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 a previous post, we showed that the mirror line for a segment $\overline{AB}$ has a special property: each point on the mirror line is equidistant from the endpoints of the segment. In this post, we’ll examine the converse of this theorem: if we have a point $O$ that is known to be equidistant from two points $A$ and $B$, does this mean that $O$ has to be on the mirror line for $A$ and $B$? It seems intuitive that the bisector of $\angle AOB$ will also act as a mirror for $A$ and $B$. Point $O$ is equidistant from $A$ and $B$, and line $OC$ bisects $\angle AOB$.

The plan for this proof is to show that $\overline{OB}$ is the mirror image of $\overline{OA}$ in line $m$. To establish this, we’ll show that these two segments 1) have a common initial endpoint, 2) have the same length, and 3) have terminal endpoints that lie on a common half-line. As long as these conditions are met, we can be sure that these two segments coincide. (The technical machinery that makes this argument work is the Ruler Postulate.)

1. Let $OC$ be the bisector of $\angle AOB$, and let $m$ be the line through $O$ and $C$; let $r$ be the reflection over line $m$. We’ll show that the endpoints of segment $\overline{AB}$ are symmetric with respect to line $m$.
2. Do $r(\overline{OA})$ and $\overline{OB}$ have a common initial endpoint?
1. Yes – since $O$ is on line $m$, $r(O) = O$.
3. Do $r(A)$ and $B$ lie on a common half-line?
1. Yes – since $m$ is the bisector of $\angle AOB$, the Side-Switching Theorem shows that $r(\overrightarrow{OA}) = \overrightarrow{OB}$, so $r(A)$ lies somewhere on ray $\overrightarrow{OB}$ (as does point $B$).
4. Do $r(\overline{OA})$ and $\overline{OB}$ have the same length?
1. Yes – we are given that $\overline{OA} = \overline{OB}$, and since reflections preserve distance, $r(\overline{OA}) = \overline{OB}$ as well.
5. It now follows that $r(\overline{OA}) = \overline{OB}$. In particular, $r(A) = B$. This implies that line $m$ is the mirror line for $A$ and $B$. Since point $O$ is on line $m$, this completes the proof that $O$ is on the mirror line for $A$ and $B$.

This proof makes use of 1) the definition of reflection, 2) the Side-Switching Theorem for angles, and 3) the fundamental assumption that reflections preserve distance.

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 proofs of the Isosceles Triangle Theorem, the Segment Congruence Theorem, and the Side-Side-Side criterion for triangle congruence.

The Equidistance Theorem is also discussed at Illustrative Mathematics, although at present the logic is problematic, as one of the commenters points out.