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Proposition 2.

Theorem. If a straight line be drawn parallel to one of the sides of a triangle, it will cut the other sides, or those sides produced proportionally; and if the sides of a triangle, or the sides produced, be cut proportionally, the straight line which joins the points of section will be parallel to the other side of the triangle.

Let DE be drawn parallel to BC, one of the sides of the triangle ABC';

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Join BE, CD; then the triangle BDE is equal to the tri- · angle CDE, because they are on the same base DE, and between the same parallels, DE, BC; and ADE is another triangle; and equal quantities have (IV. 14) the same ratio to the same quantity.

BDE: ADE :: CDE: ADE.

But (V. 1, Cor. 2)

BDE: ADE :: BD : DA,

the common altitude being the perpendicular drawn from E to AB.

Similarly,

CDE ADE:: CE: EA;

... (IV. 15), BD : DA :: CE: EA.

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also,

CE: EA :: CDE: ADE;

... (IV.15), BDE: ADE :: CDE: ADE.

Therefore BDE and CDE have the same ratio to ADE;

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Hence they are between the same parallels (I. 37); that is, DE is parallel to BC.

Corollary.—(See first figure.)

If

AD: DB:: AE: EC,

by composition, AD+DB : AD :: AE+EC : AE,

or,

AB AD: AC: AE.

:

Proposition 3.

Theorem.-If the vertical angle of a triangle be bisected by a straight line which also cuts the base, the segments of the base will have the same ratio, which the other sides of the triangle have to each other.

Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD, which meets the base at D; then

BD: DC :: BA : AC.

Through C draw CE parallel to
AD, meeting BA produced in E.
Then, because AC meets the
parallel lines AD, CE, the angles
ACE, DAC are equal (I. 27). B
But the angle CAD (Hyp.) is

D

E

equal to BAD; therefore ACE is equal to BAD. Again, because BAE meets the parallel lines AD, EC, the angles BAD, AEC are equal; but BAD was shown equal ACE, therefore the angles AEC, ACE are equal, and the side AC is equal (I. 14) to the side AE.

Now, because EC is drawn parallel to AD,

(V. 2),

But

BD: DC :: BA : AE.

AE-AC;

.. (IV. 14), BD: DC:: BA: AC.

Proposition 4.

Theorem.-The sides about the equal angles of mutually equiangular triangles are proportional; and the homologous sides are the antecedents or consequents of the ratios.

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Let the triangle DEF be so placed that the angle D will be at A, and the line DE on the line AB; and because the angle D is equal to the angle A, the line DF will fall on AC; E will be at some point in AB, as G, and F at some point in AC, as H; join GH; and because the angle AGH is equal to the angle ABC, the side GH is parallel (I. 26) to the side BC.

... (V. 2, Cor.), AB: AG :: AC: AH; alternately (IV.7), AB : AC:: AG: AH, AB AC: DE: DF.

or,

The antecedents AB and DE are opposite the equal angles C and F, and are therefore homologous, as also the consequents AC and DF.

In the same manner we may prove the sides about any other angle proportional.

Corollary.-Hence all mutually equiangular triangles are

similar.

Proposition 5.

Theorem.-If the sides of two triangles about each of their angles be proportional, the triangles will be similar.

Let the triangles ABC, DEF have their sides proportional, so that AB: DE :: BC: EF :: AC: DF;

then will the triangles be similar.

At the points E and F in the straight line EF, make (I. 13) the angles FEG, EFG equal respectively to the angles ABC, ACB; then will the remaining angle G be equal (I. 30, Cor. 4) to the remaining angle A.

Wherefore the triangles ABC, GEF are equiangular, and (V. 4),

but (Hyp.)

... (IV. 15),

.'. (IV. 14),

Similarly,

:

B

AB BC: GE: EF;
AB: BC:: DE : EF;

GE: EF:: DE: EF;

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E

D

G

And because the triangles DEF, GEF have one side, EF, common, and the other sides equal, each to each, the angles of one are equal (I. 9) to the angles of the other. But the angles of the triangle GEF are, by construction, equal to the angles of the triangle ABC. Therefore ABC, DEF are also equiangular, and therefore (V. 4) similar.

Proposition 6.

Theorem.-If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportional, the triangles will be similar.

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on AC. Let E fall at G, then will F fall at some point in

AC, as H.

And because

AB AC DE : DF,

AB AC: AG: AH.

Alternately (IV. 7), AB : AG :: AC: AH.

By division (IV. 10), BG: AG: CH: AH.

.. (V. 2), BC is parallel to GH.

Therefore the angles of the triangle A GH are equal (I. 27) to the angles of the triangle ABC. Hence ABC, DEF are equiangular, and therefore (V. 4) similar.

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