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the side DF will take the position GH, and BGH will

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or, since BG is equal to ED, and BH to EF,

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hence, the sides about the equal angles, taken in the same order, are proportional; and consequently, the triangles are similar (D. 1); which was to be proved.

Cor. If two triangles have two angles in one, equal to two angles in the other, each to each, they will be similar (B. I., P. XXV., C. 2).

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Triangles which have their corresponding sides proportional, are similar.

In the triangles ABC and DEF, let the corresponding sides be proportional; that is, let

AB : DE :: BC : EF :: :

CA: FD;

then will the triangles be similar.

For, on BA lay off BG equal to ED; on BC lay

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hence, GH is parallel to AC (P. XVI.); and consequently, the triangles BAC and BGH are equiangular, and therefore similar hence, from Prop. XVIII., we have,

BC : BH :: CA : HG.

But, by hypothesis,

Ᏼ Ꮯ : EF :: CA : FD;

hence (B. II., P. IV., C.), we have,

BH : EF :: HG : FD.

But, BH is equal to EF; hence, HG is equal to FD. The triangles BHG and EFD have, therefore, their sides equal, each to each, and consequently, they are equal in all their parts. Now, it has just been shown that BHG and BCA are similar: hence, EFD and BCA are also similar; which was to be proved.

Scholium. In order that polygons may be similar, they must fulfill two conditions: they must be mutually equiangular, and the corresponding sides must be proportional. In the case of triangles, either of these conditions involves the other, which is not true of any other species of polygons.

PROPOSITION XX. THEOREM.

Triangles which have an angle in each equal, and the including sides proportional, are similar.

In the triangles ABC and DEF, let the angle B be equal to the angle E; and suppose that

BA : ED : BC : EF;

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Ꮐ ; DF will take the position GH, and the triangle DEF will coincide with GBH, and consequently, will be equal to it.

But, from the assumed proportion, and because BG is equal to ED, and BH to EF we have,

BA : BG :: BC : BH ;

hence, GH is parallel to AC; and consequently, BAC and BGH are equiangular, and therefore similar. But, EDF is equal to BGH: hence, it is also similar to BAC; which was to be proved.

PROPOSITION XXI. THEOREM.

Triangles which have their sides parallel, each to each, or perpendicular, each to each, are similar.

1o. Let the triangles ABC and DEF have the side AB parallel to DE, BC to EF, and CA to FD:

then will they be similar.

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

the angle C is
the angle F, and the an-

gle A to the angle D;
the triangles are, therefore,
mutually equiangular, and

B

A

and BC (B. I., P.

D

E

F

consequently, are similar (P. XVIII.); which was to be

proved.

have the side

BC to EF,

and CA to

2o. Let the triangles ABC and DEF AB perpendicular to DE,

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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 GCH 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 perpendicular, 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 line be drawn parallel to the base of a triangle, and 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, lines drawn from A to points of the base: then will

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