 Hi, I'm Zor. Welcome to Unizor education. I will spend today's lecture on certain number of theorems. We can call it mini theorems. They're not really very difficult at all related to similarity. I'm planning to release another lecture devoted to different theorems. It's very, very useful for you if you spend some time trying to prove these theorems yourself. Notes for this lecture contain all the different theorems. I think there are seven or eight of them. I will prove them right now in front of you. But again, it's very important for you to spend time at least an hour for just getting into this logic, these theorems before you listen to my lecture. Now, after the lecture is finished, if you understand everything, again, go to the notes and try to prove all these theorems yourself once more just to make sure that you basically understand how it works. Okay, given two points A and B position on the leg of an angle with vertex P, point A is closer to P. Okay, so we have an angle and we have two points A and B. And two other points C and D positioned on another C and B. Segments on one leg, P A and AB, P A and AB are proportional to segments on another leg, PC and CT, which means P A to PC is AB to CT. This is given. Prove that segments AC and BD are parallel. So AC and BD must be parallel. Okay, now this is obviously about similarity of triangles PAC and PBD. Now, to prove the similarity, one of the theorems which we learned before was that if there is an angle and sides which form this angle are proportionate, then the triangles are similar. In this case, we don't have sides PB and PG proportional to P A and PC, but we do have AB and CG. So the question is, is proportionality of AB and CG to P A and PC sufficient for proportionality of PB and PB? Well, the answer is yes, and to prove it, it's actually very easy. Since this ratio is the same, let's call it, let's say K. Now, that means that P A is equal to K times PC, and AB is equal to K times CG, right? P A over PC is K, so P A is PC times K. Same thing here. AB is CG times K. Now, if P A and AB, now if I will add them together, these two, I will have on the left P A plus AB, which is P A plus AB, which is PB. Here, using distributive law, I can factor out K, and in the parentheses, I will have PC plus CD, PC plus CD, which is PD, which means that PB over PD is exactly the same K, same as this one. So now, we have this proportionality, P A over PC is equal to PB over PD and is equal to the same constant. This is already sufficient condition for similarity. P A over PC. Actually, we can write this particular equality in slightly different form. Since P A to PC is PB to PD, obviously P A to PB is equal to PC over PG. These are exactly equivalent ecologies because both of them means that P A times PG cross multiplication is equal to PD times PC, and this is exactly the same. P A times PG is equal to PB times PC. So obviously, this is exactly the same, and this is a more familiar expression for proportionality of the size. So we have a smaller triangle, P A C, and bigger, P B C. Now, we see that the corresponding sides, which form this angle, which is common for this too, P A over PB is PC over PD. So that's proportionality, that means triangles are similar, and that's why these angles are congruent to each other because they lie against proportional sides of the triangles. Okay, my kindle for whatever reason got stuck. Okay, yeah. Okay, that's it for the first problem. Now, since these angles are congruent, the lines are parallel as two lines and transversal, and these are corresponding angles. Okay, next. Two nonparallel segments, AB and CD and CD, intersecting at point X, such that AX times XB, AX times XD, is equal to CX times XD. Prove that all four endpoints of given segment lie on the same circle. And these four points lie on the same circle. That's what's necessary to prove. Okay, let's connect the dots. Now, obviously, these two angles are vertical, and for this, I will convert it into AX over XD is equal to CX over XB. Now, this is exactly the same as this. Now, we see that these two triangles have congruent angle, and sides which form this angle are proportional, AX to XD as CX to XB. So these are similar triangles. Now, since they are similar triangles, then the corresponding angles, and these are corresponding to each other, they are congruent to each other. Okay. Now, is that sufficient for staging that all four points lie on the same circle? Well, actually yes, and here is why. Let's have a circle around A, B, and C. Why point D is supposed to lie on the same circle? Well, let me put it this way. If point D lies in the same circle, then obviously, since this is an arc which supports both inscribed angle, this and this, that's the same arc, and inscribed angles which are supported by the same arc are congruent to each other. So it's kind of a reverse statement. But what if they are equal to each other? How to prove that this is actually the point D lies on the circle? Well, for a very simple reason. You know that if point D lies outside of the circle, then its measure is half the difference between two... Okay, let me put it this way. Now, this angle is measured as half of this arc minus this arc. When I'm saying arc, I mean central angle, which is supported by this arc. So this angle is equal to half the difference between this and this. But this angle is equal exactly half of this arc without any kind of a difference. Which means that if D is outside, it cannot be... And it's supported by the same arc as this angle, it cannot be equal. It will always be less than. Now, if D is inside the circle, it's also impossible