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Mathematics LibreTexts

2.3: Zero angle

  • Page ID
    23586
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    Proposition \(\PageIndex{1}\)

    \(\measuredangle AOA = 0\) for any \(A \ne O\).

    Proof

    According to Axiom IIIb,

    \(\measuredangle AOA + \measuredangle AOA \equiv \measuredangle AOA\).

    Subtract \(\measuredangle AOA\) from both sides, we get that \(\measuredangle AOA \equiv 0\).

    By Axiom III, \(-\pi < \measuredangle AOA \le \pi\); therefore \(\measuredangle AOA = 0\).

    Exercise \(\PageIndex{1}\)

    Assume \(\measuredangle AOB = 0\). Show that \([OA) = [OB)\).

    Hint

    By Proposition \(\PageIndex{1}\), \(\measuredangle AOA = 0\). It remains to apply Axiom III.

    Theorem \(\PageIndex{2}\)

    For any \(A\) and \(B\) distinct from \(O\), we have

    \(\measuredangle AOB \equiv - \measuredangle BOA.\)

    Proof

    According to Axiom IIIb,

    \(\measuredangle AOB + \measuredangle BOA \equiv \measuredangle AOA\)

    By Proposition \(\PageIndex{1}\), \(\measuredangle AOA = 0\). Hence the result.