Show that a diagonal divides a parallelogram into two triangles of equal area.

$D$and $E$are points on sides $AB$ and $AC$ respectively of $∆ABC$such that $ar(∆BCD)=ar(∆BCE).$Prove that $DE\parallel BC.$

$P$ is any point on the diagonal$AC$of a parallelogram$ABCD$. Prove that $ar(∆ADP)=ar(∆ABP)$

In the adjoining figure, the diagonals$AC$ and $BD$of a quadrilateral $ABCD$intersect at $O.$
If$BO = OD,$prove that $ar(∆ABC)=ar(∆ADC)$

The vertex $\mathrm{A}$of $∆ABC$ is joined to a point $\mathrm{D}$ on the side $\mathrm{BC}$. The midpoint of $\mathrm{AD}$ is $\mathrm{E}$.

Prove that $\mathrm{ar}(∆\mathrm{BEC})=\frac{1}{2}\mathrm{ar}(∆\mathrm{ABC}).$

$\mathrm{D}$is the midpoint of side $\mathrm{BC}$ of $∆\mathrm{ABC}$ and $\mathrm{E}$ is the midpoint of $\mathrm{BD}$. If $\mathrm{O}$is the midpoint of $\mathrm{AE},$ prove that $\mathrm{ar}(∆\mathrm{BOE}) =\frac{1}{8} \mathrm{ar}(∆\mathrm{ABC}).$

In a trapezium $ABCD,$ $AB\parallel DC$ and $M$ is the midpoint of $BC$. Through $M$, a line $PQ\parallel AD$ has been drawn which meets $AB$ in $P$ and $DC$ produced in $Q,$ as shown in the adjoining figure. Prove that $ar\left(ABCD\right)=ar\left(APQD\right).$

In the adjoining figure, $ABCD$ is a quadrilateral. A line through$D$, parallel to $AC$, meets $BC$ produced in$P$. Prove that$ar(∆ABP) =ar(\square \mathrm{ABCD}).$