Given In `DeltaABC`, L and M are points on AB and AC respectively such that `LM || BC`.
To prove `" "ar (DeltaLOB) = ar (DeltaMOC)`
Proof We know that, triangles on the same base and between the same parallels are equal in area.
Hence, `DeltaLBC` and `DeltaMBC` lie on the same base BC and between the same parallels BC and LM.
So, `" " ar (DeltaLBC) = ar (DeltaMBC)`
`rArr" "` `ar (DeltaLOB) + ar (DeltaBOC) = ar (DeltaMOC) + ar (DeltaBOC)`
On eliminating D `ar (DeltaBOC)` from both sides, we get
`ar (DeltaLOB) = ar (DeltaMOC)" "` Hence proved.