 Welcome back to our lecture series Math 3130, Modern Geometry for Students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misseldine. In the start of lecture 23, I would like to officially define what an acute and obtuse angle are. We've done this informally in the lectures before now, but this is the official place where we're going to put such a definition. Be aware that this definition takes place in a congruence geometry. Geometry satisfies the incidence between this and congruence axioms of Hilbert. We say that an angle is acute if it is less than a right angle. We know what it means for an angle to be less than another angle. We can say the angle ABC is less than the angle DEF. We also know that all right angles are congruent to each other. We proved Euclid's fourth posh as a theorem of congruence geometry. This is not an ill-defined concept here. An acute angle is an angle which is less than a right angle. Likewise, we say that angle is obtuse if it is greater than a right angle. Be aware that this definition doesn't require any notion of measure whatsoever. This can be done entirely within a congruence geometry. The notion of a neutral geometry is not necessary for these things, but it potentially could simplify things. For example, in a congruence geometry, it can be proven that the supplement of an acute angle is obtuse and vice versa. Likewise, the complement of an acute angle has to be acute while an obtuse angle doesn't have a defined complement because it's bigger than a right angle, of course. Finally, we actually have shown in congruence geometry that a triangle cannot have two right angles or triangle can't have a right angle and an obtuse angle. Now, a modification of this theorem can be used to show that no triangle can have more than one obtuse angle. Now, of course, for such triangles with multiple right angles or obtuse angles, they can't actually exist not in a congruence geometry. For example, we've seen a triple right triangle inside of elliptic geometry, and it's allowed there because we don't have the between the axioms and therefore in a congruence geometry, we need all the axioms incidence between this and congruence. So things can fall apart if we leave the setting of the congruence geometry, aka elliptic geometry, I should say as an example of that. There are some restrictions on the sizes of angles inside of a triangle. Now, each of these statements becomes very much trivial observations when you use the notion of measure like in a neutral geometry. For example, an acute angle in a neutral geometry, we could set that to just be an angle whose measure ABC is less than 90 degrees. And to be an obtuse angle, we just make it be greater than 90 degrees. And so then statements about supplements of an acute angle being obtuse is very trivial, right? Because if you have the sum of two angles that adds up to be 180 degrees and one of them is less than 90, then the other one has to be larger than 90. This is just a consequence of arithmetic. The fact that obtuse angles don't have compliments. Again, if your angle measure is already greater than 90 degrees and you subtract that from 90, you're gonna get a negative angle measure. That's not an angle. And therefore all of these observations are very, very trivial in the context of neutral geometry, okay? Now, the titular topic for this lecture 23 is the so-called Sicari-Lejeanre theorem, which is actually a distinct result in neutral geometry that doesn't have a necessary counterpart in congruence geometry. That is to say the things we're gonna prove, well, specifically the Sicari-Lejeanre theorem that we're gonna prove in this lecture is something that doesn't have a corresponding result in congruence geometry. Well, things like we can prove things about obtuse angles and acute angles. We can do it in congruence geometry without measure. It just gets a lot harder. There are gonna be some things that are impossible to prove without measure of some kind. And therefore neutral geometry is the necessary setting. And so the Sicari-Lejeanre theorem is essentially our first result that truly transcends congruence geometry and does require some notion of continuity here. And so to prepare ourselves to prove the Sicari-Lejeanre, there's two lemmas we're gonna prove first. The first lemma, this will be a lemma of neutral geometry, which tells us that the sum of measures of any two angles of a triangle is gonna be less than 180 degrees. Now, this lemma is often taken to be the exterior angle theorem for neutral geometry, which in that setting, if you have a triangle and an exterior angle like this, typically the way that the exterior angle theorem is phrased is that this angle right here, the exterior angle, is larger than the sum of the two remote interior angles like so, for which as this is an angle, it's less than 180 degrees than these two angles. There's some has to be less than 180 degrees. And so the proof of this lemma is gonna use the exterior angle theorem as a result from congruence geometry. And therefore we're basically modifying it to be the neutral version of the exterior angle theorem. That's not exactly what we're proving right here. I just wanna mention that that's essentially what we're proving here with a little bit more misogyny of the problem we get that statement as such. And so, okay, we wanna prove that sum of measures of any two angles of a triangle is always less than 180 degrees. So let's actually draw our triangle like we had on the screen a moment ago. I'm gonna try better to make it look like a triangle. Okay. And so we have our