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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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)
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)
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