In the figure above, $\angle AOB$ and $\angle A'OB'$ are vertical angles.

• How do we prove that they are congruent?
• What does it even mean to prove that two angles are congruent?
• Which proofs of this theorem are most accessible to students?
• Is it reasonable to expect students to independently produce a proof of this theorem on an assessment?
• The CCSSM call for students to use rigid motions to prove that two figures are congruent. Is this a viable approach for proving this particular theorem?

Please weigh in with your thoughts on these questions in the comments. In this post, I’ll present two proofs of this theorem. Please feel free to critique my proofs:

• Is the level of rigor appropriate for a typical high school student?
• If you feel there is too much detail, what would you leave out?
• If you feel there is not enough detail, what would you add?
• Are you aware of alternative proofs not listed here that are more accessible to students?

Approach #1: Show that $\alpha = \alpha'$ using algebra.

1. Since points $A$, $O$, and $A'$ are on the same line, $\angle AOA'=180$. It follows that $\alpha +\beta =180$.
2. Since points $B$, $O$, and $B'$ are on the same line, $\angle BOB'=180$. It follows that $\beta +\alpha' =180$.
3. From steps 1 and 2, we get $\alpha + \beta = \beta + \alpha'$, and thus $\alpha =\alpha'$.

The proof above shows that $\angle AOB$ has the same degree measure as $\angle A'OB'$. Does this prove that these two angles are congruent? It depends on how you define congruence.

• In some approaches to geometry, the link between equality of angles and congruence is established through a definition: two angles are congruent if and only if their degree measures are equal.
• In the approach taken in the CCSSM, congruence is defined in terms of rigid motions. So the idea that “two angles are congruent if and only if their degree measures are equal” is its own theorem and therefore requires a separate proof. In that sense, the argument above is not enough to prove that vertical angles are congruent.

Approach #2: Show that a rotation maps $\angle AOB$ onto $\angle A'OB'$.

1. Let $r$ be a 180-degree rotation about point $O$. We’ll show that $r(\angle AOB)=\angle A'OB'$.
2. Circle $K$ meets line $n$ at points $A$ and $A'$, so $\angle AOA'=180$ and $r(A)=A'$.
3. Circle $K$ meets line $m$ at points $B$ and $B'$, so $\angle BOB'=180$ and $r(B)=B'$.
4. A rotation about point $O$ fixes point $O$, so $r(O)=O$.
5. From steps 2, 3, and 4, we have $r(\angle AOB)=\angle A'OB'$. Since a rotation is a rigid motion, $\angle AOB \cong \angle A'OB'$, which was to be proved.

Which approach do you prefer? Does thinking about the theorem and its proof in terms of rigid motions add value? What are the strengths and limitations of the two approaches? Please add your thoughts to the comments.