triangle A, B, C. We'll call this one A, B, and C. We'll put C over here. And so let's consider starting it in three, two, one. And so here's our triangle. We'll label it A, B, and C like so. And so in order to talk about exterior angles, I'm gonna also put a point D on there such that B is between C and D on this line. And I want us to consider the angles A and the angle B, which they're not necessarily congruent to each other, but I want us to consider those two angles. And we're gonna show that the sum of angle A and angle B, which in case there's any ambiguity here when I talk about angle B, I mean angle C, B, A, that is the angle of the triangle. We wanna show that the sum of those two angles is going to be less than 180 degrees. And this is how the exterior angle theorem comes into play here. When you look at the angle D, B, A, that is an exterior angle to the triangle A, B, C. Therefore, the angle D, B, A is gonna be greater than the angle A by the exterior angle theorem, okay? But we also have that angle C, B, A and D, B, A, these are supplementary angles to each other because after all D, B, A is a exterior angle to the triangle here. And therefore the sum of the two angles, the angle A, B, C plus the angle A, D, A, D, excuse me, A, B, D is equal to 180 degrees. They're supplements to each other. And so when you combine those two observations, this and this, we get the very simple argument here that 180 degrees, that's the same thing as the measure of A, B, D, A, B, C plus A, B, D, like so, but A, B, D is less than angle A, so like so, and be aware that angle A, B, C was the measure of angle B from the triangle right here. And so we get this very quick observation that the measures of angles A and B are less than 180 degrees. And this pairing A and B together was arbitrary. A similar argument would show that if you pair A and C together, you'll get less than 180 degrees. If you pair B and C together, you'll get less than 180 degrees. And so this is essentially a consequence of the exterior angle theorem. Now, I want you to be aware that when we prove this lemma right here, we never really needed neutral geometry. That is to say, we didn't really use continuity. We didn't need measure. We talked about measure, but we didn't need it. And while measure definitely simplifies the argument, it wasn't necessary in this. And so really this lemma here could be written as a lemma of congruence geometry. And so this is what I was starting this lecture with when we talked about the acute and obtuse angles. That many theorems of congruence geometry are true, of course, because they're theorems, but in the presence of neutral geometry, the proofs get dramatically easier because measure, we're actually adding together real numbers as opposed to adding together angles or something, which is a much more complicated notion here. So yes, this lemma, this extended version of the exterior angle theorem is in fact a theorem of congruence geometry, but measure makes it so much easier. So let's prove a second lemma that we need in order to prove the Siccario-Legioner theorem. And we're also going to cast this in the setting of neutral geometry. Now for this lemma, do we need it to be neutral geometry or can we get away with it just being in congruence geometry? Well, the proof is gonna be dramatically simpler when we use the notion of measure. And since we're gonna need continuity eventually, there's really no point in not doing it right now, I suppose. Now, you could modify this proof, I do believe, to not need any notion of measure, and you could do this just in congruence geometry, but I'll leave that as an exercise to the viewer here. What we're gonna prove for this lemma here is if we have a triangle ABC, we can construct a different triangle whose vertices we'll call A1, B1, C1. Now, if you're following along here, you might wonder why do I call them A1, B1, C1? Why not like A prime, B prime, C prime? Well, here the subscript is actually suggestive of a sequence that we're gonna do when we do the proof of the securit-lajana theorem. And so this is really just the first triangle and a sequence of triangles. It really is gonna be a calculus argument here. So anyways, getting back to the statement of the lemma, if you have a triangle ABC, then there exists another triangle A1, B1, C1, such that the two triangles have the same angle measure. That is to say the sum of the angles ABC is equal to the sum of the angles A1, B1, C1. But then also we have that the measure of angle A1 is gonna be less than or equal to one half the measure of A. And I want you to be aware that it makes sense to talk about, it makes sense to say that one angle is less than another angle. We don't need measure to do that. Now, of course, since we're talking about the measures of the angles, we're talking about real numbers right now, positive real numbers. So it's easy to describe what less than means there. It gets a little bit more complicated when you talk about less than an angle, but we can do it. And then one half of an angle, well, that has to do with its bisector. So I want you to be aware that the statement, the statement of this lemma here can easily be rebranded to be inside of congruence geometry, okay? But again, we're not gonna bother with that inside of this proof here. So let us then consider the situation that we have here. Okay, so we're gonna have a triangle, ABC. And so we need to construct a new triangle, which is gonna have the same angle sum as ABC, but half the measure of angle A. And so the key to that is we're gonna use midpoints. Midpoints