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, ?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 angles||Definition of corresponding angles|
|4.||?ADE ~= ?ABC||Postulate 10.1|
|5.||?DAE ~ ?ABC||Reflexive property of ~=|
|6.||?ADE ~ ?ABC||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 ?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.
- 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 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.