prove ABCD is also a square What we know is that:
1. ABCD is a quadrilateral. 
2. The red area is a square. 
3. AH=BE=CF=DG



The question is prove that ABCD is also a square.
I have realised that the four triangles here AHG, DGF, EFC and HBE have the same length hypotenuse and also AH = DG = CF = BE, so if I can prove ∠ A, B, C, D are 90°, then four triangles are congruent. Then I will know that four sides, AB,BC,CD,DA are the same length then I can prove it. 
The problem is that I dont know how to prove angle A,B,C,D are 90 degree. 
Thanks!
If one among the angles at $A$, $B$, $C$, $D$ is a right angle, then it is easy to prove they are all right angles and $ABCD$ is a square. Suppose then none of them is a right angle: at least one of them ($\angle GAH$, for instance) must then be obtuse. I'll show that this leads to a contradiction.

Let $N$ be the foot of the perpendicular from $G$ to line $AH$. As $\angle GAH&gt;90°$ then $AH&lt;NH$.

Let $M$ be the foot of the perpendicular from $E$ to line $BH$: we have then $EM\le EB$. But triangles $EMH$ and $HNG$ are congruent (because they have $\angle NGH=\angle MHE=\pi/2-\alpha$, $\angle NHG=\angle MEH=\alpha$ and $GH=HE$), thus:
$$
AH &lt; NH = EM \le EB
$$
which is in contradiction with the hypothesis $AH=EB$.
Hint: extend side $AD$ and $EF$. Let's say they intersect at point $X$. Then you can see that $\angle{XFG}=90°$ and $\angle{DXF}\cong\angle{AGH}$. Can you take it from here?