MCQ
- A$30^{\circ}$
- B$45^{\circ}$
- C$60^{\circ}$
- ✓$90^{\circ}$
✓
Answer
Correct option: D.
$90^{\circ}$
(d) $90^{\circ}$
ABCD is a parallelogram.
$\therefore \quad \angle B=\angle D \Rightarrow \frac{1}{2} \angle B=\frac{1}{2} \angle D \Rightarrow \angle A BD=\angle A D B$
$\Rightarrow \quad A B=A D \quad$ [Sides opposite to equal angles in $\triangle A B D$ ]
$\Rightarrow \quad \triangle A B D$ is isosceles
$\Rightarrow \quad A O \perp B D \quad[\because O$ is the mid-point of $B D]$
$\Rightarrow \quad \angle A O B=90^{\circ}$
ABCD is a parallelogram.
$\therefore \quad \angle B=\angle D \Rightarrow \frac{1}{2} \angle B=\frac{1}{2} \angle D \Rightarrow \angle A BD=\angle A D B$
$\Rightarrow \quad A B=A D \quad$ [Sides opposite to equal angles in $\triangle A B D$ ]
$\Rightarrow \quad \triangle A B D$ is isosceles
$\Rightarrow \quad A O \perp B D \quad[\because O$ is the mid-point of $B D]$
$\Rightarrow \quad \angle A O B=90^{\circ}$
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