GEOMETRY

GEOMETRY.

INTRODUCTION.

THERE is reason to believe that geometry, as well as most of the other sciences, was first cultivated in Egypt; and, according to some authors, it had its origin in the necessity there was of assigning to the inhabitants every year their particular shares of land: for as the country was annually overflowed by the Nile, it has been taken for granted (perhaps without good reason), that the land-marks would be obliterated, and the possessions rendered undistinguishable from one another. Such is said to have been the origin of land-measuring, the form under which geometry was first known, and from which it has taken its name; for geometry literally signifies the measuring of the earth.

The historian Herodotus refers the origin of geometry to the time when Sesostris intersected Egypt by numerous canals, and divided the country among the inhabitants; and this account of the beginning of the science has been considered by some as very probable.

From Egypt geometry was carried into Greece by Thales of Miletus about 600 years before the Christian æra. This celebrated philosopher is said to have made numerous discoveries in geometry; and in particular to have first observed that any angle in a semicircle is a right angle; a discovery which gave him great joy, and for which he thanked the muses by a sacrifice.

Among the disciples of Thales were Anaximander and Anaxagoras: the first of these wrote an elementary treatise or introduction to geometry, the earliest of which there is any mention in history; and the last is said to have attempted the quadrature of the circle, a problem which has baffled the skill of mathematicians of every age.

Pythagorus followed Thales, and had the merit of discovering one of the most beautiful and important propositions of the whole science, namely, that the square of the hypothenuse of a right-angled triangle is equal to the squares of the two other sides. He is said to have been so transported with joy at this discovery, that he sacrificed a hundred oxen to the gods as a testimony of his gratitude. The truth of this anecdote has however been doubted, on account of the philosopher's moderate fortune and religious opinions concerning the transmigration of souls.

Zenodorus is the earliest of the geometers whose writings have reached modern times, a part of them having been preserved by Theon, in his commentary on Ptolemy.

Hippocrates of Chios cultivated geometry, and distinguished himself by the quadrature of the curvilinear

space contained between half the circumference of one circle, and the fourth part of the circumference of another circle, their concavities being both turned the same way, and the radius of the former to that of the latter as 1 to $\backslash sqrt\{2\}$. He also wrote elements of geometry which are now lost.

The founding of the school of Plato forms one of the earliest and most important epochs in the history of geometry; for to that philosopher we are said to be indebted for the discovery of the Geometrical Analysis, by which the science has been greatly extended, and which is indeed absolutely necessary for the resolution of problems of a certain degree of difficulty.

The Conic Sections, and the theory of Geometrical Loci, are commonly reckoned among the improvements which geometry received from his disciples; and there is reason to suppose that these, as well as many other important discoveries which we have not room here to enumerate, were first suggested by the attempts of the geometers of the Platonic school to resolve two celebrated problems, namely, to trisect, or divide into three equal parts, a given angle; and to construct a cube which should be the double of another cube; which last problem Hippocrates had shewn to be equivalent to the finding of two mean proportionals between two given lines. The esteem in which Plato held the science of geometry is fully evinced by the following inscription over the door of his school: Let no one enter here that is ignorant of geometry.

The science of geometry was likewise cultivated in all its branches by the philosophers of the Alexandrian school, among whom Euclid claims, in a particular manner, our attention. This celebrated mathematician lived about 300 years before the Christian æra, and probably studied geometry at Athens under the disciples of Plato. From Greece he went to Alexandria, allured thither no doubt by the fame of the celebrated school of that city, and by the favours conferred by the first Ptolemy upon learned men. He composed elements of geometry in a systematic form, comprehending in them such propositions belonging to the first principles of the science as had been discovered by mathematicians previous to his time. This work has had the singular good fortune to preserve the highest reputation in all ages and in all countries where science has been cultivated, and it has served as the groundwork of innumerable other treatises, few of which, if any, have excelled it. Many commentaries have been written on it, and it has been translated into almost all

the European and Oriental languages. Euclid is likewise known to have written other works on geometry: of these we have his Data, which may be regarded as a continuation of his elements; and an account of a work of his on porisms (see PORISMS) preserved in the writings of Pappus, but which has suffered so much from time as to be almost unintelligible.

After Euclid, lived Archimedes, who cultivated and improved all the branches of the mathematics known at that period, and in a particular manner geometry. He was the first that found nearly the ratio of the diameter of a circle to its circumference, and he squared the parabola. He likewise wrote treatises on the Sphere and Cylinder, on Spirals, on Conoids and Spheroids, besides others on mixt Mathematics. He also extended and improved the Geometrical Analysis, the principles of which had been established in the school of Plato. Many of the writings of Archimedes have been lost; but such as remain prove him to have been one of the greatest geometers that ever lived, and indeed the NEWTON of antiquity.

Apollonius of Perga was nearly contemporary with Archimedes, that is, he flourished about the end of the second century before the Christian era. He studied geometry in the Alexandrian school under the successors of Euclid, and he greatly extended the theory of the conic sections (see introduction to CONIC SECTIONS). He also composed treatises on different parts of Geometrical Analysis, but of these only one has come down to us entire; it is entitled de sectione rationis, and was discovered in the Arabic tongue, from which it has been translated into Latin by Dr Halley. Such accounts however are preserved in the mathematical collections of Pappus of his other treatises, that several of them have been restored by modern mathematicians. We may mention in particular his treatises de Locis Planis, de Sectione Spatii, de Sectione Determinata, de Tractionibus, each of which is divided into two books.

Having mentioned Archimedes and Apollonius, by far the most illustrious mathematicians of the period in which they lived, we shall pass over several others who contributed nothing to the improvement of the science, and therefore are but little known to us. We shall however briefly notice Theodosius, who lived about 50 years A. C. and who is the author of a work on Spherics, which is considered as one of the most valuable of the books on the ancient geometry.

Pappus and Theon of Alexandria deserve to be mentioned as among the most celebrated of the commentators and annotators on the ancient geometry. We are particularly indebted to Pappus (who lived about the middle of the fourth century) for our knowledge of various discoveries and treatises of the ancient geometers, which, but for the account he has given of them in his mathematical collections, would have been for ever lost to mathematicians of modern times.

Proclus, the head of the Platonic school at Athens, cultivated mathematics about the middle of the fifth century; and although it does not appear that he made any discoveries in the science, yet he rendered it some service by his example and instruction. He wrote a commentary on the first book of Euclid, which contains many curious observations respecting the history and metaphysics of mathematics.

We have now briefly noticed the principal epochs in

the history of geometry, and the most celebrated men who have contributed to its improvement from the earliest periods of history to the end of the fifth century; but long before this time the era of discovery seems to have been past, and the science on the decline. Still however the Alexandrian school existed, and it was possible that a Euclid or an Apollonius might again arise in that seminary. But the taking of Alexandria by the Arabs in the year 641 gave a death-blow to the sciences, not only in that capital, but throughout the whole Greek empire. The library, a treasure of infinite value, was burnt, and the stores of learning which had been accumulating for ages were annihilated for ever.

Although by this unfortunate event the sciences suffered an irreparable loss, it must be attributed to the fanaticism of the new religion which the conquerors had adopted, rather than to national ignorance or barbarity; for before that period, the sciences, when on the decline in Greece, had found an asylum among them, and about 120 years after the death of Mohammed they again took them under their protection.

The Arabs translated the greater part of the works of the Greek geometers, and chiefly those introductory to astronomy. They even began to study the more sublime geometry of the ancients; for Apollonius's Conic Sections became familiar to them, and some of the books of that work have only reached us in an Arabic version. They gave to Trigonometry its present simple and commodious form, and greatly simplified its operations by the introduction of sines instead of the chords of double arcs, which had been formerly used.

After geometry, as well as its kindred mathematical sciences, had remained for several centuries under the protection of the Arabs, it was again received into Spain, Italy, and the rest of Europe, about the year 1400. Among the earliest writers on the subject after this period, were Leonardus Pisanus, and Lucas Paciolus or de Burgo.

The limits within which we must necessarily confine this sketch of the history of the science, will not, however, allow us to enumerate all the improvements which it has received since the restoration of letters in Europe; for a list of the names of those who have contributed more or less to its extension, would include almost every mathematician of note from the time of Leonardus Pisanus to the present day.

The writings of the ancient geometers have been assiduously sought after, and held in great repute; for it appears that as far as they carried some of their theories, they left but little room for improvement, and of this remark we think the writings of Euclid, of Archimedes, and of Apollonius, afford remarkable instances. Euclid's elements of geometry have been considered, at least in this country, as one of the best books that could be put into the hands of the mathematical student, particularly that edition of its first six and eleventh and twelfth books which was given to the world by the late Dr Simson. An excellent system of geometry, comprehending the first six books of the illustrious ancient, together with three supplementary books, has of late years been published by Mr Professor Playfair, of the University of Edinburgh. We believe no modern system has excelled that of Euclid (as

(as restored to its original purity by Dr Simson) in respect of logical accuracy and systematic arrangement. There is one, however, which we must particularly mention on account of its great excellence, and the use

we have made of it in the system we are now to present to our readers. It is that of Mr Legendre, which we consider as the most complete and extensive that has yet appeared. First Principles.

SECT. I. THE FIRST PRINCIPLES.

DEFINITIONS.

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

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

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

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

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

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

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

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

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

X. A Plane Rectilinear 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 DAB.

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

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 every where by lines.

If the lines be straight, the space which they enclose is called a Rectilinear 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. 9.).

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. 9.); and an acute-angled triangle is that which has three acute angles (fig. 11.).

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

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

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

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 of 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 manner of 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; .

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 $\backslash times$ 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 $A\; \backslash times\; B$ 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 $m\; A$ will be used to express the product of $m$ and $A$, and so on with other expressions, agreeable 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 $\backslash sqrt\{\backslash phantom\{x\}\}$ indicates a root to be extracted; thus $\backslash sqrt\{A\; \backslash times\; B\}$ 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 7th proposition of the section in which it occurs; (4. 2.) means the 4th proposition of the 2d section; (2. cor. 28. 4.) means the 2d corollary to the 28th proposition of the 4th 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. Fig. 17

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 $ACD$, $DCB$. Fig. 18

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 Fig. 19

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.

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.

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.

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 tri-

angles shall coincide exactly, the one with the other; therefore they are equal (ax. 5.).

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

Two triangles are equal, when they have a side, Fig. 22, and the two adjacent angles of the one equal to a side, and the two adjacent angles of the other, each to each.

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 FD, 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 ED; 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.

Any two sides of a triangle are together greater Fig. 23. than the third.

