Langley's Adventitious Angles${}$ I've been running in circles and couldn't give a rigorous mathematical proof that the angle is x = 20°. 

Any idea? This is my try:
I got the answer $x=20^\circ$ using a computer program: https://www.geogebra.org/classic/qt79hpec
 A: This is a variant of the original Langley's puzzle, which has a straightforward trigonometric solution. Apply the sine rule to the triangles ADE, ADB and BDE
$$\frac{\sin x}{\sin 10}\cdot \frac{\sin 20}{\sin (30+x)}\cdot \frac{\sin 80}{\sin 60}
=\frac{DA}{DE}\cdot \frac{DE}{DB}\cdot \frac{DB}{DA} = 1$$
which simplifies to
$$2\cos^210\sin x = \sin60\sin(30+x) =\frac{\sqrt3}4\cos x + \frac{3}4\sin x$$
Solve for $\tan x$,
$$\begin{align}
\tan x  & = \frac{\sqrt3}{1+4\cos 20}  = \frac{\sqrt3\sin 20}{(\sin 20 +\sin 40 )+ \sin40} \\
& = \frac{\sqrt3\sin 20}{2\sin30\cos10 +\sin 40} = \frac{\sqrt3\sin 20}{\sin 80 +\sin 40} = \frac{\sqrt3\sin 20}{\sqrt3\cos 20}  =\tan 20 \\
\end{align}$$
Thus, $x = 20$.
A: 
Construct a line from $A$ that is $60^\circ$ off of $AB$.
It intersects $BC$ at $M$ and $BD$ at $P$
$\triangle ABP$ is equilateral.
$\triangle AMB \cong \triangle BDA$
$\triangle DMP$ is equilateral
$MP \cong DM$ 
$CP$ bisects $\angle C$
As $\angle MCA = \angle MAC = 20^\circ$ then $\triangle MCA$ is isosceles
$\triangle CMP \cong \triangle AME\\
MP\cong ME\\
DM\cong ME$
$\angle DMC = 80^\circ\\
\angle DEM = 50^\circ\\
\angle AEM = 30^\circ\\
\angle AED = 20^\circ$
A: The large triangle is isosceles. Let the left base angle be $B.$ Them in the triangle $BCE,$ we find that $B=10°,$ as given, then $C=20°$ (consider the large isosceles triangle). Thus, we have that $C\hat E D=(70+60+20)°=150°.$ Hence we have that $$150+x+10+20=180.$$ This shows us that  $x=0,$ or in other words that the configuration is impossible.
