Question

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

Answer 1

Given: In ΔABC, L and M are points on AB and AC respectively such that LM || BC.

To prove: ar (ΔLOB) = ar (ΔMOC)

Proof: We know that, triangles on the same base and between the same base between the same parallels are equal in area.

Hence, ΔLBC and ΔMBC lie on the same base BC and between the same parallels BC and LM.

So, ar (ΔLBC) = ar (ΔMBC)

⇒ ar (ΔLOB) + ar (ΔBOC) = ar (ΔMOC) + ar (ΔBOC)

On eliminating D ar (ΔBOC) from both sides, we get

ar (ΔLOB) = ar (ΔMOC)

Hence proved.