# 4.3: Transversals to Three Parallel Lines

Three Parallel Lines Theorem
Three Parallel Lines Theorem

4.3: Transversals to Three Parallel Lines

Page ID
34137

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.

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

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

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

Find $$x$$:

Solution

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

Check:

Answer: $$x = 6$$.

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

$\dfrac{a}{b} = \dfrac{c}{d}.$

Find $$x$$:

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, $$x = 2$$:

Answer: $$x = 2$$.

## Problems

1 – 6. Find $$x$$:

1.

2.

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