In triangle ABC, if L and M are the points on AB & AC respectively, such that LM is parallel to BC. If BM & CL intersect at O, prove that area of traingle LOB = area of triangle MOC.


We have LM||BC
ΔBLC and ΔCMB are on same base BC and have the same height as LM||BC.
Therefore ar(ΔBLC) = ar (ΔCMB)
 ar(ΔLOB) + ar(ΔBOC) = ar(ΔMOC) + ar (ΔBOC)
  ar(ΔLOB) = ar(ΔMOC)


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