Question

Question 155
In parallelogram MODE, the bisectors of $\angle M$ and $\angle O$ meet at Q. Find the measure of $\angle MQO$

Answer 1

Let MODE be a parallelogram and Q be the point of intersection of the bisector of $\angle M$ and $\angle O$

Since, MODE is a parallelogram,
$\therefore \angle EMO+\angle DOM={180}^{\circ }$ [$\because$ adjacent angles are supplementary]
$\frac{1}{2}\angle EMO+\frac{1}{2}\angle DOM={90}^{\circ }$ [dividing both sides by 2]
$\Rightarrow \angle QMO+\angle QOM={90}^{\circ }$ ........ (i)
Now, in $\Delta MOQ$,
$\angle QOM+\angle QMO+\angle MQO={180}^{\circ }$ [angle sum property of triangle]
$\Rightarrow {90}^{\circ }+\angle MQO={180}^{\circ }$ [from Eq. (i)]
$\therefore \angle MQO={180}^{\circ }-{90}^{\circ }={90}^{\circ }$