Question

In a $\Delta \angle ABC,\angle ACB$ and the bisectors of $\angle ABC$ and $\angle ACB$ at $O$ such that $\angle BOC={120}^0$
show that $\angle A=\angle B=\angle C={60}^0$

Answer 1

Given

In $\Delta ABC$

$\angle ABC=\angle ACB$

Divide both sides by $2$

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

$\Rightarrow \angle OBC=\angle OCB$

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

$120-90=\frac{1}{2}\angle A$

$30=\frac{1}{2}\angle A$

$\angle A=60$

$\angle A+\angle ABC+\angle ACB={180}^{\circ }$

$2\angle ABC=120$

$\angle ABC=60$

$\angle ACB=60$

$\angle A=\angle B=\angle C={60}^{\circ }$

Hence Proved