Although the mathematics received but little addition from 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 see these all we now know concerning the origin of these sciences; for it is from this source that Proclus, Simson, 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 its 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 details 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. Ioppus 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 such 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 Alexamninus 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 Heraclides; 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 Eudemus, 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 Eudemus to communicate what he had written to Philonides, also a geometer. Eudemus having died, he addressed the fourth book to Attalus, 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. Geminus 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. Posidonius 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. Diysiodorus was a skilful geometer, as appears by his solution of a difficult problem of Archimedes. Theodorus 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 Heron, 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 Thyanus, 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 scholiasts; 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 Menelaus, 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 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 day, 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 certainty 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 those 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 Menelaus and Autolycus. There is no memorial to indicate when it was written, or by whom. The other copy, marked 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 All, 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 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 anything 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 Nassirreddin, 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 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 Englishmen, such as Daniel Morlay, Robert of Reading, William Shell or William de Conchis, Clement Langton, 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 20; 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 sisters 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; Jor- nans Memorarius, 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 Navarre, 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 an approximate ratio found by Archimedes for the true exact ratio. The paralogism which he gives for a demonstration may, however, be excused in the geometers of this 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 Commedia thought worthy of publication, and had prepared 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 issues of paralogisms. It appears from the introduction to the work 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 sides 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 Mullo (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 Cindices of Serenus, besides other works on mixed mathematics; he commented on the books of Archimedes, which Eudoxus 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 Memmius, Commandine, Guido Ubald, 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 Netherlanders 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 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 Palatinum 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 1618, 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. Machinurin, 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 Veterum 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, 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 his 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 course 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 now do the same in regard to geometry. The early writers are not so much mentioned for their excellence, as in a wish to give a faithful portraiture of the state and progress of the science at different periods. The works of geometry in the last century are too numerous to be 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. a. 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. a. c. 194 Tion, Commentator on Euclid, fl. a. c. 117 Todossius, Sphericorum libri tres, printed at Oxford, 1707 Melaus, Spherics, edited by Maurolycus, fl. a. c. 100 Senus, 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 Plutarch, Commentator on Euclid, fl. a. c. 450 Taylor gave a translation in English, 1788. Verum Mathematicorum Athenaei, Bitonis, Pollidori, Heronis, Philonis, et aliorum, Opera, r. et Lat., 1693

Writers in Modern Times.

Lus de Burgo, Summa de Arithmetica, &c., 1494 Baridimus, Geometria Speculativa, 1495 Boël, L'Art et Science de Géométrie, 1514 Nicolaus Cusa, De Geometricis Transmutationibus, 1514 Albert Durer, Institutionum Geometricarum lib. iv., 1532 Olaus Finicus, Liber de Geometrica Practica, 1544 Reid, The Pathway to Knowledge, 1551 Bezo, Opuscula quaedam Geometrica, 1554 Rams, Arithmetice lib. ii. Geometricae lib. xxvi., 1580 Fickius, Geometricae Rotundae lib. iii., 1583 Stinus, Problematum Geometricorum lib. v., 1583 Rotius, Geom. Problema Tridecem Modis Dem., 1586 Vita, Opera Mathematica, 1589 Diges (Tho.), Pantometria, a geometrical treatise, 1591 Boista Porta, Elementorum Curvilineorum lib. ii., 1601 Rams, Geometria, 1604 Chrys, Geometria Practica, 1606 Ghaldus, Apollonius Redivivus, 1607 Atterton, Supplementum Apollonii Redivivi, 1612 Kepler, Nova Stereometria, &c., 1618 History. Van Ceulen, De Circulo et Adscriptis, 1619 Snellius, Cyclometricus, 1621 Metius, Arith. libri ii. et Geom. libri vi., 1626 Guidlin, De Centro Gravitatis, &c., 1635 Cavalieri, Geometria Indivisibilibus promota, 1635 Exercitationes Geometricae sex, 1647 Descartes, Geometria, 1637 Torricelli, De Sphaera et Solidis Spharalibus, &c., 1644 Gregory St Vincent, Opus Geom. quadrat. Circuli, 1647 Rudd, Practical Geometry, 1650 Wallis, Arithmetica Infinitorum, 1656 De Cycloide et Cissoidae, 1659 Pascal (A. Dettonville), Lettres (on the Cycloid), 1658 Ricci, Exer. Geom. de Maximis et Minimis, 1666 J. Gregory, Vera Circ. et Hyp. Quadratura, 1667 Geometria Pars Universalis, 1668 Slusius, Mesolabum, &c., 1668 Huygens, De Linearum Curvarum evolutione et dimensione (in Horol. Oscil.), 1673 Opera (collected by S'Gravesande), 1751 Barrow, Lectiones Geometricae, 1674 Viviani, Enodatio Problematum Gallicorum, 1677 De Omerique, Analysis Geometrica, 1698 Sharpe, Geometry Improved, 1718 Clairaut, Elémens de Géométrie, 1746 Math. Stewart, General Theorems, 1746 Propositiones Geometricae, 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, Elémens de Géométrie, 1795 Mascheroni, Géométrie du Compas, 1798 Problèmes pour les Arpenteurs, 1803 Legendre, Elémens de Géométrie (third edition), 1800 Kramp, Elémens de Géométrie, 1805 Leslie, Elements of Geometry, Geometrical Analysis, &c., 1809 L'Huillier, Elémens 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

