5. Let point $D$ be the intersection of side $\overline{AB}$ of triangle $ABC$ and the angle bisector of the triangle at vertex $C$. What are the measures of the angles of triangle $ABC$ if the incenter of triangle $ADC$ and the circumcenter of triangle $ABC$ coincide?

NATIONAL COMPETITION IN MATHEMATICSPrimošten, April 4-6, 2016.8th grade-elementary school
First method:

The line $CD$ is the angle bisector of the triangle at vertex $C$, and $\gamma=|\measuredangle ACB|=2 \cdot (180^{\circ}-2 \beta)=360^{\circ}-4 \beta$.

From the sum of the angles of triangle $ABC$, we get

$\alpha+\beta+\gamma=180^{\circ}$

$\beta+(180^{\circ}-2 \beta)+(360^{\circ}-4 \beta)=180^{\circ} \rightarrow 5 \beta=360^{\circ} \rightarrow \beta=72^{\circ}$

$\alpha=180^{\circ}-2 \cdot 72^{\circ}=36^{\circ}$

$\gamma=180^{\circ}-(72^{\circ}+36^{\circ})=72^{\circ}$

Triangle $ABC$ is isosceles with angles of $72^{\circ}, 72^{\circ}$, and $36^{\circ}$.