Question

ABCD is a parallelogram, M is the mid-point of BD and BM bisects $\angle B$. Then $\angle AMB=$

• A. ${45}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

The correct option is B: ${90}^{\circ }$

In || gm ABCD, M is the mid-point of BD and BM bisects $\angle B$

AM is joined.

We know that the diagonals of a parallelogram bisect each other.

then, AC will pass thought M.

$\therefore$ BM bisects $\angle B$

$\therefore \angle 1=\angle 2$

$AD\parallel BC,$ DB is the tranversal.

So, $\angle 2=\angle 3$ [Alternate interior angle]

$\therefore \angle 1=\angle 2=\angle 3$

In triangle ADB, we have

$\angle 1=\angle 3$ [proved above]

$AD=AB\dots \dots \left(i\right)$ [Sides opposite to equal angles are equal ]

$AB=CD\dots \dots \left(ii\right)$ [ Opposite side of parallelogram]

From eq.(i) and eq.(ii), we have

$AB=CD=AD=BC$

So, ABCD is a rhombus, and we know that the diagonals of a rhombus bisect each other at ${90}^{\circ }.$

$\therefore \angle AMB={90}^{\circ }$