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)