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2.3: Zero angle

Last updated

Sep 5, 2021
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- 2.2: Lines and half-lines
- 2.4: Straight 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 , $\measuredangle AOA = 0$ . 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.

Proposition 2.3.1\PageIndex{1}

Theorem 2.3.2\PageIndex{2}

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Proposition $\PageIndex{1}$

Theorem $\PageIndex{2}$