muse, 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 B AD, 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 BA C, 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 DA C, B A C being also similar,

DC: AC:: AC: BC;

hence each of the sides A B, A C 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 A B D, 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 B C.

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 ABC, ADE have the angle A in common; then will the triangle ABC be to the triangle A D E as ABX AC to ADX A E.

Join BE; then the triangles ABE, B

A

D

E

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 A B C, 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: ADX AE.

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

PROPOSITION XXIX.-THEOREM.

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

Let A B C, D E F be two similar triangles, and let A C, D F be homologous sides; then the triangle ABC will be to the triangle DEF as the square on AC is to the square on DF.

A

B

D

CE

F

For, the triangles being similar, they have their homologous sides proportional (Art. 210); therefore

AB: DE:: AC: DF;

and multiplying the terms of this proportion by the corresponding terms of the identical proportion,

AC: DF:: AC: DF,

we have (Prop. XIII. Bk. II.),

ABX AC: DEX DF:: A C2: D F2.

But, by reason of the equal angles A and D, the triangle ABC is to the triangle DEF as ABX AC is to DEX DF (Prop. XXVIII.); consequently (Prop. X. Bk. II.),

ABCDEF:: A C2 : D F2.

Therefore, the two similar triangles A B C, DEF are to each other as the squares described on the homologous sides A C, D F, or as the squares described on any other two homologous sides.

PROPOSITION XXX.-THEOREM.

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

to each, and similarly situated. From the homologous angles A and F, draw the diagonals A C, A D and FH, FI.

The two polygons being similar, the angles B and G, which are homologous, must be equal, and the sides A B, BC must also be proportional to FG, GH (Art. 210); that is, AB: FG:: BC: GH. Therefore the triangles ABC, FGH have an angle of the one equal to the angle of the other, and the sides containing these angles proportional; hence they are similar (Prop. XXIV.); consequently the angle BCA is equal to the angle GHF. These equal angles being taken from the equal angles BCD, GHI, the remaining angles A CD, FHI will be equal (Art. 34, Ax. 3). But, since the triangles A B C, FGH are similar, we have

AC: FH:: BC: GH;

and, since the polygons are similar (Art. 210),

BC: GH:: CD: HI;

hence (Prop. X. Cor. 1, Bk. II.),

AC: FH:: CD: HI.

But the terms of the last proportion are the sides about the equal angles ACD, FHI; hence the triangles A CD, FHI are similar (Prop. XXIV.). In the same manner, it may be shown that the corresponding triangles AD E, FIK are similar; hence the similar polygons may be divided into the same number of triangles similar each to each, and similarly situated.

276. Cor. Conversely, if two polygons are composed

of the same number of similar triangles, and similarly situated, the two polygons are similar.

For the similarity of the corresponding triangles give the angles A B C equal to F GH, BCA equal to G H F, and ACD equal to FHI; hence, B C D equal to G H I, likewise C D E equal to HIK, &c. Moreover, we have

AB: FG:: BC: GH:: AC: FH:: CD: HI, &c.; therefore the two polygons have their angles equal and their sides proportional; hence they are similar.

PROPOSITION XXXI.-THEOREM.

277. The perimeters of similar polygons are to each other as their homologous sides; and their areas are to each other as the squares described on these sides.

A B and FG, B C and GH, &c.; and their areas are to

each other as A B is to F G2, B C to GH3, &c.

First. Since the two polygons are similar, we have

AB: FG:: BC: GH:: CD: HI, &c.

Now the sum of the antecedents A B, BC, CD, &c., which compose the perimeter of the first polygon, is to the sum of the consequents FG, GH, HI, &c., which compose the perimeter of the second polygon, as any one antecedent is to its consequent (Prop. XI. Bk. II.); therefore, as any two homologous sides are to each other, or as AB is to F G.

Secondly. From the homologous angles A and F, draw