Ab,Ac = the reflections of A' in BB', CC', resp.
Bc,Ba = the reflections of B' in CC', AA', resp.
Ca,Cb = the reflections of C' in AA', BB', resp.
Na, Nb, Nc = The NPC centers of A'AbAc, B'BcBa, C'CaCb, resp.
O* = The circumcenter of NaNbNc. It is the NPC center of A'B'C'.
R* = the radical center of (Na), (Nb), (Nc)
The NPC circles (Na), (Nb), (Nc) concur on the NPC of A'B'C'. The point of concurrence, the R*, is the Poncelet point of H wrt A'B'C' and since the H of ABC is the I of A'B'C', the point is the Feuerbach point  of A'B'C' = the center of the Feuerbach hyperbola of A'B'C'.
The points R*, H, O* are collinear. The line is the INF-line of A'B'C'
Let P = point on the Euler line of ABC.
The points R*, P, O* are collinear.
P = a variable point.
Which is the locus of P such that R*,P,O* are collinear?
Euler Line + ???
Is the incenter I on the locus?
The Feuerbach point of ABC:
Let ABC be a triangle and A'B'C' the cevian triangle of I. Denote:
Ab, Ac = the reflections of A in BB', CC', resp.
Bc, Ba = the reflectuons of B in CC', AA', resp.
Ca, Cb = the reflections of C in AA', BB', resp.
The NPCs of AAbAc, BBcBa, CCaCb (and ABC) concur at Feuerbach point of ABC.
Antreas P. Hatzipolakis, 16 May 2013