Let's assume that $\angle DBC = 50^{\circ}$, $[DC]$ bisects $\angle ACB$, and that $|AC| = |BC|-|AD|$. How could we find the angle $\angle BDC$?
Applying angle bisector theorem:
$$\frac{|AD|}{|BD|} = \frac{|AC|}{|BC|}$$
Since $|AC| = |BC|-|AD|$,
$$\frac{|AD|}{|BD|} = \frac{|BC|-|AD|}{|BC|} = 1-\frac{|AD|}{|BC|}$$
$$|AD|\biggr(\frac{1}{|BD|}+\frac{1}{|BC|}\biggr) = 1$$
But this won't lead me anywhere, I believe. Could we take complex geometric approach to this problem?
A fairly simple problem. Though you ask for a method that involved the complex plane, I'll share a method relying purely on Euclidean Geometry as how you originally seemed to have attempted this question.
1.) Since we know that $AC=BC-AD$, we can rewrite this as $BC=AC+AD$. Now, for our ease, we shall label $AC=a$ and $AD=b$, this implies that $AC=a+b$. (We can also label the bisected angles as $\alpha$)
2.) Locate $E$ outside $\triangle ABC$ such that $AE=b$. Also join $B$ and $E$ via $BE$, and $D$ and $E$ via $DE$. This construction shows that $AE=AC=a+b$, therefore, $\triangle ABE$ is an isosceles triangle. Since $DC$ is an angle bisector of $\angle ACB$, we can extend $DC$ to meet $BE$ at point $F$. It is easy to see that, because $\triangle ABE$ is isosceles, that segment $CF$ is both the perpendicular and angle bisector of $\triangle ABE$
3.) Notice that, via the SAS property, $\triangle CBD$ and $\triangle CED$ are congruent. This implies that $BD=DE$ and $\angle CBD=\angle CED=50^\circ$. Via some basic angle chasing, it is easy to show that $\angle DBF=\angle DEF=40-\alpha$. This means that, via the exterior angle property, $\angle EDA=80-2\alpha$. However, we already know that $AD=b$ and $AE=b$, thus, $\angle EDA=\angle DEA=50$. This means that:
$$80-2\alpha=50$$
$$\alpha=15$$
Therefore, $\angle BDC=180-65=115$
Let $x = ∠BCD \;,\; y = ∠BDC$
We are only concerned with angles. Let $|CD|=1$
$|BC| = |AC| + |AD|$
$\displaystyle \require{cancel}\frac{\sin(y)}{\sin 50°} = \frac{\sin(y) }{\sin(y-x)} + \frac{\sin(x) }{\sin(y-x)} = \frac{\cancel{2} \sin(\frac{y+x}{2})\cancel{\cos(\frac{y-x}{2})}} {\cancel{2} \sin(\frac{y-x}{2}) \cancel{\cos(\frac{y-x}{2})}}$
$ x + y = 180°-50° = 130°$
$\displaystyle \require{cancel}\frac{\sin(y)}{\sin 50°} = \frac{\sin 65°}{\sin(y-65°)} = \frac{\sin 115°}{\sin(y-65°)} $
$→ y = 115°$
We can also solve directly for y, by cross multiply.
$\sin(y)\;\sin(y-65°) = -\frac{1}{2} \big(\cos(2y-65°) - \cos(65°)\;\big)$
$\sin(115°)\;\;\sin(50°) = -\frac{1}{2} \big(\;\;\cos(165°) \quad - \cos(65°)\;\big)$
$\cos(2y-65°) = \cos(165°)$
$\require{cancel} → y = 115°, \cancel{130°}\qquad $// 2nd solution implied $x=0°$, thus not valid
Let $X$ be the point on $[BC]$ for which $XC = AC$.
Note: this is the main move -- idea behind this being just that it "feels right" in a way that it seems like it could be a way to use the $AD + AC = BC$ information.
$\Delta XCD$ and $\Delta ACD$ are congruent triangles by SAS. So then $XD = AD$.
But $BX = BC - XC = BC - AC = AD = XD$.
