Question

In the given figure, ABCD is a parallelogram, M is the midpoint of BD and BD bisects $\angle B$ as well as $\angle D$. Then, $\angle AMB=$ ?

(a) ${45}^{\circ }$

(b) ${60}^{\circ }$

(c) ${90}^{\circ }$

(d) ${30}^{\circ }$

Answer 1

(c) 90°

ABCD is a parallelogram. BD is the diagonal and M is the mid point of BD. BD is a bisector of ∠B.

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

∴ M is the mid point of AC.

AB || CD and BD is the transversal,

∴ ∠ ABD = ∠ BDC ...(1) (Alternate interior angles)

∠ ABD = ∠ DBC ...(2) (Given)

From (1) and (2), we get

∠ BDC = ∠ DBC

In Δ BCD,

∠ BDC = ∠ DBC

⇒ BC = CD ...(3) (In a triangle, equal angles have equal sides opposite to them)

AB = CD and BC = AD ...(4) (Opposite sides of the parallelogram are equal)

From (3) and (4), we get

AB = BC = CD = DA

∴ ABCD is a rhombus.

⇒ ∠AMB = 90° (Diagonals of rhombus are perpendicular to each other)