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

If from a point O, within a triangle ABC, there Fig. 24. 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 ; that is .

Again, since , to each of these add DC and we have , but it has been shewn that , much more then is .

If two sides AB, AC of a triangle ABC are equal Fig. 24. to two sides DE, DF of another triangle DEF, Fig. 25. each to each; but if the angle BAC contained Fig. 26.

by the former is 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, (6.) and therefore GC=EF. Now there may be three cases, according as the point G falls without the triangle BAC, or on the side BC, or within the same triangle.

Fig. 24. CASE I. Because $GC\; \backslash leq\; GI\; +\; IC$, and $AB\; \backslash leq\; AI\; +\; IB$, (7.) therefore $GC\; +\; AB\; \backslash leq\; GI\; +\; AI\; +\; IC\; +\; IB$, that is $GC\; +\; AB\; \backslash leq\; AG\; +\; BC$, from each of these unequal quantities take away the equal quantities AB, AG, and there remains $GC\; \backslash leq\; BC$, therefore $EF\; \backslash leq\; BC$.

Fig. 25. 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.

Fig. 26. CASE III. Lastly, if the point G fall within the triangle BAC, then $AG\; +\; GC\; \backslash leq\; AB\; +\; BC$, (8.) therefore, taking away the equal quantities AG, AB, there remains $GC\; \backslash leq\; BC$ or $EF\; \backslash leq\; BC$.

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.

Fig. 22. Two triangles are equal, when the three sides of the one are equal to the three sides of the other, each to each.

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, as 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.

Fig. 27. 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 the others.

THEOREM XII.

If two angles of a triangle are 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, , 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 not equal to B, therefore it is greater.

THEOREM XIV.

From a given 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

produce one of them AB, so that $BF=BA$, 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, (ax. 4.) therefore it is equally impossible that two perpendiculars can be drawn from the same point to the same straight line.

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=BCA, the side CB is common, and the side $BF=BA$, therefore the third side $CF=AC$, (5.) but $AF=AC+CF$, (7.) that is ; therefore , 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 a given 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.

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.

Two right-angled triangles are equal, when the hypotenuse and a side of the one are equal to the hypotenuse and a side of the other, each to each.

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

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.

7. If two points E, G in a straight line AB are situated at unequal distances EF, GH 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.

If two straight lines AB, CD be parallel, the 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 shewn, 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 shewn that GH is perpendicular to AB.

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

Fig. 36. 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.

Fig. 37. If a straight line EF meet two parallel 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 shewn 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 shewn 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.

Fig. 38. 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.

If a side AC of a triangle ABC be produced to-wards 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.

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.

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

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.

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.

If the opposite sides of a quadrilateral ABCD are equal, so that AB = DC, and AD = BC; then the equal sides are parallel, and the figure is a parallelogram.

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.

If two opposite sides AB, DC, of a quadrilateral are equal and parallel, the two other sides are in like manner equal and parallel; and the figure is a parallelogram.

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.

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

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

Any diameter AB, divides the circle and its circumference into two equal parts.

FOR if the figure AEB be applied to AFB, so that the base AB may be common to both, the curve line AEB must fall exactly upon the curve line 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.

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 .

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.).

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

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

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

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

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

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.

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.

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.).

Since the centre C, the middle D of the chord AB, and the middle G 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.

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 those 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; 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.

Two equal chords are equally distant from the centre;

centre; and of unequal chords, that which is nearer the centre is greater than that which is more remote.

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 $AF\; (=\; \backslash frac\{1\}\{2\}AB)$ of the one is also equal to the side $DG\; (=\; \backslash frac\{1\}\{2\}DE)$ 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, the chord nearer the centre is greater than that which is farther from it.

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.).

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.

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 $A,\; D$.

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.

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.

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 $AB,\; DE$ are equal, the angles $ACB,\; DCE$ are equal.

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

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

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

Suppose in the next place that the centre is without the angle $BAD$; then, drawing the diameter $AE$, it may be demonstrated, as in the first case, that the angle $ECD$ is double of the angle $EAD$, and that the angle

angle ECB, a part of the first, is double the angle EAB a part of the other; therefore the remaining angle BCD is double the remaining angle BAD.

Fig. 57. All angles BAD, BFD in the same segment BAED
Fig. 58. of a circle are equal to one another.

Fig. 57. If the segment be greater than a semicircle, from the centre C draw CB and CD; then the angles BAD and BFD being (by last theorem) each equal to half BCD, they must be equal to one another.

Fig. 58. But if the segment BAED be less than a semicircle, let H be the intersection of BF and AD; then, the triangles ABH and FDH having the angle AHB of the one equal to FHD of the other, (4. 1.) and ABH = FDH, (by case 1.) will have the remaining angles of the one equal to the remaining angles of the other; that is the angle BAH = HFD, or BAD = BFD.

Fig. 59. The opposite angles of any quadrilateral figure ABCD described in a circle are together equal to two right angles.

DRAW the diagonals AC, BD; because the angle ABC = ACD, and CBD = CAD, (last theor.) the sum $ABD\; +\; CBD\; =\; ACD\; +\; CAD$; or $ABC\; =\; ACD\; +\; CAD$; to each of these equals add ADC, and $ABC\; +\; ADC\; =\; ACD\; +\; CAD\; +\; ADC$; but the last three angles being the angles of the triangle ADC, are taken together equal to two right angles, (24. 1.) therefore $ABC\; +\; ADC\; =$ two right angles. In the same manner, the angles BAD, BCD may be shewn to be together equal to two right angles.

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.

The angle BAC contained by AC, a tangent, and Fig. 60. 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.

I. WHEN one magnitude contains another a certain number of times exactly, the former is said to be a multiple of the latter, and the latter a part of the former.

II. When several magnitudes are multiples of as many others, and each contains its parts the same number of times, the former are said to be equimultiples of the latter, and the latter like parts of the former.

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

IV. If there be four magnitudes, or quantities, A, B, C, D, and if A contain some part of B just as often as C contains a like part of D, then, the ratio of A to B is said to be the same with (or equal to) the ratio of C to D.

It follows immediately from this definition, that if A contain B just as often as C contains D, then the ratio of A to B is equal to the ratio of C to D; for in that case it is evident that A will contain any part of B just as often as C contains a like part of D.

V. When two ratios are equal, their terms are called proportionals.

To denote that the ratio of A to B is equal to the ratio of C to D, they are usually written thus, $A\; :\; B\; ::\; C\; :\; D$, or thus, $A\; :\; B\; =\; C\; :\; D$, which is read thus; A is to B as C to D; such an expression is called an analogy or a proportion.

VI. Of four proportional quantities, the last term is called a fourth proportional to the other three taken in order.

VII. Three quantities A, B, C, are said to be proportional, when the ratio of the first A to the second B is equal to the ratio of the second B to the third C.

VIII. Of three proportional quantities, the middle term is said to be a mean proportional between the other two, and the last a third proportional to the first and second.

IX. Quantities are said to be continual proportionals, when the first is to the second, as the second to the third, and as the third to the fourth, and so on.

X. When there is any number of magnitudes A, B, C, D, of the same kind, the ratio of the first A to the last D is said to be compounded of the ratio of A

of A to B, and of the ratio of B to C, and of the ratio of C to D.

(Ax. 4.) then, if $x$ be put for one of those equal parts, Of Proportion.

XI. If three magnitudes A, B, C, be continual proportionals; that is, if the ratio of A to B be equal to the ratio of B to C; then the ratio of the first A to the third C is said to be duplicate of the ratio of the first A to the second B. Hence, since by the last definition the ratio of A to C is compounded of the ratio of A to B and of B to C, a ratio which is compounded of two equal ratios is duplicate of either of them.

XII. If four magnitudes A, B, C, D be continual proportionals, the ratio of the first A to the fourth D is said to be triplicate the ratio of the first A to the second B. Hence a ratio compounded of three equal ratios is triplicate of any one of them.

XIII. Ratio of Equality is that which equal magnitudes bear to each other.

The next four definitions explain the names given by geometers to certain ways of changing either the order or magnitude of proportionals, so that they still continue to be proportional.

XIV. Inverse Ratio is when the antecedent is made the consequent, and the consequent the antecedent. See Theor. 3.

XV. Alternate proportion is when antecedent is compared with antecedent, and consequent with consequent. See Theor. 2.

XVI. Compounded ratio is when the sum of the antecedent and consequent is compared either with the antecedent, or with the consequent. See Theor. 4.

XVII. Divided ratio is when the difference of the antecedent and consequent is compared either with the antecedent or with the consequent. See Theor. 4.

AXIOMS.

1. Equal quantities have each the same ratio to the same quantity; and the same quantity has the same ratio to each of any number of equal quantities.

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

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

4. If two quantities be divided into, or composed of parts that are equal among themselves, or all of the same magnitude, then will the whole of the one have the same ratio to the whole of the other, as the number of parts in the one has to the number of equal parts in the other.

THEOREM I.

Equimultiples of any two quantities have to each other the same ratio as the quantities themselves.

Let A and B be any two quantities, and, $m$ being put to denote any number, let $m\; A$, $m\; B$ be equimultiples of those quantities, $m\; A$ shall have to $m\; B$ the same ratio that A has to B.

Let the ratio of A to B be equal to the ratio of one number $p$ to another number $q$, that is, let A contain $p$ such equal parts as B contains $q$, VOL. IX. Part II.

and consequently, multiplying both by the same number $m$,

or, which is evidently the same,

Hence it appears that $m\; A$ contains the quantity $m\; x$ as a part $p$ times; and that $m\; B$ contains the same quantity $q$ times; therefore the ratio of $m\; A$ to $m\; B$ is the same as the ratio of the number $p$ to the number $q$ (Ax. 4.); but the ratio of A to B is also equal to the ratio of $p$ to $q$, (by hypothesis), therefore the ratio of $m\; A$ to $m\; B$ is equal to the ratio of A to B (Ax. 3.).

COR. Hence like parts of quantities have to each other the same ratio as the wholes: that is, $\backslash frac\{A\}\{m\}\; :\; \backslash frac\{B\}\{m\}\; ::$

$A\; :\; B$; for A and B are equimultiples of $\backslash frac\{A\}\{m\}$ and $\backslash frac\{B\}\{m\}$.

THEOREM II.

If four quantities of the same kind be proportionals, they shall also be proportionals by alternation.

LET A, B, C, D be four quantities, of the same kind, and let $A\; :\; B\; ::\; C\; :\; D$; then shall $A\; :\; C\; ::\; B\; :\; D$.