for it to be... This is our A and this is our D. Now, again, this angle and this angle are supported on one side by this arc. But since this is inside, there is another arc on another side. And if you remember, if this particular point is inside the circle, then its angle is measured by half a sum of these two arcs. So it's either half a difference between this arc and something else if it's outside, or half sum of the same arc and something else. So it's not equal to this angle, which is always exactly equal to half of this arc. So we don't have any other choice but to conclude that the point D is supposed to belong to this circle. That's it. Number three. Given triangle ABC and median AM from vertex A to midpoint M of the opposite side, this is median. So these two are equal. Okay, let PQ be any segment connecting point P on the side AB with Q on the side AC such that PQ is parallel. So we have some parallel segment. PQ. Okay. Prove that point of intersection X of median AM and segment PQ divides segment PQ in half. All right. So triangle, this is the median, and this is the segment which is parallel to the base BC. And we have to prove that median divides any segment which is parallel to the base in half. Okay, now, obviously since these are parallel, you can definitely say that, okay, where is my, okay, here it is. I'm losing my... All right, so let me put it down so I don't press any buttons. So how to prove that PX and XQ are equal to each other? Well, quite simply because triangle APX is obviously similar to ABM since these are too parallel. That means that all angles are equal. These are two parallel lines and transversals. So these angles are equal. These angles are equal. And this is the complement. So these two triangles are similar, which means there is a proportionality of their sides. Okay, PX relates to BM, PX to BM as AX to AM. Now, these two triangles, AXQ and AMC are similar for the same reason. And again, XQ relates to MC as AX to AM. XQ to MC relates as exactly the same, AX to AM. So what I'm interested is actually this one. I'm not interested in this anymore. I'm interested in this. These two ratios are the same. But now let's see this. BM and MC are equal to each other because this is the region. So these two halves have the same lengths, which necessitates that these two PX and XQ must also be equal to each other otherwise we would not have the equality. So this is equal to this. And that's what's necessary to prove. And the story back to my Kindle with problems. That was problem number three. Okay, number four. Given three rays with common vertex V. Three rays. One, two, three with common vertex V. Where point A be on one of these rays and the distances from point A to two other rays are X and Y. Okay, so this is our point A. Now, if we drop the perpendicular to this ray, we will have X. And if we draw perpendicular to this ray, we will have Y. So these are distances from point A to two other rays. Prove that ratio X divided by Y is constant regardless of the position of point A. So if I will take another point somewhere here and draw another perpendicular and another perpendicular, and I will have another X prime and Y prime, then this theorem actually states that ratio should stay the same. Well, now it's kind of obvious because let's think about it. This is A prime, right? Now, obviously, since these are two perpendicular to the same line, they are parallel to each other. And these two triangles, this and this, bigger and smaller, are similar. And from their similarity, we can conclude that this relates to this as this to this, right? So X to X prime is equal to D A to D A prime. X to X prime as D A to D A prime because of similarity of this pair. Now, similarity of this pair is exactly the same. Y over Y prime, this over this is equal to this over this. That's the proportionality. Again, the same thing, D A over D A. So this is the same, so these are the same. So regardless of where you point A, A prime, whatever, the ratio remains the same. It's all based on proportionality, which is the consequence of similarity. Okay, next. Given trapezoid A, B, C, G with parallel bases B, C and A, D. Okay, so we have trapezoid A, B, C, okay. Point M is midpoint of B, C and M is end point of... Okay, so we have middle of the upper base and middle of the lower base. Okay, now prove the point of intersection of diagonals. So we have two diagonals and we have point of intersection X. Y is on the segment which connects M to M. Alright, so I am not connecting M to N, but what I do right now, I will connect M to X and then X to N as two separate segments. But I will prove that M, X, N is a straight line. Now, let's think about it. How can we prove it? Well, in the triangle B, C, X, M is basically immediate, right? Since M divides in half B, C. Similarly in A, X, D, X, N is also immediate. So, let's think about the consequence from the fact that B, C, X and A, X, D are similar. Well, obviously they are similar because B, C and A, G are parallel, so these angles are equal to each other and these angles are equal to each other, as alternate interior angles with two parallel and transversal, either this or that. So B, X, C is similar to A, X, D. Now, what follows from this similarity? Well, in particular, that sides are proportional, right? So, the side which lies against this angle in a smaller triangle is B, X, and in this triangle against the same angle X, X, D. And it's equal to, let's say, B, C, the base over A, D, right? Let us now consider triangle B, M, X and X, D, N. I would like to prove that they are also similar to each other, but I will use a different theorem