Given :
Δ ABC is an isosceles with AB = AC.
Draw AP ⊥ BC,
∴ ∠APB = ∠APC = 90°

In Δ APB and Δ APC,
⇒ ∠APB = ∠APC (Each equal to 90°)
⇒ AB = AC (Since ABC is an isosceles triangle)
⇒ AP = AP (Common)
∴ Δ APB ≅ Δ APC (By R.H.S. congruence rule)
We know that,
Corresponding parts of congruent triangle are equal.
∴ ∠B = ∠C (By C.P.C.T.)
Hence, proved that ∠B = ∠C.