Let the foot of the perpendicular bisectors of $BC$ and $AB$ be $M$ and $N$ resp and $ \angle BAC=b, \angle BCA=a$. From, isosceles triangle property we must have, $XB=XC, YB=AY \implies \angle BYX=180^o-2b, \angle BXY=180^o-2a$ Now, from the triangle $AYN$ and $CMX$, we have, $\angle NYA=90^o-b, \angle MXC= 90^o-b \implies \angle OYX=\frac{1}{2} \angle BYX$, where $O$ is $(ABC)$. Hence, the circumcentre of triangle $ABC$ is incentre of triangle $BXY$ .

Notice that it suffices to show that the incenter of $\triangle BXY$ is the circumcenter of $\triangle ABC$. Restate the problem in terms of $\triangle BXY$ as follows: Restated problem wrote: Let $I$ be the incenter of $\triangle ABC$. Let the lines through $A$ perpendicular to $BI$ and $CI$ meet $BC$ at $P$ and $Q$ respectively. Show that $I$ is the circumcenter of $\triangle APQ$. Let $BI \cap AP=M,CI \cap AQ=N$. Then it suffices to show that $M$ and $N$ are the midpoints of $AP$ and $AQ$ respectively. But, as in $\triangle ABP$, we have that $BM$ is the angle bisector of $\angle ABP$, as well as the altitude from $B$ to $AP$, so $\triangle ABP$ must be isosceles, and $M$ must be the midpoint of $AP$. Similarly, we get that $N$ is the midpoint of $AQ$. Hence, done. $\blacksquare$

Let $M$ and $N$ be the foot of perpendiculars in $AB$ and $AC$ . notice that triangle $AYM$ is congruent to $MYB$ so angle $MYA$= angle $MYA$=$90-A$. And similarly triangle $BXN$is congruent to triangle $NXC$ and angle $BXN$= angle $NXC$=$90-C$. so proved

Let $P$ and $Q$ denote the foot of the perpendiculars form $O$( the circumcentre of $\triangle ABC$). onto the sides $BC$ and $AC$. Notice that $\triangle XBP \cong \triangle XCP$ (by SAS congruency). Similarly $\triangle YBQ \cong \triangle YAQ$( by SSS congruency). This gives by c.p.c.t that $\angle BYQ= \angle AYQ$ and $\angle BXP = \angle CXP$. Thus $O$ is the incentre of $\triangle XBY$. Our proof is thus complete.