed cylinders, reckoned from the bottom of the solid to the top, be denoted by V, X, Y, Z, and the inscribed cylinder by v, x, y, z, then the sums of the circumscribed and inscribed cylinder will be

\[ V + X + Y + Z \]

and

\[ v + x + y + z. \]

Now by the nature of the figure, each circumscribed cylinder is equal to the inscribed cylinder next below it; therefore \( X = v, Y = x, \) and \( Z = y, \) and hence the excess of the sum of all the circumscribed above the inscribed cylinders will be the same as the excess of the greatest circumscribed above the least inscribed cylinder; that is, it will be equal to \( V - z, \) and consequently will be less than \( V; \) but the lowest circumscribed cylinder \( V \) is less than the solid \( S, \) because it has the same base (viz. the circle having for its diameter Bb), and a less altitude KD, by construction; therefore the excess of the series of circumscribed above the series of inscribed cylinders is less than the given solid \( S. \)

Cor. The difference between the solid \( AB \) ba and either of the two series of cylinders will be less than the greatest circumscribed cylinder: For the solid \( AB \) ba is greater than the one series of cylinders and less than the other; therefore it will differ from either series by a quantity less than the difference between the two.

Theorem V.—If a cone and hemisphere have equal bases and altitudes, and if a series of cylinders be described about the cone, and another series be inscribed in the hemisphere, and the cylinders have all the same altitude; the sum of the two series will be equal to a cylinder having the same base and altitude as the hemisphere.

Let AFB be a semicircle, and CFDA, CFEB, squares described on the radius CF, and let CE be the diagonal of one of the squares BF: Let CF be divided into any number of equal parts CG, GK, KM, MF; and let perpendiculars be drawn through the points of division, meeting the diagonal CE, in the points O, P, Q; the quadrant arc BF in the points H, L, N; and the side of the square in the points R, S, T; construct the rectangles CO, GP, KQ, ME, which will circumscribe the triangle CFE; construct also the rectangles CH, GL, KN, which will be inscribed in the quadrant CFB. Suppose now the plane of the square to revolve about its side CF as an axis; the triangle CFE will then generate a cone, which will have DE for the diameter of its base, and C for its vertex; the quadrant CFB will generate a hemisphere, having for its base a circle of which AB is a diameter; and the square CBEF will generate a cylinder, having the same base and altitude as the hemisphere: also, the rectangles described about the triangle CFE will manifestly generate a series of cylinders circumscribing the cone; the rectangles inscribed in the quadrant will generate a series of cylinders inscribed in the hemisphere; and the rectangles CR, GS, KT, ME will generate a series of cylinders which will compose a cylinder having the same base and altitude as the hemisphere.

The triangles CFE, CGO are manifestly similar, and \( CF = FE; \) therefore \( CG = GO. \) In like manner, it may be proved that \( CK = KP \) and that \( CM = MQ. \)

Join CH, and because CGH is a right-angled triangle, a circle described with CH as a radius will be equal to two circles described with CG and GH as radii (2 Cor. 4, 6, Part I.), but \( CG = GO, \) and \( CH = GR; \) therefore a circle described with GR as a radius will be equal to two circles described with GO and GH as radii; hence again it follows, that the cylinder generated by the rectangle CR will be equal to both the cylinders generated by the rectangles CO and CH, for they have all the same altitude, and the base of the first is equal to the sum of the bases of the other two. It may be demonstrated in the same manner that the cylinder generated by the rectangle GS is equal to the sum of the cylinders generated by the rectangles GP and GL, and the same of all the rest; therefore the sum of the cylinders generated by the rectangles CR, GS, KT, ME is equal to the two series of cylinders, one generated by the rectangles CO, GP, KQ, ME, and the other generated by the rectangles CH, GL, KN; that is, a cylinder having the same base and altitude as the hemisphere, is equal to the sum of the two series of cylinders, one described about the cone, and the other described in the hemisphere.

Theorem VI.—Every sphere is two thirds of the circumscribing cylinder. (Fig. 154.)

Let a figure be constructed exactly as is last proposition; and to abridge, let C denote the cone, e the series of cylinders described about it, H the hemisphere, h the cylinders described in it, and K the cylinder having the same base and altitude as the hemisphere, or cone. Moreover, put \( d \) for the difference between the cone and its circumscribed cylinders, and \(d'\) for the difference between the hemisphere and its inscribed cylinders; then we have

\[ C + d = e, \quad \text{and} \quad H = h + d', \]

and adding equals to equals,

\[ C + H + d = c + h + d'. \]

But \(c + h = K\) (5); therefore, \(C + H + d = K + d'\), and \(C + H + d - d' = K\), also \(C + H = K + d' - d'\). Hence it appears that the difference between \(C + H\) and \(K\) is equal to the difference between \(d\) and \(d'\). Now \(d\) is less than the cylinder generated by the rotation of the rectangle ME (Cor. to Prop. 4), and \(d'\) is less than the cylinder generated by the rectangle CR, which is equal to ME; therefore the difference between \(d\) and \(d'\) must be less than the same rectangle; hence the difference between \(C + H\) and \(K\) is less than the cylinder generated by the revolution of the rectangle ME, or is less than a cylinder having the same base as the cone, and the line FM for its altitude. From this we may infer, that \(C + H\) is exactly equal to \(K\); for if there can be any difference, let it be a cylinder having the same base as the cone, and its altitude equal to \(FV\); then FM must be greater than \(FV\); but the number of parts into which FC is divided may be so great that FM may be less than \(FV\); therefore \(C + H\) cannot be unequal to \(K\); and since \(C + H = K\), and \(C = \frac{1}{3}K\) (3), therefore \(H = \frac{2}{3}K\); that is, the hemisphere is two thirds of its circumscribing cylinder; and taking the doubles of these, the whole sphere is two thirds of its circumscribing cylinder.

A Table showing the Theorem of the foregoing Treatise, that corresponds to each of the principal Propositions in the first Six and in the Eleventh and Twelfth Books of Euclid's Elements.

| Euclid. | Geometry. | Euclid. | Geometry. | Euclid. | Geometry. | Euclid. | Geometry. | Euclid. | Geometry. | |--------|-----------|--------|-----------|--------|-----------|--------|-----------|--------|-----------| | Book I. | Part I. Theor. Sect. | Book I. | Part I. Theor. Sect. | Book III. | Part I. Theor. Sect. | Book VI. | Part I. Theor. Sect. | Book XI. | Part II. Theor. Sect. | | Prop. 4. | 5. 1. | Pr. 41. | 2. 4. | Pr. 28. | 4. 2. | Pr. 2. | 17. 4. | 9. | 2 Cor. to | | 5. | 11. 1. | 47. | 13. 4. | 29. | 4. 2. | 18. 4. | 10. | 5. 1. | | 6. | 12. 1. | 48. | Scolium | 31. | 17. 2. | 3. | 19. 4. | 10. | 1. 1. | | 8. | 10. 1. | 15. 4. | 32. | 18. 2. | 4. | 20. 4. | 13. | 2 Cor. to | | 13. | 1. 1. | Book II. | Theor. Sect. | 35. | 28. 4. | 5. | 21. 4. | 14. | 4. 1. | | 14. | 3. 1. | 36. | 29. 4. | 6. | 22. 4. | 6. | 23. 4. | 15. | 10. 1. | | 15. | 4. 1. | Book V. | Theor. Sect. | 30. 4. | 8. | 23. 4. | 16. | 7. 1. | | 16. | 23. 1. | Pr. 4. | 10. 4. | 14. | Cor. | 17. | 15. 1. | | 17. | 24. 1. | 5. | 12. 4. | 15. | 24. 4. | 18. | 12. 1. | | 18. | 6. | 13. 1. | 7. | 11. 4. | Pr. 4. | 7. 3. | 19. | 14. 1. | | 19. | 12. | 14. 4. | 15. | 12. 3. | 17. | 8. 4. | 20. | 16. 1. | | 20. | 7. 1. | 15. 4. | 12. | 15. 4. | 19. | 25. 4. | 21. | 17. 1. | | 21. | 8. 1. | 13. | 14. 4. | 15. | 1. 3. | 20. | 26. 4. | A. | 18. 1. | | 24. | 9. 1. | Book III. | Theor. Sect. | 16. | 3. 3. | 27. | 4. | 24. | 12. 2. | | 25. | 6. 1. | Pr. 2. | 3. 2. | 17. | 5. 3. | 31. | 1 Cor. to | 25. | 9. 2. | | 27. | 22. 1. | 3. | 6. 2. | 22. | 8. 3. | 33. | 31. 4. | 28. | 4. 2. | | 28. | 21. 1. | Cor. to | 10. | 7. 2. | 24. | 10. 3. | Book XI. | Part II. Theor. Sect. | 29. | 5. 2. | | 30. | 20. 1. | Book VI. | Theor. Sect. | Pr. 1. | 1. 1. | 30. | 31. | 6. 2. | | 32. | 23. | 11. | 12. 2. | Cor. to | 2. | 32. | 9. 2. | | 33. | 24. 1. | 12. | 11. 2. | Cor. to | 3. | 33. | 12. 2. | | 34. | 25. | 14. | 8. 2. | Cor. to | 4. | 1 Cor. to | 30. | 16. 2. of Part II. | | 35. | 28. 1. | 15. | 2. 2. | Pr. 1. | 6. | 1 Cor. to | 31. | 3. 3. | | 36. | 26. 1. | 8. 2. | 16. | 9. 2. | 6. | 5. 1. | Pr. 1. | 1. 6. of Part I. | | 37. | 1. 4. | 20. | 14. 2. | 5. | 4. | 4. 6. of Part I. | | 38. | 2 Cor. to | 21. | 15. 2. | 4. | 4. | 16. 2. of Part II. |

ANALYTICAL GEOMETRY.

1. The branch of science which has obtained the name of analytical geometry owes its origin to Descartes. It consists in the application of algebra to geometry, not merely by using algebraic symbols as the representatives of magnitude, but by employing these in such a manner that the position of a point is indicated by referring it to fixed lines or planes, just as the position of a point on the earth's surface is defined by referring it to the equator and the first meridian. For example, if we suppose the positions of two lines at right angles to one another to be fixed and known, then the position of any point in the same plane is sufficiently determined by its distances from these two lines, just as the position of a place on a given map is denoted by its latitude and longitude.

This science, like that of ordinary geometry, is divided into two parts, plane and solid, or analytical geometry of two and of three dimensions.

ANALYTICAL GEOMETRY OF TWO DIMENSIONS.

2. As every line or area is supposed to be situated in one plane, we may, whenever it is convenient to do so, imagine that plane to be the paper before us, which, for convenience of phraseology, we may further suppose to be standing upright, so that any one line of print is horizontal, whilst the successive lines lie vertically under each other. The simplest way, under these circumstances, in which the position of a point can be indicated, will be by its distances from two lines at right angles to each other, the one horizontal, the other vertical. If we know the positions of these lines, then we know the position of every point in this plane whose distances from these two lines have been given us. But it is frequently necessary to adopt lines of reference inclined to one another. In this case, the position of a point is determined by its two distances from these lines respectively, each measured in a direction parallel to the other. The lines of reference are called the co-ordinate axes; their point of intersection is called the origin; the distances of any other point from the lines, measured in the way stated above are called the co-ordinates of that point; of these one is called the abscissa, the other the ordinate, the former being usually the designation of the horizontal distance.

In this figure, \(OX\), \(OY\), at right angles or not, are called the co-ordinate axes; \(O\) the origin; \(PM\), \(PN\) (or its equal \(OM\)), drawn parallel to \(OY\) and \(OX\), the co-ordinates of the point \(P\); \(OM\) the abscissa, \(PM\) the ordinate. The line \(OM\) is usually written by the letter \(x\), and \(MP\) by \(y\). We may thus form some idea of the meaning of the symbols by considering the letter \(x\) as the abbreviation for the general word longitude, and the letter \(y\) as that for latitude. Thus when the latitude and longitude are both specified, the point referred to is determined. If, on the other hand, there is stated merely a relation between them, such, for example, as that the latitude is equal to the longitude, it is clear that no one point is determined thereby, since the same property is applicable to an innumerable series of points. A little consideration will show that some line, straight or crooked, will pass through all such points. The relation, therefore, is said to determine this line, and when expressed in algebraic symbols, is called the equation to the line. Thus, for example, the relation, "that the abscissa is equal to the ordinate," or the longitude equal to the latitude, understood as above, indicates a straight line; and the corresponding equation, \(x = y\), is the equation to a straight line.

3. The requirements of analysis demand that attention should be paid to the sign as well as to the magnitude of the algebraic symbol which expresses an abscissa or an ordinate. We lay it down as a rule, deduced at once from the algebraic definitions of the symbols \(+\) and \(-\); that if one of them denotes a line measured in one direction, the other will denote a line measured in the opposite direction. Thus if \(x = +a\) be an abscissa measured to the right of \(O\), \(x = -a\) will be an equal abscissa measured to the left; if \(y = +b\) be an ordinate measured above \(O\), \(y = -b\) will be an equal ordinate measured below it; and these are the directions which we shall usually assign to the positive and negative abscissas and ordinates.

4. Besides the method of representation which we have described, there is another very unlike it in form, but equally important in the solution of mechanical and astronomical problems. In this system, the position of a point is determined by its distance from a given point, and the direction in which the line measuring that distance lies with respect to a given line. This system is called the system of polar co-ordinates. The given point to which all others are referred is called the pole, and the distance from that point the radius vector.

5. The subject before us naturally divides itself into three distinct branches,—1°, the determination of the equations to curves from the knowledge of their properties; 2°, the determination of the forms and properties of curves from their equations; and, 3°, the deduction of one property of a curve from another, or from several others. The third branch being the combination and extension of the other two, we propose briefly to illustrate the first and second, and then to include under the third a discussion of some of the more important properties of the Conic Sections in the form in which the requirements of science demand their exhibition.

SECTION I.—THE DETERMINATION OF THE EQUATIONS TO CURVES FROM THE KNOWLEDGE OF THEIR PROPERTIES.

6. a. The Straight Line.—The property of the straight line which we shall employ is this—that if two series of parallel lines be drawn from any points in it so as to form with the given line a series of triangles, the sides of these triangles will be proportional.

Let \(PQR\) be the straight line; \(OX\), \(OY\) the co-ordinate axes; \(PM\) parallel to \(OY\), and \(QE\) parallel to \(OX\): the triangles \(PQE\), \(QRO\) are similar, \(\therefore PE : EQ :: QO : OR\).

Let \(OM\) be called \(x\), or \(OM = x\), \(MP = y\), \(OQ = b\); and let the ratio \(QO : OR\) be \(a : 1\);

then \(y = b : x :: a : 1\);

or \(y = ax + b\) is the equation to the line \(PR\).

Hence it is clear that any simple equation between \(y\) and \(x\) will represent a straight line. If the axes of co-ordinates are at right angles to each other—in which case they are called rectangular axes—a is the trigonometrical tangent of the angle \(PRO\), in which the given line cuts the axis of \(x\).

7. The equation to a straight line may be written in a different form, which is frequently convenient, thus:

Let the perpendicular from \(O\) on the line \(PR\) be called \(p\), the angle \(PRM = a\); then it is evident that

\[y \cos a - x \sin a - p = 0,\]

which is the equation to the line.

This equation is abbreviated by writing for it simply \(a = 0\); or, when multiplied by a constant \(a\), by \(u = 0\); where \(u\) is a function of \(x\), \(y\), \(a\), \(p\), and a constant which is arbitrary. We shall give an instance of the use of this form of the equation in the sequel.

If the axes of co-ordinates are not at right angles, the equation becomes slightly changed in form.

Let the line make the angles \(\theta\) and \(\phi\) with the axes of \(x\) and \(y\) respectively; then

\[QO : OR :: \sin \theta : \sin \phi;\]

\[\therefore y = \frac{\sin \theta}{\sin \phi} x + b\] is the equation;

or \(y \sin \phi - x \sin \theta = b \sin \phi;\)

consequently all lines which are parallel to the given line, or for which \(\theta\) and \(\phi\) are the same, have the relative magnitudes of the co-efficients of \(x\) and \(y\) the same, and vice versa.

8. Cor. 1.—If the line pass through a given point of which the co-ordinates are \(x'\), \(y'\), we shall have \(y = ax + b\) generally,

and \(y' = ax' + b\) for the given point;

\[\therefore y - y' = a(x - x'),\]

or \(b\) is determined.

Cor. 2.—If the line pass through a second given point whose co-ordinates are \(y'\), \(x'\), we shall have \(y' - y' = a(x' - x')\), or \(a\) is determined,

and \[\therefore y - y' = \frac{y' - y'}{x' - x'} (x - x')\] is the equation. 9. To express the polar equation to the straight line.

Let O be the pole, OY perpendicular to PR; and suppose OP joined. Let OP = r, OY = p, ∠POR = θ, ∠YOR = α; then OP cos POY = OY, or r cos (θ - α) = p

is the equation required.

10. b. The Circle.—The property of the circle is, that the lines drawn from the centre to the circumference are equal.

Let us suppose the axes to be rectangular, Q the centre, QP the radius = a; OM = x, PM = y, the co-ordinates of P; ON = h, NQ = c, the co-ordinates of the centre; then PR² + QR² = PQ²; or (y - c)² + (x - h)² = a² is the equation required. If the origin be at the centre, the equation is x² + y² = a².

