[1] The locus of P is a rectangular hyperbola passing through the vertices of ABC.

[2] The hyperbola has its center at the middle M of BC and its axes parallel to the bisectors of angle A.

[3] Side BC and line (e) are conjugate directions of the hyperbola.

[4] Let A

[5] The tangent to the hyperbola at A coincides with the symmedian line of triangle ABC through vertex A.

[1, 2] From the middle M of BC draw the parallel B''C'' to the outer bisector of angle A. From the cyclicity of quadrangle BCC'B' follows that the two small (yellow) triangles BB

Project P on the parallels to the bisectors of angle A passing from M. The projections create rectangle PM'MM'' has the same area as C

[3, 4] Consider point P'=f(P). The middle of PP' is on line BC and P' is a point of the hyperbola. To see it draw P'C and P'B and their intersections B'', C'' with the sides of the triangle. It suffices to show that B''C'' is parallel to B'C', or equivalently, parallel to PP'. This is a consequence of an easy exercise on Thales theorem (see ThalesApplication.html ).

[5] Is a consequence of the fact that {P,P'} are harmonic conjugate with respect to {P

Maclaurin.html

ThalesApplication.html

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