[1] EE

[2] Triangles A*FE, B*DF, C*ED are isosceli, their equal legs being equal to the circumradius R of triangle DEF.

[3] Hexagon A*FB*DC*E is symmetric equilateral and its center is the circumcenter Q of ABC.

[4] The radical axis (ac) of the circles a, c passing respectively from points {A,A*,F

[5] The three radical axes of circles a, b, c meet at the symmedian point of DEF, which is the

[6] Diameters AA*, BB*, CC* of circles a, b, c intersect at a point coinciding with X(84).

[7] X(9), X(84) and Q are collinear.

[8] There is a conic (h) passing through the six points E

[1, 2] Since EF is bisector of angle A and FAF

[3] By the previous arguments the hexagon is equilateral, since opposite sides are parallel (f.e. FB*, EC* are both orthogonal to BC) it is symmetric. Then FE, B*C* are parallel and equal and intercept on the circumcircle of ABC (coinciding with the Euler circle of DEF) rectangle AA'A''A''' lying symmetrically with respect to the hexagon and having center Q.

[4, 5] Since EE

[6, 7] A*B*C* is symmetric with respect to Q to DEF and suffices to show it perspective to ABC. This is shown by applying Menelaus to the intersection points X, Y, Z of side pairs {BC, B*C*}, {CA,C*A*} and {AB,A*B*}. In fact XC*/XB* = C*E

[8] See the arguments in EqualCirclesConic.html for (h). The existance of (k) is a fact valid for every symmetric hexagon, see the reference below.

EqualCirclesConic.html

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