First solution (power of a point) Let $\gamma$ denote the nine-point circle of $ABC$ .

$[asy] pair A = dir(125); pair B = dir(210); pair C = dir(330); pair M = midpoint(A--B); pair N = midpoint(A--C); pair O = origin; pair H = A+B+C; draw(A--B--C--cycle, blue); pair E = foot(B, A, C); pair F = foot(C, A, B); draw(B--E, lightblue); draw(C--F, lightblue); pair R = extension(E, F, B, C); pair Q = -A+2*foot(O, A, R); filldraw(unitcircle, invisible, blue); draw(A--R--E, heavygreen); pair P = extension(M, N, A, A+dir(90)*A); draw(A--P--N, red); filldraw(circumcircle(A, M, N), invisible, red); filldraw(circumcircle(A, E, F), invisible, heavygreen); filldraw(circumcircle(M, N, E), invisible, heavycyan); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(145)); dot("$N$", N, dir(20)); dot("$O$", O, dir(315)); dot("$H$", H, dir(H)); dot("$E$", E, dir(40)); dot("$F$", F, dir(F)); dot("$R$", R, dir(R)); dot("$Q$", Q, dir(Q)); dot("$P$", P, dir(P)); /* TSQ Source: A = dir 125 B = dir 210 C = dir 330 M = midpoint A--B R145 N = midpoint A--C R20 O = origin R315 H = A+B+C A--B--C--cycle blue E = foot B A C R40 F = foot C A B B--E lightblue C--F lightblue R = extension E F B C Q = -A+2*foot O A R unitcircle 0.1 lightcyan / blue A--R--E heavygreen P = extension M N A A+dir(90)*A A--P--N red circumcircle A M N 0.1 lightred / red circumcircle A E F 0.1 lightgreen / heavygreen circumcircle M N E 0.1 lightcyan / heavycyan */ [/asy]$

Note that

$PA^2 = PM \cdot PN$ , so $P$ lies on the radical axis of $\Gamma$ and $\gamma$ .
$RA \cdot RQ = RE \cdot RF$ , so $R$ lies on the radical axis of $\Gamma$ and $\gamma$ .

Thus $\overline{PR}$ is the radical axis of $\Gamma$ and $\gamma$ , which is evidently perpendicular to $\overline{OH}$ .

Remark: In fact, by power of a point one may also observe that $R$ lies on $\overline{BC}$ , since it is on the radical axis of $(AQFHE)$ , $(BFEC)$ , $(ABC)$ . Ironically, this fact is not used in the solution.

Second solution (barycentric coordinates) Again note first $R \in \overline{BC}$ (although this can be avoided too). We compute the points in much the same way as before. Since $\overline{AP} \cap \overline{BC} = (0 : b^2 : -c^2)$ we have $P = \left( b^2-c^2 : b^2 : -c^2 \right)$ (since $x=y+z$ is the equation of line $\overline{MN}$ ). Now in Conway notation we have $R = \overline{EF} \cap \overline{BC} = (0 : S_C : -S_B) =\left( 0 : a^2+b^2-c^2 : -a^2+b^2-c^2 \right).$ Hence $\overrightarrow{PR} = \frac{1}{2(b^2-c^2)} \left( b^2-c^2 , c^2-a^2 , a^2-b^2 \right).$ On the other hand, we have $\overrightarrow{OH} = \vec A + \vec B + \vec C$ . So it suffices to check that $\sum_{\text{cyc}} a^2\left( (a^2-b^2) + (c^2-a^2) \right) = 0$ which is immediate.

Third solution (complex numbers) Let $ABC$ be the unit circle. We first compute $P$ as the midpoint of $A$ and $\overline{AA} \cap \overline{BC}$ : $\begin{align*} p &= \frac{1}{2} \left( a + \frac{a^2(b+c)-bc\cdot2a}{a^2-bc} \right) \\ &= \frac{a(a^2-bc)+a^2(b+c)-2abc}{2(a^2-bc)}. \end{align*}$ Using the remark above, $R$ is the inverse of $D$ with respect to the circle with diameter $\overline{BC}$ , which has radius $\left\lvert \frac{1}{2}(b-c) \right\rvert$ . Thus $\begin{align*} r - \frac{b+c}{2} &= \frac{\frac14(b-c)\left( \frac1b-\frac1c \right)} {\overline{\frac{1}{2}\left( a-\frac{bc}{a} \right)}} \\ r &= \frac{b+c}{2} + \frac{-\frac{1}{2} \frac{(b-c)^2}{bc}}{\frac1a-\frac{a}{bc}} \\ &= \frac{b+c}{2} + \frac{a(b-c)^2}{2(a^2-bc)} \\ &= \frac{a(b-c)^2+(b+c)(a^2-bc)}{2(a^2-bc)}. \end{align*}$ Expanding and subtracting gives $p-r = \frac{a^3-abc-ab^2-ac^2+b^2c+bc^2}{2(a^2-bc)} = \frac{(a+b+c)(a-b)(a-c)}{2(a^2-bc)}$ which is visibly self-conjugate once the factor of $a+b+c$ is deleted.

(Actually, one can guess this factorization ahead of time by noting that if $A=B$ , then $P=B=R$ , so $a-b$ must be a factor; analogously $a-c$ must be as well.)