Fermat point and 9-point Centers: What Is This About?
A Mathematical Droodle

Explanation

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Let F be the Fermat point of ΔABC. Let T_c, T_a, T_b denote the 9-point centers of triangles ABF, BCF, CAF with centroids G_c, G_a, G_b, respectively. Then

1. the Euler lines of the three triangles are parallel to CF, AF, BF, respectively,
2. they meet at G - the centroid of ΔABC,
3. ΔT_aT_bT_c is equilateral,
4. Points F, G, T_a, T_b, T_c are concyclic, with FG being a diameter of their circumcircle.

First of all, the lines through centroids G_a, G_b, G_c parallel to AF, BF, and CF, respectively, meet at G. Also, because of the choice of F, the Euler line in ΔABF is parallel to FI_c, where I_c is the incenter of ΔABF. But, the Fermat point may be constructed from the Napoleon triangle ABC', BCA', and CAB' as the intersection of AA', BB', and CC'. Each of these three lines bisects the angle formed by the other two. So it follows that the Euler line in ΔABF is parallel to CF. It also passes through G_c and, hence, through G. Triangles BCF and CAF can be treated similarly implying that the Euler lines e_a, e_b, e_c of triangles BCF, CAF, and ABF meet in G where they are equally inclined to each other.

As we know, points T_a, T_b, T_c are the feet of perpendiculars from F to e_a, e_b, e_c. This makes angles FT_aG, FT_bG, FT_cG right so that the five points F, G, T_a, T_b, T_c are indeed concyclic. In addition, the perpendiculars from F to equally inclined lines are equally inclined meaning that ΔT_aT_bT_c is equilateral.

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