Finding angle with geometric approach I would like to solve this problem just with an elementary geometric approach. I already solved with trigonometry, e.g. using the Bretschneider formula, finding that the angle $ x = 15° $. Any idea?
I edited showing how I computed the $ x $ value using the Bretschneider formula for the area of the quadrilateral $ ABDE $ and equating to the sum of the triangles' area $ ABE + EFD + BDF $
$$\begin{cases}
BC = a \\
AB = a(1/\tan(2x) - 1) \\
BD = a\sqrt{2} \\
AE = AB/\cos(2x+\pi/6) = a(1/\tan(2x) -1)/\cos(2x+\pi/6) \\
ED = a/\cos(x)
\end{cases}
$$
So I solved this equation with Mathematica, and the only solution that fit the problem is $ x = \pi/12 $
$ a^2/2+(a^2(1/\tan(2x) - 1)(1+\tan(x)))/2 + a^2 \tan(x)/2 =
((a\sqrt{2})^2 + \\
(a(1/\tan(2x)- 1)/\cos(2x+\pi/6))^2 - (a/\cos(x))^2 -(a(1/\tan(2x) - 1))^2)/4 \tan(\pi/2 -2x) $
I guess there is a simpler trigonometric approach, but I just wanted to try with that formula.

 A: Hint: Draw a circle on diameter $AD$. Extend $AE$ to meet the circle at $H$. Connect $D$ to $H$; we have:
$\overset{\frown} {DH}=60^o$
Draw diameter $HJ$ ($J$ is where diameter starting from $H$ and passing the center of circle $O$ meets the circle). Draw a perpendicular from $D$ on $HJ$ to meet the circle at $I$, we have:
$\overset{\frown} {DH}=\overset{\frown} {IH}=60^o$
therefore:
$\widehat {HDI}=\frac 12 \overset{\frown} {DH}=30^o$
Also
$\overset{\frown} {AJ}=\overset{\frown} {DH}=60^o$
Therefore:
$\overset{\frown} {AI}=60^o\Rightarrow \widehat{IDA}=30^o=2x $
which gives $x=15^o$
What remains is to show $F$ is on $DI$. For this note that the radius of circle is equal to the measure of sides of square, that is if you draw a circle centered at D and radius $r=DC$, it crosses points O(center of the first circle and of course vertex F of the square.  $DC\bot AC$ and also $FG||DC$ that means $FG\bot DI$ so $F$ is on $DI$.

A: Here I report the original figure, to which I added the segments $DH$ bisecting the angle $\widehat{FDG}$ and $EK$ orthogonal to $AD$.

Let's define
$$
d = AB, \qquad e = EF.
$$
We have, by Pythagorean theorem,
$$
AD = \sqrt{ AC^2 + CD^2 } = \sqrt{ (a+d)^2 + a^2 }.
$$
where $a$ is the known length of the side of the square.
For what follows, it would be useful to set $R=\sqrt{ (a+d)^2 + a^2 }.$
From the similar triangles $DFG$ and $ACD$
$$
\frac{FG}{CD} = \frac{DF}{AC} \quad \implies \quad FG = \frac{a^2}{a+d}.
$$
By Pythagorean theorem
$$
DG = \sqrt{ FG^2 + DF^2 } = \frac{aR}{a+d}.
$$
By the angle bisector theorem
\begin{align}
& \frac{GH}{FH} = \frac{DG}{DF}, \\[2mm]
& \frac{GH+FH}{FH} = \frac{DG+DF}{DF}, \\[2mm]
& \frac{FG}{FH} = \frac{DG+DF}{DF}, \\[2mm]
& FH = R-(a+d).
\end{align}
Note that
$$
EF = FH \quad \implies \quad e = R-(a+d)
$$
By Pythagorean theorem
$$
AE = \sqrt{ AB^2 + BE^2 } = \sqrt{ d^2 + (a+e)^2 }
$$
Being $AEK$ a 30-60-90 triangle, then
\begin{align}
& AK = \frac{\sqrt{3}}{2} AE = \frac{\sqrt{3}}{2} \sqrt{ d^2 + (a+e)^2 } \\
& EK = \frac{1}{2} AE = \frac{1}{2} \sqrt{ d^2 + (a+e)^2 }
\end{align}
We have
$$
DK = AD - AK = R - \frac{\sqrt{3}}{2} \sqrt{ d^2 + (a+e)^2 }
$$
By Pythagorean theorem
$$
DE^2 = EK^2 + DK^2 = \frac{1}{4} [ d^2 + (a+e)^2 ] + \left( R - \frac{\sqrt{3}}{2} \sqrt{ d^2 + (a+e)^2 } \right)^2
$$
but also
$$
DE^2 = DF^2 + EF^2 = a^2 + e^2
$$
Comparing these two expressions for $DE^2$, we have
$$
d^2+(a+e)^2+(a+d)^2-e^2=\sqrt{3}R\sqrt{d^2+(a+e)^2}
$$
Taking into account
\begin{align}
& e^2 = 2(a+d)^2+a^2-2(a+d)R \\
& (a+e)^2 = 2(a+d)^2-2d(R+a)
\end{align}
and squaring both sides
$$
4 \left(aR+d^2\right)^2 = 3 \left[(a+d)^2+a^2\right] \left[2 (a+d)^2+d^2-2 d \left(R+a\right)\right]
$$
by expanding the products and isolating the root $R$
$$
2 d \left(6 a^2+10 a d+3 d^2\right) R=4 a^4+16 a^3 d+32 a^2 d^2+24 a d^3+5 d^4
$$
squaring again both sides and factoring
$$
\left(2 a^2-2 a d-d^2\right) \left(8 a^6+72 a^5 d+188 a^4 d^2+208 a^3 d^3+122 a^2 d^4+50 a d^5+11 d^6\right)=0
$$
The only positive solution of this equation is
$$
d=(\sqrt{3}-1)a
$$
but this means that $AC = \sqrt{3}a$, the triangle $ACD$ is a 30-60-90 triangle, then $\widehat{CAD}=2x=30°$, so $x=15°$.
