Prove that one of the diagonals of quadrilateral EFGH is parallel to one of the sides of square ABCD if E,F,G,H are points on AB,BC,CD,DA such that.. Prove that one of the diagonals of quadrilateral $EFGH$ is parallel to one of the sides of square $ABCD$ if $E$, $F$, $G$, $H$ are points on $AB$, $BC$, $CD$, $DA$ ($E$ on $AB$, $F$ on $BC$ and so on) such that the area of $EFGH$ is half of the area of $ABCD$.
What I've figured out so far is that if the area of $EFGH$ is half of $ABCD$'s area then the area of the triangles $AEH$+$BEF$+$CFG$+$DGH$ is also half of $ABCD$, that means that $\frac{AE*AH}{2} + \frac{BE*BF}{2} + \frac{CF*CG}{2} + \frac{DG*DH}{2} = \frac{a^2}{2}$ where a is the side of square $ABCD$.
I'm not sure where to go from here, any help is appreciated.
 A: 
Suppose the opposite. Let none of the diagonals of the quadrilateral $EFGH$ be parallel to the sides of the square $ABCD$. Draw straight lines through the points $E$, $F$, $G$, $H$ parallel to the sides of the square. Since both diagonals are not parallel to the sides, these lines do not coincide. Then
$$S_{\triangle AEH} = S_{\triangle EIH}, \quad S_{\triangle EBF} = S_{\triangle EFL}, \quad S_{\triangle FCG} = S_{\triangle FKG}, \quad S_{\triangle GHD} = S_{\triangle HJG}.$$
Thus,
$$S_{EFGH} =S_{\triangle AEH} +  S_{\triangle EBF} +  S_{\triangle FCG} + S_{\triangle GHD} =  S_{\triangle EIH} + S_{\triangle EFL} + S_{\triangle FKG} + S_{\triangle HJG}.$$
On the other hand,
$$S_{\triangle EIH} + S_{\triangle EFL} + S_{\triangle FKG} + S_{\triangle HJG} = S_{EFGH} + S_{IJKL} > S_{EFGH},$$
since $\triangle EFL$ and $\triangle HJG$ intersect in a rectangle $IJKL$. We got a contradiction.
It should be noted that point $E$ may be located above point $G$, and point $H$ may be located to the right of point $F$ (not as in the picture). In these cases, the proof will be similar.
