 Hi, I'm Zor. Welcome to Unisor Education. This is the fourth lecture about construction which is using the similarity. I very strongly recommend you to go to the notes of this lecture and try to solve all these problems yourself first, then listen to the lecture and then try to solve them again by yourself. All right. So I will just start from the top. Construct the triangle by its two sides and then angle bisector between them. All right. So let's say you have a triangle and angle bisector. So you know AB, BC, and BG. These three sides we know. This one, this one, and this one. Now, what we do is, and basically it's the same way how I proved some other theorem about this bisector related to ratio of these two sides and the pieces of the base, the bisector device. I will continue this line and I draw a parallel to my bisector. Now, let's consider this triangle B, E, C. This angle obviously is equal to this one because BG and EC are parallel. That's how I constructed them. So these two angles are corresponding with these two parallel and AE as a transversal. Now, these two angles are also equal to each other because the same parallel, BC as a transversal and these being alternate inferior angles. But in between themselves, they are also equal, because BG, by definition, by the condition of this particular problem, is a bisector, it's an angle bisector. So these triangles are equal to each other, which means these triangles are equal to each other, which means this side, B, E, is exactly the same as BC, although on my drawing, which is definitely not perfect, it doesn't really seem to be this way. So consider I'm a little bit more perfect in my drawing, and B, E, B, E, C is also this triangle. Now, I would like to establish the lengths of the EC. How? Well, obviously, ABG and AEC are similar triangles. One common angle and these two also are the same. So we have similarity, and from the similarity, we can write the proportionality between BG and EC, BG and EC, which is equal to AB to AE, right? Now, what do we know in this proportion? Well, BG is a given bisector. AB is given side, we know it as well. Now, AE is a sum of AB plus BE, but B is equal to BC. So basically, AE is equal to AB plus BC. So we know it as well. It's AB plus BC. So we know three elements of this proportion, and that's why we can find the correspond EC. Now, with EC known, think about triangle BEC. We know all three sides, so we can build triangle BEC, and that's how we will start our construction. So construction starts from triangle BEC after proper finding of all components, whatever is needed. In this case, EC from this particular proportion. Now, how to reconstruct this? Well, easily, you do this, and basically from point B, you have the length AB equal to A, and that's ABC, which we need to construct. That's it. Well, what's very creative about this particular problem? This additional construction, continuation of the AB and parallel. We cannot construct ABC immediately, but we can construct something auxiliary, which we have come up with. Next problem, next problem. Construct a triangle where it's three altitudes. So we have to construct a triangle if we know it's three altitudes. Okay, well, from the first glance, it's kind of not easy. But let's consider the following thing, G, E, F. Let's consider triangles A, F, C, A, B, E. So let me maybe look in this way. This triangle and this triangle. Now, it's quite obvious that these two triangles are number one, right triangles, because these are altitudes. So angle B, E, A is 90 degrees, and C, F, A is 90 degrees. More than that, they share the same angle. B, A, C is the common angle for both of these triangles, which means they are similar. Since they are similar, I can basically write the proportionality between their sides. Okay, so they are pepotonous, which is A, B. So these are similar. So A, B over pepotonous A, C is equal to minus the catatoules, which is opposite to this common angle, which is V, E, to catatoules of this triangle opposite to the same angle, which is C, F. Now, what does it mean? These are two sides, A, B, and A, C. These are two altitudes, B, E, and C, F. So the conclusion which I came up with is that sides are basically inverse proportional to altitudes which fall on these sides, because V, E falls on A, C. You see, V, E falls on A, C. And C, F falls on A, B. So the sides are inversely proportional to the altitudes. All right, now, that's a very good observation. How can I use this particular thing? Very simply, since I know the proportionality between my altitudes and I know that sides are inverse proportional to my altitudes, I can build a triangle similar to this one. How can I do it? Well, very simply. Let me just start with A prime, C prime, any segment which I want. Now, since I know that ratio between two known altitudes and ratio between something which I will call A, B prime to my A prime, C prime is equal to this ratio, I can find out what's my A prime, B prime. I can find out the lengths of this particular segment. And similarly, from a very similar consideration about other two triangles, let's say the triangle B, E, C, triangle B, E, C, and triangle A, D, C, A, D, C. They're also similar. Once more, B, E, C shares this angle with A, D, C, which means they're also similar to each other. And I can derive from this proportionality between A, C over A, C. No, not A, C, sorry. B, C, yes, A, C is A, D, C is hypotenuse. And B, C is another hypotenuse in this triangle. They are related as the casualties opposite to this angle, which is A, D, relative to B, E. So I have exactly the same kind of thing. If I know A prime, C prime, and I know this ratio because A, D, and B, E are known to me, I can find out B, C prime, B prime, C prime. So that's how from proportionality between sides and altitudes I can actually build a triangle which is similar to the one which I need. I fixed one particular side of this triangle and using the proportionality I found other two sides. And so