A circle touches side $B C$ of triangle $A B C$ at point $M$, side $A C$ at point $N$, and intersects side $A B$ at points $K$ and $L$, such that $K L M N$ is a square. Find the angles of triangle $A B C$.
The angle between a tangent and a chord is equal to half the angular measure of the arc enclosed between them.

## Solution

Angle $B M L$ is the angle between the tangent $M B$ and the chord $M L$. It is equal to half the smaller arc $M L$ of the circle, i.e., $45^{\circ}$. Since $M L \perp A B$, $\angle B=45^{\circ}$. Similarly, we can prove that $\angle A=45^{\circ}$. Therefore, $\angle C=90^{\circ}$.

## Answer

$90^{\circ}, 45^{\circ}, 45^{\circ}$.