In a ΔABC, if L and M are points on AB and AC respectively such that LM || BC. Prove

that:

(1) ar (ΔLCM ) = ar (ΔLBM )

(2) ar (ΔLBC) = ar (ΔMBC)

(3) ar (ΔABM) ar (ΔACL)

(4) ar (ΔLOB) ar (ΔMOC)

#### Solution

(1) Clearly Triangles LMB and LMC are on the same base LM and between the same

parallels LM and BC.

∴ ar (ΔLMB) = ar (ΔLMC) ......(1)

(2) We observe that triangles LBC and MBC area on the same base BC and between the

same parallels LM and BC

∴ arc ΔLBC = ar (MBC) ..........(2)

(3) We have

ar (ΔLMB) = ar (ΔLMC) [from (1)]

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

⇒ ar (ΔABM) = ar (ΔACL)

(4) We have

ar(ΔCBC) = ar (ΔMBC) ∴ [from (1)]

⇒ ar (ΔLBC) = ar (ΔBOC) = a (ΔMBC) - ar (BOC)

⇒ ar (ΔLOB) = ar (ΔMOC)