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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/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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