\triangle ABC is a scalene triangle. K_a,L_a,M_a are the intersections of BC with the interior angle bisector, exterior angle bisector, and the median from A respectively. Circle \odot AK_aL_a intersects AM_a again at X_a. Define X_b,X_c similarly. Show that the circumcenter of triangle \triangle X_aX_bX_c lies on the euler line of triangle \triangle ABC.
the circle AK_aL_a is the A-Appolonios circle. basically it is the locus of points Z where \frac{ZB}{ZC}=\frac{AB}{AC}. prove is by the harmonic bundle (B,C,K_a,L_a).
we know now \frac{X_aB}{X_aC}=\frac{AB}{AC} and X_a is on the median meaning that X_a is the A-Humpty point. see Here for details.
now we know that \angle HX_aG = 90^\circ where H is the orthocenter and G is the centroid.
so the circle X_aX_bX_c is centered at the midpoint of HG which is obvoisly on the euler line.