Finding side-length proof in double-angle triangle. 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:


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.
