Bisector of Apex of Isosceles Triangle also Bisects Base

Theorem
Let $\triangle ABC$ be an isosceles triangle whose apex is $A$.

Let $AD$ be the bisector of $\angle BAC$ such that $AD$ intersects $BC$ at $D$.

Then $AD$ bisects $BC$.

Proof
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.