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In a $\triangle ABC, \angle ABC = \angle ACB$ and the bisectors of $\angle ABC$ and $\angle ACB$ intersect at $O$ such that $\angle BOC = 120^o$. Show that $\angle A = \angle B = \angle C = 60^o$.
Given:
In a $\triangle ABC, \angle ABC = \angle ACB$ and the bisectors of $\angle ABC$ and $\angle ACB$ intersect at $O$ such that $\angle BOC = 120^o$.
To do:
We have to show that $\angle A = \angle B = \angle C = 60^o$.
Solution:
$\angle A+\angle B+\angle C=180^{\circ}$
Dividing both sides by 2, we get,
$\frac{1}{2} \angle A+\frac{1}{2} \angle B+\frac{1}{2} \angle C=180^{\circ}$
$\frac{1}{2} \angle A+\angle O B C+\angle O B C=90^{\circ}$
$\angle O B C+\angle O C B=90^{\circ}-\frac{1}{2}A$
In $\triangle B O C$,
$\angle B O C+\angle O B C+\angle O C B=180^{\circ}$
$\angle B O C+90^{\circ}-\frac{1}{2} \angle A=180^{\circ}$
$\angle B O C=90^{\circ}+\frac{1}{2} \angle A$
This implies,
$90^o+ \frac{1}{2} \angle A = 120^o$
$\frac{1}{2} \angle A = 120^o - 90^o = 30^o$
$\angle A = 60^o$
$\angle B + \angle C = 180^o - 60^o = 120^o$
$\angle C = \angle B = 60^o$
Hence, $\angle A = \angle B = \angle C = 60^o$.
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