Abcd is a parallelogram m is midpoint of bd and bm bisects angle b find angle amb
Answers
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)
Hope its helps!!
Given : ABCD is a parallelogram. M is the midpoint of BD
BD bisects ∠B and ∠D
To Find : the measure of ∠AMB.
Solution:
ABCD is a parallelogram
=> ∠B = ∠D ( opposite angles of parallelogram are equal )
BD bisects ∠B and ∠D
=> ∠ADB = ∠ABD
=> AB = AD ( opposite sides of Equal angles)
∠ADB = ∠ABD
=> ∠ADM = ∠ABM as M lies on BD
and BM = DM as M is mid point of BD
in ΔADM and ΔABM
AD = AB ( shown above)
∠ADM = ∠ABM
DM = BM
=> ΔADM ≅ ΔABM
=> ∠AMD = ∠AMB
∠AMD + ∠AMB = 180° ( Linear Pair)
=> 2 ∠AMB = 180°
=> ∠AMB = 90°
the measure of ∠AMB is 90°
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