Question

In a triangle $ABC$, the bisectors of $\angle B$ and $\angle C$ intersects each other at a point $O$, Prove that $\angle BOC={90}^o+\frac{1}{2}\angle A$

Answer 1

Let $BM$ and $CN$ are the bisectors of the angles $B$ and $C$ intersecting at $O$.

We know that $\angle A+\angle B+\angle C=180$

$\Rightarrow \angle B+\angle C=180-\angle A...\left(1\right)$

In $\Delta BOC,\angle OBC+\angle OCB+\angle BOC=180$

$\Rightarrow \frac{\angle B}{2}+\frac{\angle C}{2}+\angle BOC=180$

$\Rightarrow \angle BOC=180-\left(\frac{\angle B+\angle C}{2}\right)$

$=180-\left(\frac{180-\angle A}{2}\right)$ using $\left(1\right)$

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

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

Hence proved.