Sarthaks Test
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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

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AP bisects ∠A 

Then, ∠AP = ∠PAB = 30o 

Adjacent angles are supplementary


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