Question:
In a triangle ABC, if L and M are points on AB and AC respectively such that LM ∥ BC. Prove that:
(i) ar(ΔLCM) = ar(ΔLBM)
(ii) ar(ΔLBC) = ar(ΔMBC)
(iii) ar(ΔABM) = ar(ΔACL)
(iv) ar(ΔLOB) = ar(ΔMOC)
Solution:
(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).