Question

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

Answer 1

In parallelogram $ABCD$, $AO$ and $BO$ are bisectors of angle $A$ and angle $B$ respectively.

We know that sum of adjacent angles in a parallelogram is ${180}^o$

$\therefore$ $\angle A+\angle B={180}^o$

$\Rightarrow$ $2x+2y={180}^o$

$\therefore$ $x+y={90}^o$ ---- ( 1 )

In $△AOB,$

$\Rightarrow$ $x+y+\angle AOB={180}^o$ [ Sum of angles in a triangles is ${180}^o$ ]

$\Rightarrow$ ${90}^o+\angle AOB={180}^o$ [ From ( 1 ) ]

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