Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is

$ABCD$ is a trapezium in which $AB \parallel DC$,$DC = 30 \, cm$ and $AB = 50 \, cm$. If $X$ and $Y$ are,respectively,the mid-points of $AD$ and $BC$,prove that $\operatorname{ar}(DCYX) = \frac{7}{9} \operatorname{ar}(XYBA)$.

In the figure,$CD \parallel AE$ and $CY \parallel BA$. Prove that $\operatorname{ar}(\triangle CBX) = \operatorname{ar}(\triangle AXY)$.

In the figure,if parallelogram $ABCD$ and rectangle $ABEM$ are of equal area,then:

In $\triangle ABC,$ if $L$ and $M$ are the points on $AB$ and $AC,$ respectively such that $LM \parallel BC.$ Prove that $\operatorname{ar}(\triangle LOB) = \operatorname{ar}(\triangle MOC).$

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.