Given Δ*ABC* 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 (July 2004)

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

3) For a proof, read my 2006 article A generalization of the Spieker circle and Nagel line in the journal *Pythagoras*.

4) Here is also an unedited proof communicated to me by John Silvester, King's College, London in 2007: Silvester Spieker Proof.

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Michael de Villiers, first created online 9 November 2007, updated 13 June 2017 with *WebSketchpad*; 2 August 2021.