$ABCD$ 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 $\operatorname{ar}(DCYX) = \frac{7}{9} \operatorname{ar}(XYBA)$.

(N/A) In $\Delta MBY$ and $\Delta DCY$,we have:
$\angle 1 = \angle 2$ [Vertically opposite angles]
$\angle 3 = \angle 4$ [Since $AB \parallel DC$ and alternate interior angles are equal]
$BY = CY$ [Since $Y$ is the mid-point of $BC$]
Therefore,$\Delta MBY \cong \Delta DCY$ [By $ASA$ Congruence Rule]
So,$MB = DC = 30 \, cm$ [$CPCT$]
Now,$AM = AB + BM = 50 \, cm + 30 \, cm = 80 \, cm$
In $\Delta ADM$,by the Mid-point Theorem,$XY = \frac{1}{2} AM = \frac{1}{2} \times 80 \, cm = 40 \, cm$
As $AB \parallel XY \parallel DC$ and $X$ and $Y$ are the mid-points of $AD$ and $BC$,the heights of the trapeziums $DCXY$ and $XYBA$ are equal. Let the equal height be $h \, cm$.
$\frac{\operatorname{ar}(DCXY)}{\operatorname{ar}(XYBA)} = \frac{\frac{1}{2}(DC + XY) \times h}{\frac{1}{2}(XY + AB) \times h} = \frac{30 + 40}{40 + 50} = \frac{70}{90} = \frac{7}{9}$
Hence,$\operatorname{ar}(DCXY) = \frac{7}{9} \operatorname{ar}(XYBA)$.