Yet another geometrical alternative. Since $\angle BAC + \angle BDC = 180^\circ$, we rotate and translate $\triangle BCD$ into $\triangle FBA$ as shown in the figure below:
(Note that $A$ lies on $CF$. However, we cannot not assume a priori that $D$ lies on $BF$, so we cannot simply observe that $\triangle BCF$ is equilateral.)
Now
$$
\angle F = \angle CBD = 180^\circ - \angle BCD - \angle BDC = 75^\circ - x.
$$
But $BC = BF$, so
$$
\angle F = \angle BCF = \angle BCD + \angle ACD = 45^\circ + x,
$$
so
$$
75^\circ - x = 45^\circ + x.
$$
Therefore, $x = 15^\circ$.
In the figure, $\Delta AED$ together point $C$ constitute the given figure.
The constructions are:
(1) Equilateral triangle $BAC$,
(2) A circle with center $B$ and radius $BA$,
(3) Note that the circle passes through $E$ because $\angle AEC=\frac{\angle ABC}{2}$
(4) Join $DB$
Note that we do not assume that $D, E, B$ are collinear.
We are going to $\color{red}{prove}$ that $\color{red}{D, E, B}$ are $\color{red}{collinear}$.
$\color{red}{Proof:}$
Note that $\angle EAB=45^o$ and $BA=BE \implies \angle ABE=90^o$
i.e. $\color{red}{BE \bot BA}$
On the other hand, $AC=CB=CD \implies \angle ABD=90^o$
i.e. $\color{red}{BD \bot BA}$
The $2$ statements implies that $B, E, D$ are collinear.
Since $CB=CD, \angle BCA=60^o $ and $BCD$ is a trianlge, $\angle CDB=30^0$
Best Answer
In this case, trigonometric solution is easier than the geometric method.
From sin rule,
$$\frac{BC}{\sin\alpha}=\frac{12}{\sin2\alpha}=\frac{12}{2\sin\alpha\cos\alpha}\implies BC\cos\alpha=6.$$
And $CD=BC\cos\alpha=6$.