There is a standard litany of theorems involving proving triangle congruence that has remained largely unchanged since my high school days. I was told that, to prove that two triangles are congruent, we need three pieces of information. The abbreviations were given as SSS, SAS, AAS, and ASA. Astute students would ask about SSA (or ASS, to be childish). Then as now, we were told that SSA was not sufficient.

Since I’ve returned to the world of high school mathematics, I’ve noticed the inclusion of HL. I don’t recall that theorem from my own high school experience. It’s possible I’ve forgotten it, but also possible it wasn’t in the curriculum. However, it got me wondering: Why does HL work, when SSA doesn’t?

For those who need a quick refresher, the basic theorems can be boiled down to this: If we know that the sides are congruent, the triangles are congruent (SSS). If we know two angles, we can figure out all three, so two angles and a side are sufficient (ASA and AAS). If we know two sides and one angle, the angle needs to be between the sides (SAS).

The exception, modern students are told, is the right triangle: If we know any two sides of a right triangle, we can prove congruence. When I first heard of this, I assumed it was because, if we know two sides of a right triangle, we can use Pythagorean’s Theorem to determine the third side. Knowing the leg lengths is effectively SAS (since the angle is 90°); knowing one leg and the hypotenuse allows us to calculate the third side and thus prove SSS congruence.

But this summer I also noticed that it’s possible to determine an angle with SSA and the Law of Sines, with some restrictions. I put this in the back of my mind and got distracted by other things.

Since I’m back to teaching SSS and its other theorems, I found myself thinking about this again. Why *does* HL work when SSA doesn’t?

In reality, we could be teaching SSA in such a way that HL is a natural consequence. Indeed, HL is not a theorem in its own right, nor is it solely because of the Pythagorean Theorem: It is due to the constraints on SSA.

Specifically, given two sides S_{1} and S_{2} and an angle A that is adjacent to S_{1} but not to S_{2}, SSA is sufficient to prove congruence if *m*S_{1} ≤ *m*S_{2}. If *m*S_{1} > *m*S_{2}, then there are two possible triangles.

If A is right or obtuse, then it must be the case that mS_{1} ≤ mS_{2}. The only way that the S_{2} can be shorter than S_{1} is if A is acute. Even when A is acute, of course, it’s possible for S_{1} to be shorter than or the same length as S_{2}.

This follows from the Law of Sines. Let A_{1} be the angle opposite S_{1}. By the Law of Sines: \[\frac{\sin A_1}{S_1} = \frac{\sin A}{S_2}\] which can be rewritten as \[\sin A_1 = \frac{S_1 \cdot \sin A}{S_2}\]

Because sine is symmetric around the 90°, this has two solutions unless A_{1} is a right angle. However, if S_{1} is shorter than S_{2}, then A_{1} has to be smaller than A. And regardless, one of the solutions will be in the range of 90° to 180°, which is not possible if A is also right or obtuse. The only condition under which A_{1} can be more than A is if S_{1} is longer than S_{2}.

Conclusion: SSA is sufficient in some conditions for proving congruence, but it’s not completely reliable, unlike the other theorems.

JoshuaReally like this and will add it to my post about little white lies in math. Only sometime in the last six months did it occur to me that the old chestnut: “angle-side-side makes an ASS of you and me” was really unhelpful. I’ve been living with that wasting a bit of memory space for over two decades!