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# In figure below, $ABCD, DCFE$ and $ABFE$ are parallelograms. Show that $ \operatorname{ar}(\mathrm{ADE})=\operatorname{ar}(\mathrm{BCF}) $.

"

Given:

$ABCD, DCFE$ and $ABFE$ are parallelograms.

To do:

We have to prove that $ar(\triangle ADE) = ar(\triangle BCF)$.

Solution:

$ABCD$ is a parallelogram.

This implies,

$AD = BC$

Similarly,

In parallelogram $ABEF$,

$AE = BF$

In parallelogram $CDEF$,

$DE = CF$

In $\triangle ADE$ and $\triangle BCF$,

$AD = BC$

$DE = CF$

$AE = BF$

Therefore, by SSS axiom,

$\triangle ADE \cong \triangle BCF$

This implies,

$ar(\triangle ADE) = ar(\triangle BCF)$ (Congruent triangles are equal in area)

Hence proved.

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