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°

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Solution

(c) 90°

Explanation:
∠B = ∠D
⇒ $\frac{1}{2}$ ∠B = $\frac{1}{2}$ ∠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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