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Mathematics LibreTexts

3.3: Same sign lemmas

  • Page ID
    23593
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    Lemma \(\PageIndex{1}\)

    Assume \(Q' \in [PQ)\) and \(Q' \ne P\). Then for any \(X \in (PQ)\) the angles \(PQX\) and \(PQ'X\) have the same sign.

    截屏2021-02-02 下午1.37.17.png

    Proof

    By Proposition 2.2.2, for any \(t \in [0, 1]\) there is a unique point \(Q_t \in [PQ)\) such that 

    \[PQ_t = (1 - t) \cdot PQ + t \cdot PQ'.\]

    Note that the map \(t \mapsto Q_t\) is continuous,

    \(Q_0 = Q\),                    \(Q_1 = Q'\)

    and for any \(t \in [0, 1]\), we have that \(P \ne Q_t\).

    Applying Corollary 3.2.1, for \(P_t = P\), \(Q_t\), and \(X_t = X\), we get that \(\angle PQX\) has the same sign as \(\angle PQ'X\).

    Theorem \(\PageIndex{1}\) Signs of angles of a triangle

    In arbitrary nondegenerate triangle \(ABC\), the angles \(ABC, BCA,\) and \(CAB\) have the same sign.

    截屏2021-02-02 下午1.47.20.png

    Proof

    Choose a point \(Z \in (AB)\) so that \(A\) lies between \(B\) and \(Z\).

    According to Lemma \(\PageIndex{1}\), the angles \(ZBC\) and \(ZAC\) have the same sign.

    Note that \(\measuredangle ABC = \measuredangle ZBC\) and 

    \[\measuredangle ZAC + \measuredangle CAB \equiv \pi.\]

    Therefore, \(\angle CAB\) has the same sign as \(\angle ZAC\) which in turn has the same sign as \(\measuredangle ABC = \measuredangle ZBC\).

    Repeating the same argument for \(\angle BCA\) and \(\angle CAB\), we get the result.

    Lemma \(\PageIndex{2}\)

    Assume \([XY]\) does not intersect \((PQ)\), then the angles \(PQX\) and \(PQY\) have the same sign.

    The proof is nearly identical to the one above.

    截屏2021-02-02 下午1.52.14.png

    Proof

    According to Proposition 2.2.2, for any \(t \in [0, 1]\) there is a point \(X_t \in [XY]\), such that

    \(XX_t = t \cdot XY.\)

    Note that the map \(t \mapsto X_t\) is continuous. Moreover, \(X_0 = X\), \(X_1 = Y\), and \(X_t \not\in (QP)\) for any \(t \in [0, 1]\). 

    Applying Corollary 3.2.1, for \(P_t = P\), \(Q_t = Q\), and \(X_t\), we get that \(\angle PQX\) has the same sign as \(\angle PQY\).