Question

In the given figure,P is any point on a diagonal AC of the parallelogram ABCD. Show that ar(ADP)=ar(ABP).

Answer 1

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)

Answer 2

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 medianar(triangle PDJ) =ar (PBJ)Adding (1) and (11)ar (JDC)+ ar  (PDJ)= ar (triangle   JBC) + ar (triangle  PBJ)

ar(ADP)= ar (ABP)