Question

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

Answer 1

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AP bisects ∠A

Then, ∠DAP=∠PAB=30  

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                      ------ ( 1 )

We know that in parallelogram adjacent angles are supplementary

∴  ∠A+∠B=180  

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⇒  60  

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+∠B=180  

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∴  ∠B=120  

o

.

BP bisects ∠B

Then, ∠PAB=∠PBC=60  

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            ---- ( 2 )

⇒  ∠PAB=∠APD=30  

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            [ Alternate angles ]   ---- ( 3 )

∴  ∠DAP=∠APD=30  

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         [ From ( 1 ) and ( 3 ) ]

∴  AD=DP               [ Since base angles are equal ]

Similarly, ∠PBA=∠BPC=60  

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              [ Alternate angles ]   --- ( 4 )

⇒  ∠PBC=∠BPC=60  

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        [ From ( 2 ) and ( 4 ) ]

∴  PC=BC                            [ Since base angles are equal  

⇒  DC=DP+PC

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

⇒  DC=AD+AD                 [ Opposite sides of parallelogram are equal ]

∴  DC=2AD

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Answer 2

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