If AD, BE and CF are 3 cevians concurrent at H, then a conic drawn through any 5 of the 3 feet of the cevians, the 3 midpoints of AH, BH and CH, and the 3 midpoints of the sides of ABC, passes through the remaining 4 points. In addition, the center of the conic N, the centroid G of ABC and H are collinear, and HN = 3GN.
This general result is therefore not only a generalization of the famous nine-point circle, but also of the Euler line.

Nine Point Conic and generalization of Euler Line

1) First click on the buttons on the left. Then drag points A, B, C, D or E. Also drag D or E onto the respective extensions of BC or CA. Drag these points until the conic becomes a hyperbola or a parabola.
2) Dynamically explore a Further generalization of the Euler line (which only involves similarity).
3) View and interact with the 'dual' of this result, which is a generalization of the famous Spieker circle (and Nagel line) to a six-point conic (and associated Nagel line) at: Spieker Conic and generalization of Nagel line.

Historical Note: The nine-point conic appears to have been first described by Maxime Bocher in 1892 in his paper On a nine-point conic, but the associated generalization of the Euler line is not mentioned.

Projective Proof: Christopher Bradley from the University of Bath, who passed away in 2013, produced the following elegant, concise proof of the nine-point conic and associated Euler line in 2010: The Nine-point Conic and a Pair of Parallel Lines.

Related Result: The nine point conic is a special case of the six point bicevian conic, which is attributed to Carnot - click on the 'Link to Generalization' button in the preceding URL to view a dynamic sketch of this 'bicevian conic'.