An Interesting Side Relation

From cosine rule , we get

$\begin{aligned} \blue{a^2} & = b^2 + c^2 - 2bc \cos A & \small \blue{\text{Given that }a^2 = b(b+c)} \\ \blue{b^2+bc} & = b^2 + c^2 - 2bc \cos A \\ b & = c-2b\cos A & \small \blue{\text{By sine rule }\frac b{\sin B} = \frac c{\sin C} = k} \\ \sin B & = \sin C - 2\sin B \cos A & \small \blue{\text{As } \sin C = \sin (180^\circ - A -B) = \sin (A+B)} \\ & = \sin (A+B) - 2\sin B \cos A \\ & = \sin A \cos B + \sin B \cos A - 2 \sin B \cos A \\ & = \sin A \cos B - \sin B \cos A \\ \sin B & = \sin (A-B) \\ B & = A-B \\ 2B & = A \\ \implies \frac {\angle A}{\angle B} & = \boxed 2 \end{aligned}$

Let us extend $\overline {CA}$ to $D$ such that $|\overline {AD}|=|\overline {AB}|=c$ . Then, since

$a^2=b(b+c)\implies \dfrac{a}{b}=\dfrac{b+c}{a}\implies \dfrac{|\overline {BC}|}{|\overline {AC}|}=\dfrac{|\overline {DC}|}{|\overline {BC}|}$ ,

therefore $\triangle {ABC}$ and $\triangle {DBC}$ are similar, with $\angle {ABC}=\angle {BDC}=\angle B$ .

So $\angle A=2\angle {BDC}=2\angle B$ and the required ratio is $\dfrac{\angle A}{\angle B}=\boxed 2$ .