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Mathematics LibreTexts

2.5: Vertical angles

  • Page ID
    23588
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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, \(\measuredangle AOA' = \measuredangle BOB' = \pi\).

    截屏2021-02-02 上午10.35.53.png

    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.