If two parallelograms ABCD and ABEF are on the same base AB and between the same parallels AB and FC, then prove that ar(ABCD) = ar(ABEF)

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Solution

Two parallelograms ABCD and ABEF on the same base AB and between the same parallels AB and FC.
To prove : ar(ABCD) = ar(ABEF)
$Proof : In △ A D F and △ B C E, we have$

$A D = B C [Opposite sides of a | |^{g m}]$
$A F = B E [Opposite sides of a | |^{g m}]$
$∠ D A F = ∠ C B E [A D | | B C and A F | | B E]$
[Angle between AD and AF = Angle between BC and BE]
$∴ △ A D F ≅ △ B C E [By SAS congruency]$
$∴ a r (A D F) = a r (B C E) . . . (i)$
$∴ a r (A B C D) = a r (A B E D) + a r (B C E)$
$= a r (A B E D) + a r (A D F) [Using (i)]$
$= a r (A B E F) .$
$Hence, a r (A B C D) = a r (A B E F) .$

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