Croatia competition problems

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.
$$\text{Finally, we have } \angle DCB = 60^\circ - \beta = \frac{1}{3}(180^\circ - 3\beta) = \frac{1}{3}\angle ACB, \text{ i.e. } \angle ACB = 3\angle DCB.$$