Brocard Points of the Triangle

Given three points A, B and C,
find the point P such that angle(PAB) = angle(PBC) = angle(PCA).
Express tan(PAB) in terms of A, B and C.
construction of P
calculation of tan(PAB)
the other point P
coordinates of P

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The trilinear coordinates of P are the 3 values (PF, PG, PE) or any 3 multiples of them by a same factor.

Therefore we have        (PF, PG, PE)  =  (PB, PC, PA)

since                    PA = (A1,A)*cos(A1,A,P)
                            = (A1,A)*sin(P,A,B)
                            = CA/sin(CAB)*sin(PAB)

therefore                (PB, PC, PA) = (AB/sin(ABC), BC/sin(BCA), CA/sin(CAB))

Simplifying notation,
The trilinears of P are          (c/sin(B), a/sin(C), b/sin(A))
And for its isogonal conjugate   (b/sin(C), c/sin(A), a/sin(B))

Remarkably simple ... with a twist.