11. The polar equation in its simplest form, when the centre is the pole, is evidently r = a.

12. c. The Parabola.—The property (Conic Sections, Part I, Def. 1) is, that the distance of any point from the focus is equal to its distance from the directrix.

Let F be the focus, DQ the directrix, then FP = PQ. Suppose the vertex A to be the origin, and AFX the axis of the parabola to be the axis of x.

Let AM = x, MP = y, AF = AD = a, then PM² + MF² = PF² = PQ² = MD²; or y² + (x - a)² = (x + a)², which gives y² = 4ax, the equation required.

13. To find the polar equation, the focus being the pole.

Let FP = r, DF = θ; then FP cos PFM = FM = DM - DF = PQ - DF = FP - DF = r - 2a

∴ r = \frac{2a}{1 - \cos PFM} = \frac{a}{\sin^2 PFM} = \frac{a}{\cos^2 \theta} is the equation required.

14. d. The Ellipse.—The property (Conic Sections, Part 2, Def. 1) is, that the sum of the two lines drawn from any point to the two foci is constant.

Let the centre C be the origin : CA which passes through the focus S, called the semi-axis major, or the semi-transverse axis, the axis of x; CB perpendicular to CA, the semi-axis minor, or the semi-conjugate axis, the axis of y; CM = x, MP = y, CA = a, CB = b; then it is evident that CP² = CS² = a² - b²; and the condition FP + SP = 2a gives FP² = (2a - SP)² = 4a² - 4a · SP + SP²,

or (FC + x)² + y² = 4a² - 4a · SP + (CS - x)² + y²;

whence 2FC · x = 4a² - 4a · SP - 2CS · x,

or a · SP = a² - CS · x;

that is, a(SM² + MP²) = a² - 2a² · CS · x + CS² · x²,

or a²(x - CS)² + a²y² = a² - 2a² · CS · x + CS² · x²;

whence a²x² + a²y² = a² + CS² · x²,

or a²b² + b²x² = a²b², the equation required.

15. Cor.—Since a · SP = a² - CS · x,

if CS = ae, we shall have SP = a - ex; and since SP + FP = 2a, FP = a + ex. Also it is evident that a²e² = a² - b², or e² = 1 - \frac{b²}{a²}.

e is called the eccentricity.

16. To find the polar equation to the ellipse, the focus being the pole.

Let SP = r, PSA = θ, then PSD = π - θ. But (Art. 15) SP = a - ex,

where x = CM = CS - SM = a - e - r cos PSM;

∴ r = a - ae² + er · cos PSM;

r = \frac{a(1 - e)}{1 - e \cos PSM} = \frac{a(1 - e)}{1 + e \cos \theta}

is the equation required.

17. e. The Hyperbola.—The property is (Conic Sections, Part 3, Def. 1), that the difference of the two lines drawn from any point to the two foci is constant.

The figure being drawn, and the same letters retained as for the ellipse, we shall have FP - SP = 2a, which equation being treated as the corresponding equation for the ellipse, will give, first, a · SP = CS · x - a²; and, finally (if CS² = a² + b²), a²y² - b²x² = -a²b² is the equation required.

18. The polar equation is r = \frac{a(e - 1)}{1 + e \cos \theta}

19. f. The Cissoid of Diocles.—This curve is generated by the following construction: AQB is a semi-circle, of which AB is the diameter; AQ is any chord; QN perpendicular to the diameter; PM is drawn parallel to QN at such a distance that AM = BN: the point P, in which the chord AQ and the line PM intersect one another, is a point in the cissoid APD: the cissoid is said to be the locus of the point P.

Let AB = 2a, AM = x, MP = y,

then y² : x² :: AN² : AN² : AN · NB : AN²

∴ NB : AN :: x : 2a - x;

∴ y² = \frac{x^3}{2a - x} is the equation required.

20. To find the polar equation.

Join BQ; let AP = r, ∠PAM = θ; then because AQB is a right angle,

AQ = AB cos θ, ∴ AN = AB cos² θ,

and AM = r cos θ, ∴ r cos θ = AB - AB cos² θ = AB sin² θ,

and r = \frac{2a \sin^2 \theta}{\cos \theta} is the equation required.

21. g. The Conchoid of Nicomedes.—This curve is generated as follows:

A line of indefinite length revolves on and also slides in a fixed pivot; whilst a constant portion of the line always projects beyond a given fixed straight line,—the extremity of the projecting line traces out a conchoid; or the conchoid is the locus of the extremity.

Let C be the pivot, AB the fixed line; PQ the constant portion of the revolving line which projects above it. Draw CA perpendicular to AB. Let PQ = a, CA = b, AM = x, MP = y.

By similar triangles,

PL² : LC² :: PM² : MQ² :: PM² : PQ² - PM²;

or (y + b)² : x² :: y² : a² - y²;

or x²y² = (y + b)²(a² - y²); which is the equation required.

22. To find the polar equation. Let C be the pole, Analytical Geometry.

23. b. The Cycloid.—This curve is described by a point in the circumference of a circle which rolls along a straight line.

Let GPD be the circle which rolls along the straight line BR; the point P in the circumference of the circle will trace out a curve BAR, which is the cycloid.

Let A be the highest point of the curve; GD = 2a the diameter of the generating circle = AC; AM = x, MP = y; O the centre of the circle; ∠POG = θ; then it is evident that the circumference of the circle coincides by succession with the straight line BR; therefore arc PD = BD, arc GPD = BC, and arc GP = DC; hence

\[ x = GL = a \cos \theta, \] \[ y = ML + LP = PG + PL = a \theta + a \sin \theta, \]

or \( y = a \text{ vers}^{-1} \frac{x}{a} + \sqrt{2ax - x^2} \) is the equation required.

24. i. The Lemniscate.—The property of this curve is that the product of the two lines drawn from any point of it to the two foci is always equal to the square of half the distance between the foci.

Let F and S be the foci, FC = CS = a, CM = x, MP = y; the property is that

\[ FP \cdot SP = a^2. \]

Hence \( \sqrt{(a + x)^2 + y^2} \times \sqrt{(a - x)^2 + y^2} = a^2; \)

or \( (y^2 + x^2 + a^2 + 2ax)(y^2 + x^2 + a^2 - 2ax) = a^4; \)

or \( (y^2 + x^2 + a^2)^2 - 4a^2x^2 = a^4; \)

or \( (y^2 + x^2)^2 + 2a^2y^2 - 2a^2x^2 = 0; \)

or \( (y^2 + x^2)^2 = 2a^2(x^2 - y^2) \) is the equation required.

25. To find the polar equation. Let CP = r, ∠PCS = θ; then

\[ FP^2 = r^2 + a^2 + 2ar \cos \theta; \] \[ SP^2 = r^2 + a^2 - 2ar \cos \theta; \] \[ FP^2 \times SP^2 \text{ or } a^4 = (r^2 + a^2)^2 - 4a^2r^2 \cos^2 \theta; \]

whence

\[ r^2 = 4a^2 \cos^2 \theta - 2a^2 = 2a^2(2 \cos^2 \theta - 1) = 2a^2 \cos 2\theta \]

is the equation required.

26. Besides the cycloid there are other curves which are generated by a point in a rolling circle. The trochoid differs from the cycloid in having the generating point within the circle at a distance b from the centre. Its equations are

\[ x = a - b \cos \theta, \] \[ y = ab + b \sin \theta. \]

The epicycloid is traced out by a point in the circumference of a circle which rolls on the convex circumference of another circle.

The hypocycloid is traced out by a point in the circumference of a circle which rolls on the concave circumference of a larger circle.

If the describing point is not in the circumference of the rolling circle, the curves become the epitrochoid and hypotrochoid respectively.

SECTION II.—THE DETERMINATION OF THE FORMS AND PROPERTIES OF CURVES FROM THEIR EQUATIONS.

27. Since very few curves can be described by a continuous motion, it is evident that, with rare exceptions, the determination of the form of a curve must resolve itself either into the expression of the numerical values of all coordinates corresponding to given abscissas, or into a general investigation of the number and nature of the branches of the curve, their flexure and mutual intersections. It is under the latter aspect that we are about to regard this subject. With respect to those few curves which can be described by a continuous motion, the modes of description of some of the most important have been already given in the treatise on Conic Sections, whilst others, such as the cycloid, involve a mechanical process so obvious that it is quite superfluous to waste words in attempting to make it plain. We proceed then at once to describe the general features of a few curves from their equations. This description we shall designate as tracing the curve.

28. a. The cubical parabola of which the equation is \( ax^3 = x^2 \).

When \( x = 0, y = 0 \), which shows that the curve passes through the origin. When \( x \) is positive, \( y \) is positive, which shows that to the right the curve lies above the axis of \( x \). When \( x \) is negative, \( y \) is negative, which shows that to the left the curve lies below the axis of \( x \).

Since \( \frac{dy}{dx} = 3x^2 \), it follows that at the origin the axis of \( x \) is a tangent to the curve (Fluxions, Art. 65, Equation 3). The form of the curve is therefore as in the figure.

29. b. The semi-cubical parabola, of which the equation is \( ay^3 = x^3 \).

When \( x = 0, y = 0 \). When \( x \) is positive \( y^2 \) is positive, or \( y \) has two equal values, one + the other −; which shows that there are two similar branches of the curve to the right of the origin, the one above and the other below the axis of \( x \). When \( x \) is negative \( y^2 \) is negative, and \( y \) is impossible; or there is no portion of the curve to the left of the axis of \( x \). The figure is like the outline of the head of a spear, or is a cusp.

30. c. The curve whose equation is \( y = \frac{x^3}{x^2 + 1} \). It is evident that its general form will be something like that of the cubical parabola already traced. But the equation may be thrown under the form \( y = x - \frac{1}{x} + &c. \); which shows that as \( x \) increases the value of \( y \) tends continually to be equal to it; which is equivalent to the fact that the curve constantly approaches a straight line whose equation is \( y = x \). This straight line is called an asymptote to the curve. An example of a similar line has already been given in the article Conic Sections (iii. 15) in the case of the hyperbola. A more instructive example will be seen in the conchoid already described, where, from the nature of the case, it is at once evident that the curve continually approaches the straight line AB, without ever reaching it.

To ascertain whether a curve has an asymptote or not, it is sufficient to expand its equation in a descending series of powers of \( x \).

If the portion which remains, after excluding that which becomes 0 on \( x \) becoming infinite, be in the form of a simple equation, that equation represents the asymptote. In addition to this there may be an asymptote parallel to the axis of \( y \), which can be ascertained by \( y \) becoming infinite for some finite value of \( x \).

31. d. There is another circumstance to be noted respecting the curve we have been discussing. Near the origin where the axis of \( x \) is a tangent, it is evidently convex to that axis; but from the nature of the case it is clear that at a considerable distance from the origin it must be convex towards its asymptote, and consequently concave towards the axis of \( x \). The point at which it changes from convex to concave is called a point of inflexion or a point of contrary flexure. To ascertain the position of this point, it is necessary to find the value of \( x \) which shall render \( \frac{d^2y}{dx^2} = 0 \). On solving this equation we shall obtain \( x = \pm \sqrt{3} \), which gives the point of inflexion. The curve is therefore as follows (fig. 11), OB being the asymptote, and P the point of inflexion.

32. e. To trace the curve of which the equation is

\[ xy^2 = (x-a)(x-b)(x-c). \]

Suppose \( a \) the least, and \( c \) the greatest of the three magnitudes. We have

\[ y = \sqrt{\frac{(x-a)(x-b)(x-c)}{x}}; \]

from which it is clear that in all cases there are two values of \( y \) equal and with opposite signs; or, in other words, there is symmetry with respect to the axis of \( x \).

When \( x = 0 \), \( y \) is infinite, or the axis of \( y \) is an asymptote.

When \( x < a \), \( y \) is impossible, or no branch of the curve lies between the origin and the distance \( a \) to the right.

When \( x = a \), \( y = 0 \); when \( x > a < b \), \( y \) is possible; and when \( x = b \), \( y = 0 \); the curve accordingly starts from the axis of \( x \) at the distance \( a \) from the origin, and returns to it again at the distance \( b \). This branch of the curve is consequently a kind of oval. If it be required to ascertain where is the broadest part of the oval, it will be necessary to find at what point \( y \) is a maximum by the method given in Fluxions, art. 60. The result is obtained by the solution of a cubic equation.

When \( x > b < c \), \( y \) is impossible.

When \( x = c \), \( y = 0 \); when \( x > c \), \( y \) is possible; and when \( x \) is infinite, \( y \) is infinite—the branch beyond the distance \( c \) consequently starts from the axis of \( x \) and extends to infinity, somewhat in the form of a parabola. When \( x \) is negative, \( y \) is always possible; and it is infinite both when \( x \) is 0 and when \( x \) is infinite—diminishing from the first point and then increasing to the second. At some point it must attain a minimum value, which can be ascertained as the maximum above. The branch to the left of the origin accordingly bears some faint resemblance to an hyperbola. The whole curve is as in the figure where O is the origin—OA = a, OB = b, OC = c.

33. f. In the last example we assumed that \( a \), \( b \), and \( c \) are all different from each other. If we suppose \( a = b \), we have the equation \( y = (x-a) \sqrt{\frac{x-c}{x}} \). The peculiarity in this equation is this—that when \( x = a \), \( y = 0 \), which gives a possible point A; but when \( x \) is a little greater or less than \( a \), \( y \) is impossible, so that no branch of the curve lies near the point. This point, an isolated point where co-ordinates satisfy the equation, but which does not belong to any branch of the curve, is called a conjugate point. By comparing this example with the last, of which it is a particular case, we shall be able to trace the origin of such a point in the oval AB, which has collapsed.

We will now give an example or two of the use of polar co-ordinates.

34. g. The Lemniscate, of which the equation (\( r \)) is

\[ r^2 = 2a^2 \cos 2\theta. \]

When \( \theta = 0 \), \( r = 2a^2 \), \( r = \pm a\sqrt{2} \); or \( r \) has two equal values, CA and CB in the figure of Art. 24.

When \( \theta < \frac{\pi}{4} \), \( r \) is possible, having two equal values, positive and negative, the former giving the upper half of the right-hand portion, and the latter the lower half of the left.

When \( \theta = \frac{\pi}{4} \), \( r = 0 \); when \( \theta > \frac{\pi}{4} < \frac{3\pi}{4} \), \( r \) is impossible.

When \( \theta = \frac{3\pi}{4} \), \( r = 0 \); when \( \theta = \pi \), \( r = \pm a\sqrt{2} \), thus giving the upper portion of the left-hand figure, and the lower portion of the right.

35. h. The Cardioid, of which the equation is

\[ r = a \cos^2 \frac{\theta}{2}. \]

When \( \theta = 0 \), \( r = a \); when \( \theta < \frac{\pi}{2} \), \( r \) is positive; when \( \theta = \frac{\pi}{2} \), \( r = 0 \); and the same is true in a reverse order between \( \frac{\pi}{2} \) and \( 2\pi \). The curve, therefore, starts at its greatest distance from the origin, and returns to the origin by having positive radii throughout a whole circumference—a portion of it accordingly lies to the left of the axis of \( y \). The curve is in the shape of a heart, thus: O being the origin; OA = a.

Many curves, whose equations are presented between linear co-ordinates, have their general form, and some of their properties, determined most readily by converting the given equation into a polar equation, as in the following example:

36. i.

\[ x^4 - ax^2y + ay^3 = 0. \]

Let \( x = r \cos \theta \), \( y = r \sin \theta \); then the equation gives

\[ r = a \sin \theta \left( \cos^2 \theta - \sin^2 \theta \right). \]

When \( \theta = 0 \), \( r = 0 \); when \( \theta < \frac{\pi}{4} \), \( r \) is positive; when \( \theta = \frac{\pi}{4} \), \( r = 0 \): hence there is an oval in the first octant, having its extremity at the origin.

When \( \theta > \frac{\pi}{4} < \frac{3\pi}{4} \), \( r \) is negative; when \( \theta = \frac{3\pi}{4} \), \( r = 0 \); hence there is no branch of the curve in the second octant, but one extending back from the origin to infinity in the sixth octant.

When \( \theta > \frac{3\pi}{4} < \pi \), \( r \) is negative; when \( \theta = \frac{3\pi}{4} \), \( r = 0 \); hence there is no branch of the curve in the third octant, but a branch returning from infinity to the origin in the seventh.

When \( \theta > \frac{3\pi}{4} < \pi \), \( r \) is positive; when \( \theta = \pi \), \( r = 0 \); hence there is an oval in the fourth octant, having its extremity at the origin. The maximum radius vector occurs at about 30. The curve is therefore in the annexed form.

37. These examples will suffice to illustrate the general determination of the form of a curve from its equation. We have next to discuss the determination of the properties of curves from their equations. This we shall do very briefly, confining ourselves to curves of the second degree. As a preliminary it will be requisite to point out the method employed to transfer our reference from one system of co-ordinates to another. This method is called... Transformation of Co-ordinates.

38. It is supposed that the equation to a curve referred to a certain system of co-ordinates \(x, y\) is given; and it is required to determine what its equation becomes when referred to another system of co-ordinates \(x', y'\). The problem obviously resolves itself into the determination of the values of \(x\) and \(y\) in terms of \(x'\) and \(y'\), and the substitution of the latter in place of the former. The formulae are thus obtained.

