ABCD is a parallelogram in which angle A = 60 degree, if the bisectors of angle A and angle B meet DC at P, prove that (i) angle APB = 90 degree (ii) AD = DP and PB = PC=BC (iii) DC = 2AD

Given: ABCD is a parallelogram in which ∠A = 60°. AP and PB are the bisector of ∠A and ∠B respectively.

∠PAB = ∠PAD = 30° (AP bisects ∠A)

and ∠ABP = ∠PBC = 60° (BP bisects ∠B)

i)

In triangle APB, using angle sum property of triangle, we get

 ∠APB + ∠PAB + ∠ABP = 180°

⇒ ∠APB + 30° + 60° = 180°

⇒ ∠APB = 90°

ii)

AB || CD and AP is the transversal.

∴ ∠PAB = ∠APD = 30° (Alternate angles)

In ΔAPD,

∠PAD = ∠APD = 30°

∴ PD = AD ...(1) (Equal sides have equal angles opposite to them)

AD || BC and AB is the transversal.

∴ ∠A + ∠B = 180° (Sum of adjacent interior angles is 180°)

⇒ 60° + ∠B = 180°

⇒ ∠B = 120°

⇒ ∠PBA = ∠PBC = 60° (PB is bisector ∠B)

∠PBA = ∠BPC = 60° (Alternate angles)

In ΔPBC,

∠BPC = ∠PBC = 60°

Also,  ∠PCB = 60°  [Using angle sum property of triangle]

So, ΔPBC is an equilateral triangle

∴ BC = PB = PC ...(2) (Equal sides of an equilateral triangle)

iii)

CD = DP + PC

∴ CD = AD + BC [Using (1) and (2)]

⇒ CD = AD + AD [AD = BC (Opposite sides of parallelogram are equal)]

⇒ CD = 2AD