In the figure, $ABCD$ is a parallelogram. $O$ is any point on $AC. PQ \parallel AB$ and $LM \parallel AD$. Prove that $ar(parallelogram\ DLOP) = ar(parallelogram\ BMOQ)$.
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Given:

$ABCD$ is a parallelogram. $O$ is any point on $AC. PQ \parallel AB$ and $LM \parallel AD$.

To do:

We have to prove that $ar(parallelogram\ DLOP) = ar(parallelogram\ BMOQ)$.

Solution:

The diagonal of a parallelogram divides it into two triangles of equal area.

Therefore,

$ar(\triangle ADC) = ar(\triangle ABC)$

$ar(\triangle APO) + ar(\|gm DLOP) + ar(\triangle OLC) = ar(\triangle AOM) + ar(\|gm BMOQ) + ar( \triangle OQC)$.....…(i)

$AO$ and $OC$ are the diagonals of parallelograms $AMOP$ and $OQCL$ respectively,

This implies,

$ar(\triangle APO) = ar(\triangle AMO)$.....…(ii)

$ar(\triangle OLC) = ar(\triangle OQC)$....…(iii)

Subtracting (ii) and (iii) from (i), we get,

$ar(\|gm DLOP) = ar(\|gm BMOQ)$.

Hence proved.

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Updated on: 10-Oct-2022

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