Problem: Let $D$ and $E$ be points in the interiors of sides $AB$ and $AC$,
respectively, of a triangle $ABC$, such that $DB = BC = CE$.
Let the lines $CD$ and $BE$ meet at $F$.
Prove that the incenter $I$ of triangle $ABC$,
the orthocenter $H$ of triangle $DEF$ and
the midpoint $M$ of the arc $BAC$ of the circumcircle of triangle $ABC$
are collinear.

Solution: \paragraph{First solution (Cynthia Du).}
Let $BI$ and $CI$ meet the circumcircle again at $M_B$, $M_C$.
Observe that we have the spiral congruence
\[ \triangle MDB \sim \triangle MEC \]
from $\dang MBD = \dang MBA = \dang MCA = \dang MCE$
and $BD = EC$, $BM = CM$.
That is, $M$ is the Miquel point of $BDEC$.

\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(240);
pair C = dir(300);

draw(A--B--C--cycle, heavycyan);
filldraw(unitcircle, opacity(0.1)+lightcyan, heavycyan);

pair I = incenter(A, B, C);
pair D = -C+2*foot(C, I, B);
pair E = -B+2*foot(B, I, C);

pair M_B = circumcenter(A, I, C);
pair M_C = circumcenter(A, I, B);
pair M = dir(90);

filldraw(M--D--B--cycle, opacity(0.1)+lightgreen, heavygreen);
filldraw(M--E--C--cycle, opacity(0.1)+lightgreen, heavygreen);
draw(B--M_B, heavycyan);
draw(C--M_C, heavycyan);

pair T = extension(M, E, B, I);
pair S = extension(M, D, C, I);

draw(circumcircle(D, I, E), lightred+dashed);
draw(D--S, heavygreen);
draw(E--T, heavygreen);
draw(D--I--E, lightred);

draw(C--D, lightgreen);
draw(B--E, lightgreen);
pair F = extension(C, D, B, E);
pair H = orthocenter(D, E, F);
draw(D--H--E, heavycyan);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(I));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$M_B$", M_B, dir(M_B));
dot("$M_C$", M_C, dir(M_C));
dot("$M$", M, dir(M));
dot("$T$", T, dir(T));
dot("$S$", S, dir(S));
dot("$F$", F, dir(F));
dot("$H$", H, dir(90));

/* TSQ Source:

A = dir 110
B = dir 240
C = dir 300

A--B--C--cycle heavycyan
unitcircle 0.1 lightcyan / heavycyan

I = incenter A B C
D = -C+2*foot C I B
E = -B+2*foot B I C

M_B = circumcenter A I C
M_C = circumcenter A I B
M = dir 90

M--D--B--cycle 0.1 lightgreen / heavygreen
M--E--C--cycle 0.1 lightgreen / heavygreen
B--M_B heavycyan
C--M_C heavycyan

T = extension M E B I
S = extension M D C I

circumcircle D I E lightred dashed
D--S heavygreen
E--T heavygreen
D--I--E lightred

C--D lightgreen
B--E lightgreen
F = extension C D B E
H = orthocenter D E F R90
D--H--E heavycyan

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

Let $T = \ol{ME} \cap \ol{BI}$ and $S = \ol{MD} \cap \ol{CI}$.
First, since $\ol{BI}$ is the perpendicular bisector of $\ol{CD}$
we have that
\[ \dang DIT = \dang CIT = \dang CIB = 90\dg - \half \angle A
  = \dang MCB = \dang MED = \dang TED \]
and so $D$, $I$, $T$, $E$ is cyclic.
Similarly $S$ lies on this circle too.
But $\dang SDE = \dang EDM = \dang MED = \dang TED$
so in fact $\ol{ST} \parallel \ol{DE}$ (isosceles trapezoid).

Then $\triangle IST$ and $\triangle HDE$ are homothetic,
so $\ol{IH}$, $\ol{DS}$, and $\ol{ET}$ concur (at $M$).

\paragraph{Second solution (Evan Chen).}
Observe that we have the spiral congruence
\[ \triangle MDB \sim \triangle MEC \]
from $\dang MBD = \dang MBA = \dang MCA = \dang MCE$
and $BD = EC$, $BM = CM$.
That is, $M$ is the Miquel point of $BDEC$.

\begin{center}
  \begin{asy}
  pair A = dir(110);
  pair B = dir(240);
  pair C = dir(300);

  draw(A--B--C--cycle, heavycyan);
  filldraw(unitcircle, opacity(0.1)+lightcyan, heavycyan);

  pair I = incenter(A, B, C);
  pair D = -C+2*foot(C, I, B);
  pair E = -B+2*foot(B, I, C);
  pair M = dir(90);
  pair F = extension(C, D, B, E);

  pair X = midpoint(B--D);
  pair Y = midpoint(C--E);
  draw(X--M--Y, red+dotted);
  pair H = orthocenter(D, E, F);

  draw(C--D, lightgreen);
  draw(B--E, lightgreen);
  draw(I--M, red+dotted);

  draw(CP(X, B), red+dashed);
  draw(CP(Y, C), red+dashed);
  filldraw(B--D--E--C--cycle, opacity(0.1)+yellow, lightgreen);

  dot("$A$", A, dir(A));
  dot("$B$", B, dir(B));
  dot("$C$", C, dir(C));
  dot("$I$", I, dir(I));
  dot("$D$", D, dir(120));
  dot("$E$", E, dir(75));
  dot("$M$", M, dir(M));
  dot("$F$", F, dir(F));
  dot("$X$", X, dir(A+B));
  dot("$Y$", Y, dir(A+C));
  dot("$H$", H, dir(90));