I have built a triangle which is proportional to mine, completely proportional, similar. Now, how to construct from this triangle the one which I need? Well, that's the easiest part. Let's say draw this particular altitude and since the triangles are similar, ratio between this altitude which is given and this altitude which I just constructed which is equal to ratio between this side which I don't know and this which is already constructed. So again, I have a force proportional between... No, this is E prime. So I have AB over A prime, B prime is equal to, let's say, BE to B prime, E prime. Now, all these except AB are known or constructed and that's how I determine AB as a force proportional. So I know this and then I'll just find all other sides very easily. Alright, so what's interesting here is to notice that sides in the triangle are inversely proportional to their altitudes and use it to construct the triangle similar to the one which we need and then scale it to the proper side. Express as algebraic formula lengths of two segments into which an angle bisector divides and opposite to this angle side of a triangle in terms of three known lengths of triangle sides. Alright, this is an explicit usage of a theorem which we have to know. So you have an angle bisector and what you have to find is lengths of two segments. So this is AB, C, D. You have to find segments X and Y in terms of three known sides. Let's say this is side A, this is side C, and this is side B. I'm using the name of the side lower case letter which is used for opposite vertex. That's how it teaches tradition for triangles. Now, this is algebra. This is non-geometry actually. And to determine X and Y, well, this is angle bisector. To determine X and Y, all you have to do is remember the theorem which I have used many times before that angle bisector divides an opposite side of a triangle in the ratio proportional to the sides it's drawn in between. So basically, C over X is equal to A over Y. Now, if you wish, I can remind you how I proved it. I think it's the same additional construction which I have to do. This is the isosceles triangle because this angle is equal to this one and this angle is equal to this one. And basically the proportionality between these sides, since triangles are similar since these are parallel lines, the proportionality between this and this, which is equal to this, is equal to proportionality between this and this. So that's how it's proven. But that's not about how to prove it. That's about an algebra which will allow me to find out exactly the values of these two segments into which Hc is divided. So I have this equation. Now, I also have another equation that X plus Y is equal to B. So these are two equations which will allow me to find out two algebraically the values of X and Y. Well, how to solve this equation? Well, for instance, from here you can put Y is equal to Yc is equal to Ax so it will be X times A over C and substitute Y into this. Y, it will be X plus XA over C is equal to B common denominator C. So it will be XC plus XI A over C equals B. X equals B times C divided by A plus C. So this is, and then similarly for the Y. So this is the answer. Now, do you remember how to construct this? Well, obviously this is the same construction which I was using many times before because it can be actually rewritten as X over B is equal to C over A A plus C. So you have force proportional between these three others which are known to us since we know A, B, and C. That's it. What's important here? Remember the theorem that a bisector divides opposite side of a triangle proportionally to the sides it's independently. All right, next. Expressed as algebraic formula lengths of two cateches are the right triangle in terms of known lengths of two segments it's a partners is divided by an altitude drawn onto it. Okay. So you have right triangle and an altitude. You have to find two cateches if you know these two segments. Basically that's it. Again, what helps here is to remember that in the right triangle H square is equal to A times B. Now, proof is really very, very elementary because these two triangles, this one and this one are obviously similar since this angle is equal to this one and this angle is equal to this one. So these are similar triangles and from the similarity of these triangles we basically derive everything including this one. Well, it's kind of obvious. In this triangle I can take A which is opposite to double-arfed angle relates to H in this triangle opposite to double-arfs as H in this triangle opposite to a single arc opposite to B. And that's where we get this particular equation. So that's simple. Now, since you know this you know the theorem Pythagorean theorem which says that X square is equal to A square plus H square, right? This is right triangle. Now, H square is AB so it's AB A square plus AB or A times A plus B from which X over A from which X square is equal to A A plus B. Now, how to construct an X square which is a product of two known segments we know. One of the previous lectures I actually explained how to do that. So basically we can reduce our problem to a known construction problem to find the segment X whose square is equal to product of two known other known segments. Now, obviously the same thing for Y Y is equal to B square plus H square which is B square plus AB which is B times A plus B which is again the same kind of a known thing. Y square is equal to B times AB product of two known segments. So that's how we found our competitive. What was important here is to again remember the theorem that the square of an altitude to a hypotenuse is equal to a product of these two segments with device hypotenuse which is an easy theorem. Next, express as an algebraic formula the lengths of the median of a triangle in terms of three known lengths of each side. All right, so in all three sides of a triangle we have to express the median. Well, when medians are involved usually what helps is to