Let the equal ratios of A to B, and of C to D, be the same as the ratio of the number $p$ to the number $q$; then A will contain $p$ such equal parts as B contains $q$, (Ax. 4.) and C will, in like manner, contain $p$ such equal parts as D contains $q$; let each of the equal parts thus contained in A and B be $x$, and let each of those contained in C and D be $y$, then

Now as $A\; =\; p\; x$, and $C\; =\; p\; y$; it is manifest that A and C are equimultiples of $x$ and $y$, therefore the ratio of A to C is equal to the ratio of $x$ to $y$, (I.) and as $B\; =\; q\; x$, and $D\; =\; q\; y$, B and D are in like manner equimultiples of $x$ and $y$; therefore the ratio of B to D is equal to the ratio of $x$ to $y$; therefore the ratio of A to C is equal to the ratio of B to D.

COR. If the first of four proportionals be greater than the third, the second is greater than the fourth; and if the first be less than the third, the second is less than the fourth.

THEOREM III.

If four quantities be proportionals, they are also proportionals by inversion.

LET $A\; :\; B\; ::\; C\; :\; D$; then shall $B\; :\; A\; ::\; D\; :\; C$.

For let the equal ratios of A to B, and of C to D, be the same as the ratio of the number $p$ to the number $q$, then as B will contain $q$ such equal parts as A contains

tains $p$ (Ax. 4.), $B$ will be to $A$ as $q$ is to $p$, and as $D$ will contain $q$ such equal parts as $C$ contains $p$, $D$ will be to $C$ also as $q$ to $p$, therefore the ratio of $B$ to $A$ is equal to the ratio of $D$ to $C$ (Ax. 3.).

If four quantities be proportionals, they are also proportionals by composition, and by division.

LET $A\; :\; B\; ::\; C\; :\; D$, then will

$A+B\; :\; A\; ::\; C+D\; :\; C$, and $A+B\; :\; B\; ::\; C+D\; :\; D$; also $A-B\; :\; A\; ::\; C-D\; :\; C$, and $A-B\; :\; B\; ::\; C-D\; :\; D$.

Let us suppose, as in the two preceding theorems, that the ratios of $A$ to $B$, and of $C$ to $D$, are each equal to the ratio of the number $p$ to the number $q$, so that $A$ contains $p$ such equal parts as $B$ contains $q$, and $C$ contains $p$ such equal parts as $D$ contains $q$; and let $x$ as before denote each of the equal parts contained in $A$ and $B$, and $y$ each of the equal parts contained in $C$ and $D$; then, since

Now as $A+B$ contains $x$ $p+q$ times, and $A$ contains the same quantity $p$ times, and $B$ contains it $q$ times, (by the 4th axiom),

$A+B\; :\; A\; ::\; p+q\; :\; p$, and $A+B\; :\; B\; ::\; p+q\; :\; q$, and as $C+D$ contains $y$ $p+q$ times, and $C$ contains it $p$ times, and $D$ contains it $q$ times,

Thus it appears that the ratios of $A+B$ to $A$, and of $C+D$ to $C$, are equal to the same ratio, namely, that of $p+q$ to $p$; therefore (Ax. 3.) $A+B\; :\; A\; ::\; C+D\; :\; C$. It also appears that the ratios of $A+B$ to $B$, and $C+D$ to $D$ are each equal to the ratio of $p+q$ to $q$; therefore (Ax. 3.) $A+B\; :\; B\; ::\; C+D\; :\; D$.

In the same manner the second part of the theorem may be proved, namely, that $A-B\; :\; A\; ::\; C-D\; :\; C$ and $A-B\; :\; B\; ::\; C-D\; :\; D$.

If four quantities be proportionals, and there be taken any equimultiples of the antecedents, and also any equimultiples of the consequents; the resulting quantities will still be proportionals.

LET $A\; :\; B\; ::\; C\; :\; D$, and $m\; A,\; m\; C$ be any equimultiples of the antecedents, and $n\; B,\; n\; D$ any equimultiples of the consequents; then $m\; A\; :\; n\; B\; ::\; m\; C\; :\; n\; D$.

The quantities $p,\; q,\; x$ and $y$ being supposed to express the same things as in the foregoing theorems; because

therefore, multiplying the antecedents by the number $m$, and the consequents by $n$,

and hence the ratio of $m\; A$ to $n\; B$ is equal to the ratio of the number $mp$ to the number $nq$, (Ax. 4.) and the ratio of $m\; C$ to $n\; D$, is equal to the same ratio of $mp$ to $nq$, therefore (Ax. 3.) $m\; A\; :\; n\; B\; ::\; m\; C\; :\; n\; D$.

If there be any number of quantities, and as many others, which, taken two and two, have the same ratio; the first shall have to the last of the first series the same ratio which the first of the other series has to the last.

FIRST, let there be three quantities $A,\; B,\; C$, and other three $H,\; K,\; L$, and let $A\; :\; B\; ::\; H\; :\; K$, and $B\; :\; C\; ::\; K\; :\; L$, then will $A\; :\; C\; ::\; H\; :\; L$.

For let the equal ratios of $A$ to $B$, and of $H$ to $K$, be the same with the ratio of a number $p$ to another number $q$, so that $x$ and $y$ being like parts of $A$ and $H$, and also like parts of $B$ and $K$, as in the former theorems,

Also let $C$ contain $q$ equal parts, each equal to $v$, and let $L$ contain $q$ equal parts, each equal to $z$, so that

then because $B\; :\; C\; ::\; K\; :\; L$, that is, $qx\; :\; qv\; ::\; qy\; :\; qz$, and $qx$ and $qy$ are equimultiples of $x$ and $y$, also $qv$ and $qz$ are equimultiples of $v$ and $z$, therefore (1. & Ax. 3.) $x\; :\; v\; ::\; y\; :\; z$, hence (by last theorem) $px\; :\; qv\; ::\; py\; :\; qz$, that is, (because $A\; =\; px,\; C\; =\; qv,\; H\; =\; py,\; L\; =\; qz$) $A\; :\; C\; ::\; H\; :\; L$.

Next, let these 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 will $A\; :\; D\; ::\; H\; :\; M$.

For, because $A\; :\; B\; ::\; H\; :\; K$, and $B\; :\; C\; ::\; K\; :\; L$; therefore, by the first case, $A\; :\; C\; ::\; H\; :\; L$; and because $C\; :\; D\; ::\; L\; :\; M$, therefore by the same case, $A\; :\; D\; ::\; H\; :\; M$. The demonstration applies in the same manner to any number of quantities.

COR. Hence it appears, that ratios compounded of the same number of like or equal ratios are equal to one another.

Note.—When four quantities are proportionals in the manner explained in this theorem, they are said to be so from equality of distance; and it is usual for mathematical writers to say that they are so, ex aequali or ex aequo.

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 shall have to the last of the first series the same ratio which the first has to the last of the other series.

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

For let the equal ratios of $A$ to $B$, and of $K$ to $L$ be equal to the ratio of the number $p$ to the number $q$, so that as before.

Also, let $C$ be supposed to contain $q$ equal parts, each equal to $x$, and let $H$ contain $p$ equal parts, each equal to $y$, so that

Then, because $B\; :\; C\; ::\; H\; :\; K$, that is, $qx\; :\; qx\; ::\; py\; :\; py$; therefore (1. & Ax. 3.) $x\; :\; x\; ::\; y\; :\; y$, and consequently (5.) $px\; :\; qx\; ::\; py\; :\; qy$, that is (because $px\; =\; A,\; qx\; =\; C,\; py\; =\; H,\; qy\; =\; L$) $A\; :\; C\; ::\; H\; :\; L$.

Next, let there be four quantities $A,\; B,\; C,\; D$, and other four $H,\; K,\; L,\; M$, such, that $A\; :\; B\; ::\; C\; :\; D$, and $B\; :\; C\; ::\; H\; :\; K$, and $C\; :\; D\; ::\; H\; :\; K$, then will $A\; :\; D\; ::\; H\; :\; M$; for because $A\; :\; B\; ::\; L\; :\; M$, and $B\; :\; C\; ::\; K\; :\; L$; by the foregoing case $A\; :\; C\; ::\; K\; :\; M$; and again because $C\; :\; D\; ::\; H\; :\; K$; therefore, by same case, $A\; :\; D\; ::\; H\; :\; M$. The demonstration applies in the same manner to any number of quantities.

Note.—In this theorem, as in the last, the four quantities $A,\; D,\; H,\; M$, are said to be proportionals from equality of distance; but because in this case the proportions are taken in a cross order, it is common to say, that they are so, ex aequali, in proportione perturbata, or ex aequo inversely.

THEOREM VIII.

If to the two consequents of four proportionals there be added any two quantities that have the same ratio to the respective antecedents, these sums and the antecedents will still be proportionals.

(where $B\text{'}$ and $D\text{'}$ denote two quantities distinct from those denoted by $B$ and $D$); then will

For since $A\; :\; B\; ::\; C\; :\; D$, by inversion, (3.) $B\; :\; A\; ::\; D\; :\; C$, but $A\; :\; B\text{'}\; ::\; C\; :\; D\text{'}$, therefore (6.) $B\; :\; B\text{'}\; ::\; D\; :\; D\text{'}$, and by composition, (4.) and inversion $B\; :\; B\; +\; B\text{'}\; ::\; D\; :\; D\; +\; D\text{'}$, and since $A\; :\; B\; ::\; C\; :\; D$; therefore (6.) $A\; :\; B\; +\; B\text{'}\; ::\; C\; :\; D\; +\; D\text{'}$.

COR. 1. If instead of two quantities $B\text{'},\; D\text{'}$, there be any number and which ta-

ken two and two have the same ratio to the antecedents $A,\; C$, that is, if

then will $A\; :\; B\; +\; B\text{'}\; +\; B\text{'}\text{'}\; ::\; C\; :\; D\; +\; D\text{'}\; +\; D\text{'}\text{'}$.

For since $A\; :\; B\; +\; B\text{'}\; ::\; C\; :\; D\; +\; D\text{'}$ (by the theor.) and $A\; :\; B\text{'}\text{'}\; ::\; C\; :\; D\text{'}\text{'}$,

therefore, by the proposition,

COR. 2. If any number of quantities of the same kind be proportionals, as one of the antecedents is to its consequent, so is the sum of all the antecedents to the sum of all the consequents.

then because $A\; :\; A\; ::\; B\; :\; B$,
and $A\; :\; C\; ::\; B\; :\; D$,
and $A\; :\; E\; ::\; B\; :\; F$;

therefore, $A\; :\; A\; +\; C\; +\; E\; ::\; B\; :\; B\; +\; D\; +\; F$; and by alternation,

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 $q\; =\; B\; -\; d$; 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.

