Question

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)

Answer 1

(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)