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

$X$ 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 $\operatorname{ar}(LZY) = \operatorname{ar}(MZYX)$.

In the figure,$ABCD$ is a parallelogram. Points $P$ and $Q$ on $BC$ trisect $BC$ into three equal parts. Prove that $\operatorname{ar}(APQ) = \operatorname{ar}(DPQ) = \frac{1}{6} \operatorname{ar}(ABCD)$.

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

In $\triangle ABC$,$AD$ is a median. $E$ is the midpoint of $BD$ and $O$ is the midpoint of $AE$. Prove that $ar(AOB) = \frac{1}{8} ar(ABC)$.

$ABCD$ is a quadrilateral whose diagonal $AC$ divides it into two parts of equal area. Then $ABCD$: