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; some for which it will not intersect at all (which means there is no such triangle in the first place); and one for which the ray will be tangent to the circle, giving a unique triangle with a right angle at C. So in general, in this case, congruency can't be guaranteed. 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/
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