Using Euclidean geometry, how to find $x$? This question comes from a friend exam that I'm helping to review.
I've been trying hard but can't find the answer.
Using Euclidean geometry, how to find the angle $x$?



I've been able to work out all the angles based on $x$ and $180^{\circ} $, but then I got stuck.
Here's my calculation:
I named the center point as $E$
$$ \angle ABD = 180^{\circ}  - 6x \\ 
   \angle ABC = 180^{\circ}  - 3x $$
$$\angle ABC = \angle ABD + \angle DBC \\
180^{\circ}  - 3x = 180^{\circ}  - 6x + \angle DBC $$
$$ \begin{align} \angle DBC &= 3x \\
\angle BEC &= 180^{\circ} - 4x\\
\angle BDC &= 180^{\circ} - 5x \end{align}$$
$$ \angle DEC = 180^{\circ} - \angle ACD - \angle BDC = 180^{\circ} - (x + 180^{\circ} - 5x) = 4x $$
Would anyone be able to help me with this question?

We have $\angle BAD=3x=\angle ADB$, hence $AB=BD$.
Let $E$ be the point on $AC$ such that $BE=AB$. Then $A,D,E$ lie on a circle with center $B$, hence $\angle EBD =2\angle EAD = 2x$.
On the other hand, $\angle AEB=\angle BAE=2x$, hence $\angle CBE=\angle AEB-\angle ACB=2x-x=x$.
We see that $\triangle CEB \sim \triangle CDA$ because these triangles have equal angles. In particular $\frac{CE}{CD}=\frac{CB}{CA}$. Since $\angle ACB=x=\angle DCE$, we have $\triangle ACB \sim\triangle DCE$ by SAS. Hence $\angle EDC =\angle BAC=2x$.
So $\angle DEA=\angle EDC+\angle DCE=2x+x=3x$. So $\angle DEB=\angle DEA+\angle AEB =3x+2x=5x$. Also $\angle BDE =\angle DEB=5x$ because $BD=BE$. We see now that the sum of angles of triangle $EBD$ equals $12x$. This leads to $12x=180^\circ$, hence $x=15^\circ$.

