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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.
Solution
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].
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