Problem: Triangle $ABC$ is inscribed in circle $\Omega$.
The interior angle bisector of angle $A$ intersects side $BC$
and $\Omega$ at $D$ and $L$ (other than $A$), respectively.
Let $M$ be the midpoint of side $BC$.
The circumcircle of triangle $ADM$ intersects sides
$AB$ and $AC$ again at $Q$ and $P$ (other than $A$), respectively.
Let $N$ be the midpoint of segment $PQ$,
and let $H$ be the foot of the perpendicular from $L$ to line $ND$.
Prove that line $ML$ is tangent to the circumcircle of triangle $HMN$.

Solution: By angle chasing, equivalent to show $\ol{MN} \parallel \ol{AD}$,
so discard the point $H$.
We now present a three solutions.

\paragraph{First solution using vectors.}
We first contend that:
\begin{claim*}
  We have $QB = PC$.
\end{claim*}
\begin{proof}
  Power of a Point gives $BM \cdot BD = AB \cdot QB$.
  Then use the angle bisector theorem.
\end{proof}
Now notice that the vector
\[ \overrightarrow{MN}
  = \half \left( \overrightarrow{BQ} + \overrightarrow{CP} \right) \]
which must be parallel to the angle bisector
since $\overrightarrow{BQ}$ and $\overrightarrow{CP}$
have the same magnitude.


\paragraph{Second solution using spiral similarity.}
let $X$ be the arc midpoint of $BAC$.
Then $ADMX$ is cyclic with diameter $\ol{XD}$,
and hence $X$ is the Miquel point of $QBPC$ and thus the midpoint of arc $BAC$.
Moreover $\ol{XND}$ collinear (as $XP=XQ$, $DP=DQ$) on $(APQ)$.

\begin{center}
\begin{asy}
pair A = dir(130);
pair B = dir(210);
pair C = dir(330);
draw(A--B--C--cycle, lightblue);
filldraw(unitcircle, opacity(0.1)+lightblue, blue);

pair L = dir(-90);
pair D = extension(A, L, B, C);
pair M = midpoint(B--C);
path w = circumcircle(A,D,M);
draw(w, dashed+orange);
pair Q = OP(A--B, w);
pair P = IP(A--C, w);
pair N = midpoint(P--Q);
pair H = foot(L, N, D);

pair X = dir(90);

draw(A--L, blue);
draw(circumcircle(H, D, M), lightblue+dotted);
draw(P--Q, blue);

draw(Q--X--C, blue);
draw(N--H, lightblue+dotted);
draw(M--L, lightblue+dotted);

filldraw(X--N--M--cycle, opacity(0.1)+lightgreen, heavygreen);
filldraw(X--P--C--cycle, opacity(0.1)+lightgreen, heavygreen);
filldraw(X--Q--B--cycle, opacity(0.1)+lightgreen, heavygreen);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$L$", L, dir(L));
dot("$D$", D, dir(D));
dot("$M$", M, dir(-90));
dot("$Q$", Q, dir(Q));
dot("$P$", P, dir(10));
dot("$N$", N, dir(130));
dot("$H$", H, dir(H));
dot("$X$", X, dir(X));

/* TSQ Source:

A = dir 130
B = dir 210
C = dir 330
A--B--C--cycle lightblue
unitcircle 0.1 lightblue / blue

L = dir -90
D = extension A L B C
M = midpoint B--C R-90
!path w = circumcircle(A,D,M);
w dashed orange
Q = OP A--B w
P = IP A--C w R10
N = midpoint P--Q R130
H = foot L N D

X = dir 90

A--L blue
circumcircle H D M lightblue dotted
P--Q blue

Q--X--C blue
N--H lightblue dotted
M--L lightblue dotted

X--N--M--cycle 0.1 lightgreen / heavygreen
X--P--C--cycle 0.1 lightgreen / heavygreen
X--Q--B--cycle 0.1 lightgreen / heavygreen

*/
\end{asy}
\end{center}
Then $\triangle XNM \sim \triangle XPC$ spirally, and
\[ \dang XMN = \dang XCP = \dang XCA = \dang XLA \]
thus done.


\paragraph{Third solution using barycentrics (mine).}
Once reduced to $\ol{MN} \parallel \ol{AB}$,
straight bary will also work.
By power of a point one obtains
\begin{align*}
  P &= \left( a^2 : 0 : 2b(b+c)-a^2 \right) \\
  Q &= \left( a^2 : 2c(b+c)-a^2 : 0 \right) \\
  \implies N &= \left( a^2(b+c) : 2bc(b+c) - ba^2 : 2bc(b+c)-ca^2 \right).
\end{align*}
Now the point at infinity along $\ol{AD}$ is $(-(b+c):b:c)$
and so we need only verify
\[
  \det \begin{bmatrix}
    a^2(b+c) & 2bc(b+c) - ba^2 & 2bc(b+c)-ca^2 \\
    0 & 1 & 1 \\
    -(b+c) & b & c
  \end{bmatrix}
  = 0
\]
which follows since the first row is $-a^2$
times the third row plus $2bc(b+c)$ times the second row.