1. 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

Proportions 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 all equal.

II. Two figures are similar, when the angles of the one are equal to the angles of the other, each to each; and the homologous sides proportional. The homologous sides are those which have the same position in the two figures; or which are adjacent to the equal angles. The angles themselves are called homologous angles.

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

III. In two different circles, similar sectors, similar arcs, similar segments, are such as correspond to equal angles at the centre. Thus the angle A being equal to the angle O, the arc BC is similar to the arc 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.

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. 1.). Now if from the quadrilateral ABED, the triangle ADF be taken away, there will remain the parallelogram ABEF; and if from the same quadrilateral ABED, the triangle CBE, equal to the former, be taken away, there will remain the parallelogram ABCD; therefore the two parallelograms ABCD, ABEF, which have the same base, and the same altitude, are equivalent.

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

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.

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. 4. 3.), that is, as the base AB to AE.

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 AF contains $q$; then shall ABCD : AEFG :: $m\; n\; :\; p\; q$.

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.)

and it was shewn that

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 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 $3\; \backslash times\; 10$ or 30; this number considered by itself has no meaning,

meaning, but if we have a second rectangle B, the base of which is twelve units, and altitude seven, this second rectangle shall be represented by the number $12\; \backslash times\; 7$ 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 $\backslash frac\{84\}\{30\}$, that is, it shall be $\backslash frac\{7\}\{5\}$ 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 30 superficial units or 30 squares, each having its side equal to an unit. To illustrate this, see fig. 72.

The area of any parallelogram is equal to the product of its base by its altitude.

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

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

The area of a triangle is equal to the product of its base by the half of its altitude.

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 $BC\; \backslash times\; AO$ (5.), therefore that of the triangle is $\backslash frac\{1\}\{2\}\; BC\; \backslash times\; AO$, or $BC\; \backslash times\; \backslash frac\{1\}\{2\}\; AO$.

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.

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

$AL\; \backslash times\; EF$. But $AL\; =\; DK$, and because the triangle IBL is equal to the triangle ICK, 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 $FE\; \backslash times\; AL$, it is also equal to $FE\; \backslash times\; \backslash left(\; \backslash frac\{AB\; +\; CD\}\{2\}\; \backslash right)$.

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 proportionals.

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

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

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

Next suppose that the rectangle ACGD = ABFE; then $ABHD\; :\; ACGD\; ::\; ABHD\; :\; ABFE$, (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 proportionals, the rectangle contained by the extremities is equal to the square of the mean; and if the rectangle contained by the extremes be equal to the square of the mean, the three straight lines are proportionals.

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

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

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

now (3.) $\backslash text\{rect.\; BP\}\; :\; \backslash text\{rect.\; BM\}\; ::\; BF\; :\; BG$, and $\backslash text\{rect.\; DN\}\; :\; \backslash text\{rect.\; DQ\}\; ::\; DH\; :\; DI$; but $BF\; :\; BG\; ::\; DH\; :\; DI$, (by hypoth.) therefore,

but

Proportions but it has been shewn that
of Figures.

rect. AF : rect. BP :: rect. CH : rect. DN,

therefore (6. 3.)

rect. AF : rect. BM :: rect. GH : rect. DQ.

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

THEOREM X.

Fig. 74. 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 those two parts: which may be expressed thus, $AC^2\; =\; AB^2\; +\; BC^2\; +\; 2AB\; \backslash times\; BC$.

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 IDGH is the square upon BC, for AC=AE, and AB=AF, therefore AC-AB=AE-AF, that is BC=EF; but BC=IG, and EF=DG, (26. 1.) therefore IGDH is the square upon BC, and the remaining two parts are the two rectangles BCGI, FEHI, which have each for their measure AB×BC, therefore the square upon AC is equal to the squares upon AB and BC, and twice the rectangle AB×BC.

THEOREM XI.

Fig. 75. 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, $AC^2\; =\; AB^2\; +\; BC^2\; -\; 2AB\; \backslash times\; BC$.

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×BC for their measure; and if these be taken from the whole figure ABILKEA, that is from AB²+BC², there will remain the square ACDE, that is, the square upon AC.

THEOREM XII.

Fig. 76. 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, $(AB+BC)\; \backslash times\; (AB-BC)\; =\; AB^2\; -\; BC^2$.

CONSTRUCT upon AB and AC the squares ABIF, ACDE; produce AB, so that BK=BC, and complete the rectangle AKLE. The base AK of the rectangle is the sum of the two lines AB, BC; and its altitude AE is the difference of the same lines; therefore, the rectangle AKLE=(AB+BC)(AB-BC); but the same rectangle is composed of two parts ABHE+BHLK, of which, BHLK is equal to the rectangle EDGF,

for BH=DE, and BK=FE; therefore, AKLE=ABHE+EDGF; but these two parts constitute the excess of the square ABIF above the square DHIG, the former of which is the square upon AB, and the latter the square upon BC, therefore $(AB+BC)\; \backslash times\; (AB-BC)\; =\; AB^2\; -\; BC^2$.

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 is 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×CD; that is,

FIRST. Suppose AD to fall within the triangle, then BD=BC-CD, and consequently (11.) BD²=BC²+CD²-2BC×CD; to each of these equals add AD²; then, observing that BD²+DA²=BA², and CD²+DA²=CA²,

Next, suppose AD to fall without the triangle, so that BD=CD-BC, and therefore BD²=CD²+BC²-2BC×CD, (11.) to each of these add AD² as before,

before, and we get

In a triangle ABC, if the angle C is obtuse, the square of the opposite side AB is greater than the sum of the squares of the sides which contain the angle C; and if AD a perpendicular be drawn to BC from the opposite angle, the difference shall be equal to twice the rectangle $BC\; \backslash times\; CD$, that is,

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

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 squares 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.

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

therefore, by adding equals to equals, and observing that $BE\; =\; CE$, and therefore $BE^2\; =\; CE^2$, and $2BE\; \backslash times\; ED\; =\; 2CE\; \backslash times\; ED$,

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.)

COR. Hence by composition $AB\; :\; AD\; ::\; AC\; :\; AE$, and $AB\; :\; BD\; ::\; AC\; :\; CE$.

Conversely, if two of the sides AB, AC of a triangle are 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, $AE\; :\; BD\; ::\; AC\; :\; CE$, 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.

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

THROUGH the point C draw CE parallel to AD, so as to meet BA produced. In the triangle BCE, the line AD is parallel to one of its sides CE, therefore $BD\; :\; DC\; ::\; BA\; :\; AE$; now the triangle 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 $BD\; :\; DC\; ::\; BA\; :\; AC$.

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 $BC\; :$

Proportions of Figures. $BC\; :\; CE\; ::\; BA\; :\; AF$; or since $AF\; =\; CD$, $BC\; :\; CE\; ::\; BA\; :\; CD$. Again, in the same triangle, because $CD$ is parallel to the side $BF$, $BC\; :\; CE\; ::\; FD\; :\; DE$, or, since $FD\; =\; AC$, $BC\; :\; CE\; ::\; AC\; :\; DE$; having now shewn that $BC\; :\; CE\; ::\; BA\; :\; CD$, and that $BC\; :\; CE\; ::\; AC\; :\; DE$, it follows that $BA\; :\; CD\; ::\; AC\; :\; DE$; therefore the equiangular triangles $BAC$, $CDE$ have their homologous sides proportional, and hence (def. 2.) the triangles are similar.

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

Fig. 33. 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 hypothesis $BC\; :\; EF\; ::\; AB\; :\; DE$, therefore $GE\; =\; DE$ (Ax. 2. 3.). In like manner, because by the same theorem $BC\; :\; EF\; ::\; CA\; :\; FG$; and by hypothesis $BC\; :\; EF\; ::\; CA\; :\; FD$; therefore $FG\; =\; FD$; but it was shewn that $EG\; =\; ED$, therefore, the triangles $GEF$, $DEF$, having the sides of the one equal to those of the other, each to each, are equal, but, by construction, the triangle $GEF$ is equiangular to $ABC$ therefore also the triangles $DEF$, $ABC$ are equiangular and similar.

Fig. 35. Two triangles which have an angle of the one equal to an angle of the other, and the sides about these angles proportional, are similar.

LET the angle $A\; =\; D$, and let $AB\; :\; DE\; ::\; AC\; :\; DF$, 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 $ABC$.

Fig. 36. In a right-angled triangle, if a perpendicular $AD$ be drawn from the right angle upon the hypotenuse, then,
1. The triangles $ABD$, $CAD$ on each side of the perpendicular are similar to the whole triangle $BAC$, and to one another.
2. Each side $AB$ or $AC$ is a mean proportional between the hypotenuse $BC$, and the adjacent segment $BD$ or $DC$.

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

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

2. Because the triangle $BAD$ is similar to the triangle $BAC$, their homologous sides are proportional. Now the side $BD$ of the lesser triangle is homologous to the side $BA$ of the greater, because they are opposite to the equal angles $BAD$, $BCA$; in like manner $BA$, considered as a side of the lesser triangle, is homologous to the side $BC$ of the greater, each being opposite to a right angle; therefore, $BD\; :\; BA\; ::\; BA\; :\; BC$. In the same manner it may be shewn 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 similar triangles $ABD$, $ACD$, it appears that $BD\; :\; DA\; ::\; DA\; :\; DC$; therefore the perpendicular is a mean proportional between the segments of the hypotenuse.

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 contain the equal angles; that is, the triangle $ABC$ is to the triangle $ADE$, as the rectangle $AB\; \backslash times\; AC$ to the rectangle $AD\; \backslash times\; AE$.

JOIN $BE$; because the triangles $ABE$, $ADE$ have a common vertex $E$, they have the same altitude, therefore $ABE\; :\; ADE\; ::\; AB\; :\; AD$, (Cor. to 6.); but $AB\; :\; AD\; ::\; AB\; \backslash times\; AE\; :\; AD\; \backslash times\; AE$, (3.) therefore,

In the same manner it may be demonstrated that

Therefore (6. 3.) $ABC\; :\; ADE\; ::\; AB\; \backslash times\; AC\; :\; AD\; \backslash times\; AE$.

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

What has been proved of triangles is also true of parallelograms, they being the doubles of such triangles.

Two similar triangles are to each other as the squares of their homologous sides.

Proportions LET the angle $A=D$, the angle $B=E$, and therefore the angle $C=F$,

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

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 similarly situated.

