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.
 A: No constructions needed.
ADY and ABX have half the base and the same altitude (to different sides) so their areas are 1/4 of ABCD.
XYC has half the base and half the altitude so its area is 1/8 of ABCD.
Together they make 1/4+1/4+1/8=5/8, so what remains is 3/8.
A: Use $$S_{ABC}=\frac12 a\cdot h_a$$ and $$S_{ABCD}= a\cdot h_a$$
Let $Ar(ABCD)=S$, then $Ar(\triangle ABX)=\frac14S$, $Ar(\triangle ADY)=\frac14S$, $Ar(\triangle CXY)=\frac18S$
Then $Ar(\triangle AXY)=S-\frac14S-\frac14S-\frac18S=\frac38S$
A: 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.
