AAA, ASS, SSA Theorems

Date: 11/16/2001 at 20:32:34
From: Zakiyah
Subject: SSA, AAA or ASS postulates

Can you please tell me in detail why the ASS, SSA, and AAA postulates 
can't be used to determine triangle congruence?

Thanks.

Date: 11/17/2001 at 08:45:33
From: Doctor Jubal
Subject: Re: SSA, AAA or ASS postulates

Hi Zakiyah,

Thanks for writing to Dr. Math.

Just a side note: the SSS, SAS, and ASA triangle congruency theorems 
are theorems, not postulates. A postulate is something you just state 
and assume to be true. A theorem is something you can prove, based on 
your postulates.

Let's start with the angle-angle-angle or AAA Congruency Theorem. 
Think of two equilateral triangles, one with side length 1 and one 
with side length 2. They have the same angles as each other (each 
angle is 60 degrees), but they're not congruent, because they have 
different side lengths.

Actually, there is an AAA theorem, the AAA Similarity Theorem, but 
it's not a congruency theorem. We say that two triangles are similar 
if all their angles are the same. This means that they are exactly the 
same shape, but are different sizes. Similarity comes in quite handy 
when trying to prove two objects have certain proportions (like 
proving that one is exactly twice the size of the other).

As for angle-side-side (ASS) and side-side-angle (SSA), they are the 
same theorem. The two triangles have two congruent sides, and one 
congruent angle that is not the angle between the two congruent sides.  
To demonstrate why this isn't a congruency theorem, I'm going to ask 
you to draw some figures.

First, draw a line segment and label its ends points A and B. This is 
one of the known sides, the "middle" S in ASS or SSA.  At one end of 
it, we know the length of the next side, but we don't know what angle 
it comes off at. So draw a circle centered at point A - the radius of 
the circle is the length of the the other line segment, and the third 
vertex of the triangle must lie on this circle.

There are two possibilities: either the radius of the circle is longer 
than segment AB, or it is shorter. Draw one figure for each case.

 

In the case where the circle's radius is longer than AB, point B lies 
inside the circle. So if you draw a ray originating at point B, it 
will only intersect the circle in one place. Let's say that the angle 
between that ray and segment AB is the congruent angle a, and the 
third vertex of the triangle must line on that ray. Since there is 
only one point that lies on both the circle and the ray, there is only 
one triangle that has sides with the two shared lengths and also has 
the shared angle, and an SSA congruency theorem would hold in this 
case.

But now go to the case where the circle's radius is smaller than AB.  
You can draw a ray originating at B that intersects the circle in two 
places, so there are two possibilities for the third vertex of the 
triangle. Both of these triangles have sides S and S and angle A, but 
they are not congruent. So SSA congruency fails here.

We can summarize SSA congruency this way: SSA congruency fails if the 
middle S is equal to or longer than the other S, and holds if the 
middle S is the shorter side.

Does this help?  Write back if you'd like to talk about this some
more, or if you have any other questions.

- Doctor Jubal, The Math Forum

  http://mathforum.org/dr.math/ 
  

Date: 11/17/2001 at 19:10:04
From: Zakiyah
Subject: Re: SSA, AAA or ASS postulates

This is exactly what I wanted to know. Thank you so much; I now 
understand.

Date: 05/29/2009 at 04:08:13
From: Mike
Subject: SSA or ASS triangle Congruency

I read your answer and completely understand until this point...

"We can summarize SSA congruency this way: SSA congruency fails if
the middle S is equal to or longer than the other S, and holds if the 
middle S is the shorter side."

I believe that if the middle S is congruent to the other S, you will
have only one possible triangle.

If the middle S is equal to the other S you will have an isosceles 
triangle.  An isosceles triangle has two congruent angles.  You 
could prove the last pair of angles are also going to be congruent 
by the No-Choice Theorem.  At this point, it appears to me the 
triangles are congruent by SAS (because you have two congruent sides 
and all three angles are congruent).  Unless I am missing something, 
it appears you are incorrect.  Shouldn't it say the only way SSA 
fails is if the middle S is longer than the other S?

Date: 05/29/2009 at 21:35:55
From: Doctor Peterson
Subject: Re: SSA or ASS triangle Congruency

Hi, Mike.

I think you're right. Dr. Jubal's reasoning, filled out a little 
more, is this:

Suppose that you are given AB = c and AC = b, along with angle B.

In the case where b < c (the left picture), there are SOME angles B
for which the ray BC will intersect the circle at two points, so
congruency will not be guaranteed:

 

There are also some for which it will not intersect at all, which means
there is no such triangle in the first place:

 

Finally, there is one for which the ray will be tangent to the circle,
giving a unique triangle with a right angle at C:

 

So in general, congruency can't be guaranteed here.

In the case where b > c (the right picture), for ANY angle B the ray
AC will intersect the circle in one place, so congruency is guaranteed:

 

He doesn't show the case where b = c, which you are considering. In
that case, the circle passes through B, and if the given angle B is
acute, the ray will again intersect the circle in only one place, so
congruency will be guaranteed:

 

But if angle B is obtuse, there will be no such triangle; so that case
will not arise:

 

So it would be proper to make the following SSA theorem:

If two triangles have one congruent angle B = B', and two congruent
sides AB = A'B' and AC = A'C', and if AB <= AC, then the triangles are
congruent. If AB > AC, the triangles may or may not be congruent.
Other special conditions may hold that would determine congruency, 
however.

If you have any further questions, feel free to write back.


- Doctor Peterson, The Math Forum

  http://mathforum.org/dr.math/