Is there any pure geometric proof for this primary geometry problem? The original problem can be found here:
Nick's Mathematical Puzzle 62: Four squares on a quadrilateral : Squares are constructed externally on the sides of an arbitrary quadrilateral. Show that the line segments joining the centers of opposite squares lie on perpendicular lines and are of equal length.
$ABCD$ is an arbitrary quadrilateral; $E,F,G$ and $H$ are centers of squares outside the quadrilateral. Prove $EF\bot GH$ and $\overline{EF}= \overline{GH}$.

The solution presented is using the geometric meaning of complex numbers.
Is there any pure geometric approach to prove it?
 A: Obviously, we can assume that $ABCD$ is oriented counterclockwise as in the figure. (Otherwise reflect through some line.)
For any point $M$ we write $\rho_M$ for the counterclockwise $90^{\circ}$ rotation about $M$, and $\sigma_M$ for the symmetry centred at $M$.
Lemma Given two points $M$ and $N$, we have $\rho_N \circ \rho_M = \sigma_O$ for some point $O$. Moreover $\rho_O(N) = M$.
Proof It is well-known that a composite of two rotations is a rotation whose angle is the sum of the oriented angles of the individual rotations (or a translation if this sum is $0$.) Thus there is a point $O$ for which $\rho_N \circ \rho_M = \sigma_O$.
Now write $M' = \sigma_O(M)$ and $N' = \sigma_O(N)$. $MNM'N'$ is a parallelogram with centre $O$. But since 
$$\rho_N(M) = \rho_N \circ \rho_M(M) = \sigma_O(M) = M',$$
the parallelogram $MNM'N'$ is in fact a clockwise-oriented square. The lemma follows immediately from this.
Proof of the theorem
Let $O$ be the midpoint of the segment $AC$. 
We have $\rho_H \circ \rho_E = \sigma_{O'}$ for some point $O'$ as in the lemma. But since $\sigma_{O'}(A) = \rho_H \circ \rho_E (A) = C$, the point $O'$ must in fact be $O$. Thus $\rho_O(H) = E$.
Considering $\rho_G \circ \rho_F$ and exchanging the roles of $A$ and $C$, we similarly prove $\rho_O(G) = F$.
Putting these facts together, we see that $\rho_O$ transforms the segment $HG$ into $EF$. This proves what we want.
A: This was not intended to be an answer; I posted it as an answer so that I could attach a figure for illustration. But now it solves the problem.
I would prefer to prove a lemma that for any $\Delta ABC$ (in the figure below), $\Delta MEF$ obtained is an isosceles right-angled triangle, where $M$ is the mid-point of $AC$. $E$ and $F$ are centers of squares. 
Update 1
Now it seems the lemma can be proved by adding auxiliary conductors like below:

Then $\Delta ABG\cong\Delta BPQ \cong \Delta CGB$, this can easily lead to:
$\overline{AM}=\overline{BN}, \overline{BM}=\overline{PN}$ and therefore:
$\Delta ABM\cong \Delta BPN$; similarly $\Delta BCM\cong\Delta QBN$
Then $\angle MEN$ and $\angle MFN$ are both right angles; and $\Delta MEN$ and $\Delta MFN$ are isosceles; $ENFM$ is a square.
Update 2
By applying lemma above it is easy to conclude that: $\Delta EOF $ and $ \Delta HOG$ are congruent to each other and one can be obtained by rotating another around $O$ by 90°.

