# 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.

Statements | Reasons | |
---|---|---|

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.

Statements | Reasons | |
---|---|---|

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.

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**See also:**