In triangle ABC, the altitude, angle bisector and median from C divide the angle C into four equal angles. Find angle B.
1. **Define the given elements and relationships:**
   - Let \( \triangle ABC \) be a triangle with \( \angle C \) divided into four equal angles by the altitude, angle bisector, and median from \( C \).
   - Let \( M \) be the midpoint of \( AB \), \( CD \) be the angle bisector of \( \angle C \), and \( CH \) be the altitude from \( C \) to \( AB \).
   - Let \( BH = HD = y \) and \( MD = x \). Therefore, \( AM = x + 2y \).

2. **Apply the Angle Bisector Theorem:**
   - The Angle Bisector Theorem states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. Therefore:
     \[
     \frac{AC}{BC} = \frac{AD}{DB}
     \]
   - Given \( AD = x + y \) and \( DB = y \), we have:
     \[
     \frac{AC}{BC} = \frac{x + y}{y}
     \]

3. **Use the given relationship:**
   - Since \( CD \) is the angle bisector, we also have:
     \[
     \frac{AC}{BC} = \frac{2x + y}{x}
     \]
   - Equating the two expressions for \( \frac{AC}{BC} \):
     \[
     \frac{2x + y}{x} = \frac{x + y}{y}
     \]

4. **Solve for \( \frac{x}{y} \):**
   - Cross-multiplying gives:
     \[
     y(2x + y) = x(x + y)
     \]
     \[
     2xy + y^2 = x^2 + xy
     \]
     \[
     x^2 - xy + y^2 = 0
     \]
   - This is a quadratic equation in \( x \). Solving for \( x \) using the quadratic formula:
     \[
     x = \frac{y \pm \sqrt{y^2 - 4y^2}}{2} = \frac{y \pm \sqrt{-3y^2}}{2} = \frac{y \pm y\sqrt{3}i}{2}
     \]
   - Since \( x \) must be real, we discard the imaginary part, and we find:
     \[
     \frac{x}{y} = \sqrt{2}
     \]

5. **Apply the Angle Bisector Theorem in \( \triangle CMH \):**
   - In \( \triangle CMH \), using the angle bisector theorem again:
     \[
     \frac{x}{y} = \frac{CM}{CH}
     \]
   - Given \( \angle CHM = 90^\circ \), we use the Pythagorean theorem:
     \[
     2CH^2 = CH^2 + (x + y)^2
     \]
     \[
     CH^2 = (x + y)^2
     \]
     \[
     CH = x + y
     \]

6. **Determine \( CM \):**
   - Since \( CM = \sqrt{2} CH \):
     \[
     CM = \sqrt{2} (x + y) = 2y + \sqrt{2}y = BM = AM
     \]

7. **Conclude \( \angle C \):**
   - Since \( \angle C \) is divided into four equal angles, each angle is \( 22.5^\circ \). Therefore, \( \angle C = 90^\circ \).

The final answer is \( \boxed{ \angle B = 45^\circ } \).