 I'm Zorf. Welcome to Unisor Education and today is lecture number eight in a series of different problem-solving lectures related to triangles, quadrangles, parallel lines. These are all elementary problems which we are solving here. There are much more complex and difficult problems but I will probably put it a little bit later in the course after we will talk about circles and maybe some other elements. So these are, as I was saying in the previous lecture, your tools basically. You have to have a toolbox how to solve certain elementary problems so when more complex ones come along you will not be actually concentrated on constructing some elementary stuff. You probably will just have it like automatically basically. Just as an example I'm not talking about how to drop a perpendicular from a point to a line. I assume you have already covered it and this is in your toolbox. And other problems basically are all the same type. So let me go through this last series of problems here. Construct the trapezoid by base two legs and one angle formed by that base with one of its legs. So trapezoid you have two bases, no sorry one base, right? Yeah, one base, two legs and one angle formed by that base with, okay, how can we solve this problem? Well it's actually relatively easy. When you have a trapezoid it's kind of common to bring one of the legs close to another one. So from B we will draw a parallel line E, B E parallel to C G which means it's also the same, the congruent to C G. So we know it basically. So we know A B, we know B E, we know this angle. Now since we know this angle and these two sides in the triangle A B E we can actually construct it. So we have an angle first then we measure A B and then using the ranges equal to B E we just mark where the B E is. Now I understand that if it's a circle it can actually intersect in two points this line or in one point or in no points at all. So anyway let's just assume we have some point E. Now if we do, so this is what we have meanwhile, right? Now how to reconstruct the whole trapezoid? Well you know that A G has a known length. So from A you just extend beyond point A or point E and just measure the segment A G. Now from here you draw a parallel line to this one and from B you have parallel line to this one and here is your trapezoid. Now as far as intersection of this circle ranges of E E with the line, obviously if you have two different intersections then you will have a second solution like in this particular case it will also intersect here as well because this will be a circle like this, right? So it will intersect here, this line will also be exactly the same size and basically again you can just have the same E G from here and then you will have not parallel to this, parallel to this and of the C prime. So A B C prime D would be also a trapezoid which has the same base, the same lag, the same lag and the same angle. So there are a number of solutions like in this case two solutions. If this lag is too short obviously we will not have any intersection and the original solution or if it's exactly equal to the distance between B and the straight line A D it will be a perpendicular here so it will be the right trapezoid which means one of the angles will be right. It will be both C D and C prime D would be actually close to each other and perpendicular, it will be one line so it will be one solution. So depending on usual relationship between the sizes of different components we will have different number of solutions from zero to two. Construct a trapezoid by a difference between bases diagonal and two legs. Okay again trapezoid difference between two bases diagonal and two legs. Okay now when you have a difference between two bases then it's customary again as I was saying in the previous problem bring one of the lines, one of the legs closer to another. Since these two are congruent to each other because this is parallelogram, bases are parallel and this we have constructed as a parallel. The AE exactly is equal to the difference between bases in this case right. So besides the difference between the bases we have diagonal and two legs. So this leg is congruent to this now this we also have right. So we have basically AE we have this triangle by three sides right. Now the only thing which remains we do have a diagonal, one of two diagonals. So how to find points C and G if we have a diagonal? Well very simple if you have this for instance if you have this diagonal from B to G so what you do is you take a compass center at B the radius is equal to diagonal and basically mark the arc of this radius and crosses in D. From D you draw a parallel to BE and equal in size and you have a Z, you have point C. So that's the end of it. So most importantly is to shift CG to the left so you will have a triangle ABE and all three sides of this triangle are equal. Now whatever happens after this is trivial you just have the radius or whatever else. This is this is already simple stuff. Important is to understand that if you have a difference between two bases then the proper way is just to build this line BE parallel to CE and that helps you to build piece of this drawing and then expand it anywhere you want. Construct a trapezoid by its four sides. Okay so again you have a trapezoid you have all four sides. Yeah but look since we have four sides that means we can subtract the bases from each other and do exactly the same thing as in the previous problem. Build triangle ABE by side, another side and the difference between the bases. So it's two X and difference between the bases. Okay after that we don't have a diagonal as in the previous case but we do have size of AD and size of BC separately right? So we just extend this line and then using the size of AD basically mark where the point D is supposed to be because we have this segment. Now parallel from point B parallel to AD and again we measure the upper base and the point C. Important thing about this again reduce the problem to a simple one to build this triangle using both legs and difference between the bases. Everything else is not so interesting and very simple. Construct a trapezoid by its base altitude onto it and two diagonals. Okay trapezoid by base so we have a base now we have an altitude onto it so basically the distance between these two so if you