<s>[INST] Finding the angle $\angle BDC$ 
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?
 [/INST] 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:
$\displaystyle 2\left(\frac{\sin 50°}{\sin 2x}\right) 
= 1 + (\tan25°)(\tan x)
= \frac{(\cos 25°)(\cos x) + (\sin 25°)(\sin x)}{(\cos 25°)(\cos x)}$
$\displaystyle \require{cancel} \frac{\sin 50°}{(\sin x)\cancel{(\cos x)}}
= \frac{\cos(x-25°)}{(\cos 25°){\cancel{(\cos x)}}}$
$(\sin 50°)(\cos 25°) = \sin(x)\cos (x-25°) \quad → x=50°$
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