
# 2.3: Zero angle


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.