Question

In a parallelogram $PQRS$, the bisectors of $\angle P$ and $\angle Q$ meet at $O$. Find $\angle POQ$.

Answer 1

Given that:

$\angle SPO=\angle OPQ$ and $\angle RQO=\angle OQP.$

$\angle P+\angle Q={180}^{\circ }$ [$SincePS\parallel QR$]

$\frac{1}{2}(\angle P+\angle Q)=\frac{{180}^{\circ }}{2}$

$\angle OPQ+\angle OQP={90}^{\circ }$ ......(1)

Now,

In $△PQO,$

$\angle OPQ+\angle OQP+\angle POQ={180}^{\circ }$

$\angle POQ={90}^{\circ }$ [From equation (1)]

$\angle O={90}^{\circ }.$