Question

In the adjecent figure, angle bisectors of angles A and B of parallelogram ABCD meet at M.
where M lies on CD. Prove that M is mid point CD.​

Answer 1

◻prove that M is the midpoint of CD⬇⬇

ABCD is a parallelogram,

in which ∠A = 60° ⇒ ∠B = 120° [Adjacent angles of a parallelogram are supplementary]

∠C = 60° = ∠A [Opposite angles of a parallelogram are equal]

∠D = ∠B = 120° [Opposite angles of parallelogram are equal]

AM bisects ∠A ⇒ ∠DAM = ∠MAB = 30°

BM bisects ∠B ⇒ ∠CBM = ∠MBA = 60°

◽ In ΔMAB, ∠AMB = 90° [Angle sum property]

◽In ΔMBC, ∠BMC = 60° [Angle sum property]

◽In ΔADM, ∠AMD = 30° [Angle sum property]

◽In ΔMBC,

∠BMC = ∠CBM = 60° [Linear angles are supplementary]

⇒ BC = MC [Sides opposite to equal angles of a triangle are equal] -------- (1)

◽ In ΔADM,

∠APD = ∠DAM = 30° ⇒ AD = DM [Sides opposite to equal angles of a triangle are equal]

But AD = BC [Opposite sides of parallelogram are equal]

So,

BC = DM -------- (2)

From (1) and (2),

◽we get,

DM = MC ⇒ M is the midpoint of CD.

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