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
This is a variant of the original Langley's puzzle and it does have a geometric solution, albeit a bit involved.  Below is 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 $\tan x$,

$$\begin{align}
\tan x  &amp; = \frac{\sqrt3}{1+4\cos 20}  = \frac{\sqrt3\sin 20}{(\sin 20 +\sin 40 )+ \sin40} \\
&amp; = \frac{\sqrt3\sin 20}{2\sin30\cos10 +\sin 40} = \frac{\sqrt3\sin 20}{\sin 80 +\sin 40} = \frac{\sqrt3\sin 20}{\sqrt3\cos 20}  =\tan 20 \\
\end{align}$$

Thus, $x = 20$.
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.