If two lines be cut by three or more parallels, the intercepts on one are proportional to the corresponding intercepts on the other. PROP. III.—Theorem. If a line (AD) bisect any angle (A) of a triangle (ABC), it divides the opposite side (BC) into segments proportional to the adjacent sides. Conversely, If the segments (BD, DC) into which a line (AD) drawn from any angle (A) of a triangle divides the opposite side be proportional to the adjacent sides, that line bisects the angle (A). Dem.—1. Through C draw CE parallel to AD, to meet BA produced in E. Because BA meets the parallels AD, EC, the angle BAD [I. xxix.] is equal to AEC; and because AC meets the parallels AD, EC, the angle DAC is equal to ACE; but the angle BAD is equal to DAC (hyp.); therefore the angle ACE is equal to AEC; therefore AE is equal to AC [I. vi.]. Again, because AD is parallel to EC, one of the sides of the triangle BEC, BD : DC :: BA : AE [ii.]; but AE has been proved equal to AC. Therefore BD : DC :: BA : AC. 2. If BD : DC :: BA : AC, the angle BAC is bisected. Dem.—Let the same construction be made. Because AD is parallel to EC, BA : AE :: BD : DC [ii.]; but BD : DC :: BA : AC (hyp.). Therefore [V. xi.] BA : AE :: BA : AC, and hence [V. ix.] AE is equal to AC; therefore the angle AEC is equal to ACE; but AEC is equal to BAD [I. xxix.], and ACE to DAC; hence BAD is equal to DAC, and the line AD bisects the angle BAC. Exercises. 1. If the line AD bisect the external vertical angle CAE, BA : AC :: BD : DC, and conversely. Dem.—Cut off AE = AC. Join ED. Then the triangles ACD, AED are evidently congruent; therefore the angle EDB is bisected; hence [iii.] BA : AE :: BD : DE; or BA : AC :: BD : DC. 2. Exercise 1 has been proved by quoting Proposition iii. Prove it independently, and prove iii. as an inference from it. 3. The internal and the external bisectors of the vertical angle of a triangle divide the base harmonically (see Definition, p. 191). 4. Any line intersecting the legs of any angle is cut harmonically by the internal and external bisectors of the angle. 5. Any line intersecting the legs of a right angle is cut harmonically by any two lines through its vertex which make equal angles with either of its sides. 6. If the base of a triangle be given in magnitude and position, and the ratio of the sides, the locus of the vertex is a circle which divides the base harmonically in the ratio of the sides. 7. If a, b, c denote the sides of a triangle ABC, and D, D′ the points where the internal and external bisectors of A meet BC; prove 8. In the same case, if E, E′, F, F′ be points similarly determined on the sides CA, AB, respectively; prove + + = 0, and + + = 0. PROP. IV.—Theorem.