Instead of Steiner points, if we use X(110) centers, fixed perspector of ABC and A1B1C1 is X(54)-Kosnita point.
The other points satisfyind same property are X(107) and X(476)-Tixier point.
If we use X(5)-NPC centers, AA1, BB1,CC1 are parallel lines.
Let H=X(4)-Orthocenter of ABC. A 1 B 1 C 1 cevian triangle of H. B1 A , C1 A are reflections of B1, C1 on A. L A , line through B1 A , C1...
Instead of Steiner points, if we use X(110) centers, fixed perspector of ABC and A1B1C1 is X(54)-Kosnita point.
ReplyDeleteThe other points satisfyind same property are X(107) and X(476)-Tixier point.
ReplyDeleteIf we use X(5)-NPC centers, AA1, BB1,CC1 are parallel lines.
ReplyDelete