Question

In the given figure, $ABCD$ is the parallelogram, $AO$ and $BO$ are the angle bisectors of $\angle A$ and $\angle B$.

find $\angle AOB$.

• A. Greater than ${90}^o$
• B. Less than ${90}^o$
• C. ${90}^o$
• D. ${60}^o$

Answer 1

The correct option is C ${90}^o$


As $AO$ and $BO$ are the angle bisectors of $\angle A$ and $\angle B$
Also $\angle DAB+\angle ABC={180}^o\text{(Co interior angles)}$
$\angle 1+\angle 1+\angle 2+\angle 2={180}^o\Rightarrow 2(\angle 1+\angle 2)={180}^o\Rightarrow \angle 1+\angle 2=\frac{{180}^o}{2}\Rightarrow \angle 1+\angle 2={90}^o$

Now, in $△AOB$
Sum of all interior angles of a triangle is ${180}^o$.
So, $\angle 1+\angle 2+\angle AOB={180}^o$
${90}^o+\angle AOB={180}^o$
$\angle AOB={180}^o-{90}^o$
$\angle AOB={90}^o$

Hence, option $\left(c\right)$ correct.