4A. Determine the angles in a triangle where the height and the median drawn from the same vertex divide the angle at that vertex into three equal parts.
Solution. Let $ABC$ be a triangle such that the altitude $CH$ and the median $CT$ divide the angle at vertex $C$ into three equal parts. $\triangle AHC \cong \triangle CHT$ because $CH$ is common, $\measuredangle AHC=\measuredangle CHT=90^{\circ}$, and from the condition of the problem, $\measuredangle ACH=$ $\measuredangle HCT$. Therefore,

$$
\overline{AH}=\overline{HT}=\frac{1}{2} \overline{TB}=\frac{1}{4} \overline{AB}
$$

$CT$ is the angle bisector of the angle at vertex $C$ in $\triangle CHB$, so from the property of the angle bisector, it follows that $\overline{CH}: \overline{CB}=\overline{HT}: \overline{TB}$. From this, we get

$$
\overline{CH}=\frac{\overline{CB} \cdot \overline{HT}}{\overline{TB}}=\overline{CB} \cdot \frac{1}{\frac{2}{TB}} \overline{TB}=\frac{1}{2} \overline{CB}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_e95f2dbd17df8adc2c7cg-02.jpg?height=272&width=505&top_left_y=1008&top_left_x=959)

Triangle $CHB$ is a right triangle, so one of the legs is half the hypotenuse. It follows that $\measuredangle HBC=30^{\circ}$ and $\measuredangle BHC=60^{\circ}$. Therefore, the angles in triangle $ABC$ are $30^{\circ}, 60^{\circ}$, and $90^{\circ}$.