In which of the following figures,do you find two polygons on the same base and between the same parallels?

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 $\operatorname{ar}(\triangle BPQ) = \frac{1}{2} \operatorname{ar}(\triangle ABC)$.

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

In the given figure,$ABCD$ is a trapezium with $AB \parallel DC$. $E$ is a point on extended $BC$. Prove that $ar(BDE) = ar(ACED)$.

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides $8 \, cm$ and $6 \, cm$ is:

$ABCD$ is a trapezium with parallel sides $AB = a \text{ cm}$ and $DC = b \text{ cm}$. $E$ and $F$ are the mid-points of the non-parallel sides. The ratio of $\operatorname{ar}(ABFE)$ and $\operatorname{ar}(EFCD)$ is