My solution : Let $XY$ cut $\odot (O)$ again at $Z$ . Since $\angle XBA=\angle ZAC=\angle ZBC, \angle BAX=\angle BCZ$ , so $\triangle BAX \sim \triangle BCZ \Longrightarrow \triangle BAC \sim \triangle BXZ \Longrightarrow XZ:AC=BZ:BC$ . $(\star )$ Similarly, we can prove $\triangle CAY \sim \triangle CBZ \Longrightarrow AY:BZ=AC:BC$ , so combine $(\star)$ we get $AY=XZ \Longrightarrow$ the perpendicular bisector of $XY$ pass through $O$ . i.e. $OX=OY$ Q.E.D

An easy extension Let $ABC$ be a triangle with circumcenter $O$. $E,F$ lie on $CA,AB$. $EF$ cuts circles $(AEB),(AFC)$ again at $M,N$. Prove that $OM=ON$.

buratinogigle wrote: An easy extension Let $ABC$ be a triangle with circumcenter $O$. $E,F$ lie on $CA,AB$. $EF$ cuts circles $(AEB),(AFC)$ again at $M,N$. Prove that $OM=ON$. My solution : Let $T=BM \cap CN$ . Since $\angle TMN=\angle BAC=\angle TNM$ , so we get $TM=TN$ and $\angle MTN=180^{\circ}-2\angle BAC=180^{\circ}-\angle BOC \Longrightarrow T \in \odot (BOC)$ , hence from $\angle CTO=\angle CBO=90^{\circ}-\angle BAC \Longrightarrow TO \perp MN \Longrightarrow O$ lie on the perpendicular bisector of $MN$ . i.e. $OM=ON$ Q.E.D

Would it be useful to invert around A?

Soph11 wrote: Would it be useful to invert around A? Yes. Notice the problem is equivalent to $$OX^{2}-R^{2}=XZ.XA=YZ.YA=OY^{2}-R^{2}$$ (Power of a point). Invert around $A$ and use Thales' Theorem and distance formula.