In Figure, `A B C D` is a parallelogram in which `/_A=60^@` . If the bisectors of `/_A` and `/_B` meet at `P ,` prove that `A D=D P ,P C=B C` and `D C=2A Ddot` Figure

Explanation

"given,AP bisects`/_A`
Then,
`/_DAP=/_PAB=30^@`------ ( 1 )
We know that in parallelogram adjacent angles are supplementary
so, `/_A+/_B=180^@`
`60^@+/_B=180^@`
`/_B=120^@`
BP bisects angle B

Then,
`/_PAB=/_PBC=60^@`---- ( 2 )
`/_PAB=/_APD=30^@`[ Alternate angles ]---- ( 3 )
`/_DAP=/_APD=30^@`[ From ( 1 ) and ( 3 ) ]
so,
AD=DP[Since base angles are equal ]
Similarly, `/_PBA=/_BPC=60^@`[ Alternate angles ]--- ( 4 )
`/_PBC=/_BPC=60^@`[ From ( 2 ) and ( 4 ) ]
so, 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 ]
so,
`DC=2AD`"