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

In trapezium $ABCD$,$AB || CD$ and diagonals $AC$ and $BD$ intersect at point $O$. Prove that $ar(AOD) = ar(BOC)$.

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

$ABCD$ is a trapezium in which $AB \parallel DC$,$DC = 30 \, cm$ and $AB = 50 \, cm$. If $X$ and $Y$ are,respectively,the mid-points of $AD$ and $BC$,prove that $\operatorname{ar}(DCYX) = \frac{7}{9} \operatorname{ar}(XYBA)$.

$ABCD$ is a parallelogram and $BC$ is produced to a point $Q$ such that $AD = CQ$. If $AQ$ intersects $DC$ at $P$,show that $\operatorname{ar}(BPC) = \operatorname{ar}(DPQ)$.

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 $ar(\triangle BDE) = \frac{1}{4} ar(\triangle ABC).$