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.

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.