Finding side-length proof in double-angle triangle.

Question

In triangle $ABC$, $|AC| = b$ and $|AB| = c$. Angle $A$ is twice angle $B$. Prove that $$|BC| = \sqrt{b\cdot \left(b+c\right)}$$

I understand how to apply laws such as the cosine and sine law to this triangle, but I have no idea how to end up at the desired expression.

Here's a diagram of the triangle:

Answer 1

An elementary solution:

Let's draw $[AD]$ angle biector, $|DB|=x$ and $|DC|=a-x$. Therefore, $\triangle ABC \sim \triangle DAC$ and by properties of ratio

$$ \dfrac ba = \dfrac {a-x}b = \dfrac xc =\dfrac a{b+c}$$

Then $$a^2=b(b+c).$$

Answer 2

Applying the cosine rule in two ways ... we have \begin{eqnarray*} a^2=b^2+c^2-2bc \cos(B) \\ c^2=a^2+b^2-2ab \cos(2B) \end{eqnarray*} Now reacll the double angle formula $\cos(2B)=2 \cos^2(B)-1$ & after a little algebra \begin{eqnarray*} (a^2-bc-c^2)(a+b-c)(a-b+c)(b+c)b=0 \end{eqnarray*} The latter four factors can be rejected as either the triangle inequality or quantities that are clearly positive ... so we have $a^2-bc-c^2=0$ & the required result follows.