Question

In a parallelogram $PQRS$ of the given figure, the bisectors of $\angle P$ and $\angle Q$ meet $SR$ at $O$. Show that $\angle POQ={90}^o$

Answer 1

$PQRS$ is a parallelogram.

$PO$ is angle bisector of $\angle P$

$\therefore$ $\angle SPO=\angle OPQ$ --- ( 1 )

$QO$ is an angle bisector of $\angle Q$

$\therefore$ $\angle RQO=\angle OQP$ ---- ( 2 )

$\therefore$ $PS\parallel QR$

$\Rightarrow$ $\angle SPQ+\angle PQR={180}^o$ [ Sum of adjacent angles are supplementary ]

$\Rightarrow$ $\angle SPO+\angle OPQ+\angle OQP+\angle OQR={180}^o$

$\Rightarrow$ $2\angle OPQ+2\angle OQP={180}^o$ [ From ( 1 ) and ( 2 ) ]

$\Rightarrow$ $\angle OPQ+\angle OQP={90}^o$ ---- ( 3 )

Now, in $△POQ,$

$\Rightarrow$ $\angle OPQ+\angle OQP+\angle POQ={180}^o.$

$\Rightarrow$ ${90}^o+\angle POQ={180}^o$ [ From ( 3 ) ]

$\Rightarrow$ $\angle POQ={90}^o.$