This suggests to find a point P which is the centroid of its antipedal triangle as to ABC.

Let LMN be the antipedal triangle of P and A′, B′, C′ the inverse of A′, B′, C′ with respect to a circle Γ having its centre at P. If J is the intersection of AP and B′C′, the inverse J′ of J as to Γ is the point where the parallel to MN drawn through L meets again the circle BPCL. Hence the ratio PJ: A′J equals PA: J′A, and this means that the barycentric coordinates of P are the same in the triangle LMN as in the triangle A′B′C′ Therefore, if P is the centroid of its antipedal triangle, it is also the centroid of A′B′C′