# ABCD is a parallelogram and ∠DAB=60  , If the bisectors AP and BP of the angles A and B respective meet at P on CD. Prove that P is the mid-point of CD.

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## ABCD is a parallelogram, in which ∠A=60o.

∠A+∠B=180o            [ Sum of adjacent angles of parallelogram are supplementary ]

60o+∠B=180o.

∠B=120o.

∠D=∠B=120o         [ Opposite angles of parallelogram are equal ]

∠A=∠C=60o         [ Opposite angles of parallelogram are equal ]

AP bisects ∠A

∠DAP=∠PAB=30o.

BP bisects ∠B

∠CBP=∠PBA=60o

In △PAB,

∠PAB+∠PBA+∠APB=180o.

30o+60o+∠APB=180o

∠APB=90o.

In △PBC,

∠PBC+∠PCB+∠BPC=180o.

60o+60o+∠BPC=180o.

∠BPC=60o

30o+120o+∠APD=180o.

∠APD=30o.

In △PBC,

∠BPC=∠CBP=60o.    [ linear angles are supplementary ]

BC=PC      [ Sides opposite to equal angles of a triangle are equal ]     ----- ( 1 )

∠APD=∠DAP=30o.

AD=DP          [ Sides opposite to equal angles of triangle are equal ]

But AD=BC            [ Opposite sides of parallelogram are equal ]

So, BC=DP          ----- ( 2 )

From ( 1 ) and ( 2 ), we get

DP=PC

P is the mid-point of CD 