Congruency Theorems for Triangles
Date: 13 Mar 1995 17:20:13 -0500 From: Roger Gillies Subject: Re: Congruency Theorums for Triangles Dear Dr. Math, Given the three methods of proving two triangles congruent: SSS SAS ASA Using SAS, the angle must be a contained angle, Correct? Then, two triangles, one which has two sides that are of equal length to the second triangle, and both having an angle (not contained) equal, cannot be proved congruent. It seems to me that they are congruent, though. Any thoughts on this? Thanks. Roger
Date: 13 Mar 1995 20:58:08 -0500 From: Dr. Ethan Subject: Re: Congruency Theorums for Triangles Hey Roger, What you are looking for is an SSA postulate. I am afraid that it just isn't there. It is interesting to see why it can't be; and there are a few special cases in which it actually works. First let's look at one special case and then we'll look at why it doesn't work in general. Are you familiar with the HL postulate of congruency? It is for right triangles, and it says that if the hypotenuse and legs of two right triangles are congruent, then the triangles are congruent. That essentially is an SSA postulate, except that we require that the A be a right angle. I know that that is a pretty exclusive restriction, but we actually can get a little broader. Before I tell you how broad, let's look at a triangle for which it won't work. Here is an experiment for you to try. This should at least give you one example of a one place that it won't work, and then you can write back with what you find out and I'll give you another hint about what you can try. Get three sticks, one that is three inches long and two that are five inches long. Use them to make an isosceles triangle. It will look something like this: /\ / \ / \ / \ 5/ \5 / \ / \ / \ /________________\ 3 Label them like this: /\ / \ / \ / \ a/ \c / \ / \ / \ /________________\ b Now I want you to keep the angle fixed between a and c and move the stick of length 3 in until it forms a triangle again. If you have followed these instructions, you have constructed two triangles that have congruencies SSA but are clearly not congruent triangles. I hope that this helps some. Please write back if this doesn't make sense and I will try again. Ethan, Doctor On Call
Date: 01/05/99 at 11:22:17 From: Nathan Smith Subject: The SSA theorem in geometry I can't figure out how you could move the stick with a length of 3 to make it touch the side again. It seems to me that it only hits in one spot. Move it one way or the other, and it becomes farther and farther away from the other stick. I have gone to other sites that use trigonometry, but I am only in geometry, so I don't understand. Thanks for help, Nathan Smith
Date: 01/05/99 at 12:39:56 From: Doctor Rob Subject: Re: The SSA theorem in geometry Try this diagram: B o ,/|\ ,'/ | \ ,' / | \ 15,' / | \13 ,' / 12| \ ,' 13/ | \ ,' 4 / 5 | 5 \ A o'------o-------o-------o C 14 P D Triangles ABC and ABD have sides BC = BD = 13, and they share side AB = 15 and <BAC = <BAD. That gives you the conditions for S.S.A. Sides AC = 4 and AD = 14, however, so the triangles are not congruent. This is constructed out of right triangles ABP, BCP, and BDP. ABP has sides 9, 12, and 15, and 9^2 + 12^2 = 15^2. BCP and BDP are congruent and have sides 5, 12, and 13, and 5^2 + 12^2 = 13^2. A host of other examples could be constructed. All you need is three numbers x, y, and z with 0 < x < y < z. Make AB = z BP = x BC = BD = y Then AP = sqrt(z^2-x^2) CP = DP = sqrt(y^2-x^2) AC = AP - CP = sqrt(z^2-x^2) - sqrt(y^2-x^2) AD = AP + DP = sqrt(z^2-x^2) + sqrt(y^2-x^2) In the example above the numbers are x = 12, y = 13, and z = 15, which were carefully chosen to make all the lengths whole numbers. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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