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?