Question

If the bisectors of the angles $\angle ABC$ and $\angle ACB$ of a triangle ABC meet at a point O, then Prove that $\angle BOC={90}^o+\frac{1}{2}\angle BAC$.

Answer 1

Given:
A triangle ABC and the bisectors of $\angle ABC$ and $\angle ACB$ meeting at point O.

To prove:
$\angle BOC={90}^o+\frac{1}{2}\angle BAC$.

Proof :
In triangle BOC we have

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

In triangle ABC, we have $\angle A+\angle B+\angle C={180}^o$. Since BO and CO are bisectors of $\angle ABC$ and $\angle ACB$ respectively.

We have
$\angle B=2\angle 1$ and $\angle C=2\angle 2$.

We therefore get $\angle A+2\left(\angle 1\right)+2\left(\angle 2\right)={180}^o$. Dividing by 2, we get $\frac{\angle A}{2}+\angle 1+\angle 2={90}^o$. This gives
$\angle 1+\angle 2={90}^o-\frac{\angle A}{2}$ (2)

From (1) and (2), we get

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

$\therefore \angle BOC={90}^o+\frac{1}{2}\angle BAC$. $\left[henceproved\right]$