Problem: Acute triangle $ABC$ is inscribed in circle $\omega$.
Let $H$ and $O$ denote its orthocenter and circumcenter, respectively.
Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively.
Rays $MH$ and $NH$ meet $\omega$ at $P$ and $Q$, respectively.
Lines $MN$ and $PQ$ meet at $R$.
Prove that $\ol{OA} \perp \ol{RA}$.

Solution: Let $MH$ and $NH$ meet the nine-point circle again at $P'$ and $Q'$, respectively.
Recall that $H$ is the center of the homothety between
the circumcircle and the nine-point circle.
From this we can see that $P$ and $Q$ are the images of this homothety, meaning that
\[ HQ = 2HQ' \quad\text{and}\quad HP = 2HP'. \]
Since $M$, $P'$, $Q'$, $N$ are cyclic, Power of a Point gives us
\[ MH \cdot HP' = HN \cdot HQ'. \]
Multiplying both sides by two, we thus derive
\[ HM \cdot HP = HN \cdot HQ. \]
It follows that the points $M$, $N$, $P$, $Q$ are concyclic.

\begin{center}
  \begin{asy}
    size(11cm);
    defaultpen(fontsize(8pt));
    pair A = dir(110);
    pair B = dir(210);
    pair C = dir(330);
    pair H = A+B+C;
    pair O = circumcenter(A, B, C);
    pair M = midpoint(A--B);
    pair N = midpoint(A--C);
    pair U = -C;
    pair V = -B;
    pair P = -U+2*foot(O, U, H);
    pair Q = -V+2*foot(O, V, H);
    pair R = extension(M, N, P, Q);

    draw(A--B--C--cycle, blue);
    draw(unitcircle, blue);
    draw(M--P);
    draw(N--Q);
    draw(circumcircle(M, N, midpoint(B--C)), green);
    draw(circumcircle(P, Q, M), red+dashed);
    draw(P--R--A);
    draw(R--N, dotted);
    draw(circumcircle(A, M, N), dashed);

    pair Qp = midpoint(H--Q);
    pair Pp = midpoint(H--P);

    dot("$A$", A, dir(A));
    dot("$B$", B, dir(250));
    dot("$C$", C, dir(C));
    dot("$H$", H, dir(-90));
    dot("$O$", O, dir(315));
    dot("$M$", M, 1.4*dir(150));
    dot("$N$", N, dir(20));
    dot("$P$", P, dir(P));
    dot("$Q$", Q, dir(Q));
    dot("$R$", R, dir(R));
    dot("$Q'$", Qp, dir(255));
    dot("$P'$", Pp, dir(Pp));

    /* Source generated by TSQ */
  \end{asy}
\end{center}

Let $\omega_1$, $\omega_2$, $\omega_3$ denote the circumcircles of
$MNPQ$, $AMN$, and $ABC$, respectively.
The radical axis of $\omega_1$ and $\omega_2$ is line $MN$, while the
radical axis of $\omega_1$ and $\omega_3$ is line $PQ$.
Hence the line $R$ lies on the radical axis of $\omega_2$ and $\omega_3$.

But we claim that $\omega_2$ and $\omega_3$ are internally tangent at $A$.
This follows by noting the homothety at $A$ with ratio $2$ sends
$M$ to $B$ and $N$ to $C$.
Hence the radical axis of $\omega_2$ and $\omega_3$ is a line tangent to
both circles at $A$.

Hence $\ol{RA}$ is tangent to $\omega_3$.
Therefore, $\ol{RA} \perp \ol{OA}$.