## 21128Re : [EMHL] Re: Which angle ?

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• Aug 1, 2012
Dear Jean-Louis,

I still owe you the proof of this old question.
I will summarize your question first.
1. let ABCD a convex quadrilateral
2. E, F, G the points of intersection of AC and BD, AD and BC, AB and CD
3. You want to proof synthetically the equality <FXG=<FEG.
Since FEG is the Diagonal Triangle of the complete quadrangle ABCD you describe, it actually is enough to proof that X lies on the circumcircle of the complete quadrangle, because then <FXG=<FEG.
Later you called this point yourself the Euler-Poncelet Point in http://jl.ayme.pagesperso-orange.fr/vol8.html
This point X is called QA-P2 in the Encyclopedia of Quadri-Figures (EQF).

Let me first restate the problem in other words:
Given a complete quadrangle with defining points P1.P2.P3.P4 (A,B,C,D in your example),
proof synthetically that the Euler-Poncelet point X lies on the circumcircle of the diagonal triangle of the complete quadrangle.

There are enough synthetic proofs that the Euler-Poncelet Point X in a complete quadrangle P1.P2.P3.P4 is the point with these properties:
1. The center of the orthogonal hyperbola through P1, P2, P3, P4 and the orthocenters of the 4 Component Triangles (P1.P2.P3, P2.P3.P4, P3.P4.P1, P4.P1.P2) of the complete quadrangle.
2. The common point of the nine-point circles of the Component Triangles of the complete quadrangle.

Here is the synthetic proof that this point X lies on the circumcircle of the diagonal triangle of the complete quadrangle.
1. Let P1.P2.P3.P4 be the defining points of a complete quadrangle.
Let S1=P1.P2^P3.P4, S2=P1.P3^P2.P4, S3=P1.P4^P2.P3.
S1.S2.S3 is called the Diagonal Triangle of P1.P2.P3.P4.

2. Because of the construction of S1, S2, S3 the following is true:
S1.S2.S3 is the P1-cevian triangle of triangle P2.P3.P4,
S1.S2.S3 is the P2-cevian triangle of triangle P3.P4.P1,
S1.S2.S3 is the P3-cevian triangle of triangle P4.P1.P2,
S1.S2.S3 is the P4-cevian triangle of triangle P1.P2.P3.

3. Stated in another way: the diagonal triangle of a complete quadrangle is the Pi-cevian triangle wrt triangle Pj.Pk.Pl, for all permutations of i,j,k,l in (1,2,3,4).

4. According to the theorem described in Hyacinthos message #20282 the center of the orthogonal hyperbola of some triangle ABC through some point P lies on the circumcircle of the cevian triangle of P wrt ABC (proved by Poncelet).

5. Since X is the center of the orthogonal hyperbola through some vertice Pi of the complete quadrangle and the triangle Pj.Pk.Pl formed by the other vertices of the component quadrangle it now is evident that X lies on the circumcircle of the Pi-cevian triangle wrt Pj.Pk.Pl, which is the diagonal triangle of the complete quadrangle.
q.e.d.

Best regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, Jean-Louis Ayme <jeanlouisayme@...> wrote:
>
> DearÂ Francisco and Hyacinthists,
> perhaps I wasn't to precise.
> I want to find another angle which is equal to <FEG.
> For example the point of intersection X of the Euler's circles of the four triangles ABC, BCD, CDA, DAB verify <FXG=<FEG.
> So, I want to search another point(s) than X, so that I will prove the equality before.
> Sincerely
> Jean-Louis
>
>
> --- En date deÂ : Lun 13.9.10, Francisco Javier <garciacapitan@...> a Ã©critÂ :
>
>
> De: Francisco Javier <garciacapitan@...>
> Objet: [EMHL] Re: Which angle ?
> Ã: Hyacinthos@yahoogroups.com
> Date: Lundi 13 septembre 2010, 8h30
>
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> Â
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> I know that this is not what you asked, but you may wish to know when this angle is right.
>
> Take a point D' on the hyperbola K through the traces of the internal and external angle bisectors of angles A and C, and having the point B as focus (the other focus is on the B-altitude).
>
> Let D be the isogonal conjugate of D'
>
> Then the angle FEG for the quadrilateral ABCD is right.
>
> Best regards, Francisco Javier.
>
> --- In Hyacinthos@yahoogroups.com, Jean-Louis Ayme <jeanlouisayme@> wrote:
> >
> > Dear Hyacinthists,
> > let ABCD a convex quadrilateral
> > E, F, G the points of intersection of AC and BD, AD and BC, AB and CD
> > Question : to which angle is equal <FEG?
> > Sincerely
> > Jean-Louis
> >
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> >
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