Triangle inscribed in Steiner Circumellipse
 

Let P a point interior to triangle ABC and DEF its cevian triangle. Draw parallels to BC through E and F, intersecting cevian AP at EA and FA. We call MBA the intersection point of DF and BFA, and MCA the intersection point of DE and CEA,

1. The points MBA, MCA and P are collinear. Call ra the line which these points lies on. Call rb, rc the corresponding lines to other cevians.
2. If we construct in a similar way the points MAB, MCB, MAC, MBC, the six points M?? lie on the same conic GM.
3. The lines ra, rb, rc cut the sides of ABC at six points N?? lying on the same conic GN.
4. Let XYZ the triangle bounded by the lines joining points N?? nearest to the vertices of ABC. Then XYZ is inscribed in the Steiner circumellipse of ABC and XYZ is perspective with ABC, with perspector P.