But the triangles BDE, DEC are equivalent; therefore,

we have, (11. 3,)

AD: DB:: AE : EC. Hence,

If a line be drawn parallel to the base of any triangle, it will divide the other sides proportionally.

Cor. 1. Hence by composition, (16. 3.) we have AD+ DB : AD :: AE+EC: AE, or AB: AD :: AC: AE; and also AB : BD :: AC : CE.

Cor. 2. If between two straight lines AB, CD, any number of parallels AC, EF, GH, BD, &c., be drawn, those straight lines will be cut proportionally, and we shall have AE CF : EG: FH :: GB: HD.

OF, those two proportions give AE: CF :: EG: FH. (11. 3.) It may be proved in the same manner, that EG: FH :: GB: HD, and so on; hence the lines AB, CD are cut proportionally by the parallels AC, EF, GH, &c. Therefore, If a line, &c.

Conversely, if the sides AB, AC, are cut proportionally by the line DE, so that we have AD: DB:: AE: EC, the line DE will be parallel to the base BC.

For if DE is not parallel to BC, suppose that DO is. Then, by the preceding theorem we shall have AD: BD :: AO: OC.

But

A

by hypothesis, we have AD: DB:: AE: EC; DA hence we must have AO: OC :: AE: EC, or AO: AE :: OC: EC; an impossible result, B since AO, the one antecedent, is less than its consequent AE, and OC, the other antecedent, is greater than its consequent EC. Hence the parallel to BC, drawn from the point D, cannot differ from DE; therefore DE is that parallel. Hence,

Conversely, if the sides of a triangle be cut proportionally by a straight line, this line will be parallel to the base or third side.

Scholium. The same conclusion would be true, if the proportion AB : AD: : AC: AE, were the proposed one. For this proportion would give us, AB-AD: AD: : AC-AE: AE, or BD: AD::CE: AE.

The line AD, which bisects the angle BAC of a triangle, divides the base BC into two segments BD, DC proportional to the adjacent sides, AB, AC; so that we have

But the triangle ACE is isosceles: for since AD, CE are parallel, we have the angle ACE DAC, and the angle AEC=BAD; (24. 1;) but by hypothesis DAC=BAD; hence the angle ACE AEC, and consequently AE=AC. (12. 1.) In place of AE in the above proportion, substitute AC, and we shall have BD: DC :: AB: AC. Therefore,

The line which bisects any angle of a triangle, divides the base into two segments, which have the same ratio as the adjacent sides have to each other.

Two equiangular triangles ABC, CDE have their homologous sides proportional, and are similar.

In the given triangles, let the angle BAC CDE, ABC=DCE and ACB DEC; then the homologous sides, or the sides adjacent to the equal angles, will be proportional; that is,

B

C

E

BC: CE:AB: CD :: AC: DE. Place the homologous sides BC, CE in the same straight line; and produce the sides BA, ED till they meet in F.

Since BCE is a straight line, and the angle BCA is equal to CED, it follows that AC is parallel to DE. (21. 1.) In like manner, since the angle ABC is equal to DCE, the line AB is parallel to DC. Hence the figure ACDF is a parallel

ogram.

In the triangle BFE, the line AC is parallel to the base FE; hence, (15.4,) BC: CE:: BA: AF; or, putting CD in the place of its equal AF,

BC: CE: BA: CD.

In the same triangle BEF, BF may be considered as the

base, CD is parallel to it; then we have the proportion BC: CE: FD: DE;

and putting AC in the place of its equal FD,

BC CE: AC: DE.

And finally, since both these proportions contain the same ratio BC CE, we have, (11. 3,)

:

AC DE : BA: CD.

Thus the equiangular triangles BAC, CDE have their homologous sides proportional. But two figures are similar when they have their angles respectively equal, and their homologous sides proportional; (Def. 3. 4;) consequently the equiangular triangles BAC, CDE, are similar. Hence,

Any two equiangular triangles have their homologous sides proportional, and are similar.

Cor. For the similarity of two triangles, it is enough that they have two angles equal each to each; since the third will also be equal in both, and the two triangles will be equiangular.

Scholium. It should be observed that, in similar triangles, the homologous sides are opposite to the equal angles; thus the angle ACB being equal to DEC, the side AB is homologous to DC; in like manner, AC and DE are homologous, because opposite to the equal angles ABC, DCE. When the homologous sides are determined, it is easy to form the proportions: AB: BC: AC: DE:: BC: CE.

PROPOSITION XIX. THEOREM.

Two triangles, ABC, DEF, which have their homologous sides proportional, are equiangular and similar.

At the point E, make the angle FEG=B, and at F, the angle EFG C; the third G will be equal to the third A, and the two triangles ABC, EFG will be equiangular. Therefore, by the last proposition, we shall have

BC EF: : AB: EG;

but by hypothesis, we have BC EF:: AB : DE; hence EG DE. By the same proposition, we also have BC: EF :: AC: FG; and by hypothesis, we have BC : EF : : AC: DF; hence FG-DF. Therefore the triangles EGF, DEF, having their three sides respectively equal, are themselves equal. (10. 1.) But by construction, the triangles EGF and ABC are equiangular: therefore DEF and ABC are also equiangular and similar. Hence,

Any two triangles which have their homologous sides proportional, are equiangular and similar.

D

A

E

B

Scholium. By the last two propositions, it appears that in triangles, equality among the angles is a consequence of proportionality among the sides, and conversely; so that one of these conditions sufficiently determines the similarity of two triangles. But in quadrilaterals, the proportion between the sides may be altered without altering the angles, or the angles be altered without altering the proportion between the sides; and thus proportionality among the sides cannot be a consequence of equality among the angles of two quadrilaterals, or vice versa. For example, by drawing EF parallel to BC, the angles of the quadrilateral AEFD, are made equal to those of ABCD, though the proportion between the sides is different; and, in like manner, without changing the four sides AB, BC, CD, AD, we can make the point B approach D or recede from it, which will change the angles.