If E and F are any two points lying on the sides DC andAD, respectively of a parallelogram ABCD, then

B F + C F = A E + B E

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A E = B F

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A r (Δ A E B) = A r (Δ B F C)

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A r (Δ A D E) = A r (Δ B E C)

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Solution

The correct option is D $A r (Δ A E B) = A r (Δ B F C)$
We know that area of triangle and parallelogram on same base and same set of parallel line in half the area of parallelogram

In triangle AEB and parallelogram ABCD
$a r e a Δ A E B = \frac{1}{2} a r e a A B C D$ .............................(1)

In triangle AEB and parallelogram ABCD
$a r e a Δ B F C = \frac{1}{2} a r e a A B C D$ .............................(2)

As per (1) and (2) we get
$a r e a Δ A E B = a r e a Δ B F C$

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