(a) Prove that if, in a triangle $ABC$ , $\angle A = 2 \angle B$ , then $|BC|^2=|AC|^2+ |AC|\cdot |AB|$
(b) Is the converse true?
1. Let \(AB = c\), \(AC = b\), and \(BC = a\). We are given that \(\angle A = 2 \angle B\).
2. Construct the bisector of \(\angle BAC\) which meets \(BC\) at \(D\). By the angle bisector theorem, we have:
   \[
   \frac{BD}{DC} = \frac{AB}{AC} = \frac{c}{b}
   \]
3. Let \(BD = x\) and \(DC = y\). Then, \(x + y = a\) and \(\frac{x}{y} = \frac{c}{b}\). Solving for \(x\) and \(y\), we get:
   \[
   x = \frac{ac}{b+c} \quad \text{and} \quad y = \frac{ab}{b+c}
   \]
4. Since \(\angle A = 2 \angle B\), \(\triangle ABD \sim \triangle CAD\) by the angle-angle similarity criterion. Therefore, we have:
   \[
   \frac{AB}{AD} = \frac{AD}{AC} \quad \Rightarrow \quad \frac{c}{AD} = \frac{AD}{b}
   \]
5. Solving for \(AD\), we get:
   \[
   AD^2 = bc \quad \Rightarrow \quad AD = \sqrt{bc}
   \]
6. Using the Law of Cosines in \(\triangle ABD\) and \(\triangle ACD\), we have:
   \[
   BD^2 = AB^2 + AD^2 - 2 \cdot AB \cdot AD \cdot \cos(\angle BAD)
   \]
   \[
   DC^2 = AC^2 + AD^2 - 2 \cdot AC \cdot AD \cdot \cos(\angle CAD)
   \]
7. Since \(\angle BAD = \angle CAD = \angle B\), we have:
   \[
   BD^2 = c^2 + bc - 2c \cdot \sqrt{bc} \cdot \cos(\angle B)
   \]
   \[
   DC^2 = b^2 + bc - 2b \cdot \sqrt{bc} \cdot \cos(\angle B)
   \]
8. Adding these two equations, we get:
   \[
   BD^2 + DC^2 = c^2 + b^2 + 2bc - 2 \sqrt{bc} (c + b) \cos(\angle B)
   \]
9. Since \(BD + DC = a\), we have:
   \[
   a^2 = BD^2 + DC^2
   \]
10. Therefore, we get:
    \[
    a^2 = b^2 + bc
    \]

The final answer is \( \boxed{ a^2 = b^2 + bc } \)