# 2.4: Straight angle

- Page ID
- 23587

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.

**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.