Consider a triangle ABC and its altitudes intersecting at the orthocenter H. There is an interesting figure created by considering the additional line PP' and its intersection point O with the base BC of the triangle.
[1] There is a circle c(O,r) orthogonal to the circumcircles c1, c2 of quadrangles BCPP' and AP'HP. c1, c2 being also orthogonal to each other.
[2] Altitude AD coincides with the radical axis of c and c1 and the circle with diameter EE' is also orthogonal to c2.
[3] x=|DH| and y = |AD| satisfy x*y = |DE|2.
Everything follows trivially from the properties of the orthocenter. In particular F, the middle of AE' and center of the circumcircle of quadrangle AP'HP is on the radical axis of c and c1, since it is simultaneously orthogonal to these circles.
One can see the figure from another viewpoint:
[1] Starting with two orthogonal circles c(O,r) and c1, consider the inversion P' = G(P) with respect to c' (G preserves circle c1).
[2] Consider the good parametrization (stereographic projection) which to each point P of c1 corresponds the intersection point A of CP with line EE'.
[3] x*y = |DE|2 and DE is uniquely from c and c1 defined. x and y being the coordinates of P and P' this shows that the inversion G coincides with an involutive homography G' on the points of circle c1. The Fregier point of the involution being O, having two fixed points E and E' and homography axis the line EE'.
![[alogo]](alogo.png){kind=small}
Inversion as involution ![[0_0]](InversionAsInvolution_files/InversionAsInvolution_0_0_0.png)
![[0_1]](InversionAsInvolution_files/InversionAsInvolution_0_0_1.png)
![[0_2]](InversionAsInvolution_files/InversionAsInvolution_0_0_2.png)
![[1_0]](InversionAsInvolution_files/InversionAsInvolution_0_1_0.png)
![[1_1]](InversionAsInvolution_files/InversionAsInvolution_0_1_1.png)
![[1_2]](InversionAsInvolution_files/InversionAsInvolution_0_1_2.png)