Question

In parallelogram $ABCD$, the bisectors of $\angle A$ and $\angle B$ intersect at $M$.If $\angle A={80}^o$, then $\angle AMB=$.........

• A. ${40}^o$
• B. ${50}^o$
• C. ${80}^o$
• D. ${90}^o$

Answer 1

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

Given in parallelogram $ABCD,AM$ and $BM$ are bisectors of $\angle A$ and $\angle B$ respectively.

As we know that sum of adjacent angles in a parallelogram is $180°$.

$\therefore \angle A+\angle B=180°$

$\Rightarrow 2\angle BAM+2\angle ABM=180°$

$\Rightarrow \angle BAM+\angle ABM=\frac{180°}{2}=90°.....\left(1\right)$

Now, in $△AMB$,

$\angle BAM+\angle AMB+\angle ABM=180°$

$\Rightarrow \angle AMB+90°=180°$

$\Rightarrow \angle AMB=180°-90°=90°$