In a Δ ABC, if L and M are points on AB and AC respectively such tha

1.	In a Δ 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)
Answer» (i) Clearly, Triangles LMB and LMC are on the same base LM and between the same parallels LM and BC. Therefore, Area (ΔLMB) = Area (ΔLMC) ...(i) (ii) We observe that, Triangles LBC and MBC are on the same base BC and between the same parallels LM and BC. Therefore, Area (ΔLBC) = Area (ΔMBC) ...(ii) (iii) We have, Area (ΔLMB) = Area (ΔLMC) [From (i)] Area (ΔALM) + Area (ΔLMB) = Area (ΔALM) = Area (ΔLMC) Area (ΔABM) = Area (ΔACL) (iv) We have, Area (ΔLBC) = Area (ΔMBC) [From (ii)] Area (ΔLBC) - Area (ΔBOC) = Area (ΔMBC) - Area (ΔBOC) Area (ΔLOB) =Area (ΔMOC)

In a Δ 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)

Answer»

(i) Clearly,

Triangles LMB and LMC are on the same base LM and between the same parallels LM and BC.

Therefore,

Area (ΔLMB) = Area (ΔLMC) ...(i)

(ii) We observe that,

Triangles LBC and MBC are on the same base BC and between the same parallels LM and BC.

Therefore,

Area (ΔLBC) = Area (ΔMBC) ...(ii)

(iii) We have,

Area (ΔLMB) = Area (ΔLMC) [From (i)]

Area (ΔALM) + Area (ΔLMB) = Area (ΔALM) = Area (ΔLMC)

Area (ΔABM) = Area (ΔACL)

(iv) We have,

Area (ΔLBC) = Area (ΔMBC) [From (ii)]

Area (ΔLBC) - Area (ΔBOC) = Area (ΔMBC) - Area (ΔBOC)

Area (ΔLOB) =Area (ΔMOC)