Finding angle with geometric approach I would like to solve the problem in this picture:

with just an elementary geometric approach. I already solved with trigonometry, e.g. using the Bretschneider formula, finding that the angle $ x = 15° $. Any idea?
This is 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: I do not know if this is elementary enough, though the exercise could fit in an Olympiad style easily. So here is another analytic way. In the figure, let the point $D$ be the center $(0;0)$ and assume that the sides of the square are of unit length. So you get $C(1;0)$, $B(1;1)$, $A(1;\tan(X))$, $G(\frac{1}{\tan(X)};1)$.
Here we should bound $X$ so the problem has a meaning. If $E'$ is the symmetric of $E$ with respect to $F(0;1)$, your problem is equivalent to the following equation in $X$ (which characterises the angle bisector $(DE')$ in the triangle $DFG$):
$$(1)\quad\dfrac{DG}{DF}=\dfrac{E'G}{E'F}.$$
Finding the line $(EA)$ as $$y=\tan(X-\frac{\pi}{6})x+\tan(X)-\tan(X-\frac{\pi}{6}),$$ then $E\left(\dfrac{1+\tan(X-\frac{\pi}{6})-\tan(X)}{\tan(X-\frac{\pi}{6})};1\right).$
Putting this in $(1)$, squaring both sides expanding and simplifying, you get the equation $$(2\sqrt{3}+1)t^4- (2\sqrt{3}+4)t^3-4t^2+ (6\sqrt{3}-4)t-2\sqrt{3}+3=0,$$
where $t=\tan(X)$. A root is $t=\sqrt{3}$, the other roots are not acceptable from the figure condition on $X$, for example $x_E<0$ etc. (I guess this may be also made elementary).