39. 1°. If the new axes are parallel to the old, the origin alone being changed—it is evident that nothing is required but to write \(x' + a\) in place of \(x\), and \(y' + b\) in place of \(y\); where \(a\) and \(b\) are the co-ordinates of the new origin referred to the old.

40. 2°. Suppose the origin unchanged, and both the old and new axes rectangular, the axis of \(x'\) being inclined to that of \(x\) by the angle \(\theta\).

Let \(OM = x, MP = y, ON = x', NP = y'\). Draw \(NR, NQ\) parallel to \(Oy\) and \(Ox\) respectively;

then

\[x = OR - QN = x' \cos \theta - y' \sin \theta.\]

\[y = NR + PQ = x' \sin \theta + y' \cos \theta.\]

41. 3°. Suppose the origin to be unchanged; and neither the old nor the new axes to be necessarily rectangular.

Let \(OM = x, MP = y, ON = x', NP = y'\). Draw \(ME, NF\) perpendicular to \(Oy\), and let \(NG\) be parallel to \(Oy\) and meet \(EM\) produced in \(G\); then \(ME = NF - GM,\)

or

\[x' \sin x Oy = x' \sin x' Oy - y' \sin y' Oy.\]

Similarly, by drawing perpendiculars from \(P\) and \(N\) on \(Ox\), we get

\[y \sin y Ox = x' \sin x' Ox + y' \sin y' Ox.\]

42. 4°. When both the origin and the direction of the axes are to be changed, we have only to unite the processes of No. 1 and Nos. 2 and 3 to effect the transformation.

43. 5°. To obtain a polar equation, let \(OP = r\);

then

\[x = r \frac{\sin OPM}{\sin PMx} = \frac{\sin POy}{\sin x Oy},\]

\[y = r \frac{\sin POM}{\sin PMx} = \frac{\sin POx}{\sin x Oy}.\]

Properties of Curves of the Second Degree, as deduced from the general equation.

The general equation of the second degree is

\[Ay^2 + 2 Bxy + Cx^2 + 2 Dy + 2 Ex + F = 0.\]

44. Prop. I.—No straight line can cut a curve of the second degree in more than two points.

Let \(y = mx + n\) be the equation to a straight line, then by substituting this value of \(y\) in the given equation, we have a quadratic for the determination of the values of \(x\) at the points where the co-ordinates are common to it and to the straight line. A similar process gives a quadratic equation in \(y\). Consequently there are but two solutions, and therefore but two points of intersection.

By referring to the equations to the conic sections given above, it will be seen that they are curves of the second degree.

45. Prop. II.—Admitting that the above equation represents a conic section, it is required to determine the particular one.

If we solve the equation with respect to \(y\), we obtain

\[Ay = -(Bx + D) \pm \sqrt{(B^2 - AC)x^2 + 2(BD - AE)x + D^2 - AF}.\]

Now, I. If \(B^2 - AC\) be negative, this value of \(y\) is impossible when \(x\) is infinite; which shows that the curve does not extend to an infinite distance. It is therefore a circle or an ellipse, according as \(C\) is equal to \(A\) or not. The ellipse, however, may be simply a point, and the equation may be impossible.

II. If \(B^2 - AC\) be positive, \(y\) has two real roots when \(x\) is infinite, whether \(x\) be positive or negative. The curve has consequently four infinite branches as in the hyperbola or in two intersecting straight lines.

III. If \(B^2 - AC = 0\); then if \(BD - AE\) is positive, \(y\) will be possible when \(x\) is positive and infinite, but impossible when \(x\) is negative, and vice versa. The curve is consequently a parabola. If \(BD - AE = 0\), the equation represents two straight lines or one, or is impossible.

IV. If the quantity under the radical is a complete square, i.e., if \((B^2 - AC)(D^2 - AF) = (BD - AE)^2\) the equation represents two straight lines; for it is the product of two simple equations.

46. Prop. III.—To find the centre of the curve.

The centre is that point which bisects every chord which can be drawn through it. Let \(a, b\) be the co-ordinates of the centre; \(x', y'\), the co-ordinates when the centre is the origin; then (Art. 39) we have \(x = x' + a, y = y' + b\); whence

\[A(y' + b)^2 + 2B(x' + a)(y' + b) + C(x' + a)^2 + 2D(y' + b) + 2E(x' + a) + F = 0.\]

Now when \(y = 0\), this equation must give two values of \(x'\) equal and with opposite signs, because the chord of \(x'\) is bisected at the centre. Hence their sum is equal to 0;

or

\[Bb + Ca + E = 0;\]

similarly,

\[Ab + Ba + D = 0;\]

whence

\[a = -\frac{BD - AE}{B^2 - AC}, b = -\frac{BE - CD}{B^2 - AC},\]

which are the co-ordinates of the centre.

In the case of the parabola these co-ordinates are infinite, for (Art. 45) \(B^2 - AC = 0\); hence we may say that the parabola has no centre. In analysis, however, it is desirable to regard the centre as existing at an infinite distance.

47. Prop. IV.—To find the locus of the middle points of parallel chords.

Let \(a, b\) be the co-ordinates of one of those middle points, and let the origin be transferred to that point by writing \(x' + a, y' + b\) for \(x\) and \(y\). Let also \(y = mx\) be the equation to this chord; then the points in which it intersects the curve will be those for which \(x'\) and \(y'\) are the same in the equation to the curve and to the chord. Hence

\[A(mx' + b)^2 + 2B(mx' + a)(mx' + b) + C(mx' + a)^2 + 2D(mx' + b) + 2E(mx' + a) + F = 0\]

is the equation for determining the values of \(x'\) at the points of section; but since these are by hypothesis equal and with opposite signs, the co-efficient of \(x'\) must be equal to 0;

therefore

\[Amx + Bb + Bma + Ca + Dm + E = 0.\]

Now \(m\) remains the same for all chords which are parallel. This equation is therefore a simple equation between \(a\) and \(b\); consequently if \(a\) and \(b\) be considered as the co-ordinates of middle points generally, the equation which connects them is the equation to a straight line. The locus of the middle points of all parallel chords is therefore a straight line. 48. Every straight line which bisects a system of parallel chords is called a diameter, and the chords themselves are called ordinates to that diameter.

49. Cor.—From the definition of a centre it is evident that every diameter passes through it.

In the case of the parabola, which has no finitely situated centre, we observe that \( B^2 = AC \). Now the equation to the diameter (Art. 47) is

\[ Bb + Ca + m(Ab + Ba) + E + mD = 0; \]

or

\[ Bb + Ca + m\left(\frac{B^2}{C}b + Ba\right) + E + mD = 0; \]

or

\[ \left(1 + \frac{mB}{C}\right)(Bb + Ca) + E + mD = 0; \]

i.e.

\[ Bb + Ca + \frac{C}{1 + mB}(E + mD) = 0; \]

in which the co-efficients of the co-ordinates \(a\) and \(b\) are independent of \(m\), which shows (Art. 8) that all diameters of the parabola are parallel to one another.

50. Prop. V.—If two diameters of a curve of the second degree be such that one of them bisects all chords parallel to the other, then the latter also will bisect all chords parallel to the former.

By Art. 47 the equation to the line which bisects all chords parallel to the line whose equation is \(y = mx\), is

\[ (Am + B)b + (Bm + C)a + Dm + E = 0, \]

or

\[ y + \frac{Bm + C}{Am + B}x + \frac{Dm + E}{Am + B} = 0, \]

by writing \(y\) and \(x\) for the symbols which stand for the general phrases latitude and longitude of a point in the line. Let this be abbreviated by \(y = mx + &c.\),

where

\[ m' = -\frac{Bm + C}{Am + B} \]

In the same manner the diameter which bisects all chords parallel to \(y = mx\) is \(y = m'x\);

where \(m'' = -\frac{Bm' + C}{Am' + B}\)

\[ = -\frac{Bm + BC}{Am + B} + C \]

\[ = -\frac{ABm + AC}{Am + B} + B \]

\[ = m \]

i.e. (Art 8), this diameter is parallel to the original chord; hence the truth of the proposition. Diameters so related to each other are called conjugate diameters.

51. Prop. VI.—If through any point \(O\) two chords be drawn, meeting the curve in the points \(P\), \(Q\), and \(R\), \(S\), respectively; then, the ratio of the rectangle \(OP.OQ\) to the rectangle \(OR.OS\), is a ratio which is the same whatever be the point \(O\), provided the direction of the chords remain unchanged.

Let the co-ordinates of \(O\) be \(a\) and \(b\); the polar radius through \(O = r\); then (Art. 43) we have

\[ x = r\frac{\sin POy}{\sin xOy} + a, \quad y = r\frac{\sin POx}{\sin xOy} + b \]

\(mr + a\) suppose \(mr + b\),

where \(m\) and \(n\) depend on the direction of the radius or chord.

The general equation becomes, by substitution,

\[ A(nr + b)^2 + 2B(nr + a)(nr + b) + C(nr + a)^2 + 2D(nr + b) + 2E(nr + a) + F = 0; \]

which gives two values of \(r\), viz. OP, OQ. Their product is the last term of the equation, viz.

\[ \frac{Ab^2 + 2Bab + Ca^2 + 2Db + 2Ea + F}{An^2 + 2Bmn + Cm^2} \]

Similarly the product \(OR.OS\) is

\[ \frac{Ab^2 + 2Bab + Ca^2 + 2Db + 2Ea + F}{An^2 + 2Bmn + Cm^2} \]

Consequently \(OP.OQ : OR.OS : : An^2 + 2Bmn + Cm^2 : An^2 + 2Bmn + Cm^2\);

a ratio which is independent of \(a\) and \(b\), and therefore of the position of the point \(O\). If \(O'\) be any other point, and \(O'P\) be parallel to \(OP\), \(O'R'\) to \(OR\), &c.; then

\[ OP.OQ : OR.OS : : O'P.O'Q : O'R'.O'S'. \]

52. Cor. I.—If \(O'\) be the centre; \(O'P = O'Q\), \(O'R' = O'S'\); then \(OP.OQ : OR.OS : : O'P^2 : O'R'^2\). That is, if two chords intersect one another, the rectangles by their segments are to one another as the squares of the diameters parallel to them respectively.

53. Cor. 2.—\(O'\) being still the centre, let \(O'O\) be the diameter of which \(PQ\) is the ordinate; and let \(RS\) and also \(RS'\) pass through \(O'\) and coincide with one another; then \(OP^2 : OR.OS : : O'P^2 : O'R'^2\); or the square of the ordinate is to the rectangle by the abscissas, as the square of the diameter which is parallel to the former to the square of that which passes through the latter.

54. In the case of the parabola this proposition is inapplicable; for the diameter and the chord parallel to it are both infinite. To determine the corresponding ratio for this case, let the axes be transformed to \(x', y'\) such that the axis of \(x'\) is parallel to the diameter, and the axis of \(y'\) to a bisected chord. Let also the origin be on the curve—the equation will then assume the form \(Ay'^2 + 2Ex' = 0\); for \(C = 0\); because when \(y = 0\) there must be only one value of \(x'\), viz., \(x' = 0\); but \(B^2 = 4AC\); \(B = 0\); and since the two values of \(y\) are equal and with opposite signs we must have \(D = 0\); lastly, since the origin is on the curve we must have \(F = 0\). The equation in this form shows that the abscissas are to one another as the squares of the ordinates.

The theorems contained in this second corollary constitute the fundamental properties of the conic sections, and are the same as demonstrated in the treatise on that subject. (Arts. i. 12, ii. 13, iii. 12.)

55. Prop. VII.—To find the equation to the tangent.

The tangent is that line to which a line cutting the curve in two points continually approaches as its limit, as the points of section approach to each other.

The equation to a line which meets the curve in the two points \(x'y', x'y'\) is

\[ y - y' = \frac{y' - y}{x' - x}(x - x') \quad (\text{Art. 8, Cor. 2}) \]

Now \(Ay'^2 + 2Bx'y' + Cx'^2 + 2Dy' + 2Ex' + F = 0\)

and \(Ay'^2 + 2Bx'y' + Cx'^2 + 2Dy' + 2Ex' + F = 0\);

\[\begin{align*} &\text{by subtraction,} \\ &A(y'^2 - y^2) + B(x'(y' - y') + y'(x' - x') + x'(y' - y') \\ &+ y'(x' - x')) + C(x'^2 - x^2) + 2D(y' - y') \\ &+ 2E(x' - x') = 0; \end{align*}\]

whence

\[ \frac{y' - y}{x' - x} = -\frac{B(y' + y) + C(x' + x') + 2E}{A(y' + y) + B(x' + x') + 2D}; \]

and the equation to the cutting line is

\[ y - y' = -\frac{B(y' + y) + C(x' + x') + 2E}{A(y' + y) + B(x' + x') + 2D}(x - x'). \]

The equation to the tangent is derived from this by writing \(y'\) for \(y'\) and \(x'\) for \(x'\). It is therefore

\[ y - y' = -\frac{By' + Cy' + Ex'}{Ay' + By' + Dx'}(x - x'). \]

Those who are familiar with the differential calculus will obtain this result at once from the equation

\[ y - y' = \frac{dy}{dx}(x - x'). \]

If we multiply out equation (2), and substitute for \(Ay'^2 + Analytical Geometry.

2 Bx' y' + Cx^2 + Dy' + Ex' - F from equation (1), shall value - Dy' - Ex' - F from

Ayy' + B (xy + x'y) + Cxx' + D (y + y') + E (x + x') + F = 0

as the equation to the tangent. The equation to the tangent is therefore to be got from the equation to the curve by writing xx' for x^2, yy' for y^2, x + x' for 2x, y + y' for 2y, and xy + x'y for 2xy; or by changing one of the x's into x', and one of the y's into y'.

56. Cor.—Since the equation to the tangent is symmetrical in x and x', y and y', if we abbreviate it by ϕ(x, y, x', y'), we shall have the equation ϕ(x, y, x', y') = ϕ(x, y, x, y); x and y being interchangeable with x' and y'.

57. Prop. VIII.—To find the equation to the straight line which joins the points of contact of two tangents, real or imaginary.

Abbreviate the equation to the curve by f(x, y) = 0, and to the tangent by ϕ(x, y, x', y') = 0.

Let X, Y be the co-ordinates of the point of intersection of two tangents; then the equations f(x', y') = 0, ϕ(x', y', X, Y) = 0 give the values of x', y', the co-ordinates of the point of contact of the one tangent.

Also the equations f(x', y') = 0, ϕ(x', y', X, Y) = 0, give the co-ordinates x', y' of the point of contact of the other tangent which passes through the point X, Y.

Now, of the equations constituting these respective pairs, one is a simple equation, and the other a quadratic. Each pair will therefore give two values of the variables. But each pair is the same as the other; therefore the two roots of the one are the roots of both; or x', y', x", y" are the roots of the equations f(x, y) = 0, ϕ(x, y, X, Y) = 0.

Again, let ψ(x, y, X, Y) = 0 be the equation to the chord which passes through the points (x', y', x", y") of contact; then will x', y', x", y" be the roots of the equations f(x, y) = 0, ψ(x, y, X, Y) = 0.

But it has been already proved that

f(x, y) = 0, ϕ(x, y, X, Y) = 0

have the same roots. Moreover, one of the equations, f(x, y) = 0, is common to both pairs; therefore the other two equations must be identical; or ψ(x, y, X, Y) = 0 is the same as ϕ(x, y, X, Y) = 0; i.e., the equation to the line which joins the points of contact of two tangents is of the same form as the equation to the tangent; or if the equation to the tangent at the point x', y' in the curve is ϕ(x, y, x', y') = 0; the equation to the line which joins the points of contact of two tangents which meet in a point x', y', out of the curve is ϕ(x, y, x', y') = 0.

58. The line which joins the points of contact of two tangents is called the polar of their point of intersection; and that point accordingly is called the pole of the line.

59. Prop. IX.—If the point B is in the polar of A, then is A also in the polar of B.

Let x', y' be the co-ordinates of A,

x', y', ... ... B.

The equation to the polar of A is ϕ(x, y, x', y') = 0; but by hypothesis B (x', y') is a point in this line; therefore ϕ(x', y', x', y') = 0. Hence also (Art. 56) ϕ(x', y', x', y') = 0.

But ϕ(x', y', x', y') = 0 is the equation to the polar of B; the equation ϕ(x', y', x', y') = 0 consequently shows that x', y' are the co-ordinates of a point in that line; i.e., A is a point in the polar of B.

60. Cor. 1.—If any number of points be taken in the polar of A, their polars will all pass through A; i.e., the polars of every point in a straight line all pass through the pole of that line.

61. Cor. 2.—If A and B are the poles of two lines PQ and RS which meet in O, the line AB is the polar of the point O.

For the polar of O passes through A (by the Proposition) and through B; it is therefore the straight line AB.

62. Prop. X.—To find the condition that the straight line whose equation is mx + nx + p = 0 may touch a given conic section.

