Geometric "proof"

[plasmid]

First some background on geometry for those who have either never taken it or have forgotten it. Two triangles are congruent if they have the same shape and size. They may be reflected or rotated or translated relative to each other, but by applying these transformations it is possible to make the two triangles superimposable. Given that there are two triangles, it is well known that if one triangle has sides of length {S1, S2, S3} and the other also has sides of length {S1, S2, S3} then the two triangles are congruent (this is sometimes called the SSS or side-side-side theorem in geometry). Similarly, if one triangle has two sides of length {S1, S2} and the angle between those two sides is {A3}, then any other triangle that also has two sides of length {S1, S2} and an angle in between them of {A3} must be congruent with it (this is called the SAS or side-angle-side theorem). There is a similar ASA theorem if you know the length of one of the sides and the angles at each end of that side.

In grade school, I was quite distraught that there was no theorem stating that if you knew that two of the sides and one of the angles that does NOT fall between the two sides are equal then the triangles are congruent (that's because it's not true!), which would naturally be called the A$$ theorem. So I made up a proof for such a theorem on my own. The proof is below, see if you can spot the flaw in it.

Proof: Take two triangles where you know the lengths of sides S1 and S2 and you know the angle A1 that is opposite from side S1. Position the two triangles so that their sides with length S1 are overlapping, the sides of length S2 for each triangle are both emerging from the same end of the common S1, and the triangles are pointing in opposite directions, as shown in the attached figure.

Now imagine a line passing from the A1 of one triangle to the A1 of the other triangle. Consider the triangle formed by this line and the two lines of length S2. We next take advantage of a theorem on isosceles triangles: if a triangle has two sides of the same length, then the two angles opposite of those sides must have the same angle. We therefore know that the two angles labeled A1a are the same.

Since the two angles A1 are the same, and the two angles A1a are the same, then the remaining angles labeled A1b in the figure (which span from the imaginary line to the side of unknown length for each triangle) must also be the same: just A1 minus A1a. Considering the triangle formed by the imaginary line and the two sides of unknown length, we can again use a theorem on isosceles triangles: if two angles of a triangle have the same angle, then the sides opposite those angles must have the same length. That proves that the two sides that were originally of unknown length must actually have the same length. Then the two triangles must be congruent, by the SSS theorem.

[Guest]

The biggest flaw that I can find is that there are still too many variables. Since the angle between s1 and s2 is unknown (and cant be found, even from the two angles A1a), the third angle is unknown, and the remaining side is unknown the triangle could have any number of combinations of remaining sides and angles, and the other triangle could be something totally different. The figure is tricky because it assumes (at least the way it is drawn) that you already know the angle between s1 and s2 and that it is the same in both triangles, in which case you could just use the SAS theorem.

[Guest]

The triangle(s) formed by the dotted line in the diagram will only be isosceles triangles if the angles between S1 and S2 on both triangles are the same. Assuming that they are isosceles triangles brings about the proof that the original triangles are congruent-a bit of circular reasoning.

[plasmid]

The triangle(s) formed by the dotted line in the diagram will only be isosceles triangles if the angles between S1 and S2 on both triangles are the same. Assuming that they are isosceles triangles brings about the proof that the original triangles are congruent-a bit of circular reasoning.

An isosceles triangle is a single triangle where two of its sides have the same length. The single triangle formed by the two S2 lines and the dotted line would be an isosceles triangle (regardless of what the angle between S1 and S2 is -- S1 is not part of the isosceles triangle).

[HoustonHokie]

Your mistake is in assuming that A1 > A1a. If A1 = A1a, then A1b = 0, and that changes the look of the diagram (imagine keeping only the part of the diagram above the dotted line). S1 would not extend below the dotted line and the cyan line could "swivel" about the intesection of the two red S2 lines, creating a whole range of triangles where the given conditions are met, but are not congruent. They would have an angle A1, sides S1 & S2, but A2 on one triangle would equal A3 on the other, and vice versa. Most importantly, S3 would not be equal for the two triangles thus created.

Edited March 30, 2009 by HoustonHokie

[Guest]

An isosceles triangle is a single triangle where two of its sides have the same length. The single triangle formed by the two S2 lines and the dotted line would be an isosceles triangle (regardless of what the angle between S1 and S2 is -- S1 is not part of the isosceles triangle).

Yeah, you're right

[plasmid]

Your mistake is in assuming that A1 > A1a. If A1 = A1a, then A1b = 0, and that changes the look of the diagram (imagine keeping only the part of the diagram above the dotted line). S1 would not extend below the dotted line and the cyan line could "swivel" about the intesection of the two red S2 lines, creating a whole range of triangles where the given conditions are met, but are not congruent. They would have an angle A1, sides S1 & S2, but A2 on one triangle would equal A3 on the other, and vice versa. Most importantly, S3 would not be equal for the two triangles thus created.

Yep, that's it!

That is exactly the only condition where the two triangles can be non-congruent: they have to have the angle across from S2 on one triangle plus the angle across from S2 on the other triangle be equal to 180 degrees. Then putting the two triangles together would only make the upper half of that figure. Incidentally, the proof would still work if A1 were less than A1a, then you would just have the bottom half of the triangle (as it's shown in the figure) pointing up into the top half instead of pointing down below the dashed line.

[Guest]

Torcher pure torcher... WHy bring up the past horrors of forced math. Honestly I dont remember any of it but good for you who does. Interesting problem but just cant do it.

[Guest]

Draw a line of a specific length, then draw cirle of a specific radius at one end and a second line at the other end at a specific angle. If the radius of the circle is less than the the length of the first line than the second line will intersect the circle twice showing two possible triangles. This is kinda hard to visualized but I can't illustrate it while at work.

Edited March 30, 2009 by burninator777

[Guest]

Proof: Take two triangles where you know the lengths of sides S1 and S2 and you know the angle A1 that is opposite from side S1. Position the two triangles so that their sides with length S1 are overlapping, the sides of length S2 for each triangle are both emerging from the same end of the common S1, and the triangles are pointing in opposite directions, as shown in the attached figure.

Now imagine a line passing from the A1 of one triangle to the A1 of the other triangle. Consider the triangle formed by this line and the two lines of length S2. We next take advantage of a theorem on isosceles triangles: if a triangle has two sides of the same length, then the two angles opposite of those sides must have the same angle. We therefore know that the two angles labeled A1a are the same.

Since the two angles A1 are the same, and the two angles A1a are the same, then the remaining angles labeled A1b in the figure (which span from the imaginary line to the side of unknown length for each triangle) must also be the same: just A1 minus A1a. Considering the triangle formed by the imaginary line and the two sides of unknown length, we can again use a theorem on isosceles triangles: if two angles of a triangle have the same angle, then the sides opposite those angles must have the same length. That proves that the two sides that were originally of unknown length must actually have the same length. Then the two triangles must be congruent, by the SSS theorem.

The angles A1 in the figure are not necessarily equal...

[Guest]

That is exactly the only condition where the two triangles can be non-congruent: they have to have the angle across from S2 on one triangle plus the angle across from S2 on the other triangle be equal to 180 degrees.

That is incorrect. It depends only on which of the two known sides you "attach" the known angle two. In your figure, you have made A1 the angle between the unknown side and S2 in both triangles, which is enough to make them congruent. However, you could just as easily have made A1 the angle between the unknown side and S1 in one of those triangles, in which case they would be non-congruent.