Let $O_1, O_2$ be centers of $(ABO), (ACO)$. 1) $X\in OO_1$ because they are all on perp bisector of $AB$. Similarly $Y\in OO_2$. 2) $X\in (AOC)$ because $\angle BXO=90^\circ -\angle B=\angle CAO$. Similarly $Y\in (ABO)$. 3) Note that $AO_1Y\sim AOC$. So by spiral similarity $O_1Y\mapsto OC$ we have $A\in (XYO_1)$. Similarly $A\in (XYO_2)$.

Denote by $O_{1},O_{2}$ centers of $(AOB),(AOC).$ Obviously $O_{1}\in OX,O_{2}\in OY.$ Then $$\measuredangle AO_{1}X=\frac{1}{2}\measuredangle AO_{1}B=\measuredangle AOB=2\measuredangle ACB=\measuredangle AYX\implies O_{1}\in (AXY).$$Analogously $O_{2}\in (AXY).$