95. Determine the angles of a triangle in which the median, bisector, and altitude divide an angle into four equal parts.
95. Let $ABC$ be a given triangle, $BH$, $BD$, and $BM$ the altitude, bisector, and median of this triangle, respectively, which divide angle $B$ into four equal parts (Fig. 219). Describe a circle $\mathcal{S}$ around triangle $ABC$ and extend the bisector of the triangle to intersect the circle at point $D_{1}$; connect point $D_{1}$ with the midpoint $M$ of side $AC$. Since $D_{1}$ is the midpoint of arc $AD_{1}C$, $D_{1}M$ is perpendicular to chord $AC$, i.e., $D_{1}M \parallel HB$. From this, we conclude that $\angle BD_{1}M = \angle DBH$, and since by condition $\angle DBH = \angle DBM$, then

$$
\angle BD_{1}M = \angle DBM \quad \text{and} \quad BM = MD_{1}
$$

From this, it follows that the perpendicular to segment $BD_{1}$ at its midpoint, on which lies the center of circle $\mathcal{S}$, passes through point $M$. Since the perpendicular $D_{1}M$ to chord $AC$ at its midpoint should also pass through the center of the circumscribed circle $S$, point $M$ is the center of the circumscribed circle around $ABC$ and angle $ABC$ is a right angle.

Now it is quite simple to determine the other angles of triangle $ABC$:

$$
\begin{aligned}
& \angle BCA = \angle ABH = \frac{1}{4} d = 22^{\circ} 30' \\
& \angle BAC = \angle CBH = \frac{3}{4} d = 67^{\circ} 30'
\end{aligned}
$$