Let PQ be the diameter of a circle. Let A and B be points on the circle, and let the tangents to the circle through A and B intersect at C. Let the tangent to the circle at Q intersect PA, PB, and PC at A', B', and C', respectively. Show that C' is the midpoint of A'B'.

While I have done a solution using similar triangles and symmedian properties, it is encouraged to solve this problem using inversion and I am having a lot of trouble doing that. I think I should draw a circle with radius PQ centered at P and possibly invert a circle centered at C with radius CA but I haven't been able to do much other than that. Could someone please explain the inversion solution?

Guest Aug 21, 2020