Write True or False and justify your answer:
In the figure,$ABCD$ and $EFGD$ are two parallelograms and $G$ is the mid-point of $CD.$ Then $\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(EFGD).$

(FALSE) The statement is False.
Justification:
Since $\triangle DPC$ and parallelogram $ABCD$ are on the same base $DC$ and between the same parallels $AB$ and $DC$,we have:
$\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(ABCD)$.
Given that $G$ is the midpoint of $DC$,the base of parallelogram $EFGD$ is $DG = \frac{1}{2} DC$.
Since parallelogram $EFGD$ and parallelogram $ABCD$ are between the same parallels,their areas are proportional to their bases.
$\operatorname{ar}(EFGD) = \frac{DG}{DC} \times \operatorname{ar}(ABCD) = \frac{1}{2} \operatorname{ar}(ABCD)$.
Comparing the two results:
$\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(ABCD)$ and $\operatorname{ar}(EFGD) = \frac{1}{2} \operatorname{ar}(ABCD)$.
Therefore,$\operatorname{ar}(\triangle DPC) = \operatorname{ar}(EFGD)$.