Question

In the 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

Fastest and the correct answer will be marked as brainliest
Wrong answers will be reported



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Answer 1

AP bisects ∠A

Then, ∠DAP=∠PAB=30

o

------ ( 1 )

We know that in parallelogram adjacent angles are supplementary

∴ ∠A+∠B=180

o

⇒ 60

o

+∠B=180

o

∴ ∠B=120

o

.

BP bisects ∠B

Then, ∠PAB=∠PBC=60

o

---- ( 2 )

⇒ ∠PAB=∠APD=30

o

[ Alternate angles ] ---- ( 3 )

∴ ∠DAP=∠APD=30

o

[ From ( 1 ) and ( 3 ) ]

∴ AD=DP [ Since base angles are equal ]

Similarly, ∠PBA=∠BPC=60

o

[ Alternate angles ] --- ( 4 )

⇒ ∠PBC=∠BPC=60

o

[ 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 ]

Answer 2

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