Question

In Fig., below, 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.

Answer 1

AP bisects `∠`A

Then, `∠`AP = `∠`PAB = 30°

Adjacent angles are supplementary

Then,`∠`A + `∠`B =180°

`∠`B + 60° =180°             ÐA = 60°

`∠`B = 180° - 60°

`∠`B = 120°

BP bisects `∠`B

Then, `∠`PBA  `∠`PBC = 30°

`∠`PAB = `∠`APD = 30°                  [Alternative interior angles]

∴AD = DP      [  ∵  Sides opposite to equal angles are in equal length]

Similarly

`∠`BA = `∠`BPC = 60°                  [Alternative interior angle]

∴ PC = BC

DC = DP + PC

DC = AD + BC     [ ∵ DP = AD, PC = BC ]

DC = 2AD     [  ∵  AD = BC  Opposite sides of a parallelogram are equal].