 Hi, I'm Zor. Welcome to Unisor Education. I'm continuing talking about similarity. In this case, we will talk about similarity of triangles. Well, as I'm sure you've noticed, we are trying to consider more and more complicated geometrical figures. We considered angles, segments, now we are combining them together into triangles. Next will be polygons, obviously. Do you understand? Alright, so today is about triangles. What we have learned from the previous lectures is that transformation of scaling, homoththy, as they call it, preserves the angles and converts, transforms all segments into other segments, parallel to the original ones, having the lengths equal to the lengths of the original, multiplied by the factor of scaling. So if there is a scaling, which is a center, and a factor, then the image of every angle is angle congruent to the one which is given, and image of every segment is the one which is parallel and lengths of which is multiplied by the factor. Therefore, if you have a triangle and some center of scaling, let's say scaling by factor of two, it goes this way and that would be our new triangle. So this new triangle is similar to the original triangle and what's important is angles are exactly the same and sides are proportional to the sides of the original, proportional in terms of the ratio between the lengths of any corresponding sides is exactly the factor of the scaling. Now the question is whether the reverse is true, which means if two triangles have three angles equal to each other and sides are proportional, which means there is the same ratio between the lengths of all corresponding sides then they are similar to each other, which means they can be transformed one into another using some scaling. Well, the answer is yes. So the proportionality of the sides and the quality of the lengths of the angles is not only necessary but also a sufficient condition for similarity of triangles. Now actually there are some much simpler statements you don't really have to have all angles equal to each other and all sides proportional to each other to get to the similarity. You can suffice with a simpler statement actually three different simpler statements, for instance it's sufficient to have only two angles equal to each other. Well, obviously because the third angle will also be equal because the sum should be equal to 180 degrees but it's also sufficient to have two angles to have sides proportional to each other and I'm going to prove it. Another simplification is you don't have to have all three angles equal and all three sides proportional, you can have only one angle equal to another and sides which are making these angles proportional to each other. That is also sufficient for similarity. And the third again you don't have to have three angles equal and three sides proportional, it's enough to have three sides proportional. It's sufficient for similarity. Now when I'm saying it's sufficient for similarity it means that there is a scaling which would transform one triangle into another. Maybe combined with some parallel shift or anything like that, that's obvious. So we have three different theorems. Each one of them states some kind of a condition which is sufficient for similarity and I'm going to prove one after another. However before doing that I will prove auxiliary theorem called lemma. Lemma is auxiliary theorem and using this lemma I will prove each one of those three theorems. So what is this lemma? If you have a triangle and a point somewhere on one side and the line parallel to the base and n is point of intersection of this line which is parallel to base with another side, then triangle Bm and is similar to triangle B1, ABC. Now how to prove the similarity? Well to prove the similarity we have to actually find the scaling which transforms one into another. So scaling means we have to define what's the center and what's the factor of scaling. So I choose point B as a center and I choose the factor am over mb as a scaling factor. So what I have to prove is that if I use exactly the same scaling factor and apply to segment Bm, I will hit the point C. So how can we do it? Well I went in a similar fashion to the way how I was proving that during this transformation of scaling all segments are changing their lengths proportionately. Let's assume that this factor is some kind of a rational number, P over Q. Now what does it mean that the ratio between lengths of this and lengths of this is some rational number PQ? Well it means that there is some kind of a unit segment which would fit P times in am and Q times in Bm. So let's say we will do it this way. So it fits 1, 2, 3, 4 times and 5, 6, 7, 8. Let's make it a little bit less obvious. So it will be 4 times here, 5 times here, so it will be Oh, sorry that's wrong. Not am to mb, ab to mb. So it will be 9, 5. So what does it mean that this ratio is equal to PQ? Like in this case 9, 5. It means that there is some unit segment which fits 9 times in the big one and 5 times, no this is 4. Again my mistake. And 4 times in this case. 1, 2, 3, 4 and 5 here so it's 9 all together. So 9 over 4. So this unit segment, this small one fits 9 times in ab and 4 times in mb. That's what basically factorization means. Alright. What do we do now? Now since we have this unit segment we basically divide the whole ab into 9 partitions, divide the whole ab into 9 segments and draw parallel lines. 1, 2, 3, 4. So the fourth point of division starting from B to A will fall onto point M and the 9s would fall onto point A. Or if you wish the point number Q in a general case M and point number, division point number P will fall onto A. Now since I have drawn all lines parallel to the base AC line MM will be among them because M is one of the division points. So they will coincide. So I will have whatever number of lines here but the line number Q would coincide with MM. Now obviously as we know and again I mentioned this in the previous lecture about segments. If you have equal segments on one side of the ankle and parallel lines which are intersecting to another you will have equal segments here. Which means the number of segments will be exactly the same. Which means P in this case or 9 in our particular case and the number which fits into this part will be also the same as this which is Q or for another case. Which means that the ratio of these two would be exactly the same as the ratio of these two which is P to Q or 9 to 4. So this actually proves for any rational factor that the ratio between BM and BA is the same as BM to BC. So all I need to do is for any point M draw a parallel line and now I have a top triangle which is similar to the whole triangle as long as these lines are parallel. Now I have proved that only in case BM to AB is a rational number. And again as in the previous lecture I'm not going to oversimplify saying okay it's obviously that for any real number that's true. It's not out this because irrational numbers are not easy to define and it needs significant amount of efforts to really rigorously prove it for irrational numbers. However since any irrational number can be approximated to any degree of precision with rational numbers you can probably say that it's kind of obvious that this theorem can be proven and it is. Yes it can be proven so let's just take it for granted that for irrational numbers the theorem can be expanded and that's why I'm basically staging that this particular lemma, this auxiliary theorem that if you take any point on the triangle's side draw a parallel line you will cut the triangle similar to the big