A geometry problem (how to find angle x) The solution is $x=50^{\circ}$.

How to prove $x=50^{\circ}$ without trigonometry?
Find $K$ on $\overline{BD}$ such that $\angle KAB = 10^\circ$. Then $\angle AKB = 160^\circ = 2\cdot\angle ACB$; this makes $K$ the center of the circumcircle of $\triangle ABC$. Thus, $\overline{KB}\cong\overline{KC}$, whereupon $\angle KCB = 20^\circ$. Note also that $\triangle ACK$ must be equilateral. (Why?)

We also have that $\angle DKA = 20^\circ = \angle DAK$. This makes $\triangle DAK$ isosceles, so that $\overline{CD}$ bisects $\angle ACK = 60^\circ$. The result follows. $\square$
This is incorrect and just left for reference

The strategy to take appears to be to find all the angles in the centre at $D$  in terms of $x$ and then equate to $360$ but I don't get $x=50$

$\angle BDC=180-20-x=160-x$

$\angle ADB=180-30-10=140$

Use the largest triangle to find $\angle ACD$ 

$\angle ACD =180-70-30-x=80-x$

This means $\angle ADC =180-40-(80-x)=60+x$

At the centre you have $160-x+140+60+x=360$ 

The $x$ cancel and I briefly checked that $x=60$ is a solution as well.

See comments below as to why this fails