 Hi, I'm Zor. Welcome to Unisor Education. Today we will talk about very important geometrical property of triangles about triangles being congruent. Now this course, this lecture actually is part of the course called Math Plus and Problems, which is on Unisor.com. I suggest you to watch this lecture from Unisor.com website because every lecture including this one has very important notes, which basically are like a good textbook, very very detailed. Then there is a prerequisite course in the same website. It's called Maths for Teens, which basically is the foundation of school mathematics, high school mathematics, algebra, geometry, logic, combinatorics, etc. All the topics actually. So this particular course, Math Plus and Problems, is a continuation in in certain respect. It's not about standard problems, which basically are checking if you have correctly mastered the theory. These are problems, mostly problems, which are not exactly standard. They are forcing you to think about to be creative, to find out something which is not exactly within the framework of whatever you have been taught. So that's the purpose of the whole course and in this particular case we will talk about one particular problem of geometry. Geometry of triangles when triangles are congruent to each other. Now let's talk about congruency, geometrical figures. What is congruent? Basically congruent means that you have two geometrical figures. Let's say it's on the plane. So you have, let's say, two triangles. You have this triangle and you have this triangle. We are talking about congruency in what kind of respect? Well, we are talking about process of transformation of one into another when they will completely coincide. Now, what kind of transformation do we allow to call these triangles congruent? Well, first of all, it's translation, which means we're just shifting the whole thing along some line, straight line. Then it's rotation because we can actually rotate and they still be, they will still be congruent in our in our sense, in our definition of congruency. And they can be symmetrically reflected along some kind of line. So these three components of general transformation are allowed to be used to make these two figures coincide and if using one of the or three of them in some combination, these transformation, if we can make these two coincide, then we can say that these geometrical figures are congruent. In all times, actually, we used the word equal. Now it's more traditional to call them congruent. Equal means that the lengths of the each site in this in this particular case corresponding site is equal. Now now let's talk about something which is closer to problem at hand. Now when two triangles can be congruent to each other. Now, regular course of geometry actually we used side-side-side, which means three sides of one are of the same lengths as three sides of another triangle. The corresponding sides are supposed to be equal lengths. Now there is a side angle side. So we have two sides of equal lengths and the angle in between them of one triangle correspondingly equal to another. And we have angle side angle. So we have in this case all three. In this case two and angle. In this case, it's line and two angles are equal to each other. Okay, so it makes an impression that you see three elements of a triangle are sufficient to be correspondingly equal to have triangles congruent to each other. Is it any three elements? Can I say, let's say instead of side angle side, I can put side-side angle. Which means side and then side and this angle. Like this, something like this. Is it also a good theorem to prove that triangles which have two sides and angle not in between them, but they're supposed to be congruent triangle. Or some other combination, whatever is not covered by these three. Well, the answer is no. So this is not the right theory. Now angle, angle, angle. This is obviously not the right one because you can just enlarge triangle, scale it. So the new triangle will be similar to the original one. With three angles equal, but sides be proportionally bigger. So also not good. So my point is that these are the good theorems and general theorem about, okay, let's say just any three elements and it will be equal. Now that's not true. More than that. What I will prove is that you have, you can have two sides and all three angles of one triangle be equal to two sides and three angles of another triangle and triangles be completely different. Non-congrant. And that's the purpose of this particular lecture. Okay, so let's do it now. I will suggest to you exactly an example when two sides and three angles are not sufficient to be equal between two triangles for these triangles to be congruent. Here's an example. So let's draw something like this. This would be my A, B and C somewhere here. Now, I will put D, E in such a way that A, C is equal to C, E and D, E this one is equal to BC. So let's assume that I can build such a triangle. Now as you see triangles are obviously similar. D, E is parallel to this one. So all angles are the same. This is common and this is obviously equal to this. This is equal to this. Now speaking about symbols, I'm usually preferring to call the side which is across the capital letter, let's say D, lower case B. Now, I will have it for a bigger triangle. So a bigger triangle will be with index one. Across A will be A1 and across C will be C1. Now inside C, E as I said by definition by my construction is supposed to be equal to A, C. So C, E would be B also B because they are equal but I will put index two and D, E again, I assume that I can construct such a way that D, E is equal to BC. It would be A 2, A because it's equal to this, A but 2 because it's a smaller triangle. And this one whatever it happens to be, I will put X on it. So I assume that this triangle is two triangles actually are constructed as I said. So A1 is equal to A2 and B1 is equal to B2. Now, A, C is actually B1 and D, E is equal to A1. Now, so I assume that I can construct this triangle with these properties. Now, if I can and I will prove that I can, so if I can, what follows from here? Because if I can construct it, I already have two sides are equal to each other. This one is equal to this and this one is equal to this, the big one, A, C to C, E. All angles are obviously equal, so we have five elements of one triangle equal to five elements of another