Inversion in a circle of two points => similar triangles
Inversion in a circle, K of radius R, with center q. Point a is inverted to A, and b to B. An amazing fact is that triangles qab and qBA are similar! This is easy to show (a little algebra), using the fact that both triangles share angle aqb, and the definition of inversion of points in K. Denote distance from q to a as qa, q to b as qb, q to A as qA, q to B as qB. The definition of inversion is that qA=R*R/qa, and qB=R*R/qb. (Think of what this is if R=1.) Dividing equations gives qA/qB=qb/qa, showing the triangles are similar.