A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Then,

A r e a (Δ A B C D) = A r e a (Δ A C F)

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A r e a (Δ A D F) = A r e a (Δ A B F C)

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A r e a (Δ A B F) = A r e a (Δ A C F)

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Solution

The correct option is B $A r e a (Δ A D F) = A r e a (Δ A B F C)$
Since, ABCD is a parallelogram and diagonal AC divides it into two triangles of equal area, we have,
$a r (Δ A D C) = a r (Δ A B C)$ . . . . . (1)
As DC||AB, so CF||AB
Since, triangles on the same base and between the same parallels are equal in area, so we have,
$a r (Δ A C F) = a r (Δ B C F)$ . . . . . (2)
Adding (1) and (2), we get,
$a r (Δ A D C) + a r (Δ A C F)$
$= a r (Δ A B C) + a r (Δ B C F)$
$\Rightarrow a r (Δ A D F) = a r (A B F C)$

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