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

Because the polygons are similar, the angle $ABC$ is equal to its homologous angle $FGH$ (Def. 2.) also the sides $AB$, $BC$ are proportional to $FG$, $GH$, so that $AB\; :\; FG\; ::\; BC\; :\; GH$, therefore the triangles $ABC$, $FGH$ are similar (22.); therefore the angle $BCA\; =\; GHF$, and these being taken from the equal angles $BCD$, $GHI$, the remainders $ACD$, $FHI$ are equal; but the triangles $ACD$, $FHI$ being similar, $AC\; :\; FH\; ::\; CD\; :\; HI$, besides, because of the similarity of the polygons, $BC\; :\; GH\; ::\; CD\; :\; HI$; therefore $AC\; :\; FH\; ::\; CD\; :\; HI$; now it has been already shewn 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.

For, since by the nature of similar figures $AB\; :\; FG\; ::\; BC\; :\; GH\; ::\; CD\; :\; HI$, &c. therefore, (2. cor. 8. 3.) $AB\; +\; BC\; +\; CD$, &c. the perimeter of the first figure, is to $FG\; +\; GH\; +\; HI$, &c. the perimeter of the second, as the side $AB$ to its homologous side $FG$.

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

By the same manner of reasoning,

and so on if there be more triangles; hence, from this series of equal ratios, it follows (2. cor. 8. 3.) that of Figures. $ABC\; +\; ACD\; +\; ADE$, 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. 11. 3.) $AB$ has to $L$ the duplicate ratio of $AB$ to $FG$; but $AB\; :\; L\; ::\; AB^2\; :\; AB\; \backslash times\; L$ (3.), or, since $AB\; \backslash times\; L\; =\; FG^2$, (cor. to 8.) $AB\; :\; L\; ::\; AB^2\; :\; FG^2\; ::\; ABCDE\; :\; FGHIK$, therefore the figure $ABCDE$ has to the figure $FGHIK$, the duplicate ratio of $AB$ to $FG$.

THEOREM XXVIII.

The segments of two chords $AB$, $CD$, which cut each other within a circle, are reciprocally proportional, that is $AO\; :\; DO\; ::\; CO\; :\; OB$. Fig. 88.

JOIN $AC$ and $BD$; and because the triangle $AOC$, $BOD$ 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 $AO\; \backslash times\; BO\; =\; CO\; \backslash times\; DO$, (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 $OB\; :\; OC\; ::\; OD\; :\; OA$. Fig. 90.

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.) $OA\; \backslash times\; OB\; =\; OC\; \backslash times\; OD$, 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

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 $ACO$ is equal to $DAO$ (18. 2.), therefore the triangles are similar (20.), and consequently $CO\; :\; OA\; ::\; OA\; :\; OD$.

COR. Therefore (cor. to 8.) $CO\; \backslash times\; OD\; =\; OA^2$, 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.

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 $A\; g,\; g\; h,\; h\; B$, the number of which is $p$, the arch $DF$ shall contain $q$ equal parts $D\; k,\; k\; l,\; l\; m,\; m\; n,\; n\; F$, 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 centres, observing, however, that the radii of the circles must be equal.

To bisect a given straight line $AB$; that is, to divide it into two equal parts.

FROM the points $A$ and $C$ as centres, with any radius greater than the half of $AB$, describe arches, cutting each other in $D$ and $D\text{'}$ on each side of the line $AB$. Draw a straight line through the points $D,\; D\text{'}$, cutting $AB$ in $C$; the line $AB$ is bisected in $C$.

For the points $D,\; D\text{'}$, 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 $DCD\text{'}$ is that perpendicular, and consequently $C$ is the middle of $AB$.

To draw a perpendicular to a given straight line $BC$, from a given point $A$ in that line.

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

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 $E$; draw a straight line through the points $A$ and $E$, meeting $BD$ in $C$; the line $AC$ is the perpendicular required.

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

At a given point $A$, in a given line $AB$, to make an angle equal to a given angle $K$.

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

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

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

Problems. D as before, and the line CD will manifestly bisect the angle C, as required.

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.

Fig. 5. 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.).

Fig. 6. 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.

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

Fig. 7. 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.

To find the centre of a given circle, or of a circle of which an arch is given. Fig. 101.

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. 2.

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 ABC.

To draw a tangent to a given circle through a given point A. Fig. 102, 103.

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

To inscribe a circle in a given triangle ABC. Fig. 104.

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 angles ODB = OEB, and the angle OBD = OBE; therefore the remaining angles BOD, BOE, are equal; and as the side DB 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.

Upon a given straight line AB, to describe a segment. Fig. 105.

ment 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.).

Fig. 106. To divide a straight line, AB, into any proposed
Fig. 107. 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.

Fig. 108. To find a fourth proportional to three given lines,
A, B, C.

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

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

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

Fig. 109. 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.).

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 $BF\; :\; AF\; ::\; AF\; :\; AB$.

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

To make a square equivalent to a given parallelo-
gram or to a given triangle.

FIRST, Let ABCD be a given parallelogram, (fig. 112.) 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, $XY^2\; =\; AB\; \backslash times\; DE$ (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 $\backslash frac\{1\}\{2\}BC\; :\; XY\; ::\; XY\; :\; AD$; therefore (8. 4.) $XY^2\; =\; \backslash frac\{1\}\{2\}BC\; \backslash times\; AD\; =\; \backslash text\{triangle\; \}\; ABC$ (6. 4.).

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.) $EF\; \backslash times\; EX\; =\; AB\; \backslash times\; AD$, that is, the rectangle EFGX is equal to the rectangle ABCD.

To make a triangle equivalent to a given polygon ABCDE.

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

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

In like manner, if the diagonal CA be drawn into 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.

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

Upon a given line FC to construct a polygon similar to a given polygon ABCDE.

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 FHI, 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 FGHKI similar to ABCDE (26. 4.).

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.

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 chord 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.

As the form of reasoning by which it has been shewn that an equilateral hexagon inscribed in a circle is also equiangular, will apply alike to any equilateral polygon: it may be inferred that every equilateral polygon inscribed in a circle is also equiangular.

To inscribe a regular pentagon in a given circle. Fig. 117.

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 shewn that the sides about the 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 arc 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.).

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

angles $TOH=NOH$; 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 the chord $AB$ (6. 2.), therefore $GH$ is parallel to $AB$ (9. 2. and 18. 1.), and $HI$ to $BC$, therefore the angle $GHO=ABO$, and $IHO=CBO$, and hence $GHI=ABC$; and in like manner it appears, that $HIK=BCD$, &c, therefore the angles of the circumscribed polygon are equal to those of the inscribed polygon. And because of the parallels, $GH:AB::OH:OB$, and $HI:BC::OH:OB$, therefore $GH:AB::HI:BC$; but $AB=BC$; therefore $GH=HI$. For the same reason $HI=IK$, &c, therefore, the polygon is regular, and similar to the inscribed polygon.

Fig. 120. 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 $ABC$.

Fig. 125. COR. Hence the perimeter of any polygon described about a circle, is greater than the circumference of the circle.

Fig. 119. Equilateral polygons, $ABCDEF$, $GHIKLM$, of the same number of sides inscribed in circles are similar, and are to one another as the squares of the radii of the circles.

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

Fig. 121. 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 cir-

cle, 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, $ODH:OAK::OH^2:OK^2$ (25. 4.), and on account of the similar triangles $OAK$, $LAB$, $OA^2$ or $OH^2:OK^2::LA^2:LB^2$ (20. and 9. 4.); therefore, $P:p::LA^2:LB^2$, and by division and inversion, $P:P-p::LA^2:LA^2-LB^2$, or $AB^2$; but $LA^2$, 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 $LA^2\; >\; P$, and as it has been proved that $P:P-p::LA^2:AB^2$, of which proportion, the first term $P$ is less than the third $LA^2$; therefore (2. 3.) the second $P-p$ is less than the fourth $AB^2$, but , therefore .

COR. I. 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,

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.

The area of any circle is equal to a rectangle contained by the radius, and a straight line equal to half the circumference.

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 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 $n\; \backslash times\; \backslash frac\{1\}\{2\}DE\; =\; \backslash frac\{1\}\{2\}P$, $n\; \backslash times\; \backslash frac\{1\}\{2\}DE\; \backslash times\; OH\; =\; \backslash frac\{1\}\{2\}P\; \backslash times\; OH$, but $n\; \backslash times\; \backslash frac\{1\}\{2\}DE\; \backslash times\; OH\; =\; n\; \backslash times\; \backslash text\{triangle\; \}\; DOE\; =\; \backslash text\{polygon\; \}\; DEF$, &c. therefore, $\backslash frac\{1\}\{2\}P\; \backslash times\; OH\; =\; \backslash text\{polygon\; \}\; DEF$, &c.; and in like manner it appears, that $\backslash frac\{1\}\{2\}p\; \backslash times\; OK\; =\; \backslash text\{polygon\; \}\; ABC$, &c. Now let $Q$ denote the circumference of the circle, then, because $\backslash frac\{1\}\{2\}Q\; =\; \backslash frac\{1\}\{2\}p$, and $OH\; =\; OK$, therefore $\backslash frac\{1\}\{2\}Q\; \backslash times\; OH\; =\; \backslash frac\{1\}\{2\}p\; \backslash times\; OK$, that is $\backslash frac\{1\}\{2\}Q\; \backslash times\; OH$ is greater than the inscribed polygon. Again, because $\backslash frac\{1\}\{2\}Q\; \backslash leq\; \backslash frac\{1\}\{2\}P$ (axiom), therefore $\backslash frac\{1\}\{2\}Q\; \backslash times\; OH\; \backslash leq\; \backslash frac\{1\}\{2\}P\; \backslash times\; OH$, that is, $\backslash frac\{1\}\{2\}Q\; \backslash times\; OH$ is less than the circumscribed polygon: Thus it appears that $\backslash frac\{1\}\{2\}Q\; \backslash times\; OH$ is greater than any polygon inscribed in the circle, but less than any polygon described about it; therefore, $\backslash frac\{1\}\{2\}Q\; \backslash times\; OH$ is equal to the circle (2.).

The areas of circles are to one another as the squares of their radii.

