 In a congruence geometry, we assume side-angle-side as an axiom. In previous videos, we've proven angle-side angle as a triangle criterion. We actually implied that these two are logically equivalent to each other. It doesn't matter which one you take. In the last video, we proved side-side-side as a triangle congruence. In this video, we're going to prove our last triangle congruence, that is side-angle-angle. Now, it's very tempting to say that, oh, side-angle-side-angle-angle follows immediately from angle-side-angle. The difference, of course, is the location of the side. We have two congruent angles. Triangles have three angles, of course. Is the congruent angle between the two angles, or is it exterior to the two angles? That does make a difference. Now, in a trigonometry class, which is probably about Euclidean geometry, oftentimes in that Euclidean geometry setting, we use the notion of angle sums, and we say that, oh, a triangle's angles add up to be 180 degrees. I want you to be aware that in this lecture series, we have not yet introduced the idea of an angle measure. We can't use angle measure to prove this right now, because we don't have it. Then the statement that angle measure adds up to 180 degrees is actually equivalent to the Euclidean parallel posh. That is to assume your angle measure is always 180 degrees. It's to assume your Euclidean. Anything that uses an angle sum of 180 degrees only applies to Euclidean geometry. We are trying to stay neutral with regard to the parallel postulates, and therefore we can't make any assumptions about that, because typically, like I said in trigonometry, what you do is something like the following. You know about this angle, you know about this angle, and you know about this side. Well, since the angle measures add up to be 180 degrees, if you know two of the angle measures, then you actually know the third one, and so you really have a, you can turn a side angle situation into an angle side angle situation. So you can always convert this into that one, but that uses angle measure in Euclidean geometry. What we're gonna do in this video is we're gonna prove it just in congruent geometry. So this is neutral geometry without any notion of continuity. We do have incidents between this and congruence, and in this situation, we can still prove side angle side, thus providing a proof for Euclidean geometry, but also for hyperbolic geometry as well. So how are we gonna do that? Well, well, let's first make sure we understand the statement of the problem here. So we have two triangles, ABC, and another triangle, A prime, B prime, C prime. So we assume that the side ABC is congruent to the side A prime, B prime, and that the angle A is congruent to the angle A prime, and the angle C is congruent to the angle C prime. Under those situations, we then can infer that the triangle ABC is congruent to the triangle A prime, B prime, C prime. And so I wanna sketch the picture of our triangle for us as we work through this thing, okay? So we have our first triangle, like so. We will call this triangle ABC, of course. And so we have ABC, like so. And then we have our other triangle. We'll put it down below here, like so. And so by corresponds, we'll call this one C prime, B prime, and A prime, like so. And so what did we assume about these triangles? We assumed that the segment A prime was congruent to the segment A prime, B prime. We also assumed that the angle A and A prime were congruent. So CAB was congruent to C prime, A prime, B prime. We're also assuming that angle C are congruent to each other. So angle ACB is congruent to A prime, C prime, B prime, like so. So these are the assumptions we have going on so far. Now, in the Euclidean sense, I just alluded to it, in the Euclidean sense, if we could prove that the other angles are congruent, then we could use angle side angle and we would be done. And that's kind of how we're gonna approach this one. What we're gonna do is we're going to then, we want that B, the angle B is congruent to B prime. So we're gonna accomplish that by basically supposing they're not what happens in that situation. So suppose that the two angles are not congruent to each other. The angle B is not congruent to the angle B prime. Now, if that's the case, then one of the angles is larger to the other. And without the Lawson Generality, we can assume that angle ABC is larger than angle A prime, B prime, C prime. Now, if that's the case, if angle B is larger than angle B prime, that means we can translate angle B prime into angle B and it'll sit inside the interior of the angle. So that gives us that there's gonna be some point, let's call that point D. There's gonna be some point inside the angle ABC, and I'm gonna inform the Ray right here, the Ray BD. We're gonna get some point D interior to the angle ABC such that the angle ABD is congruent to the angle A prime, B prime, C prime, like so. So this angle ABD is congruent to the angle A prime, B prime, C prime, like so. So with three marks there to indicate those are congruent to each other. That's what it means for one angle to be larger than the other. Then since D is an interior point to the angle ABC, the cross bar theorem applies and gives us a point of intersection between the line segment AC and the Ray BD. And so let's call that point of intersection E like so. So if look at the triangle ABE versus the triangle A prime, B prime, C prime, notice that in these triangles we have an angle, we have a side, we have an angle that are congruent to each other. And so angle side angle applies to show that those two triangles are congruent to each other. So in particular, the triangle ABE is congruent to the triangle A prime, B prime, C prime. So these triangles are congruent, okay? So next, again, that followed by angle side angle, just say that again. So as corresponding parts of congruent triangles are congruent, that gives us that the angle ABE is congruent to the angle A prime, C prime, B prime. So let's label those angles there. The angle ABE is congruent to the angle A prime, C prime, B double prime. So that's this angle right here, okay? Now by transitivity of congruence, these two angles are congruent. These two angles are congruent. So these two angles are congruent to each other, all right? So in particular, the angle ABE is congruent to the angle ACB. But look at the triangle ECB right here. The angle ABE is an exterior angle to the triangle ECB. But the angle ECB is then a remote interior angle associated to that exterior angle right there. By the exterior angle theorem, that means that ECB should be smaller than the angle ABE, right? But we just mentioned a moment ago that these two angles are actually congruent to each other so they can't be smaller than one another. So we end up with this contradiction. What was the thing that was contradicted of course? The contradiction was that angle ACB was greater than A prime, the angle A prime, B prime, C prime. And the other direction is basically the same thing, right? If you assume the angle ACB is less than the angle A prime, B prime, C prime, by changing the appropriate parts you get the same contradiction. So therefore it must in fact be the case that angle ACB is congruent to angle A prime, B prime, C prime. And then by the assumptions we have here, since angles A and A prime are congruent, side AB is congruent to A prime, C prime. And since angle B is congruent to angle B prime, you get by angle side angle that the two triangles are congruent to each other. So I wanted you to point out here that the proof of this theorem was very similar to how you do it in Euclidean geometry. You argue that since you know about these two angles, you argue that the third angle also is congruent and then it follows by angle side angle. You turn into an angle side angle situation. But we didn't use angle sums here. We actually used the exterior angle theorem which does compare the sizes of angles. So we are able to avoid this issue of angle measure for the moment being at least so that this becomes a theorem of neutral geometry.