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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

In △ADP,

  ∠PAD+∠ADP+∠APD=180o.

  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 )

In △ADP,

  ∠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

solution

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