The equation to the conic section may, by properly selecting the co-ordinates, be thrown into one or other of the two forms,

\[ \frac{x^2}{a} + \frac{y^2}{b} = 1 \quad \text{or} \quad \frac{2x}{a} + \frac{y^2}{b} = 1. \]

The tangents to these have for their respective equations (Art. 55),

\[ \frac{x'}{a} + \frac{y'}{b} = 1, \quad \text{and} \quad \frac{x'^2}{a} + \frac{y'^2}{b} = 1. \]

In order that the former of these may coincide with the equation mx + nx + p = 0, we must have

\[ -\frac{n}{p} = \frac{x'}{a}, \quad -\frac{m}{p} = \frac{y'}{b}. \]

But \( \frac{x'^2}{a} + \frac{y'^2}{b} = 1 \); whence \( a^2 n^2 + b^2 m^2 = p^2 \)..............(1)

is the resulting condition.

In order that the latter of the above equations may coincide with the equation mx + nx + p = 0, we must have

\[ -\frac{n}{p} = \frac{1}{a}, \quad -\frac{m}{p} = \frac{1}{b}. \]

\[ \therefore \quad \frac{x'}{a} = 1 + \frac{p}{an}, \quad \frac{y'}{b} = \frac{m}{an}; \]

which being substituted in the equation

\[ \frac{2x}{a} + \frac{y^2}{b} = 1, \]

there results

\[ a^2 n^2 + 2 an p + b^2 m^2 = 0 \].................(2)

as the condition required.

63. Prop. XI.—Given that chords to a curve of the second degree always touch a conic section, it is required to prove that the locus of their poles is a conic section.

Let A be the conic section to which the chords are always tangents; B the curve of the second degree (a conic section) to which the chords are drawn; C the curve which is the locus of the poles of the chords to B.

Let x', y' be the co-ordinates of a point in C, the equation to the polar of this point is

\[ \phi(x, y, x', y') = 0 \quad \text{(Art. 57)} \quad \text{or} \quad (Ay' + Bx' + D) y + (By' + Cx' + E) x + Dy' + Ex' + F = 0 \quad \text{(Art. 55)}. \]

In order that this line may be a tangent to the conic section A, whose equation we shall suppose to be either

\[ \frac{x^2}{a} + \frac{y^2}{b} = 1 \quad \text{or} \quad \frac{2x}{a} + \frac{y^2}{b} = 1, \]

we have to satisfy one or other of the two conditions obtained in the last proposition, viz. Analytical Geometry.

\[a(By' + Cx' + E)^2 + b(Ay' + Bx' + D)^2 = (Dy' + Ex' + F)^2\] or \[a(By' + Cx' + E)^2 + 2a(By' + Cx' + E)(Dy' + Ex' + F) + b(Ay' + Bx' + D)^2 = 0\]

as necessary to be fulfilled.

If the requisite one of these relations between \(x'\) and \(y'\) hold true, the polar to every point in \(C\) will touch \(A\). But these conditions are both expressed in the form of a quadratic equation between \(x'\) and \(y'\). Hence the curve \(C\) is a curve of the second degree or a conic section.

64. Prop. XII.—Given that \(C\) is generated by \(A\) through the intervention of \(B\), as in the last proposition, to prove that \(A\) would be generated by \(C\) through the intervention of the same curve \(B\).

Let two tangents be drawn to \(A\), forming chords to \(B\), and let the poles of those chords be \(C\) and \(C'\) in the curve \(C\); then (Art. 61) the point of intersection of those tangents is the pole of \(CC'\) (a chord in \(C\)). But as the tangents become indefinitely near each other, their point of intersection tends to become the point \(A\) in the curve \(A\); and the points \(C, C'\) also tend to become indefinitely near each other, or the chord \(CC'\) tends to become a tangent. Hence the point \(A\) on the curve \(A\) is the pole of the tangent to a point \(C\) on the curve \(C\).

65. The curves \(A\) and \(C\) are therefore reciprocal, and the process which employs this method of demonstration is termed the method of reciprocal polars. We shall confine ourselves to a single application of this method, that of connecting two theorems relative to the conic sections which are known as Pascal's and Brianchon's theorem respectively.

Pascal's Theorem.

66. Prop. XIII.—If any hexagon be inscribed in a conic section, and the opposite sides be produced to meet in three points, those points are in a straight line.

Let \(ABCDEF\) be a hexagon inscribed in a conic section, and let the opposite sides produced meet in the points \(G, H,\) and \(K\); viz., \(AB\) and \(ED\) in \(G\); \(BC\) and \(EF\) in \(H\); \(CD\) and \(AF\) in \(K\); then the points \(G, H,\) and \(K\) are in a straight line. Let \(u = 0, v = 0, w = 0\), be three simple equations between \(x\) and \(y\), each multiplied by an arbitrary constant, as in Art. 7; and let \(u = 0\) (1) be the equation to the line \(AB, v = 0\) (2) to the line \(CD,\) and \(w = 0\) (3) to the line \(EF\).

Now since \(u, v, w\) involve \(x\) and \(y, u^2, v^2, w^2,\) &c., will involve \(x^2, y^2,\) &c., consequently

\[u^2 + v^2 + w^2 - (\lambda + \frac{1}{\lambda})uv - (\mu + \frac{1}{\mu})uw - (\nu + \frac{1}{\nu})vw = 0\]

is an equation between \(x, y, x^2, y^2,\) &c., containing six arbitrary constants; and is therefore the general equation of the second degree which represents any conic section.

At the points \(A\) and \(B\), where the straight line represented by equation (1) intersects the conic section represented by (4), the same values of \(x\) and \(y\) render both equations identical, or \(u\) is the same in fact as well as in form in both equations:

We must have, therefore,

\[v^2 + w^2 - (\lambda + \frac{1}{\lambda})vw = 0,\]

or \(v = \lambda w\) and \(v = \frac{1}{\lambda} w\)

at those points respectively. Similarly at the points \(D\) and \(C\) we must have \(u = \mu v,\) \(u = \frac{1}{\mu} v\) respectively, and at the points \(E\) and \(F, u = \nu v\) and \(u = \frac{1}{\nu} v.\) Hence the points are thus determined, viz.:

\[A \text{ by the equations } u = 0, v = \lambda w \quad \ldots \quad (5)\] \[B \quad \ldots \quad u = 0, v = \frac{1}{\lambda} w \quad \ldots \quad (6)\] \[C \quad \ldots \quad v = 0, u = \mu w \quad \ldots \quad (7)\] \[D \quad \ldots \quad v = 0, u = \frac{1}{\mu} w \quad \ldots \quad (8)\] \[E \quad \ldots \quad w = 0, u = \nu w \quad \ldots \quad (9)\] \[F \quad \ldots \quad w = 0, u = \frac{1}{\nu} w \quad \ldots \quad (10)\]

Let \(u = Av + Bw\) be the equation to the line which passes through the points \(B\) and \(C\); then, by (6) we must have

\[0 = \frac{A}{\lambda} + B;\] and by (7) \(\frac{1}{\mu} = B:\) consequently

\[u = \frac{1}{\mu} (-\lambda v + w)\]

is the equation to the line \(BC\). Similarly \(u = \nu v + \mu w\) (12) is the equation to the line \(DE\), and \(u = \frac{1}{\nu} (v - \lambda w)\) (13) is the equation to the line \(FA\).

Let \(Au + Bv + Cw = 0\) be the equation to \(GK\).

This equation and the equations to \(AB\) and \(ED\) must be simultaneously satisfied at the point \(G\). Combining then (1) and (12) with this equation we get \(\frac{\mu}{C} = \frac{\nu}{B}\).

In like manner the equation must be satisfied simultaneously with those to \(AF\) and \(CD\), at the point \(K\). The equations (2) and (13) combined with it give \(\frac{\nu}{A} = \frac{\lambda}{C}\), whence

\[A = \frac{1}{\lambda}, \quad B = \frac{1}{\mu}, \quad C = \frac{1}{\nu};\] and the equation to the line \(GK\) becomes

\[\frac{u}{\lambda} + \frac{v}{\mu} + \frac{w}{\nu} = 0 \quad \ldots \quad (14).\]

Again, let \(Au + Bv + Cw = 0\) be the equation to the line \(GH\). The point \(G\) gives as before \(\frac{\mu}{C} = \frac{\nu}{B}\); at the point \(H\) the equation must be satisfied simultaneously with the equa- 67. Con.—It is evident that the demonstration does not require any order of position amongst the points, or, in other words, that the points are interchangeable. Consequently, if for hexagon we read six points, and admit every variety of modes of joining them, we shall obtain a great number (sixty) of different straight lines in which three points of intersection of joining lines lie.

Briancon's Theorem.

68. Prop. XIV.—If any hexagon be described about a conic section, the three diagonals which join opposite angles meet in a point.

Let two tangents be drawn to a conic section A as in Art. 63, meeting in the point P, and let their poles in the conic section C be a, b; then (Art. 61) P is the pole of the line ab. The same is true of any other pair of tangents; consequently, if six tangents be drawn to A forming a hexagon about it, their points of intersection will be the poles of a hexagon in C.

Let us designate the angles of the hexagon described about A by P, Q, R, S, T, U; and the sides of the hexagon described in C to which they are respectively poles by p, q, r, s, t, u.

Let p and s meet in G; q and t in H, and r and u in K; then, by the last proposition, G, H, and K are in a straight line.

Now (Art. 61) G is the pole of PS, H of QT, and K of RU; and since G, H, and K are in a straight line, their polars PS, QT, RU pass through the pole of that line (Art. 60); or meet in a point.

SECTION III.—DEDUCTION OF PROPERTIES OF CURVES FROM THEIR EQUATIONS AND KNOWN PROPERTIES.

We propose in this section to deduce some of the more important properties of the conic sections from their equations, combined with their known geometrical figures. The axes of co-ordinates will be supposed rectangular unless otherwise expressed.

1. The Straight Line.

69. Prop. I.—To find the equation to a straight line which shall be perpendicular to a given straight line.

Let \( y = ax + b \) be the equation to the given straight line, \( y = px + q \) that of the required line.

If \( \theta, \phi \) be the angles in which those lines respectively cut the axis of \( x \); it is evident that the condition imposed requires that

\[ \phi = \frac{\pi}{2} + \theta; \]

\[ \tan \phi = -\cot \theta = -\frac{1}{\tan \theta}; \]

or \( p = -\frac{1}{a} \) (Art. 6);

hence \( y = -\frac{1}{a} x + q \) is the equation required.

70. Prop. II.—To find the length of the perpendicular from a given point on a given line.

Let \( x', y' \) be the co-ordinates of the given point A;

\( y = ax + b \) the equation to the given line PR.

Draw AP perpendicular to the given line, and let \( y' \) cut RP in E;

then \( AE = AD - ED = y' - (ax' + b) \)

and \( AP = AE \sin AEP = AE \cos PRM \)

\[ = \frac{AE}{\sqrt{1 + \tan^2 PRM}} = \frac{y' - (ax' + b)}{\sqrt{1 + a^2}} \]

71. We may exhibit this result in a very different form by adopting the equation to a straight line given in Art. 7.

Let \( AP = P \); then the equation to a line through A parallel to RP is \( y \cos a - x \sin a - (p + P) = 0 \); and since A is a point in this line,

\[ y' \cos a - x' \sin a - (p + P) = 0; \]

or \( P = y' \cos a - x' \sin a - p \);

\( = a \) (Art. 7) when \( x', y' \) are written for \( x \) and \( y \).

72. Prop. III.—If \( a = 0, \beta = 0 \) be the equations to two straight lines, agreeably to the abbreviated notation of Art. 7; then \( la + m\beta = 0 \), or \( a - k\beta = 0 \) will represent a line which passes through their point of intersection.

For the co-ordinates which render both \( a = 0 \) and \( \beta = 0 \) render \( la + m\beta = 0 \), or \( a - k\beta = 0 \); and are therefore co-ordinates of a point in the last line as in the other two.

73. Prop. IV.—If \( a = 0 \) be the equation to the line QA, \( \beta = 0 \) to the line QB, and \( a - k\beta = 0 \) to the line QC;

then \( k = \frac{\sin AQC}{\sin BQC} \)

Let \( x', y' \) be the co-ordinates of C.

Draw CA, CB perpendicular to QA, QB; then (Art. 71)

\( CA = y' \cos a - x' \sin a - p \),

\( = a \), when \( x', y' \) are written in place of the co-ordinates.

CB = \( \beta \), when \( x', y' \) are written for the co-ordinates.

But since \( x', y' \) is a point in the line \( a - k\beta = 0 \), we have

\( k = \frac{a}{\beta} \) when \( x', y' \) are written for the co-ordinates,

\[ CA = \frac{QC \sin AQC}{QC \sin BQC} = \frac{\sin AQC}{\sin BQC} \]

74. Cor. 1.—The equation to the line QC is therefore \( a \sin BQC - \beta \sin AQC = 0 \).

75. Cor. 2.—The straight line which bisects the angle between two straight lines whose equations are \( a = 0, \beta = 0 \) is \( a - \beta = 0 \).

76. Prop. V.—If \( a = 0, \beta = 0, \gamma = 0 \) be the equations to three straight lines, and it be possible to find three numerical quantities, \( l, m, n \), such that \( la + m\beta + n\gamma \) shall be equal to 0; the three straight lines shall meet in one point.

At the point where two of them whose equations are \( a = 0, \beta = 0 \) intersect, the co-ordinates render \( la + m\beta = 0 \). But, by hypothesis, \( la + m\beta + n\gamma = 0 \) for all co-ordinates; Analytical Geometry.

77. Prop. VI.—The straight lines which bisect the angles of a triangle meet in a point.

Let \(a = 0\), \(\beta = 0\), \(\gamma = 0\) be the equations to the sides of the triangle; \(a - \beta = 0\), \(a - \gamma = 0\), \(\beta - \gamma = 0\) (Art. 75) are the equations to the lines which bisect the angles. But \(a - \beta - (\alpha - \gamma) + \beta - \gamma = 0\); consequently (Art. 76) these lines meet in a point.

78. Prop. VII.—The perpendiculars from the angles of a triangle on the opposite sides meet in a point.

Let the angle between the lines whose equations are \(a = 0\), \(\beta = 0\) be denoted by \((a\beta)\), &c.; then (Art. 74) the equation to the perpendicular from the angle in which these two meet is

\[u = a \cos (\beta y) - \beta \cos (ay) = 0.\]

Similarly the equations to the other two perpendiculars are

\[v = y \cos (a\beta) - a \cos (\gamma\beta) = 0,\] \[w = \beta \cos (\gamma a) - \gamma \cos (\beta a) = 0.\]

Hence we have \(u + v + w = 0\), therefore (Art. 76) the three lines meet in a point.

79. Prop. VIII.—If there be two triangles such that the perpendiculars from the angles of the one on the sides of the other meet in a point, then the perpendiculars from the angles of the latter on the sides of the former will also meet in a point.

Let the equations to the sides of the former triangle be \(a = 0\), \(\beta = 0\), \(\gamma = 0\); and of the latter \(a' = 0\), \(\beta' = 0\), \(\gamma' = 0\). Let also \((a\beta)\) denote the angle between the lines \(a = 0\), \(\beta = 0\); \((a'\beta')\) between the lines \(a' = 0\), \(\beta' = 0\), &c. Then (Art. 74) the equation to the perpendicular from the point \((a\beta)\) on the line \(\gamma = 0\) is \(a \cos (\beta y) - \beta \cos (ay) = 0\); and the equation to the perpendicular from the point \((a'\beta')\) on the line \(\gamma' = 0\) is \(a' \cos (\beta' y) - \beta' \cos (a'y) = 0\); and from \((\beta y)\) on \(a = 0\), \(\beta \cos (\gamma a) - \gamma \cos (\beta a) = 0\).

Now at the point where these three lines meet, \(a\) is the same in (1) and (2), \(\beta\) in (1) and (3), and \(\gamma\) in (2) and (3). Hence by elimination between (1) and (3), we get

\[a \cos (\beta y) \cos (\gamma a) - \gamma \cos (\beta a) \cos (ay) = 0;\]

which, when combined with (2), gives

\[\cos (\beta y) \cos (\gamma a) \cos (a\beta) = \cos (\beta a) \cos (ay) \cos (\gamma\beta).\]

Now, if we change \(a\) into \(a'\), \(\beta\) into \(\beta'\), \(\gamma\) into \(\gamma'\), and conversely, we obtain the equations to the perpendiculars from the angles of the latter triangle on the sides of the former. At the point where the first and third of these lines intersect, we shall have, as before,

\[a' \cos (\beta' y) \cos (\gamma' a) - \gamma' \cos (\beta' a) \cos (a'y) = 0;\]

and the equation to the second line is

\[a' \cos (\gamma' \beta) - \gamma' \cos (a'\beta) = 0.\]

But because, by equation (4),

\[\cos (\beta y) \cos (\gamma a) \cos (a\beta) = \cos (\beta a) \cos (ay) \cos (\gamma\beta),\]

the same values of \(a'\) and \(\gamma'\) satisfy both these equations. Hence the three lines meet in one point.

2. The Circle.

We shall confine ourselves to two properties only of the circle, both because of their more simple geometric deduction, and because they are included in the properties of the ellipse.

80. Prop. I.—If TPQ be a tangent to a circle; CT, BQ perpendiculars to the diameter from its extremities; then the rectangle PT.PQ = rad².

When the centre is the origin, the equation to the circle is \(x^2 + y^2 = a^2\) (Art. 10), and that of the tangent is \(ax + yy' = a^2\) (Art. 55).

Hence

\[ax + BQ.y' = a^2\]

\[ax + CT.y' = a^2(a^2 - x^2)\]

\[= a^2y'^2\]

or

\[CT.BQ = a^2\]

but \(CT = PT\), \(BQ = PQ\), \(\therefore PT.PQ = a^2\).

