Which of the following figures lie on the same base and between the same parallels? Write the common base and the two parallels for the figure for which the answer is affirmative.

If in the figure,$PQRS$ and $EFRS$ are two parallelograms,then $\operatorname{ar}(MFR) = \frac{1}{2} \operatorname{ar}(PQRS)$. State whether the statement is True or False.

Write True or False and justify your answer:
$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. Then $ar(\triangle BDE) = \frac{1}{4} ar(\triangle ABC).$

The mid-points of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to:

$(1)$ If two figures are congruent,they must have $\ldots \ldots$ areas.
$(2)$ Area of figure $A$ is denoted as $\ldots \ldots$ symbolically.

In $\triangle ABC$,$D$ is the mid-point of $AB$ and $P$ is any point on $BC$. If $CQ \parallel PD$ meets $AB$ in $Q$,then prove that $\operatorname{ar}(\triangle BPQ) = \frac{1}{2} \operatorname{ar}(\triangle ABC)$.