Let the secants FB, FC be drawn from the point F; then will

FB: FC: FD : FA. For, drawing AC, BD, the triangles FAC, FBD have the angle at F common, and the angle at C equal to the angle at B, since each is measured by half the arc AD (B. III, Prop. vII;) there

B

fore these triangles are similar, and we have FB FC: FD: FA.

D

Cor. If we take the products of the means and extremes of the above proportion, we shall have

THEOREM. If either angle of a triangle is bisected by a line terminating in the opposite side, the rectangle of the sides, including the bisected angle, is equal to the square of the bisecting line, together with the rectangle contained by the segments of the third side.

Let AD bisect the angle A: then will

ABX AC AD2+ BD × DC.

Describe a circumference through the three points B, A, C, (B. III, Prop. 11;) produce AD till it meets this circumference at F, and join CF.

B

D

The triangle BAD is similar to the triangle FAC; for, by hypothesis, the angle BAD=FAC; also the angle B=F, each being measured by half the arc AC. Hence these triangles are similar, and we have

BA: AF :: AD: AC; which gives
BAX AC AF × AD;

=

or, using AD+DF for AF, we have

BAX AC AD2+ AD × DF.

=

But AD× DF=BD × DC, [B. IV, Prop. xxvIII, Cor. ;] therefore we finally obtain

BA× AC=AD2+BD × DC.

PROPOSITION XXXI.

THEOREM. In every triangle, the rectangle contained by any two sides is equal to the rectangle contained by the diameter of the circumscribing circle, and the perpendicular drawn to the third side from the opposite angle.

other at A; also the angle B=F, each being measured by half the arc AC, (B. III, Prop. VIII.) Hence the triangles are similar, and we have AB: CF : : AD : AC; consequently, ABX AC ADX CF.

PROPOSITION XXXII.

THEOREM. If a point be taken on the radius of a circle, and this radius be then produced, and a second point be taken on it without the circumference, these points being so situated that the radius of the circle shall be a mean proportional between their distances from the centre, then, if lines be drawn from these points to any points of the circumference, the ratio of such lines will be constant.

Let D be the point within the circumference, and F the point without; then, if CD: CA :: CA : CF, the ratio of GD to GF will be the same for all positions of the point G.

A

For, by hypothesis, CD CA: CA: CF, or substituting CG for CA, CD: CG :: CG : CF; hence the triangles CDG and CFG have each an equal angle C contained by proportional sides, and are therefore equiangular and similar, (B. IV, Prop. XIV,) and the third side GD is to the third side GF as CD to CG or CA. And since the ratio of CD to CA is constant, it follows that the ratio of GD to GF is also constant.

BOOK FIFTH.

DEFINITIONS.

1. ANY polygonal figure is said to be equilateral, when all its sides are equal; and it is equiangular, when all its angles are equal.

2. Two polygons are said to be mutually equilateral, when their corresponding sides, taken in the same order, are equal. When this is the case with the corresponding angles, the polygons are said to be mutually equiangular.

3. A regular polygon has all its sides and all its angles equal. If all the sides and all the angles are not equal, the polygon is irregular.

4. A regular polygon may have any number of sides not less than three. The equilateral triangle (Def. XV,) is a regular polygon of three sides. The square (Def. XIX,) is also a regular polygon of four sides.

PROPOSITION I.

THEOREM. Two regular polygons of the same number of sides, are similar figures.

Suppose we have, for example, the two regular hexagons ABCDFG, abcdfg; then will these two polygons be similar figures.

For, the sum of all the angles is the same in the one as the other, (B. I, Prop. XXIV.) In this case the sum of all the angles is eight right-angles; the angles are

therefore each equal to one-sixth of eight right-angles, and hence the two polygons are mutually equiangular. Again, since AB, BC, CD, etc., are equal, and ab, bc, cd, etc., are also equal, we have

AB: ab: : BC : bc : : CD : cd, etc. It therefore follows that two regular polygons of the same number of sides have equal angles, and the sides about those equal angles proportional; consequently they are similar, (B. IV, Def. 3.)

Cor. The perimeters of two regular polygons having the same number of sides, are to each other as their homologous sides; and their areas are to each other as the squares of those sides, (B. IV, Prop. xvI.)

(92.) It has been shown, (B. I, Prop. xxiv,) that the sum of all the interior angles of a polygon is found by multiplying two rightangles, or 1800, by a number which is two less than the number of sides of the polygon; or, if n denote the number of sides, then the sum of all the angles will be (n-2.) 1800. But, since all the angles of a regular polygon are equal, the magnitude of one of these angles may be found by dividing the sum of all by the number of angles, or, which is the same, by the number of sides in the polygon; therefore

n- -2
n

each angle is equal to 1800, or 1800

360°

n