Conjecture. Let P be triangle center of ABC on K001 Neuberg cubic of ABC. P1, P2 are two points on the K001-Neuberg Cubic of ABC such that line P1P2 passes through P. Circle with diameter [P1P2] intersects the Neuberg Cubic of ABC at P3, P4. * P3, P4 and P* (Isogonal Conjugate of P) are collinear points.
Conjecture. Let P be triangle center of ABC on K001 Neuberg cubic of ABC. P1, P2 are two points on the K001-Neuberg Cubic of ABC such that line P1P2 passes through P.
ReplyDeleteCircle with diameter [P1P2] intersects the Neuberg Cubic of ABC at P3, P4.
* P3, P4 and P* (Isogonal Conjugate of P) are collinear points.