# 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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Updated on: 10-Oct-2022

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