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Mathematics LibreTexts

2.4: Straight angle

  • Page ID
    23587
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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.