[1] A*A'B'B*, B*B'C'C*, C*C'A'A* are cyclic quadrilaterals.

[2] The triangle A''B''C'' formed by the centers of the three circumcircles of the previous quadrilaterals has its medians coinciding with the altitudes of ABC.

[3] The symmedian point of A''B''C'' coincides with the circumcenter of ABC.

[1] follows from angle(A*A'C)=angle(BAC)=angle(BB*A*) and analogous relations.

[2] A''A is orthogonal on the middle of B'C'. Analogous relations hold for B''B and C''C.

[3] angle(B''A''A)=angle(C*C'A)=angle(C*CB)=angle(C*B*B)=angle(OA''C''), O being the circumcenter of ABC. This proves that A''O is the symmedian from A'' of triangle A''B''C''. Analogous arguments are valid for the other symmedians.

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