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1807. Circles Through Isodynamic Points
Let S=X(15)-1st Isodynamic point of ABC.
Sa,Sb,Sc are X(100) points of SBC, SCA, SAB resp.*S,Sa,Sb,Sc lie on same circle.
** If Sa,Sb,Sc are X(101) points of SBC, SCA, SAB resp. again S,Sa,Sb, Sc lie on same circle.
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1987. Circumcenter On Euler Line
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...