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 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 them 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, Figures 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 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 one 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 D; 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, 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, 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, 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; triangle ABC shall be equal to DEF. The proposition will evidently be true (10), if the remaining sides BC, F are equal. Now, if it be possible to suppose that they unequal, let BC be the greater, take BG = EF, and AG; then the triangles ABG, DEF, having the side B = DE, BG = EF, and the angle B = E, will be equal one another (5), and will have the remaining side G = DF; but by hypothesis DF = AC; therefore G = AC; but AG cannot be equal to AC (15), therefore 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, GF 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, 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; then, the equal sides are parallel, and the figure is a parallelogram. (Fig. 42.)

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 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 line AB may be common to both, the curve line AEB must fall exactly upon the curve line AFB; 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.

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 \( DCE \); thereby 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... Theorem XVII.—In a circle, the angle BAD in a semicircle 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 ABDE 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 Figures: 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. THEOREM II.—If four quantities be proportionals, they are proportionals 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 the same parts as A contains two; and D will contain the same 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 proportionals, they will also be proportionals when taken alternately.

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

Or, 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

\[ \begin{align*} A &= 3X \\ B &= 4X \\ C &= 3Y \\ D &= 4Y. \end{align*} \]

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 \( B : D = X : Y \) \(\{\) Prop. I.

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

OR. 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 proportionals, they shall also be proportionals 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 each of the equal parts contained in C and D be Y. Then, because

\[ \begin{align*} A &= 5X \\ B &= 3X \\ C &= 5Y \\ D &= 3Y, \end{align*} \]

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 proportionals, they will also be proportionals 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 \( A = 5X \), \( B = 3X \), \( C = 5Y \), \( D = 3Y \), 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 proportionals, they shall also be proportionals by conversion.

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

For, making the same supposition as in the last proposition, because

\[ \begin{align*} A &= 5X \\ B &= 3X \\ C &= 5Y \\ D &= 3Y, \end{align*} \]

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

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 \( A = 2X \), \( B = 3X \), \( C = 2Y \), \( D = 3Y \).

Then \( mA = 2mX \), \( nB = 3nX \), \( mC = 2mY \), \( nD = 3nY \).

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 \( mA : B = mC : D \),

and \( A : nB = C : nD \).

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

\[ \begin{align*} A &= 2X \\ B &= 3X \\ C &= 7X \\ H &= 2Y \\ K &= 3Y \\ L &= 7Y. \end{align*} \]

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 æquali or ex æquo.

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 æquali in proportione perturbata, or ex æquo 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 æquali (Prop. VIII), A : E = C : F; and, by composition, A + E : E = C + F : F. Again, by hypothesis, E : B = F : D, therefore ex æq. 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 æq. 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

\[ A = 2X, \quad B = 3X, \] \[ C = 2Y, \quad D = 3Y, \] \[ E = 2Z, \quad F = 3Z. \]

Hence, by addition,

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

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; 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 AF = 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, I). 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

\[ \frac{ABCD}{AEFG} = \frac{mn}{pq}. \]

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)

\[ \frac{ABCD}{AEHD} = \frac{AB}{AE} = \frac{m}{p}, \]

but \( m : p = nm : np \) (1, 3),

therefore \( \frac{ABCD}{AEHD} = \frac{nm}{np}. \)

Again, \( \frac{AEHD}{AEFG} = \frac{AD}{AG} = \frac{n}{q}; \)

but \( n : q = pn : pq; \)

therefore \( \frac{AEHD}{AEFG} = \frac{pn}{pq}; \)

and it was shown that

\[ \frac{ABCD}{AEHD} = \frac{nm}{np} \text{ or } \frac{pn}{pq}, \]

therefore (8, 3) \( \frac{ABCD}{AEFG} = \frac{mn}{pq}. \)

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. I); but the measure of the rectangle is BC × AO (Schol. 4), therefore the area of the parallelogram is BC × 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 × AO (5), therefore that of the triangle is $\frac{1}{2}BC \times AO$, or BC × $\frac{1}{2}AO$.

