In a ΔABC, ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that

Given,

In ΔABC,

∠ABC = ∠ACB

Dividing both sides by '2'

$\frac{1}{2} \angle \mathrm{ABC}=\frac{1}{2} \angle \mathrm{ACB}$

⇒ ∠OBC = ∠OCB [ ∴ OB, OC bisects ∠B and ∠C]

Now,

$\angle \mathrm{BOC}=90^{\circ}+\frac{1}{2} \angle \mathrm{A}$

$\Rightarrow 120^{\circ}-90^{\circ}=\frac{1}{2} \angle \mathrm{A}$

⇒ 30°∗ (2) = ∠A

⇒ ∠A = 60°

Now in ΔABC

∠A + ∠ABC + ∠ACB = 180° (Sum of all angles of a triangle)

⇒ 60° + 2∠ABC = 180° [∴ ∠ABC = ∠ACB]

⇒ 2∠ABC = 180° − 60°

⇒ ∠ABC = 120°/2 = 60°

⇒ ∠ABC = ∠ACB

∴ ∠ACB = 60°

Hence Proved.