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!
 A: 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 $\angle GAH>90°$ then $AH<NH$.
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:
$$
AH < NH = EM \le EB
$$
which is in contradiction with the hypothesis $AH=EB$.

A: HINT: prove that the triangles $$FCE=EBH=HAG=GDF$$ are congruent
A: 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?
A: Scheme of the proof:


*

*If $ABCD$ is rectangle, then it is easy to show that it is a square too (congruent triangles).

*If $ABCD$ isn't rectangle, then one of angles $A,B,C,D$ is obtuse one.
W.l.o.g., $A>90^{\circ}$.


*

*Then the distance from the line $AB$ to point $E$ is greater than $|AH|$.

*So, $|BE|>|AH|$; contradiction.




More detailed:
first, draw the case with right angle $\angle A$:

note that:


*

*point $A$ is on the semicircle $GAH$ (with diameter $GH$);

*line $GA$ is tangent to orange circle;

*line $AB$ is tangent to blue circle;

*$|AH|=|BE|$.


Now, build point $A'$ such that $|AH|=|A'H|$ and $\angle GA'H>90^{\circ}$:

 Point $A'$ belongs to orange circumference, and is inside the semicircle $GAH$ (inside $\triangle GAH$ too).
 So $\angle A'HG<\angle AHG$, hence ray $A'H$ cannot intersect blue circle.

So any segment $B'E$ (where $B'$ belongs to the ray $A'H$) has length greater than $|BE|$.
Here we came to contradiction: if $\angle GA'H$ is obtuse, then $|B'E|>|A'H|$.
