[Geometry] Not Samuraiwarm's Theorem

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 ABCABC (not necessary acute), midpoint of sides BC,CA,AB\overline{BC},\overline{CA},\overline{AB} called D,E,FD,E,F respectively, and symmedians of a triangle ABCABC called AD,AE,AF\overline{AD'},\overline{AE'},\overline{AF'} BC,CA,AB\overline{BC},\overline{CA},\overline{AB} called D,E,FD,E,F respectively. Construct a line from A,B,CA,B,C, pass through circumcenter OO, and intersect BC,CA,AB\overline{BC},\overline{CA},\overline{AB} at point P1,P2,P3P_{1},P_{2},P_{3} respectively.

Theorem: 6 points, D,E,F,P1,P2,P3D,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 X1X^{-1} of an object XX in a triangle ABCABC is constructed by reflecting AX,BX,CX\overline{AX},\overline{BX},\overline{CX} about the angle bisectors of A,B,CA,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.

Note by Samuraiwarm Tsunayoshi
4 years, 1 month ago

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Well, this result is actuallyu true for any two points. In other words, any two cevian triangles of ABCABC are conconic. You can prove this using crossratio.

Xuming Liang - 4 years, 1 month ago

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