How to find the angle in this trapezoid?

Maybe this will help : By trial and error(https://www.geogebra.org/classic/yvwyjx59), we find $45^\circ$.

For $45^\circ$, we have in $\mathbb C$, with $A=0, \, D=1, \, C=e^{i\frac{\pi}{6}}$,$$E=\left(\frac{3-\sqrt{3}}{2},\frac{\sqrt{3}-1}{2}\right) \quad \text{ and } \quad DE=CD=\frac{\sqrt{6}-\sqrt{2}}{2}.$$ So, we have one solution.

In general, you can formulate the problem in algebraic form with $\tan(x)$ as unknown. So set a system of coordinates centered at $A(0,0)$, $D(1,0)$, and you basically want the ordinates of $B$ and $C$ to be equal. So, if $(\vec{AD},\vec{AC})=\alpha=2x$ we solve $$\dfrac{\tan(\frac{\pi}{2}-3x)\tan(2x+\frac{\pi}{12})}{\tan(\frac{\pi}{2}-3x)+\tan(2x+\frac{\pi}{12})}=\dfrac{\tan(\frac{\pi}{2}-x)\tan(2x)}{\tan(\frac{\pi}{2}-x)+\tan(2x)}$$ Expanding everything (With GeoGebra for example) for $0<\alpha<\frac{\pi}{2}$, you get to solve $\tan^4(x)+(-8\sqrt{3}-16)\tan^3(x)-6\tan^2(x) +1=0.$ Or $X^4+(-8\sqrt{3}-16)X^3-6X^2+1=0.$ This algebraically has only one real solution in $(0,1)$ for $X=\tan(x)=2-\sqrt{3}$ and $x=\frac{\pi}{12}.$

In picture the extension of AD meets circumcircle d of triangle ABC at S. We connect S to B. SB inter sects SB at U. Circle e is circumcircle of triangle ABU and intersects AS at D'. In circle d angles CBS and SAC are opposite to arc SC so we have:

$\angle CBS=15^o$

In circle e angles D'AE and D'BU are opposite to arc D'U so we have:

$\angle D'BU=15^o$

Which results in :

$\angle EBC=\angle D'BU+\angle CBS=30^o$

Since $BE=BC$ and BU is bisector of D'BC, then $BU\bot EC$ that is AB is diameter of circle e, hence $AD'B=90^o$.Also we have:

$\angle BEC=\angle ECB=\angle ABC= \frac{180-30}2=75^o$

so we have :

$\angle EBC=75-30 = 45^o$

Which results in:

$\angle D'AB=45^o\Rightarrow \angle CAB=45-15=30^o$

BF is the reflection of AC over D'Q where D' is the mid point of arc AD'B. we have :

$\overset{\Large\frown}{BC}=\overset{\Large\frown}{AF}\Rightarrow \angle FCA=\angle CAB$

That is $FC||AB$, so FC passes through D whic means D' is coincident on D which means circle e passes through D. Hence $\angle ADB=90^o$.