The CCSSM calls for students to learn to recognize line-symmetric figures as early as Grade 4:

Recognize a line of symmetry for a two-dimensional figure as a line across the figure such that the figure can be folded along the line into matching parts. Identify line-symmetric figures and draw lines of symmetry.

In high school, students can be expected to use logical reasoning to demonstrate that a given figure is line-symmetric. In order to do this, we need to take the intuitive idea about “folding the figure along a line into matching parts” and turn it into a statement that is mathematically precise.

Definition: let line $m$ be given. A figure $S$ is line-symmetric with respect to $m$ if the reflection over line $m$ maps $S$ onto itself. That is, $r_m(S)=S$.

An angle is an extremely simple example of a line-symmetric figure. Students can explore this phenomenon by tracing an angle on a piece of patty paper, then folding the paper so that the two sides coincide.

In this post, we’ll explore the logic underlying what some texts refer to as “The Side-Switching Theorem.” The aim of the proof is to show that there is a line in which the sides of the angle are mirror images of one another. That is, we’ll find a suitable reflection $r$ with $r(\overrightarrow{OA}) = \overrightarrow{OB}$ and $r(\overrightarrow{OB}) = \overrightarrow{OA}$. It seems intuitively correct to say that the mirror should be the line that bisects $\angle AOB$. How can we prove that this is so?

Line $OC$ bisects $\angle AOB$.

Click here to interact with the figure in GeoGebra.

The plan for this proof is to show that $\angle BOC$ is the mirror image of $\angle AOC$ in line $m$. To establish this, we’ll show that these two angles 1) have a common initial side, 2) have the same degree measure, and 3) have terminal sides that lie in a common half-plane. As long as these conditions are met, we can be sure that these two angles coincide. (The technical machinery that makes this argument work is the Protractor Postulate.)

1. Let $m$ be the line through $O$ and $C$, and let $r$ be the reflection over line $m$. We’ll show that the sides of $\angle AOB$ are symmetric with respect to line $m$.
2. Do $r(\overrightarrow{OA})$ and $\overrightarrow{OB}$ lie in the same half-plane relative to $m$?
• Since $\overrightarrow{OA}$ lies to the left of line $m$, its mirror image must lie to the right of $m$, as does $\overrightarrow{OB}$.
3. Do $r(\angle AOC)$ and $\angle BOC$ have a common initial side?
• Since points $O$ and $C$ are on line $m$, $r(\overrightarrow{OC}) = \overrightarrow{OC}$.
4. Do $r(\angle AOC)$ and $\angle BOC$ have the same degree measure?
• Since line $OC$ bisects $\angle AOB$, we have $\angle AOC = \angle BOC$. And since reflections preserve angles, we can be sure that $r(\angle AOC) = \angle BOC$ as well.
5. It now follows that $r(\angle AOC)$ coincides with $\angle BOC$. In particular, $r(\overrightarrow{OA}) = \overrightarrow{OB}$, which implies that $r(\overrightarrow{OB}) = \overrightarrow{OA}$ also. This completes the proof that $\angle AOB$ is line-symmetric with respect to its angle bisector $OC$.

The proof above makes use of three important ideas: 1) the definition of reflection, 2) the fundamental assumption that reflections preserve angles, and 3) the Protractor Postulate.

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