Area of a parallelogram $=$ .........

In the figure,$ABCD$ and $AEFD$ are two parallelograms. Prove that $\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.

In $\triangle ABC$,$AD$ is a median. $E$ is the midpoint of $BD$ and $O$ is the midpoint of $AE$. Prove that $ar(AOB) = \frac{1}{8} ar(ABC)$.

The diagonals of a parallelogram $ABCD$ intersect at a point $O$. Through $O$,a line is drawn to intersect $AD$ at $P$ and $BC$ at $Q$. Show that $PQ$ divides the parallelogram into two parts of equal area.

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

If a triangle and a parallelogram are on the same base and between the same parallels,then the ratio of the area of the triangle to the area of the parallelogram is: