any solution?

Solved with v4913. Our solution was motivated by generalizing this post. Let $\measuredangle$ denote directed angles modulo $180^\circ$. We prove the following much more general statement. Quote: In $\triangle ABC$ with orthocenter $H$, choose $X$ on $\overline{AB}$ and $Y$ on $\overline{AC}$ such that $XBH \sim YCH$. Let $M_A$ be the midpoint of $\overline{BC}$, $K$ be a point on the perpendicular bisector of $\overline{BC}$, and $P$ be the point with \[\measuredangle AXP=\measuredangle PYA=\measuredangle M_AKB=\theta.\]Prove that $P$ lies on $\overline{HK}$. Notice that as $X$ and $Y$ move linearly, $P$ moves linearly, so it suffices to check two cases of $(X,Y)$. Case 1: Suppose $X$ and $Y$ satisfy $\measuredangle AXH=\measuredangle HYA=\theta$. Then, $\measuredangle BXH=\measuredangle HYC$ and $\measuredangle XBH=\measuredangle YCH$ give $XBH \sim YCH$, so $X$ and $Y$ satisfy the conditions of the above statement. In this case, $P=H$ lies on $\overline{HK}$, as desired. Case 2: Suppose $(X,Y)=(B,C)$. Let $H'$ be the point such that $BHCH'$ is a parallelogram. Notice that since \[\measuredangle CBK=90^\circ+\theta=90^\circ+\measuredangle ABP=\measuredangle H'BP\]and analogously $\measuredangle BCK=\measuredangle H'CP$, we obtain that $K$ and $H'$ are isogonal conjugates in $PBC$, so $\overline{PK}$ and $\overline{PH'}$ are isogonal in $\angle BPC$. However, notice that since $\measuredangle PBH=\measuredangle ABH-\theta=\measuredangle HCP$, the first isogonality lemma implies that $\overline{PH}$ and $\overline{PH'}$ are isogonal in $\angle BPC$. Thus, $P$ lies on $\overline{HK}$. By setting $(X,Y)=(M,N)$ and $K=D$, we obtain that the antipode of $H$ in the circumcircle of $HMN$ lies on $\overline{HD}$, so $O$ lies on $\overline{HD}$, as desired. $\square$