about similarity. Now, in case of B, C, X and A, X, D, I was using the theorem that if two angles are correspondingly equal to other two angles, then triangles are similar. In this case, I will use another theorem. An angle and sides which form this angle are proportional. If this is true, then I will have the similarity. Now, the B, X to X, D, I know is equal to B, C over A, D. But now, since this is a ratio, I can always multiply a ratio by one half, B, C, A, D. Why did they do that? I mean, obviously I didn't change the ratio since the numerator and denominator are multiplied by the same number. But half of B, C is B, M. That's why I did it. And half of A, D is N, D. So, what I have is, let's just wipe out this piece. I have this, B, M over N. And this is exactly the proportionality which I need. B, X over X, D is the same as B, M over N, D. So, B, M, X and X, D, M have equal angle. And sides which form this angle are proportion. So, it means that we have similarity of triangles B, M, X and X, D, N. From similarity follows the angles which are supposed to be equal to each other. But now, let's think about B, X, D is a straight line. Therefore, M, X, N also must be a straight line. And this is basically the situation. If this is a straight line, if I will put it any other way, not as a straight line, then this angle will not be equal to this. Since I can always continue this line and have a vertical one and it will be smaller. So, if these angles are equal to each other, it necessitates that since B, X, D is a straight line, M, X, N also supposed to be a straight line. So, if I connect M to N, it will intersect both diagonals exactly at the point X of their own intersection. Alright, next. Given triangle A, B, C, segment A, L is a bisector of angle B, A, C. Okay, A, B, C, A, B, C, and this is angle bisector. Prove that point L divides side B, C proportionally to corresponding sides A, B, and A, C. That form angle B, C. In other words, I have to prove that A, B over A, C is equal to B, L over L, C. This over this is equal to this over this. Well, actually, this is one of those interesting moments in mathematics when it's a very interesting and in some way aesthetically appealing result which is not really obvious. I mean, there are many very plain theorems which are kind of obvious. This is definitely not an obvious theorem. This length over this, if this is an angle bisector, relates the same as B, L to L, C. However, this is true, so let's try to prove. I'll put it on the top. A, B over A, C, B, L over L, C. However, considering that this is not really an obvious thing, the proof is not really very difficult. Here it is. Let's draw a line from C parallel to A, L until it intersects at point D with continuation of A, B. So, A, L is parallel to D, C, D. Well, since they are parallel, obviously these angles as corresponding are equal to each other. At the same time, this angle and this angle are also equal to each other because we have again parallel, transversal, and these are alternate interior. But, since A, L is an angle bisector, these two are equal to each other. So, I will use just one R rather than two. And same thing here. Now, what do we have now? We have these two angles are equal to each other and this one is congruent to this and this one is congruent to this and these are congruent to each other since A, L is an angle bisector. Okay, that helps. Now, why is this helping? Here is why. In this particular ecology which I have to prove, I will exchange A, C with A, D. Since these two angles are equal to each other, A, C, D is a socialist triangle which means that this and this, these sides are the same. So, A, C is equal to A, D. But now, if you look at the triangle D, B, C, obviously just similar to A, B, L. Right? Because these are two parallel lines, this one and this one. So, these triangles are similar to each other. All angles are equal, etc., etc. Now, from their similarity, we see that A, B and A, D, this ratio, this over this is equal to this over this. This is an angle, if you remember actually we had a theorem about this, that if you have an angle in parallel lines, then they cut proportional segments. So, the relationship between these segments is exactly the same as the relationship between them. Or you can use similarity of triangles which is exactly the same thing. Which basically is what we have to prove except this is A, D and this is A, C. But as we have already pointed out, they are equal to each other. And that proves the original theorem. That the y-sector, angle-by-sector divides an opposite side proportionally to the sides which form this angle. Very interesting theorem. And in some way unusual, I would say, or unexpected. There are some theorems which are unexpected and look very aesthetically appealing, as I was saying. That was number six. Given triangle A, B, C with unequal sides A, B and B, C. Segment A, M unequal sides A, B and A, C. Okay. A, B and A, C unequal. Okay, fine. Now A, M is a median of side B, C. So this is M midpoint. Segment A, L is a bisector of the same angle. So this is L. A, L is a bisector. So A, M is a median. A, L is angle-by-sector. Finally, A, H is an altitude. The lengths of these three segments are related as A, H less than A, L less than A, M. So if these two sides are not equal to each other, then altitude is smaller than angle-by-sector and angle-by-sector is smaller than the median. Well, obviously these sides are equal to each other, then they all coincide and they're all equal to each other. But if sides are different, then there is such a ratio between, not ratio, there is such a relationship between