10.2: Relación cruzada
- Page ID
- 114453
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)El siguiente teorema da algunas cantidades expresadas en distancias o ángulos que no cambian después de la inversión.
Dejar\(ABCD\) y\(A'B'C'D'\) ser dos cuadriláteros tales que los puntos\(A',B',C'\), y\(D'\) son los inversos de\(A,B,C\), y\(D\) respectivamente.
Entonces
a)
\(\dfrac{AB \cdot CD}{BC \cdot DA} = \dfrac{A'B' \cdot C'D'}{B'C' \cdot D'A'}\).
b)
\(\measuredangle ABC + \measuredangle CDA \equiv -(\measuredangle A'B'C' + \measuredangle C'D'A')\).
(c) Si el cuadrilátero\(ABCD\) está inscrito, entonces así es\(\square A'B'C'D'\).
- Prueba
-
a). \(O\)Sea el centro de la inversión. Según Lemma 10.1.1,\(\triangle AOB \sim \triangle B'OA'\). Por lo tanto,
\(\dfrac{AB}{A'B'} = \dfrac{OA}{OB'}.\)
Análogamente,
\(\dfrac{BC}{B'C'} = \dfrac{OC}{OB'}\),\(\dfrac{CD}{C'D'} = \dfrac{OC}{OD'}\),\(\dfrac{DA}{D'A'} = \dfrac{OA}{OD'}\).
Por lo tanto,
\(\dfrac{AB}{A'B'} \cdot \dfrac{B'C'}{BC} \cdot \dfrac{CD}{C'D'} \cdot \dfrac{D'A'}{DA} = \dfrac{OA}{OB'} \cdot \dfrac{OB'}{OC} \cdot \dfrac{OC}{OD'} \cdot \dfrac{OD'}{OA}.\)
De ahí que a) sigue.
b). Según Lemma 10.1.1,
\[\begin{array} {l} {\measuredangle ABO \equiv -\measuredangle B'A'O, \measuredangle OBC \equiv -\measuredangle OC'B',} \\ {\measuredangle CDO \equiv -\measuredangle D'C'O, \measuredangle ODA \equiv -\measuredangle OA'D'.} \end{array}\]
Por Axioma IIIb,
\(\measuredangle ABC \equiv \measuredangle ABO + \measuredangle OBC\),\(\measuredangle D'C'B' \equiv \measuredangle D'C'O + \measuredangle OC'B'\),
\(\measuredangle CDA \equiv \measuredangle CDO + \measuredangle ODA\),\(\measuredangle B'A'D' \equiv \measuredangle B'A'O + \measuredangle OA'D'\),Por lo tanto, sumando las cuatro identidades en 10.2.1, obtenemos que
\(\measuredangle ABC +\measuredangle CDA \equiv -(\measuredangle D'C'B' + \measuredangle B'A'D')\).
Aplicando el Axioma IIIb y el Ejercicio 7.4.5, lo conseguimos
\(\begin{array} {rcl} {\measuredangle A'B'C' + \measuredangle C'D'A'} & \equiv & {-(\measuredangle B'C'D' + \measuredangle D'A'B') \equiv} \\ {} & \equiv & {\measuredangle D'C'B' + \measuredangle B'A'D'.} \end{array}\)
De ahí que siga el inciso b).
c). Se desprende de (b) y Corolario 9.3.2.