 In this video, we are going to prove the so-called angle opposite side triangle relation or we'll often just call this AOS for short, angle opposite side. Now, when I do these three letter acronyms like SAS or, you know, like AAS, things like this, this often makes us think of triangle congruences. The angle opposite side relation is not a congruence criterion. It's a relation, but in many ways the AOS does sort of, you can use it in similar ways to triangle congruences, but let me say exactly what it says. So AOS tells us the following for any triangle ABC. If the angle A is greater than the angle B, this happens if and only if the side length BC is greater than the side length AC. So this tells us that the bigger angles coincide with the bigger side, but also the bigger sides coincide with the bigger angles. And so that's what we mean by this angle opposite side relationship. The big side goes with the big angle. The small side goes with the small angle and things like that. And so this observation is extremely useful, extremely critical when we analyze triangles in a congruence geometry. So as this is an if and only if statement, we have to go, we have to do both directions. So in one situation we make an assumption about a side that we prove something about the angle. And in the other direction we have to assume something about the angle and prove something about the side. The direction that we are going to go is this one here. We are going to suppose that the side length BC is greater than the side length AC. So before I go any farther, let me draw a triangle on the screen for us so we can consider that as we go forward. So just some hunky-dory triangle, something like this. And let me label some sides. We'll call this one A, we'll call this one B, we'll call this one C. And so we are going to suppose that BC is longer than AC, which maybe that's not accurate for my drawing here, but really doesn't matter. It's not such a big deal. So we're going to assume that AC, BC is larger than AC. So to accomplish that, we're going to consider the bisector Cp to the angle C, which is going to intersect the line segment AB at some point D according to the crossbar theorem. So there's a couple of things going on there. So we're going to take the bisector Cp, so we're going to take angle C and cut it in half. We know that angle bisectors exist by what we've proven previously. Let me try that again. So the angle bisector would look something like that. Okay. And so then we know that this angle bisector, since it's an interior angle, excuse me, it's an interior ray to the angle ACB. So by the crossbar theorem, this will intersect some point. We'll call that point D. Like so. Cp was the ray, but we really could have called that ray CD. No big deal there. Now, because this is an angle bisector, we know that the angle ACD is congruent to the angle BCD. That's the whole point of using the bisector right there. So then we're going to play around with segment translation. So by segment translation, there exists some point E on the ray CB such that CE is congruent to AC. And we know that point by the original assumption is going to have to sit in between B and E. Because after all, we said that AC is shorter than BC. So we can translate the segment AC onto the ray CB. But because AC is shorter, it's going to have to do the case that E sits in between B and C. So we get that this segment is congruent to that segment right there. And so then if I were just to add a few more lines here, we're going to look at the triangle ACD and then the triangle ECD like so. So I want you to consider these triangles that are on the screen right here. There's the triangle ACD and there's the triangle ECD like so. So looking at those triangles, the segment EC and AC are congruent to each other. The angles C are congruent to each other. And then you also have that the segment DC is shared by both triangles. It's clearly congruent to itself. And so we have right now a side angle side situation. And so by side angle side, these two triangles are congruent to each other. Side angle side gives that to us. So that tells us because the triangles are congruent to each other, corresponding parts of congruent triangles are congruent. So angle A is then going to be congruent to its corresponding side, which is going to be angle CED. So we get that those two angles are congruent to each other. Like we said earlier, E is between B and C because of how we by original assumption that BC was longer than AC. As such, this angle CED that we just talked about is an exterior angle to the triangle BED triangle bed right there. Now, since that's an exterior angle, this tells us that angle CED is larger than both of the remote interior angles. That's a consequence of the exterior angle theorem. One of those interior angles is in fact angle B. So we get that our angle here CED is larger than angle B by the alternating exterior angle theorem. Now, notice in this situation that these angles right here, right? Angle A and angle CED are congruent to each other. That's larger than this angle right here. And so by putting those two statements together, we get that angle A was greater than angle B. And that's then the first direction, right? If BC was greater than AC, we then get that angle A was greater than BC. So that's the first direction. And so I want to give some more space for another drawing right here. But let's go in the other direction. Let's suppose that angle A is now larger than angle B. Now, in that situation, if we do our triangle again, something like the following, and we'll say this time we have A right here, we have B right here, and we have C at the bottom, like so. And this situation suppose that angle A is congruent to angle B. Now, assume for the sake of contradiction that the segment BC is congruent to the segment AC. This would be an Asosceles triangle. So by the Asosceles triangle theorem, that would say that these two angles are congruent, which then gives us a contradiction. So that's not possible. Let's also assume the other situation. So we're assuming that angle A is greater than angle B. And then let's suppose that BC is less than AC. Well, by the very previous part that we just did up here, if BC is less than AC, that would imply that angle A is greater, excuse me, angle A is less than angle B, which is then another contradiction with our assumption angle A is greater than angle B. Now, there's only three possibilities. The segment BC right here in the segment AC, they're either congruent, or one is bigger than the other. They can't be congruent. We took care of that. A can't be less than B. So we actually get that angle A is bigger than angle B. Take care of it. So this is a very interesting proof because what we did here is it's an if and only if statement, right? So we have something like statement X, if and only if statement Y. So it's like, okay, we assumed Y and then we get X. We did that. But then we're like, oh, we're going to assume X, but then we use this statement to prove the other direction. And so we then get Y. So it's interesting because since one of the directions held, you actually can use the direction that you're going against to prove the direction you have. And the reason we are able to do that is because we had this total ordering on line segments and angles. So we are able to go through cases. And many of those cases led to contradictions because of the previous case. It's a very fascinating result where an if and only if statement where one direction actually proves both directions. You don't see that often. But the angle opposite side relation is exactly a situation where you can use that.