Proving that a square inscribed in a rhombus (with unequal diagonals) has sides parallel to the rhombus' diagonals 
How would you prove that a square inscribed in a Rhombus (with unequal diagonals) has its sides parallel to the diagonals of the rhombus using elementary geometry?

Actually I have spent a LOT of time doing this, but couldn't prove it so please provide a mathematically rigorous proof.
 A: 
Take a look at triangles $AEI$ and $DGH$. Angles $\angle AIE$ and $\angle DGH$ are obviously equal (parallel sides). In the same way angles $\angle AEI$ and $\angle DHG$ are also equal. And EI=GH so by ASA triangles $AEI$ and $DHG$ are congruent. Therefore:
$$AE=DH$$
...and consequentially:
$$BE=CH$$
Denote with $F$ intersection of rhombus' diagonals. Now take a look at triangles $DHF$ and $AEF$: $\angle EAF=\angle HDF, AE=DH,AF=DF$, so by SAS triangles $DHF$ and $AEF$ are congruent and $\angle HFD=\angle EFA,FE=FH$. So points $E,F,H$ are colinear and line $EFH$ is a diagonal of the square. You can easily prove in the same way that $IFG$ is also a digalonal of the square.
Now draw perpendicular lines from $F$ to sides $AB$ and $CD$. Triangles $EFJ$ and $GFK$ are congruent by SSA ($FE=FG,FJ=FK,\angle EJF=\angle GKF$). Therefore:
$$EJ=KG$$
Triangles $BJF$ and $BKF$ are also congruent by ASA (all angles are equal and triangles share a common side $BF$). Consequentially:
$$JB=BK$$
It means that:
$$EJ+JB=BK+KG\iff BE=BG$$
Because of $FE=FG,BE=BG$ line $BF$ is the median of side $EG$. In rhombus $ABCD$, $BF$ is also median of diagonal $AD$. Because of that, lines $EG$ and $AD$ are parallel (perpendicular to the same median BF). It's trivial to show that all other square sides must be parallel to rombus' diagonals.
EDIT: You can use the same proof in case of inscribed rectangle, not just square. Reducing the problem to a square wasn't necessary.
