## 5. In a given triangle ABC, the altitude and the median drawn from vertex $C$ divide angle $C$ into three equal parts. Determine the angles of the triangle.
Solution: Let $\overline{\mathrm{CM}}=\mathrm{t}_{\mathrm{c}}$ and $\overline{\mathrm{CN}}=\mathrm{h}_{\mathrm{c}}$ divide angle C into three equal parts. If $P \in A C$, such that $M P \perp A C$, then $\triangle \mathrm{CMP} \cong \triangle \mathrm{CMN}\left(\triangle \mathrm{P}=\Delta \mathrm{N}=90^{\circ}\right.$, CM is a common side and $\triangle \mathrm{PCN}=\triangle \mathrm{MCN}=\varphi$ ). It follows that $\overline{\mathrm{PM}}=\overline{\mathrm{MN}}=\frac{\mathrm{c}}{4}$ ( $\triangle \mathrm{MBC}$ is isosceles, hence

![](https://cdn.mathpix.com/cropped/2024_06_05_3e201888496d77f8c9ddg-2.jpg?height=327&width=452&top_left_y=959&top_left_x=990)
$\overline{\mathrm{MN}}=\overline{\mathrm{NB}}=\frac{\mathrm{c}}{4}$ ). Since the hypotenuse $\mathrm{AM}$ in $\triangle \mathrm{AMP}$ is twice the length of the leg PM, it follows that $\alpha=30^{\circ}$. $\triangle \mathrm{PMN}=180^{\circ}-60^{\circ}=120^{\circ}$, and from the congruence of the triangles, it follows that $\triangle \mathrm{PMC}=\triangle \mathrm{CMN}=60^{\circ}$, i.e., $\varphi=30^{\circ}$, so $\triangle \mathrm{C}=90^{\circ}$, and $\triangle B=60^{\circ}$.

## VIII section