# Proof 4.20

## Prove Theorem 4.5 about the radical axis of three circles.

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.

[Construction of Assumptions] Construct three circles named for their center point, , that are not collinear. Note that each pair of circles creates a radical axis. and overlap and the radical axis is the line through those intersection points, . and also overlap and form a similar radical axis through their intersection points . To create the third radical axis, construct a line segment between points . The perpendicular bisector of this segment is the radical axis between these non-intersecting circles by definition.
Proof: Assume there are three noncollinear circles and three radical axes as described above. Now consider a triangle that is created by connecting where is the point of concurrency of the radical axes created by and . Notice, and by the definition of a perpendicular bisector. By Common Notion 1, we know that equal things are equal, so . By Proposition 4 (SAS), we can conclude that . Therefore, the radical axis of and must be concurrent with the other two radical axes at point by construction.

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