 Hi, I'm Zor. Welcome to Unizor Indication, the kingdom of knowledge. Today we will talk about triangle inequality. I will explain what the triangle inequality is in a simple case, first of all. And it actually states the following, that any two sides of a triangle, if you add together their lengths, it will be longer than the third side. So any triangle, whatever the triangle is, acute angle, obtuse angle, right angle, longer sides, shorter sides, whatever it is, for any triangle, some of the lengths of two sides is always greater than the lengths of the third side. Alright, now, how to prove it? It's actually based on the previous lecture, I was talking about if in the triangle you have one angle greater than another angle, then the opposite side will be longer than the opposite side corresponding. So let me just remind it here. Okay, here's our original triangle ABC. So what I'm saying is that if one side, if one angle, let's say this angle is bigger than this angle, then the corresponding opposite side, this will be bigger than this one. Alright, so based on this theory, I will prove that in the triangle ABC, sum of AB plus BC will be greater than AC. I can always assume that AC is the longest side of the triangle. It's just easier this way. Why? Because if two sides of a triangle in sum gives you a greater length than the longest side, then obviously in the same triangle, any other sum of any other two sides will be longer than the third side. So two smaller side added together will be greater than the bigger side. Alright, how to prove this particular theory. Let's extend side AB to a segment BD equal to BC. So now BDC is an equilateral triangle with these two sides equal to each other. What's interesting is that sum of two sides, which I'm considering AB plus BC, is exactly the same as ABD. Why? Because AB is the same piece and instead of BC, we just turn it around and now we have a bigger segment AD, which has the length equal to sum of lengths of AB plus BC. Since BC and BD are equal. Okay, so our theorem says that sum of two sides AB plus BC should be greater than AC. Now instead, I will prove that AG is greater than AC. Since sum of these sides, AB plus BC is the same as segment AG's sides. Alright, so now consider the triangle ADC and I would like to prove that AD is greater than AC. Now, you remember from the previous lecture that in the triangle against a bigger angle lies the bigger side. Now, what's the angle opposite to AC? It's this one. What's the angle opposite to AD? It's this one. But here is a very interesting consideration. Instead of proving that this angle is bigger than this, which would prove that the whole side AG is bigger than AC, which we need. I will use the fact that BGC is equilateral triangle and that's why the angle BGC is equal to angle BCG. So these two angles, this one and this one, are equal to each other because BCG is an equilateral triangle and angles at the base opposite to equal sides are equal. And now it becomes obvious that this angle or this one, which is equal, is smaller than this because, again, we go back to this consideration that one angle lies inside another angle. So BCG is inside ACG. Why? Because the point B is inside the angle ACG. Again, these considerations might actually seem not exactly rigorous, but it can be proven more rigorously. We just don't want to spend so much time on this. But using Hilbert axioms, it can be proved more rigorously. On the level of Euclid, that's enough, basically, and intuitively just obvious that if this point B is inside the angle, then the whole angle BCG is inside angle ACG. Since it's inside, it's part of it, it's smaller. So angle BCG and, therefore, BGC equal to it, smaller than angle ACACB. And that's why this side is less, this side, which is supposed to be proven. Okay, so any two sides of a triangle have combined lengths greater than the third side. Now, let's do some more interesting logical considerations. Let's consider you have two points on the plane, and they are connected with a chain of segments. This is how some not exactly smart guy would travel from point A to point B, and the smarter people on the plane would take the direct route from A to B. Now, we intuitively know that direct route is always shorter than any kind of segmented and turning direction, etc., broken in different directions. Now, how to prove that the sum of all the lengths of segments which constitute this chain of segments, chain I mean the end of the first segment coincides with the beginning of the second segment, end of the second segment coincides with the beginning of the third segment, etc., etc., and the end of the last segment is our point B. So how to prove that this chain of segments is longer than direct segment from A to B? Well, it's actually quite simple based on the theory which I can just prove it. And here is how. Let's consider first two segments from here to here, and from here to here. I can always replace it with a shorter way. I will go directly from this point to this point, and it will be shorter because of the triangle inequality. So instead of these two segments, I will use just this one. Now, instead of one, two, three segments, again, I will go directly to the very good picture. Let's consider our segment is a little bit more broken down. So instead of going from this to this, and from this to this, I can go directly, and this segment would be shorter than some of these two. So I have shortened my way again. So I will repeat this process instead of going from here to here, and then to here. I will go directly, and this segment is shorter than some of these. So I am shortening my lengths on every step I am making, and I am straightening the way how I go from A to B. So the next one will be the replacement of this. So every time I am replacing two segments with one, I am shortening my lengths. Now from this to this, I will shorten the gate by going straight, and finally, instead of going in two segments, I can go straight from A to B. So basically, what I wanted to say is that using this process, we can always shorten our way from A to B on each step. Since my total lengths has been shorter and shorter, shortened and shortened on every step, my