Let $N$ be the midpoint of $OH$, let $K'$ be the reflection of $K$, and let $A'$ be the reflection of $A$ with respect to $OH$, and let $O'$ be the reflection of $O$ with respect to $BC$. It is well known that $AHO'O$ is a parallelogram with center $N$. Therefore, $\Delta A'OH =\Delta AOH=\Delta O'OH$, which implies that $A'OH$ and $O'OH$ have the same circumcircle, and thus the same circumcenter $K'$. Since $K'$ is the circumcenter of $OO'H$, it must lie on the axis of $OO'$, which by definition is $BC$.

Complex bash Let $(ABC)$ be unit circle , so that $o=0, h=a+b+c$ Since $o=0$ we can easily compute $k$: $k=\frac{ha(\bar{h}-\bar{a})}{\bar{h}a-h\bar{a}}=\frac{a(a+b+c)(\frac{ab+bc+ca}{abc}-\frac{1}{a})}{\frac{ab+bc+ca}{bc}-\frac{a+b+c}{a}}=\frac{a^2(a+b+c)}{a^2-bc} \wedge \bar{k}=\frac{ab+bc+ca}{a(bc-a^2)}$ Then $n=\frac{h\bar{k}}{\bar{h}}=\frac{\frac{abc(a+b+c)(ab+bc+ca)}{a(bc-a^2)}}{ab+bc+ca}=\frac{bc(a+b+c)}{bc-a^2} \wedge \bar{n}=\frac{a(ab+bc+ca)}{bc(a^2-bc)} $ Now we just need to show $\begin{vmatrix} \frac{bc(a+b+c)}{bc-a^2} & \frac{a(ab+bc+ca)}{bc(a^2-bc)} & 1 \\ b & \frac{1}{b} & 1 \\ c & \frac{1}{c} & 1 \end{vmatrix}=0 \iff \frac{bc(a+b+c)}{bc-a^2}(\frac{1}{b}-\frac{1}{c})-\frac{a(ab+bc+ca)}{bc(a^2-bc)}(b-c)+(\frac{b}{c}-\frac{c}{b})=0 \iff 0=0$ so we are done

Let $\omega$ denote the reflection of circle $AHO$ with respect to $OH$. It is enough to show that the center of $\omega$ lies on $BC$. If $A'=AH\cap \odot ABC$ then $A'$ must lie on $\omega$. This is because $HO$ must subtend an angle of $\angle HAO$ in $\omega$ as well, and we know $\angle HAO=\angle OA'H$. We know that the center of $\omega$ must lie on the perpendicular bisector of $HA'$, which is $BC$.