Option : (C)
Given,
ABCD is a parallelogram,
M is mid point of BD
BM bisects ∠B
∵ ∠A+∠B = 180°
[adjacent angles of parallelogram are supplementary]
In ∆AMB ,
∠AMB + \(\frac{1}{2}\)∠A + \(\frac{1}{2}\)∠B = 180°
= ∠AMB + \(\frac{1}{2}\)(∠A + ∠B) = 180°
=∠AMB + \(\frac{1}{2}\) x 180° = 180°
= ∠AMB = 180° - 90° = 90°