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4.3: Transversals to Three Parallel Lines

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    34137
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    In Chapter 1 we defined a transversal to be a line which intersects two other lines, We will now extend the definition to a line which intersects three other lines. In Figure \(\PageIndex{1}\), \(AB\) is a transversal to three lines.

    屏幕快照 2020-11-17 下午7.05.15.png
    Figure \(\PageIndex{1}\): \(AB\) is a transversal to three lines.

    If the three lines are parallel and we have two such transversals we may state the following theorem:

    Theorem \(\PageIndex{1}\)

    The line segments formed by two transversals crossing three parallel lines are proportional.

    In Figure \(\PageIndex{2}\), \(\dfrac{a}{b} = \dfrac{c}{d}\).

    屏幕快照 2020-11-17 下午7.09.40.png
    Figure \(\PageIndex{2}\): The line segments formed by the transversals are proportional.
    Example \(\PageIndex{1}\)

    Find \(x\):

    屏幕快照 2020-11-17 下午7.12.17.png

    Solution

    \[\begin{array} {rcl} {\dfrac{x}{3}} & = & {\dfrac{8}{4}} \\ {4x} & = & {24} \\ {x} & = & {6} \end{array}\]

    Check:

    屏幕快照 2020-11-17 下午7.12.37.png

    Answer: \(x = 6\).

    Proof of Theorem \(\PageIndex{1}\)

    Draw \(GB\) and \(HC\) parallel to \(DF\) (Figure \(\PageIndex{3}\)). The corresponding angles of the parallel lines are equal and so \(\triangle BCH \sim \triangle ABG\). Therefore

    \[\dfrac{BC}{AB} = \dfrac{CH}{BG}.\]

    Now \(CH = FE = c\) and \(BG = ED = d\) because they are the opposite sides of a parallelogram. Substituting, we obtain

    \[\dfrac{a}{b} = \dfrac{c}{d}.\]

    屏幕快照 2020-11-17 下午7.16.41.png
    Figure \(\PageIndex{3}\): Draw \(GB\) and \(HC\) parallel to \(DF\).
    Example \(\PageIndex{2}\)

    Find \(x\):

    屏幕快照 2020-11-17 下午7.18.35.png

    Solution

    \(\begin{array} {rcl} {\dfrac{x}{3}} & = & {\dfrac{2x + 2}{4x + 1}} \\ {(x)(4x + 1)} & = & {(3)(2x + 2)} \\ {4x^2 + x} & = & {6x + 6} \\ {4x^2 - 5x - 6} & = & {0} \\ {(x - 2)(4x + 3)} & = & {0} \end{array}\)

    \(\begin{array} {rcl} {x - 2} & = & {0} \\ {x} & = & {-2} \end{array}\) or \(\begin{array} {rcl} {4x + 3} & = & {0} \\ {4x} & = & {-3} \\ {x} & = & {-\dfrac{3}{4}} \end{array}\)

    We reject \(x = -\dfrac{3}{4}\) because \(BC = x\) cannot be negative.

    Check, \(x = 2\):

    屏幕快照 2020-11-17 下午7.27.25.png

    Answer: \(x = 2\).

    Problems

    1 - 6. Find \(x\):

    1.

    Screen Shot 2020-11-17 at 7.30.19 PM.png

    2.

    Screen Shot 2020-11-17 at 7.30.40 PM.png

    3.

    Screen Shot 2020-11-17 at 7.30.57 PM.png

    4.

    Screen Shot 2020-11-17 at 7.31.11 PM.png

    5.

    Screen Shot 2020-11-17 at 7.31.31 PM.png

    6.

    Screen Shot 2020-11-17 at 7.31.50 PM.png


    This page titled 4.3: Transversals to Three Parallel Lines is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Henry Africk (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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