continue medium by the same lengths and convert our triangle to a parallelogram. So this is A, this is B, this is C, this is X, and this is also X. All right? A, B, C. So you know AB is equal to C, AC is equal to B, and BC is equal to A. The median is X, this is also X. I'm not going to prove that this is a parallelogram. It's an easy stuff which we have already covered in previous lectures dedicated to triangles and parallelograms. And now let's recall that in the parallelogram again that's something which we have learned before. Sum of squares of diagonals is equal to sum of squares of all sides. This is a theorem which I have proven before. Now, this diagonal is 2X, this diagonal is C, which means 2X squared plus C squared is equal to sum of squares of all sides. So side A is twice and side B is twice. So this is an equation. Again, this is algebraic formula from which we can derive our X. Now, how to construct this particular X? And the formula is trivial. I'm not going to write it. But how can we construct the formulas like that? Well, actually very easily. Step by step. How can we construct, let's say, D squared which is equal to 2A squared? How to construct this? Well, easy. That's basically 2A times 2. And we know, sorry, times 8. And we know how to construct a segment square of which is equal to a product of two given segments. So if we know this, we can substitute instead of 2A squared we can put D squared. So we know how to construct B. We know how to construct similarly E. Then we can always construct F squared which is equal to D squared plus E squared. This is the Pythagorean theory. So instead of this, we can put F squared, right? Then next construction, move C over there. So it's 2X squared is equal to F squared minus C squared. How to construct this? Again, Pythagorean theory. This is the hypogenesis. This is the calculus. After F, we have G, right? So we can replace it with G squared. Well, now we can actually do a square root from both sides. 2X is equal to G. And X is equal to G divided by 2. So by sequentially constructing D, E, F, G, we will construct X. So the algorithm is using this formula construct each individual component and then you derive with X. So from algebra we move to geometry. Express it as an algebraic formula in terms of an altitude of a triangle in terms of three known lengths of its sides. Again, it's kind of similar. So if you have three sides, how to express this particular thing? Well, very briefly, this is A, this is B, this is C. So let's call it Y, this particular segment. And just use the Pythagorean theorem. X squared plus Y squared is equal to B squared, right? This is the right triangle. Now, this is also the right triangle. So X squared plus, if this is C, this is Y, then this is C minus Y, right? C minus Y squared is equal to A squared. So you have two equations with two unknowns. The easiest way right now is to subtract one from another. So X will be out and we will have, now, this is a very simple equation. I just opened the parenthesis and Y goes out from which we have C is, let's see, Y. So Y is equal to minus A squared plus B squared plus C squared, right? This is minus, so it goes to the plus, divided by two C. So that's how we can construct Y. Now, again, algebraically it's very easy to construct. How? First, we replace B squared plus C squared with a D squared using the Pythagorean theorem where D is a hypotenuse and D and C are cateches. Now, this can be replaced again with E squared, where D is a hypotenuse, A is a catechus, so this will be, E squared would be another catechus. And this is a simple fourth proportion as we did many times before. So that's how you build Y. That's how you build Y. This is Y, right? So from Y and B, you build X. So, again, first from geometry to algebra using a couple of Pythagorean theorems and system of two equations, solve the equations, get the values of one of those things, and then you construct another. Calculate lengths of the side of a square inscribed in a triangle. Two vertices of a square are in the base. By known triangles, base lengths and then L2P. Okay. So you have a triangle with inscribed square in such a way that two vertices of a square are on the base of a triangle. And what you do know about the whole thing is the base of the triangle and its altitude. So how can we find the side of the square which is inscribed into this particular triangle? Well, again, algebra. Let's say this is X. This piece. We talk from the vertex to the beginning. And this is Y which is the side of a square. So obviously, X plus Y is H, where H is known, right? Now, we also know that this triangle is similar to the big one, right? So it's altitude X divided by an altitude of an entire triangle is equal to its side, which is Y. This is the square, right? Relates to the base of the big triangle. Okay. Simple system of two equations which actually is very easily resolved. X is equal to, from this one, Y times H over A. Substitute to this one. So you have Y times H over A plus Y is equal to H from which Y times H H plus A is equal to HA. So Y is equal to HA divided by H plus A. Simple thing. How to construct if you know A and H and that's how you get the side of a square. Well, that's it. That concludes this particular lecture. It seems to be interesting at least to me that you can always use algebraic methodology to solve geometric problems. And it actually goes both ways. I mean, from geometry, you jump to algebra. You do certain calculations and the results of the calculation you can convert back into geometry by constructing whatever the formula you have come up with. All right. That's it. I do recommend you to go through all these problems again and I do recommend you to register and take exams. Very important. What's also very important is for parents and supervisors to get involved into this and also register so they can actually see the results and actively participate in the educational process. That's it for today. Thank you very much.