In Figure, ABCD is a parallelogram in which ∠A = 60°.
Question:

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

Solution:

AP bisects ∠A

Then, ∠DAP = ∠PAB = 30°

Adjacent angles are supplementary

Then, ∠A + ∠B = 180°

∠B + 600 = 180°

∠B = 180° − 60°

∠B = 120°

BP bisects ∠B

Then, ∠PBA = ∠PBC = 30°

∠PAB = ∠APD = 30° [Alternate interior angles]

Therefore, AD = DP   [Sides opposite to equal angles are in equal length]

Similarly

∠PBA = ∠BPC = 60° [Alternate interior angles]

Therefore, PC = BC

DC = DP + PC

DC = AD + BC   [Since, DP = AD and PC = BC]

DC = 2AD    [Since, AD = BC, opposite sides of a parallelogram are equal]

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