In a triangle ABC, if L and M are points on AB and AC respectively such that LM ∥ BC. Prove that:

(i) Clearly triangles LMB and LMC are on the same base LM and between the same parallels LM and BC.

∴ ar(ΔLMB) = ar(ΔLMC)               ... (1)

(ii) We observe that triangles LBC and MBC are on the same base BC and between same parallels LM and BC.

∴ ar(ΔLBC) = ar(ΔMBC)          ... (2)

(iii) We have,

ar(ΔLMB) = ar(ΔLMC)            [From 1]

⇒ ar(ΔALM) + ar(ΔLMB) = ar(ΔALM) + ar(ΔLMC)

⇒ ar(ΔABM) = ar(ΔACL)

(iv) we have,

ar(ΔLBC) = ar(ΔMBC)             [From 1]

⇒ ar(ΔLBC) − ar(ΔBOC) = ar(ΔMBC) − ar(ΔBOC)

⇒ ar(ΔLOB) = ar(ΔMOC).