Let $P$ be the midpoint of $\overline{BC}$. It is well-known that $P$ lies on $\overline{EH}$ and obvious that $P$ lies on $\overline{MN}$. As the circumcircles of $BHC$ and $ABC$ are reflections over $BC$, it follows that $PM \cdot PN = \text{pow}(P, (BHC))$. But if $A'$ is the antipode of $A$ on $(ABC)$, then $\text{pow}(P, (ABC)) = PA' \cdot PE = PH \cdot PE$ by orthocenter reflections, so as $P$ lies on the radical axis of $(BHC)$ and $(ABC)$ we're done.

After realizing $PM\cdot PN= pow(P,(BHC))= PB\cdot PC= PB^2$, then we must prove $PB$ is tangent to $(EHB)$ at $B$ which is equivalent to $\angle HEB=\angle HBC$. But this follows from the fact that $EH$ lies on the diameter of $(ABC)$

Solution by complex numbers is possible...

electrovector wrote: After realizing $PM\cdot PN= pow(P,(BHC))= PB\cdot PC= PB^2$, then we must prove $PB$ is tangent to $(EHB)$ at $B$ which is equivalent to $\angle HEB=\angle HBC$. But this follows from the fact that $EH$ lies on the diameter of $(ABC)$ Nice solution!

WolfusA wrote: Solution by complex numbers is possible... Indeed! Toss on the complex plane with $a = x^2$, $m = -yz$ etc. and $(ABC)$ the unit circle, then $h = a+b+c = x^2+y^2+z^2$. Since $(ABC)$ and $(BHC)$ are symmetric wrt to $BC$, the point $N$ is symmetric with the centre of the arc $BC$ containing $A$ on $(ABC)$. This point is represented by $yz$, so $n = b+c-bc\overline{(yz)} = y^2+z^2-yz$. Furthermore, it is pretty well-known that $E,H$, the midpoint of $BC$ and the antipode of $A$ on $(ABC)$ are collinear, so $$e = \frac{h-(-a)}{-a \overline{h} - 1} = -\frac{y^2 z^2 (2x^2+ y^2+z^2)}{x^2 y^2+x^2 z^2+2 y^2 z^2}.$$Now it remains to compute $\frac{e-m}{e-h} : \frac{n-m}{n-h}$. Firstly $n-m = y^2+z^2$ and $n-h = -(x^2+yz)$. The rest is computation: $$e - m = -\frac{y^2 z^2 (2x^2+ y^2+z^2)}{x^2 y^2+x^2 z^2+2 y^2 z^2} - yz = yz \frac{yz(2x^2+y^2+z^2) + (x^2 y^2+x^2 z^2+2 y^2 z^2)}{x^2 y^2+x^2 z^2+2 y^2 z^2} =- yz \frac{(y+z)^2 (x^2 + yz)}{x^2 y^2+x^2 z^2+2 y^2 z^2},$$$$e - h = -\frac{y^2 z^2 (2x^2 + y^2+z^2) -(x^2+y^2+z^2)(x^2 y^2+x^2 z^2+2 y^2 z^2)}{x^2 y^2+x^2 z^2+2 y^2 z^2} = - \frac{(x^2+y^2)(x^2+z^2)(y^2+z^2)}{x^2 y^2+x^2 z^2+2 y^2 z^2},$$so: $$\frac{e-m}{e-h} : \frac{n-m}{n-h} = \frac{yz(y-z)^2 (x^2+ yz)}{(x^2+y^2)(x^2+z^2)(y^2+z^2)} : \frac{(y^2+z^2)}{(x^2+yz)} $$$$=\frac{yz (y-z)^2 (x^2+yz)^2}{(x^2+y^2)(x^2+z^2) (y^2+z^2)^2},$$which is clearly self-conjugate as required. 'twas not even that painful

Easy problem my solution was the same as @above