$a^2 - b^2 = bc$, prove $2\alpha = \beta$

Question

In a triangle with sides $a$, $b$, $c$ and angles $\alpha$, $\beta$, $a^2 - b^2 = bc$. Prove $2\beta = \alpha$.

I've been trying to solve this for a while, but can't come up with anything useful. I've tried using the sine and cosine rules, to no avail. It seems like it's extremely simple, but I'm missing something.

Answer 1

we have $$a^2=b^2+c^2-2bc\cos(\alpha)$$ with our condition we get $$1+2\cos(\alpha)=\frac{c}{b}=\frac{\sin(\gamma)}{\sin(\beta)}=\frac{\sin(\alpha+\beta)}{\sin(\beta)}$$ using the addition formuals we obtain $$1+2\cos(\alpha)=\sin(\alpha)\cot(\beta)+cos(\alpha)$$ now we have $$2\cos^2\left(\frac{\alpha}{2}\right)=2\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)\cot(\beta)$$ dividing by $$\cos\left(\frac{\alpha}{2}\right)\ne 0$$ we get $$\tan\left(\frac{\alpha}{2}\right)=\tan(\beta)$$

Answer 2

An elementary proof of $\alpha = 2 \beta$

$a^2=b(b+c)$ is given. Let's condider the following figure. Since $|BD|=b+c$, we find that

$$ |BC|^2=|BA|\cdot |BD|$$

Therefore $\triangle DBC \sim \triangle CBA$ (side-angle-side) and $\angle BDC = \angle BCA$. Thus

$$\alpha = 2 \beta $$

Note: Furthermore, converse of this proposition proved at here