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 $m$ is equidistant from $A$ and $B$. 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 $\overline{AB}$ be given, and let line $m$ be the perpendicular bisector of $\overline{AB}$. Let $r$ be a reflection over line $m$, and let $P$ be an arbitrary point on $m$. To prove that $P$ is equidistant from $A$ and $B$, we essentially just need to answer two questions: where does $r$ take point $P$, and where does it take point $A$?

1. Since $P$ is on $m$, $r(P)=P$.
2. Since $m$ is the perpendicular bisector of $\overline{AB}$, $r(A)=B$.
3. Since reflections preserve distance, the distance from $r(P)$ to $r(A)$ is equal to the distance from $P$ to $A$. That is, the distance from $P$ to $B$ is equal to the distance from $P$ to $A$. 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?