Question

AB is parallel to CD, EF intersects them at M and N. The bisectors of $\angle BMNand\angle MND$ meet at Q. If $\angle AME={80}^{\circ }$, then $\angle MQN$ is:

• A. ${90}^{\circ }$
• B. ${70}^{\circ }$
• C. ${80}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

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

Given: $\angle AME={80}^{\circ }$
So, $\angle BMF=\angle AME={80}^{\circ }$
($\because$ Vertically opposite angles)
$\therefore \angle QMN=\frac{1}{2}\angle BMF=\frac{1}{2}\times {80}^{\circ }={40}^{\circ }$
($\because$ QM bisects $\angle BMF$)
Also, $\angle CNE=\angle AME={80}^{\circ }$
($\because$ Corresponding angles)
So, $\angle CNE+\angle END={180}^{\circ }$
($\because$ CD is a straight line)
$\Rightarrow {80}^{\circ }+\angle END={180}^{\circ }$
$\Rightarrow \angle END={100}^{\circ }$
$\therefore \angle QNM=\frac{1}{2}\angle END=\frac{1}{2}\times {100}^{\circ }={50}^{\circ }$
($\because$ QN bisects $\angle END$)
In $\Delta QMN$,
$\angle MQN+\angle QMN+\angle QNM={180}^{\circ }$
$\Rightarrow \angle MQN+{40}^{\circ }+{50}^{\circ }={180}^{\circ }$
$\therefore \angle MQN={90}^{\circ }$