Find $\angle BDC$ Quadrilateral $ABCD$ , $\angle ABD = 17^{\circ}, \angle DBC = 34^{\circ}, \angle ACB = 43^{\circ}, \angle ADB = 13^{\circ}$, Find $\angle BDC$.
 A: 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 $x=39^\circ$.
A: We can consider in $\mathbb{C}$ : $A=0$, $B=1$
Denote $AB=a=1, \,AC=c,\, AD=d$. Then:


*

*$\frac{a}{\sin 43^{\circ}}=\frac{c}{\sin 51^{\circ}}$, so $c=\frac{\sin 51^{\circ}}{\sin 43^{\circ}}$

*$C=B\cdot c(\cos{86^{\circ}}+i\sin 86^{\circ}) = \cos{86^{\circ}}\frac{\sin 51^{\circ}}{\sin 43^{\circ}} + i \sin 86^{\circ}\frac{\sin 51^{\circ}}{\sin 43^{\circ}}$

*$\frac{a}{\sin 13^{\circ}}=\frac{d}{\sin 17^{\circ}}$, so $d=\frac{\sin 17^{\circ}}{\sin 13^{\circ}}$

*$D=(-1)\cdot d(\cos(-30^{\circ})+i \sin(-30^{\circ}))=-\frac{\sqrt{3}}{2} \frac{\sin 17^{\circ}}{\sin 13^{\circ}} +\frac{i}{2}\frac{\sin 17^{\circ}}{\sin 13^{\circ}}$


Now we know that $\angle DBC = \arg(C-D)+\arg(B-D)$. $\arg(B-D)=17^{\circ}$, so we need only to compute $\alpha = \arg(C-D)$:
$\tan \alpha = \frac{\mathfrak{Im }(C-D)}{\mathfrak{Re }(C-D)}=\frac{\sin 86^{\circ}\frac{\sin 51^{\circ}}{\sin 43^{\circ}}-\frac{1}{2}\frac{\sin 17^{\circ}}{\sin 13^{\circ}}}{\cos{86^{\circ}}\frac{\sin 51^{\circ}}{\sin 43^{\circ}}+\frac{\sqrt{3}}{2} \frac{\sin 17^{\circ}}{\sin 13^{\circ}}}$
Well, now that looks a little creepy, but it wil surely get less complicated, if we will use some trigonometrical formulas (sine and cosine of sum, sine and cosine of doubled angle... and so on). See, that:


*

*$86^{\circ}=2\cdot 43^{\circ}$

*$43^{\circ}=30^{\circ}+13^{\circ}$

*$17^{\circ}=30^{\circ}-13^{\circ}$

*$51^{\circ}=90^{\circ}-3\cdot 13^{\circ}$


Unfortunately I don't have now enough time to deal with this without calculator, but in some days I'll manage some time to do it. 
Edit:
Wolfram Alpha [1] tells, that $\alpha=22^{\circ}$, so we obtain $\angle DBC = 39^{\circ}$.
[1] https://www.wolframalpha.com/input/?i=ArcTan((Sin(86Deg)Sin(51Deg)%2F(Sin(43+Deg))+-+1%2F2+Sin(17Deg)%2F(Sin(13+Deg)))%2F(Cos(86Deg)Sin(51Deg)%2F(Sin(43+Deg))+%2B+Sqrt(3)%2F2+Sin(17Deg)%2F(Sin(13+Deg))))
A: See the image for a correct solution.

The original and, incorrect answer follows.
Don't read what follows, it's wrong. 
The problem you presented has no unique solution.  This is easy to prove.
First, only one angle of the triangle BDC is specified, therefore the angles BCD and CDB are free to change.  In other words, this is a triangle with 2 unknown angles.  Knowing one angle is not sufficient to determine the value of the other two, additional information is needed.
The other triangles are totally superfluous.  The location of point A is irrelevant as it does not affect the shape of the triangle BDC.  The two angles in the BDC triangle remain unknown ** and unaffected ** by any movement of point A.
Dynamic geometry software, such as GeoGebra can tell you the value of the angle because it is internally using a lot more information than the one presented in your figure.  For instance, it knows the equations of every line, from that, calculating any angle is trivial.  
If you wish to convince yourself, in your figure delete the segments AB and AD (don't delete the segment AC, keep it for visualization purposes.)  After doing that you can move AC whereever you want, which will change the angle measure of 43 degrees but, the problem remains identical, one triangle with 2 unknowns.  Knowing a portion (43 in your figure) of one unknown is useless.
If more information was given, such as, DA = DC then that would be sufficient to determine the value of the angles.
The problem you've presented is equivalent to presenting one triangle with one known angle, no other information and asking for the value of the other 2 angles.  The answer is: there is no unique solution.  It is undetermined.
Hope that helps.
