Euler line ABC is X(3)X(74) line of A'''B'''C'''. So ınstead of X(3) circumcenter of A'''B'''C''', any point on the line X(3)X(74) of A'''B'''C''' will be on the euler line of ABC.
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...
Euler line ABC is X(3)X(74) line of A'''B'''C'''. So ınstead of X(3) circumcenter of A'''B'''C''', any point on the line X(3)X(74) of A'''B'''C''' will be on the euler line of ABC.
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