  /* TSQ Source:

  A = dir 110
  B = dir 240
  C = dir 300

  A--B--C--cycle heavycyan
  unitcircle 0.1 lightcyan / heavycyan

  I = incenter A B C
  D = -C+2*foot C I B R120
  E = -B+2*foot B I C R75
  M = dir 90
  F = extension C D B E

  X = midpoint B--D RA+B
  Y = midpoint C--E RA+C
  X--M--Y red dotted
  H = orthocenter D E F R90

  C--D lightgreen
  B--E lightgreen
  I--M red dotted

  CP X B red dashed
  CP Y C red dashed
  B--D--E--C--cycle 0.1 yellow / lightgreen

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

Let $X$ and $Y$ be the midpoints of $\ol{BD}$ and $\ol{CE}$.
Then $MX = MY$ by our congruence.

Consider now the circles with diameters $\ol{BD}$ and $\ol{CE}$.
We now claim that $H$, $I$, $M$ all lie on the radical axis
of these circles.
Note that $I$ is the orthocenter of $\triangle BFC$
and $H$ is the orthocenter of $\triangle DEF$,
so this follows from the so-called Steiner line of $BCDE$
(perpendicular to Gauss line $\ol{XY}$).
For $M$, we observe $MX^2 - XB^2 = MY^2 - YC^2$ thus completing the proof.

\paragraph{Third solution (homothety, official solution).}
Extend $DH$ and $EH$ to meet $BI$ and $CI$ at $D_1$ and $E_1$.
Note $DD_1 \perp BE$, $CI \perp BE$, so $DD_1 \parallel CI$.
Similarly $EE_1 \parallel BI$.
So $HE_1ID_1$.

\begin{center}
  \begin{asy}
    size(9cm);
    pair A = dir(100);
    pair B = dir(240);
    pair C = dir(300);
    pair I = incenter(A, B, C);
    pair D = -C+2*foot(C, B, I);
    pair E = -B+2*foot(B, C, I);
    pair F = extension(B, E, C, D);
    pair H = orthocenter(D, E, F);
    draw(A--B--C--cycle);
    draw(B--E);
    draw(C--D);

    pair D_1 = extension(D, H, B, I);
    pair E_1 = extension(E, H, C, I);
    pair D_2 = dir(20);
    pair E_2 = dir(170);
    draw(unitcircle);

    pair M = D_2+E_2-I;

    draw(circumcircle(B, C, D_1));
    draw(B--D_2--M--E_2--C);
    draw(E_1--E, dotted);
    draw(M--I, dotted);
    draw(D_1--D, dotted);


    dot("$A$", A, dir(A));
    dot("$B$", B, dir(B));
    dot("$C$", C, dir(C));
    dot("$I$", I, dir(I));
    dot("$D$", D, dir(D));
    dot("$E$", E, dir(E));
    dot("$F$", F, dir(F));
    dot("$H$", H, dir(90));
    dot("$D_1$", D_1, dir(100));
    dot("$E_1$", E_1, dir(80));
    dot("$D_2$", D_2, dir(D_2));
    dot("$E_2$", E_2, dir(E_2));
    dot("$M$", M, dir(M));

    /* Source generated by TSQ

    A = dir 100
    B = dir 240
    C = dir 300
    I = incenter A B C
    D = -C+2*foot C B I
    E = -B+2*foot B C I
    F = extension B E C D
    H = orthocenter D E F R90
    A--B--C--cycle
    B--E
    C--D

    D_1 = extension D H B I R100
    E_1 = extension E H C I R80
    D_2 = dir 20
    E_2 = dir 170
    unitcircle

    M = D_2+E_2-I

    circle B C D_1
    B--D_2--M--E_2--C
    E_1--E dotted
    M--I dotted
    D_1--D dotted

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

Angle chase to show that $B$, $E_1$, $F$, $C$ are cyclic --
$\angle DCE_1 = \angle DCI$ is computable in terms of $ABC$
and \[ \angle E_1BF = \angle E_1BE = \angle E_1EB = \angle HEF = \angle HDF = \angle HDC = \angle DCE_1 = \angle FCE_1. \]
Thus $B$, $D_1$, $F$, $C$ are also cyclic.
So $B$, $D_1$, $E_1$, $C$ are cyclic.

Extend $BI$ and $CI$ to meet the circumcircle again at $D_2$ and $E_2$.
Direct computation gives that $ME_2ID_2$ is also a parallelogram.
We also get $E_1D_1$ is parallel to $E_2D_2$
(both are antiparallel to $BC$ through $\angle BIC$).
So we have homothetic paralellograms and that finishes the problem.