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.