# In the given figure,p is any point on the diagonal ac of parallelogram abcd.show that ar(adp)=ar(abp).

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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)