2.4: Straight angle

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If $\measuredangle AOB = \pi$, we say that $\angle AOB$ is a straight angle. Note that by Proposition 2.3.2, if $\angle AOB$ is a straight, then so is $\angle BOA$.

We says that the point $O$ lies between points $A$ and $B$, if $O \ne A$, $O \ne B$, and $O \in [AB]$.

Theorem $\PageIndex{1}$

The angle $AOB$ is straight if and only if $O$ lies between $A$ and $B$.

Proof

By Proposition 2.2.2, we may assume that $OA = OB = 1$.

$截屏2021-02-02 上午10.04.24.png$

"If" part. Assume $O$ lies between $A$ and $B$. Set $\alpha = \measuredangle AOB$.

Applying Axiom IIIa, we get a half-line $[OA')$ such that $\alpha = \measuredangle BOA'$. By Proposition 2.2.2, we can assume that $OA' = 1$. According to Axiom IV,

$\triangle AOB \cong \triangle BOA'$.

Suppose that $f$ denotes the corresponding motion of the plane; that is, $f$ is a motion such that $f(A) = B$, $f(O) = O$, and $f(B) = A'$.

$截屏2021-02-02 上午10.04.45.png$

Then

$O = f(O) \in f(AB) = (A'B)$.

Therefore, both lines $(AB)$ and $(A'B)$ contain $B$ and $O$. By Axiom II, $(AB) = (A'B)$.

By the definition of the line, $(AB)$ contains exactly two points $A$ and $B$ on distance 1 from $O$. Since $OA' = 1$ and $A' \ne B$, we get that $A = A'$.

By Axiom IIIb and Proposition 2.3.1, we get that

\[\begin{array} {rcl} {2 \cdot \alpha} & = & {\measuredangle AOB + \measuredangle BOA' =} \\ {} & = & {\measuredangle AOB + \measuredangle BOA \equiv} \\ {} & equiv & {\measuredangle AOA =} \\ {} & = & {0} \end{array}\]

Therefore, by Exercise 1.8.1, $\alpha$ is either 0 or $\pi$.

Since $[OA) \ne [OB)$, we have that $\alpha \ne 0$, see Exercise 2.3.1. Therefore, $\alpha = \pi$.

"Only if" part. Suppose that $\measuredangle AOB = \pi$. Consider the line $(OA)$ and choose a point $B'$ on $(OA)$ so that $O$ lies between $A$ and $B'$.

From above, we have that $\measuredangle AOB' = \pi$. Applying Axiom IIIa, we get that $[OB) = [OB')$. In particular, $O$ lies between $A$ and $B$.

A triangle $ABC$ is called degenerate if $A, B$, and $C$ lie on one line. The following corollary is just a reformulation of Theorem 2.4.1.

Corollary $\PageIndex{1}$

A triangle is degenerate if and only if one of its angles is equal to $\pi$ or 0. Moreover in a degenerate triangle the angle measures are 0, 0, and $\pi$.

Exercise $\PageIndex{1}$

Show that three distinct points $A, O$, and $B$ lie on one line if and only if

$2 \cdot \measuredangle AOB \equiv 0$.

Hint: Apply Proposition 2.3.1, Theorem 2.4.1 and Exercise 1.8.1.

Exercise $\PageIndex{2}$

Let $A, B$ and $C$ be three points distinct from $O$. Show that $B, O$ and $C$ lie on one line if and only if

$2 \cdot \measuredangle AOB \equiv 2 \cdot \measuredangle AOC$.

Hint: Axiom IIIb, $2 \cdot \measuredangle BOC \equiv 2 \cdot \measuredangle AOC - 2 \cdot \measuredangle AOB = 0$. By Exercise 1.8.1, it implies that $\measuredangle BOC$ is either 0 or $\pi$. It remains to apply Exercsie 2.3.1 and Theorem 2.4.1 respectively in these two cases.

Exercise $\PageIndex{3}$

Show that there is a nondegenerate triangle.

Answer

Fix two points $A$ and $B$ provided by Axiom I.

Fix a real number $0 < \alpha < \pi$. By Axiom IIIa there is a point $C$ such that $\measuredangle ABC = \alpha$. Use Proposition 2.2.1 to show that $\triangle ABC$ is nondegenerate.

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