Bulgaria competition problems, 2022

Choose a point $M_1$ on the ray $CA$ such that $AM_1 = AM$ and $CM_1 = BC$. It follows that triangle $M_1MA$ is isosceles and $\angle AM_1M = \angle AMM_1 = \frac{1}{2} \angle BAC = 40^\circ$.

Triangles $CM_1M$ and $CBM$ are congruent ($CM_1 = CB$, $CM = CB$, $\angle CMM_1 = \angle CMB = 110^\circ$). Therefore $\angle ABC = \angle CM_1M = 40^\circ$, $\angle ACB = 60^\circ$.

The equality $a^2 = b(b+c)$ follows from $\alpha = 2\beta$. Choose a point $B_1$ on the ray $CA$ such that $AB_1 = AB$ or $CB_1 = b+c$. The triangle $\triangle B_1BA$ is isosceles and thus $\angle AB_1B = \angle B_1BA$. The sum of these angles is equal to $\angle BAC = 2\beta$, i.e. $\angle AB_1B = \angle B_1BA = \beta$ and $\angle B_1BC = \angle B_1BA + \angle ABC = 2\beta$.

From $\triangle ABC \sim \triangle BB_1C$ we obtain $\frac{BC}{B_1C} = \frac{AC}{BC}$, $\frac{a}{b+c} = \frac{b}{a}$, so $a^2 = b(b+c)$.