In a $\triangle ABC$, if $L$ and $M$ are points on $AB$ and $AC$ respectively such that $LM \| BC$.
Prove that $ar(\triangle LCM) = ar(\triangle LBM)$.


Given:

In a $\triangle ABC$, $L$ and $M$ are points on $AB$ and $AC$ respectively such that $LM \| BC$.

To do:

We have to prove that $ar(\triangle LCM) = ar(\triangle LBM)$.

Solution:

Join $LM, LC$ and $MB$.


$L$ and $M$ are the mid points of $AB$ and $AC$.

This implies,

$LM \| BC$

$\triangle LBM$ and $\triangle LCM$ are on the same base $LM$ and between the same parallels.

Therefore,

$ar(\triangle LBM) = ar(\triangle LCM)$......…(i)

$ar(\triangle LCM) = ar(\triangle LBM)$

Hence proved.

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Updated on: 10-Oct-2022

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