altitude, angle-by-sector and median. Median is the longest. Now, the fact that altitude is the shortest is obvious because altitude is perpendicular and we have learned many, many lectures ago that the perpendicular from a point to a line is the shortest distance. So I'm not talking about this. This is obvious. Let's think about this. So we have, we don't really need this. I actually added it just for completeness. So we have a median and an angle-by-sector. And we have to prove that angle-by-sector is smaller than the median. Now, here is how we can do it. Think about the previous theorem, which I have just proven. It stated that AB over AC equals to BL over LC, right? An angle-by-sector divides the opposite side proportionally to sides which form the angle. Now, let's assume for definiteness that AB is smaller than AC. So we have been given information that they are not equal. So one of them is smaller, another is bigger. So let's say AB is smaller. Since AB is smaller than AC, this is less than one. AB is smaller than AC, which means BL over LC is less than one. So BL is less than LC, right? Now, since BL, BL is less than LC, then the point M, the point L is shifted towards B from the midpoint. Because M is a midpoint, BM is equal to MC. Because AM is a median. So as we see, the proportionality of these segments, BL and LC to unequal AB and AC, causes the fact that point L is shifted from the middle towards point B. Now, what does it actually mean? Well, it means that if you again remember our old theorems, we were talking about that the closer the point is to the base of the perpendicular, the shorter it is. The shorter the segment is. Basically, that's exactly what's causing AL to be less than AM. Since AL is shorter, here is our H again, since L is closer to H than AM, AL is shorter than AM. There is one consideration which I silently omitted here. Is it possible that AL, while being closer to B, will actually overshoot the H and be on this side? Well, the answer is no. It's not possible. However, its proof should be slightly different. And it depends actually on the fact that point L, let's just wait for my telephone to come down, it actually is related to the fact that since this is a perpendicular, AH is a perpendicular, and we have that AB is smaller than AC, it means that point B is closer to H than point C. And since this is an angle bisector, it should be closer to the C which is much farther from the H than B. So that's basically kind of consideration which I'm not really pretending that this is 100% robust. But still, that's where you should really maybe continue to research yourself and try to prove it a little bit more rigorously. This is an interesting point, but nevertheless, what is important is to always base the fact that the lengths of the AL should be shorter, the closer it is to the base of the perpendicular H. Alright, and I think I still have one more theorem to prove, which is number 8. Given triangle and three altitudes, ABC and three altitudes, D, E and F. Okay, connect points D, E and F, connect points D, E and F to form a triangle, and prove that each of these small triangles is similar to the original triangle ABC. This is also quite an interesting theorem, and again, I would say unexpected. So A, F, E is similar to ABC. Okay, how can we prove that? Well, let's consider two triangles, A, B, E, which is the right triangle, and A, F, C. These are right triangles, right, because B is perpendicular and C, F is perpendicular. And they share the same angle, B, A, C, right? So this triangle and this triangle, both are right triangles, they share the same acute angle, which means they are similar, because another acute angle also will be equal to each other. So they are similar. Now, from their similarity, we can draw a proportionality of their sides, right? So let's consider, let's say, a smaller catatose of A, A, B, E, which is A, E. Now, it's related to, it's against this angle, these two arcs. Now, in this triangle, against these two arcs, a smaller catatose, A, F. Now, how about hypogenesis? They're supposed to be proportional, too. So A, B in A, B, E, A, B, hypogenesis, two A, C, hypogenesis in A, F, C. Now, let's consider this triangle. Let me just wipe out everything except one. So, now let's consider A, F, E, and A, B, C. Well, again, the angle is common. And now we have the proportionality of the sides. A, E, we can exchange these, right? A, B, A, F, A, C. So A, E, relative to A, B, is like A, F, relative to A, C. So basically, if you would like these two to be parallel, you have to turn the touch side down. But anyway, this proves actually that these triangles are similar to each other because there is a common angle and proportional sides. Well, exactly the same way we can prove that the triangle on, on small triangles at other vertices exactly, similarly in exactly the same fashion. Well, that basically proves the point that these triangles are similar. Again, something which is, I would say, unexpected, but interesting. Well, that concludes my today's lecture about different theorems related to similarity. I will probably have another one. And don't forget, go back to notes. Go on this site, Unisor.com, go to the notes for this lecture, and try to repeat basically all these proves yourself. And if you can't, go back to the lecture, listen to the game, and, well, finish it up completely. I mean, it's very, very important for you to be able to prove on your own. And not just understand what I'm saying. It should be inculcated in your mind. Because again, as you remember, the whole purpose of this site is to develop your mind. Thank you very much.