Given ΔABC and A'B'C' as its median triangle, A'D, B'E & C'F as 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.

Spieker Conic and generalization of Nagel line

1) Drag points A, B, C, D or E. Also drag D or E onto the extensions of the sides. Can you drag until the conic becomes a hyperbola or parabola?

2) View and interact with the 'dual' of this result, which is a generalization of the famous nine-point circle (and Euler line) to a nine-point conic (and associated Euler line) at: Nine Point Conic and generalization of Euler Line.