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  • In ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC)

In \triangle ABC, if L and M are the points on AB and AC, respectively such that LM \parallel BC. Prove that ar (LOB) = ar (MOC)

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Solution.

Given: \triangle ABC with L and M points on AB and AC respectively such that LM \parallel BC.

To prove: ar (\triangle LOB) = ar(\triangle MOC)
Proof:
We know that, triangles on the same base and between the same parallels are equal in area.
Hence, \triangle LBC and \triangle MBC lie on the same base BC and between the same parallels BC and LM.
So, ar(\triangle LBC) = ar(\triangle MBC) …(1)
ar(\triangle LBC) = ar(\triangle LOB) + ar(\triangle BOC)
ar(\triangle MBC)=ar(\triangle MOC)+ar(\triangle BOC)
Using these values in equation 1, we get:
\Rightarrow ar(\triangle LOB) + ar(\triangle BOC) = ar(\triangle MOC) + ar(\triangle BOC)
On eliminating ar(\triangle BOC) form both sides, we get
ar (\triangle LOB) = ar(\triangle MOC)
Hence proved.

 

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