Question

In Fig. 13.35, ABCD is a parallelogram in which angle A = 60°. If the bisectors of
angle A and angle B meet at P, prove that AD = DP, PC = BC and DC = 2AD.​

Answer 1

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Answer 2

Step-by-step explanation:

AP bisects ∠A

Then, ∠DAP=∠PAB=30  ------ ( 1 )

We know that in parallelogram adjacent angles are supplementary

∴  ∠A+∠B=180

⇒  60 +∠B=180

∴  ∠B=120

BP bisects ∠B

Then, ∠PAB=∠PBC=60  ---- ( 2 )

⇒  ∠PAB=∠APD=30  [ Alternate angles ]   ---- ( 3 )

∴  ∠DAP=∠APD=30

 [ From ( 1 ) and ( 3 ) ]

∴  AD=DP               [ Since base angles are equal ]