In the given figure, ABCD is a parallelogram, M is the mid-point of BD and BD bisects ∠B as well as ∠D.


In the given figure, ABCD is a parallelogram, M is the mid-point of BD and BD bisects B as well as ∠D. Then, ∠AMB = ?
(a) 45°
(b) 60°
(c) 90°
(d) 30°



(c) ​90°

∠B = ∠D  

$\Rightarrow \frac{1}{2} \angle B=\frac{1}{2} \angle D$

⇒​ ∠ADB = ∠​ABD
∴ ∆ABD is an isosceles triangle and M is the midpoint of BD. We can also say that M is the median of ∆ABD.
∴ AM ⊥ BD and, hence, ​∠AMB =​​ 90°


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