**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 \(ABC\) (not necessary acute), midpoint of sides \(\overline{BC},\overline{CA},\overline{AB}\) called \(D,E,F\) respectively, and symmedians of a triangle \(ABC\) called \(\overline{AD'},\overline{AE'},\overline{AF'}\) \(\overline{BC},\overline{CA},\overline{AB}\) called \(D,E,F\) respectively. Construct a line from \(A,B,C\), pass through circumcenter \(O\), and intersect \(\overline{BC},\overline{CA},\overline{AB}\) at point \(P_{1},P_{2},P_{3}\) respectively.

Theorem: 6 points, \(D,E,F,P_{1},P_{2},P_{3}\) lie on the same conic.

Details: 3 symmedians of a triangle are isogonal conjugate of 3 medians of a triangle.

Details: The isogonal conjugate \(X^{-1}\) of an object \(X\) in a triangle \(ABC\) is constructed by reflecting \(\overline{AX},\overline{BX},\overline{CX}\) about the angle bisectors of \(A,B,C\).

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.

## Comments

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TopNewestWell, 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.

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