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# Geometry: Parallel Segments and Segment Proportions

## Parallel Segments and Segment Proportions

Suppose you have a segment ¯AB. You've talked about breaking up a segment into two equal pieces (using the midpoint). You can also break up a segment into thirds, quarters, or whatever fraction you want.

Suppose you have two segments, ¯AB and ¯RS, as shown in Figure 14.3. These segments can have the same length, or they can have different lengths. You can break each segment up into a variety of pieces. One specific way you can divide up your segments is proportionally. When two segments, ¯AB and ¯RS, are divided proportionally, it means that you have found two points, C on ¯AB and T on ¯RS, so that

• AC/RT = CB/TS.

You are now ready to prove the following theorem:

• Theorem 14.2: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides these sides proportionally.
• Example 1: Write a formal proof of Theorem 14.2.
• Solution: This theorem is illustrated in Figure 14.4.
• Given: In Figure 14.4, ΔABC has ↔DE ‌ ‌ ¯BC , with ↔DE intersecting ¯AB at D and ¯AC at E.
• Proof: In order to show that D and E divide the segments ¯AB and ¯AC proportionally, you will need to show that ΔADE ~ ΔABC and then use CSSTAP. To show that ΔADE ~ ΔABC, you will use the AA Similarity Theorem. To determine the angle congruencies, you will use our postulate about corresponding angles and parallel lines.
StatementsReasons
1. ΔABC has ↔DE ‌ ‌ ¯BC , with ↔DE intersecting ¯AB at D and ¯AC at E Given
2.↔DE ‌ ‌ ¯BC cut by a transversal ¯AB Definition of transversal
3. ∠ADE and ∠ABC are corresponding anglesDefinition of corresponding angles
5. ∠DAE ~ ∠ABC Reflexive property of ~=
6. ΔADE ~ ΔABCAA Similarity Theorem
8. AB - AD/AD = AC - AE/AE Property 3 of proportionalities
10. AD/BD = AE/EC Property 2 of proportionalities

You can also use similar triangles to show that two lines are parallel. For example, suppose that ΔADE ~ ΔABC in Figure 14.5. You can prove that ¯DE ‌ ‌ ¯BC.

• Example 2: If ΔADE ~ ΔABC as shown in Figure 14.5, prove that ¯DE ‌ ‌ ¯BC.
• Solution: Your game plan is quite simple. Because ΔADE ~ ΔABC, you know that ∠ADE ~= ∠ABC. Because ∠ADE and ∠ABC are congruent corresponding angles, you know that ↔DE ‌ ‌ ↔BC by Theorem 10.7.
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