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 CIK = BIL, 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 $AL \times EF$. 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 $FE \times \frac{(AB + CD)}{2}$.

**Theorem VIII.**—If four straight lines AB, AC, AD, AE be proportional; 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 proportional.

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,

$$\frac{ABHD}{ACGD} = \frac{AB}{AC} (3)$$

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

$$\frac{ABHD}{ABFE} = \frac{AD}{AE}$$

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

Next suppose that the rectangle ACGD = ABFE; then

$$\frac{ABHD}{ACGD} = \frac{ABHD}{ABFE} (Ax.1,3)$$

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

Cor. If three straight lines be proportional, 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 proportional.

**Theorem IX.**—If four straight lines be proportional, and also other four; the rectangles contained by the corresponding terms shall be proportional; 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),

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

$$\text{and rect. CH : rect. DN = CD : DE;}$$

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 proportional.

**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, EF = BC = IG, and EF = DG (26, 1); therefore IGDH is the square upon BC; and the remaining parts are the two rectangles BCGI, FEHI, which hire 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 ABIF, 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 ABLIKEA, 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 = DI and BK = FE; therefore AKLE = ABHE + EDGF; but these two parts constitute the excess of the square AIF 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 \( AD \) 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 \( AD \), 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 \( CAE \) 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 \( BAC = CDE \), \( ABC = DCE \), and \( ACB = DEC \); 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 \( FD = AC, BC : CE = AC : DE \); 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 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 = D (Ax. 2, 3). In like manner, because by the same theo- rem BC : EF = CA : FG; and by hypothesis BC : EF = C : FD; therefore FG = FD; but it was shown that E = ED, therefore, the triangles GEF, DEF, having the sides of the one equal to those of the other, each to each, unequal; 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 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,

then the triangle ABC is similar to DEF. Take AG = DE, and draw GH parallel to BC, then the angle AGH = ABC (21, 1), 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.

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, the 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.

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,

\[ \frac{ABE}{ADE} = \frac{AB \times AE}{AD \times AE}. \]

In the same manner, it may be demonstrated that

\[ \frac{ABC}{ABE} = \frac{AB \times AC}{AB \times AE}; \]

therefore (8, 3) \( \frac{ABC}{ADE} = \frac{AB \times AC}{AD \times AE}. \)

Con. Therefore the two triangles are equivalent, if the rectangle \( AB \times AC = AD \times AE \), 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) \( \frac{AB}{DE} = \frac{AC}{DF}; \) now \( \frac{AB}{DE} = \frac{AB}{DE}, \) for the two ratios are identical, therefore (9),

\[ \frac{AB^2}{DE^2} = \frac{AB \times AC}{DE \times DF} \] but \( \frac{ABC}{DEF} = \frac{AB \times AC}{DE \times DF} \) (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 poly- gon 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 \( \frac{AB}{FG} = \frac{BC}{GH}, \) 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 = AC² : FH² (25), in like manner ACD : FHI = AC² : FH²; therefore,

\[ \text{ABC : FGH} = \text{ACD : FHI}. \]

By the same manner of reasoning,

\[ \text{ACD : FHI} = \text{ADE : FIK}, \]

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

\[ \frac{AB}{L} = \frac{AB^2}{AB \times L} \quad (3), \]

or since \(AB \times L = FG^2\),

\[ \text{(Cor. to 8)} \quad \frac{AB}{L} = \frac{AB^2}{FG^2} = \text{ABCDE : FGHIK}; \]

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.

Cor. 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 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.

Cor. 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.

Cor. 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, AB, the number of which is p, the arch DF shall contain q equal parts Dh, hl, 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.

Cor. 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 centre; 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.

Fig. 93.

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.

Fig. 94.

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 F 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.

Fig. 96.

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, AD'C 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, 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 XIV.**—To find a fourth proportional to three given lines A, B, C.

Draw two straight lines DE, DF, containing any angle; 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 : IK (17, 4), that is, A : B = C : DX; therefore DX is a fourth proportional to A, B, and C.

Con. The same construction serves to find a third proportional to two lines A and B; for it is the same as 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, meeting the circle in D; make AF = AD, and AB shall be divided at F in the manner required.

For since AB is perpendicular to the radius, it is a tangent to the circle (9, 2); and if AC be produced to meet the circle in E, AB : AF = AE : AB (30, 4), and by division AB — AF : AF = AE — AB : AB; but AB — AF = BF; and since DE = 2BC = AB, therefore AE — AB = AD = AF, therefore BF : AF = AF : AB.

