[ $\underline{\text { Mutual }} \underline{\text { position of altitudes, medians, angle bisectors, etc. ] }}$

It is known that in a certain triangle, the median, angle bisector,

and altitude drawn from vertex $C$ divide the angle into four equal

parts. Find the angles of this triangle.
Prove that the given triangle is a right triangle.

## Solution

Let's denote each of these angles as $\alpha$. Since the bisector divides the angle between the height and the median in half, $4 \alpha=90^{\circ}$. Therefore, $\alpha=22.5^{\circ}$.

## Answer

$90^{\circ} ; 22.5^{\circ} ; 67.5^{\circ}$.