Let $\angle BAC = \alpha$, so $\angle ABC = 2\alpha$.

In triangle $ABC$, the sum of angles is $180^\circ$:
$$\alpha + 2\alpha + \angle BCA = 180^\circ$$
So $\angle BCA = 180^\circ - 3\alpha$.

Let $|AB| = c$, $|BC| = a$, $|CA| = b$.

By the Law of Sines:
$$\frac{a}{\sin \alpha} = \frac{b}{\sin 2\alpha} = \frac{c}{\sin(180^\circ - 3\alpha)} = \frac{c}{\sin 3\alpha}$$

We want to prove $b < 2a$.

From the Law of Sines:
$$\frac{b}{a} = \frac{\sin 2\alpha}{\sin \alpha}$$
Recall $\sin 2\alpha = 2\sin \alpha \cos \alpha$:
$$\frac{b}{a} = \frac{2\sin \alpha \cos \alpha}{\sin \alpha} = 2\cos \alpha$$
So $b = 2a \cos \alpha$.

We want $b < 2a$, i.e. $2a \cos \alpha < 2a$, or $\cos \alpha < 1$.

Since $0 < \alpha < 60^\circ$ (otherwise $\angle BCA$ would be negative), $\cos \alpha < 1$ is always true for $0 < \alpha < 60^\circ$.

Therefore, $|AC| < 2|BC|$.