Nice x in geometry

$|\overline {CD}|=2|\overline {AD}|\sin 20\degree$ .

In $\triangle {BCD}, \dfrac{|\overline {CD}|}{\sin 20\degree}=\dfrac {|\overline {BD}|}{\sin 120\degree}$

$\implies |\overline {BD}|=\sqrt 3|\overline {AD}|$ .

In $\triangle {ABD}, \dfrac {|\overline {AD}|}{\sin (110\degree-x)}=\dfrac {|\overline {BD}|}{\sin (40\degree+x)}$

$\implies \tan x=\dfrac {\sqrt 3\cos 20\degree-\sin 40\degree}{\cos 40\degree-\sqrt 3\sin 20\degree}$

$=\dfrac {\sin 20\degree+\sqrt 3\cos 20\degree}{\cos 20\degree-\sqrt 3\sin 20\degree}$

$=\dfrac{\cos (20\degree-\tan^{-1} \frac{1}{\sqrt 3})}{\cos (20\degree+\tan^{-1} \sqrt 3)}$

$=\dfrac {\cos 10\degree}{\cos 80\degree}=\tan 80\degree$

$\implies x=\boxed {80\degree}$ .

Nice solution from @Nibedan Mukherjee . To show that the solution works, we have to show that $AB=AC=AD$ . Then $A$ is the center of a circle of radius $AB$ and $\triangle ABD$ is isosceles, and $x = 180^\circ - 2(30^\circ) - 40^\circ = \boxed{80}^\circ$ . Let $AB=AC=AD = 1$ . We need to show $AB=1$ .

Since $\triangle ACD$ is isosceles, $AC=AD=1$ . Then $CD=2\sin 20^\circ$ . Let the diagonals of $ABCD$ intersect at $E$ . We note that $\triangle CDE$ is isosceles and $CE=4\sin^2 20^\circ$ . Then $AE=AC-CE = 1-4\sin^2 20^\circ=2\cos 40^ circ -1$ .

By sine rule , $\dfrac {BE}{\sin 50^\circ} = \dfrac {CE}{\sin 20^\circ} \implies BE = 4\sin 20^\circ \sin 50^\circ = 4\sin 20^\circ \cos 40^\circ$ .

By cosine rule ,

$\begin{aligned} AB^2 & = AE^2 + BE^2 - 2AE\cdot BE \cos \angle AEB \\ & = (2\cos 40^\circ-1)^2 + (4\sin 20^\circ \cos 40^\circ)^2 - 8(2\cos 40^\circ-1)\sin 20^\circ \cos 40^\circ \blue{\cos 70^\circ} \quad \quad \small \blue{\text{As }\cos 70^\circ = \sin 20^\circ} \\ & = 4\cos^2 40^\circ - 4\cos 40^\circ + 1 + \cancel{16\sin^2 20^\circ \cos^2 40^\circ} - \cancel{16\sin^2 20^\circ \cos^2 40^\circ} + 8 \sin^2 20^\circ \cos 40^\circ \\ & = 4\cos^2 40^\circ + 1 - 4\cos 40^\circ (1-2\sin^2 20^\circ) \\ & = \cancel{4\cos^2 40^\circ} + 1 - \cancel{4\cos^2 40^\circ} \\ & = 1 \\ \implies AB & = 1 \end{aligned}$