Geometry: The Big Five
The Big Five
In order to make use of the fact that corresponding parts of congruent triangles are congruent you will need to be able to prove that the two triangles are, in fact, congruent without using the definition of triangle congruence. I'll give you some shortcuts (in the form of postulates and theorems) that will help you do this. Most of these postulates and theorems will be known by their abbreviations: S stands for side and A stands for angle.
Eureka!
To come up with a game plan for proving triangles congruent, here is some helpful advice:
- 1. Mark the figures systematically. Use a square in the opening of a right angle, use the same number of dashes on congruent sides and use the same number of arcs on congruent angles.
- 2. Trace the triangles suspected to be congruent in different colors.
- 3. If the triangles overlap, draw them separately.
The SSS Postulate
As the name implies, you can conclude that two triangles are congruent based on just the lengths of the sides of two triangles.
- Postulate 12.1: SSS Postulate. If the three sides of one triangle are congruent to the three sides of a second triangle, then the triangles are congruent.
Let's practice using this postulate. Take a look at Figure 12.2. Suppose that ¯AB and ¯CD bisect each other at M and that ¯AC ~= ¯DB. Prove that ΔAMC ~= ΔBMD.
Figure 12.2¯AB and ¯CD bisect each other at M, and ¯AC ~= ¯DB.
- Example 1: Given the situation shown in Figure 12.2, if ¯AB and ¯CD bisect each other at M, and ¯AC ~= ¯DB, write an informal proof that ΔAMC ~= ΔBMD.
- Solution: You can skip the formalities of rewriting the statement you are trying to prove. You already have a picture (Figure 12.2) to help you visualize the situation. You are given that ¯AB and ¯CD bisect each other at M, and ¯AC ~= ¯DB, and you need to show that ΔAMC ~= ΔBMD. To make use of the SSS Postulate you will need to explore the relationships between the sides of the two triangles. Because ¯AB and ¯CD bisect each other at M you know that ¯AM ~= ¯MB and ¯CM ~= ¯MD. That should be enough to nail down the details of the proof.
| Statements | Reasons | |
|---|---|---|
| 1. | ¯AB and ¯CD bisect each other at M, and ¯AC ~= ¯DB | Given |
| 2. | ¯AM ~= ¯MB and ¯CM ~= ¯MD | Definition of segment bisector |
| 3. | ΔAMC ~= ΔBMD | SSS Postulate |
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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