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In a $\triangle ABC$, if $L$ and $M$ are points on $AB$ and $AC$ respectively such that $LM \| BC$.
Prove that $ar(\triangle LBC) = ar(\triangle MBC)$.
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 $ar(\triangle LBC) = ar(\triangle MBC)$.
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)
$\triangle LBC$ and $\triangle MBC$ are on the same base $BC$ and between the same parallels.
Therefore,
$ar(\triangle LBC) = ar(\triangle MBC)$......…(ii)
Hence proved.
Updated on: 10-Oct-2022
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