do exist in any congruence geometry, yes. But in the presence of neutral geometry, it's much easier to prove the existence. But we already have the existence. We don't need to prove that again or anything like that. So we're gonna let D be the midpoint of the segment BC. So in particular, BD is congruent to DC like so. And we're then going to take our line here. We're gonna take the line segment AD. And we're gonna extend that to some point E over here, so that these segments AD and DE are congruent to each other as well. So the existence of such a point D and E is the consequence we have in congruence geometry, but in fact, neutral geometry, it's a much easier thing to say here. Also, I'm gonna add the segment BE to my picture here because I'm actually gonna talk about some triangle congruences right now. So in particular, there's a couple angles that we have that are congruent to each other, or at least one that's worth noting here. Right now, we have that angle ADC and EDB are congruent to each other because these are vertical angles like so. And so now at this moment, we can then say that the triangles ADC and EDB are congruent to each other by a side angle side. So which triangles are we talking about? So you have ADC, which is this triangle right here, ADC. And you have the triangle EDB, which is this triangle right here. Why they congruent to each other? Well, we have side angles side. And so those triangles are in fact, congruent to one another. This triangle is congruent to that one. And as such, corresponding parts of congruent triangles are congruent. And so this is gonna give us the congruence of angles that I was referring to. So we have angle CAD is congruent to BED. So where are those angles in our diagram here? You have CAD right here, it's congruent to BED, which would be right here. And then likewise, we have angle EBD is congruent to ACD. So ACD is right here and EBD is right here. So we have those angles are congruent to each other. After all, these two triangles are congruent to each other. So all three of their angles are congruent to each other. That's gonna be useful to us as we go forward here. Now I want us to consider the triangle. We're now gonna define triangle A1, B1, C1. We're gonna define that to be the triangle ABE. So notice we started with the triangle ABC, which is this triangle right here. Our triangle ABE is this triangle right here, okay? And I claim that this triangle ABE has the same angle sum that the original triangle ABC has and let's see that. So if we start off with the angle measure ABC of the original triangle, let's prove that this is the same thing as ABE right here. So when we talk about angles AB and C less B specific, which angles are we talking about? Cause with the extra line segments we've added here, there's some ambiguity what one means, but angle A, angle B, et cetera right here. So by angle A, we're referring to this one. This is the original angle A right here. So this is the angle BAC. The original angle B was this one right here. So that'd be angle ABC. And then while there isn't necessarily any confusion with angle C with our diagram, this is our angle over here, BCA, like so. So we have those three angles. So what can we say about these angles here? Now the angle BAC, which is this angle right here, we can dissect it into the two pieces. One of them right here is CAD. The other one is BAD like so. So we can break up this angle A into these two angles, leave the other two alone for a moment. So note here that angle CAD, we already provisely showed was congruent to BED. So these two angles because they're congruent to each other, their measures are equal to one another. So we can make that substitution. We also have that angle EBD is congruent to angle ACD. Which remember, ACD is this angle right here, but that's the same thing as angle ACB or BCA, which you prefer. And so that tells us that angle BCA and angle EBD are congruent to each other. So their angle measures are equal. So we can make that substitution. No big deal there whatsoever. So now we're gonna rearrange things a little bit. So we have this sum right here. So I'm gonna move angle BAD to the front. We're gonna then move, we're gonna put the EBD and the ABC together. And then we're gonna move angle BED to the end right there. Again, just some rearrangements why we're doing that. Well, because these angles are gonna combine together. If we take angle ABC, which is this angle right here and combine it with angle EBD, that's this angle right here, this together combines to become angle ABE. Which notice ABE, this is the angle B in the triangle ABE, right? Which is what we mean by angle B1, right? Then what about angle BAD? Angle BAD was this angle right here. That's exactly angle A1, note there. And then BED, where is that on our diagram? That's over here, BED. In this triangle, the vertex E coincides with C1. And so making that substitution, we get exactly that BED will become angle C1 right there. And so just by a little bit of rearrangement, the congruences we have already, we then get that the angle measures are identical to each other, okay? And so that's a very important observation here. So when we look at our triangle one more time here, ABC has the same measure, angle some measure as ABE right here, okay? And so now let's fixate on angle A for a moment, okay? Now when you look at this angle, angle, the measure of angle A, it's the sum of the measure of angle A1 right here and C1 right here. Cause after all, this angle DAC is congruent to the angle DEB that we talked about earlier. And so this