have a distance you draw a perpendicular and draw a parallel line so that's where the upper base is supposed to be located but now you have two diagonals right? One and two so what you do from A using the radius of AC you find the C and from D using the radius of being D you draw a B. Begin there are certain variations based on whether we cut in one point or in two points with this circle etc so but these are variations you can just examine them yourself but basically it's enough for this construction. This is the start this is the beginning bottom base and the parallel line which is distance on the altitude. In every problem actually there is something important and something I would say technical there is a creative moment in the very beginning and then the technicality which you just you know use your regular scales basically. So in this case there's not much basically you just use altitude as the distance between two parallel lines that that's the most important part. Construct the trapezoid by its two bases and two diagonals. Trapezoid again two bases and two diagonals well if you have two bases and two diagonals I think the best way would be from this point C draw a parallel line to a diagonal to a BD. Now think about B, G, E, C it's a parallelogram right? So these are equal. Now look at the triangle A, C, E. In this triangle we know A, C because it's one of the diagonals C, E because it's congruent to another diagonals and we know this side A, E which is sum of two bases right? We do have every base two bases and two diagonals right? So we have two bases and two diagonals which means just combine these two bases together and you have A, E. So that's the end of the creative part. This triangle A, C, E can be built by three sides diagonal diagonal and sum of two bases. Now from here how can we basically how from this particular triangle A, C, E how can we reproduce the rest of this? So we start it from A, we do A, E, we have point C because this is one diagonal this is another diagonal all right? Now we do have two bases right? Which means we know separately A, G and G, E. So that's how from point A we can find point G. We know A, D by lines okay? So we got this point. What else? Now obviously from here we have to draw a parallel line because that's where B is supposed to be. Now from D we can always use B, D as a radius and just put an R crossing this parallel line using the radius B, D and D as a center and that's how you get the B. That's it. Well I'd like you to notice that basically when we are dealing with trapezoids most of the problems were related to some kind of a shift either a shift a side closer to a side or diagonal outside of the of the trapezoid. It's kind of a typical trapezoid so most of the problems related to trapezoids can be solved using one or another methods by shifting left or right diagonals or sides. That's the purpose of solving many problems of relatively close conditions because it kind of develops the skill okay these are typical methods to solve these problems all right? Okay construct parallelogram by two diagonals and an altitude okay parallelogram two diagonals and an altitude. There are again many ways of solving this particular problem. As an example let's think about this one A, B, C, D, E. Now obviously A, C, E is a right triangle and we know hypotenuse and the casualties so we can build it. Now knowing this we know this but also notice two diagonals and an altitude okay so we have used one diagonal but we have another one so we can draw a parallel line here that's where the B, C would be right? Now this diagonal is intersecting with another diagonal in the midpoint so we basically know what B, P is equal B, P is half of this diagonal so using B, P is a radius now the center of A, C the midpoint is P obviously you can just have an arc here which basically gives you the point B so since you know this well obviously my drawing is not exactly correct that would be and this would be A and this would be C that would be the parallel area so important thing is A, C, E is a triangle which we can build because it's a right triangle with all hypotenuse and the casualties and what's important thing is how to find the point B having B, D as a diagonal we know the B, P which is half of it and using the half of it we just find this point now I wonder what happens if you find another point here would that be the same thing then diagonal would be this one and it would be this type of parallelogram something like this well it has exactly the same two diagonals this one and this one and exactly the same type so in this case there are two solutions so number of solutions can be more than one as you see it's only in the triangle when you have three sides you have only one solution and not always by the way because there is a triangle inequality like if sides are A, B and C then A plus B should be greater than C so there are certain conditions there are certain triangles you cannot build but in this case not only there are cases when you cannot build something there are cases when you can build more than one something in this case parallelogram okay construct the parallelogram by side some of two diagonals and angle between them parallelogram side some of two okay let's have two diagonals so we have a side we have some of two diagonals and angle between them right but let's think about this way let me reduce the problem to a simpler one you have a side you have an angle you have some of two diagonals but diagonals are crossing in the midpoint which means some of two diagonals is twice as big as A, P plus B, D so let's consider triangle A, P, D triangle which has one side known some of two sides also known and an angle between these two sides so let's just completely change the problem and let's just solve this one how to construct a triangle by side and sum of two other sides and angle between them okay to do that is usually if you have sum of two sides you have to extend one side by the length equal to this one now what do we know about let me just use similar letters or rather this is C and this is G now what do we know about ADD ADC sorry ADC well we know AD we know AC because this is given and actually we know that this triangle BCD is isosceles by construction we know this angle which is exterior angle which means these are