Let $N$ be the midpoint of $BC$. Then reflecting about $N$ sends $M$ to $A$ and $H$ to the $A$ antipode, proving the claim.

$a, b, c$ on unit circle, then $h = a + b + c, d = 2 \cdot b - a, e = 2 \cdot c - a, m = \frac{d+e}{2} = b + c - a, h - m = 2a$.

Solution by Sir Jawad(TanR314) IMO gold orz orz orz $I$ be the midpoint of $BC.$ Reflect $H$ over $I$ to $H'$ and Reflecting $A$ over $I$ sends to $M$. Hence $AHMH'$ is a parallelogram. $$\angle XAH+\angle XHA=\angle XAH +\angle HAH'=90° \blacksquare$$ Another solution is by noting $BHCM$ is the nine-point circle of $\triangle ADC$ then showing $\angle XHA=|\angle D -\angle E|$

idk why i couldn't see the synthetic solution.... Computations left to reader. Let $N$ be midpoint of $BC$. Take $N$ as the origin. $N=(0,0), B=(-1,0), C=(1,0), A=(a,b)$. From $NA=NM$ $\implies$ $M=(-a,-b)$. From $AO=BO$ $\implies$ $O=(0,\frac{a^2+b^2-1}{2b})$. Take altitudes from $B$ and $C$ to $AC$ and $AB$ respectively and get equations of lines and their intersection which is $H$. So, $H=(a,\frac{a^2-1}{-b})$. We need to prove $AO$ is parallel to $HM$. Which means their slopes should be equal. Which is true since, $\frac{b-\frac{a^2+b^2-1}{2b}}{a}=\frac{\frac{a^2-1}{-b}+b}{2a}$.