We consider the configuration as shown in the figure. Other configurations proceed analogously. Let $D$ be the second intersection of $A H$ with the circumcircle of $\triangle A B C$. Let $S$ be the second intersection of the circumcircle of $A B C$ and the circumcircle of $A H O$. (Since $\triangle A B C$ is acute, $O$ and $H$ lie inside $\triangle A B C$ and thus also inside its circumcircle, so $D$ and $S$ both exist.)

We have

$$\angle O S H=\angle O A H=\angle O A D=\angle O D A=\angle O D H,$$

where we use that $|O A|=|O D|$. Furthermore, we have

$$\angle O H D=180^{\circ}-\angle O H A=180^{\circ}-\angle O S A=180^{\circ}-\angle O A S=\angle O H S,$$

where we use that $|O A|=|O S|$. We conclude that $\triangle O H S \cong \triangle O H D$ (SAS). It follows that $D$ and $S$ are each other's reflections in $O H$.

If we reflect the center $K$ of the circumcircle of $\triangle O H S$ in $O H$, we get the center $L$ of the circumcircle of $\triangle O H D$. We need to prove that $L$ lies on $B C$.

Point $D$ is the reflection of $H$ in $B C$. This is a known fact, which we can prove as follows: $\angle D B C=\angle D A C=\angle H A C=90^{\circ}-\angle A C B=\angle H B C$ and similarly $\angle D C B=\angle H C B$, so $\triangle D B C \cong \triangle H B C(\mathrm{SAS})$. Therefore, $D$ is indeed the reflection of $H$ in $B C$, which means that $B C$ is the perpendicular bisector of $H D$. Since $L$ lies on the perpendicular bisector of $H D$, $L$ lies on $B C$, and that is what we wanted to prove.