[alogo] Isogonals of circumcircle points

Let P be a point of the circumcircle c0 of triangle ABC and consider its reflexions {A',B',C'} on the bisectors of the angles of the triangle.
[1] Lines {CC', BB', AA'} are parallel.
[2] The direction of these parallel lines is orthogonal to the direction of the Simson line of P.
[3] Triangle A'B'C' is always similar to the pedal triangle A0B0C0 of the incenter I of ABC.
[4] The circumcircle of A'B'C' is centered at I and passes through P.
[5] The isogonal conjugate P* of P is on the line at infinity.
[6] Triangles ABC and A'B'C' are point perspective with respect to P*. Their line of perspectivity passes through the incenter I of ABC.

[0_0] [0_1]
[1_0] [1_1]

[1,5] follows by comparing angles, at A and B say (figure).
[2] follows similarly by comparing angles at P of the cyclic quadrilateral APPBPC created by the projections of P on the sides of ABC.
[3,4] follow by noticing that, by their definition, I is on the medial lines of {PA', PB', PC'}. Then comparing the angles of A'B'C' with those formed at P.
[6] follows from an easy calculation with trilinears.

Remark-1 The previous arguments give a geometric proof of the (trivial) fact that the circumcircle is the isogonal conjugate of the line at infinity. In fact in trilinears this line is represented by equation
ax+by+cz=0, ({a,b,c} being the side-lengths of the triangle), whereas
is the equation of the circumcircle (see CircumcircleInTrilinears.html ).

Remark-2 From the previous remark follows that every conic intersecting the line at infinity in two different points (hyperbola) has isogonal transform a line intersecting the circumcircle. Every conic tangent to the line at infinity (parabola) has isogonal transform a line touching the circumcircle and finally every conic not intersecting the line at infinity (ellipse) has isogonal transform a line not intersecting the circumcircle.
In the file IsogonalOfParabola.html I discuss the special case of parabolas, resulting by taking the isogonal image of a tangent to the circumcircle.

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