Question

In figure, $OP$ bisects $\angle BOC$ and $OQ$ bisects $\angle AOC$. Show that $\angle POQ={90}^{\circ }$

Answer 1

According to question, $OP$ is bisector of $\angle BOC$
$\therefore \angle 1=\angle 2$
or $\angle 1=\frac{\angle BOC}{2}$
and $OQ$ is bisector of $\angle AOC$
$\therefore \angle 3=\angle 4$
or $\angle 4=\frac{\angle AOC}{2}$
$\therefore \angle 1+\angle 4=\frac{\angle BOC}{2}+\frac{\angle AOC}{2}$
$=\frac{\angle BOC+\angle AOC}{2}=\frac{180}{2}=90°$