## Nine Point Conic and Generalization of Euler Line

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.

#### .sketch_canvas { border: medium solid lightgray; display: inline-block; } 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'.

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Michael de Villiers, 27 Oct 2007; modified June 2017 using WebSketchpad; further modified 6 January 2023.