**Scholium.** When a line is divided in this manner, it is said to be divided in extreme and mean ratio.

**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 Lines and equivalent to the triangle CDE (2 Cor. 2, 4); to each add Figures the polygon ABCE, and the polygon ABCDE shall be upon a Plane.

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 FGH, 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 (15, 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 because OT bisects the arch AB; it is perpendicular to chord AB (6, 2); therefore GH is parallel to AB (2, 18, 1), and HI to BC, therefore the angle IHO = ABO, and IHO = CBO, and hence GHI = ABC; 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 = CI : 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.

**ECT. 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, GHKLM, 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.

**Fig. 120.**

If each of the polygons is by hypothesis equilateral, it will also be equangular (Schol. 23, 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 A BO, GP, HP to the centres of the circles; then, because the angle AOB is the same part of four right angles as 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 = O : GP; therefore (9 and 27, 4) polygon ABCDEF = polygon GHKLM = 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 a certain number of times exactly, its chord AB is the side of a regular figure inscribed in the circle; apply lines in the circle, each equal to AB, thus forming the regular figure ABC, &c., and describe a regular figure DEF, &c. of the same number of sides about the circle. Then the excess of the circumscribed figure above the inscribed figure shall be less than the square upon AG. For draw lines from D and E to O the centre, these lines will pass through A and B (24, 5); also, a line drawn from O, to H the point of contact of the line DE, will bisect AB, and be perpendicular to it; and AB will be parallel to DE. Draw the diameter AL, and join BL, which will be parallel to HO (18, 4). Put P for the circumscribed polygon, and p for the inscribed polygon; then, because the triangles ODH, OAK are evidently like parts of P and p, P : p = ODH : OAK (1, 3); but the triangles ODH, OAK being similar, ODH : OAK = OK² : OA² (25, 4); and on account of the similar triangles OAK, LAB, OA² : OK² = LA² : LB² (20, and 9, 4); therefore P : p = LA² : LB², and by division and inversion, P : P — p = LA² : LA² — LB², or AB²; but LA², that is, the square described about the circle, is greater than the equilateral polygon of eight sides described about the circle, because it contains that polygon, and for the same reason the polygon of eight sides is greater than the polygon of sixteen sides, and so on; therefore LA² > P, and as it has been proved that P : P — p = LA² : AB², of which proportion the first term P is less than the third LA²; therefore (2, 3) the second P — p is less than the fourth AB²; but AB² > AG², therefore P — p < AG².

Cor. 1. Because the polygons P and p differ from one another more than either of them differs from the circle, the difference between each of them, and the circle, is less than the given space, viz. the square of AG. And therefore, however small any space may be, a polygon may be inscribed in the circle, and another described about it, each of which shall differ from the circle by less than the given space.

Cor. 2. A space which is greater than any polygon that can be inscribed in a circle, but which is less than any polygon that can be described about it, is equal to the circle itself.

**Prop. III. Theorem.**—The area of any circle is equal to a rectangle contained by the radius, and a straight line equal to half 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 \( n \times \frac{1}{2}DE \times OH = \frac{1}{2} \times \text{triangle DOE} = \text{polygon DEF}, \ldots \); therefore, \( \frac{1}{2}P \times OH = \text{polygon DEF}, \ldots \); and in like manner it appears that \( \frac{1}{2}p \times OK = \text{polygon ABC}, \ldots \).

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 GHJKLM 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 \( AO^2 : GP^2 = \text{polygon ABCDEF : polygon GHJKLM} \), and \( AO^2 : GP^2 = \text{circle ABE : Q} \), therefore polygon ABCDEF = polygon GHJKLM = circle ABE : Q; but circle ABE > polygon ABCDEF, therefore Q > polygon GHJKLM; 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 \( AO^2 : GP^2 = \text{circle ABE : circle GHL} \), therefore \( AO^2 : GP^2 = \text{circle ABE : circle GHL} \).

Cor. I. 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 \( \frac{1}{2}M \times AO : AO^2 = \frac{1}{2}N \times GP : GP^2 \), therefore (3, 4), \( \frac{1}{2}M : AO = \frac{1}{2}N : GP \); and again, by alternation, \( \frac{1}{2}M : N = AO : GP \), 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 are by hypothesis known, and it is required to find \( a \) and \( b \).

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 \( a = \sqrt{AB} \).

