We consider the configuration as in the figure. Other configurations are treated analogously. Denote by $D$ the second intersection of $AH$ with the circumcircle of $\triangle ABC$. Denote by $S$ the second intersection of the circumcircles of $ABC$ and $AHO$. (Because $\triangle ABC$ is acute, both $O$ and $H$ lie in the interior of $ABC$ and also in the interior of the circumcircle, hence $D$ and $S$ both exist.)
We have
$$\angle OSH = \angle OAH = \angle OAD = \angle ODA = \angle ODH,$$
where we use that $|OA| = |OD|$. Moreover, we have
$$\angle OHD = 180^\circ - \angle OHA = 180^\circ - \angle OSA = 180^\circ - \angle OAS = \angle OHS,$$
where we use that $|OA| = |OS|$. Now we conclude that $\triangle OHS \cong \triangle OHD$ (SAA). This yields that $D$ and $S$ are each others reflection images in $OH$.

Therefore, if we reflect the circumcentre $K$ of $\triangle OHS$ in $OH$, we get the circumcentre $L$ of $\triangle OHD$. Now we must prove that $L$ lies on $BC$.
Point $D$ is the reflection of $H$ in $BC$. This is a known fact, which we can prove as follows: $\angle DBC = \angle DAC = \angle HAC = 90^\circ - \angle ACB = \angle HBC$ and analogously $\angle DCB = \angle HCB$, hence $\triangle DBC \cong \triangle HBC$ (ASA). Hence, $D$ is indeed the reflection of $H$ in $BC$, from which we get that $BC$ is the perpendicular bisector of $HD$. Because $L$ lies on the perpendicular bisector of $HD$, we get that $L$ lies on $BC$, which is what we wanted to prove. $\square$