Nine Point Conic and generalization of Euler Line
If AD, BE and FG are 3 cevians concurrent at F, then a conic drawn through any 5 of the 3 feet of the cevians, the 3 midpoints of AF, BF and CF, 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 O of ABC and F are collinear, and FN = 3NO. (On the left, scroll down to see the measurement 'o' = FN/NO at the bottom).
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Drag points A, B, C, D or E. Also drag D onto the extension of BC. What do you notice?
Read my article A generalization of the nine-point circle and the Euler line
Michael de Villiers, 27 Oct 2007, Created with GeoGebra