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.
AB/BC = EF/FG
Since AB≅BC, we have that AB=BC. Let's now find ABBC.
AB=BC ⇒ AB/BC = AB/AB=1
This gives us the following relationship between EF and FG.
1 = EF/FG ⇒ FG = EF
Segments EF and FG are the same length. By definition of congruent segments, we conclude that EF≅FG.
Paragraph Proof
We can now summarize our proof in one paragraph.
Given: & AE∥ BF∥ CG andAB≅BC
Prove: & EF≅FG
Proof: By applying the Proportional Parts of Parallel Lines Corollary, we get that ABBC = EFFG. Since AB≅BC, we have that AB=BC. Thus, ABBC =1. By substitution, 1= EFFG which implies that EF=FG. Finally, by the definition of congruent segments, EF≅FG.
