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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)
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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)
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