4.3: Transversales a Tres Líneas Paralelas
- Page ID
- 114597
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)En el Capítulo 1 definimos una transversal para ser una línea que intersecta otras dos líneas, ahora extenderemos la definición a una línea que intersecta otras tres líneas. En la Figura\(\PageIndex{1}\),\(AB\) es una transversal a tres líneas.

Si las tres líneas son paralelas y tenemos dos de esas transversales podemos exponer el siguiente teorema:
Los segmentos de línea formados por dos transversales que cruzan tres líneas paralelas son proporcionales.
En la Figura\(\PageIndex{2}\),\(\dfrac{a}{b} = \dfrac{c}{d}\).

Encuentra\(x\):
Solución
\[\begin{array} {rcl} {\dfrac{x}{3}} & = & {\dfrac{8}{4}} \\ {4x} & = & {24} \\ {x} & = & {6} \end{array}\]
Comprobar:
Respuesta:\(x = 6\).
Dibujar\(GB\) y\(HC\) paralelo a\(DF\) (Figura\(\PageIndex{3}\)). Los ángulos correspondientes de las líneas paralelas son iguales y así\(\triangle BCH \sim \triangle ABG\). Por lo tanto
\[\dfrac{BC}{AB} = \dfrac{CH}{BG}.\]
Ahora\(CH = FE = c\) y\(BG = ED = d\) porque son los lados opuestos de un paralelogramo. Sustituyendo, obtenemos
\[\dfrac{a}{b} = \dfrac{c}{d}.\]

Encuentra\(x\):
Solución
\(\begin{array} {rcl} {\dfrac{x}{3}} & = & {\dfrac{2x + 2}{4x + 1}} \\ {(x)(4x + 1)} & = & {(3)(2x + 2)} \\ {4x^2 + x} & = & {6x + 6} \\ {4x^2 - 5x - 6} & = & {0} \\ {(x - 2)(4x + 3)} & = & {0} \end{array}\)
\(\begin{array} {rcl} {x - 2} & = & {0} \\ {x} & = & {-2} \end{array}\)o\(\begin{array} {rcl} {4x + 3} & = & {0} \\ {4x} & = & {-3} \\ {x} & = & {-\dfrac{3}{4}} \end{array}\)
Rechazamos\(x = -\dfrac{3}{4}\) porque\(BC = x\) no puede ser negativo.
Comprobar,\(x = 2\):
Respuesta:\(x = 2\).
Problemas
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