Spieker Conic and generalization of Nagel line

Given A'B'C' as the median triangle, A'D, B'E & C'F are 3 concurrent cevians at N, and L, J and K are the respective midpoints of A'N, B'N and C'N, and X, Y and Z are the midpoints of the sides of A'B'C' as shown. If S is constructed as the common midpoint of XL, YJ and ZK, and the intersections of AN, BN and CN with the sides of the median triangle are P, Q and R as shown, and their respective reflections through S are P', Q' and R', and a conic is drawn through any 5 of P, Q, R, P', Q' and R', then the conic cuts through the 6th point, and is inscribed in the median triangle. The conic is also inscribed in the triangle obtained from the median triangle through a half-turn around S. In addition, the center of the conic (S), the centroid of ABC (G) and N (Ceva point) are collinear, and NS = 3 SG. (On the left, scroll down to see the measurement 'w' = NS/SG at the bottom).

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Drag points A, B, C, D or E. Also drag D or E onto the extensions of the sides. What do you notice?

Read my article A generalization of the Spieker circle and Nagel line

Michael de Villiers, 9 November 2007, Created with GeoGebra