In a $△ A B C,$ if L and M are points on AB and AC respectively such that LM || BC. Prove that:
$(i) a r (△ L C M) = a r (△ L B M) (i i) a r (△ L B C) = a r (△ M B C) (i i i) a r (△ A B M) = a r (△ A C L) (i v) a r (△ L O B) = a r (△ M O C) .$

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Solution

Given : In $△ A B C,$
L and M are mid points on AB and AC
LM, LC and MB are joined

To Prove:
$(i) a r (△ L C M) = a r (△ L B M) (i i) a r (△ L B C) = a r (△ M B C) (i i i) a r (△ A B M) = a r (△ A C L) (i v) a r (△ L O B) = a r (△ M O C) .$
Proof: $∵$ L and M are the mid points of AB and AC
$∴ L M | | B C$
(i) Now $△ L B M a n d △ L C M$ are on the same base LM and between the same parallels
$∴ a r (△ L B M) = a r (△ L C M) \Rightarrow a r (△ L C M) = a r (△ L B M)$
(ii) $∵ △ L B C a n d △ M B C$ are on the same base BC and between the same parallels
$∴ a r (△ L B C) = a r (△ M B C)$
$(i i i) a r (△ L M B) = a r (△ L M C) \Rightarrow a r (△ A L M) + a r (△ L M B) = a r (△ A L M) + a r (△ L M C) \Rightarrow a r (△ A B M) = a r (△ A C L)$
(iv) $∵ a r (△ L B C) = a r (△ M B C) \Rightarrow a r (△ L B C) - a r (△ B O C) = a r (△ M B C) - a r (△ B O C) a r (△ L B O) = a r (△ M O C)$

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