one. Okay we will use this particular lemma to prove our three theorems each one of them basically states a sufficient condition for two triangles to be similar. Okay theorem number one it's angle one and angle two are congruent to angle one and angle two. Now what does it mean? It means that you have two triangles and they have this angle and this congruent to each other it's sufficient for these two triangles to be similar. I think I have to correct it slightly. Alright how can we prove it? Okay let's say this is ABC and this is A' B' C' so I know that these angles and these angles are congruent to each other. Now what do I do? Okay let's take this segment A' B' and find the point M in such a way that these two segments have the same lengths. So AM as a segment is congruent to A' B' and I draw a line parallel to my side BC. So MM is parallel to BC. Well since these are parallel lines these angles are corresponding and that's why they're equal. Now this angle and this are equal to each other and that's why since these are equal these two angles are equal. Now if you consider these two triangles the line A' B' and AM are the same by construction this angle is again equal by condition of the theorem this we have just proved that this angle is also equal so what do we have? Triangles are congruent to each other by angle, side and angle. So these are two congruents to each other but these two the small one and the big one are similar to each other because of the angle we have just proven. We took the point and draw a line parallel to an opposite side. That's the proof that this triangle being congruent to this and this being similar to this that's why this is similar to the ABC and the proof. So this is angle and angle. So if you remember for congruency we need like three elements. Angle, side, angle or side, angle, side or three sides for similarity we need slightly different conditions. The first condition we were just facing was two angles. That's sufficient for similarity. Now the second condition is side well it is side angle side but it's a proportional side angle and another proportional side. So if you have one triangle and then another triangle and you know that these angles are congruent to each other and side a prime, b prime, c prime, a, b, c and a, b over a prime, b prime is equal to a, c over a prime, c prime. So if you know that the sides which are forming this angle are proportional then triangles are similar. Compare it with congruence. Compare it with two triangles are congruent if two sides and angle in between are equal to each other. In this case sides must be proportional and angle equal. So whenever you're replacing equality with proportionality you are basically shifting from congruence to similarity. Because similarity again is something which is proportional as far as the sizes of the sides but the angles are supposed to be the same. So that's why we have this particular difference. Whatever was equal for congruence proportional for similarity. Okay, how can we prove it? We'll do exactly the same way as in the previous theorem which means I take this length and find the point m and draw a parallel line. So this is equal to this. Now, what do I know? I know that a m is the same as a prime b prime. Now in which case I can replace this with a b divided by instead of a prime b prime I will put a m. I know that triangle a m m is similar to triangle a b c. Which means that a b to a m is supposed to be equal to a c to a m. Right? Which means a c to a m. a b to a m, a b to a m like a c to a m. So it's exactly the same thing. But now look at this. This is a c and this is a c. Which means a c a prime c prime should be equal to a m. So this should be equal to this. Now what we have triangle a prime b prime c prime and a m m. It's obviously that they are congruent because this is the same by construction. The angle is given as equal and this we have just proven are equal. So we have a side angle side equal to side angle side. So triangles are congruent. This one is similar to the big one. Which means this one is similar to the big one a b c. And to prove. So we are just using this proportionality to expand proportionality of these sides to basically equality of these segments. Now the third theorem will be again similar to this one. Third is proportionality of sides. Let's do exactly the same thing. So we have a prime b prime and c prime. And I know that all sides are proportional. Which means a prime b prime relates to a b s. a prime c prime to a c equals b prime c prime to b c. Right? Okay. Do exactly the same thing. A prime b prime find the point m here. And draw a parallel line. Always the same thing. Now we know that this triangle a m m is similar to a b c. Which means that a m which is by the way equal to a prime b prime relates to a b as this to this and this to that. So in this particular case on the left, since a prime b prime is equal to a m we can try this. So a m to a b instead of a prime b prime to a b. But a m to a b a m to a b is exactly like m n to b c. Now using this as you see we conclude that m n and b prime c prime are equal to each other. And again the same ratio is equal to a m to a c. Using this you see a c are the same so a prime c prime should be equal to a m. So this one we construct it as equal. And we have just proven that these lines, these segments are also equal to each other. So these two triangles are equal to three sides. So since these two triangles are congruent and these two triangles are similar this is similar to that. So we have proven the third theory. So there are three theorems about similar triangles. In a way it is similar to three theorems about congruence of triangles. So for triangles we have what? Side angle side. For similarity we have proportional sides equal angle and proportional side. For triangles we have two angles and side in between for congruence. For similarity two angles are sufficient. And finally for congruence of triangles we have three sides supposed to be equal to each other. For similarity it's sufficient that they are proportional to each other. So in a way it is basically similar. Okay I think I've covered what I wanted about different theorems of similarity. Yeah. So for triangles you basically have to remember these three theorems and the way how they are all proved is basically the same way. I find one of the sides of the bigger triangle if you wish. I find the point to cut the same lengths as the side of the smaller triangle and then we're just using proportionality of the sides to basically prove the theorem. All based on the same lemma which is if triangle has a line parallel to its base it cuts from the top at triangle similar to the big one. That's the most important part of it. And the most difficult actually. That's it for triangles. Everything should be on the Unisor.com. I will load this lecture as well. Please use the site to basically go through the whole course of algebra or geometry etc. Because the lectures are basically supposed to contain a concise course when one of the later statements are always based on something which was done before that. So consider this not as a reference material, this site Unisor.com, but as a textbook which means you have to have chapter after chapter that are logically related to each other. And that's the way how I suggest you to study. And that's exactly how I suggest parents and supervisors and teachers to use this site for educational process of their students. They can enroll students, they can check this course on exams etc. So that's it for today. Thank you very much.