Let $ ABC$ be a triangle such that \[ \frac{BC}{AB \minus{} BC}\equal{}\frac{AB \plus{} BC}{AC}\] Determine the ratio $ \angle A : \angle C$.
1. Given the equation:
   \[
   \frac{BC}{AB - BC} = \frac{AB + BC}{AC}
   \]
   Let's denote the sides of the triangle as follows: \( BC = a \), \( AC = b \), and \( AB = c \). Substituting these into the given equation, we get:
   \[
   \frac{a}{c - a} = \frac{c + a}{b}
   \]

2. Cross-multiplying to eliminate the fractions, we obtain:
   \[
   a \cdot b = (c - a)(c + a)
   \]
   Simplifying the right-hand side using the difference of squares:
   \[
   ab = c^2 - a^2
   \]

3. Rearranging the equation to isolate \( c^2 \):
   \[
   c^2 = ab + a^2
   \]

4. Using the Law of Cosines in triangle \( ABC \):
   \[
   c^2 = a^2 + b^2 - 2ab \cos C
   \]
   Equating the two expressions for \( c^2 \):
   \[
   ab + a^2 = a^2 + b^2 - 2ab \cos C
   \]

5. Simplifying the equation:
   \[
   ab = b^2 - 2ab \cos C
   \]
   Rearranging to solve for \( \cos C \):
   \[
   ab + 2ab \cos C = b^2
   \]
   \[
   ab (1 + 2 \cos C) = b^2
   \]
   \[
   1 + 2 \cos C = \frac{b}{a}
   \]

6. Using the Law of Sines, we know:
   \[
   \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
   \]
   Since \( C = 2A \), we can use the double-angle identity for sine:
   \[
   \sin 2A = 2 \sin A \cos A
   \]

7. Substituting \( C = 2A \) into the Law of Sines:
   \[
   \frac{a}{\sin A} = \frac{b}{\sin 2A} = \frac{b}{2 \sin A \cos A}
   \]
   Simplifying:
   \[
   \frac{a}{\sin A} = \frac{b}{2 \sin A \cos A}
   \]
   \[
   a = \frac{b}{2 \cos A}
   \]

8. Since \( \cos A \) must be a positive value less than or equal to 1, we conclude that:
   \[
   \cos A = \frac{b}{2a}
   \]

9. Given that \( C = 2A \), the ratio of the angles is:
   \[
   \angle A : \angle C = 1 : 2
   \]

The final answer is \(\boxed{1:2}\)