one right here is what we called this is the measure of angle A1 like we said earlier and this is the measure of SC1 that we talked about beforehand. Now the sum of the measures of angle A1 and angle C1 gives us the original angle A. And so we have this and this is a real number for which if two numbers add up together to give us a whole, that means that one of the numbers has to be less than or equal to half of the whole, right? Cause if both the measure of A1 and the measure of C1, if both of those were at least half of the measure of A, then their sum would give us something larger than the measure of A, that doesn't make sense. So basically when you partition these things, one of them has to be less than or equal to half of the whole thing. Now, if that was A1 that we labeled it great, no big deal. If it turns out that C1 was the smaller one, it was less than or equal to half, then we can actually swap the labels A1, C1, B1 can stay where it's at, right? We can swap these labels, we can call that one A1, we can call this one C1 if we have to, it doesn't matter. But one of those two angles will be less than or equal to half the measure of angle A. And that's what we need in this situation. Thus proving the lima. So now we're in the situation where we're ready to prove the Sicieri-Lejeune theorem, for which the statement here in neutral geometry, the angle sum of any triangle is less than or equal to 180 degrees. Now, in this argument, we really are gonna use continuity. Like I said, in the previous two lemmas, we used measure, but that was more of a convenience. Could we have done it? Probably, like the first lemma, we were very explicit on how we could change it. In the second lemma, you know, we probably could have gotten away with it because we talked about half of angles, those are bisectors, yada, yada. It makes the argument a lot harder, but we could have probably done it. With the Sicieri-Lejeune theorem, we do need a notion of continuity. In particular, the Archimedean principle is gonna come into play here. So if we don't have the Archimedean principle, then this theorem might fall apart because the proof we're gonna provide does use the Archimedean principle. Now, what that does tell us is that sure, we don't necessarily need full-blown neutral geometry. If we just have an Archimedean congruence geometry, such as the rational plane, right? That's a congruence geometry that has the Archimedean principle. It doesn't have circle continuity, or it doesn't have full continuity like the Dedekin axiom that we take for neutral geometry, but this proof would be valid in that situation as well because it's really the Archimedean principle that's gonna come into play here. All right, so in neutral geometry, the sum, the angle sum of any triangle is less than or equal to 180 degrees. And so you can kind of see why someone might be interested in such a theorem here. To give you some historical background, the point of this theorem was to try to prove the Euclidean parallel postulate as a theorem of neutral geometry. That was what many people like Sikeri were very, very interested in proving that the Euclidean parallel postulate was a theorem of neutral geometry. Now, we will see in the future that the statement that the angle sum of a triangle was equal to 180 degrees always in neutral geometry's equivalent to the Euclidean parallel postulate. So Sikeri was trying to tackle this idea of the triangle sum theorem here, that he wanted to prove that the angle sum of every triangle was equal to 180 degrees, because then that proves by equivalence, again, we'll prove that later, that the Euclidean parallel postulate is a theorem of neutral geometry. But Sikeri was not able to do it. The Sikeri-Lijonov theorem has a very important escape hatch here. The angle sum is less than or equal to 180 degrees. So we will see soon that the angle sum of a triangle equaling 180 degrees is equivalent to the Euclidean parallel postulate. But what about less than, okay? Well, if the angle sum is strictly less than 180 degrees, so it's not equal, but it's less than 180 degrees, we're gonna see that in neutral geometry, that's equivalent to the hyperbolic parallel postulate. Now, with regard to elliptic geometry, you might guess that in elliptic geometry, the angle sum of a triangle is greater than 180 degrees, and in that appropriate setting, the two notions are equivalent to each other. Now, the proof of the Sikeri-Lijonov theorem falls apart in elliptic geometry because we don't have the exterior angle theorem. I should point out that the original, the first lemma we proved in this lecture 23, used the exterior angle theorem, which is not valid in, it's not valid in elliptic geometry because notions of betweenness are falling apart there. So that's the main kicker in such a situation. So right now, we're in this neutral geometry, we're remaining neutral on our parallel postulates. We're not assuming the hyperbolic or Euclidean, elliptic we know is not the case because the elliptic geometry is incompatible with congruent geometry, it's incompatible with neutral geometry here. So let's get to the proof of the Sikeri-Lijonov theorem. Why is the angle sum of a neutral triangle always less than or equal to 180 degrees? Well, turns out we're gonna have a geometric proof where I'm not even gonna draw a picture this time because this is honestly just a calculus argument. We're gonna assume, well, I should say we're gonna assume the opposite. So for the sake of contradiction, suppose that the sum of angles and some neutral triangle ABC exceeds 180 