81. Prop. II.—If through a fixed point chords are drawn to a circle, and tangents are drawn at the extremities of those chords, the locus of the points of intersection of the tangents is a straight line.

Let \(b, c\) be the co-ordinates of the fixed point, \(x', y'\) the co-ordinates of a point in which two tangents to the circle intersect one another; then (Art. 57) the equation to the chord which joins the points of contact is \(xx' + yy' = a^2\). But, by hypothesis, \(b, c\) are co-ordinates of a point in this line, \(\therefore bx' + cy' = a^2\). Now this is a simple equation between \(x'\) and \(y'\). Hence the truth of the proposition.

3. The Parabola.

82. Prop. I.—To find the latus rectum of the parabola.

The latus rectum is the ordinate to the axis drawn through the focus.

By Art. 12 the equation to the parabola is \(y^2 = 4ax\). Now, when \(x = a\), \(y\) is half the chord through the focus; \(\therefore\) that half-chord² = \(4a^2\), or the latus rectum = \(4a\).

83. Prop. II.—To find the subtangent of the parabola.

The subtangent is the portion of the diameter included between the ordinate and the tangent.

Let PT be the tangent meeting the axis in T; MT is the subtangent.

The equation to the tangent PT is (Art. 55) \(yy' = 2a(x + x')\).

Now, at the point T, \(y = 0\), \(\therefore x = -x'\), or AT = AM; hence the subtangent is double the abscissa.

84. Cor.—Since AT = AM and AF = AD; by addition, FT = DM = PQ = FP; \(\therefore \angle FPT = \angle FTP = \angle QPT\), or the tangent bisects the angle between two lines drawn from the point of contact, the one to the focus, the other perpendicular to the directrix.

This is the property of the tangent employed in the treatise on Conic Sections, to which the reader is referred for the geometrical properties to which it leads.

85. Prop. III.—To find the equation to the normal of the parabola.

The normal is the straight line which is perpendicular to the tangent at the point of contact.

Let PT be the tangent at the point P, PG perpendicular to PT is the normal.

The equation to PT is

\[yy' = 2a(x + x'), \text{ or } y = \frac{2a}{y'}x + \frac{2ax'}{y'};\]

and the equation to a line perpendicular to this is (Art. 69)

\[y = -\frac{y'}{2a}x + q.\] Now the point \( P \), whose co-ordinates are \( x', y' \), is a point in this line; therefore

\[ y' = -\frac{y'}{2a}x' + q, \]

and by subtraction \( y - y' = -\frac{y'}{2a}(x - x') \),

which is the equation to the normal.

86.—Cor. At the point \( G, y = 0, x = AG \). Hence \( AG = x' + 2a = AM + 2a \); \( MG = 2a \).

The line \( MG \) is called the subnormal; and it follows that the subnormal is equal to half the latus rectum.

87. Prop. IV.—The perpendicular from the focus on the tangent intersects the tangent in the axis of \( y \).

The equation to the tangent is

\[ yy' = 2a(x + x'), \]

or

\[ y'\left(y - \frac{y'}{2}\right) = 2ax; \]

and the equation to the perpendicular from \( F \) on the tangent is

\[ y = -\frac{y'}{2a}(x - a) \]

or

\[ y - \frac{y'}{2} = -\frac{yx}{2a}. \]

At their point of section \( x \) and \( y \) are the same in both.

Now if \( x \) be not 0, \( y - \frac{y'}{2} \) is positive in equation (1) and negative in equation (2), or vice versa, which is impossible; \( x = 0 \), or the point of intersection is in the axis of \( y \).

88. Cor.—FP : FY :: FY : FA,

or

\[ FP = \frac{FY^2}{FA} = \frac{FY^2}{FA^2} = \frac{FA}{\sin^2 T}. \]

89. Prop. V.—To find the locus of the intersection of the tangent with the perpendicular on it from the vertex.

The equation to the tangent is \( yy' = 2a(x + x') \), \( \ldots \) (1)

and the equation to the perpendicular on it from the vertex is \( y = -\frac{y'}{2a}x \); \( \ldots \) (2)

Also the equation to the parabola is \( y'^2 = 4ax \); \( \ldots \) (3)

From equation (2), \( y' = -2ay/x \), and hence, from equation (1)

\[ 2ax' = -\frac{2ay^2}{x} - 2ax. \]

Substituting these values in (3), there is obtained

\[ \frac{4a^2y^2}{x^2} = -\frac{4ay^2}{x} - 4ax \]

or

\[ y^2 = -\frac{x^3}{a + x}; \]

which shows (Art. 19) that the locus is a cissoid, the diameter of whose generating circle is \( AD \).

90. Prop. VI.—To find the equation to the parabola when the axes are any diameter and the tangent at its extremity.

Let \( PN \) be the diameter parallel to \( AM \); \( QN \) the ordinate parallel to the tangent at \( P \); \( x', y' \) the co-ordinates of \( P \); \( x, y \) of \( Q \) referred to the axis; \( x', y' \) the co-ordinates of \( P \), \( NQ \) referred to the diameter \( PN \) and the tangent \( dP \).

Hence, \( x = x' + x' + y' \cos T \),

\( y = y' + y' \sin T \);

therefore the equation \( y'^2 = 4ax \) becomes

\[ (y' + y'\sin T)^2 = 4a(x' + x' + y'\cos T), \]

or

\[ y'^2 + 2y'y'\sin T + y'^2\sin^2 T = 4ax' + 4ax' + 4ay'\cos T. \]

Now tan \( T = \frac{2a}{y} \) (Art. 6); \( \ldots \) \( 2y'y'\sin T = 4ay'\cos T \); and \( y'^2 = 4ax' \); hence the equation is reduced to \( y'^2\sin^2 T = 4ax' \).

But \( \frac{4a}{\sin^2 T} = 4FP \) (Art. 88); hence \( y'^2 = 4FPx^2 \) is the equation required.

The equation when the curve is referred to any diameter and its bisected chords is consequently of the same form as the equation when the curve is referred to the axis. The conclusions arrived at in the one case are very easily adapted to the other.

91. Prop. VII.—The semi-latus-rectum is an harmonic mean between the segments of any chord drawn through the focus.

Let \( PFR \) be a chord through the focus; then (Art. 13),

\[ \frac{1}{FP} = \frac{1}{2a}(1 - \cos PFM); \]

similarly

\[ \frac{1}{FR} = \frac{1}{2a}(1 - \cos RFM). \]

Now, \( \cos PFM + \cos RFM = 0 \); \( \ldots \) \( \frac{1}{FP} + \frac{1}{FR} = \frac{2}{2a} \), which proves the proposition.

4. The Ellipse.

92. Prop. I.—If a circle be described about the axis major, then ordinates to the ellipse and the circle to the same abscissa, have to one another the proportion of the axis minor to the axis major.

Let \( MPQ \) be an ordinate to the abscissa \( CM \); then

\[ MP^2 = \frac{b^2(a^2 - x^2)}{a^2} \text{ (Art. 14)}, \]

and

\[ MQ^2 = a^2 - x^2 \text{ (Art. 10)}; \]

\( \therefore \)

\[ MP : MQ :: b : a; \]

or

\[ MP : MQ :: b : a. \]

93. Prop. II.—To find the subtangent of the ellipse.

The equation to the tangent \( PT \) is \( ax'y' + bx'x = a'b' \).

At the point \( T, y = 0 \); \( \ldots \) \( xx' = a^2 \), or \( CT.CM = CA^2 \); whence the subtangent

\[ MT = CT - CM = \frac{CA^2}{CM} - CM = \frac{CA^2 - CM^2}{CM}. \]

94. Cor.—If a tangent be drawn to the circle at \( Q \), it will cut the axis in the same point \( T \), for \( CT \) is independent of \( b \). 95. Prop. III.—To find the value of the subnormal.

The equation to the tangent being \(a^2y' + b^2x' = a^2b\), the equation to the normal \(PG\) will be

\[y - y' = \frac{a^2}{b^2} (x - x')\] (Art. 69).

Now, at the point \(G\) we have \(y = 0\);

\[\therefore CG = x' - \frac{b^2x'}{a^2} = e^2x' = e^2CM.\]

96. Prop. IV.—The normal bisects the angle between the two focal distances.

For, retaining the letters placed at the foci in Art. 14, we have

\[FG = FC + CG = ae + e^2x';\] \[SG = SC - CG = ae - e^2x'.\]

Also

\[SP = a - ex'\] (Art. 15), and therefore

\[FP = a + ex';\] whence

\[FG : SG :: FP : SP,\]

and (Euc. vi. 3) the angle \(FPS\) is bisected by the line \(PG\).

97. Cor.—Hence also the tangent bisects the angle between one of the focal distances and the other produced.

98. Prop. V.—To find the locus of the intersection of the tangent with the perpendicular on it from the focus.

If we write \(m = \frac{b^2x'}{a^2y'}\), we shall have as the equation to the tangent, \(y + mx = \frac{b^2}{a^2}\).

Now

\[\left(\frac{b^2}{a^2}\right)^2 = m^2 \left(\frac{a^2}{x'}\right)^2 = \frac{m^2a^2}{1 - \frac{y'^2}{b^2}};\]

hence \(\left(\frac{b^2}{a^2}\right)^2 - b^2 = m^2a^2\), and \(y + mx = \sqrt{b^2 + m^2a^2}\) is the equation to the tangent.

Also the equation to the perpendicular on it from the focus is, \(y = \frac{1}{m}(x - ae)\); or \(my - x = -ae\). Squaring the equations and adding them, we get

\[(1 + m^2)(x^2 + y^2) = b^2 + m^2a^2 + a^2e^2,\]

\[= (1 + m^2)a^2;\]

\[\therefore x^2 + y^2 = a^2;\]

or the locus required is the circle described about the major axis.

99. Prop. VI.—The rectangle by the two perpendiculars from the foci on the tangent is equal to the square of the semi-axis minor.

The length of the perpendicular from a point whose co-ordinates are \(x_1, y_1\) on a straight line whose equation is \(y = -mx + c\) is (Art. 70)

\[y_1 + mx_1 - c = \sqrt{1 + m^2};\]

Now, the co-ordinates of \(S\) are \(x_1 = ae, y_1 = 0\); hence the perpendicular from \(S\) on the tangent is equal to

\[\frac{mae - c}{\sqrt{1 + m^2}},\] where \(c = \sqrt{b^2 + m^2a^2}.\)

In the same way the length of the perpendicular from \(F\) on the tangent is

\[-\frac{mae - c}{\sqrt{1 + m^2}};\] consequently their rectangle

\[\frac{e^2 - m^2a^2}{1 + m^2} = b^2.\]

100. Prop. VII.—To find the equation to the ellipse when referred to two diameters as axes, of which the one is parallel to a tangent at the extremity of the other.

Let \(CP\) be one diameter, and \(CD\) parallel to the tangent at \(P\) the extremity of the other.

Let the co-ordinates of \(P\), referred to the axes as before, be \(x', y'\); of \(Q, x_2, y_2\), and let the co-ordinates of \(Q\) referred to \(CP\) and \(CD\) be \(x\) and \(y\), or \(CV = x,\)

\(QV = y\); also let the angle \(PCA = \theta, DCO = \phi\); then tan \(\phi = \frac{b^2x'}{a^2y'}\) and tan \(\theta = \frac{y'}{x'};\)

\[\therefore \tan \theta \tan \phi = \frac{b^2}{a^2};\] i.e., \(a^2 \sin \theta \sin \phi = b^2 \cos \theta \cos \phi.\)

And (Art. 41)

\[x_1 = x \cos \theta - y \cos \phi,\] \[y_1 = x \sin \theta + y \sin \phi.\]

Substituting these values in the equation \(a^2x_1^2 + b^2y_1^2 = a^2b^2\), there results

\[a^2(x \cos \theta + y \sin \phi)^2 + b^2(x \cos \theta - y \sin \phi)^2 = a^2b^2,\]

or \(a^2(x^2 \sin^2 \theta + y^2 \sin^2 \phi) + b^2(x^2 \cos^2 \theta + y^2 \cos^2 \phi) = a^2b^2,\)

or \(a^2y^2 + b^2x^2 = a^2b^2\), where

\[a^2 = \frac{a^2b^2}{a^2 \sin^2 \theta + b^2 \cos^2 \phi},\] \[b^2 = \frac{a^2b^2}{a^2 \sin^2 \phi + b^2 \cos^2 \phi}.\]

101. Cor. 1.—Since only the squares of \(x\) and \(y\) appear in the equation, the line \(CP\) bisects all chords parallel to \(CD\), or (Art. 50) the diameters \(CP, CD\) are conjugate diameters, and \(CP\) is parallel to the tangent at \(D\).

102. Cor. 2.—By making successively \(x = 0\) and \(y = 0\), we obtain \(CP = a, CD = b\). Hence the equation is exactly the same as when referred to the principal axes; and the results obtained in that case are easily adapted to the present.

103. Cor. 3.—We have

\[a^2 \sin^2 \theta : b^2 \cos^2 \phi :: a^2 \tan^2 \phi : b^2 : a^2 + b^2 \cot^2 \theta,\]

\[\therefore PM : CN :: b : a.\]

Similarly

\[DN : CM :: b : a,\]

\[\therefore PM : CN :: DN : CM.\]

104. Cor. 4.—\(CP^2 + CD^2 = CM^2 + MP^2 + CN^2 + ND^2\)

\[= CM^2 + MP^2 + \frac{a^2}{b^2} MP^2 + \frac{b^2}{a^2} CM^2\]

\[= (a^2 + b^2) \left(\frac{CM^2}{a^2} + \frac{MP^2}{b^2}\right)\]

\[= a^2 + b^2;\]

or the sum of the squares of any two conjugate axes is the same as the sum of the squares of the principal axes.

105. Cor. 5.—\(CD^2 = \frac{a^2}{b^2} y^2 + \frac{b^2}{a^2} x^2\),

\[= a^2 \left(1 - \frac{x^2}{a^2}\right) + \frac{b^2}{a^2} x^2\]

\[= a^2 - e^2 x^2\]

\[= FP \cdot SP\] (Art. 15).

106. Prop. VIII.—All parallelograms circumscribing the ellipse are equal.

If tangents be drawn at the extremities of conjugate diameters, they will form a parallelogram of which the Analytical Geometry.

107. Prop. IX.—The ellipse is the curve generated by a point whose distance from the focus is to its distance from the directrix as \( e : 1 \).

Let EQ be the directrix at a distance from the centre CE = \( \frac{a}{e} \); then

\[ PQ = ME = CE - CM = \frac{a}{e} - x = \frac{a - ex}{e} = \frac{FP}{e} \]

(Art. 15); \( \therefore FP : PQ :: e : 1 \).

5. The Hyperbola.

The equation to the hyperbola differs from that to the ellipse only in having \( -b^2 \) in place of \( b^2 \). With this limitation, the properties given above and their demonstrations are common to both curves, and it would be superfluous to repeat them. We shall, accordingly, confine ourselves to that property of the hyperbola which has no parallel in the ellipse—the asymptote and its consequences.

108. Prop. I.—To find the equation to the asymptote of the hyperbola.

We have already (Art. 30) shown that if the values of \( y \) be expressed in a descending series of powers of \( x \), the equation to the asymptote may be obtained from it by omitting all negative powers.

Now in the hyperbola (Art. 17)

\[ y^2 = \frac{b^2}{a^2} \left( x^2 - a^2 \right) \]

\[ = \frac{b^2}{a^2} x^2 \left( 1 - \frac{a^2}{x^2} \right) \]

\[ \therefore y = \pm \frac{b}{a} x \left( 1 - \frac{1}{2} \frac{a^2}{x^2} + &c. \right) \]

Hence the straight line whose equation is \( y = \pm \frac{bx}{a} \) is an asymptote to the curve.

If the curve be referred to conjugate diameters, the same reasoning will show that the equation to the asymptotes is

\[ y = \pm \frac{b}{a} x. \]

It is evident that the asymptotes are the diagonals of a parallelogram of which the conjugate diameters are the lines joining the points of bisection of opposite sides.

109. Prop. II.—If any chord of the hyperbola be terminated by the asymptotes, the rectangle by its segments is equal to the square of the semi-axis parallel to that chord.

Let RPQH (fig. 27) be a chord terminated by the asymptotes; then RP . PH = \( b^2 \).

Let the curve be referred to conjugate diameters, of which one is parallel to RH; its equation is

\[ y^2 = \frac{b^2}{a^2} \left( x^2 - a^2 \right); \]

and the equation to the asymptote

\[ y^2 = \frac{b^2}{a^2} x^2; \]

\[ \therefore y^2 - y^2 = \frac{b^2}{a^2} a^2 = b^2; \]

i.e. \( MR^2 - MP^2 = b^2 \).

Now MP = MQ being respectively \( \pm \frac{b}{a} \sqrt{x^2 - a^2} \), and

MR = MH being respectively \( \pm \frac{b}{a} x; \)

the above equation gives RP . PH = \( b^2 \).

110. Prop. III.—To find the equation to the curve when referred to the asymptotes as axes.

Draw NP parallel to CR, and let CN = \( x' \), NP = \( y' \).

