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

Nikos Dergiades

> 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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