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