Question

The bisector of $\angle B$ and $\angle C$ of triangle ABC intersect each other at the point O. Prove that $\angle BOC={90}^0+\frac{1}{2}\angle A$

Answer 1

Given :

A $△$ ABC such that the bisectors of $\angle$ ABC and $\angle$ ACB meet at a point O.

To prove :

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

Proof :

In $△$ BOC, we have

$\angle 1+\angle 2+\angle BOC={180}^o$ ....(1)

In $△$ ABC, we have,

$\angle A+\angle B+\angle C={180}^o$

$\angle A+2\left(\angle 1\right)+2\left(\angle 2\right)={180}^o$

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

$\angle 1+\angle 2={90}^o-\frac{\angle A}{2}$

Therefore, in equation 1,

${90}^o-\frac{\angle A}{2}+\angle BOC={180}^o$

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