Let $O$ be the circumcenter of $(ABC)$, then $S$, $M$, $O$ lie on the perpendicular bisector of side $AB$. Let $w$ and $(ABC)$ intersect at $S'$, $C$. Since $AS'BC$ is cyclic and $\angle$ $ACS'$ $=$ $\angle$ $BCS'$, it's easy to show that $S'$ lies on the perpendicular bisector of side $AB$, therefore $S$ $\equiv$ $S'$ and $S$ $\in$ $(ABC)$. Let $D$, $E$ be the feet of the altitude from $A$, $B$ of $\bigtriangleup$ $ABC$ and $J$ be the midpoint of side $CH$. We have $M$ is the center of the circle passing through $A$, $B$, $D$, $E$ and $J$ is the center of the circle passing through $D$, $H$, $F$, $E$, $C$ and so, $ME$ $=$ $MD$, $JE$ $=$ $JD$. Since $DFEC$ is cyclic and $\angle$ $FCE$ $=$ $\angle$ $FCD$, it's easy to show that $FE$=$FD$. $M$, $F$, $J$ lie on the perpendicular bisector of side $DE$. We know that $CH$ $=$ $2OM$ and $CH$ $\parallel$ $OM$, so $MJCO$ is an parallelogram. $MF$ $\parallel$ $OC$ and $OS$ $=$ $OC$, so $MF$ $=$ $MS$.

We use complex numbers, letting $A = u^2, B = v^2, C = w^2$ on the unit circle. We know that $S$ is the midpoint of arc $AB$ not containing $C,$ so $s = -uv.$ On top of that, we know that $m = \frac{u^2 + v^2}{2}$ and $h = u^2 + v^2 + w^2.$ This means that $m - s = \frac{u^2 + v^2}{2} + uv = \frac{u^2 + 2uv + v^2}{2} = \frac{(u+v)^2}{2},$ so $$|m-s| = \frac{|u+v|^2}{2}.$$Now, by the complex foot formula, we get \begin{align*} f &= \frac{1}{2} (w^2 - uv + u^2 + v^2 + w^2 - \overline{u^2 + v^2 + w^2} \cdot w^2 (-uv)) \\ &= \frac{1}{2} \left(2w^2 - uv + u^2 + v^2 + \left(\frac{1}{u^2} + \frac{1}{v^2} + \frac{1}{w^2}\right) \cdot w^2 uv\right) \\ &= \frac{1}{2} \left(2w^2 - uv + u^2 + v^2 + \frac{w^2 uv(u^2v^2 + v^2w^2 + w^2u^2)}{u^2v^2w^2}\right) \\ &= \frac{1}{2} \left(2w^2 - uv + u^2 + v^2 + \frac{u^2 v^2 + v^2 w^2 + w^2 u^2}{uv}\right). \end{align*}Therefore, \begin{align*} f-m &= \frac{1}{2} \left(2w^2 - uv + u^2 + v^2 + \frac{u^2 v^2 + v^2 w^2 + w^2 u^2}{uv} - u^2 - v^2\right) \\ &= \frac{1}{2} \left(2w^2 - uv + \frac{u^2 v^2 + v^2 w^2 + w^2 u^2}{uv}\right) \\ &= \frac{1}{2} \left(\frac{2w^2 uv - u^2 v^2 + u^2 v^2 + v^2 w^2 + w^2 u^2}{uv}\right) \\ &= \frac{1}{2} \left(\frac{w^2 v^2 + 2w^2 uv + w^2 u^2}{uv}\right) \\ &= \frac{w^2 (u+v)^2}{2uv}. \end{align*}This means that $$|m-f| = \left|\frac{w^2 (u+v)^2}{2uv}\right| = \frac{|w^2| |(u+v)^2|}{2|u||v|}.$$Noting that $|w^2| = |u| = |v| = 1,$ we get that $$|m-f| = \frac{|u+v|^2}{2} = |m-s|,$$as desired.

It is well-known that $S$ is the midpoint of the arc of circle $ABC$ which does not contain $A$. Reflect $H,F$ in $M$ to get $H', F'$. It is well-known that $H'$ is the antipode of $A$ in $ABC$. Additionally, $HF || H'F'$ implies $\angle(HF, w) = 90 = \angle(H'F', w) = \angle(HS, w)$ which implies that $H'F'S$ are collinear. This gives us $\angle F'SF = 90$ and since $M$ is the midpoint of $FF'$, we have $MF = MS$ by Thales' theorem.