Let $BX\cap AC\equiv E, CY\cap AB\equiv F, H\equiv BX\cap CY $. Let $M,N $ be the midpoints of $XY,O_BO_C $ respectively. We shall show that both $AM,AN $ pass through $O $, the circumcentre of $\triangle ABC $, which will finish the problem. Claim : $AM $ passes through $O$. Proof : $\angle XYA=\angle XBA=90^{\circ}-\angle BAC=\angle YCA=\angle YXA $, so $AM\perp XY\implies \angle YAM=90^{\circ}-\angle XYA=\angle BAC $. Also $\angle YAB=\angle YCB=90^{\circ}-\angle ABC $. Hence $\angle BAM=\angle YAM-\angle YAB=90^{\circ}-\angle ACB=\angle BAO $, so $AM $ passes through $O $ which proves the claim. Claim : $AN $ passes through $O $. Proof : Extend $BA $ to a point $S $ such that $AS=AB $. $\triangle AO_BB\sim \triangle AO_CC $ using a simple angle chase $\implies \tfrac {AO_B}{AO_C}=\tfrac {AB}{AC}=\tfrac {AS}{AC}$. Also $\angle O_BAO_C=180^{\circ}-\angle BAC=\angle SAC $, so $\triangle O_BAO_C \sim\triangle SAC $. Let $T $ be the midpoint of $CS $. Then the similarity implies that $\angle O_BAM=\angle SAT $. Since $A,T $ are midpoints of $BS,CS $ respectively, Midpoint Theorem implies that $AT||BC\implies \angle SAT=\angle ABC\implies \angle O_BAM=\angle ABC$. So we see that $\angle BAN=\angle O_BAN-\angle O_BAB=\angle ABC-(90^{\circ}-\angle BAC)=90^{\circ}-\angle ACB=\angle BAO$, so $AN $ passes through $O $ which proves the claim. These two claims finish the problem as both $M,N $ lie on $AO $. EDIT : Completely simplified the proof of Claim $2$. Initially I was using the properties of symmedians to prove it.

My solution is similar as above. Let the midpoint of $XY$ and $O_BO_C$ be $M,N$ respectively. I will prove that $M,N$ both lie of $AO$ where $O$ is the circumcenter of $ABC$. $AO_BOO_C$ is a parallelogram, So, $N$ lies on $AO$. In quadrilateral $AYOX$, $AX=AY$ and $OX=OY$, so $M$ lies on $AO$ and we are done.

SergeyKrakowska wrote: Given an acute triangle $ABC$. $\Gamma _{B}$ is a circle that passes through $AB$, tangent to $AC$ at $A$ and centered at $O_{B}$. Define $\Gamma_C$ and $O_C$ the same way. Let the altitudes of $\triangle ABC$ from $B$ and $C$ meets the circumcircle of $\triangle ABC$ at $X$ and $Y$, respectively. Prove that $A$, the midpoint of $XY$ and the midpoint of $O_{B}O_{C}$ is collinear. Please see what you write. I think that you mean \(\gamma_B\) passes through \(B\). Please edit