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

COR. 1. The circumferences of circles are to one another as their radii. Put $M$ for half the circumference of the circle ABE, and $N$ for half the circumference of GKL; then, circle ABE : circle GHL :: $AO^2\; :\; GP^2$; but $\backslash frac\{1\}\{2\}M\; \backslash times\; AO\; =\; \backslash text\{circle\; \}\; ABE$, also $\backslash frac\{1\}\{2\}N\; \backslash times\; GP\; =\; \backslash text\{circle\; \}\; GHL$, (3.) therefore $\backslash frac\{1\}\{2\}M\; \backslash times\; AO\; :\; \backslash frac\{1\}\{2\}N\; \backslash times\; GP\; ::\; AO^2\; :\; GP^2$, and by alternation $\backslash frac\{1\}\{2\}M\; \backslash times\; AO\; :\; AO^2\; ::\; \backslash frac\{1\}\{2\}N\; \backslash times\; GP\; :\; GP^2$, therefore (3. 4.) $\backslash frac\{1\}\{2\}M\; :\; AO\; ::\; \backslash frac\{1\}\{2\}N\; :\; GP$, and again by alternation $\backslash frac\{1\}\{2\}M\; :\; \backslash frac\{1\}\{2\}N\; ::\; AO\; :\; GP$, therefore $M\; :\; N\; ::\; AO\; :\; GP$.

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

because $A\; :\; H\; ::\; a^2\; :\; h^2$
and $B\; :\; H\; ::\; b^2\; :\; h^2$,
therefore $A\; +\; B\; :\; H\; ::\; a^2\; +\; b^2\; :\; h^2$ (8. 3.)
but $a^2\; +\; b^2\; =\; h^2$ (13. 4.), therefore $A\; +\; B\; =\; H$.

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. Fig. 111.

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 2 PM 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 polygon of which AM is a side, and $a$ 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 $a\; =\; \backslash sqrt\{A\; \backslash times\; B\}$.

II. The triangles CPM, CPE, having the same altitude CM, are to one another as PM to PE. But as CP bisects the angle MCE, $PM\; :\; PE\; ::\; CM\; :\; CE$ (19. 4.) :: $CD\; :\; CA\; ::\; A\; :\; a$; therefore, $CPM\; :\; CPE\; ::\; A\; :\; a$; and consequently $CPM\; +\; CPE$, or $CME\; :\; CPM\; ::\; A\; +\; a\; :\; A$, and $CME\; :\; 2\; CPM\; ::\; A\; +\; a\; :\; 2\; A$; but CME and 2 CPM, or CMPA, are to one another as the polygons B and $b$, of which they are like parts; therefore, $A\; +\; a\; :\; 2\; A\; ::\; B\; :\; b$. Now the polygon $a$ has been already found, therefore by this last proportion the polygon $b$ is determined; that is,

To find nearly the ratio of the circumference of a circle to its diameter.

LET the radius of the circle $=\; r$, then, the sides of the inscribed square being the hypotenuse of a right-angled triangle of which the radii are the sides, (see fig.)

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 formulae

the area of the inscribed octagon, and $b\; =\; 3.3137085$, &c. the area of the circumscribed octagon. By substituting these numbers in the formulae, 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:

Hence it appears that the areas of a regular polygon of 32768 sides inscribed in the circle, and of a simi-

lar polygon described about it, differ so little from each other, that the numbers which express them are the same as far as the eighth decimal place. And as the circle is greater than the one polygon, and less than the other, its area will be nearly 3.1415926. But the area is the product of the radius and the half of the circumference; therefore, the radius being unity, half the circumference is 3.1415926 nearly; and the radius is to half the circumference, or the diameter is to the circumference, nearly as 1 to 3.1415926.

In this way the ratio of the diameter to the circumference may be found to any degree of accuracy; but neither by this, nor any other method yet known, can the ratio be exactly determined.

ARCHIMEDES, by means of inscribed and circumscribed polygons of 96 sides, found that the diameter is to the circumference as 7 to 22, nearly, which ratio is nearer to the truth than can be expressed by any smaller numbers; and METIUS found it to be more nearly as 113 to 355. Both of these expressions are convenient on account of the smallness of the numbers, but later mathematicians have carried the approximation to a much greater degree of accuracy. Thus, it has been found that the diameter being 1, the circumference is greater than 3.1415926535897932, but less than the same number having its last figure increased by unity; and some have even had the patience to carry the approximation as far as the 150th place of decimals.

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

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

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

IV. It will be demonstrated (Theor. 3.) that the 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 one perpendicular being drawn in the one plane, and the other in the other plane.

This angle may be either acute or obtuse.

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

VI. A solid angle is that which is made by the meeting of more than two plane angles, which are not in the same plane, in one point. Thus the solid angle $S$ is formed by the plane angles $ASB$, $BSC$, $CSD$, $DSA$.

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

FOR from the definition of a plane (7. def. 1.) it is manifest that if a straight line coincide with a plane in two points it must be wholly in the plane.

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.

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 $AB$,

(7. def. 1.) and the same line must also be in the same plane CD, therefore it must be the common section of them both.

If a straight line AP is 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,

and in the like manner, in the triangle PBC,

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

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.) and consequently AP is perpendicular to the plane MN (Def. 1.).

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.

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. 4. 1.) and because AP is perpendicular to the plane MN, so that APB, APC are right angles, AB=AC, (cor. 5. 1.) therefore ABC is an isosceles triangle; and since its base BC is bisected at D, BC is perpendicular to AD; (schol. 11. 1.) 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. 1.) 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 are 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.

Two planes MN, PQ, perpendicular to the same straight line AB, are parallel to each other. Fig. 127.

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.

The intersections EF, GH of two parallel planes MN, PQ with a third plane FG, are parallel. Fig. 128.

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.

Any straight line AB, perpendicular to MN, one of two parallel planes MN, PQ, is also perpendicular to PQ the other plane. Fig. 127.

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. 1.) hence (Def. 1.) the line AB is perpendicular to the plane PQ.

Parallel straight lines EG, FH, comprehended between two parallel planes MN, PQ, are equal. Fig. 128.

LET a plane pass through the lines EG, FH, and meet

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 EF=GH.

COR. Hence two parallel planes are everywhere at the same distance from each other. For, if EF and GH are perpendicular to the two planes, they are parallel, (1. cor. 5.) therefore they are equal.

THEOREM X.

Fig. 125. If two straight lines CA, EA, meeting one another, be parallel to two other lines DB, FB, that meet one another, though not in the same plane with the first two; the first two and the other two shall contain equal angles, and the plane passing through the first two shall be parallel to the plane passing through the other two.

TAKE AC=BD, AE=BF, and join CE, DF, AB, CD, EF. Because AC is equal and parallel to BD, the figure 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 (2 cor. 5. and 28. 1.); therefore the triangles CAE, DBF are equal, (10. 1.) hence the angle CAE=DBF.

In the second 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 shewn to be equal; therefore, CD=GD, and FH=EF, which is absurd, therefore the plane ACE is parallel to BDF.

THEOREM XI.

Fig. 130. 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 XII.

Fig. 130. If the plane AB is perpendicular to the plane MN; and 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.

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, 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 XIII.

If two planes AB, AD are perpendicular to a third, Fig. their common intersection AP is perpendicular to the third plane.

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 XIV.

If two straight lines be cut by parallel planes, they Fig. 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, are parallel (7.), and in like manner it appears, that AC, GF are parallel; therefore AE : EB (:: AG : GD) :: CF : FD.

THEOREM XV.

If a solid angle be contained by three plane an- Fig. gles, 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.), therefore BE=BC; now AB < AC + BC, therefore, taking BD from the first of these unequal quantities, and BC from the second, we get AD < AC; and as the triangles ASD, ASC have SD=SC, and SA common to both, and AD < AC, therefore (9. 1.) the angle ASD < ASC; and, adding DSB to the one, and CSB to the other, ASB < ASC + BSC.

THEOREM XVI.

If each of two solid angles be contained by three Fig. plane

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 shewn that $AB\; =\; DE$, and bounded by 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 besides 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.

SECT. VIII. 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: 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 $ABCDEFHIK$ 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 face 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 $A$, and terminate in a polygonal plane $BCD$.

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

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

THEOREM I.

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

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

For let the base $ABCDE$ be applied to its equal the base $abcde$, so that they may coincide with each other; then, as 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 as these angles are similarly situated, the solid angles $B$ and $b$ are equal (15. 7.) 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 shewn, that $GH$ falls upon $gh$, therefore the upper base $FGHIK$ coincides entirely with its equal $fg\; h\; i\; k$, and the two solids coincide with each other, or occupy the same space, therefore the prisms are equal.

SCHOLIUM.

A prism is entirely determined, when its base $ABCDE$ is known, and its edge $BG$ is given in magnitude and position; for if through the point $G$, $GF$ be drawn equal and parallel to $AB$, and $GH$ equal and parallel to $BC$, and the polygon $FGHIK$ be described equal to $ABCDE$ (20. 5.), it is evident that the

Of Solid points FKI will have determinate positions; therefore bearded by any two prisms constructed with the same data cannot be unequal.

Fig. 135. In any parallelopiped the opposite planes are equal and parallel.

FROM the nature of the solid (4. def.) the bases ABCD, EFGH are equal parallelograms, and their sides are parallel, therefore the planes AC, EG are parallel; and because AD is equal and parallel to BC, and AE is equal and parallel to BF, the angle DAE = CBF, and the plane DAE is parallel to the plane CBF, (10. 7.) therefore also the parallelogram DAEH is equal to the parallelogram CBFG. It may in like manner be demonstrated, that the opposite parallelograms ABFE, DCGH are equal and parallel.

COR. Hence, in a parallelopiped, any one of the six planes which contain it may be taken for its base.

Fig. 136. The plane BDHF, which passes through two parallel opposite edges BF, DH, of a parallelopiped AG, divides it into two triangular prisms ABDHEF, GHFBCD, equal to one another.

FOR the triangles ABD, EFH, having their sides equal and parallel, are equal, and the lateral faces ABFE, ADHE, BDHF are parallelograms; therefore the solid ABDHEF is a prism; for like reasons the solid GHFBCD is a prism. Again, because the plane angles which contain the solid angle at G are equal to those which contain the solid angle at A, viz. the angle FGH = DAB, FGC = DAF, and HGC = BAE, the planes in which these angles are have the same inclination to one another, (16. 7.); as, however, these angles are not disposed in the same order, but in a contrary order, the solid angles cannot be made to coincide with one another, and consequently the prisms cannot be proved equal by superposition, as in Theorem I. Their equality may however be established by reasoning thus:

The inclination of each of any two adjacent faces of a prism to the base, and the length of an edge being given, the prism is evidently restricted to one determinate magnitude; and it will evidently have the same magnitude whichever of the two sides of the base it may stand upon; that is, whether it be constructed above or below the base. Now if the upper base FGH of the one prism be applied to the lower base DAB of the other, so that the sides FG, GH, FH may be upon the sides DA, AB, DB equal to them, then the prism GHFBCD will have the position ABDHEF; and the two faces ABFE' of ADHE' of the prism below the base will have each the same inclination to it, as the equivalent faces ABFE, ADHE of the prism above the base; and the edge AE' is equal to the edge AE; therefore the conditions which determine the magnitude of both prisms are identical, and consequently the prisms are equal.

