2.3: Zero angle

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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.