Question

In the figure given alongside, the bisector of the angles $\angle B$ and $\angle Cof\Delta ABC$ meet at $O$. Prove that $\angle BOC=90+\frac{1}{2}\angle A$

Answer 1

In $\Delta ABC$, by angle sum property we have

$2x+2y+\angle A=180$°

$\Rightarrow x+y+\frac{\angle A}{2}=90$°

$\Rightarrow x+y=90°–\frac{\angle A}{2}$

In $\Delta BOC$, we have

$x+y+\angle BOC=180$°

$90°–\frac{\angle A}{2}+\angle BOC=180$°

$\angle BOC=180°–90°+\frac{\angle A}{2}$

$\angle BOC=90°+\frac{\angle A}{2}$