$A$ point $E$ is taken on the side $BC$ of a parallelogram $ABCD$. $AE$ and $DC$ are produced to meet at $F$. Prove that $\operatorname{ar}(\triangle ADF) = \operatorname{ar}(ABFC)$.

(N/A) Given: $ABCD$ is a parallelogram. $A$ point $E$ is taken on the side $BC$. $AE$ and $DC$ are produced to meet at $F$.
Proof: Since $ABCD$ is a parallelogram and diagonal $AC$ divides it into two triangles of equal area,we have
$\operatorname{ar}(\triangle ADC) = \operatorname{ar}(\triangle ABC) \quad ....(1)$
As $DC \parallel AB$,so $CF \parallel AB$.
Since triangles on the same base $CF$ and between the same parallels $AB$ and $DF$ are equal in area,we have
$\operatorname{ar}(\triangle ACF) = \operatorname{ar}(\triangle BCF) \quad ....(2)$
Adding $(1)$ and $(2)$,we get
$\operatorname{ar}(\triangle ADC) + \operatorname{ar}(\triangle ACF) = \operatorname{ar}(\triangle ABC) + \operatorname{ar}(\triangle BCF)$
$\Rightarrow \operatorname{ar}(\triangle ADF) = \operatorname{ar}(ABFC)$
Hence,proved.