Vertical Angles

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Three proofs.

Below are three different proofs that vertical angles are congruent. Please consider them separately.

Point P is the intersection of lines $m$ and $n$ . Prove that $\angle 1\cong \angle 3$ .

Fix note: When students write equations about linear pairs, they often write two equations for non-overlapping linear pairs—which doesn’t help. The figure above is intended to help avoid that dead end, but it would be worthwhile to discuss that dead end anyway.

Proof: Using adjacent angles, $\angle 1\cong \angle 3$ because they are both to $\angle 2$ . $\blacksquare$

Additional detail: First write down equations about linear pairs of angles: $m\angle 1 + m\angle 2 = \answer {180} \textrm { degrees}$ $m\angle 3 + m\angle 2 = \answer {180} \textrm { degrees}$ By comparing the two equations, one might see that $m\angle 1=m\angle 3$ . Alternatively, one may do some algebra to conclude that $m\angle 1 = 180^\circ - m\angle 2 = m\angle 3$ , which is essentially what the one-sentence proof says.

Point P is the intersection of lines $m$ and $n$ . Prove that $\angle 1\cong \angle 3$ .

Proof: A rotation of about $P$ maps $m$ onto itself, maps $n$ onto itself, and swaps $\angle 1$ and . Because rotations preserve angle measures, it must be that $\angle 1\cong \angle 3$ . $\blacksquare$

Additional detail: Line $m$ is the union of two opposite $\answer [format=string]{rays}$ with endpoint $P$ . The rotation about $P$ swaps these opposite rays, and the same idea holds for line $n$ . That rotation maps the sides of $\angle 1$ onto the sides of $\angle 3$ and vice versa.

Point P is the intersection of lines $m$ and $n$ . Prove that $\angle 1\cong \angle 3$ .

Proof: Reflecting about the of $\angle 2$ swaps the sides of $\angle 2$ and therefore lines $m$ and $n$ . Thus, that reflection swaps $\angle 1$ and . Because reflections preserve angle measures, it follows that $\angle 1\cong \angle 3$ . $\blacksquare$

Additional detail: Because reflections take lines to $\answer [format=string]{lines}$ , the reflection that swaps the sides of $\angle 2$ must swap not just the rays but lines $m$ and $n$ , which contain the rays.