Then $\Delta BXD$ is isosceles. And so $\angle BDX = 50^{\circ}$. And because $\Delta XCD$ and $\Delta ACD$ are congruent, $\angle XDC = \angle ADC = 65^{\circ}$. And so $\angle BDC = 115^{\circ}$.
Let $\displaystyle x = \frac{∠A}{2} \;,\; y = \frac{∠C}{2}\;$
Given $∠B=50° \quad → 2x+2y = 130°$
Law of Tangent, on ΔACD, given $AD = (a-b)$
$\displaystyle \frac{b-(a-b)}{b+(a-b)} = \frac{2b}{a} - 1 = \frac{\tan \left(\frac{(180-2x-y)-y}{2}\right)}{\tan \left(\frac{(180-2x-y)+y}{2}\right)} = \frac{\tan 25°}{\tan (90°-x)} = (\tan 25°)(\tan x) $
Law of Sine, on on ΔABC
$\displaystyle \frac{b}{a} = \frac{\sin 50°}{\sin 2x}$
Let $t_1 = \tan 25° \;,\; t_2 = \tan x\;$, combine the two expressions:
$\displaystyle 2 \left(\frac{2t_1}{1+t_1^2} \right) ÷ \left(\frac{2t_2}{1+t_2^2} \right) - 1 = t_1\,t_2 $
$\displaystyle → t_2 = \frac{1}{t_1} \;\lor\; \frac{2\,t_1}{1-t_1^2} = \tan(65°) \;\lor\; \tan(50°)$
$x=65° ⇒ ∠C = 0°$, not a triangle, thus not a solution.
$x=50° ⇒ ∠C = 30° ⇒ ∠BDC = \left(180 - 50 - \frac{30}{2}\right)\!° = 115°\;$
Perhaps simpler way to solve for x:
$\displaystyle 2\left(\frac{\sin 50°}{\sin 2x}\right) = 1 + (\tan25°)(\tan x) = \frac{(\cos 25°)(\cos x) + (\sin 25°)(\sin x)}{(\cos 25°)(\cos x)}$
$\displaystyle \require{cancel} \frac{\sin 50°}{(\sin x)\cancel{(\cos x)}} = \frac{\cos(x-25°)}{(\cos 25°){\cancel{(\cos x)}}}$
$(\sin 50°)(\cos 25°) = \sin(x)\cos (x-25°) \quad → x=50°$
Law of Sines works nice. We know the following three facts.
$$\frac{\sin(50+x)}{|AC|}=\frac{\sin(x)}{|AD|}$$
$$\frac{\sin(130-2x)}{|BC|}=\frac{\sin(2x)}{|AD|+|BD|}$$ $$\frac{\sin(x)}{|BD|}=\frac{\sin(130-x)}{|BC|}$$ These three equations imply that $$\frac{|AC|}{|AD|}=\frac{\sin(50+x)}{\sin(x)}$$ $$\frac{|AD|}{|BC|}=\frac{\sin(2x)}{\sin(130-2x)}-\frac{\sin(x)}{\sin(130-x)}$$ It follows that $$\frac{\sin\left(50+x\right)}{\sin\left(x\right)}-\left(\frac{1}{\frac{\sin\left(2x\right)}{\sin\left(130-2x\right)}-\frac{\sin\left(x\right)}{\sin\left(130-x\right)}}\right)=\frac{|AC|}{|AD|}-\frac{|BC|}{|AD|}=-1$$ The above implies $x=15^{\circ}$ and the angle measure you seek is $130-x=115^{\circ}$
Let $2\alpha=\angle ACB$. Then the law of sines in $\triangle ADC$ and $\triangle ABC$ gives respectively $$\frac{AD}{AC}=\frac{\sin\alpha}{\sin(\alpha +50^\circ)}$$ $$\frac{BC}{AC}=\frac{\sin(2\alpha+50^\circ)}{\sin50^\circ}$$ Since $\frac{BC}{AC}=\frac{AD+AC}{AC}=\frac{AD}{AC}+1$ we get
$$\frac{\sin(2\alpha+50^\circ)}{\sin50^\circ}=\frac{\sin\alpha}{\sin(\alpha +50^\circ)}+1$$
Thanks to Albert Chan's comment, we can see that $\alpha= 15^\circ$ and thus $\angle BDC= 115^\circ.$
