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.

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 $Ar(\triangle AXY)=S-\frac14S-\frac14S-\frac18S=\frac38S$
