P is any point on the diagonals AC of a parallelogram ABCD.Prove that $a r (△ A D P) = a r (△ A B P)$ .

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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)
⇒ar△ADO-ar△DPO=ar△ABO-ar△BPO
Thus, ar(∆ADP) = ar(∆ABP)
Case II:

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

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Similar questions

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

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F

are points on diagonals

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of a parallelogram

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. Show that

B F D E

is a parallelogram.

Q. In the adjoining figure, ABCD is a parallelogram and O is any point on the diagonal AC. Show that ar(∆AOB) = ar(∆AOD).

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

P is any point on the diagonals AC of a parallelogram ABCD.Prove that $a r (△ A D P) = a r (△ A B P)$ .