The circle $\omega$ passes through the vertices $A$ and $B$ of a unit square $A B C D$. It intersects $A D$ and $A C$ at $K$ and $M$ respectively. Find the length of the projection of $K M$ onto $A C$.
Let $T$ be the point of intersection of $\omega$ with $B C$. Then, as $\angle A B T$ a right angle, $A T$ is a diameter, and $\angle A M T$ is also a right angle. Therefore the projections of $K M$ and $K T$ on $A C$ coincide. But the length of the projection of $K T$ is $\frac{\sqrt{2}}{2}$ because the length of $K T$ is one, and the angle between $K T$ and $A C$ is 45 .