In triangle $A B C$, the median $B D$ is equal to half of side $A C$. Find the angle $B$ of the triangle.
The median $BD$ divides the given triangle into two isosceles triangles.

## Solution

First method. Let $\angle A=\alpha, \angle C=\gamma$. Since $AD=BD$ and $CD=BD$, then $\angle ABD=\alpha, \angle DBC=\gamma$, and since the sum of the angles in triangle $ABC$ is $180^{\circ}$, we have $2\alpha + 2\gamma = 180^{\circ}$.

Therefore, $\angle B=\alpha+\gamma=90^{\circ}$.

Second method. The circle with center at point $D$ and radius $BD$ passes through points $A$ and $C$, and $AC$ is its diameter. Therefore, $\angle B=90^{\circ}$.

## Answer

$90^{\circ}$.