Geometry: Find angle x in triangle I have not been able to find a euclidean geometry solution to this, but any other solutions are also appreciated.

Let ABC be a triangle with AB=CD and angles as marked in the diagram. Find the measure in degrees of angle $x$.
Answer: $15^o$.
 A: let $AB = DC = 1, AD = a.$  the by the rule sine applied to the triangle $ABD,$ we have $$\frac a{\sin x} = \frac 1{\sin 3x}\to a = \frac{\sin x}{\sin 3x} = \frac{\sin x}{\sin x \cos 2x + \sin 2x \cos x} = \frac{1}{\cos 2x + 2\cos^2x} $$
apply the rule of sine to the triangle $ABC,$ then $$\frac{1+a}{\sin 7x} =\frac1{\sin 9x}$$  so we need to solve $$\sin 9x =(\sin 7x - \sin 9x)(\cos 2x + 2\cos^2 x),\, 0 < x < 20^\circ.$$ 
this has one positive solution in the $x = 15^\circ$ in the interval $0 < x < 20^\circ.$
A: Here is purely synthetic approach.
We know that $AB=DC$ so there exists rotation $\varphi$ such that $\varphi(D)=A$ and $\varphi(C)=B$. Denote $E=\varphi(A)$ and $F=\varphi(B)$.
We have $EF=AB=DC$ and $\angle AEF=\angle DAB=2x$. Moreover $E \in AB$ and $AE=AD$. Thus $\angle DEA=\frac 12 \angle DAB=x=\angle EBD$. So $DE=BD$.
We claim that $\triangle DEF \equiv \triangle BDC$. We have $DE=BD$, $EF=DC$ and $\angle DEF = \angle DEA+\angle AEF=2x+x=3x=\angle CDB$, so by SAS these triangles indeed are congruent.
In particular $DF=BC$. 
Angle chasing gives $\angle DAF = \angle DAB + \angle BAF = 2x+\angle CDB = 2x+3x = 6x-x = \angle DBC - \angle DEA = \angle FDE - \angle ADE = \angle FDA$ therefore $AF=DF$. 
So $DB=AF=DF=BC$ which implies that $\angle CDB = \angle BCD$. Suming up angles of $\triangle DBC$ we get $180^\circ = \angle DBC + \angle BCD + \angle CDB = 6x+3x+3x=12x$. Thus $x=15^\circ$.

