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

B

by half the arc AD (B. III, Prop. vIII);
therefore these triangles are similar, and we have
FB FC: FD: FA.

:

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

FB.FA = FC.FD.

PROPOSITION XXVI.

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.

=

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

:

BA X AC = AF × AD;

or, using AD + DF for AF, we have

BA × AC = AD2 + AD × DF.

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

.

BA X AC = AD2 + BD × DC.

PROPOSITION XXVII.

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.

In the triangle ABC, let AD be drawn perpendicular to BC, and let CF be the diameter of the circumscribed circle; then will

For, joining AF, the triangles ABD, AFC are right

angled, the one at D, the

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; con

sequently,

AB X AC

AD × CF.

PROPOSITION XXVIII.

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.

the ratio of FG to GD will be the same for all positions of the point G.

For, by hypothesis, stituting CG for CA,

:

CD CA: CA: CF, or subCD: CG :: CG: CF; hence the triangles CDG and CFG have each an equal angle C contained by proportional sides, and are therefore similar (B. IV, Prop. x), and the third side GD is to the third side GF as CD to CG or CA. But, by division, the proportion CD: CA :: CA: CF gives

CD
GD

CA: AD: AF; therefore,

GF AD : AF; and since the

ratio of AD to AF 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.

M

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 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).