Problem: Let $ABC$ be a scalene triangle. Let $K_a$, $L_a$, and
$M_a$ be the respective intersections with $BC$ of the internal
angle bisector, external angle bisector, and the median from
$A$. The circumcircle of $AK_aL_a$ intersects $AM_a$ a second time
at a point $X_a$ different from $A$. Define $X_b$ and $X_c$
analogously. Prove that the circumcenter of $X_aX_bX_c$ lies on
the Euler line of $ABC$.

Solution: The main content of the problem:
\begin{claim*}
  $\angle H X_a G = 90\dg$.
\end{claim*}
This implies the result,
since then the desired circumcenter is the midpoint of $\ol{GH}$.
(This is the main difficulty; the Euler line is a red herring.)

In what follows, we abbreviate $K_a$ $L_a$, $M_a$, $X_a$
to $K$, $L$, $M$, $X$.

\begin{proof}
  [First proof by Brocard]
  To do this, it suffices to show that $M$ has the
  same power with respect to the circle
  with diameter $\ol{AH}$ and the circle with diameter $\ol{K L}$.
  In fact I claim both circles are orthogonal
  to the circle with diameter $\ol{BC}$!
  The former follows from Brocard's theorem, noting that $A$
  is on the polar of $H$,
  and the latter follows from the harmonic bundle.

  \begin{center}
  \begin{asy}
  size(11cm);
  pair A = dir(150);
  pair B = dir(200);
  pair C = dir(340);
  filldraw(A--B--C--cycle, opacity(0.1)+lightred, red);

  pair M = midpoint(B--C);

  pair H = orthocenter(A, B, C);
  pair G = centroid(A, B, C);
  pair X = foot(H, M, A);

  pair K = extension(A, incenter(A, B, C), B, C);
  pair L = extension(A, A+dir(90)*(K-A), B, C);

  draw(K--A--L--B, red);
  draw(circumcircle(A, K, L), orange+dashed);
  draw(A--M, red);
  draw(X--H, orange);

  dot("$A$", A, dir(A));
  dot("$B$", B, dir(270));
  dot("$C$", C, dir(270));
  dot("$M$", M, dir(M));
  dot("$H$", H, dir(270));
  dot("$G$", G, dir(10));
  dot("$X$", X, dir(10));
  dot("$K$", K, dir(K));
  dot("$L$", L, dir(L));

  /* TSQ Source:

  !size(11cm);
  A = dir 150
  B = dir 200 R270
  C = dir 340 R270
  A--B--C--cycle 0.1 lightred / red

  M = midpoint B--C

  H = orthocenter A B C R270
  G = centroid A B C R10
  X = foot H M A R10

  K = extension A incenter A B C B C
  L = extension A A+dir(90)*(K-A) B C

  K--A--L--B red
  circumcircle A K L orange dashed
  A--M red
  X--H orange

  */
  \end{asy}
  \end{center}
  Then $\ol{AM}$ is the radical axis, so $X$ lies on both circles.
\end{proof}

\begin{proof}
  [Second proof by orthocenter reflection, Bendit Chan]
  As before, we know $MX \cdot MA = MK \cdot ML = MB \cdot MC$,
  but $X$ lies inside segment $AM$.
  Construct parallelogram $ABA'C$.
  Then $MX \cdot MA' = MB \cdot MC$,
  so $XBA'C$ is concyclic.

  However, it is well-known the circumcircle of $\triangle BA'C$
  (which is the reflection of $(ABC)$ across $\ol{BC}$)
  passes through $H$ and in fact has diameter $\ol{A'H}$.
  So this gives $\angle HXA' = 90\dg$ as needed.
\end{proof}

\begin{proof}
  [Third proof by barycentric coordinates]
  Alternatively we may just compute $X = (a^2:2S_A:2S_A)$.
  Let $F = (0 : S_C : S_B)$ be the foot from $H$.
  Then we check that $X H F M$ is cyclic,
  which is power of a point from $A$.
\end{proof}