In the figure, CD || AE and CY || BA.
(i) Name a triangle equal in area of $△ C B X .$
(ii) Prove that $a r (△ Z D E) = a r (△ C Z A) .$
(iii) Prove that ar (BCZY) = $a r (△ E D Z)$

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Solution

Given: IN the figure,
CP || AE and CY || BA

Proof:
(i) $△ C B X a n d △$ CYX are on the same base BY and between same parallels.
$∴ a r (△ C B X) = a r (△ C Y X)$
(ii) $△ A D E a n d △ A C E$ are on the same base and between the same parallels (AE || CD)
$∴ a r (△ A D E) = a r (△ A C E)$
Substracting $a r (△ A Z E)$ from both sides
$\Rightarrow a r (△ A D E) - a r (△ A Z E) = a r (△ A C E) - a r (△ A Z E) \Rightarrow a r (△ Z D E) = a r (△ A C Z) \Rightarrow a r △ Z D E = a r △ C Z$
(iii) $∵$ $△ x A C Y$ and BCY are on the same base CY and between the same parallels
$∴ a r (△ A C Y) = a r (△ B C Y) N o w a r (△ A C Z) = a r (△ Z D E) \Rightarrow a r (△ A C Y) + a r (△ C Y Z) = a r (△ E D Z) \Rightarrow a r (△ B C Y) + a r (△ C Y Z) = a r (△ E D Z) ∴ a r (B C Z Y) = a r (△ E D Z)$

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In the figure, CD || AE and CY || BA. (i) Name a triangle equal in area of△CBX. (ii) Prove that ar(△ZDE)=ar(△CZA). (iii) Prove that ar (BCZY) = ar(△EDZ)

In the figure, CD || AE and CY || BA.
(i) Name a triangle equal in area of $△ C B X .$
(ii) Prove that $a r (△ Z D E) = a r (△ C Z A) .$
(iii) Prove that ar (BCZY) = $a r (△ E D Z)$