the perpendicular AD is a mean proportional between BD and DC, the two segments of the diameter; that is,

Two triangles, having an angle in the one equal to an angle in the other, are to each other as the rectangles of the sides which contain the equal angles.

Let the two triangles ABC, ADE have the angle A in common; then will the triangle ABC be to the triangle ADE as the rectangle ABX AC is to the rectangle AD XAE.

Join BE. Then the two triangles ABE, ADE, having the common vertex E, have the same altitude, and are to each other as their bases AB, AD (Prop. VI., Cor. 1); therefore

ABE ADE::AB: AD.

B

D

A

E

Also, the two triangles ABC, ABE, having the common vertex B, have the same altitude, and are to each other as their bases AC, AE; therefore

ABC: ABE :: AC: AE.

Hence (Prop. XI., Cor., B. II.).

ABC: ADE:: ABXAC: AD XAE.

Therefore, two triangles, &c.

Cor. 1. If the rectangles of the sides containing the equal angles are equivalent, the triangles will be equivalent.

Cor. 2. Equiangular parallelograms are to each other as the rectangles of the sides which contain the equal angles.

PROPOSITION XXIV. THEOREM.

Similar triangles are to each other as the squares described on their homologous sides.

Let ABC, DEF be two similar triangles, having the angle A equal to D, the angle B equal to E, and C equal to F; then the triangle ABC is to the triangle DEF as the square on BC is to B the square on EF.

A

By similar triangles, we have (Def. 3)
AB: DE:: BC: EF.

D

CE

F

Multiplying together the corresponding terms of these pro portions, we obtain (Prop. XI., B. II.),

ABX BC: DE XEF:: BC: EF2.

But, by Prop. XXIII.,

ABC: DEF:: AB×BC: DE×EF;

hence (Prop. IV., B. II.)

ABC: DEF :: BC2 :

EF2.

Therefore, similar triangles, &c.

PROPOSITION XXV. THEOREM.

Two similar polygons may be divided into the same number of triangles, similar each to each, and similarly situated.

polygon FGHIK, the angle B is equal to the angle G (Def. 3), and AB BC:: FG: GH. And, because the triangles ABC, FGH have an angle in the one equal to an angle in the other, and the sides about these equal angles proportional, they are similar (Prop. XX.); therefore the angle BCA is equal to the angle GHF. Also, because the polygons are similar, the whole angle BCD is equal (Def. 3) to the whole angle GHI; therefore, the remaining angle ACD is equal to the remaining angle FHI. Now, because the triangles ABC FGH are similar,

AC: FH:: BC: GH.
And, because the polygons are similar (Def. 3),

whence

BC: GH:: CD: HI;

AC: FH CD: HI;

that is, the sides about the equal angles ACD, FHI are proportional; therefore the triangle ACD is similar to the triangle FHI (Prop. XX.). For the same reason, the triangle ADE is similar to the triangle FIK; therefore the similar polygons ABCDE, FGHIK are divided into the same number of triangles, which are similar, each to each, and similarly situated.

Cor. Conversely, if two polygons are composed of the same number of triangles, similar and similarly situated, the polygons are similar.

For, because the triangles are similar, the angle ABC is equal to FGH; and because the angle BCA is equal to GHF, and ACD to FHI, therefore the angle BCD is equal to GHI. For the same reason, the angle CDE is equal to HIK, and so on for the other angles. Therefore the two polygons are mutually equiangular.

Moreover, the sides about the equal angles are proportional. For, because the triangles are similar, AB : FG :: BC : GH. Also, BC: GH:: AČ: FH, and AC: FH :: CD: HI; hence BC: GH:: CD: HI. In the same manner, it may be proved that CD: HI :: DE: IK, and so on for the other sides. Therefore the two polygons are similar.

PROPOSITION XXVI. THEOREM.

The perimeters of similar polygons are to each other as their homologous sides; and their areas are as the squares of those sides.

Let ABCDE, FGHIK be two similar polygons, and let AB be the side B homologous to FG; then the perimeter of ABCDE is to the perimeter of A

FGHIK as AB is to FG;

and the area of ABCDE

E

C

is to the area of FGHIK as AB is to FG2

D

F

K

H.

