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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) - Mathematics

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)

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 28 | Page 48
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