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.

In the previous post, we showed that all of the points on the perpendicular bisector of a segment are equidistant from the endpoints of the segment. But are these the only points with this equidistance property? Let’s use the properties of reflections to show that they are.

Let segment $\overline{AB}$ be given, and let $O$ be a point in the plane that is equidistant from $A$ and $B$. To show that $O$ is on the perpendicular bisector of $\overline{AB}$, we’ll find a line $l$ through $O$ in which $B$ is the mirror image of $A$. That is, we’ll show that $B$ is the image of $A$ under the transformation that reflects the plane over line $l$.

First let’s construct a circle through $A$ centered at $O$. Since $A$ and $B$ are the same distance from $O$, this circle passes through $B$ as well as $A$.

Which line through $O$ will act as a mirror for the endpoints of $\overline{AB}$?

The entire figure appears to be line-symmetric, but we need some way of describing the line of symmetry. To that end, let $C$ be the point that divides the arc from $A$ to $B$ into two equal parts, and let $m$ be the line through $O$ and $C$. Let $r$ be the reflection of the plane over line $m$ – we’ll show that $r$ maps $A$ onto $B$.

Let $A'=r(A)$. To show that $A' = B$, we need to answer two questions: How can we be sure that $A'$ is on the same circle as $A$ and $B$? How can we be sure that $A'$ is located precisely at point $B$ on this circle?

1. Since $r(O)=O$ and reflections preserve distance, $A'$ and $A$ must be equidistant from $O$. That is, $A'$ and $A$ are on the same circle centered at $O$.
2. Let $\alpha$ represent the measure of arcs $AC$ and $BC$. Since $r(C) = C$, $r(O) = O$, and reflections preserve angles, the measure of $\angle COA'$ must be $\alpha$ as well. Since there is only one counter-clockwise arc from $C$ with degree measure $\alpha$, it follows that $A'$ coincides with $B$.
3. Having shown that $r(A) = B$, it follows from the definition of reflection that line $m$ is the perpendicular bisector of segment $\overline{AB}$. But point $O$ is on $m$, which means that $O$ is on the perpendicular bisector of $\overline{AB}$ – this was to be proved!

This proof relies on three very important ideas:

We are now in a position to completely characterize the points on the perpendicular bisector of a segment, thus fulfilling the final requirement of Standard G-CO.9:

The points on the perpendicular bisector of a segment are exactly those that are equidistant from the segment’s endpoints.

This theorem has many applications. In particular, it will play a critical role in our proof of the Side-Side-Side criterion for triangle congruence.