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 \( a \) 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: \( a = \sqrt{AB} \),

\[ b = \frac{2A \times B}{A + a} \] give us \( a = 28284271 \), &c. the area of the inscribed octagon, and \( b = 313137085 \), &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 | 20000000 | 40000000 | | 8 | 28284271 | 33137085 | | 16 | 30614674 | 31825979 | | 32 | 31214451 | 31517249 | | 64 | 31365485 | 31441184 | | 128 | 31403311 | 31422236 | | 256 | 31412772 | 31417504 | | 512 | 31415138 | 31416321 | | 1024 | 31415729 | 31416025 | | 2048 | 31415877 | 31415951 | | 4096 | 31415914 | 31415133 | | 8192 | 31415923 | 31415928 | | 16384 | 31415925 | 31415927 | | 32768 | 31415926 | 31415926 |

Hence it appears that the area of a regular polygon of 32768 sides inscribed in the circle, and of a similar poly- PART II.—GEOMETRY OF SOLIDS.

SCT. I.—OF PLANES AND SOLID ANGLES.

Definitions.

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

2. 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.

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

4. It will be demonstrated (Theor. 3) that the common section of two planes is a straight line; this being premised, the inclination of two planes is the angle contained by two straight lines drawn perpendicular to the line which is their common section, from any point in it, the perpendicular being drawn in the one plane, and neither in the other plane.

This angle may be either acute or obtuse.

If it be a right angle, the two planes are perpendicular to each other.

5. A solid angle is that which is made by the meeting 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.

It is evident, from the definition of a plane (Sect. I, Part I), 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; therefore 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 a 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 intersection; 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, meeting 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 as 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 I), 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 I); 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. 11, 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, to the same straight line AB, which is impossible.

Theorem VII.—The intersections GE, 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.1) 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 (1 Cor. 5); therefore they are equal.

Theorem X.—If two straight lines CA, EA, meeting one another, be parallel to two other lines DB, EB, 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 ABDC 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 PH = 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.) 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 \(MN\).

Cor. 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 \(AVB\), \(BVC\), &c., formed about the vertex \(V\) is equivalent to the sum of the angles of a like number of triangles \(AOB\), \(BOC\), &c., 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 behg; 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 FGHK coincides entirely with its equal fghk, 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 BE, the In the same manner it may be demonstrated that the Geometry oblique prism BCD-FGH is equal to the right prism of Solids Bed-Fgh. But the two prisms are equal (Cor. to 1), since they have the same altitude, BF and their bases are equal, 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, I). 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 plane angles are equal each to each, and similarly situated, 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 BIL (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 \(AG\), \(AK\), 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,

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

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, \textit{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, \textit{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 \(AC\), \(AN\), 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 parallelopiped 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 parallelopipeds 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 parallelopipeds, 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 parallelopipeds \( P \) 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 parallelopiped 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 base 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 \( BCD, bed \). 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, I), 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),

\[ AH \times \text{base } BCD : ah \times \text{base } BCD = BC^3 : bc \times BC^2, \]

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

therefore, ex aequo,

\[ AH \times \text{base } BCD : ah \times \text{base } bcd = BC^3 : bc^2. \]

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

\[ \text{prism } P : \text{prism } p = BC^3 : bc^2. \]

Cor. 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 \( S^2s \) 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 \( AFG \), and the triangle \( ABD \) 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.) Draw AKE, ake perpendicular to the bases BCD, bed, 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 = bc : fg; therefore BC : FG = be : fg; and hence BC² : FG² = bc² : fg² (23, 4, Part I); but because of the similar triangles BDC, FHG, BC² : FG² = trian. BDC : trian. FHG, and in like manner bc² : fg² = trian. bed : trian. fgh (25, 4, Part I), therefore

trian. BCD : trian. FGH = trian. bed : trian. fgh.

Now, trian. BCD = trian. bed (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 CP, 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, GRS, 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 KBCDLF 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 k; 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; wherefore, 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.

**Proposition XVI.**—Pyramids that have equal bases and altitudes are equal to one another.

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

For if they are not equal, let Z represent the solid which is equal to the excess of one of them, a-bed, 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-bed. 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, gf, 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 denote the pyramids A-BCD, and a-bed 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 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. I.), 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. I.), and which therefore will be greater than... Geometry LIM. Let this be the polygon BECED, and upon BECFD of Solids; 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 ba 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 \( ABba \) and either of the two series of cylinders will be less than the greatest circumscribed cylinder: For the solid \( ABba \) 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, CFEF, 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 quadrantial 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 CF. 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, c 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

\[ C + d = c, \quad \text{and} \quad H = h + d', \]

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 \(C = \frac{1}{2}K\) (3), therefore \(H = \frac{1}{2}K\); 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.

Table showing the Theorem of the foregoing Treatise, 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. |

GEORGE I. II. III. and IV. kings of Great Britain.