Problem: Let $ABC$ be a scalene triangle with circumcircle $\Omega$ and incenter $I$.
Ray $AI$ meets $\ol{BC}$ at $D$ and $\Omega$ again at $M$;
the circle with diameter $\ol{DM}$ cuts $\Omega$ again at $K$.
Lines $MK$ and $BC$ meet at $S$, and $N$ is the midpoint of $\ol{IS}$.
The circumcircles of $\triangle KID$ and $\triangle MAN$
intersect at points $L_1$ and $L_2$.
Prove that $\Omega$ passes through the midpoint of either $\ol{IL_1}$ or $\ol{IL_2}$.

Solution: Let $W$ be the midpoint of $\ol{BC}$,
let $X$ be the point on $\Omega$ opposite $M$.
Observe that $\ol{KD}$ passes through $X$,
and thus lines $BC$, $MK$, $XA$ concur at
the orthocenter of $\triangle DMX$, which we call $S$.
Denote by $I_A$ the $A$-excenter of $ABC$.

Next, let $E$ be the foot of the altitude from $I$ to $\ol{XI_A}$;
observe that $E$ lies on the circle centered at $M$
through $I$, $B$, $C$, $I_A$.
Then, $S$ is the radical center of $\Omega$
and the circles with diameter $\ol{IX}$ and $\ol{II_A}$;
hence line $SI$ passes through $E$;
accordingly $I$ is the orthocenter of $\triangle XSI_A$;
denote by $L$ the foot from $X$ to $\ol{SI_A}$.

\begin{center}
\begin{asy}
size(10cm);
pair A = dir(160);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A, B, C);
pair D = extension(A, I, B, C);
pair M = circumcenter(B, I, C);
pair I_A = 2*M-I;
pair X = -M;
pair K = foot(M, D, X);
pair S = extension(M, K, B, C);
pair N = midpoint(I--S);

pair W = midpoint(B--C);
pair E = foot(I, X, I_A);

draw(A--B--C--cycle, orange);
filldraw(unitcircle, opacity(0.1)+lightred, orange);
filldraw(CP(M, I), opacity(0.1)+yellow, olive);
draw(A--I_A, orange);
draw(M--X, orange);
draw(S--E, olive);
draw(X--I_A, olive);
draw(B--S, orange);

pair L = foot(X, I_A, S);
draw(M--S--X--cycle, heavyred+1);
draw(X--K, heavyred+1);
draw(A--M, heavyred+1);
draw(S--W, heavyred+1);

draw(S--I_A, orange);

draw(circumcircle(K, I, D), palered+1);
filldraw(circumcircle(M, A, N), opacity(0.1)+yellow, palered+1);
draw(X--L, orange+1);

pair T = midpoint(I--L);
draw(T--M, orange);

dot("$A$", A, dir(150));
dot("$B$", B, dir(B));
dot("$C$", C, dir(-10));
dot("$I$", I, dir(I));
dot("$D$", D, dir(D));
dot("$M$", M, dir(M));
dot("$I_A$", I_A, dir(I_A));
dot("$X$", X, dir(X));
dot("$K$", K, dir(K));
dot("$S$", S, dir(S));
dot("$N$", N, dir(135));
dot("$W$", W, dir(-45));
dot("$E$", E, dir(45));
dot("$L$", L, dir(L));
dot("$T$", T, dir(T));

/* Source generated by TSQ:

!size(10cm);
A = dir 160 R150
B = dir 210
C = dir 330 R-10
I = incenter A B C
D = extension A I B C
M = circumcenter B I C
I_A = 2*M-I
X = -M
K = foot M D X
S = extension M K B C
N = midpoint I--S R135

W = midpoint B--C R-45
E = foot I X I_A R45

A--B--C--cycle orange
unitcircle 0.1 lightred / orange
CP M I 0.1 yellow / olive
A--I_A orange
M--X orange
S--E olive
X--I_A olive
B--S orange

L = foot X I_A S
M--S--X--cycle heavyred+1
X--K heavyred+1
A--M heavyred+1
S--W heavyred+1

S--I_A orange

circumcircle K I D palered+1
circumcircle M A N 0.1 yellow / palered+1
X--L orange+1

T = midpoint I--L
T--M orange

*/
\end{asy}
\end{center}

We claim that this $L$ lies on
both the circumcircle of $\triangle KID$ and $\triangle MAN$.
It lies on the circumcircle of $\triangle MAN$
since this circle is the nine-point circle of $\triangle XSI_A$.
Also, $XD \cdot XK = XW \cdot XM = XA \cdot XS = XI \cdot XL$,
so $KDIL$ are concyclic.
%For the other, note that $\triangle MWI \sim \triangle MIX$,
%according to which $\angle IWM = \angle MIX = 180^{\circ} - \angle LIM = 180^{\circ} - \angle MLI$,
%enough to imply that quadrilateral $MWIL$ is cyclic.
%But lines $IL$, $DK$, and $WM$ meet at $X$, implying the conclusion.

