Proof

Suppose two angles are vertical. Prove that they are congruent.

Fig. 1: Two intesecting lines segments form vertical angles, shown in pink.

To prove that two vertical angles are congruent, we need to prove two things :

1. \(\angle AOB\cong \angle COD\Rightarrow m\angle AOB = m\angle COD\)

2. \(m\angle AOB = m\angle COD\Rightarrow \angle AOB\cong \angle COD\)

\(\underline{\text{Proof}}\):

1. \(\angle AOB\cong \angle COD\Rightarrow m\angle AOB = m\angle COD\)

Suppose \(\angle AOB\cong \angle COD\).

Then there is an isometry that superimposes \(\angle AOB\) onto \(\angle COD\).

Let \(f\) be that isometry : \(f(\angle AOB)=\angle COD\).

Isometries preserve angle.

The angle measure of \(f(\angle AOB)\) equals the angle measure of \(\angle COD\).

So by substitution \(m\angle AOB=m\angle COD\).

2. \(m\angle AOB = m\angle COD\Rightarrow \angle AOB\cong \angle COD\)

Suppose \(m\angle AOB = m\angle COD\).

Since by definition of vertical angles, \(\angle AOB\) and \(\angle COD\) are across from each other, there is a reflection, \(f(m\angle AOB)\) about the origin (O).

We also know that there are four pairs of supplementary angles, two of which are:

• \(m\angle AOB+m\angle BOD=180^\circ\)

• \(m\angle BOD+m\angle COD=180^\circ\)

So, \(m\angle AOB+m\angle BOD=m\angle BOD+m\angle COD\).

And if we subtract \(m\angle BOD\) from both sides we get, \(m\angle AOB=m\angle COD\).

It would then make sense that \(f(O)=O\), \(f(A)=C\), and \(f(B)=D\).

Then by definition of reflection \(f(m\angle AOB)=m\angle COD\).

Therefore by the Congruence and Angle measure theorem \(\angle AOB\cong \angle COD\).

\(\square\)