If two parallelopipeds AG, AL have a common base ABCD, and have their upper bases in the same plane, and between the same parallel straight lines EK, HL, the two parallelopipeds are equivalent to each other.

BECAUSE 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 AEIDHM be applied to the prism BFKCGI, so that their bases AEI, BFK, which are equal, may coincide with each other, then, because the solid angle E is equal to the solid angle F, the side EH shall fall upon FG, and this is all that is necessary to prove that the two prisms coincide entirely, for the base AEI and the edge EH determine the prism AEM, and the base BFK and the edge FG determine the prism BFL; therefore the prisms are equal. But 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.

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 parallel to IK, (cor. 2. 5. 7.) for a similar reason GF is parallel to LK. Let the sides EF, HG, as also the sides IK, LM, 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 parallelopipeds AL; therefore the two parallelopipeds AG, AL, which have the same base and the same altitude, are equivalent to one another.

Any parallelopiped AG is equivalent to a rectangular parallelopiped, having the same altitude, and an equivalent base.

AT the points A, B, C, D, let AI, BK, CL, DM, be drawn perpendicular to the plane ABCD, and terminating in the plane of the upper base; then, IK, KL,

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 is also a rectangle, the solid AL will be a rectangular parallelopiped equivalent to the parallelopiped AG. But if ABCD is not a rectangle, (fig. 140.) draw AO and BN perpendicular to CD, and OQ and NP perpendicular to DC, meeting MI 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. 139.) and they have the same altitude, and the base ABNO of the former is equivalent to the base ABCD of the latter.

Any section NOPQR of a prism, made by a plane parallel to its base ABCDE, is equal to the base.

For the parallels AN, BO, CP contained between the parallel planes ABC, NOP are equal (9. 7.); and thus all the figures ABON, BCPO, &c. are parallelograms; hence the side ON=AB, OP=BC, PQ=CD, &c. also, the equal sides are parallel, therefore, the angle ABC=NOP, the angle BCD=OPQ, &c. therefore the two polygons ABCDE, NOPQR, have their sides and angles equal, each to each; therefore, they are equal.

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 $p$ and $q$, so that AE will contain $p$ such equal parts as AI contains $q$. Let AE and AI be divided into $p$ and $q$ 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 $p$ solids, which will also be parallelopipeds having equal bases (7.) and equal altitudes, therefore they will be equal among themselves; and in like manner the parallelopiped AL will be divided into $q$ equal solids; and as each of the solids in AG is equal to each of the solids in AL, the parallelopiped AG will contain $p$ such equal parts as the parallelopiped AL contains $q$; therefore the parallelopiped AG will be to the parallelopiped AL as the number $p$ to the number $q$, that is, as AE the altitude of the former to AI the altitude of the latter.

Two rectangular parallelopipeds AG, AK, which have the same altitude AE, are to each other as their bases ABCB, AMNO.

LET the two solids be placed, the one by the side of the other, as represented in the figure, and let the plane ONKL be produced, so as to meet the plane DUGH

in PQ, thus forming a third parallelopiped AQ, which may be compared with each of the parallelopipeds AG, AK. The two solids AG, AQ, having the same base ADHE, are to each other as their altitudes AB, AO, (8.) and, in like manner, the two solids AQ, AK, having the same base AOLE, are to each other as their altitudes AD, AM; that is,

but $\backslash text\{AB\}\; :\; \backslash text\{AO\}\; ::\; \backslash text\{base\; AC\}\; :\; \backslash text\{base\; AP\}$ (3. 4.)
and $\backslash text\{AD\}\; :\; \backslash text\{AM\}\; ::\; \backslash text\{base\; AP\}\; :\; \backslash text\{base\; AN\}$,

therefore,

therefore (7. 3.)

Rectangular parallelopipeds are to each other as the products of the numbers proportional to their bases and altitudes, or as the products of the numbers proportional to their three dimensions.

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

and by the last theorem but one,

but, considering the bases AC, AN as measured by numbers, as also the altitudes AE, AX,

therefore,

therefore, (7. 3.)

which proportion, by substituting for the bases AC, AN their numerical values $AB\; \backslash times\; AD$ and $AO\; \backslash times\; AM$ becomes

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 for its numerical measure;

Of Solids bounded by solids are estimated. When two parallelopipeds are compared together 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 contains 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, $P\; :\; Q\; ::\; 3\; \backslash times\; 2\; :\; 4\; \backslash times\; 5\; ::\; 6\; :\; 20$.

If all the dimensions of each solid are used in comparing them together, 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 four, three, and six linear units, respectively; and those of Q, seven, two, and five of the same unit; then $P\; :\; Q\; ::\; 4\; \backslash times\; 3\; \backslash times\; 6\; :\; 7\; \backslash times\; 2\; \backslash times\; 5\; ::\; 72\; :\; 70$.

As lines are compared together 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 we have just now supposed, 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 (6.); and it has been shewn, that the solidity of such a parallelopiped is equal to the product of its base and 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 the polygon which forms its base can be divided into triangles, but the solidity of each of these 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.

COR. 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.

THEOREM XII.

Similar prisms are to one another as the cubes of their homologous sides.

LET AG, IP be two similar prisms, of which AB, IK are two homologous sides, the prism AG is to the prism IP as the cube of AB to the cube of IK. Let E and N be two homologous angles of the prisms, and ES, NV perpendiculars to the planes of their bases; join IV; take IR=AE, and in the plane INV draw RT perpendicular to IV; then RT shall be perpendicular to the plane IL (11. and 12. of 7.), also RT shall be equal to ES; for if the solid angles A and I were applied the one to the other, the planes which contain them would coincide (schol. 16. 7.), and the point E would fall upon the point R, and therefore the perpendicular ES would coincide with the perpendicular RT (2. cor. 4. 7.). Now the content of a prism being the product of its base by its altitude (11.), it follows that $\backslash text\{prism\; AG\}\; :\; \backslash text\{prism\; IP\}\; ::\; ES\; \backslash times\; \backslash text\{base\; AC\}\; :\; NV\; \backslash times\; \backslash text\{base\; IL\}$; but $\backslash text\{base\; AC\}\; :\; \backslash text\{base\; IL\}\; ::\; AB^2\; :\; IK^2$ (27. 4.) and therefore considering the lines expressed by numbers, $ES\; \backslash times\; \backslash text\{base\; AC\}\; :\; NV\; \backslash times\; \backslash text\{base\; IL\}\; ::\; RT\; \backslash times\; AB^2\; :\; NV\; \backslash times\; IK^2$ (5. 3.) therefore $\backslash text\{prism\; AG\}\; :\; \backslash text\{prism\; IP\}\; ::\; RT\; \backslash times\; AB^2\; :\; NV\; \backslash times\; IK^2$; but $RT\; :\; NV\; ::\; RI\; \backslash text\{\; or\; \}\; AE\; :\; NI$ (20. 4.) :: $AB\; :\; IK$ (def. of sim. figs.), and consequently $RT\; \backslash times\; AB^2\; :\; NV\; \backslash times\; IK^2\; ::\; AB^3\; :\; IK^3$ (5. 3.); therefore $\backslash text\{prism\; AG\}\; :\; \backslash text\{prism\; IP\}\; ::\; AB^3\; :\; IK^3$.

COR. Similar prisms are to one another in the triplicate ratio of the homologous sides. For let Y and Z be two such lines that $AB\; :\; IK\; ::\; IK\; :\; Y\; ::\; Y\; :\; Z$, then the ratio of AB to Z is triplicate the ratio of AB to IK (12. def. 3.). Now, since $AB\; :\; IK\; ::\; IK\; :\; Y$, therefore $AB^2\; :\; IK^2\; ::\; IK^2\; :\; Y^2$ (9. 4.) and, multiplying the antecedents by AB, and consequents by IK, $AB^3\; :\; IK^3\; ::\; AB\; \backslash times\; IK^2\; :\; IK\; \backslash times\; Y^2\; ::\; AB\; \backslash times\; IK\; :\; Y^2$, but $Y^2\; =\; IK\; \backslash times\; Z$ (8. 4.); therefore $AB^3\; :\; IK^3\; ::\; AB\; \backslash times\; IK\; :\; IK\; \backslash times\; Z\; ::\; AB\; :\; Z$, but $\backslash text\{prism\; AG\}\; :\; \backslash text\{prism\; IP\}\; ::\; AB^3\; :\; IK^3$, therefore $\backslash text\{prism\; AG\}\; :\; \backslash text\{prism\; IP\}\; ::\; AB\; :\; Z$, which last ratio is triplicate the ratio of AB to IK.

THEOREM XIII.

If a triangular pyramid ABCD be cut by a plane $bcd$ parallel to its base, the section $bcd$ is similar to the base BCD.

FOR because the planes $bcd$, BCD are parallel, their intersections $bc$, BC with a third plane BAC are parallel (7. 7.); and, for a like reason, $cd$ is parallel to CD, and $db$ to DB; therefore the angle $bcd\; =\; BCD$, $cd\; =\; CDB$, and $db\; =\; DBC$ (10. 7.); hence the triangles $bcd$, BCD are equiangular, and consequently similar.

COR. 1. If two triangular pyramids ABCD, EFGH, which have equal bases, and equal altitudes, be cut by planes $bcd$, $fgh$ that are parallel to the bases, and at equal distances from them, the sections are equal. For conceive the bases of the pyramids to be in the same plane, then their vertices will be in a plane parallel to their bases, and the sections, $bcd$, $fgh$ will also be in a plane parallel to their bases, therefore, $AB\; :\; AE\; ::\; EF\; :$

Solids EF : Ef (14. 7.), but because the triangles ABC, FGH are similar, AB : A b :: BC : b c, and, in like manner EF : E f :: FG : f g, therefore, BC : b c :: FG : f g, and BC² : b c² :: FG² : f g² (9. 4.); but BC² : b c² :: triangle BCD : trian. b c d, and FG² : f g² :: trian. FGH : trian. f g h (25. 4.); therefore, trian. BCD : trian. b c d :: trian. FGH : trian. f g h, but trian. BCD = trian. FGH (by hyp.) therefore trian. b c d = trian. f g h.

SCHOLIUM.

It is easy to see that what is here demonstrated of triangular pyramids, is equally true of polygonal pyramids having equal bases and altitudes.

THEOREM XIV.

145. 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 be equal to 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 CE, 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 h; and let the line MQ be produced till it meet DC in g. Join hg, then hCgQFP is a prism; and is equal to the prism PM (cor. 11.). 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.

