Question

In figure ABCD is a parallelogram in which angle D=60 degree if the the bisector of Angle B and angle C meet ab at C prove that CE D angle equal to 90 degree, AD=AE,AB=2AD, triangle BCE in an equilateral triangle​

Answer 1

Answer:

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 ]

∴ DC=2AD