 In this video, we are going to state and prove the triangle inequality as a theorem of congruence geometry, which remember in congruence geometry, we accept Hilbert's four axioms of incidence, four axioms of betweenness, and the six axioms of congruence. And so the statement of the triangle inequality is the following. If ab and c are three distinct non-colonial points in this congruence geometry, then the segment ac will be less than the segment ab plus bc. Now I should mention that this statement itself has a lot of jargon built into it. What does it mean to take ab plus bc? It wasn't mean to add together two segments, which in a congruence geometry, we don't necessarily have the notion of measure. So these aren't numbers we're adding together. These are segments we're adding together. Of course, the previous video in lecture 20 made sense of what it means to add together two segments. The idea is you take the segment bc and you translate it onto the ray ab at the point b and thus extending the segment. So we can talk about their sum here. What does it mean for one segment to be less than the other? Well, the idea is this segment, when translated onto this segment, the translation lies in between the endpoint of a and c in that situation. So in a congruence geometry, these three non-colonial points always satisfy the relation ac is less than ab plus bc. So why does it get the name triangle inequality? Inequality makes sense because it is an actual inequality with regard to this inequality on segments here. It's called a triangle inequality, the triangle quality, because really we think of these three distinct non-colonial points as vertices of a triangle. So you have a segment a, a segment b, excuse me not segment, you have the vertex a, the vertex b, vertex c of your triangle here. And so the segments ac is in this segment right here. And the idea here then of the triangle inequality is if you take any two sides of a triangle, their total length is longer than the sum of the other sides. Now in this situation, ac does feel like it's the shortest side here. But if you were to take the side length ab, which kind of looks like it's the biggest length in this triangle, it still is true that if you take ac plus bc, it's going to have to be bigger than ab. And then same thing here, if you take cb, maybe cb is the longest, it's hard to tell here which one's longer in my diagram. But nonetheless, if you take ac plus ab, that's longer than the segment bc. You know, if you were to flatten these two sides into a single segment, it's going to be longer than that one. And so the moral of the story when it comes to the triangle inequality is that if you take a detour, it's always longer than if you take the straight path between any points. So we get this very important triangle inequality here. Now in this statement, I am assuming that the triangle or that the three points are non-colinear. Some people rephrase the triangle inequality to allow for a collinearity in that situation. So in that situation, I should mention that if ab and bc were in fact collinear, then as they're collinear, there's going to have to be some type of betweenness statement between the three points, ab and c. This is a result of trichotomy. And so without the loss of generality, you can assume that b is between a and c in that situation. Then it would then follow that ab, the segment, plus the segment bc would equal the segment ac, right? The way we've defined segment addition, we get that ab plus bc is equal to ac by the segment addition axiom. These are the exact same things. We have equality in that situation. Therefore, the triangle inequality is sometimes written in the following way that if ab and c are three distinct points with no assumptions about collinearity or non-colinearity, then the segment ac is less than or equal to ab plus bc, where equality only happens in the case where ab and c are collinear. If they're non-colinearity, then it'll be a strict inequality. And in fact, I should say that the equality only happens when b is between a and c in that situation. Alright, because notice b is the point that we're acting like is in the middle. Now we're going to prove this situation where it's a strict inequality because then we don't have to worry about all this equality or congruence of the betweenness of things. So we're just going to consider the case where these points are non-colinear, so we get a strict inequality. But it generalizes the other case very quickly by what we talked about just a moment ago. Alright, so let's then look at the proof of the triangle inequality. I think I can leave the statement on the screen while we also look at the proof. So consider the following situation. So let's take the ray ab, like so, and so we'll label our ray. We have a, we have b, and this is some ray, but we also should think of this as the line, as a triangle, right? So we have the vertex c also here as well. So we have our triangle abc here and then think of the ray ab. There is going to exist a unique point d on the ray ab such that ad is congruent to ab plus bc, like so. So there's some point over here, like we called it d. And so we're assuming there's a congruence between the segment, the segment bd is then going to be congruent to bc, like so. You know, imagine we just took this, we took this point and rotated this segment over here. So in particular, ad is going to equal the segment ab plus bc as we define segment addition beforehand. So we need to prove that the segment ad is greater than the segment ac, right? So we want to show that this one is longer than that one. Because if ad is longer than ac, then that shows that since ad is equal to ab plus bc, we get exactly what we're trying to prove right here, okay? I guess we should remember that these segments are congruent. That'll probably come in handy right now, right? Because by construction, the segment bd is congruent to bc. We also have, of course, that the point b is between the segment, is between the points ad. I want us next to consider the triangle bcd, which is not currently drawn here, but if I were to connect it here. The triangle bcd is an Asosceles triangle because we have these two congruent segments. By the Asosceles triangle theorem, their corresponding angles are also going to be congruent to each other as well. So the angle bcd is congruent to the angle bdc. Also, by the between cross lima, we have that b is an interior point to the angle acd. So you have this angle right here, b is an interior point. And that's of course, because b is between ad, which are these bounded values on the angle acd like so. So between cross lima gives us that. So this shows us the following. We have that the angle acd, which let's draw that on our screen right here. So acd, we're considering this angle right here. This angle is larger than the angle bcd. And that's because b is an interior point and that's exactly what that inequality means for angles. But this angle is congruent to this angle because we have an Asosceles triangle. Therefore, looking at these angles here, we have this angle right here. We have this angle right here. We know that this is the big angle and this is the small angle. This is now the time we're going to use the angle opposite side relation, or the AOS for short, which tells us that the big angles coincide with the big sides and vice versa. So if angles, looking at the triangle abd there, excuse me, adc, c is a bigger angle than d. Therefore, their opposite sides, the opposite side to c, which is ad, will be larger than the opposite side to angle d, which is ac, which then proves the triangle inequality.