In this series of posts, we are taking a look at how to develop the congruence theorems found in the CCSSM using a transformations-based approach.

What is special about points that lie on the perpendicular bisector of a segment? In the figure below, it looks as if each point on is equidistant from and . How can we prove this?

The figure appears to be symmetric with respect to line m, so let’s try to develop an argument based on the properties of reflections. (Click here to interact with this figure in GeoGebra.)

Let segment be given, and let line be the perpendicular bisector of . Let be a reflection over line , and let be an arbitrary point on . To prove that is equidistant from and , we essentially just need to answer two questions: where does take point , and where does it take point ?

- Since is on , .
- Since is the perpendicular bisector of , .
- Since reflections preserve distance, the distance from to is equal to the distance from to . That is, the distance from to is equal to the distance from to . This completes the proof!

Key points: to establish the theorem, we used the **definition of reflection** (Standard G-CO.4) and the fundamental assumption that **a reflection is a rigid motion **(and thus it preserves distance).

Next up: having shown that the points on the perpendicular bisector of a segment have a particular equidistance property, it’s natural to wonder – are these the only points in the plane with this property?

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