2. Determine the angles of triangle $ABC$ if the median, the angle bisector, and the altitude from vertex $C$ divide $\measuredangle ACB$ into four equal parts.
Solution. Let $D, E, F$ be the points of intersection of the altitude, the angle bisector, and the median drawn from vertex $C$ to side $AB$ of triangle $ABC$, and let $G$ be the intersection of the angle bisector of $\measuredangle BCA$ and the circumcircle of triangle $ABC$, as shown in the diagram below. Then point $G$ lies on the perpendicular bisector of side $AB$. By assumption, we have

$$
\begin{aligned}
\measuredangle ACD & =\measuredangle DCE \\
& =\measuredangle ECF=\measuredangle FCB.
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_778c473e5a351e479833g-1.jpg?height=497&width=528&top_left_y=1275&top_left_x=940)

Furthermore, from $CD \| FG$ it follows that

$$
\measuredangle FCG=\measuredangle DCE=\measuredangle CGF
$$

so triangle $CGF$ is isosceles and point $F$ lies on the perpendicular bisector of segment $CG$. Since $F$ is the midpoint of segment $AB$, we conclude that $F$ is the center of the circumcircle of triangle $ABC$. Therefore,

$$
\measuredangle BCA=90^{\circ}, \measuredangle CAB=\measuredangle DCB=67^{\circ} 30' \text{ and } \measuredangle ABC=22^{\circ} 30'.
$$