In the adjoining figure, AB = AC, D is a point in the interior of △ABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC of △ABC.

Given, AB = AC.

∴ ∠ABC = ∠ACB (As angles opposite to equal sides of isosceles triangle are equal.)

Given, ∠DBC = ∠DCB

∴ DB = DC. (Sides opposite to equal angles are equal.)

In △ABD and △ACD,

AB = AC (Given)

DB = DC (Proved)

AD = AD (Common)

Thus, △ABD ≅ △ACD by SSS axiom.

We know that corresponding parts of congruent triangle are equal.

∴ ∠BAD = ∠CAD.

Hence, proved that AD bisects ∠BAC.