Question

In a parallelogram $ABCD,$ the bisectors $\angle A$ and $\angle B$ meet at $O.$ Find $\angle AOB.$

Answer 1

$ABCD$ is a parallelogram

$\Rightarrow$ $\angle A+\angle B={180}^o$ [ Sum of adjacent angles of parallelogram are supplementary ]

$\Rightarrow$ $\frac{1}{2}\angle A+\frac{1}{2}\angle B=\frac{{180}^o}{2}$

$\Rightarrow$ $\angle OAB+\angle OBA={90}^o$

In $△AOB$

$\Rightarrow$ $\angle OAB+\angle AOB+\angle OBA={180}^o$

$\Rightarrow$ $\angle AOB={180}^o-(\angle OAB+\angle OBA)$

$\Rightarrow$ $\angle AOB={180}^o-(\angle OAB+\angle OBA)$

$\Rightarrow$ $\angle AOB={180}^o-{90}^o$

$\therefore$ $\angle AOB={90}^o$