In a parallelogram $ABCD$, any point $E$is taken on the side $BC$. $AE$ and $DC$when produced meet at a point $M.$Prove that
$ar(∆ADM)=ar\left(ABMC\right)$

$\mathrm{PQRS}$ are respectively the midpoints of the sides $\mathrm{AB}, \mathrm{BC}, \mathrm{CD}$ and $\mathrm{DA}$ of ${\parallel }^{gm}$ $\mathrm{ABCD}$. Show that $\mathrm{PQRS}$ is a parallelogram and also show that

$\mathrm{Ar}({\parallel }^{gm} \mathrm{PQRS}) = \mathrm{x} \mathrm{ar}({\parallel }^{gm} \mathrm{ABCD}).$

The base $\mathrm{BC}\mathit{ }$of $∆\mathrm{ABC}$is divided at $\mathrm{D}$such $\mathrm{BD} = \frac{1}{2}\mathrm{DC}$. Prove that $\mathrm{ar}(∆\mathrm{ABD}) = \frac{1}{3}\times \mathrm{ar}(∆\mathrm{ABC}).$

In the adjoining figure, $BD\parallel CA$, $E$ is the midpoint of$CA$and$BD=\frac{1}{2}CA$ . Prove that $ar(∆ABC)= 2ar(∆DBC).$

The given figure shows a pentagon$\mathit{ }\mathrm{ABCDE}, \mathrm{EG}$ drawn parallel to $\mathrm{DA}$ meets $\mathrm{BA}$ produced at $\mathrm{G}, \mathrm{and} \mathrm{CF}$ drawn parallel to $\mathrm{DB}$ meets $\mathrm{AB}$produced at $\mathrm{F}.$

Show that $\mathrm{ar}(\mathrm{pentagon} \mathrm{ABCDE}) = \mathrm{ar}(∆\mathrm{DGF}).$

In the adjoining figure,$CE \left|\right| AD$ and$CF \left|\right| BA$. Prove that $ar(∆CBG)=ar(∆AFG).$

In the adjoining figure, the points $\mathrm{D}$ divides theSide $\mathrm{BC}$ of $∆\mathrm{ABC}$ in the ratio $\mathrm{m}:\mathrm{n}.$ prove that $\mathrm{area}(∆\mathrm{ABD}): \mathrm{area}(∆\mathrm{ABC})=\mathrm{m}:\mathrm{n}$

In a trapezium $ABCD$,$AB \left|\right| DC, AB=a \mathrm{cm},$ and $DC=b \mathrm{cm}$. If $M$and $N$ are the midpoints of the nonparallel sides, $AD$ and $BC$ respectively then find the ratio of $ar\left(DCNM\right) \mathrm{and} ar\left(MNBA\right).$