All that remains to show is that the
midpoint $T$ of $\ol{IL}$ lies on $\Omega$.
But this follows from the fact that
$\ol{TM} \parallel \ol{LI_A} \implies \angle MTX = 90^{\circ}$,
thus the problem is solved.

\begin{remark*}
  Some additional facts about this picture:
  the point $T$ is the contact point of the $A$-mixtilinear incircle
  (since it is collinear with $X$ and $I$),
  while the point $K$ is such that $\ol{AK}$ is an $A$-symmedian
  (since $\ol{KD}$ and $\ol{AD}$ bisect $\angle A$ and $\angle K$, say).
\end{remark*}

\begin{center}
\begin{asy}
pair I_A = dir(115);
pair I_B = dir(215);
pair I_C = dir(325);
pair A = foot(I_A, I_B, I_C);
pair B = foot(I_B, I_C, I_A);
pair C = foot(I_C, I_A, I_B);
pair X = midpoint(I_B--I_C);
pair S = extension(B, C, I_B, I_C);
pair I = incenter(A, B, C);
pair L = extension(X, I, S, I_A);

pair D = extension(A, I, B, C);
pair M = midpoint(I--I_A);
pair K = foot(M, D, X);
pair N = midpoint(I--S);

draw(I_A--I_B--I_C--cycle, heavycyan);
filldraw(circumcircle(I_A, I_B, I_C), opacity(0.1)+palered, heavycyan);
filldraw(circumcircle(A, B, C), opacity(0.1)+orange, orange);
draw(A--B--C--cycle, orange);

draw(S--I_A, orange);
draw(S--I, olive);
draw(S--I_B, heavycyan);
draw(A--I_A, heavycyan);

filldraw(circumcircle(K, I, D), opacity(0.07)+yellow, palered+1);
filldraw(circumcircle(M, A, N), opacity(0.07)+yellow, palered+1);
draw(X--L, orange+1);

pair T = midpoint(I--L);
draw(M--X, olive);
draw(K--X, olive);
draw(M--S, olive);
draw(X--I_A, heavycyan);
draw(S--C, orange);

filldraw(circumcircle(I, B, C), opacity(0.1)+palecyan, heavycyan);

dot("$I_A$", I_A, dir(I_A));
dot("$I_B$", I_B, dir(I_B));
dot("$I_C$", I_C, dir(I_C));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$X$", X, dir(X));
dot("$S$", S, dir(S));
dot("$I$", I, dir(I));
dot("$L$", L, dir(L));
dot("$D$", D, dir(D));
dot("$M$", M, dir(M));
dot("$K$", K, dir(K));
dot("$N$", N, dir(80));
dot("$T$", T, dir(T));

/* TSQ Source:

I_A = dir 115
I_B = dir 215
I_C = dir 325
A = foot I_A I_B I_C
B = foot I_B I_C I_A
C = foot I_C I_A I_B
X = midpoint I_B--I_C
S = extension B C I_B I_C
I = incenter A B C
L = extension X I S I_A

D = extension A I B C
M = midpoint I--I_A
K = foot M D X
N = midpoint I--S R80

I_A--I_B--I_C--cycle heavycyan
circumcircle I_A I_B I_C 0.1 palered / heavycyan
circumcircle A B C 0.1 orange / orange
A--B--C--cycle orange

S--I_A orange
S--I olive
S--I_B heavycyan
A--I_A heavycyan

circumcircle K I D 0.07 yellow / palered+1
circumcircle M A N 0.07 yellow / palered+1
X--L orange+1

T = midpoint I--L
M--X olive
K--X olive
M--S olive
X--I_A heavycyan
S--C orange

circumcircle I B C 0.1 palecyan / heavycyan

*/
\end{asy}
\end{center}

\begin{remark*}
  In fact, the point $L$ is the Miquel point of cyclic quadrilateral
  $I_B I_C B C$ (inscribed in the circle with diameter $\ol{I_B I_C}$).
  This implies many of the properties that $L$ has above.
  For example, it directly implies that $L$ lies on the circumcircles
  of triangles $I_A I_B I_C$ and $BCI_A$,
  and that the point $L$ lies on $\ol{SI_A}$
  (since $S = \ol{BC} \cap \ol{I_B I_C}$).
  For this reason, many students found it easier to think about
  the problem in terms of $\triangle I_A I_B I_C$ rather than $\triangle ABC$.
\end{remark*}