Bisector of Apex of Isosceles Triangle also Bisects Base
Then $AD$ bisects $BC$.
By definition of isosceles triangle, $AB = AC$.
By definition of bisector, $\angle BAD = \angle CAD$.
By construction, $AD$ is common.
Thus by Triangle Side-Angle-Side Equality, $\triangle ABD = \triangle ACD$.
Thus $AD = DC$.
The result follows by definition of bisection.