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.