Funny problem, resembles an extremely similar trick to the one use in ISL 2017 G3 and some other problems from previous years. Let $K,K'$ midpoints of mayor and minor arc $BC$ in $(ABC)$ respectively, let $AK \cap BC=J$, $AK' \cap BC=I$, let $N,L$ midpoints of $AC,AB$ respectively and let $OL,ON \cap AI=P,Q$ respectively. Now let $H'$ the orthocenter of $\triangle KBC$. By the 90 degrees we get $JAMK'$ cyclic so $\angle KJI=\angle IK'M=\angle IH'M$ (orthocenter reflection lemma) therefore $JIH'K$ is cyclic as well. Now the killing blow, let $AO \cap (OPQ)=R$ and notice that $\angle RQP=ROP=\angle ANL=\angle ACB$ and similar for the other, we get that $\triangle RPQ \cup O \sim \triangle ABC \cup K$ so in this similarity we have $A \to J$ which means that $\angle HAP=\angle H'JI=\angle IKM=\angle IAM$ (last one is true because $AKMI$ is trivially cyclic), therefore $A,H,M$ colinear as desired thus we are done .

One can use complex bashing to solve it as well. Notice the circumcenter will be one of the vertices of this new triangle, then after computing the two other vertices (easy because the perpendicular bisector intersects the midpoint of both the major and minor arcs), one can just use the formula for the circumcenter of an arbitrary triangle to calculate the value of the circumcenter (not time consuming as one of the vertices is the circumcenter). After that one uses the Euler line to find the orthocenter and then boom just show A, the orthocenter, and M are colinear (which is not a difficult computation).

Invert at $O$. Note that $H'$ is the intersection of tangents at $D',E'$ in $(AD'E')$. Similarly $M'$ is intersection of tangents at $B,C$ to $(ABC)$. We claim that a spiral similarity maps the whole configuration off $\triangle ABC$ to the configuration off $\triangle AD'E'$, because then we have $\angle AH'O=\angle AM'O$ as $O$ is midpoint of arc $D'AE'$ (Because $ODE$ isosceles from $\angle BAD=\angle EAC$, take complements) which maps to midpoint of arc $BAC$ that is on $OM'$, so $AH'M'O$ cyclic. This spiral similarity follows from similar switch where \[\angle D'AB=\angle ADD'=\angle EDO=\angle OED=\angle E'AC\]and isosceles implies $\triangle D'AB\sim E'AC$ which gives $\triangle AD'E'\sim \triangle ABC$. We are done.