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

(N/A) Given: $ABCD$,$DCFE$,and $ABFE$ are parallelograms.
$1$. Since $ABCD$ is a parallelogram,its opposite sides are parallel and equal. Therefore,$AD = BC$ and $AD \parallel BC$. Also,$AB \parallel DC$....$(1)$
$2$. Since $DCFE$ is a parallelogram,its opposite sides are parallel and equal. Therefore,$DC \parallel EF$....$(2)$
$3$. From $(1)$ and $(2)$,we have $AB \parallel DC$ and $DC \parallel EF$,which implies $AB \parallel EF$.
$4$. Now,consider $\Delta ADE$ and $\Delta BCF$.
- We have $AD = BC$ (opposite sides of parallelogram $ABCD$).
- We have $DE = CF$ (opposite sides of parallelogram $DCFE$).
- We have $AE = BF$ (opposite sides of parallelogram $ABFE$).
- Thus,$\Delta ADE \cong \Delta BCF$ by $SSS$ congruence criterion.
$5$. Since the triangles are congruent,their areas must be equal.
- Therefore,$\operatorname{ar}(ADE) = \operatorname{ar}(BCF)$.