The bisectors of ∠B and ∠C of an isosceles ΔABC with AB = AC intersect each other at a point O.

Given: In an isosceles ΔABC, AB = AC, BO and CO are the bisectors of ∠ABC and ∠ACB, respectively.

To prove: ∠ABD = ∠BOC

Construction: Produce CB to point D.

Proof:

In ΔABC,

∵">∵∵ AB = AC                    (Given)
∴">∴∴ ∠ACB = ∠ABC            (Angle opposite to equal sides are equal)

$\Rightarrow \frac{1}{2} \angle A C B=\frac{1}{2} \angle A B C$

$\Rightarrow \angle O C B=\angle O B C \quad \ldots$. (i)

(Given, $B O$ and $C O$ are angle bisector of $\angle A B C$ and $\angle A C B$, respectively)

$\ln \triangle B O C$

$\angle O B C+\angle O C B+\angle B O C=180^{\circ} \quad$ (By angle sum property of triangle)

$\Rightarrow \angle O B C+\angle O B C+\angle B O C=180^{\circ} \quad[$ From (i) $]$

$\Rightarrow 2 \angle O B C+\angle B O C=180^{\circ}$

$\Rightarrow \angle A B C+\angle B O C=180^{\circ} \quad(B O$ is the angle bisector of $\angle A B C) \quad \ldots \ldots$ (ii)

Also, DBC is a straight line.

So, $\angle A B C+\angle D B A=180^{\circ} \quad$ (Linear pair) $\ldots$ (iii)

From (ii) and (iii), we get

$\angle A B C+\angle B O C=\angle A B C+\angle D B A$

$\therefore \angle B O C=\angle D B A$