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.


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 $DI$.

