Line BcCb is perpendicular bisector of BC.
let A1B1C1 be the CircumCevian triangle of X(15). A',B',C' are the antipodes of A1,B1,C1 on the circumcircle respectively
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...
Line BcCb is perpendicular bisector of BC.
ReplyDeletelet A1B1C1 be the CircumCevian triangle of X(15). A',B',C' are the antipodes of A1,B1,C1 on the circumcircle respectively
ReplyDelete