In triangle $A B C$, the altitudes $A A_{1}$ and $C C_{1}$ intersect at point $H$, which lies inside the triangle. It is known that $H$ is the midpoint of $A A_{1}$, and $C H: H C_{1}=2: 1$. Find the measure of angle $B$.
Mark the midpoint of segment $C H$.

## Solution

Let $M$ be the midpoint of segment $C H$. Then $A C_{1} A_{1} M$ is a parallelogram. Therefore, $\angle A_{1} M H=\angle A C_{1} H=90^{\circ}$. Segment $A_{1} M$ is the height and median of the right triangle $H A_{1} C$, which means this triangle is isosceles. Therefore, $\angle C_{1} C B=45^{\circ}$. Consequently, $\angle B=90^{\circ}-\angle C_{1} C B=45^{\circ}$.

## Answer

$45^{\circ}$.