In the figure, ABCD is a ||gm . O is any point on AC. PQ || AB and LM || AD. Prove that ar (||gm DLOP) = ar(||gm BMOQ).

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Solution

Given : In || gm ABCD, O is any point on Diagonal AC. PQ || AB and LM || BC
To Prove : ar (|| gm DLOP ) = ar(||gm BMOQ)
Proof : $∵$ Since a diagonal of a parallelogram divides it into two triangles of equal area.
$∴ a r (△ A D c) = a r (△ A B C) \Rightarrow a r (△ A P O) + a r (| | g m D L O P) + a r (△ O L C) = a r (△ A O M) + a r (| | g m B M O Q) + a r (△ O Q C)$
Since, AO and OC are diagonals of parallelograms AMOP and OQCL respectively,
$∴ a r (△ A P O) = a r (△ A M O)$
and $a r (△ O L C) = a r (△ O Q C)$

$∴ a r (| | g m D L O P) = a r (| | g m B M O Q)$

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