Note: If you have seen this theorem before, please tell me.
Also if you have a proof, feel free to post it. I'm working on it too!
With nine-point circle
The small black circle is nine-point circle, and the red conic is what my theorem says.
Construct a triangle (not necessary acute), midpoint of sides called respectively, and symmedians of a triangle called called respectively. Construct a line from , pass through circumcenter , and intersect at point respectively.
Theorem: 6 points, lie on the same conic.
Details: 3 symmedians of a triangle are isogonal conjugate of 3 medians of a triangle.
Details: The isogonal conjugate of an object in a triangle is constructed by reflecting about the angle bisectors of .
Idea: I got this idea from a nine-point circle, where the feet of altitudes, midpoints, and the midpoints between vertices and orthocenter all lie on the same circle. I also think that the isogonal conjugates are also the same.
(Orthocenter and circumcenter are isogonal conjugate to each other. Also centroid and symmedian point are isogonal conjugate to each other.)
Side notes: If you appear to see 3 other points (the similar to the midpoint of each 3 vertices and orthocenter), please tell me too.
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Well, this result is actuallyu true for any two points. In other words, any two cevian triangles of ABC are conconic. You can prove this using crossratio.