degrees. Therefore the angle sum of our triangle is 180 degrees plus X something and that X is a positive number. So this X is measuring the excess beyond 180 degrees. So by the limit we just proved, which in our lecture that was lima 353 here, if we have this triangle ABC, then we can construct a triangle A1B1C1 whose angle sum is the same as the original triangle ABC for which we can consider this the triangle A0B0C0 like so. This new triangle will have the same angle sum as the original triangle. So its angle sum will be 180 degrees plus X, but the measure of angle A1 will be at most one half of the measure of angle A0. Okay, then if we apply that lima again, so we applied the lima to triangle ABC to get triangle A1B1C1, but we can then apply that same lima to triangle A1B1C1 to construct a triangle A2B2C2, which what do we know about that triangle? Well, the angle sum of triangle A2B2C2 will be the same as the triangle A1B1C1, which I'm just gonna call that one triangle one. We'll come and call this one triangle two. The original triangle we'll call that triangle zero, okay. So triangle zero, its angle sum is 180 degrees plus X, so triangles one also had that angle sum. Triangle two will also have that angle sum. It'll be 180 degrees plus X, but the measure of angle two, excuse me, A2 in triangle two will be at most one half the measure of angle A1 in triangle one, which what do we know about the measure of angle A1? It was at most one half the measure of angle A, so therefore the measure of angle A2 will be at most one fourth the measure of angle A in the original triangle. And so then by induction, we can continue to construct these triangles. By induction, there exists a triangle whose vertices are given as A, N, B, N, C, N. For simplicity, we'll just call this triangle N right here. And what can we say about triangle N? Its angle sum will be the same as the original triangle zero. Its angle sum will be 180 degrees plus X, but the measure of angle AN will be less than or equal to one over two to the N times the measure of the original measure angle A, like so. And so this inequality is really what we're looking for here. The idea is that we can take triangles whose angle measure is getting smaller and smaller and smaller, that that is the measure of a specific angle. The sum of all the angles always retain to be 180 degrees plus that sum value X. But angle A is getting smaller and smaller and smaller and each iteration we have in this sequence of triangles. Okay, this is now where the Archimedean principle comes into play here. By the Archimedean principle, there exists a sufficiently large number N, we're gonna call that capital in here, such that the measure of angle AN is less than or equal to one over two N times the measure of angle A, right? Which that is then less than X. So we can, since these things are getting smaller and smaller and smaller, there's gotta be a point that our angle measure gets smaller than the measure of this X, this XS. And I don't care how small it is. You know, if X is 0.0000000001, whatever, then if we take sufficiently large N, eventually the measure of angle A will be smaller than that value X right there. This is because of the Archimedean principle here. If you have a known Archimedean geometry, it's very possible this might fall apart because it could be that, you know, you maybe have some infinitesimal value or something you can't get smaller than that, but I'm not gonna delve too much into a non-Archimedean geometry right here. The Archimedean principle does guarantee that eventually we can make this angle get smaller than X. And so then look at the angle measures here. So the measure of angle AN plus BN plus CN by the induction statement, this has to equal 180 degrees plus X. Well, then what happens if we subtract the measure of angle AN from both sides, the measure of AN length? So, well, then it'll cancel on the left-hand side, you're just left the sum of BN plus CN. But then on the right-hand side, we're gonna put that together with the X here. So you have 180 degrees plus X minus the measure of AN. But remember, by construction here, the measure of AN is less than X. So when you subtract the two, this is a positive number. This is greater than zero. And so you have 180 degrees plus something positive. Therefore, the sum of those two things is still larger than 180 degrees. But this is where then the first lemma comes into play here. So this sequence of triangles, we use the second lemma repeatedly. But then when you look at the first lemma, the first lemma we proved in this lecture says that the sum of any two angles in a triangle is less than 180 degrees. But we have constructed a triangle that the sum of two angles is greater than 180 degrees. That's a contradiction. What was the contradiction? Well, the contradiction was that the sum of the three angles was less than, excuse me, was greater than 180 degrees. And so that's what's kind of cool about the Archimedes principle in this situation is that we were able to push the sum of three angles into the sum of two angles because again, because the Archimedes principles, we were able to take smaller and smaller and smaller angle sums. And therefore, if a sum of three angles of triangles greater than 180 degrees, it must have in fact been that the sum of two of the angles was larger than 180 degrees because the third angle is actually negligible. We can make it as small as we want. Before we close this lecture here, I wanted to provide one very small, short application of the Sicari-Lijonar theorem. And this is to the notion of a