Nice problem!
Let $X$ be a point symmetric to $A$ with respect to $BD$. Let $Y$ be a point symmetric to $A$ with respect to $BX$. Let $Z$ be a point symmetric to $A$ with respect to $DX$.
Then $DA=DX=DZ$, $BA=BX=BY$, and $AX=XY=XZ$. Angle chasing gives
$$\angle ZXY=360^\circ - \angle AXZ - \angle YXA = 360^\circ - 2\angle AXD - 2\angle BXA = \angle XDA + \angle ABX = 2\angle BDA + 2 \angle ABD = 2\cdot 13^\circ+2\cdot 17^\circ = 60^\circ$$
which along with $XZ=XY$ implies that $XYZ$ is equilateral. Thus
$$\angle BZA = \angle BZX + \angle XZA = 30^\circ + 13^\circ = 43^\circ.$$
We also have
$$\angle DBZ = \angle DBX + \angle XBZ = 2\angle ABD = 34^\circ.$$
This means that $Z=C$. Therefore
$$\angle CDB = \angle ZDB = 3\angle BDA = 39^\circ.$$
Below is a trigonometric solution. Let $\angle CDB = x$. We use Snellius' theorem thrice:
\begin{align}
\frac{AC}{DC} & = \frac{\sin(13^\circ + x)}{\sin 64^\circ}, \\
\frac{DC}{CB} & = \frac{\sin 34^\circ}{\sin x}, \\
\frac{CB}{AC} & = \frac{\sin 86^\circ}{\sin 51^\circ}.
\end{align}
Multiplying yields
$$1=\frac{\sin(13^\circ + x)\sin 34^\circ \sin 86^\circ}{\sin 64^\circ \sin x \sin 51^\circ}$$
so $$\sin(13^\circ + x)\sin 34^\circ \sin 86^\circ = \sin 64^\circ \sin x \sin 51^\circ.$$
Using $2\sin A \sin B = \cos(A-B) - \cos(A+B)$ twice we get
$$\sin(13^\circ + x)\left(\cos 52^\circ - \cos 120^\circ \right) = \sin x \left(\cos 13^\circ + \cos 65^\circ\right).$$
Since $\cos 120^\circ = -\frac 12$, we have
$$\sin(13^\circ + x) \cos 52^\circ + \frac 12 \sin(13^\circ + x) = \sin x \cos 13^\circ + \sin x \cos 65^\circ.$$
Multiplying by two and using $2\sin A \cos B = \sin(A+B) + \sin(A-B)$ we infer
$$\sin(65^\circ + x) + \sin(x-39^\circ) + \sin(13^\circ + x) = \sin(x+13^\circ) + \sin(x - 13^\circ) + \sin(x+65^\circ) + \sin(x-65^\circ).$$
Therefore
$$\sin(x - 39^\circ) = \sin(x-13^\circ) + \sin(x-65^\circ).$$
We use now $\sin A + \sin B = 2 \sin \frac{A+B}2 \cos\frac{A-B}2$:
$$\sin(x-39^\circ) = 2\sin(x-39^\circ)\cos 26^\circ$$
or
$$\sin(x-39^\circ)(1-2\cos 26^\circ)=0.$$
Since $\cos 26^\circ \neq \frac 12$, we have $\sin(x-39^\circ)=0$ and so $x=39^\circ$.
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, $x = 15^\circ$.
Best Answer
Here is a proof that an angle inscribed in an arc is half the intersepted arc. That is, $m\angle APB=\frac{1}{2}m(arc \widehat{ACB})$
Following the figure, let $\mathrm{M}$ be the center of the circle, and $\angle APB=\alpha , \angle APM=\beta\\ \mathrm{Now, \: in \: \Delta MPB,}\\ \mathrm{MP}=\mathrm{MB}\\ \implies \angle MPB= \angle MBP= \alpha + \beta\\ \implies \angle CMB= 2 ( \alpha +\beta)\\ \: \\ \mathrm{In \: \Delta MPA,}\\ \mathrm{MP} =\mathrm{MA} \\ \angle MPA= \angle MAP= \beta \\ \implies \angle CMA=2\beta\\ \mathrm{Since \:} \angle CMB= \angle CMA+ \angle AMB\\ 2(\alpha+ \beta)= 2\beta + \angle AMB\\ \implies 2\alpha=\angle AMB\\ \: \\ \mathrm{And, \: since\:} \angle APB=\alpha, \mathrm{we \: have \: shown \:that} \\ \boxed{\angle APB=\frac{1}{2} \angle AMB} $
Now, your statement, that is, angles inscribed in the same arc are congruent, is a direct consequence of the previous theorem.
For in your case, $\angle BAC= \frac{1}{2}\angle BOC\\ \angle BDC=\frac{1}{2}\angle BOC\\ \implies \angle BAC=\angle BDC$