If $P$ is any point on the median $AD$ of a $\triangle ABC$,then $\operatorname{ar}(ABP) = \operatorname{ar}(ACP)$. State whether this statement is True or False.

In the given figure,$PQM$ is a line and $SQ || RM$. Prove that $ar(PQR) = ar(PMS)$.

In the figure,$BD \parallel CA$,$E$ is the mid-point of $CA$,and $BD = \frac{1}{2} CA$. Prove that $\operatorname{ar}(ABC) = 2 \operatorname{ar}(DBC)$.

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

$D, E$ and $F$ are the midpoints of the sides $BC, CA$ and $AB$ respectively of $\Delta ABC$. Show that:
$(i)$ $BDEF$ is a parallelogram.
$(ii)$ $ar(DEF) = \frac{1}{4} ar(ABC)$
$(iii)$ $ar(BDEF) = \frac{1}{2} ar(ABC)$

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.