In the figure given below, ABC is an isosceles triangle in which AB=AC. Also D is a point such that BD=CD. How do I prove that AD bisects angle A & angle D?

This is what I have done:

Consider the triangles ABD & ACD.

AB=AC (given)

BD=CD (given)

AD=DA (common side)

Thus the triangles ABD & ACD are congruent by Side-Side-Side (SSS) congruence rule but how do I prove that AD bisects angle A & angle D?

Thanks,

Ron