The map f: D ---> E is the

f is involutive i.e. f

[1] The concurrence of lines at E can be proved easily using Ceva's theorem. For the trace D

D

The corresponding trace E

E

Later because of the symmetry on the bisector. By Ceva the product of the analogous ratios for all sides k

[2] The six projections of D and its isogonal E on the sides lie on a circle of radius r (the circumcircle of the triangle with vertices I, J, H, the projections of D on the sides of t). This is seen easily by considering the middle U of DE and identifying it with the center of the circle.

[3] Triangle t(D) = IJH is called the

[4] The symmetry of D, E on U implies that P*, diametral of P, has DP* parallel and equal to PE.

[5] This implies that |DI||EP| = r² - |DU|² = k (the

The isogonal conjugation, given in trilinears by f : (x,y,z) -->(k/x, k/y, k/z), is not properly defined on the vertices of the triangle. Outside the sides (and their prolongations) f is one-to-one (invertible). f sends all the points of a side of the triangle to the opposite vertex.

The isogonal conjugation is a special case of the more general

There are several applications of this transform. A non standard one can be found in the file: Isogonal_3TangentCircles.html .

Some remarkable

The periods in the definitions of coordinates mean a cyclic permutation of {a,b,c}.

Isogonal_3TangentCircles.html

IsogonalGeneralized.html

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