Answer: The segment $XY$ such that $AXYB$ is a rectangle and the midpoint of $XY$ coincides with $C$.

(Solution of E. Dauhiala, B. Gilevich, K. Kostevich.) Let $O$ be the point in the same half-plane as $C$ with respect to the line $MN$ such that $\angle MON = 90^\circ$ and $OM = ON$. Then $OM = ON = \dfrac{MN}{\sqrt{2}} = CH$ (the altitude of the triangle $ABC$). Since $\angle MON = \angle MCN = 90^\circ$, the quadrilateral $MOCN$ is cyclic and hence $\angle OCM = \angle ONM = 45^\circ$. It follows that $OC \parallel AB$, so the distance between $O$ and $AB$ equals $OK = CH = OM = ON$. Therefore $O$ is the center of the circle passing through $M$, $N$ and touching $AB$. Hence the needed locus is contained in the segment $XY$ described in the answer.

Conversely, one can verify that any point of this segment is a center of some circle which touches $AB$ and intersects $CA$ and $CB$ at points $M$ and $N$ such that $MN = BC$.