Constructions connected with Euclid VI., 3 and A, and the Circle of Apollonius

Extract

Let ABC be a triangle and AD the bisector of the angle A meeting BC in D. (Fig. 8.)

For simplicity let us suppose AB > AC.

From AB cut off AE = AC, and join DE. Then by Euclid I. 4 it follows that the triangles AED, ACD are equal.

This proof of VI. 3 has the merit of depending only on I. 4 and VI. 1. A similar proof for VI. A can be got by cutting off AE′ = AC from BA produced.