<s>[INST] $\angle DAB = 10^\circ$, $\angle CAD = 20^\circ$, $\angle CDB = 140^\circ$, $CD = BD$. Find $\angle ABD$ $\angle DAB = 10^\circ$
$\angle CAD = 20^\circ$
$\angle CDB = 140^\circ$
The triangle $\triangle BCD$ is isosceles. $CD = BD$
Find the angle $x$

The answer is $30º$.
Playing with GeoGebra, I found that $AC = AD$. But I don't know how to prove it.
 [/INST] Set $\angle ACD= y, x+y=110^\circ$
For $\triangle ADC$ and $\triangle ADB$:
\begin{align}\frac{DB}{AD}=\frac{\sin 10^\circ}{\sin x}=\frac{CD}{AD}=\frac{\sin 20^\circ}{\sin (110^\circ -x)}\end{align}
\begin{align}\frac{\sin 10^\circ}{\sin 20^\circ}=\frac{sin x}{\cos(x-20^\circ)}\end{align}
\begin{align}\tan x=\frac{\cos 20^\circ}{2\cos 10^\circ -\sin 20^\circ}=\frac{\sqrt 3}{3}^*, x=30^\circ\end{align}
EDIT:
\begin{align}\frac{\cos 20^\circ}{2\cos 10^\circ -\sin 20^\circ}=\frac{\cos 20^\circ}{\cos 10^\circ+\cos 10^\circ-\sin 20^\circ}=\frac{\cos 20^\circ}{\cos 10^\circ+\sin 80^\circ-\sin 20^\circ}=\frac{\cos 20^\circ}{\cos 10^\circ+\cos 50^\circ}=\frac{\cos 20^\circ}{2\cos 20^\circ\cos 30^\circ}=\frac{\sqrt 3}{3}\end{align}
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