Exercise 29 Page 497

We are asked to write a paragraph proof of the following corollary.

If three or more lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal.

Let's consider three parallel lines and two transversals such that the parallel lines divide one transversal into congruent segments.

We want to show that the parallel lines divide the line passing through points $E$ and $G$ into congruent segments. To do this, we apply the following property.

Proportional Parts of Parallel Lines Corollary
If three or more parallel lines intersect two transversals, then they cut off the transverals proportionally.

We can write the following proportions.

\frac{A B}{B C} = \frac{E F}{F G}

Since

\overline{A B} ≅ \overline{B C},

we have that

A B = B C .

Let's now find

\frac{A B}{B C} .

A B = B C \Rightarrow \frac{A B}{B C} = \frac{A B}{A B} = 1

This gives us the following relationship between

E F

and

F G .

1 = \frac{E F}{F G} \Rightarrow F G = E F

Segments

E F

and

F G

are the same length. By definition of congruent segments, we conclude that

\overline{E F} ≅ \overline{F G} .

Paragraph Proof

We can now summarize our proof in one paragraph.

$Given : Prove : \overline{A E} ∥ \overline{B F} ∥ \overline{C G} and \overline{A B} ≅ \overline{B C} \overline{E F} ≅ \overline{F G}$
Proof: By applying the Proportional Parts of Parallel Lines Corollary, we get that $\frac{A B}{B C} = \frac{E F}{F G} .$ Since $\overline{A B} ≅ \overline{B C},$ we have that $A B = B C .$ Thus, $\frac{A B}{B C} = 1 .$ By substitution, $1 = \frac{E F}{F G}$ which implies that $E F = F G .$ Finally, by the definition of congruent segments, $\overline{E F} ≅ \overline{F G} .$