Find an angle in the given quadrilateral In the following problem, I want to find the angle marked as $x$. It seems so simple and yet I am out of ideas. It is very easy to get all angles except two of them: angle ADB and angle CBD. 

Is there a calculation for the angle $x$ that only uses parallel lines and no circles?

Edit: A nice solution by using a circle is given below. In the book, this question was asked at the start of the chapter after introducing parallel lines and all the angles in that setup. Sum of angles of a triangle was proved and this question was asked in the exercise.

From geogebra, I know the answer is 60 degrees. But I do not know how to argue that the answer is 60 degrees. 


Given:
  1) $\angle ABC=30^\circ$
  2) $\angle BAD=80^\circ$
  3) $\angle DAC=20^\circ$
  4) $\angle ACB=50^\circ$
  5) $\angle BCD=50^\circ$ 


1) (the key point of this solution) Let $F\in AB$ such, that $\angle BCF$ $=\angle FBC=30^\circ$, $\angle ACF$ $=\angle CAD=20^\circ$
2) $\angle ADC $ $= 180^\circ-\angle DAC-\angle ACB-\angle BCD$ $=180^\circ-20^\circ-50^\circ-50^\circ$ $=60^\circ$
3) $\angle AFC $ $= \angle FCB+\angle FBC$ $=30^\circ+30^\circ$ $=60^\circ=\angle ADC$ $\Rightarrow$
4) $ACDF$ is cyclic $\Rightarrow$
5) $\angle CFD=\angle CAD=20^\circ$
6) $\angle CDF$ $=180^\circ-\angle CFD-\angle FCB-\angle BCD$ $=180^\circ-20^\circ-30^\circ-50^\circ$ $=80^\circ$ $\Rightarrow$
7) $\angle FCD=\angle FDC$ $\Rightarrow$
8) $DF=CF$
9) $CF=FB$ $\Leftarrow$ $\angle FBC=\angle FCB=30^\circ$
10) (8-9$\Rightarrow$)$DF=FB$ $\Rightarrow$ $\angle FDB$ $=\angle FBD$ $=\frac{180^\circ-\angle BFD}{2}$ $=\frac{180^\circ-(\angle BFC-\angle CFD)}{2}$
$=\frac{\angle BCF+\angle FBC+\angle CFD}{2}$ $=\frac{30^\circ+30^\circ+20^\circ}{2}=40^\circ$
11) $\angle ADF=\angle ACF$ $\Leftarrow$ (4)
12) $\angle ADB=\angle ADF+\angle FDB=20^\circ+40^\circ=60^\circ$
13) Bonus: $\angle DFB=\angle CAB$ $\Rightarrow$ $AC||DF$

I can't see any parallel lines or circles as well.

So, construct them)
Seriously, $\angle BCF=30^\circ$ is the first thing that it wants to construct.
If needed with no circles, $\triangle AEF$ and $\triangle CED$ are congruent, thus $\triangle DEF$ is equilateral and similar to $\triangle CEA$, but I think it's a bit longer.
