Question

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

Answer 1

Given : In$△ABC$ ,BO and CO are the bisectors of $\angle B$ and $\angle C$ respectively and $\angle BOC={120}^{\circ }$, $\angle ABC=\angle ACB$

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

Proof: $\angle BOC={90}^{\circ }+\frac{1}{2}\angle A$
$\text{B}ut\angle BOC={120}^{\circ }$
$\therefore {90}^{\circ }+\frac{1}{2}\angle A={120}^{\circ }$
$\Rightarrow \frac{1}{2}\angle A={120}^{\circ }-{90}^{\circ }={30}^{\circ }$
$\therefore \angle A={60}^{\circ }$
$\because \angle A+\angle B+\angle C={180}^{\circ }$
(Angles of a triangle)
$\therefore \angle B+\angle C={180}^{\circ }-{60}^{\circ }={120}^{\circ }$
$\text{a}nd\angle B=\angle C$
$\therefore \angle B=\angle C=\frac{{120}^{\circ }}{2}={60}^{\circ }$
$\angle A=\angle B=\angle C={60}^{\circ }$