final result, which is the direct line from A to B, the segment which contains along the direct line, is shorter than the original chained way. Okay, now is it as strictly a rigorous proof? Well, not exactly, not exactly, because you see the process which I have just described, it's more of a human process, I would say. It's not a rigid mathematical logic. Who knows, maybe this process after segment number 1000 in my chain would not actually be the same type of process which I started. So I can always talk about certain finite number of steps, and I can prove whatever I just did with whatever number of points I had, like five, for instance, five segments, and for a chain of segments which contains, let's say, 25 segments, I have to basically repeat this logic again, because the logic has certain steps. Now, you probably remember there is a wonderful mechanism in mathematics, it's called method mathematical induction, and this method can help us to deal with the problem of any chain of segments which contains any number of segments. Well, how to do it? Well, it's actually quite easy, you remember that mathematical induction contains three different steps. Step number one, we check for n equals some initial value. Now, number two, we assume that the theorem is true for some number n, or we are usually saying not just for n, but for n equals k. And then three, based on this, we prove for n equals k plus one. I hope you remember it, if you don't, go to the corresponding lecture which I have, among basic principles of mathematical induction, and repeat it. But in theory, it's quite obvious, if I have proven that some statement where n participates, certain number of something, if this statement is true for some initial value of n, let's say for n equals to one, or for n equals to 10. And then I have proved a theorem that from the validity of this statement for certain n equals to k follows the validity for n equals k plus one. It makes sense then to state that this is basically the end of the proof for any n, because we can start from this initial point which we have basically checked that the statement is true, let's say for n equals 10. And then considering that we already know for n equals 10, and we have proven that from n equals k follows n equals k plus one, then from 10 we can basically say that if it's true for n equals 10, it means it's true for n equals 11. Then, since n equals 11 is proven as well, it follows for n equals 12, et cetera, et cetera, and that's why the theorem is true for any finite number. Now, we will use this approach in this particular case to be a little bit more rigorous with our proof. So, let's consider we have a chain which contains any number of segments, and the number of segments would be n. Now, for a chain which contains only two segments, the theorem that the sum of lengths of these segments is greater than the erect way from a to b is basically a triangle in equality which is proved in before. So, for n equals 2, we have proven it. For initial value of n equals 2, when the number of segments in the chain is 2, we have proven our theorem. So, its statement is true. Now, what if you have any number of segments n? Then, how to prove it for n equals to k plus one if for n equals k is already proven? Well, very easy. Let's just connect, let's say, point A with the end of the second segment. Draw this particular line. Now, what have we done right now, converging this chain into this chain? Well, we have reduced the number of segments. Instead of these two, we have one. Now, no matter how many segments here we have, let's say it's k plus one segments, then reducing by one, we will have n equals k segments. So, this particular logic works fine because if we assume that the theorem is true for n equals k, that means that this particular chain of segments is actually longer than this line. But this chain of segments is longer because it's exactly the same in this particular part, but in this particular part is longer than one particular segment. We have replaced two segments with one segment, so this way will always be longer than this. So, if this theorem is true for n equals k, which means if the length of this segment, the chain of segments, is longer than the direct line, then it will definitely be true that for k plus one segments, it's even bigger because we have just changed one segment to a longer combination of two segments. So, if we assume that for n equals k, the k-based chain of segments is longer, then it's obvious, using the triangle in ecology, that k plus one segment will be even longer. So, that's why we have proven, basically, the theorem relatively rigorously that any chain of segments would be longer than the direct segment correcting two points. So, smart people go directly from A to B on the plane and going along any kind of a chain of segments would be longer. No matter how you try to do it, you cannot reduce the lengths of a direct way, which obviously every people knows. Okay, now, what if you go not using these direct segments, a chain of segments, what if you want to go by some kind of curved line? This is more difficult now to supply. And I promise that I will not be exactly rigorous. So, how to prove that any smooth line would be still longer than the direct line? Well, now we have to define what is longer. So, we have to define what is the length of a curve, which is not easy. On an intuitive level, we understand that if we will take a certain number of points on this curve and replace this curve with a chain of segments connecting these two points, it will approximate the curve. The more densely we position these points on the curve, the more number of segments, obviously, we will have in this chain of segments which approximates the curve. The closer will be the approximation to the real curve. So, basically, we can say, we can continue actually this logic, talking about certain limits. If you add lengths of all these segments which constitute the chain approximating the curve, well, the more densely populated with these points curve is, the closer it is, some of these lengths of these segments to the lengths of the curve itself. And then the limit of this with the density going to infinity