solution of all problems; since the general properties of triangles include, by implication, those of all figures.

PROPOSITION XXIV.-THEOREM.

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

Let the two triangles A B C, DEF have the angle A equal to the angle D, and the sides containing these angles proportional, so that AB DE:: AC: DF; then the triangles are similar.

G

A

D

H

B

C E

F

Take A G equal D E, and draw GH parallel to BC. The angle AGH will be equal to the angle A B C (Prop. XXII. Bk. I.); and the triangles AGH, ABC will be equiangular; hence we shall have AB: AG:: AC: AH.

But, by hypothesis,

AB:DE::AC: DF;

and, by construction, A G is equal to D E; hence AH is equal to D F. Therefore the two triangles A G H, DEF, 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 themselves equal (Prop. V. Bk. I.). But the triangle AGH is similar to ABC; therefore DEF is also similar to AB C.

PROPOSITION XXV. - THEOREM.

265. Two triangles, which have their sides, taken two and two, either parallel or perpendicular to each other, are similar.

Let the two triangles A B C, DEF have the side A B parallel to the side DE, BC parallel to EF, and A C

parallel to DF; these triangles will then be similar.

A D

E

F

For, since the side AB is parallel to the side D E, and BC to EF, the angle ABC is equal to the angle DEF (Prop. XXVI. Bk. I.). Also, since AC is parallel to DF, the angle ACB is equal to the angle D F E, and the angle BAC to EDF; therefore the triangles ABC, DEF are equiangular; hence they are similar (Prop. XXII.).

Again, let the two triangles

B

A

ABC, DEF have the side

DE perpendicular to the side

AB, DF perpendicular to AC, and EF perpendicular to BC; these triangles are similar.

E

G

D

B

F

Produce FD till it meets AC at G; then the angles DGA, DE A of the quadrilateral AEDG are two right angles; and since all the four angles are together equal to four right angles (Prop. XXIX. Cor. 1, Bk. I.), the remaining two angles, EDG, EAG, are together equal to two right angles. But the two angles EDG, EDF are also together equal to two right angles (Prop. I. Bk. I.); hence the angle EDF is equal to EAG or BAC.

The two angles, G F C, GCF, in the right-angled triangle FGC, are together equal to a right angle (Prop. XXVIII. Cor. 5, Bk. I.), and the two angles GFC, GFE are together equal to the right angle EFC (Art. 34, Ax. 9); therefore GFE is equal to GCF, or DFE to BCA. Therefore the triangles ABC, DEF have two angles of the one equal to two angles of the other, each to each; hence they are similar (Prop. XXII. Cor.).

266. Scholium. When the two triangles have their sides. parallel, the parallel sides are homologous; and when they have them perpendicular, the perpendicular sides a

D

B F G H C

DI is parallel to BF, the triangles ADI equingular; and we have (Prop. XXII.), DI: BF::AI: AF;

is parallel to F G, we have in like manner, AI:AF::IK: FG;

se these two propositions contain the same ratio, we shall have (Prop. X. Cor. 1, Bk. II.),

DI: BF::IK: FG.

same manner, it may be shown that

IK:FG::KL: GH::LE: HC.

Dre the line D E is divided at the points I, K, L, as the base BC is, at the points F, G, H.

98. Or FBC were divided into equal parts at the

the parallel DE would also be divided the points I, K, L.

TION XXVIL-THEOREM.

ngled triangle, if a perpendicular is rtex of the right angle to the hypothe

I

nuse, the triangle will be divided into two triangles similar to the given triangle and to each other.

A

B

D

In the right-angled triangle A B C, from the vertex of the right angle BAC, let AD be drawn perpendicular to the hypothenuse BC; then the triangles BAD, DAC will be similar to the triangle A B C, and to each other. For the triangles BAD, BAC have the common angle B, the right angle B D A equal to the right angle BA C, and therefore the third angle, BAD, of the one, equal to the third angle, C, of the other (Prop. XXVIII. Cor. 2, Bk. I.); hence these two triangles are equiangular, and consequently are similar (Prop. XXII.). In the same manner it may be shown that the triangles DAC and BAC are equiangular and similar. The triangles BAD and DA C, being each similar to the triangle BAC, are similar to each other.

270. Cor. 1. Each of the sides containing the right angle is a mean proportional between the hypothenuse and the part of it which is cut off adjacent to that side by the perpendicular from the vertex of the right angle. For, the triangles BAD, BAC being similar, their homologous sides are proportional; hence

BD: BA:: BA: BC;

and, the triangles D A C, B A C being also similar,

DC: AC:: AC: BC;

hence each of the sides A B, AC is a mean proportional between the hypothenuse and the part cut off adjacent to that side.

271. Cor. 2. The perpendicular from the vertex of the right angle to the hypothenuse is a mean proportional between the two parts into which it divides the hypothenuse.

For, since the triangles ABD, ADC are similar, by comparing their homologous sides we have

BD: AD::AD: DC;

hence, the perpendicular A D is a mean proportional between the parts D B, D C into which it divides the hypothenuse BC.

PROPOSITION XXVIII.—THEOREM.

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

Let the two triangles A B C, ADE have the angle A in common; then will the triangle ABC be to the triangle A D E as ABX AC to AD X A E.

B

A

D

E

Join BE; then the triangles A BE, ADE, having the common vertex E, and their bases in the same line, AB, have the same altitude, and are to each other as their bases (Prop. VI. Cor.); hence

ABE: ADE::AB: A D.

In like manner, since the triangles ABC, ABE have the common vertex B, and their bases in the same line, A C, we have

ABC: ABE:: AC: A E.

By multiplying together the corresponding terms of these proportions, and omitting the common term ABE, we have (Prop. XIII. Bk. II.),

ABC: ADE:: ABX AC: ADXA E.

273. Cor. If the rectangles of the sides containing the equal angles were equivalent, the triangles would be equivalent.