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.
- 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 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.
|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 angles||Definition of corresponding angles|
|4.||∠ADE ~= ∠ABC||Postulate 10.1|
|5.||∠DAE ~ ∠ABC||Reflexive property of ~=|
|6.||ΔADE ~ ΔABC||AA Similarity Theorem|
|7.||AB/AD = AC/AE||CSSTAP|
|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 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.
|1.||ΔADE ~ ΔABC||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 angles||Definition 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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