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