- Aug 1, 2003Dear Darij
in Hyacinthos 7408, you wrote
> Let ABC be a triangle and AA', BB', CC' its altitudes.Note that if an angle of ABC is obtuse, the incenter of A'B'C' is
> We have the following theorems:
> (1a) The Euler lines of triangles HB'C', HC'A', HA'B' concur
> at one point, namely the Schiffler point of triangle
> A'B'C'. It lies on the Euler line of triangle A'B'C'.
> (1b) The Euler lines of triangles AB'C', BC'A', CA'B' concur
> at one point on the nine-point circle of triangle ABC.
> Note that (1a) follows from the definition of the Schiffler
> point (H is the incenter of triangle A'B'C')
not H, but the corresponding vertex of ABC.
> and (1b) is a result of Victor Thebault.Notethat the common point is the center of Jerabek hyperbola.
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