Suppose CM in the direction of the transverse* axis = \( x \), MP perpendicular to it = \( y \), the angle RCM = \( \theta \);

then \( x = x' \cos \theta + y' \cos \theta; \)

\[ y = y' \sin \theta - x' \sin \theta; \]

\[ \therefore a^2 \left( y' - x' \right)^2 \sin^2 \theta - b^2 \left( x' + y' \right)^2 \cos^2 \theta = -a^2 b^2. \]

Now tan \( \theta = \frac{b}{a}; \)

\[ \sin^2 \theta = \frac{b^2}{a^2 + b^2}; \]

\[ \cos^2 \theta = \frac{a^2}{a^2 + b^2}; \]

hence

\[ \frac{a^2 b^2}{a^2 + b^2} \left( \left( y' - x' \right)^2 - \left( x' + y' \right)^2 \right) = -a^2 b^2; \]

or

\[ x'y' = \frac{a^2 + b^2}{4}, \] the equation required.

111. Prop. IV.—If any tangent be produced to meet the two asymptotes, the area of the triangle contained by the tangent and the two asymptotes is always the same.

Let UP'T be the tangent; its equation (Art. 55);

\[ xy' + yx' = \frac{a^2 + b^2}{2}. \]

Now, when \( y = 0, x = CT; \)

\[ \therefore CT = \frac{a^2 + b^2}{2y}; \]

and when \( x = 0, y = CU; \)

\[ \therefore CU = \frac{a^2 + b^2}{2x}. \]

Hence

\[ CT . CU = \frac{(a^2 + b^2)^2}{4xy} = a^2 + b^2; \]

and as the angle TCU is constant, the area TCU is always the same, whatever be the point P.

We cannot close this portion of our subject without acknowledging our obligations to Mr Salmon, whose masterly treatise can hardly be too highly recommended.

ANALYTICAL GEOMETRY OF THREE DIMENSIONS; OR SOLID CO-ORDINATE GEOMETRY.

1. As in plane co-ordinate geometry, the position of a point in a plane is denoted by its distances from two given lines, so in solid geometry the position of a point in space

* The terms transverse and conjugate are more appropriate than major and minor in the case of the hyperbola. is denoted by its distances from three given planes. The direction in which the distance from any one plane is measured is parallel to the line of intersection of the other two. The planes of reference are called the co-ordinate planes, and their lines of intersection the co-ordinate axes; the point in which the three intersect one another being the origin.

SECTION I.—PROPERTIES OF THE PLANE AND STRAIGHT LINE.

We shall at present confine ourselves to rectangular axes.

2. Prop. I.—To find the distance between two points in terms of their co-ordinates.

Let Ox, Oy, Oz be the axes of co-ordinates; P, Q the two points; x, y, z the co-ordinates of P; x', y', z' the co-ordinates of Q. Through P, Q draw planes parallel to the co-ordinate planes of xy, xz, yz; it is evident that they will form a parallelepipedon, of which PQ is the diagonal. Let QM be the edge of this parallelepipedon parallel to x, MN parallel to y, and NP to z; then it is evident that QM = x - x', MN = y - y', NP = z - z'; and QM being perpendicular to the plane PMN, the angle QMP is a right angle.

\[ \therefore \quad PQ^2 = QM^2 + MP^2 = QM^2 + MN^2 + NP^2 = (x - x')^2 + (y - y')^2 + (z - z')^2. \]

3. Cor. 1.—If α, β, γ be the angles which PQ makes with the axes of x, y, z, we shall have

\[ x - x' = QM = PQ \cos \alpha, \quad y - y' = PQ \cos \beta, \quad z - z' = PQ \cos \gamma; \]

\[ \therefore \quad PQ^2 = PQ^2 \cos^2 \alpha + PQ^2 \cos^2 \beta + PQ^2 \cos^2 \gamma, \]

and \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1; \)

consequently the angles which a straight line makes with the axes are not all three arbitrary, but where two are given, the third is determined by the above equation.

4. Cor. 2.—If we take the three equations,

\[ x - x' = PQ \cos \alpha, \quad y - y' = PQ \cos \beta, \quad z - z' = PQ \cos \gamma; \]

and multiply the first by \( x - x' \), the second by \( y - y' \), and the third by \( z - z' \), and add the results, we shall get

\[ (x - x') \cos \alpha + (y - y') \cos \beta + (z - z') \cos \gamma = PQ. \]

5. Prop. II.—To find the angle between two straight lines in terms of the angles which the lines make with the axes.

Draw through the origin two lines OP, OQ parallel to the given lines, and let OP make with the axes the angles α, β, γ, and OQ the angles α', β', γ'.

Take OP = 1, OQ = 1, and let the angle POQ be θ; then

\[ x = OP \cos \alpha = \cos \alpha, \quad y = OP \cos \beta = \cos \beta, \quad z = OP \cos \gamma = \cos \gamma. \]

Now \( PQ^2 = OP^2 + OQ^2 - 2 OP \cdot OQ \cos \theta, \)

\[ = 2 - 2 \cos \theta. \]

But \( PQ^2 = (x - x')^2 + (y - y')^2 + (z - z')^2 \)

\[ = (\cos \alpha - \cos \alpha')^2 + (\cos \beta - \cos \beta')^2 + (\cos \gamma - \cos \gamma')^2 \]

\[ = \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \alpha' + \cos^2 \beta' + \cos^2 \gamma' - 2 (\cos \alpha \cos \alpha' + \cos \beta \cos \beta' + \cos \gamma \cos \gamma'); \]

\[ \therefore \quad \cos \theta = \cos \alpha \cos \alpha' + \cos \beta \cos \beta' + \cos \gamma \cos \gamma'. \]

6. Prop. III.—To find the equation to a plane.

Let ABC be the plane, P any point in it, of which the co-ordinates are x, y, z; OQ perpendicular to the plane = p, making with the axes of x, y, and z the angles α, β, and γ.

Let also OP make with the axes the angles α', β', γ': then, as in Art. 3, we have \( x = OP \cos \alpha, \)

\[ y = OP \cos \beta, \quad z = OP \cos \gamma. \]

Now \( OQ = OP \cos POQ \)

\[ = OP (\cos \alpha \cos \alpha' + \cos \beta \cos \beta' + \cos \gamma \cos \gamma') \] (Art. 5),

or \( p = x \cos \alpha + y \cos \beta + z \cos \gamma. \)

7. Cor. 1.—If OA = a, OB = b, OC = c, we have

\[ \cos \alpha = \frac{a}{p}, \quad \cos \beta = \frac{b}{p}, \quad \cos \gamma = \frac{c}{p}; \]

\[ \therefore \quad x \cos \alpha + y \cos \beta + z \cos \gamma = \frac{xp}{a} + \frac{yp}{b} + \frac{zp}{c}; \]

hence \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1, \) is another form of the equation.

8. Cor. 2.—Any simple equation in x, y, z of which the general form is \( Ax + By + Cz = D \) will represent a plane.

9. Cor. 3.—If the equation be written in the form exhibited in the last corollary, we must have

\[ \cos \alpha = \frac{A}{D}, \quad \cos \beta = \frac{B}{D}, \quad \cos \gamma = \frac{C}{D}; \]

\[ \therefore \quad \text{(Art. 3)} \quad \frac{1}{p^2} = \frac{A^2 + B^2 + C^2}{D^2}, \]

or \( p = \sqrt{A^2 + B^2 + C^2}. \)

\[ \cos \alpha = \frac{A}{\sqrt{A^2 + B^2 + C^2}}, \quad \cos \beta = \frac{B}{\sqrt{A^2 + B^2 + C^2}}, \quad \cos \gamma = \frac{C}{\sqrt{A^2 + B^2 + C^2}}. \]

10. Cor. 4.—If z = 0 in Cor. 1, we get \( \frac{x}{a} + \frac{y}{b} = 1, \) which is obviously the equation to the line AB. This line is called the trace of the plane on the plane of xy. In like manner, AC, BC are the traces of the plane or the planes of xz and yz.

11. Cor. 5.—If the plane be parallel to the axis of z or perpendicular to the plane of xy, the line OQ is in the place of xy; \( \therefore \quad \gamma = \frac{\pi}{2} \) and \( \cos \gamma = 0; \) hence the equation to the plane becomes \( x \cos \alpha + y \cos \beta = p, \) or the equation to the plane is the same as the equation to its trace on the plane of xy.

12. Prop. IV.—To find the equations to a straight line.

A straight line is the intersection of two planes, consequently the equations to those planes will be satisfied simultaneously for the line. These are therefore the equations to the line.

If the two equations be \( Ax + By + Cz = D \) and \( A'x + B'y + C'z = D', \) we may obtain by elimination

\[ x = az + p, \quad y = bz + q \]

as the form of the equations required.

13. The equations are sometimes obtained in the following manner:

Let planes pass through the line respectively perpendicular to the planes of xz and yz. Their equations will have the form (Art. 11) \( x = az + p, y = bz + q \), which are the equations to the line.

14. Cor. 1.—It is evident that if perpendiculars be drawn from every point in the given line to the plane of \( xx \), they will intersect that plane in the line whose equation is \( x = az + p \). This latter line is called the projection of the given line on the plane of \( xx \). Consequently, the equations to a line are the equations to its projections on two of the co-ordinate planes.

15. Prop. V.—To express the angles which a straight line makes with the co-ordinate axes in terms of \( a \) and \( b \).

Let \( PQ \), fig. of Art. 2, be the line, \( x = az + p, y = bz + q \) its equations; then because the point \( Q \), of which the co-ordinates are \( x', y', z' \), is a point in the line, we have

\[ x - x' = a(z - z'), \quad y - y' = b(z - z'), \]

and \( \cos \alpha = \frac{x - x'}{PQ}, \quad \cos \beta = \frac{y - y'}{PQ}, \quad \cos \gamma = \frac{z - z'}{PQ}; \]

i.e.

\[ \cos \alpha = \frac{a}{\sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}} = \frac{a}{\sqrt{a^2 + b^2 + 1}}; \]

\[ \cos \beta = \frac{b}{\sqrt{a^2 + b^2 + 1}}, \quad \cos \gamma = \frac{1}{\sqrt{a^2 + b^2 + 1}}. \]

16. Cor. 1.—\( a = \frac{\cos \alpha}{\cos \gamma} = \frac{\cos \beta}{\cos \gamma} \).

17. Cor. 2.—\( \frac{x - x'}{a} = \frac{y - y'}{b} = \frac{z - z'}{c} \);

or \( \frac{x - x'}{\cos \alpha} = \frac{z - z'}{\cos \gamma} \) and \( \frac{y - y'}{\cos \beta} = \frac{z - z'}{\cos \gamma} \);

which are the equations to a straight line in terms of the angles which it makes with the co-ordinate axes.

18. Cor. 3.—If \( x = az + p, y = bz + q \) be the equations to another straight line; then, if \( \theta \) be the angle between them, we shall have

\[ \cos \theta = \cos \alpha \cos \alpha' + \cos \beta \cos \beta' + \cos \gamma \cos \gamma' (\text{Art. 5}) \]

\[ = \frac{ad + bb' + 1}{\sqrt{a^2 + b^2 + 1} \cdot \sqrt{a'^2 + b'^2 + 1}}. \]

19. Cor. 4.—In order that the two straight lines may be at right angles to one another, we must have \( \cos \theta = 0 \), or \( ad + bb' + 1 = 0 \).

20. Prop. VI.—To find the conditions that a straight line may be at right angles to a plane.

It is evident that the angles determined in Art. 9, which the perpendicular to the plane makes with the axes, must be equal to those determined in Art. 15, which the straight line makes with them; or

\[ \frac{A}{\sqrt{A^2 + B^2 + C^2}} = \frac{a}{\sqrt{a^2 + b^2 + 1}}, \quad \frac{B}{\sqrt{A^2 + B^2 + C^2}} = \frac{b}{\sqrt{a^2 + b^2 + 1}}, \]

\[ \frac{C}{\sqrt{A^2 + B^2 + C^2}} = \frac{1}{\sqrt{a^2 + b^2 + 1}}; \]

\[ \therefore A = ac, B = bc \text{ are the conditions required.} \]

21. Prop. VII.—To find the conditions that a straight line may coincide with a plane.

Let \( x = az + p, y = bz + q \) be the equations to the line; \( Ax + By + Cz = D \) the equation to the plane; then, since they coincide, we must have \( A(az + p) + B(bz + q) + Cz = D \), whatever be \( z \); hence \( Aa + Bb + C = 0 \) and \( Ap + Bq = D \) are the conditions required.

22. Cor.—The condition that the straight line may be parallel to the plane is \( Aa + Bb + C = 0 \); being the con-

dition that a line and plane parallel to these through the origin shall coincide.

23. Prop. VIII.—To find the length of the perpendicular from a given point on a given plane.

Let \( P \) be the point of which the co-ordinates are \( x', y', z' \); \( PR \) perpendicular to the plane whose equation is \( Ax + By + Cz = D \).

Draw \( PQ \) parallel to the axis of \( z \), meeting the plane in \( Q \); then \( PR = PQ \cos QPR \)

\[ = PQ \frac{C}{\sqrt{A^2 + B^2 + C^2}} (\text{Art. 9}); \]

But \( PQ = z' - D - Ax' - By' \),

\[ \therefore PR = \frac{Ax' + By' + Cz' - D}{\sqrt{A^2 + B^2 + C^2}}. \]

24. If from every point in the boundary of a given surface, straight lines be drawn parallel to a given line and meeting a fixed plane, they form the boundary of the projection of the surface on that plane. When the lines are drawn perpendicular to the plane, the projection is called an orthogonal projection. It is with this class alone that we are concerned here.

Prop. IX.—The projection of a plane area is to the area itself, as the cosine of the angle between the two planes is to unity.

Suppose the area divided into triangles, indefinitely small if necessary. Let \( ABC \) be one of these triangles, \( DEF \) its projection; and let the line of intersection of the planes of \( ABC, DEF \) be \( GK \). Through \( A, B, C \) draw planes at right angles to \( GK \), and therefore passing through \( D, E, F \); each of the angles \( AGD, PHQ, CKF \) is equal to the angles between the planes.

Now the triangles \( ABP, DEQ \) have \( BP, EQ \) as their bases, and \( GH \) as their common altitude;

\[ \therefore DEQ : ABP :: EQ : BP :: \cos PHQ : 1. \]

Similarly of the other triangles. Hence the truth of the proposition.

25. Prop. X.—The square of any plane area is equal to the sum of the squares of its projections on the three co-ordinate planes.

Let \( A_x, A_y, A_z \) represent the projections of \( A \) on the planes of \( yz, zx, \) and \( xy \); \( \alpha, \beta, \gamma \), the angles which the perpendicular on \( A \) makes with the axes of \( x, y, \) and \( z \); then \( a \) is the angle which the area makes with the plane of \( yz \).

\[ \therefore A_x = A_y \cos \alpha, \quad A_y = A \cos \beta, \quad A_z = A \cos \gamma. \]

Now \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1; \quad \therefore A_x^2 + A_y^2 + A_z^2 = A^2. \)

26. Prop. XI.—To find the equation to a plane in terms of the area and its projections on the co-ordinate planes.

As in the last proposition \( A = A_x \cos \alpha, \&c. \)

\[ \therefore \cos \alpha = \frac{A_x}{A}, \quad \cos \beta = \frac{A_y}{A}, \quad \cos \gamma = \frac{A_z}{A}. \] Now (Art. 6) the equation to the plane is

\[ x \cos \alpha + y \cos \beta + z \cos \gamma = p; \]

hence it becomes

\[ x \Lambda_x + y \Lambda_y + z \Lambda_z = pA. \]

28. Prop. XII.—The projection of a straight line on a plane is to the straight line, as the cosine of the angle between them is to unity.

Let \( AB \) be the straight line meeting the plane in \( E \); \( CD \) the projection; then \( AC, BD \), which are perpendicular to \( ED \), are in the plane \( BED \) at right angles to the plane of projection; and \( AEC \) or its equal \( BAF \) is the angle between the straight line and the plane; hence

\[ CD : AB :: \cos E : 1. \]

29. Prop. XIII.—The projection of a straight line on another straight line is to the straight line itself, as the cosine of their angle of inclination is to unity.

When two lines are not in the same plane, their angle of inclination is the angle which one of them makes with a straight line drawn through it parallel to the other.

Let \( AB \) be the line to be projected; \( CD \) the line on which it is to be projected. Draw \( AE \) parallel to \( CD \); then \( BAE \) is the angle of inclination of the lines. Through \( B \) draw the plane \( BFE \) perpendicular to \( CD \), and therefore to \( AE \); and draw \( AC \) parallel to \( EF \). Then \( EFC, BFC, AEB \) are right angles, \( \therefore AF \) is a parallelogram and \( AE = CF \) the projection of \( AB \) on \( CD \). But

\[ AE : AB :: \cos BAE : 1; \]

\[ \therefore CF : AB :: \cos BAE : 1. \]

30. Prop. XIV.—If any two points be connected both by a straight and by a broken line, the projection of the former on a given line is equal to the sum of the projections of every part of the latter.

The projection of \( PQ \) on the line \( AB \) is equal to the sum of the projections of \( PM, MN, NR, RQ \). Through \( P, M, N, R, Q \) draw planes perpendicular to \( AB \); the intercepted portions of \( AB \) are the projections of \( PM, MN, \&c.; \) hence the proposition is evidently true provided none of the intercepted portions lap over each other or are negative, as would be the case with the projection of \( MN \) in the following figure.

