[alogo] On the Circum-medial triangle

Consider the triangle A*B*C* (sometimes called circum-medial triangle of ABC) formed by the intercepts of the medians of ABC with the circumcircle (circumcevian w.r. to the centroid, see Circumcevian.html ). Consider also the anticomplementary triangle A'B'C'.
[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.

[0_0] [0_1] [0_2] [0_3]
[1_0] [1_1] [1_2] [1_3]
[2_0] [2_1] [2_2] [2_3]
[3_0] [3_1] [3_2] [3_3]
[4_0] [4_1] [4_2] [4_3]

[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.

Remark 1 The construction of triangle A*B*C* from ABC can be applied to A*B*C* itself and produce A**B**C** and also a whole sequence of triangles produced each from the previous by the method A*B*C* is produced from ABC. The resulting sequence of triangles converges rapidly to an equilateral triangle inscribed in the circumcircle of ABC and having its sides parallel to the Napoleon triangle of ABC.
Remark 2 The previous remark could be applied to a basic triangle ABC and a fixed triangle center to produce a triangle A*B*C* defined by the intercepts of the cevians through the triangle center and the circumcircle. The answer to the question of convergence of the resulting sequence is unknown to me.

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