Proof 4.20

Theorem 4.5: Suppose three circles are given for which the centers are not collinear. Each pair of circles determines a radical axis, and these three radical axes are concurrent.
Proof: Consider three circles without collinear centers and their radial axes. Create . Let G be the point of concurrency of  and . We see by construction that  and  act like perpendicular bisectors of the . Previously (Proof 3.42), we showed that the perpendicular bisectors are concurrent. In that argument, we show that the three smaller triangles within  can be broken into smaller congruent triangles with perpendicular angles and equal sides that all meet at the point of concurrency (in this case G). Consider  because the radial axis of two circles is perpendicular to the line created by connecting the two centers of the circles. Consider . Since , we know that  to  is congruent in  and  by Common Notion 1. Since they share side lengths  and angles, we can conclude that  by Proposition. Similarly we can show that  on line . Using the same arguments, we can conclude that  on line . Since the  shares sides with  and  we know that the point G is shared by all three triangles. Therefore, the radial axes of three triangles is concurrent.