G

tinue diminishing, and would soon
elude our comparisons by their mi-
nuteness, this would be but an imper-
fect mechanical method, from which
no conclusion could be obtained to
determine whether the lines AC, CB,
have or have not a common measure.
There is a very simple way, however,
of avoiding these decreasing lines, A F
and obtaining the result, by operating

B

E

only upon lines which remain always of the same magnitude. The angle ABC being a right angle, AB is a tangent, and AE a secant drawn from the same point; so that AD: AB :: AB AE (Prop. XXX.). Hence in the second operation, when AD is compared with AB, the ratio of AB to AE may be taken instead of that of AD to AB; now AB, or its equal CD, is contained twice in AE, with the remainder AD; the result of the second operation is therefore the quotient 2 with the remainder AD, which must be compared with AB.

Thus the third operation again consists in comparing AD with AB, and may be reduced in the same manner to the comparison of AB or its equal CD with AE; from which there will again be obtained 2 for the quotient, and AD for the remainder.

Hence, it is evident that the process will never terminate; and therefore there is no common measure between the diagonal and the side of a square: a truth which was already known by arithmetic, since these two lines are to each other :: √2:1 (Prop. XI. Cor. 4.), but which acquires a greater degree of clearness by the geometrical investigation.

BOOK V.

REGULAR POLYGONS, AND THE MEASUREMENT OF THE CIRCLE.

Definition.

A POLYGON, which is at once equilateral and equiangular, is called a regular polygon.

Regular polygons may have any number of sides: the equilateral triangle is one of three sides; the square is one of four.

PROPOSITION I. THEOREM.

Two regular polygons of the same number of sides are similar

Suppose, for example, that ABCDEF, abcdef,

figures.

are two regular hexagons. The sum of all the

angles is the same in both figures,being in each equal

to eight right angles (Book I. Prop. XXVI. Cor. 3.). The angle A is the sixth part of that sum; so is the angle a: hence the angles A and a are equal; and for the same reason, the angles B and b, the angles C and c, &c. are equal.

Again, since the polygons are regular, the sides AB, BC, CD, &c. are equal, and likewise the sides ab, bc, cd, &c. (Def.); it is plain that AB : ab :: BC: bc :: CD: cd, &c.; hence the two figures in question have their angles equal, and their homologous sides proportional; consequently they are similar (Book IV. Def. 1.).

Cor. The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their surfaces are to each other as the squares of those sides (Book IV. Prop. XXVII.).

Scholium. The angle of a regular polygon, like the angle of an equiangular polygon, is determined by the number of its sides (Book I. Prop. XXVI.).

K

PROPOSITION II. THEOREM.

Any regular polygon may be inscribed in a circle, and circumscribed about one.

Let ABCDE &c. be a regular polygon: describe a circle through the three points A, B, C, the centre being O, and OP the perpendicular let fall from it, to the middle point of BC: draw AO and H OD.

B

F

If the quadrilateral OPCD be placed upon the quadrilateral OPBA, they will coincide; for the side OP is common; the angle OPC OPB, each being a right angle; hence the side PC will apply to its equal PB, and the point C will fall on B: besides, from the nature of the polygon, the angle PCD= PBA; hence CD will take the direction BA; and since CD= BA, the point D will fall on A, and the two quadrilaterals will entirely coincide. The distance OD is therefore equal to AO; and consequently the circle which passes through the three points A, B, C, will also pass through the point D. By the same mode of reasoning, it might be shown, that the circle which passes through the three points B, C, D, will also pass through the point E; and so of all the rest: hence the circle which passes through the points A, B, C, passes also through the vertices of all the angles in the polygon, which is therefore inscribed in this circle.

Again, in reference to this circle, all the sides AB, BC, CD, &c. are equal chords; they are therefore equally distant from the centre (Book III. Prop. VIII.): hence, if from the point O with the distance OP, a circle be described, it will touch the side BC, and all the other sides of the polygon, each in its middle point, and the circle will be inscribed in the polygon, or the polygon described about the circle.

Scholium 1. The point O, the common centre of the inscribed and circumscribed circles, may also be regarded as the centre of the polygon; and upon this principle the angle AOB is called the angle at the centre, being formed by two radii drawn to the extremities of the same side AB.

Since all the chords AB, BC, CD, &c. are equal, all the angles at the centre must evidently be equal likewise; and therefore the value of each will be found by dividing four right angles by the number of sides of the polygon.

Scholium 2. To inscribe a regular polygon of a certain number of sides in a given circle, we have only to divide the circumference into as many equal parts as the polygon H has sides: for the arcs being equal, the chords AB, BC, CD, &c. will also be equal; hence likewise the triangles AOB, BOC, COD, must

E

A

B

C

be equal, because the sides are equal each to each; hence all the angles ABC, BCD, CDE, &c. will be equal; hence the figure ABCDEH, will be a regular polygon.

PROPOSITION III. PROBLEM.

To inscribe a square in a given circle.

Draw two diameters AC, BD, cutting each other at right angles; join their extremities A, B, C, D: the figure ABCD will be a square. For the angles AOB, BOC, &c. being equal, the A chords AB, BC; &c. are also equal: and the angles ABC, BCD, &c. being in semicircles, are right angles.

B

D

Scholium. Since the triangle BCO is right angled and isosceles, we have BC: BO :: √2 : 1 (Book IV. Prop. XI. Cor. 4.); hence the side of the inscribed square is to the radius, as the square root of 2, is to unity.

PROPOSITION IV. PROBLEM.

In a given circle, to inscribe a regular hexagon and an equilateral triangle.

Suppose the problem solved, and that AB is a side of the inscribed hexagon; the radii AO, OB being drawn, the triangle AOB will be equilateral.

For, the angle AOB is the sixth part of four right angles; therefore, taking the right angle for unity, we shall have AOB==

and the two other angles ABO, BAO, of the same triangle, are together equal to 2-3

; and being mutually equal,

A

each of them must be equal to ; hence the triangle ABO is equilateral; therefore the side of the inscribed hexagon is equal to the radius.

Hence to inscribe a regular hexagon in a given circle, the radius must be applied six times to the circumference; which will bring us round to the point of beginning.

And the hexagon ABCDEF being inscribed, the equilateral triangle ACE may be formed by joining the vertices of the alternate angles.

Scholium. The figure ABCO is a parallelogram and even a rhombus, since AB-BC-CO-AO; hence the sum of the squares of the diagonals AC2+BO2 is equivalent to the sum of the squares of the sides, that is, to 4AB2, or 4BO (Book IV. Prop XIV. Cor.): and taking away BO2 from both, there will remain AC2-3BO2; hence AC2: BO2:: 3 : 1, or AC BO :: √3 1; hence the side of the inscribed equilateral triangle is to the radius as the square root of three is to unity.

PROPOSITION V. PROBLEM.

In a given circle, to inscribe a regular decagon; then a pentagon, and also a regular polygon of fifteen sides.