P is any point on the diagonal AC of a parallelogram ABCD. Prove that ar(∆ADP) = ar(∆ABP).

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Solution

Join BD.
Let BD and AC intersect at point O.
O is thus the midpoint of DB and AC.
PO is the median of $△$ DPB,
So,
$ar (△ DPO) = ar (△ BPO) . . . . . (1) ar (△ ADO) = ar (△ ABO) . . . . . (2) Case 1 : (2) - (1) \Rightarrow ar (△ ADO) - ar (△ DPO) = ar (△ ABO) - ar (△ BPO)$
Thus, ar(∆ADP) = ar(∆ABP)

Case II:

$ar (△ ADO) + ar (△ DPO) = ar (△ ABO) + ar (△ BPO)$
Thus, ar(∆ADP) = ar(∆ABP)

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