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:

In trapezium $ABCD$,$AB || CD$ and diagonals $AC$ and $BD$ intersect at point $O$. Prove that $ar(AOD) = ar(BOC)$.

In $\Delta ABC$,$\angle B = 90^{\circ}$,$BC = 8 \, \text{cm}$,and $AC = 17 \, \text{cm}$. $BE$ is a median of the triangle and $M$ is the midpoint of $BE$. Find the area of $\Delta BMC$ in $\text{cm}^2$.

$X$ and $Y$ are points on the side $LN$ of the triangle $LMN$ such that $LX = XY = YN$. Through $X$,a line is drawn parallel to $LM$ to meet $MN$ at $Z$ (See figure). Prove that $\operatorname{ar}(LZY) = \operatorname{ar}(MZYX)$.

$ABCD$ 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,$ABCDE$ is any pentagon. $BP$ is drawn parallel to $AC$ and meets $DC$ produced at $P$,and $EQ$ is drawn parallel to $AD$ and meets $CD$ produced at $Q$. Prove that $\operatorname{ar}(ABCDE) = \operatorname{ar}(APQ)$.