triangle. Not exactly correspondingly, so the angle is not in between equal sides, etc. But nevertheless, my point was that you can specify five equal elements of one and five equal to corresponding elements of another and still triangles will be different in this case. Okay, so what I will do is I will start constructing certain equations based on similarity of these two triangles. Now, they're obviously similar because I'm assuming that the E is parallel. So all angles are the same, so they're similar. Now, in similar triangles, now I will have the ratio of corresponding sides. Let's say X, which is CD to AC, which is B1 supposed to be equal to this to this, A2 to C1, and this smaller one is B2 to A1. Now, from this pair, we have X is equal to A2, sorry, A2B1 divided by A1, sorry, C1, C2, C1. From this one, the same X is equal to B1 times B2 divided by A1. Okay, now I'm interested in this. So B1 can be cancelled to each other and I will have A2 divided by C1 is equal to B2 divided by A1. But let's not forget that instead of A2, I can put A1. Instead of B2, I can put B1 because I assume that this is equal. That's how I actually made it from which follows that A1 square is equal to B1 times C1. So the A is supposed to be A is supposed to be in between B and C to satisfy this. So B is the smallest one, C is the biggest one, the biggest side, and A1 is supposed to be the middle one in between. So from existence of this these two triangles built the way how I did follows this particular equation which the sides are supposed to satisfy. So this is a necessary condition for existence of this construction. Now I will prove that this is sufficient condition as well, and if we do have this triangle which satisfies this particular equation, then I will be able to build the second, the smaller one. How do I do it? Okay. Easy. Let's assume that I have ADC with this particular property and B1 is smaller than A1, it's smaller than C1. So I will just take this B1 and put my point E on the distance from C equals to the distance of A. So A and E are in the same distance from C. So if this is B1, then this is also B1. So B2 is equal to B1. Then I will build a parallel line. I don't know that this line would be equal to A, so I'll put Y here. Okay, but since it's a parallel line, triangles are obviously similar, and I can put the correspondence Y divided by C1 should be equal to B2, which basically I constructed as being equal to B1, to A1. So I can put actually instead of B2, I can put B1 because it's the same thing. From which follows that Y is equal to B1, C1 divided by A1. But I assumed that my triangle, my initial triangle has this property. So instead of Bc, I can put A2 divided by A, it would be A. So this is equal to this. Okay, now the only thing is are there any triangles which satisfy this particular equality? Well, yes. For example, four, nine and six. Six square is equal to four times nine, so it is six. So triangles do exist and this condition is necessary and sufficient for being able to make such a construction. So the original question, if I have five elements, two sides and all three angles of one triangle, correspondingly, well, not correspondingly exactly, but equal, somehow equal to two sides and three angles of another triangle, does it mean that triangles are congruent? The answer is no. They will be similar, but not congruent. And there is an example. So there is a definite conclusion of this that original theorems, which I presented in the very beginning, side, side, side, side, angle, side, angle, these are good theorems. But to say that any three or even any five elements, as we have just seen, are sufficient to make triangles congruent, no. It's not exactly any. You have to have it in proper order, like side, angle, side. Not side, side, angle or something like this. Because this is an example when two sides and all three angles of one triangle is equal to two corresponding sides of another, but triangles are similar, but not congruent. So this is an example. Six, four and nine. And one more little thing. If this is six, four and nine, what is exactly? Now we have this. Okay, if this is four, this is six, and this is nine. So this is four. This is six. On the smaller one. So what is x? Well, we can derive x from, let's say, from here. So it would be x divided by b, which is what, four equals to a. A is what? Six divided by nine, right? Let's see. So what do we have? 24 times equals x, which is eight-thirds. This is eight-thirds. So two triangles. One of them has four, six and nine. Another has eight-thirds, four and six. Have two sides equal, four equals four, six equals six, and all angles are equal to each other. This is an example. Now the last thing which I wanted to touch is what about this angle? Between the biggest and the smallest sides of triangle ABC. Now we have come out with this necessary and sufficient condition. I would like to also derive a condition on this angle. That's very interesting. It's not just from this condition, follows some condition on angle alpha. How? Well, let's drop the indices ABC, so it will be ABC, and we know that a square is equal to d square plus c square minus 2 dc cosine of alpha, right? This is the law of cosines. Now I can substitute BC here. So for these triangles, which satisfy this, which satisfy this particular condition, we can say this. Now 2 BC will be on the left. This BC will be on the right. So what I will have is 2 BC equals b square plus c square minus minus BC. I will put minus 2 BC and plus BC. Why? For a simpler reason, because this is b minus c square. Positive value, which means that 2 BC since this is positive, this is greater than just simple BC. Because it's equal to BC plus something positive. b and a and c are all different. We have just made this sure in the very beginning. From which follows that not to BC, I'm sorry, it's supposed to be cosine alpha here. Cosine alpha. From which BC we can cancel, which means cosine alpha greater than one half. Or alpha is less than p naught over 3, which is 60 degrees. So angle alpha should be less than 60 degrees. Okay, that's it. I recommended to read the notes for this lecture. There is a little better picture and again, I'm using indices 1 and 2 here in the notes for this lecture and use different colors, red and blue. So that might actually be a little better. I don't know. So read the notes and well, good luck.