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.