Re: [EMHL] Nine-point circles concurring at Feuerbach point (was: Cyclic quadrilateral)
- Dear Darij:
"Darij Grinberg " wrote:
>Yes, Coolidge mentions Fontene several times in the
> >> "If four finite points be given, whereof no three
> >> are collinear, which are not the vertices and the
> >> orthocenter of a triangle, the four nine-point
> >> circles which they determine three by three, and
> >> the pedal circle of each with regard to the
> >> triangle of the other three, are concurrent."
> Somehow this must be connected with the famous Fontene
neighborhood of that theorem. For example in a footnote:)
"Fontene speaks of this as a well-known theorem."
From the above Coolidge derives Fontene's theorem about the
pedal circle of a point on a line through the circumcenter.
The isogonal conjugate of such a point will trace a conic
through the vertices and the orthocenter of the triangle.
So that the conic is bound to be a rectangular hyperbola.
And then refer to Jean-Pierre's message #6252.
> but I have no time to look further at it. IThis is what I hope for.
> am sure that it can be proven elementarily
> What I further know on the four pedal triangles is thatBoth are indeed very nice results. Thank you.
> they are all similar (this can be proven easily by
> calculation of lengths).
> Here is a nice theorem I have discovered:
> Let I be the incenter of ABC. Then, the Feuerbach
> point of ABC (i. e. the point of tangency of the
> incircle with the nine-point circle) lies on the
> nine-point circles of triangles BIC, CIA, AIB.
All the best,