In the given figure,$ABED$ is a parallelogram and $DE = EC$. Prove that $\operatorname{ar}(ABF) = \operatorname{ar}(BEC)$.

(N/A) Given: $ABED$ is a parallelogram and $DE = EC$.
To prove: $\operatorname{ar}(ABF) = \operatorname{ar}(BEC)$.
Proof:
$1$. Since $ABED$ is a parallelogram,$AB \parallel DE$ and $AB = DE$.
$2$. In $\triangle ABF$ and $\triangle BEC$,the base $AB$ is parallel to the base $DC$ (since $AB \parallel DE$ and $F, E$ lie on $DC$).
$3$. The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
$4$. Since $AB \parallel DC$,both $\triangle ABF$ and $\triangle BEC$ lie between the same parallel lines $AB$ and $DC$,so they have the same height $h$.
$5$. $\operatorname{ar}(ABF) = \frac{1}{2} \times AB \times h$.
$6$. $\operatorname{ar}(BEC) = \frac{1}{2} \times EC \times h$.
$7$. Since $AB = DE$ (opposite sides of parallelogram $ABED$) and $DE = EC$ (given),it follows that $AB = EC$.
$8$. Substituting $AB = EC$ into the area formula: $\operatorname{ar}(ABF) = \frac{1}{2} \times EC \times h = \operatorname{ar}(BEC)$.
Thus,$\operatorname{ar}(ABF) = \operatorname{ar}(BEC)$.