3.3: Same sign lemmas
- Page ID
- 23593
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Assume \(Q' \in [PQ)\) and \(Q' \ne P\). Then for any \(X \in (PQ)\) the angles \(PQX\) and \(PQ'X\) have the same sign.
- 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 \(\angle PQ'X\).
In arbitrary nondegenerate triangle \(ABC\), the angles \(ABC, BCA,\) and \(CAB\) have the same sign.
- 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.
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.
- 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 \(\angle PQY\).