Woahh nicemax indeed . Also, I just came to know that this $H'$ is nothing but the De Longchamps point lol . Let $B_1, C_1$ denote the foot of the $B, C-$altitudes respectively. $H$ denote the orthocenter, $O$ the circumcenter, and $E,F$ denote the midpoints of $B,C-$altitudes respectively. Furthermore, $H'$ denote the reflection of $H$ over $O$ and $M$ be the midpoint of $BC$. Also, $T$ denote the reflection of $H$ over $BC$. Firstly, as we are given that $H'$ lies on $BC$, so, $\angle HA_1H' = 90^{\circ}$, and so, $O$ becomes the circumcenter of $\odot(HA_1 H')$ as $O$ is the midpoint of $HH'$. This implies, $\boxed{OH = OA_1} = OH'$, and also, $O$ being the circumcenter $\boxed{OT = OA}$. So, $$\angle OA_1 T = 180^{\circ} - \angle OA_1H \stackrel{OA_1 = OH}{=} 180^{\circ} - \angle OHA_1 = \angle OHA,$$and moreover, $$\angle OTA \stackrel{OT = OA}{=} \angle OAT.$$ Thus, $\angle A_1OT = 180^{\circ} - \angle OA_1T - \angle OTA_1 = 180^{\circ} - \angle OHA - \angle OTA$ $= 180^{\circ} - \angle OHA - \angle OAT=180^{\circ} - \angle OHA - \angle OAH= \angle AOH \implies \boxed{\angle A_1OT=\angle AOH}$. These all together by SAS criterion imply, $\triangle AOH \cong \triangle TOA_1 \implies AH = A_1T$. But $T$ being reflection of $H$ over $BC \implies HA_1 = A_1T = AH \implies H$ is the midpoint of $AA_1$, i.e., $H$ is the midpoint of $A-$altitude. Now, by midpoint theorem, we get $ME \parallel CB_1 \implies \measuredangle MEH = \measuredangle CB_1B = 90^{\circ}$ and similarly, $\measuredangle MFH = \measuredangle BC_1C = 90^{\circ}$, moreover, $\measuredangle MA_1H = 90^{\circ} \implies \measuredangle MA_1H = \measuredangle MEH = \measuredangle MFH = 90^{\circ} \implies MA_1HEF$ is cyclic and we are done.

The only key observation is that $H$(orthocentre) is the midpoint of altitude passing through $A$.

One liner? Consider the anticomplementary triangle of $ABC$ since the orthocenter (since de-longchamps point is orthocenter of anticomplimentary triangle) lies on midline so same should be the case with $ABC$, so $H$ is midpoint of altitude from $A$, and therefore they are cyclic with diameter $MH$.

Nice Let $M,N,K$ be midpoints of $BC$ and altitudes of $B,C$. Note that $O$ lies on perpendicular bisector of $A_1H$ so $HA_1 = 2OM = AH$ so $H$ is the midpoint of $AA_1$. Now since $\angle MKH = \angle MNH = \angle MA_1H = 90$ we have that $A_1NHKM$ is cyclic.

lol headsolved Dilate at $H$, orthocentre, so the perpendicular bisector of $HA_1$ passes through $O$. Extending $AA_1$ to the circumcircle at $D$, we see that $A_1D=AH$ as $OA=OD,OH=OA_1$. Yet well known $A_1D=A_1H$, hence $H$ is midpoint of $AA_1$. Further if $M$ is midpoint of $BC$, easy to see that the midpoints of altitude are foot of $M$ onto the altitudes, now Thales tells us that the circle is just $(MH)$.