2.3: Zero angle
- Page ID
- 23586
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\(\measuredangle AOA = 0\) 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.
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.