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DE

K

H

B

Through I, the middle of the side BC, draw KL, parallel to the opposite side AD, and produce DC until it meet KL in K. In the triangles IBL, ICK, the side IB is equal to IC, and the angle B equal to C (74), the angle BIL equal to CIK; therefore the triangles are equal (54), and the side CK equal to BL. Now the parallelogram ALKD is the sum of the polygon ALICD; and the triangle CIK, and the trapezoid ABCD, is the sum of the same polygon and the triangle BIL; therefore, the trapezoid ABCD is equal to the parallelogram ALKD, and has, for its measure, AL EF. Now AL is equal to DK, and BL equal to CK; and CD=DK – CK; but DK is equal to AL, and CK equal to BL;

Therefore, CD=AL-BL.

But...... AB = AL +BL.
Therefore, AB+CD=2AL
Consequently I (AB+CD) = AL.
It follows that .. AL REF=} (AB+CD) EF.

THEOREM 55.

D

152. A straight line, DE, drawn parallel to the side of a triangle ABC, divides the other sides AB, AC, proportionally, or so that AD : DB :: AE : EC.

Join BE and DC: the two triangles BDE, CDE, have the same base, DE, and they have also the same altitude, because BC is parallel to DE; and, consequently, the triangles DBE and DCE are equal. Since the triangles BED and AED have the same altitude, they are to one another as their bases; and since the triangles CED and AED have the same altitude, they are to one another as their bases.

Therefore, the triangle BDE: ADE :: BD: DA. But since the triangle BDE is equal to the triangle CED, therefore the triangle CED : ADE :: BD : DA.

But the triangle CED: ADE :: CE: EA.
It follows that, BD : DA :: CE : EA.

THEOREM 56.

D

153. If the two sides, AB, AC, of a triangle be cut proportionally by the line DE, so that AD : DB :: AE : EC, the line DE shall be parallel to the remaining side of the triangle.

For, if DE be not parallel to BC, some other line, DO, will be parallel to BC; then, by the preceding proposition,

AD: DB :: AO: OC.
And, by hypothesis, AD : DB :: AE : EC.
Therefore, ...... AO: OC :: AE : EC.

And, by addition, AC: OC :: AC : EC.
And hence OC must be equal to EC; which is impossible, unless the point
O coincide with E; therefore no line besides DE can be parallel to BC.

B

THEOREM 57.

154. A line, BD, which bisects any angle, ABC, of a triangle, will divide the opposite side AC into two segments, AD, DC, which shall have the same ratio as the other two sides, AB, BC, of the triangle.

From C, one extremity of the base, draw CE, parallel to BD, meeting AB produced in E. Then the angle ABD is equal to the angle BEC (75), and the angle CBD equal to BCE (74); but, by hypothesis, the angle ABD is equal to CBD; therefore the angle BEC is equal to BCE: hence the side BC is equal to BE (63). Now, because ACE is a triangle, and BD is drawn parallel to one of its sides, AD : DC :: AB : BE (152); but, since BE is equal to BC ; therefore AD : DC :: AB : BC.

B

THEOREM 58.

155. Two equiangular triangles have their sides proportional, and are similar to each other.

D

B

Let ABC, DCE, be two triangles, which have their angles equal, each to each ; viz. BAC equal to CDE, ABC equal to DCE, and ACB equal to DEC; the homologous sides, or the sides adjacent to the equal angles, shall be proportionals; that is, BC : CE :: BA : CD, and BA : CD :: AC : DE.

Place the homologous sides BC, CE, in a straight line; and, because the angles B and E are together less than two right angles, the lines BA and ED shall meet, if produced (76) : let them meet in F. Then, since BCE is a straight line, and the angle BCA equal to E, AC is parallel to EF. In like manner, because the angle DCE is equal to B, the straight line CD is parallel to BF; therefore ACDF is a parallelogram.