quadrilateral, which we have informally talked about quadrilaterals before, but we should formally define it right now. And that's sort of like an appropriate place to end because that's how we started this lecture. We formally defined acute and obtuse angles, even though we've talked about them before, and therefore in the lecture in the same way, what is a quadrilateral? So this is a definition that makes sense in ordered geometry. A quadrilateral, A, B, C, D, is gonna be the union of two triangles. In particular, the quadrilateral, A, B, C, D is the union of the triangles, A, B, D, and C, B, D, which the intersection of the two triangles is only gonna be the side, B, D. That's how they overlap. So what I mean is something like the following. If we have a quadrilateral, maybe we have something like this, right? The end, so we call this A, B, C, D. It's a quadrilateral because it can be dissected into two triangles. There's the triangle A, B, D, and there's the triangle C, B, D. And the overlap between the two triangles is just the segment B, D. So if we have triangles like this, this is not considered a quadrilateral because their intersection is more than just a side. And likewise, something like this is not a quadrilateral either because the intersection needs to be exactly the common side between the two. You can't have one being longer than the other. But I do wanna mention that something like the following is an acceptable quadrilateral by definition. So you have something like A, B, C, D, like so. Therefore, this quadrilateral is a union of two triangles. This is perfectly fine. This first example would be an example of a convex quadrilateral as opposed to this one we have some concave quadrilateral. That is accepted as part of this definition. So one has to be very cautious when you prove things about quadrilaterals. Are you assuming convexity or does your proof also apply to concave quadrilateral? Oftentimes by default, we prove things about convex quadrilateral and thus limiting our theory unnecessarily. Now, when it comes to quadrilaterals, we do have to be a little bit more careful about angles. What do we mean by angles here? So in a quadrilateral, when you talk about the angle A, you're referring to the angle BAD, which would be this angle right here, no big deal. When it comes to angle C, that's also easy to describe. You're talking about the angle BCD, which would be this angle here. And whether you're convex or concave, there's no confusion what we mean about angles A or C right here. What does it mean to talk about angle B or angle D? So when it comes to a quadrilateral, angle B of the quadrilateral is gonna be the union of the angle ABD with the angle CBD. So we put together this angle right here and this angle right here. Its union is what we mean by angle B, all right? And that's also what we do for angle D here. It's gonna be the union of angles ABD and CBD right here. So angle D is then the union of those two. Okay, that kind of makes sense. But again, one has to be cautious when you're in the setting of a concave quadrilateral. Because sure, with our diagram here, here's angle D, but look at angle B. Angle B would be angle ABD plus CBD, like so. So when you take the union of those angles, you actually get an angle that is larger than 180 degrees with regard to its measure, right? And so that's something we do have to allow for inside of a quadrilateral, particularly concave quadrilattles, do allow angle measures to be 180 degrees, greater than 180 degrees in this setting right here. But nonetheless, what we can prove is the following. A corollary to the securely genre theorem is that the angle sum of any quadrilateral will be less than or equal to 360 degrees, which you'll notice 360 degrees is two times 180 degrees. It's the angle sum of two triangles. And when you look at it in this perspective, it's not too hard to see a quadrilateral as the union of two triangles. If we take the sum of these angles that has to be at most 180 degrees, if you take the sum of those three angles, that's at most 180 degrees. And when you add those two numbers together, you get something that's at most 360 degrees, which in particular, if you come to a concave triangle, the sum of these three angles is at most 180. The sum of these three angles is at most 180. So the sum is gonna be at most 360. But in particular, this angle right here, well, boom, is a huge angle larger than 180 degrees. But nonetheless, the sum of all the angles is gonna be less than 360 degrees. Analog statements can be said for higher polynomials, not polynomials, polygons, excuse me, for which we can define pentagons and hexagons, where what's a pentagon? It's gonna be a quadrilateral union, another triangle for which those triangles and basically you take three triangles together that only share an edge. You can do that for hexagons. You add another triangle, heptagons, octagons, nonagons, decagons, you get the idea. We can just keep on adding more and more triangles. Just make sure their intersection is only ever an edge when you join these things together. Every polygon can be dissected into triangles. And so that's how we can define these things for higher number of sides. And analogous statements can be there as well. So the idea here is that a quadrilateral has four sides and so two of course is four minus two. So for an n-gon, that is a polygon with n sides, we can expect that their angle sum is going to be less than or equal to n minus two times 180 degrees. And that's a statement that follows by induction using the Sicari-Legionner theorem.