Question
In the figure, given below, $AM$ bisects angle $A$ and $DM$ bisects angle $D$ of parallelogram $ABCD.$ Prove that: $\angle A D=90^{\circ}$.
✓
Answer
From the given figure we conclude that
$\angle A +\angle D =180^{\circ} \ldots[$ Since consecutive angles are supplementary $]$
$\frac{\angle A}{2}+\frac{\angle D}{2}=90^{\circ}$
Again from the $\triangle A D M$
$\frac{\angle A}{2}+\frac{\angle D}{2}+\angle M =180^{\circ}$
$\Rightarrow 90^{\circ}+\angle M =180^{\circ} \ldots\left[\sin c e \frac{\angle A}{2}+\frac{\angle D}{2}=90^{\circ}\right]$
$\Rightarrow \angle M =90^{\circ}$
Hence $\angle AMD =90^{\circ}$
$\angle A +\angle D =180^{\circ} \ldots[$ Since consecutive angles are supplementary $]$
$\frac{\angle A}{2}+\frac{\angle D}{2}=90^{\circ}$
Again from the $\triangle A D M$
$\frac{\angle A}{2}+\frac{\angle D}{2}+\angle M =180^{\circ}$
$\Rightarrow 90^{\circ}+\angle M =180^{\circ} \ldots\left[\sin c e \frac{\angle A}{2}+\frac{\angle D}{2}=90^{\circ}\right]$
$\Rightarrow \angle M =90^{\circ}$
Hence $\angle AMD =90^{\circ}$
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