2.5: Vertical angles

A pair of angles $AOB$ and $A'OB'$ is called vertical if the point $O$ lies between $A$ and $A'$ and between $B$ and $B'$ at the same time.

Proposition $\PageIndex{1}$

The vertical angles have equal measures.

Proof

Assume that the angles $AOB$ and $A'OB'$ are vertical. Note that $\angle AOA'$ and $\angle BOB'$ are straight. Therefore, $\measuredangle AOA' = \measuredangle BOB' = \pi$.

It follows that

$$\begin{array} {rcl} {0} & = & {\measuredangle AOA' - \measuredangle BOB' \equiv} \\ {} & equiv & {\measuredangle AOB + \measuredangle BOA' - \measuredangle BOA' - \measuredangle A'OB' \equiv} \\ {} & \equiv & {\measuredangle AOB - \measuredangle A'OB'.} \end{array}$$

Since $-\pi < \measuredangle AOB \le \pi$ and $-\pi < \measuredangle A'OB' \le \pi$, we get that $\measuredangle AOB = \measuredangle A'OB'$.

Exercise $\PageIndex{1}$

Assume $O$ is the midpoint for both segments $[AB]$ and $[CD]$. Prove that $AC = BD$.

Hint

Applying Proposition 2.5.1, we get that $\measuredangle AOC = \measuredangle BOD$. It remains to apply Axiom IV.