Let P be a point, A'B'C' its pedal triangle. Draw squares on AC', BA', CB', outwards ABC. The sidelines of these squares oppsite to the sidelines of ABC bound a triangle homothetic to ABC. If we do the same with AB', BC', CA', we obtain a triangle congruent with the first. The areas of these triangles are independent of P and the midpoint between the two homothetic centers is the symmedian K.