Let $AO_{1}\cap (O_{1})=D$ and $AO_{2}\cap (O_{2})=E$. We know that $\angle ABD=\angle ABE=90 \rightarrow D,B,E$ are collinear. The distances from $A$ and $B$ to the line $O_{1}O_{2}$ are equal but so is the distance from $C$ since the A-altitude of the triangle $AO_{1}O_{2}$ is equal to the C-altitude of the triangle $CO_{2}O_{1}$, since both triangles are congruent, then $BC\parallel O_{1}O_{2}$ or also $D,B,C,E $ are collinear. It is easy to see that $C$ is the midpoint of $DE$ (O1 and O2 are midpoints of AD and AE) $\angle AXD=\angle AYE=90$, by Reim's theorem $XD\parallel YE$ then the perpendicular bisector of $XY$ passes through $C$.

Attachments: