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.

Figure 14.3¯AB and ¯RS are divided proportionally, 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.

Figure 14.4ABC has DE ¯BC , with DE intersecting ¯AB at D and ¯AC at E.

  • Given: In Figure 14.4, ABC has DE ¯BC , with DE intersecting ¯AB at D and ¯AC at E.
  • Prove: AD/DB = AE/EC.
  • Proof: In order to show that D and E divide the segments ¯AB and ¯AC proportionally, you will need to show that ADE ~ ABC and then use CSSTAP. To show that ADE ~ ABC, you will use the AA Similarity Theorem. To determine the angle congruencies, you will use our postulate about corresponding angles and parallel lines.
1. ABC 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
4. ADE ~= ABCPostulate 10.1
5. DAE ~ ABC Reflexive property of ~=
6. ADE ~ ABCAA Similarity Theorem
8. AB - AD/AD = AC - AE/AE Property 3 of proportionalities
9. BD/AD = EC/AE Segment Addition Postulate
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 ADE ~ ABC in Figure 14.5. You can prove that ¯DE ¯BC.

  • Example 2: If ADE ~ ABC as shown in Figure 14.5, prove that ¯DE ¯BC.

Figure 14.5ADE ~ ABC.

  • Solution: Your game plan is quite simple. Because ADE ~ ABC, you know that ADE ~= ABC. Because ADE and ABC are congruent corresponding angles, you know that DE BC by Theorem 10.7.
1. ADE ~ ABC Given
2. ADE ~= ABC Definition of ~
3.DE and BC are two lines cut by a transversal AB Definition of transversal
4. ADE and ABC are corresponding anglesDefinition of corresponding angles
5. ¯DE ¯BC Theorem 10.7

Excerpted from The Complete Idiot's Guide to Geometry © 2004 by Denise Szecsei, Ph.D.. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.

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