For P=X(15), OAOBOC is equilateral.
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...
For P=X(15), OAOBOC is equilateral.
ReplyDelete