In parallelogram $ABCD$,$P$ is the midpoint of $CD$. Then,$\operatorname{ar}(ABCD) : \operatorname{ar}(PBC) = \dots$

If $E, F, G$ and $H$ are respectively the midpoints of the sides of a parallelogram $ABCD$,show that $ar(EFGH) = \frac{1}{2} ar(ABCD)$.

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 quadrilateral whose diagonal $AC$ divides it into two parts of equal area. Then $ABCD$:

In the figure,$X$ and $Y$ are the mid-points of $AC$ and $AB$ respectively,$QP \parallel BC$ and $CYQ$ and $BXP$ are straight lines. Prove that $\text{ar}(ABP) = \text{ar}(ACQ).$

$AC$ is one of the diagonals of quadrilateral $ABCD$. $BM$ and $DN$ are altitudes on $AC$ from $B$ and $D$ respectively. If $AC = 18 \, cm$,$BM = 10 \, cm$,and $DN = 6 \, cm$,then $ar(ABCD) = \dots \dots \, cm^2$.