Question

In parallelogram ABCD, bisectors of angles A and B intersect at O. Find the value of $\angle$AOB.

Answer 1

$\begin{array}{c}In\Delta \\ \Rightarrow \angle AOB+\frac{1}{2}\angle A+\frac{1}{2}\angle B={180}^{\circ }\to \left(i\right)\\ \Rightarrow \angle A+\angle B={180}^{\circ }\{propertyofparallelogram\to \left(ii\right)\\ \Rightarrow \angle AOB+\frac{1}{2}\left\{180\right\}={180}^{\circ }\{from\left(i\right)\&\left(ii\right)\\ \angle AOB={90}^{\circ }\end{array}$