Question

ABCD is a parallelogram. The angle bisectors of $\angle Aand\angle D$ meet at O. Find the measure of $\angle AOD.$

• A. ${120}^{\circ }$
• B. ${35}^{\circ }$
• C. ${40}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is D ${90}^{\circ }$


Let $\angle BAD=x^{\circ }$
$\Rightarrow$ $\angle ADC={180}^{\circ }-x^{\circ }$
[Sum of adjacent angles of a parallelogram is ${180}^{\circ }$]
$\therefore \frac{\angle BAD}{2}=\frac{x}{2}=\angle OAD$ and
$\frac{\angle ADC}{2}=90-\frac{x}{2}=\angle ADO$
[AO and DO are angle bisectors]
$\therefore \angle AOD=180-(\frac{x}{2}+90-\frac{x}{2})={90}^{\circ }$