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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 now summarize our proof in one paragraph.
Given: Prove: AE∥BF∥CG and AB≅BCEF≅FG
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Proof: By applying the Proportional Parts of Parallel Lines Corollary, we get that BCAB=FGEF. Since AB≅BC, we have that AB=BC. Thus, BCAB=1. By substitution, 1=FGEF which implies that EF=FG. Finally, by the definition of congruent segments, EF≅FG. |