Saturday, March 5, 2022

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.
-
Labels: , , ,

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)

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...