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Let the diagonal  DB   and CA    intersect at  j.
 Since the diagonai of a parallelogram bisect  each other.
therefore  j is the mid point of both AC  and  BD . Since a median of a triangle of equal area.
 In triangle  DCB,   CJ   is  a  median

ar(triangle  JDC) =ar(JBC)   ...........(1)

In triangle  PDB  ,PJ  is a median
ar(triangle PDJ) =ar (PBJ)
Adding (1) and (11)
ar (JDC)+ ar  (PDJ)= ar (triangle   JBC) + ar (triangle  PBJ)

ar(ADP)= ar (ABP)
0 0 0
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