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# 2.5: Vertical angles

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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.