ABCD is a parallelogram M is the mid-point of BD and BM bisects ∠B. Then, ∠AMB=
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
Therefore,
M is the mid-point of AC
AB || CD and BD is the transversal,
Therefore,
∠ ABD = ∠ BDC (i) (Alternate interior angle)
∠ ABD = ∠ DBC (ii) (Given)
From (i) and (ii), we get
∠ BDC = ∠ DBC
In Δ BCD,
∠ BDC = ∠ DBC
BC = CD (iii) (In a triangle, equal angles have equal sides opposite to them)
AB = CD and BC = AD (iv) (Opposite sides of the parallelogram are equal)
From (iii) and (iv), we get
AB = BC = CD = DA
Therefore,
ABCD is a rhombus
∠AMB = 90° (Diagonals of rhombus are perpendicular to each other)