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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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1988. Circumcenters On Euler Line
Let ABC be a triangle with P, Q are two points on Euler line of ABC. A',B',C' are inverse images of A,B,C through circle diame...