Problem: Let $ABC$ be an acute-angled triangle with $AB\neq AC$.
The circle with diameter $BC$ intersects the sides $AB$ and $AC$
at $M$ and $N$ respectively.
Denote by $O$ the midpoint of the side $BC$.
The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$.
Prove that the circumcircles of the triangles $BMR$ and $CNR$
have a common point lying on the side $BC$.

Solution: By Miquel's theorem it's enough to show $AMRN$ is cyclic.
\begin{center}
\begin{asy}
pair A = dir(130);
pair B = dir(210);
pair C = dir(330);
pair H = A+B+C;
pair M = foot(C, A, B);
pair N = foot(B, A, C);
pair O = midpoint(B--C);
pair R = extension(A, incenter(A, B, C), O, midpoint(M--N));
pair T = extension(H, R, B, C);
filldraw(A--B--C--cycle, opacity(0.1)+lightblue, blue);
draw(B--N, blue);
draw(C--M, blue);
filldraw(circumcircle(A, M, N), opacity(0.1)+lightgreen, heavygreen);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$H$", H, dir(H));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$O$", O, dir(O));
dot("$R$", R, dir(R));
// dot("$T$", T, dir(T));

/* TSQ Source:

A = dir 130
B = dir 210
C = dir 330
H = A+B+C
M = foot C A B
N = foot B A C
O = midpoint B--C
R = extension A incenter A B C O midpoint M--N
T = extension H R B C
A--B--C--cycle 0.1 lightblue / blue
B--N blue
C--M blue
circumcircle A M N 0.1 lightgreen / heavygreen
R--T--B blue

*/
\end{asy}
\end{center}
In fact, since the bisector of $\angle MON$
is just the perpendicular bisector of $\ol{MN}$,
the point $R$ is actually just the arc midpoint
of $\widehat{MN}$ of $(AMN)$ as desired.