Problem: Let $O$ and $H$ be the circumcenter
and the orthocenter of an acute triangle $ABC$.
Points $M$ and $D$ lie on side $BC$
such that $BM = CM$ and $\angle BAD = \angle CAD$.
Ray $MO$ intersects the circumcircle of triangle $BHC$ in point $N$.
Prove that $\angle ADO = \angle HAN$.

Solution: Let $P$ and $Q$ be the arc midpoints of $\widehat{BC}$,
so that $ADMQ$ is cyclic (as $\dang QAD = \dang QMD = 90\dg$) .
Since it's known that $(BHC)$ and $(ABC)$ are reflections across line $BC$,
it follows $N$ is the reflection of the arc midpoint $P$ across $M$.

\begin{claim*}
  Quadrilateral $ADNO$ is cyclic.
\end{claim*}
\begin{proof}
  Since $PN \cdot PO = \half PN \cdot 2PO = PM \cdot PQ = PD \cdot PA$.
\end{proof}

\begin{center}
\begin{asy}
pair A = dir(130);
pair B = dir(220);
pair C = dir(320);
filldraw(unitcircle, opacity(0.2)+lightcyan, lightblue);

pair P = dir(-90);
pair Q = dir(90);
pair D = extension(A, P, B, C);
pair O = origin;
pair M = extension(B, C, O, P);
pair N = 2*M-P;

draw(A--B--C--cycle, lightblue);
draw(A--P--Q, lightblue);
draw(A--N--D--O--A, lightblue);

filldraw(A--D--N--O--cycle, opacity(0.1)+yellow, red);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$D$", D, dir(225));
dot("$O$", O, dir(315));
dot("$M$", M, dir(315));
dot("$N$", N, dir(315));

/* TSQ Source:

A = dir 130
B = dir 220
C = dir 320
unitcircle 0.1 lightcyan / lightblue

P = dir -90
Q = dir 90
D = extension A P B C R225
O = origin R315
M = extension B C O P R315
N = 2*M-P R315

A--B--C--cycle lightblue
A--P--Q lightblue
A--N--D--O--A lightblue

A--D--N--O--cycle 0.1 yellow / red

*/
\end{asy}
\end{center}
To finish, note that $\dang HAN = \dang ONA = \dang ODA$.

\begin{remark*}
  The orthocenter $H$ is superficial
  and can be deleted basically immediately.
  One can reverse-engineer the fact that $ADNO$ is cyclic
  from the truth of the problem statement.
\end{remark*}
\begin{remark*}
  One can also show $ADNO$ concyclic by just
  computing $\dang DAO = \dang PAO$
  and $\dang DNO = \dang DPN = \dang APQ$
  in terms of the angles of the triangle, or even more directly just because
  \[ \dang DNO = \dang DNP = \dang NPD = \dang OPD = \dang ONA = \dang HAN. \]
\end{remark*}