Given triangle ABC, consider the triangle DEF of the excenters. Project the excenters on the sides of ABC to obtain points DB, DC, EC, EA, FA, FB.
[1] EEC and FFB intersect at A* symmetric of the circumcenter O of DEF with respect to side EF. Analogously FFA and DDC intersect at B*, symmetric of O with respect to side FD, DDB and EEA intersect at C* which is symmetric of O with respect to DE.
[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*,FB,EC} and {C,C*,DB,EA} is the symmedian through E of triangle DEF.
[5] The three radical axes of circles a, b, c meet at the symmedian point of DEF, which is the Mittenpunkt of triangle ABC (X(9) in the enumeration of Kimberling).
[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 EC, FB, FA, DC, DB, EA and a conic (k) circumscribing the symmetric hexagon A*FB*DC*E.
[1, 2] Since EF is bisector of angle A and FAFB, AEEC are right-angled EFA* is isosceles. Measuring its lengths in terms of angle(A) we see easily that they are equal to the circumradius R of DEF.
[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 EEC*EA = EEA*EC* E belongs to the radical axis ac. Consider the middle M of AC. ADB has length half the perimeter of ABC. The same is true for CFB. Hence AFB = CDB and M belongs to the radical axis of ac. Thus AM is the radical axis of circles a and c. Since AC is antiparallel to FD (A,C,F,D are on a circle) M belongs to the symmedian of DEF from E.
[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*EA/B*FA, YA*/YC* = A*FB/C*DB , ZB*/ZA* = B*DC/A*EC and the product (XC*/XB*)*(YA*/YC*)*(ZB*/ZA*) = (C*EA/B*FA)*(A*FB/C*DB)*(B*DC/A*EC) = ((R-EEA)/(R-FFA))*((R-FFB)/(R-DDB)*((R-DDC)/(R-EEC)). Since D, E, F are on the bisectors every factor appearing in the nominator participates also in the denominator (f.e. (R-DDC) cancels with (R-DDB)). The collinearity of Q, X(9), X(84) can be shown easily using trilinear coordinates.
[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.
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