IK: FG :: KL : GH :: LE : HC;

which was to be proved.

Cor. If BC is divided into equal parts at F, G, and H, then will DE be divided into equal parts, at I, K, and L.

PROPOSITION XXIII. THEOREM.

If, in a right-angled triangle, a perpendicular be drawn from the vertex of the right angle to the hypothenuse:

1°. The triangles on each side of the perpendicular will be similar to the given triangle, and to each other:

2o. Each side about the right angle will be a mean proportional between the hypothenuse and the adjacent segment:

3°. The perpendicular will be a mean proportional between the two segments of the hypothenuse.

1o. Let ABC be a right-angled triangle, A the vertex

of the right angle, BC the hypo

thenuse, and AD perpendicular to BC: then will ADB and ADC

be similar to ABC, and consequently, similar to each other.

The triangles ADB and ABC have the angle B

B

A

D

C

B common, and the angles ADB and

BAC equal, because both are right angles; they are, therefore, similar (P. XVIII., C). In like manner, it may be shown that the triangles ADC and ABC are similar; and since ADB and ADC are both similar to ABC, they are similar to each other; which was to be proved.

3o. AD will be a mean proportional between BD and DC. For, the triangles ADB and ADC being similar,

their homologous sides are proportional; hence,

Cor. 2. If from any point A, in a semi-circumference BAC, chords be drawn to the

A

B D

extremities B and C of the diameter BC, and a perpendicular AD be drawn to the diameter: then will ABC be a right-angled triangle, right-angled at A; and from what was proved above, each chord will be a mean proportional between the diameter and the adjacent segment; and, the perpendicular will be a mean proportional between the segments of the diameter.

PROPOSITION XXIV. THEOREM.

Triangles which have an angle in each equal, are to each other as the rectangles of the including sides.

Let the triangles GHK and ABC have the angles G and A equal: then will they be to each other as the rectangles of the sides about these angles.

The triangles ADE and ABE have their bases in the same line AB, and a common vertex E; therefore, they have the same altitude, and consequently, are to each other as their bases; that is,

multiplying these proportions, term by term, and omitting the common factor ABE (B. II., P. VII.), we have,

ADE : ABC
ABC :: AD × AE AB × AC;

:

and for AD × AE,

GHK : ABC :: : ×
GH × GK AB X AC;

which was to be proved.

:

Cor. If ADE and ABC are similar, the angles D and B being homologous, DE will be parallel to BC, and we shall have,

that is, ABE is a mean proportional between ADE and ABC.

B

PROPOSITION XXV. THEOREM.

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

Let the triangles ABC and DEF be similar, the angle A being equal to the angle D, B to E, and C to F: then will the triangles be to each other as the squares of any two homologous sides.

Because the angles A and D XXIV.),

are equal, we have (P.

2

AB × AC DE × DF :: AC2 : DF2;

:

F

combining this, with the first proportion (B. II., P. IV.), we have,

ABC : DEF :: ÃỠ2 : DF.

In like manner, it may be shown that the triangles are to each other as the squares of AB and DE, or of BC! and EF; which was to be proved.