Question

If ABCD is a parallelogram, then the angle bisectors of $\angle A$ and $\angle D$ meet at ‘O’. The measure of $\angle AOD$ is ${45}^{\circ }$​

• A.
• B.

Answer 1

The correct option is B

Let$\angle A=x^{\circ }$
then $\angle D=(180-x{)}^{\circ }$
(Since the adjacent angles of a parallelogram sum up to ${180}^{\circ })$

$\therefore \frac{\angle A}{2}=\frac{x}{2}=\angle DAO$
and$\frac{\angle D}{2}=\frac{(180-x)}{2}=90-\frac{x}{2}=\angle ADO$

In $\Delta AOD,$

$\angle AOD+\angle ADO+\angle DAO={180}^{\circ }$ [Angle sum property of a triangle]
$\angle AOD={180}^{\circ }-[\angle ADO+\angle DAO]$
$\angle AOD={180}^{\circ }-[\frac{x}{2}+90-\frac{x}{2}]$
$or\angle AOD=180-90={90}^{\circ }$
$\therefore$ The measure of $\angle AOD$ is ${90}^{\circ }$