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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 this post, we prove one of the most classic results in elementary geometry: the Isosceles Triangle Theorem. Standard G-CO.9 demands that students be able to prove theorems about triangles, including the fact that the base angles of isosceles triangles are congruent.

This is an interesting moment for this series, because this proof brings two key ideas to the forefront: symmetry and congruence. In fact, it is this very theorem that gets mentioned in the Introduction to HS Geometry:

Reflections and rotations each explain a particular type of symmetry, and the symmetries of an object offer insight into its attributes—as when the reflective symmetry of an isosceles triangle assures that its base angles are congruent.

Since this is the first post in this series that explicitly uses the idea of congruence, let’s remind ourselves what this means in the context of the CCSSM:

In the approach taken here, two geometric figures are defined to be congruent if there is a sequence of rigid motions that carries one onto the other. This is the principle of superposition.

The idea of this theorem is simple: an isosceles triangle has a fold line, and when the figure is folded along this line, the base angles are made to match up.

The angle bisector has been useful in our discussion of symmetry up until now, and this figure is no exception.

Triangle $\triangle AOB$ is isosceles with $\overline{OA}=\overline{OB}$. Line $OC$ bisects $\angle AOB$.

We’ve already done the heavy lifting to establish the Side-Switching Theorem and the Equidistance Theorem – this proof will be a cinch by comparison. The essence of the argument is that angles $\angle OAB$ and $\angle OBA$ are mirror images in line $m$. To prove this, we’ll show that the mirror images of points $O$, $A$, and $B$ are $O$, $B$, and $A$, respectively.

1. Let $m$ be the line through $O$ and $C$, and let $r$ be the reflection over line $m$. We’ll show that $r(\angle OAB) = \angle OBA$.
2. What is the mirror image of point $O$?
• Since $O$ is on line $m$, $r(O)=O$.
3. What is the mirror image of point $A$? of point $B$?
• We are given that $OA=OB$. The Equidistance Theorem tells us that the bisector of $\angle AOB$ is a mirror for $A$ and $B$, so $r(A) = B$ and $r(B) = A$.
4. What is the mirror image of $\angle OAB$?
• Since $r(O)=O$, $r(A)=B$, and $r(B)=A$, it follows that $r(\angle OAB) = \angle OBA$.
5. How does this establish that the base angles are congruent?
• Since a reflection is a rigid motion, it follows that $\angle OAB \cong \angle OBA$. This completes the proof that the base angles of an isosceles triangle are congruent.

Key ideas for this proof include 1) the definition of reflection, 2) the Equidistance Theorem, and 3) the definition of congruence.

As an added bonus, we are now in a position to establish the following result about the symmetry line for an isosceles triangle:

• The line of symmetry bisects the vertex angle.
• The line of symmetry is the perpendicular bisector of the base of the triangle.
• The line of symmetry contains the median to the base of the triangle.

Proofs are left to the imagination.

Next up: can a triangle have more than one line of symmetry? What can we say about such triangles?

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.

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 $\angle AOB$.

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.

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.

What do these figures have in common?

• the isosceles triangle
• the equilateral triangle
• the isosceles trapezoid
• the rectangle
• the kite
• the rhombus
• the square
• the circle.

Answer: they all have reflection symmetry! Being able to predict the effect of a reflection, to recognize figures that have reflection symmetry, and to make inferences about the attributes of line-symmetric figures are all key skills for students of geometry. But in order to reason about reflections, we first need to formulate a precise definition of this term. The CCSSM call for students to be able to do this in Standard G-CO.4.

Let line $m$ be given, and let $r$ be the reflection of the plane over line $m$. Intuitively, line $m$ acts like a mirror, and the reflection over $m$ sends each point $A$ to its “mirror image” $A'$. Another way to think about reflections is to represent the plane with a sheet of paper, then crease the paper along line $m$. Folding the paper in half should superpose $A$ onto its image $A'$.

The definition below captures our intuitions about reflections in a mathematically precise way.

1. If point $P$ is on line $m$, then $r$ fixes $P$. That is, $r(P) = P$.
2. Let $A$ be a point that is not on line $m$. In this case, $r$ takes $A$ to a point $A'=r(A)$ for which $m$ is the perpendicular bisector of the segment $\overline{AA'}$.

What kinds of inferences are available from this definition?

• If line $m$ is the perpendicular bisector of segment $\overline{PQ}$, then $r(P) = Q$.
• A reflection interchanges pairs of points: if $r(P) = Q$, then $r(Q) = P$.
• If the reflection over line $j$ takes point $X$ to point $Y$, then $j$ must be the perpendicular bisector of segment $\overline{XY}$.

The task of defining a reflection is described further at Illustrative Mathematics.

Homework: which of these theorems can be proved by performing a single reflection?

By my count, 4 of the 15 theorems listed above have a very simple and clear link to reflections. (Although technically, they can all be expressed in terms of reflections!! But that’s a story for another day.)

Next up: let’s use reflections and line-symmetry to prove some theorems.

In the Introduction to High School Geometry in the CCSSM, we read this:

The concepts of congruence, similarity, and symmetry can be understood from the perspective of geometric transformation. Fundamental are the rigid motions: translations, rotations, reflections, and combinations of these, all of which are here assumed to preserve distance and angles (and therefore shapes generally).

This excerpt raises several questions: what exactly is a transformation? What is a rigid motion? And what does it mean to preserve distance and angles? Let’s take these questions one at a time.

What is a transformation of the plane? Standard G-CO.2 speaks directly to this question: we read that students must be able to “describe transformations as functions that take points in the plane as inputs and give other points as outputs.” This same standard also calls for students to be able to “compare transformations that preserve distance and angle to those that do not.” In other words, students should be able to determine which transformations are rigid motions and which are not. Let’s give this term a precise definition.

Definition: a rigid motion is a transformation of the plane that preserves distance and angles.

What does it mean to say that a transformation preserves distance?

Let $T$ be a transformation, and let $A$ and $B$ be any two points in the plane. $T$ is a distance-preserving function (or isometry) if the distance between $T(A)$ and $T(B)$ is the same as the distance between $A$ and $B$. That is, $A$ and $B$ remain the same distance apart even after being transformed.

Which of the familiar transformations are rigid motions?

Axiom 1: reflections are rigid motions.

Axiom 2: rotations are rigid motions.

Axiom 3: translations are rigid motions.

What does it mean to say that a transformation preserves angles?

Let $T$ be a transformation, and let $\alpha$ be any angle in the plane. $T$ preserves angles if the measure of $T(\alpha)$ is the same as the measure of $\alpha$. That is, the measure of $\alpha$ remains the same even after being transformed.

There you have it! The facts outlined above will be used repeatedly as we develop theorems about congruent figures. Together with the definitions of reflection, rotation, and translation, the axioms above form the backbone of the theory of congruence by rigid motions.

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?