But the cosines employed will rectify this, provided we attend to the directions of the lines as indicated in the figure by the arrows.

For the projection of \( MN \) will be negative, because the angle is greater than a right angle, whilst all the others will be positive, including that of \( RQ \), for which the angle is negative, but its cosine positive.

31. By a similar process, and retaining the same restriction as to direction, we may show that if an area be connected with any number of areas, having common edges with the former area and with each other, the projection of the former on any plane is equal to the sum of the projections of all the others.

SECTION II.—EQUATIONS TO SURFACES OF THE SECOND DEGREE DEDUCED FROM THEIR PROPERTIES.

32. a.—The Sphere.

The characteristic property is that every point is equally distant from the centre.

Let \( a, b, c \) be the co-ordinates of the centre, and \( r \) the radius. The equation is evidently

\[ (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2. \]

33. b.—Oblique Cylinder on a Circular Base.

We shall give the simplest form of the equation, by supposing the circular base to be in the plane of \( xy \) with its centre at the origin, so that its equation is \( x^2 + y^2 = r^2 \). This circle is called the directrix. A straight line moves parallel to itself in such a way as always to pass through the circumference of this circle. This line is called the generatrix. Let \( x = az + p, y = bz + q \) be its equation. Then \( a, b, r \) are known constants. Now \( p, q \) are the values of \( x \) and \( y \) when \( z = 0 \); they are therefore the co-ordinates of a point in the circumference of the directrix, or \( p^2 + q^2 = r^2 \), i.e., \( (az - ex)^2 + (bz - cy)^2 = r^2 \), which is the equation required.

34. Con.—A similar process will determine the equation to the cylinder whatever be the equation to the directrix.

35. c.—Oblique Cone on a Circular Base.

We shall suppose the circle to be the same as in Art. 33. The generatrix is now required to pass through a fixed point called the vertex of the cone. Let \( a, b, c \) be the co-ordinates of this point; \( x = Az + p, y = Bz + q \), the equations to the generatrix; then we must have \( a = Ae + p, b = Be + q \), from which equations we obtain

\[ p = \frac{az - ex}{z - e}, \quad q = \frac{bz - cy}{z - e}, \]

and, since \( p^2 + q^2 = r^2 \), the equation required is

\[ (az - ex)^2 + (bz - cy)^2 = r^2 (z - e)^2. \]

36. d.—The Ellipsoid.

The directorices are two ellipses at right angles to each other, having a common axis. We shall suppose their planes to be those of \( xy \) and \( zx \). The generatrix is a variable ellipse, such that its semi-axes are the ordinates of the directorices.

Let

\[ \frac{x^2}{a^2} + \frac{MQ^2}{b^2} = 1 \quad (1) \text{ be the equation to } \Delta OB, \]

\[ \frac{x^2}{a^2} + \frac{MR^2}{c^2} = 1 \quad (2) \text{ the equation to } \Delta OC, \]

the two directorices; so that \( OA = a, OB = b, OC = c \). The equation to the generatrix RPQ will be

\[ \frac{y^2}{MQ^2} + \frac{z^2}{MR^2} = 1. \]

But by equations (1) and (2) \( MQ^2 = b^2 \left( 1 - \frac{x^2}{a^2} \right), MR^2 = c^2 \left( 1 - \frac{x^2}{a^2} \right) \). Hence (3) becomes

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1. \]

37. Con.—If \( b = c \), the ellipsoid becomes a spheroid, prolate or oblate according as \( b \) is less or greater than \( a \). 38. e.—Paraboloid of Revolution.

The directrix is a parabola, which we shall suppose to be in the plane of \(xz\), the axis of \(x\) being its axis, so that its equation is \(z^2 = 4ax\). The generatrix is a variable circle in the plane of \(yz\), such that its radius is always the ordinate of the parabola. The equation to the generatrix is \(y^2 + z^2 = \text{radius}^2\); hence \(y^2 + z^2 = 4ax\) is the equation required.

39. f.—Hyperboloid of One Sheet.

This surface differs from the ellipsoid only in having the directorices two hyperbolas, with a common conjugate axis. Hence \(a^2\) is negative, and the equation is therefore

\[ \frac{y^2}{b^2} + \frac{z^2}{c^2} - \frac{x^2}{a^2} = 1. \]

40. h.—Hyperboloid of Two Sheets.

This surface differs from the ellipsoid only in having the directorices hyperbolas with their transverse axis common. Hence \(b^2\) and \(c^2\) are negative, and the equation is therefore

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1. \]

41. h.—Elliptic Paraboloid.

The directorices are here two parabolas placed as the two ellipses were in the case of the ellipsoid. The generatrix is a variable ellipse as before.

In the figure of last Article, if we suppose \(AB, AC\) to be two parabolas whose equations are \(MQ^2 = 4ax, MR^2 = 4bx\), the origin being at \(A\), we shall have as the equation to the ellipse \(RPQ\)

\[ \frac{y^2}{MQ^2} + \frac{z^2}{MR^2} = 1. \]

Hence the equation to the surface is

\[ \frac{y^2}{4a} - \frac{z^2}{4b} = x. \]

42. i.—Hyperbolic Paraboloid.

The directorices are the same as in the last Article, but the generatrix is a variable hyperbola, whose semi-axes are the ordinates of the two parabolas. The equation to the hyperbola is therefore

\[ \frac{y^2}{MQ^2} - \frac{z^2}{MR^2} = 1. \]

Hence the equation to the surface is

\[ \frac{y^2}{4a} - \frac{z^2}{4b} = x. \]

SECTION III.—TRANSFORMATION OF CO-ORDINATES.

Hitherto we have supposed the system of co-ordinates to be rectangular. This is indeed the fundamental system; but for the solution of certain problems it is necessary to pass to an oblique system. The investigation of the requisite formulae will constitute the subject of the present section.

43. Prop. I.—To find the distance between two points in terms of their oblique co-ordinates.

Retaining the figure of Art. 2, and supposing the axes not to be at right angles, we have

\[ PQ^2 = QM^2 + MP^2 - 2QM \cdot MP \cos QMP \\ = QM^2 + MN^2 + NP^2 - 2MN \cdot NP \cos MNP \\ = 2QM \cdot MP \cos QMP. \]

Now \(MP \cos QMP\) is the projection of \(MP\) on the line \(MQ\), and is therefore (Art. 29) equal to the sum of the projections of \(MN\) and \(NP\) on the same line, i.e., to \(MN \cos NMQ + NP \cos AMQ\). But if \(a, b, c\) be the angles \(yOz, xOz, xOy\), it is evident that \(MNP, AMQ,\) and \(NMQ\) are their respective supplements; hence

\[ PQ^2 = QM^2 + MN^2 + NP^2 - 2MN \cdot NP \cos \beta \\ + 2QM \cdot MN \cos \gamma \\ = (x-a)^2 + (y-b)^2 + (z-c)^2 + 2(y-b)(z-c) \cos \alpha \\ + 2(x-a)(z-c) \cos \beta + 2(x-a)(y-b) \cos \gamma. \]

44. Prop. II.—To change the origin without altering the directions of the axes.

It is evident that this is effected by writing \(x' + a, y' + b, z' + c\) for \(x, y, z\).

45. Prop. III.—To pass from a rectangular system to any other, the origin remaining unchanged.

Let \(OM = x, MN = y, NP = z; OM' = x', MN' = y', NP' = z'\). If the line \(OP\) be projected on the line \(Ox\), the projection is \(OM\) or \(x\); and if the broken line \(OM', MN', NP'\) be projected on \(Ox\), the projection is equal to

\[ x' \cos \alpha' + y' \cos \beta' + z' \cos \gamma', \]

\[ x \cos \alpha + y \cos \beta + z \cos \gamma, \]

care being taken, as in Art. 30, to take the angles \(x', y', z'\) between the positive axis of \(x'\) and the positive axis of \(x\).

Similarly

\[ y' \cos \alpha' + y' \cos \beta' + z' \cos \gamma', \]

\[ z' \cos \alpha' + z' \cos \beta' + z' \cos \gamma'. \]

Of the nine angles \(x', y', z'\) employed above, six only are independent of one another; since by Art. 3 we have the three following relations between them, as conditions to be satisfied, viz.:

\[ \cos^2 x' + \cos^2 y' + \cos^2 z' = 1, \]

\[ \cos^2 y' + \cos^2 y' + \cos^2 z' = 1, \]

\[ \cos^2 z' + \cos^2 z' + \cos^2 z' = 1. \]

46. Cor.—If the new axes are rectangular as well as the old, we have three other equations of condition amongst the angles, to be satisfied, viz.: \(-\cos x'y = 0, \cos x'z = 0, \cos y'z = 0\), which give (Art. 5)

\[ \cos x' \cos y' + \cos x' \cos y' + \cos x' \cos z' = 0, \]

\[ \cos x' \cos z' + \cos x' \cos y' + \cos x' \cos z' = 0, \]

\[ \cos y' \cos z' + \cos y' \cos y' + \cos y' \cos z' = 0. \]

These make up the six equations of condition.

47. Prop. IV.—To pass from one rectangular system to another in a form which shall involve only three angles.

To avoid the six equations of condition exhibited above, Euler devised the following beautiful method of transformation, which is of frequent use in the solution of Mechanical Problems. Let the angle between the planes of \(xy\) and \(x'y'\) be called \(\theta\); let also these two planes intersect in the line \(Ox_1\), which makes with the axis of \(x\) the angle \(xOx_1 = \phi\), and with the axis of \(x'\) the angle \(x'Ox_1 = \psi\).

(1.) Transform the co-ordinates to \(xy_1\) in the plane of \(xy\), the axis of \(z\) remaining unchanged; there results (Plane Geometry, Art. 40)

\[ \begin{align*} x &= x_1 \cos \phi - y_1 \sin \phi, \\ y &= x_1 \sin \phi + y_1 \cos \phi. \end{align*} \]

2.) Transform the co-ordinates to \(yz'\) in the plane \(yz\), the axis of \(x_1\) remaining unchanged; there results

\[ \begin{align*} y' &= y_1 \cos \theta - z_1 \sin \theta, \\ z' &= y_1 \sin \theta + z_1 \cos \theta. \end{align*} \]

3.) Transform the co-ordinates to \(x'y'\) in the plane \(xy\), the axis of \(z'\) remaining unchanged; there results

\[ \begin{align*} x' &= x_1 \cos \psi - y_1 \sin \psi, \\ y' &= x_1 \sin \psi + y_1 \cos \psi. \end{align*} \]

By substitution, we obtain

\[ \begin{align*} x &= (x' \cos \psi - y' \sin \psi) \cos \phi - (x' \sin \psi + y' \cos \psi) \sin \phi \cos \theta \\ &+ z' \sin \phi \sin \theta, \\ y &= (x' \cos \psi - y' \sin \psi) \sin \phi + (x' \sin \psi + y' \cos \psi) \cos \phi \cos \theta \\ &+ z' \cos \phi \sin \theta, \\ z &= (x' \sin \psi + y' \cos \psi) \sin \theta + z' \cos \theta, \end{align*} \]

the requisite formulae.

48. Cor.—If \(\psi = 0\),

\[ \begin{align*} x &= x_1 \cos \phi - y_1 \cos \theta - z_1 \sin \theta \sin \phi, \\ y &= x_1 \sin \phi + y_1 \cos \theta - z_1 \sin \theta \cos \phi, \\ z &= y_1 \sin \theta + z_1 \cos \theta, \end{align*} \]

formulae which are of frequent use.

49. Prop. V.—To transform the co-ordinates from rectangular to polar.

Let OP (figure of Art. 45) \(= r\), POZ \(= \theta\), NOx \(= \phi\); then \(x = ON \cos \phi = r \sin \theta \cos \phi\), \(y = r \sin \theta \sin \phi\), and \(z = r \cos \theta\) are the requisite formulae.

SECTION IV.—TANGENT PLANES AND NOMALS TO SURFACES.

50. Prop. I.—To find the equation to the tangent plane to a surface.

If such a plane exists, it must have the property that if a section of it and of the surface be made by a plane parallel to any of the co-ordinate planes, the section of the plane will be a tangent line to the section of the surface.

Let \(x, y, z\) be the co-ordinates of the point of contact, \(x', y', z'\) those of any point in the tangent plane; and let the equation to that plane be

\[ z' - z = p(x' - x) + q(y' - y). \]

Put \(y' = y\) in the plane and in the surface; then, since the resulting line is a tangent to the resulting curve, we have

\[ z' - z = \frac{dz}{dx}(x' - x), \quad y \text{ being constant}; \]

\[ \therefore p = \frac{dz}{dx} \text{ the partial differential co-efficient}. \]

Similarly \(q = \frac{dz}{dy}\).

51. Cor.—If the equation to the surface be written under the form \(u = 0\); and \(\frac{du}{dx}, \frac{du}{dy}, \frac{du}{dz}\) represent the partial differential co-efficients of \(u\) with respect to \(x, y, z\) respectively; since the total differential co-efficient of \(u\) with respect to \(x\), when \(y\) is constant, must \(= 0\), we shall have

\[ \frac{du}{dx} + \frac{du}{dy} \frac{dz}{dx} = 0. \quad \text{Similarly} \quad \frac{du}{dy} + \frac{du}{dz} \frac{dz}{dy} = 0. \]

Hence \(\frac{dz}{dx} = -\frac{du}{dx}, \quad \frac{dz}{dy} = -\frac{du}{dy}, \quad \frac{dz}{dz} = -\frac{du}{dz}\),

and the equation to the tangent plane becomes

\[ (x' - x) \frac{du}{dx} + (y' - y) \frac{du}{dy} + (z' - z) \frac{du}{dz} = 0. \]

52. Prop. II.—To find the equation to the normal to a surface.

Retaining the notation of Art. 51; since the normal is perpendicular to the surface, and therefore to the tangent plane; if its equations are \(x' - x = a(z' - z), y' - y = b(z' - z)\), we must have (Art. 20)

\[ \frac{du}{dx} - a \frac{du}{dz} = 0, \quad \frac{du}{dy} - b \frac{du}{dz} = 0; \]

whence the equations to the normal are,

\[ (x' - x) \frac{du}{dz} = (z' - z) \frac{du}{dx}, \quad (y' - y) \frac{du}{dz} = (z' - z) \frac{du}{dy}. \]

53. Cor.—The equations may also be written

\[ x' - x + \frac{dz}{dx}(z' - z) = 0, \quad y' - y + \frac{dz}{dy}(z' - z) = 0. \]

54. Example.—To find the equation to the tangent plane to the ellipsoid.

We have

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1; \]

\[ \therefore \quad \frac{dz}{dx} = -\frac{c^2 x}{a^2 z}, \quad \frac{dz}{dy} = -\frac{c^2 y}{b^2 z}, \]

and the equation to the tangent plane becomes

\[ z' - z + \frac{c^2 x}{a^2 z}(x' - x) + \frac{c^2 y}{b^2 z}(y' - y) = 0; \]

or \(\frac{zz'}{c^2} + \frac{yy'}{b^2} + \frac{xx'}{a^2} = \frac{z^2}{c^2} + \frac{y^2}{b^2} + \frac{x^2}{a^2} = 1.\)

55. If \(p\) be the perpendicular from the centre on the tangent; we have (Art. 23)

\[ p = \frac{1}{\sqrt{\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}}}. \]

56. Prop. III.—To find the locus of the intersection of the tangent plane of the ellipsoid with the perpendicular on it from the centre.

The equation to the tangent plane is

\[ \frac{xx'}{a^2} + \frac{yy'}{b^2} + \frac{zz'}{c^2} = 1. \tag{1} \]

Consequently the equations to the perpendicular on it from the centre are (Art. 20)

\[ x' = \frac{c^2 x}{a^2 z}, \quad y' = \frac{c^2 y}{b^2 z}; \tag{2} \]

also \(\frac{zz'}{c^2} + \frac{yy'}{b^2} + \frac{xx'}{a^2} = 1. \tag{3}\)

From equations (2), \(\frac{xx'}{a^2} = \frac{zz'}{c^2} x'\) and \(\frac{yy'}{b^2} = \frac{zz'}{c^2} y'\);

\[ \therefore \quad \frac{xx'}{a^2} + \frac{yy'}{b^2} + \frac{zz'}{c^2} = \frac{zz'}{c^2} (x'^2 + y'^2 + z'^2), \] Analytical Geometry.

which, by equation (1), gives

\[ z = \frac{c^2z}{x^2 + y^2 + z^2} \]

Similarly \( y = \frac{b^2y}{x^2 + y^2 + z^2} \);

\[ x = \frac{a^2x}{x^2 + y^2 + z^2} \]

and these values being substituted in equation (3), there results \( a^2x^2 + b^2y^2 + c^2z^2 = (x^2 + y^2 + z^2)^2 \), an equation which occurs in the undulatory theory of light.

SECTION V.—ON THE CURVATURE OF SURFACES.

57. If through the normal to a surface different planes be drawn, their intersections with the surface will be plane curves, which will generally have different degrees of curvature depending on the direction of the tangent line; and, further, if other sections than those through the normal are drawn, they will determine curves which have different degrees of curvature for the same direction of the tangent line. We proceed to determine the radius of curvature of any such section.

