
# 2.4: Straight angle


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  $$OA = OB = 1$$.

"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 $$f(B) = A'$$.

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.

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.