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