156. In the triangle BFE, the straight line AC is parallel to FE; wherefore BC : CE :: BA : AF (152). Again, in the same triangle, BFE, CD is parallel to BF; therefore, BC : CE :: FD : DE; but, by substituting CD for its equal AF, in the first set of proportionals, and AC for its equal FD in the second set, we have....

BC : CE :: BA : CD by the first, and ..

BC : CE :: AC : DE by the second, therefore, by equality, BA : CD :: AC : DE; therefore the homologous sides are proportionals; and, because the triangles are equiangular, they are similar.

SCHOLIUM.—It may be remarked that the homologous sides are opposite to the equal angles.

THEOREM 59.

157. Two triangles, which have their homologous sides proportionals, are equiangular and similar.

Suppose that BC: EF :: AB : DE

and that.... AB : DE :: AC : DF the triangles ABC, DEF, have their angles equal : viz. A equal to D, B equal to E, and C equal to F. At the point E make the angle FEG equal to B, and at the

E

B

point F make the angle EFG equal to C, then G shall be equal to A (80), and the triangles GEF, ABC, shall be equiangular; therefore,

by the preceding prop. BC : EF :: AB : EG and, by hypothesis, .. BC : EF :: AB : DE; therefore EG is equal to DE. In like manner. ..BC : EF :: AC : FG and, by hypothesis, ..BC : EF :: AC : DF; therefore FG is equal to DF.

Thus, it appears that, the triangles DEF, GEF, have their three sides equal, each to each, therefore they are equal (58). But, by construction, the triangle GEF is equiangular to the triangle ABC; therefore, also, the triangles DEF, ABC, are equiangular and similar.

THEOREM 60.

D

H

158. Two triangles which have an angle of the one equal to an angle of the other, and the sides about them proportionals, are similar.

Let the angle A equal D, and suppose that AB : DE :: AC : DF, the triangle ABC is similar to DEF.

Take AG equal to DE, and draw GH parallel to BC, the angle AGH shall be equal to ABC (75), and the triangle AGH equiangular to the triangle ABC; therefore AB : AG :: AC : AH; but AG is equal to DE;

therefore........ AB : DE :: AC : AH,

but, by hypothesis, AB : DE :: AC : DF; therefore AH is equal to DF. The two triangles AGH, DEF, have therefore an angle of the one equal to an angle of the other, and the sides containing these angles equal ; therefore they are equal (53); but the triangle AGH is similar to ABC.

B

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THEOREM 61.

159. A perpendicular, AD, drawn from the right angle, A, of a right-angled triangle, upon the hypothenuse, or longest side, BC, will divide that triangle The triangles BAD and BAC have the common angle B; and, besides, the right angle BDA is equal to the right angle BAC ; therefore, the third angle BAD of the one is equal to the third angle C of the other (80); therefore the B two triangles are equiangular and similar. In like manner it may be demonstrated that the triangle DAC is equiangular and similar to the triangle BAC; therefore the three triangles are equiangular and similar to one another.

THEOREM 62.

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A

B

160. The square described upon the hypothenuse, or longest side, is equal to the squares described upon the other two sides.

From the right angle C draw CD, perpendicular to the hypothenuse AB; then the triangle ABC is divided into two triangles, ADC, CDB, which are similar to one another, and to the whole triangle ABC (159);

therefore, by the similar triangles, ABC, CBD... AB : BC :: BC : BD; again, by the similar triangles, BAC, CAD.... AB : AC :: AC : AD; therefore, reducing the first to an equation, ... AB x BD=BC? and, reducing the second analogy to an equation, AB x AD=AC?

then, adding these two equations . AB (AD +BD)= AC? + BC? but, since AB is equal to the sum of the two lines AD, DB, therefore AB2= AC? + BC?,

THEOREM 63.

161. Two triangles, which have an angle of the one equal to an angle of the other, are to each other as the rectangle of the sides about the equal angles.

A

D

B

Suppose the two triangles joined, so as to have a common angle, and let the two triangles be ABC, ADE. Draw the straight line BE.

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