Let P be a triangle center on K001-Neuberg Cubic of ABC.
Any line through P intersect the K001-Neuberg Cubic at P1 and P2.
*As P1, P2 varies on K001-Neuberg Cubic, the conic {A,B,C,P1,P2} intersect the K001-Neuberg Cubic at a fixed point.
**If P=X(1)-Incenter of ABC, fixed point is X(7164).
If P=X(3)-Circumcenter of ABC, fixed point is X(1138).
If P=X(4)-Orthocenter of ABC, fixed point is X(8431).
If P=X(13)-1st Fermat-Toricelli point of ABC, fixed point is X(8445).
If P=X(399) of ABC, fixed point is X(4).