Area of a triangle whose vertices lie on a parallelogram In the parallelogram $ABCD$, $X$ and $Y$ are the midpoints of $BC$ and $CD$. Then prove that $$Ar(\triangle AXY) = \frac {3}{8} Ar(ABCD)$$

My Attempt :
Construction; Joining $BY$ and $AC$, I got 
$$\triangle AYC=\triangle BCY$$.
But I couldn't move further from here.

Draw the diagonal $AC$. We see that $\triangle ACY$ has the same area as $\triangle ADY$, by choosing $CY$ and $DY$ to be the bases, since they have the same height. In the same way, $\triangle ACX$ has the same area as $\triangle ABX$. Therefore, the quadrilateral $AYCX$ takes up half of the parallelogram.
It remains to show that $\triangle CXY$ has area one eighth that of the parallelogram. Draw two midlines through the parallelogram, from $X$ to the midpoint of $AD$, and from $Y$ to the midpoint of $AB$. This divides the parallelogram into four congruent parts, and the segment $XY$ divides one of them in half. Thus we are done.
