Solution:
Answer: $1 : 2$.

Denote $|BC| = a$, $|AC| = b$, $|AB| = c$. The condition $\frac{a}{c - a} = \frac{c + a}{b}$ implies $c^2 = a^2 + ab$ and
$$\frac{c}{a + b} = \frac{a}{c}.$$
Let $D$ be a point on $AB$ such that $|BD| = \frac{a}{a + b} \cdot c$ (see Figure 5). Then
$$\frac{|BD|}{|BC|} = \frac{c}{a + b} = \frac{a}{c} = \frac{|BC|}{|BA|}$$
so triangles $BCD$ and $BAC$ are similar, implying $\angle BCD = \angle BAC$. Also, $\frac{|AC|}{|BC|} = \frac{|AD|}{|BD|}$ yields $\frac{|BC|}{|BD|} = \frac{|AC|}{|AD|}$, and hence by the bisector theorem $CD$ is the bisector of $\angle BCA$. So the ratio asked for is $1 : 2$.

Figure 5