First. Because the polygon ABCDE is similar to the polygon FGHIK (Def. 3),

AB: FG:: BC: GH:: CD: HI, &c.; therefore (Prop. IX., B. II.) the sum of the antecedents AB +BC+CD, &c., which form the perimeter of the first figure, is to the sum of the consequents FG+GH+HI, &c., which form the perimeter of the second figure, as any one antecedent is to its consequent, or as AB to FG.

Secondly. Because the triangle ABC is similar to the tri. angle FGH, the triangle ABC: triangle FGH :: AC2: FH' (Prop. XXIV.).

And, because the triangle ACD is similar to the triangle FHI,

ACD: FHI :: AC2: FH'.

Therefore the triangle ABC : triangle FGH :: triangle ACD triangle FHI (Prop. IV., B. II.). In the same manner, it may be proved that

ACD: FHI :: ADE: FIK.

Therefore, as the sum of the antecedents ABC+ACD+ ADE, or the polygon ABCDE, is to the sum of the conse quents FGH+FHI+FIK, or the polygon FGHIK, so is any one antecedent, as ABC, to its consequent FGH; or, as AB1 to FG. Therefore, similar polygons, &c.

PROPOSITION XXVII. THEOREM.

If two chords in a circle intersect each other, the rectangle contained by the parts of the one, is equal to the rectangle contained by the parts of the other.

Let the two chords AB, CD in the circle ACBD, intersect each other in the point E; the rectangle contained by AE, EB is equal to the rectangle contained by DE, EC.

Join AC and BD. Then, in the triangles ACE, DBE, the angles at E are equal, being vertical angles (Prop. V., B. Î.); the angle A is equal to the angle D, being in

scribed in the same segment (Prop. XV., Cor. 1., B. III.); therefore the angle C is equal to the angle B. The triangles are consequently similar; and hence (Prop. XVIII.)

or (Prop. I., B. II.),

AE: DE EC: EB,

AEXEB-DEXEC.

Therefore, if two chords, &c.

Cor. The parts of two chords which intersect each other in a circle are reciprocally proportional; that is, AE: DE:: EC: EB.

If from a point without a circle, a tangent and a secant be drawn, the square of the tangent will be equivalent to the rectangle contained by the whole secant and its external segment.

Let A be any point without the circle. BCD, and let AB be a tangent, and AC a secant; then the square of AB is equivalent to the rectangle ADX AC.

Join BD and BC. Then the triangles ABD and ABC are similar; because they B have the angle A in common; also, the angle ABD formed by a tangent and a chord is measured by half the arc BD

A

D

(Prop. XVI., B. III.); and the angle C is measured by half the same arc, therefore the angle ABD is equal to C, and the two triangles ABD, ABC are equiangular, and, consequently, similar; therefore (Prop. XVIII.)

AC AB AB : AD;

Therefore, if from a point, &c.

Cor. 1. If from a point without a circle, a tangent and a stcant be drawn, the tangent will be a mean proportional between the secant and its external segment.

Cor. 2. If from a point without a circle, two secants be drawn, the rectangles contained by the whole secants and their external segments will be equivalent to each other; for each of these rectangles is equivalent to the square of the tangent from the same point.

Cor. 3. If from a point without a circle, two secants be drawn, the whole secants will be reciprocally proportional to their external segments.

If an angle of a triangle be bisected by a line which cuts the base, the rectangle contained by the sides of the triangle, is equivalent to the rectangle contained by the segments of the base, together with the square of the bisecting line.

Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; the rectangle BAXAC is equivalent to BD XDC together with the of AD.

square

C

D

E

Describe the circle ACEB about the triangle, and produce AD to meet the circumference in E, and join EC. Then, because the angle BAD is equal to the angle CAE, and the angle ABD to the angle AEC, for they are in the same segment (Prop. XV., Cor. 1, B. III.), the triangles ABD, AEC are mutually equiangular and similar; therefore (Prop. XVIII.)

BA: AD: EA: AC; consequently (Prop. I., B. II.),

BAXAC ADXAE.

But AE=AD+DE; and multiplying each of these equals by AD, we have (Prop. III.) ADXAE-AD2+ADDE. But ADXDE=BDX DC (Prop. XXVII.); hence

BAXAC BD × DC+AD'.

=

Therefore, if an angle, &c.