and distance between the points going to zero, the limit can be called the lengths of the curve itself. But this is something which is significantly higher if you want to do it on a really rigorous way than the way we would like to be on right now. However, it still gives you some kind of a sense of how the whole theory might be extended towards curve. Since we have proven our theorem that the total lengths of all the chain segments are greater, is greater than the direct segment between A and B. And considering that any curve can be as good as possible approximated with the chain of linked segments, then it obvious that, well, again, it's obvious on certain intuitive level. It's not obvious on the rigorous mathematical level that the lengths of the curve will also be greater than the lengths of the direct line. So this is some kind of a glimpse into a little bit more difficult pieces of logic. It's related to the theory of limits, because we have to prove actually that the limit exists depending on how we populate these points in the curve, etc. So we don't go into the more details, but please understand that this is the way how to approach the same kind of logic in the curve world. All right. What else? We still have one more theory which I would like to prove. Let's consider two triangles, which have one pair of sides congruent to each other and another pair of sides congruent to each other. But these angles are different. Now, the point of this theory is to prove the following. If you have a situation like this, when this particular triangle has two sides congruent to each other, AB congruent to ED and BC congruent to EF, but angles are different, then the statement of this theory is that if this angle is bigger than this, then the opposite side is bigger than the opposite side here. It's kind of obvious from the drawing, but now we have to prove it rigorously. Okay, to prove it, we'll do the following. We will basically duplicate this triangle inside this. This side, GE, will coincide with AB. This EF will coincide with BM and angle will be equal to angle. Okay, now it looks more or less close. So triangle ABM, ABM is congruent to GEF. Angle is equal, side AB is congruent and side DM is congruent to EF. So now, what I have to prove is that, well, side DF is obviously congruent to AM. So I need to prove that AM is shorter than AC. Okay, what is important in this case? Since this angle is smaller than this one, then BM is inside the angle ABC. Again, we are referring to this concept of being inside, smaller and being inside, kind of very much related to each other. So BM is inside AC, angle ABC. All right, now to prove that AC is bigger than AM is really quite easy. What we do, we are using actually this concept of being inside. Since it's inside, we can consider angle MBC and draw a bisector of this angle and connect M to M. What's very important right now is to understand that this segment MN is congruent to segment NC. Why? Well, that's actually very simple because the triangle MBN and CBN are congruent. Why? Because this side is exactly the same. We built it congruent to this triangle and this side is also the same by premises of this particular theorem. And since these two angles are congruent because it's a bisector, we have side, angle, and side which they share in a triangle MBN and NBC. Side, angle, side, side, angle, side. So we have the congruence between these two triangles and that's why MN and NC congruent to each other. And that's why the length of AC is equal to sum of AM plus NM. That's obvious too, right? So we just took this particular piece and turned it this way. But now it is also obvious that AM plus NM, since these are two sides of a triangle AMM, some of them is greater than AM. So AM plus NM is greater than AM. Now, since AM is the same as NC, we have the same thing for AM plus NC. But this is AC. So instead of this part of this, we consider this. So AC is greater than AM. And that's exactly what needs to be proved. So if this angle is smaller than this angle, then we have constructed the congruent triangle to this, which is inside. And that's why we can build this little construction to help us to understand that the length AC is greater than AM. That's the end of this particular proof. Now, there is a converse theory that in this particular situation of two triangles with equal pairs of sides, what I said was that if this angle is smaller than this, then the side is smaller than this. Same thing can be done in reverse. If this side is smaller than this, then the angle is also smaller than this. And to prove it, we will repeat the same logic as as I was using before. Let's consider opposite. Now, what is opposite? Opposite is that this angle is not bigger than this, which means equal or smaller. But if it's equal, then the triangles would be congruent by side angle side, right? Side angle side, side angle side. So if these two angles are the same, then these two sides would be congruent, which is not the case. We have already the premise of our converse theory is that this is bigger than this. Okay. So angles cannot be the same. Now, if it's not the same, not bigger, which means it's smaller. But if this is smaller than this, then using the direct theory, this side will be smaller than this, which also contradicts our premise. Well, basically that's it. That proves that if you have a situation with two pairs of sides of two different triangles are congruent to each other, then bigger angles correspond to bigger opposite sides. And bigger opposite sides correspond to bigger angles. That's the end of this particular proof. Well, that's it for today. I would like to refer you to Unisor.com website where this and many other lectures are recorded and there are notes for the lectures. And also it's important parents can supervise the educational process of their children using this website by enrolling them into particular courses and basically checking how the score, an exam actually, what kind of score students have in their exams. It gives you a really very good supervision of the whole educational process. Good for homeschooling, by the way, of your students. Well, that's it for today. Thank you very much. Good luck.