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For, since the side AB is parallel to DE, to EF, the angle B is equal to the angle E XXIV.) ; in like manner,

the angle C is equal to

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and BC

(B. I., P.

Да

consequently, are similar (P. XVIII.); proved.

2°. Let the triangles ABC AB perpendicular to DE,

FD:

which was to be

and

DEF have the side

and CA to

BC to EF,

then will they be similar. For, prolong the sides of the triangle DEF till they meet the sides. of the triangle ABC. The sum of the interior angles of the quadrilateral BIEG is equal to four right angles (B. I., P. XXVI.); but, the angles EIB and EGB are each right

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angles, by hypothesis; hence, the sum of the angles IEG IBG is equal to two right angles; the sum of the angles IEG and DEF is equal to two right angles, because they are adjacent; and since things which are equal to the same thing are equal to each other, the sum of the angles IEG and IBG is equal to the sum of the angles IEG and DEF; or, taking away the common part IEG, we have the angle IBG equal to the angle DEF. In like manner, the angle GCII may be proved equal to the angle EFD, and the angle HAI to the angle EDF; the triangles ABC and DEF are, therefore, mutually equiangular, and consequently similar; which was to be proved.

Cor. 1. In the first case, the parallel sides are homolo

gous; in the second case, the perpendicular sides are homologous.

Cor. 2. The homologous angles are those included by sides respectively parallel or perpendicular to each other.

Scholium. When two triangles have their sides perpenlicular, each to each, they may have a different relative position from that shown in the figure. But we can always construct a triangle within the triangle ABC, whose sides shall be parallel to those of the other triangle, and then the demonstration will be the same as above.

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If a straight line be drawn parallel to the base of a triangle, and straight lines be drawn from the vertex of the triangle to points of the base, these lines will divide the base and the parallel proportionally.

Let ABC be a triangle, BC its base, A its vertex, DE parallel to BC, and AF, AG, AH,

from A to points of the base: then will

lines drawn

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

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DI : BF :: 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.

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

1o. 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 common,

B

D

and the angles ADB and

BAC equal, because both are right angles; they are, there fore, similar (P. XVIII., C). In like manner, it may be shown that the triangles ADC and 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.

2o. AB will be a mean proportional between BC and BD; and AC will be a mean proportional between CB and CD.

For, the triangles ADB and B BAC being similar, their homologous sides are proportional: hence,

D

BC AB
:

:: AB : BD.

In like manner,

BC : AC :: AC DC;

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,

BD : AD :: AD : DC;

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whence, by addition,

AB2 + AC2 = BC (BD + DC);

or,

AB2 + AC2 = BC2 ;

as was shown in Proposition XI.

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

extremities B and C of the diam

diameter : then

B D

eter BC, and a perpendicular AD be drawn to the 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.

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