r538

Let P be a point of the Kiepert hyperbola of the triangle having as vertices the centers of the Vecten squares of the antimedial triangle of ABC. We take A", B", C" orthogonally to the sides and so that PA" = a, PB" = b, PC" = c. Then AA", BB", CC" concur at a point in the Kiepert hyperbola of ABC.


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