Croatia_2018

Let $E$ be the intersection of the bisector of the segment $\overline{AB}$ with the segment $\overline{BC}$. Denote $\beta =\angle ABC$ and notice that $\angle ACB=180^{\circ }-3\beta$. Since $E$ lies on the bisector of the segment $\overline{AB}$, we have $|AE|=|BE|$. Therefore, $\angle BAE=\angle ABC=\beta$. This implies $\angle EAC=\angle CAB-\beta =2\beta -\beta =\beta$. Let $F$ be the other intersection of the line $AE$ with the circle of radius $\overline{CA}$ centred at $C$. Since $CAF$ is an isosceles triangle ($\overline{CA}$ and $\overline{CF}$ are both radii of the same circle), we get $\angle CFA=\beta$. From $\angle CFA=\angle BAF$ we get $CF\parallel AB$ (these are the angles of the transversal). This also means that $\angle BCF=\angle CBA=\beta$, which shows that $CEF$ is an isosceles triangle. From $CF\parallel AB$ and the fact that $E$ is equidistant to $C$ and $F$, we conclude that the line $DE$ is the bisector of the segment $\overline{CF}$ as well. Thus, $|DF|=|DC|=|CF|$, i.e. the triangle $DFC$ is equilateral. $$\text{Finally, we have }\angle DCB=60^{\circ }-\beta =\frac{1}{3}(180^{\circ }-3\beta )=\frac{1}{3}\angle ACB,\text{ i.e. }\angle ACB=3\angle DCB.$$