In the adjoining figure, $D E ∥ B C$ . Prove that
i) $a r (Δ A C D) = a r (δ A B E)$
ii) $a r (Δ O C E) = a r (δ O B D)$

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Solution

$B C | | D E$
$△ D E C$ and $△ D E B$ lies on the same base and between the same parallel lines
So $a r (△ D E C) = a r (△ D E B)$
$(1)$ on adding $a r (△ A D E)$ in both sides of equation $(1)$ we get
$a r (△ D E C) + a r (△ A D E) = a r (△ A D E) + a r (△ A D E)$
$a r (△ A C D) = a r (△ A B E)$
$(2)$ on substracting $a r (O D E)$ from both sides of equation $(1)$ we get
$a r (△ D E C) - a r (△ O D E) = a r (△ D E B) - a r (△ O D E)$
$\Rightarrow a r (△ O C E) = a r (O B D)$

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