58. Prop. I.—The sum of the reciprocals of the radii of curvature of any two normal sections at right angles to each other is constant.

Let the normal be taken as the axis of \( z \), and consequently the tangent plane as the plane of \( xy \).

Let also the angle which a section makes with the plane of \( xz \) be \( \theta \).

It is evident that \( \frac{dz}{dx} = 0, \frac{dz}{dy} = 0 \). Put \( r \) for \( \frac{d^2z}{dx^2}, s \) for \( \frac{d^2z}{dy^2} \); and let \( OM = k \),

\( MN = K \); then if \( R \) be the radius of curvature of this section,

\[ R = \frac{1}{2} \text{ limit of } \frac{ON^2}{PN} = \frac{1}{2} \text{ limit of } \frac{k^2 + K^2}{\frac{r^2}{2} + skK + tK^2 + \text{&c.}} \]

\[ = \frac{1 + \tan^2 \theta}{r + 2s \tan \theta + t \tan^2 \theta}, \]

and \( \frac{1}{R} = r \cos^2 \theta + 2s \sin \theta \cos \theta + t \sin^2 \theta \).

Let \( \frac{1}{R} \) be the radius of curvature of the section at right angles to this, or for which the angle is \( 90^\circ + \theta \); then

\[ \frac{1}{R'} = r \sin^2 \theta - 2s \sin \theta \cos \theta + t \cos^2 \theta; \]

\[ \therefore \frac{1}{R} + \frac{1}{R'} = r + t, \text{ which is independent of } \theta. \]

59. Prop. II.—To determine the sections of greatest and least curvature.

Retaining the notation of the last article, we have

\[ \frac{1}{R} = r \cos^2 \theta + 2s \sin \theta \cos \theta + t \sin^2 \theta, \]

which will be a maximum or a minimum, when

\[ -r \sin \theta \cos \theta + s (\cos^2 \theta - \sin^2 \theta) + t \sin \theta \cos \theta = 0, \]

or \( \tan 2\theta = \frac{2s}{r-t} \); an equation which gives two values of \( \theta \), viz., \( \theta = a, \theta = 90^\circ + a \), or the maximum and minimum sections are at right angles to each other. These are called the principal sections.

60. Prop. III.—The curvature of any normal section (measured by the reciprocal of the radius of curvature) is equal to the sum of the products of the curvatures of the two principal sections by the cosines of the angles which they respectively make with it.

Let one of the principal sections be the plane of \( xz \), or let the two values of \( \theta \) be \( \theta = 0, \theta = 90^\circ \). Let also \( p, p' \) be the two principal radii of curvature;

then when \( \theta = 0 \), we have \( \frac{1}{p} = r, \)

and when \( \theta = 90^\circ \), \( \frac{1}{p'} = t. \)

Also, since \( \tan 2\theta = 0 \), we have \( s = 0, \)

\[ \therefore \frac{1}{R} = \frac{1}{p} \cos^2 \theta + \frac{1}{p'} \sin^2 \theta. \]

61. Prop. IV.—If through a given tangent line at any point, both a normal and an oblique section be made, the radius of curvature of the oblique section is equal to the projection on its plane of the radius of curvature of the normal section.

Let the axis of \( x \) be the given tangent, the plane of \( xz \) the normal section, that of \( xz' \) the oblique section.

Let \( OM = h, MP = z, MN = K, NP' = z', \angle zOz' = \theta, R \), the radius of curvature in the normal section \( zz', R' \) in the oblique section \( zz'x \).

Then \( R : R' :: \text{ limit of } \frac{OM^2}{MP} : \frac{OM^2}{MP'}, \)

\[ :: \text{ limit of } \frac{MP'}{MP} :: \text{ limit of } \frac{z'}{\cos \theta} : z, \]

\[ :: \text{ limit of } \frac{(rh^2 + 2shk + tk^2)}{rh^2 \cos \theta} :: 1 : \cos \theta, \text{ because the limit of } \frac{k}{h} = 0; \]

therefore \( R' = R \cos \theta. \)

This is called Meunier's Theorem, and it establishes the fact, that if the diameter of a circle in the plane of \( yz \) be that of the circle of curvature of the normal section, the chord of the same circle in the direction of the oblique section will be the diameter of the circle of curvature of that section; and consequently if the diameter of a sphere be the diameter of curvature of any normal section, the chord of the same sphere made by an oblique section through the same tangent line will be the diameter of the circle of curvature of that section; or the sections of the sphere by planes through the same tangent line will all be circles of curvature of the corresponding sections of the surface; and conversely.

62. Prop. V.—To find the conditions that a section of a surface may have a contact of the second order with a circle.

We shall suppose the touching circle to be determined by the section of a sphere, the cutting plane which is common to it and the given surface being supposed to pass through their ordinate \( z \), so as to have for its equation \( y = m(x-x') \); or if \( x+h, y+k \) are the co-ordinates of a point in the section of either surface near their point of contact, the increments are connected by the equation \( k=mh. \)

Let \( f(x,y,z)=0 \) be the equation to the given surface; \((X-a)^2+(Y-b)^2+(Z-c)^2=R^2\) the equation to the sphere of which the section has a contact of the second order with the given surface at the point whose co-ordinates are \(x, y, z\).

Let \(z_1, Z_1\) be the ordinates of points in the common section of the two surfaces near their point of contact; then at the point in the curve of which the co-ordinates are \(x + h, y + k\), we have

\[ z_1 = z + \frac{dz}{dx} h + \frac{dz}{dy} k + \frac{1}{2} \left( \frac{d^2z}{dx^2} h^2 + 2 \frac{d^2z}{dx dy} hk + \frac{d^2z}{dy^2} k^2 \right) + \text{etc.} \]

and at the corresponding point in the circle,

\[ Z_1 = Z + \frac{dZ}{dX} h + \frac{dZ}{dY} k + \frac{h^2}{2} \left( \frac{d^2Z}{dX^2} + 2m \frac{d^2Z}{dX dY} + m^2 \frac{d^2Z}{dY^2} \right) + \text{etc.} \]

But the requirement for a contact of the second order is, that to the extent of \(h^2\) these two expressions shall be coincident.

The conditions to be fulfilled are, therefore,

\[ \frac{dZ}{dx} = \frac{dz}{dx}, \quad \frac{dZ}{dy} = \frac{dz}{dy}, \quad \frac{d^2Z}{dx^2} + 2m \left( \frac{d^2Z}{dX dY} - \frac{d^2z}{dx dy} \right) + m^2 \left( \frac{d^2Z}{dY^2} - \frac{d^2z}{dy^2} \right) = 0. \]

63. Prop. VI.—To find the radius of curvature and the co-ordinates of the centre of a normal section.

By differentiation, we obtain from the equation to the sphere of which the section of the normal plane is the circle of curvature

\[ (X-a)^2 + (Y-\beta)^2 + (Z-\gamma)^2 = R^2, \]

\[ X - a + (Z - \gamma) \frac{dZ}{dX} = 0, \]

\[ Y - \beta + (Z - \gamma) \frac{dZ}{dY} = 0, \]

\[ 1 + \left( \frac{dZ}{dX} \right)^2 + (Z - \gamma) \frac{d^2Z}{dX^2} = 0, \]

\[ \frac{dZ}{dX} \frac{dZ}{dY} + (Z - \gamma) \frac{d^2Z}{dX dY} = 0, \]

\[ 1 + \left( \frac{dZ}{dY} \right)^2 + (Z - \gamma) \frac{d^2Z}{dY^2} = 0. \]

Now the conclusion arrived at in Art. 61 enables us to apply the conditions of the last Article; for if the section of the sphere by one plane is a section of curvature, every section is so. Hence if we write \(dz/dx = p, dz/dy = q, dz/dz = r, dz/dy = s, dz/dy = t\), we shall have, by equations (1) and (2) of the last Article compared with (4), (5), and (6) of the present,

\[ \frac{d^2Z}{dX^2} = -\frac{1+p^2}{Z-\gamma}, \quad \frac{d^2Z}{dX dY} = -\frac{pq}{Z-\gamma}, \quad \frac{d^2Z}{dY^2} = -\frac{1+q^2}{Z-\gamma}, \]

and consequently equation (3) of the last Article gives

\[ \left( \frac{1+p^2}{Z-\gamma} + 2m \left( \frac{pq}{Z-\gamma} + s \right) + m^2 \left( \frac{1+q^2}{Z-\gamma} + t \right) = 0. \]

Whence \(Z - \gamma = -\frac{1+p^2 + 2mpq + m^2(1+q^2)}{r + 2ms + m^2t}\) ....(7)

Also by equations (2) and (3) we have

\[ (X-a)^2 = p^2(Z-\gamma)^2, \quad (Y-\beta)^2 = q^2(Z-\gamma)^2. \]

Equations (7), (8), and (9) determine the co-ordinates of the centre; and equation (10) gives the value of the radius of curvature of the section.

SECTION VI.—THE GENERATION OF SURFACES BY THE MOTION OF LINES.

We propose in the present section to determine the ordinary and differential equations to cylindrical, conical, and conoidal surfaces, and to surfaces of revolution.

Cylindrical Surfaces.

64. To determine the ordinary equation, we adopt the definition, that cylindrical surfaces are generated by the motion of a straight line which always moves parallel to itself.

Let \(x = ax + p, y = bx + q\) be the equations to the generating line, where \(a\) and \(b\) are the same for all the lines; and let \(x = F(z), y = \varphi(z)\) be the equations to the directing curve: then at the points where the straight line meets the curve we have

\[ x = ax + p, \quad y = bx + q, \quad x' = F(z'), \quad y' = \varphi(z'); \]

from which equations we may eliminate \(x', y'\), and \(z'\); and obtain a relation between \(p\) and \(q\) and the constants \(a\) and \(b\).

Let this relation be \(p = f(q)\); i.e.

\[ x - az = f(y - bz), \]

the equation required.

65. To obtain the differential equation, we observe that a tangent plane always touches the curve along one of the generating lines. Hence the line whose equation is \(x = ax + p, y = bx + q\) is parallel to the plane whose equation is (Art. 50)

\[ z' - z = \frac{dz'}{dx}(x' - x) + \frac{dz}{dy}(y' - y). \]

The condition is (Art. 22) \(a \frac{dz}{dx} + b \frac{dz}{dy} = 1\), which is the differential equation required.

Conical Surfaces.

66. A conical surface is generated by a straight line which passes through a given point, and always meets a given curve.

Let \(a, b, c\) be the co-ordinates of the given point; \(x = F(z), y = \varphi(z)\), the equations to the given curve, the directrix; the equations to the generatrix are \(x - a = m(z - c), y - b = n(z - c)\); also at the point where the line meets the directrix we shall have \(x - a = m(z - c), y - b = n(z - c), y' = F(z'), y = \varphi(z')\), from which equations \(x', y', z'\) may be eliminated, and there will remain

\[ m = f(n), \quad \text{or} \quad \frac{x-a}{z-c} = f\left(\frac{y-b}{z-c}\right), \]

the equation required.

67. To find the differential equation, it is only necessary to observe, that the tangent plane passes through the given point; hence \(z - e = \frac{dz}{dx}(x - a) + \frac{dz}{dy}(y - b)\) is the differential equation required. Right Conoidal Surfaces.

68. A right conoidal surface is generated by a straight line which passes through a given curve, and is at right angles to a fixed axis.

We shall suppose the fixed axis to be that of \( z \), so that the equations to the generatrix are \( z = p, y = mx \). Let the equations to the directrix be \( x = F(z), y = \phi(z) \); then at the point where the line meets the directrix we must have \( z' = p, y' = mx', x' = F(z'), y' = \phi(z') \), from which if \( x', y', z' \) be eliminated, there will remain

\[ p = f(m), \quad \text{or} \quad z = f\left(\frac{y}{x}\right) \]

is the equation required.

69. To find the differential equation, we observe that the tangent plane always contains the generating line. Retaining the same notation, the equation to the tangent plane is

\[ z' - z = \frac{dz}{dx}(x' - x) + \frac{dy}{dx}(y' - y). \]

Now when \( x' = 0, y' = 0 \), we must have \( z' = z \),

\[ \therefore \quad x \frac{dz}{dx} + y \frac{dz}{dx} = 0 \]

is the requisite condition, or the equation required.

Surfaces of Revolution.

70. A surface of revolution is generated by a variable circle whose plane is always perpendicular to a given fixed line through its centre; and whose circumference passes through a given curve.

Let \( Ax + By + Cz = D \) be the equation to the plane in which the circle lies; where \( \frac{A}{C} \) and \( \frac{B}{C} \) are known (Art. 20); \( a, b, c \) the co-ordinates of any known point in the axis in which the centre of the generating circle lies; the equation to a sphere whose centre is in this line will be \((x-a)^2 + (y-b)^2 + (z-c)^2 = R^2\); and the intersection of the plane \( Ax + By + Cz = D \) with this sphere will give the generating circle, provided \( D \) and \( R \) be properly selected.

Let also the equations to the directrix be \( x = F(z), y = \phi(z) \); then at the point of junction of the generatrix and directrix we must have

\[ Ax' + By' + Cz' = D \\ (x-a)^2 + (y-b)^2 + (z-c)^2 = R^2 \\ x = F(z') \\ y = \phi(z'); \]

whence by eliminating \( x', y', z' \), we obtain the relation

\[ R^2 = f(D) \]

or \((x-a)^2 + (y-b)^2 + (z-c)^2 = f(Ax + By + Cz)\), the equation required.

71. To find the differential equation, we observe that the normal always passes through the axis of revolution.

Let \( x' - a = m(z' - c) = 0, y' - b = n(z' - c) = 0 \) be the equation to the axis of revolution;

\[ x' - x + \frac{dz}{dx}(z' - z) = 0, \quad y' - y + \frac{dz}{dy}(z' - z) = 0, \]

(Art. 53), the equations to the normal; then since \( x', y', z' \) are the same in all the four equations; we have, by elimination,

\[ y - b + nc - nz = \left(\frac{dz}{dy} + n\right)(z' - z), \]

\[ x - a + mc - mz = \left(\frac{dz}{dy} + m\right)(z' - z); \]

and therefore

\[ \left(\frac{dz}{dy} + m\right)(y - b + nc - nz) = \left(\frac{dz}{dy} + n\right)(x - a + mc - mz), \]

the equation required.

For further details, the reader is referred to Hymers's Analytical Geometry of Three Dimensions, Gregory's Solid Geometry, or Leroy's Géométrie des Trois Dimensions.

Geometric Progression. See Progression.

George I., II., III., and IV., kings of Great Britain. See Britain.

George, St., one of the Bermudas, and also a town there. See Bermuda.

George of Cappadocia, or St George, the patron saint of England, was born about the beginning of the fourth century—according to some accounts, at Epiphania, in Cilicia; according to others, in Cappadocia. His father was a fuller; and the future saint himself had long a severe struggle to maintain against the disadvantages of his humble birth. According to Gregory of Nazianzen, George distinguished himself in his early career as a parasite of so mean a type that he would sell himself for a cake. By these arts he obtained the contract for supplying bacon to the troops; but he fulfilled its terms so ill, that he with difficulty escaped death at the hands of the indignant soldiers. He then fled to Alexandria, where he entered the public service, embraced Christianity, and finally became bishop of the city. Arianism was at that time rampant, and George became an Arian. Indeed, he owed his episcopate to the pliancy of his conscience, and the readiness with which he lent himself to further the political views of the court. On taking possession of his see, he found a fierce persecution going on against the Trinitarians. Instead of mitigating this evil, he carried it to such an extreme as at length to raise a rebellion against him. He fled for his life; but being soon after re-instated by the court, he returned to Egypt and signalized himself by redoubled cruelties. As might have been expected, his conduct raised up numerous enemies against him, even among his own partizans. His downfall could not be long delayed. A tyrannical act which he perpetrated towards the heathens in his diocese, shortly after the accession of Julian, irritated the people so keenly that they rose up en masse, dragged him out of the prison to which he had retired for safety, paraded him through the streets on the back of a camel, and after tearing him to pieces, burnt his remains. It is a mystery how this George was ever admitted into the calendar at all. Some writers, such as Papenbroche and Heylyn, deny altogether that the patron saint of England is the same person as the bishop of Alexandria. Other stories, however, give a version of George's history which explain to a certain extent the honour in which he is now held. One of these makes him a soldier in the service of Diocletian, in whose reign he suffered martyrdom along with many other thousands of Christians. Among the Greeks St George became known as the Victorious; and in England his renown gradually increased after the era of the Crusades to such an extent, that by the time of Edward III. he had become the patron saint of the kingdom. (Acta Sanctorum; Heylyn, History of St George; Gibbon, Decline and Fall, chap. xxii., &c.)

Cross of St George, a red cross in a field argent, forming part of the British standard. See Heraldry.

Knights of St George. See Garter. There have been various other orders of this denomination, most of which are now extinct. Among these may be noticed one founded by the Emperor Frederick III., in 1470, to guard the frontiers of Bohemia and Hungary against the Turks; that of St George of Alfama, founded by the kings of Religious Orders of St George. Of these there were different orders and congregations, particularly canons regular of St George in Alga, at Venice, established by authority of Pope Boniface IX, in 1404. The foundation of this order was laid by Bartolomeo Colonna, who preached in 1396 at Padua and some other places in the state of Venice. Pope Pius V., in 1570, gave these canons precedence over all other religious orders.