Langley's Adventitious Angles${}$

Question

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

Answer 1

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.

Answer 2

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, $x = 20$.

Answer 3

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

$\triangle CMP \cong \triangle AME\\ MP\cong ME\\ DM\cong ME$

$\angle DMC = 80^\circ\\ \angle DEM = 50^\circ\\ \angle AEM = 30^\circ\\ \angle AED = 20^\circ$