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Here is proof they are not congruent - an example of it not working:

The lines 'A' & 'B' are both the same length (picture not perfect, but its close), and are also the same length as the side on the other triangle. The other marked sides are the same length, and the hyp passed through this 'middle' point between triangles.

The (circle) arc is drawn to represent "possible" solutions. As you can see, there are TWO solutions (lines 'A' & 'B') where the lines are the correct length AND join to the hypt.

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When the shared angle between the triangles is the largest angle in the triangles (i.e. greater than 90deg), there is only 1 possible solution. This forms the case where an SSA solution does allow for congruency.

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As various people have mentioned above, this example forms the "classic" case where the triangles are not congruent (and its now time for me to go to bed... 4am :P).



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All of this is ASSUMING that the lines crossing in the middle are straight, and that the shared angles in the middle are equal. Not an assumption you can make based on the diagram.



ihuerta said:

This discussion makes my eyes bleed and sort of makes me realiZe that people really don't know much math... like at all.

Let's call the sides marked with one line as A, and those marked with two lines as B. Now, ASSUMING the lines in the diagram are straight, the two angles at the center of the figure have to be identical, as they are cuased by the interseccion of two straight lines. Lets say these angles have a value of ALPHA. Now, the Sin Rule says that A/sin(angle opposite A) = B/(sin opposite B) in a triangle. Therefore, since we know the anlge opposite both A's are the same (ALPHA), we therefore conclude that the angles opposibe B are also the same, let's call them BETA.


ihuerta, before you go saying "people really don't know much math", you should really double-check your argument to be sure it holds up.

Your sine rule argument is a noble attempt, but it doesn't work. You know that (sin a)/A = (sin b)/B, but this does not prove that a = b even if A = B. For A = B and assuming that 0 < a < 180 degrees, and 0 < b < 180 degrees (since it's a triangle), for a given A, B, and a, there are exactly two values of b which satisfy (sin a)/A = (sin b)/B. One is an acute angle, the other is an obtuse angle (or it could be the degenerate case where they are both 90 degrees).

Think about it, if the sine rule argument worked, then side-side angle would always be a valid proof of congruency. It's been shown repeatedly here that side-side angle is not a valid proof of congruency.

 

I think this thread does show one very important thing about society: People will continue to argue back and forth even when the correct answer can be mathematically proven.

Looks like shams has drawn a very nice illustration of what I'm talking about. :)



txags911 said:
All of this is ASSUMING that the lines crossing in the middle are straight, and that the shared angles in the middle are equal. Not an assumption you can make based on the diagram.

In geometry it is commonly accepted that you can assume that intersecting lines are straight, so that gives you the opposite angles.  It is not commonly accepted to assume that lines or angles which look equal are equal, though.



Yeah graphs like these come with an attached "Figure is not drawn to scale" assumption. By the way shams, you can't prove the triangles are not congruent from the information either, so that statement is somewhat misleading. All you know, is that they can be either congruent or not congruent. :)



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Sadly this is the first time i've ever had to think about congruent angles since learning about them about 8 years ago