THEOREM XV.

Pyramids that have equal bases and altitudes are equal to one another. Fig. 146.

LET ABCD, EFGH be two pyramids that have equal bases BCD, FGH, and also equal altitudes; the pyramid ABCD is equal to the pyramid EFGH.

If they are unequal, let the pyramid EFGH exceed the pyramid ABCD by the solid Z. Let a series of prisms of the same altitude be circumscribed about the pyramid ABCD that shall exceed it by a solid less than Z, (14.) and let another series equal in number to the former, and having all the same altitude, be described about the pyramid EFGH; then, because the pyramids have equal altitudes, the altitude of each of the prisms described about the one pyramid is equal to the altitude of each of the prisms described about the other pyramid; therefore the sections of the pyramids which are the bases of the corresponding prisms will be at equal distances from the bases of the pyramids, and hence these sections will be equal; (1. cor. 13.) and because the prisms have all the same altitude, the corresponding prisms will be equal, and the sum of the prisms described about the pyramid ABCD will be equal to the sum of the prisms described about the pyramid EFGH. Let the pyramid EFGH be denoted by P, and the pyramid ABCD by p, and put Q for the sum of the prisms described about P, and q for the prisms described about p: Then by hypothesis Z = P - p, and by construction Z = q - p, therefore P - p = q - p, and consequently P = q; but it has been shewn that q = Q, therefore P = Q, that is, the pyramid EFGH is greater than the sum of the prism described about it, which is impossible, therefore the pyramids ABCD, EFGH are not unequal, that is, they are equal.

THEOREM XVI.

Every prism having a triangular base may be divided into three pyramids that have triangular bases, and that are equal to one another. Fig. 147.

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, (15.) 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, DEFC are equal to one another; but these pyramids make up the whole prism ABCDEF; therefore, the prism ABCDEF is divided into three equal pyramids.

COR. 1. From this it is manifest that every pyramid

Of Cylinders, Cones, base and the same altitude with it; for if the base and the Sphere. may be divided into prisms having triangular bases.

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

SECT. IX. OF CYLINDERS, CONES, AND THE SPHERE.

DEFINITIONS.

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

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

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

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

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

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

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

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

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

The Diameter of a sphere is any straight line which passes through the centre, and is terminated both ways by the superficies of the sphere.

IV. Similar cones and cylinders are those which have their axes and diameters of their bases proportional.

THEOREM I.

Fig. 143. If from any point E in the circumference of the base of a cylinder ABCD, a perpendicular EF be drawn to the plane of the base AEB, the straight line EF is wholly in the cylindric superficies.

LET HG be the axis, and AGHD the rectangle, which by its revolution describes the cylinder. Because HG is perpendicular to AG in every position of the revolving rectangle, it is perpendicular to the plane of the circle described by AG; and because AD, the line which describes the cylindric superficies, is parallel to GH, it is also perpendicular to the plane of that circle. (5. 7.) Now when by the revolution of the rectangle AGHD the point A coincides with the point E, the line EF will coincide with AD, and thus will be wholly in the cylindric superficies; for otherwise two perpendiculars might be drawn to the same plane, from the same point, which is impossible (2 cor. 4. 7.).

THEOREM II.

Fig. 149. A cylinder and a parallelopiped having equivalent bases and the same altitude are equal to one another.

LET ABCD be a cylinder, and EF a parallelopiped having equivalent bases, viz. the circle AGB and the parallelogram EH, and having also equal altitudes; the cylinder ABCD is equal to the parallelopiped EF. If not, let them be unequal; and first let the cylinder be less than the parallelopiped EF; and from the parallelopiped EF let there be cut off a part EQ by a plane PQ parallel to NF, equal to the cylinder ABCD. In the circle AGB inscribe the polygon AGKBLM that shall differ from the circle by a space less than the parallelogram PH, (1 cor. 2. 6.) 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 shewn 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.

Fig. 150. 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.) and which therefore will be greater than LIM. Let this be the polygon BECFD; and upon BECFD let there be constituted the pyramid ABECFD, and the prism BCFKHG. Because the polygon BECFD is greater than the circle LIM, the prism BCFKHG is greater than the cylinder LMNO, for they have the same altitude, but the prism has the greater base. But the pyramid ABECFD is the third part of the prism BCFKHG (16. 8.); therefore it is greater

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 in 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 shewn, 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.

If a hemisphere and cone have equal bases and altitudes, a series of cylinders may be inscribed in the hemisphere, and another series may be circumscribed about the cone, having all the same altitudes with one another, and such that their sum shall differ from the sum of the hemisphere and the cone by a solid, less than any given solid.

LET ADB be a semicircle, of which the centre is C, and let CD be at right angles to AB; let DB and DA be squares described on DC, draw CE, and let the figure thus constructed revolve about DC: then the quadrant BCD will describe a hemisphere having C for its centre, and the triangle CDE will describe a cone having its vertex at C, and having for its base the circle described by DE, equal to that described by BC, which is the base of the hemisphere. Let W be a given solid, a series of cylinders may be described in the hemisphere ADB, and another described about the cone ECI, so that their sum shall differ from the sum of the hemisphere and cone, by a solid less than the solid W.

Upon the base of the hemisphere let a cylinder be constituted equal to W, and let its altitude be CX. Divide CD into such a number of equal parts, that each of them shall be less than CX; let these be CH, HG, GF and FD. Draw FN, GO, HP parallel to CB, meeting the circle in K, L, and M, and the straight line BE in Q, R, and S. Draw Kf, Lg, Mh, perpendicular to GO, HP, and CB; and draw Qq, Rr, Ss, perpendicular to the same lines. It is evident that the figure being thus constructed, if the whole revolve about CD, the rectangles Ff, Gg, Hh, will describe cylinders that will be circumscribed by the hemisphere BDA; and that the rectangles DN, Fg, Gr, Hs will also describe cylinders that will circumscribe the cone ICE. Now it may be demonstrated, as was done of the prisms inscribed in a pyramid (14. 8.), that the hemisphere exceeds the sum of all the cylinders described within it, by a solid less than the cylinder generated by the rectangle HB, that is, by a solid less than W. In the same manner it may be demonstrated, that the sum of the cylinders circumscribing the cone ICE is greater than the cone by a solid less than the cylinder generated by the rectangle DN, that is, by a solid less than W. Therefore, since the sum of the cylinders inscribed in the hemisphere together with a solid less than W, is equal to the hemisphere; and

since the sum of the cylinders described about the cone is equal to the cone together with a solid less than W; adding equals to equals, the sum of all the cylinders together with a solid less than W is equal to the hemisphere and cone together with a solid less than W; therefore, the difference between the whole of the cylinders, and the sum of the hemisphere and the cone, is equal to the difference of two solids, each of which is less than W; but this difference must also be less than W; therefore the difference between the two series of cylinders, and the sum of the hemisphere and cone is less than the given solid W.

The same things being supposed as in last theorem, Fig. 151. the sum of all the cylinders inscribed in the hemisphere, and described about the cone, is equal to a cylinder having the same base and altitude with the hemisphere.

FOR, the same construction being supposed as in last theorem, let L be the point in which GO meets the circle ADB, then because CGL is a right angle, if CL be joined, the circles described with the radii CG and GL are equal to the circle described with the radius CL or GO (2. cor. 4. 6.). Now CG=GR, because CD=DE, therefore, the circles described by the revolution of the radii GR and GL about the point G are together equal to the circle described by the revolution of the radius GO about the same point G; therefore also the cylinders that stand upon the two first of these circles having the common altitude GH are equal to the cylinder which stands upon the remaining circle, and which has the same altitude GH. The cylinders described by the revolution of the rectangles Gg and Gr are therefore equal to the cylinder described by the rectangle GP. And as the same may be shewn of all the rest, the cylinders described by the rectangles Hh, Gg, Ff, and by the rectangles Hs, Gr, Fg, DN, are together equal to the cylinder described by DB, that is, to the cylinder having the same base and altitude with the hemisphere.

Every sphere is two-thirds of the circumscribing cylinder.

LET the figure be constructed as in the two last theorems, and if the hemisphere described by the quadrant BDC be not equal to two-thirds of the cylinder described by the rectangle BD, let it be greater by the solid W. Then as the cone described by CDE is one-third of the cylinder described by BD, the cone and the hemisphere together will exceed the cylinder by W. But that cylinder is equal to the sum of all the cylinders described by the rectangle Hh, Gg, Ff, Hs, Gr, Fg, DN; therefore, the hemisphere and the cone added together exceed the sum of all these cylinders by the solid W, which is absurd; for it has been shewn (4.) that the hemisphere and the cone together differ from the sum of these cylinders by a solid less than W. The hemisphere is therefore equal to two-thirds of the cylinder described

Of Cylinders, cones, sphere is two thirds of the cylinder described by twice the rectangle BD, that is, to two-thirds of the circumference describing cylinder.

We here conclude the Elements of Geometry. Their application, constituting what is sometimes called Practical Geometry, will be given under the article MENSURATION.

A TABLE shewing the Theorem of the foregoing Treatise, that corresponds to each of the most material Propositions in the first six, and in the eleventh and twelfth, books of Euclid's Elements.

GEORGE I. II. and III. kings of Great Britain.
—George I. the son of Ernest Augustus, duke of Brunswick Lunenburg, and elector of Hanover, succeeded to the throne of Great Britain in 1714, in virtue of an act of parliament, passed in the latter part of the reign of King William III. limiting the succession of the crown, after the demise of that monarch, and Queen Anne (without issue), to the princess Sophia of Hanover, and the heirs of her body, being Protestants.—George II. the only son of the former, succeeded him in 1727, and enjoyed a long reign of glory; dying amidst the most rapid and extensive conquests in the 77th year of his age. He was succeeded by his grandson George III. our late sovereign. For particulars, see BRITAIN, N^o 374—701.

GEORGE, or Knights of St GEORGE, has been the denomination of several military orders, whereof that of the Garter is one of the most illustrious. See GARTER, and St GEORGE, below.

King GEORGE'S Islands, are two islands in the South sea, lying in W. Long. 144. 56. S. Lat. 14. 28. They were first discovered by Commodore Byron in 1765, and were since visited by Captain Cook in 1774. Commodore Byron's people had an encounter with the inhabitants, which proved fatal to some of the natives; but Captain Cook was more fortunate. A lieutenant and two boats well-armed were sent on shore by Captain Cook; and landed without opposition. As soon as the gentlemen landed, the islanders embraced them by touching noses, a mode of civility used in New Zealand,

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