## 15399Θέμα: Re: [EMHL] Re:Cevian Trace Equal Area Points

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• Aug 2, 2007
Dear Francois,
I've sent you a message
and now I saw your message that you sent it
before mine.
As you see we use almost the same symbolism.
You conclude the harmonic relation using vectors
and I conclude it because my lines La+, La-, AB, AC
are parallel to the sides and diagonals of a
parallelogram.
Now I have an interesting result.
I investigated pedal triangles of
isogonal conjugate points.
If abc is the pedal triangle of P and
a'b'c' is the pedal triangle of P' isogonal
conjugate of P then the areal point Q of these
triangles lies on the circumcircle of ABC.
This is from a proof.
The following is from observation:
The point Q is the fourth intersection
of circumcircle with the rectangular circumhyperbola
of the circumcevian triangle of P, that passes
through P.
Similarly Q lies on the rectangular circumhyperbola
of the circumcevian triangle of P', that passes
through P'.

Best regards

> Dear Nikos
> I give you my own construction of the equal area
> axis based on vectors.
> So you could compare it with yours.
> Given 4 points A, B, A', B' such that line AB and
> A'B' are through point O,
> let L+ be the line locus of points M such that:
> Area(MAB) = Area(MA'B')
> and L- be the line locus of points M such that:
> Area(MAB) = -Area(MA'B')
> where Area means signed area.
> Then L+ is the line through O directed by the
> vector V(AB) - V(A'B')
> and L- is the line through O directed by the vector
> V(AB) + V(A'B').
> That's why the four lines AB, A'B', L+, L- are in an
> harmonic range.
> Friendly
> Francois

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