halves of this angle so knowing this angle we know half of these angle so we know this one that's important so ADC we have an angle we have a side and we have another side now we can build a triangle using two sides and angle opposite to one of these sides using whatever methodology we used before we did solve this problem many times before so now we know how to build this triangle ADC now after building ADC we draw a perpendicular bisector to to to cg to get the point B our have a point B and now we have built ADC so my point was that knowing AP plus Pd angle between them and AD we can build APD triangle and let's just use it since we can build APD everything else is trivial just extend AP by the same length and extend Pd by the same length and that's how you get to other points of the parallel okay construct a triangle by side altitude on it and medium to another side triangle by side altitude altitude onto this side and the medium to another side so this is altitude BG is perpendicular to AC and AE is medium so these two pieces are congruent to each other so we need base height or altitude and and medium to the side all right so what can we do about it okay here is something which we can consider let's just draw perpendicular from here sorry so we have this now this is half of the altitude obviously because these are equal pieces and that's why if you draw this line it will divide BD in half so EF is half of BD if BD is known EF is known as well so what should we do is we can draw AC let's consider AEC what do we know about AEC we know side which is medium in the big triangle another side AC and we know it's hot so using AC we start with this using this altitude you draw a parallel line and using AE a medium known to you this is a center you just find the point E just using the compass now since you have these three points everything else is easy again the constructive part is finished so you connect C to E you extend it to the same lengths to get the B and that's it that's your triangle so what was important here to basically understand that AEC is a triangle where you also have three elements you remember that for triangles you need three elements you have the same base AC as a big triangle now AE in the medium becomes a side of it but the third element and that's what you have to really you know kind of think about why it's important the height of this the altitude is equal to half of the given altitude only because E is a midpoint and that's why every line parallel to it cuts in half every other line connecting B to any point on the base which we also by the way proved before all right there was a little bit to kind of guess here about this altitude that it's equal to half but okay construct right triangle by a hypotenuse and sum of two pages I think we did something similar as a auxiliary problem for a previous problem but let's do it again so you have a right triangle you have hypotenuse and sum of two cartridges all right so as usual the extend one side and just have this one turned by 90 degree to this one so now if this is ABC our initial triangle what do we know about ADB well we know a lot actually we know about this side we know this side now this is original hypotenuse now this side is sum of two cartridges so we know it as well and we know an angle angle is 45 degrees which we know where it is so by side and another side and an angle opposite to one of these two sides again we solved this problem before we can build a triangle ABG and having ABG we just have BG and perpendicular bisector gives us the point C construct the right triangle by hypotenuse and the difference between two cartridges okay it was the sum now it's difference it's very similar actually so now instead of turning this particular cartridges down to extend another cartridges we have to turn it up why to have the difference between two cartridges now what can we say about this situation so this is original triangle ABC what can we say about ADB what's known hypotenuse difference between two cartridges which is AD and beneath the third element right so we have hypotenuse and difference between two cartridges right so the third element is probably some kind of an angle now we don't have an angle we have 45 degrees here which is sum of these two angles now does it help us let me think about it what can we do about it so in this case we have this angle is 45 degrees in this oh yes absolutely this angle is 135 degrees it's supplemental to 45 right so again same situation we have triangle ADB with one known side another known side and an angle opposite to one of these two sides same thing we can always build it and from this triangle ADB we can find the point C by exactly the same technique perpendicular bisector to BD will give us the C since this is the collateral triangle BDC all right and I think the very last problem given a straight line and two points A and B on one side of the line even also a segment mn segment mn we have to find two points x and y on the line such that segment x y is congruent to mn so they have the same lengths and some of these three segments ax plus xy plus yb is minimal okay let me just remind you another problem if you have line and two points A and B and you have to find a point P let's say such that this AP plus PB is minimal what you're doing in this case you reflect B relative to this line so you draw a perpendicular and extend it by the same length then you connect this is B prime then you connect A and B prime with a straight line and AB prime actually crosses the line at point P and APB would be minimal why because if you will take any other point let's say this one P prime and you connect AP now if you reflect again the P prime B to P prime B prime the lengths will be equal so some of these two will be exactly some of these two but this is not a straight line AB prime is a straight line a straight line is shorter than anything else between two points so by reflecting B relative to this line to B prime and connecting to I you will get the point P to minimize AP plus PB any other point would be greater because AP prime plus P prime B prime is definitely greater than AP prime direct line okay so with a point we know how to solve this problem now what if there is a a segment here instead of a point well we will just reduce this problem to this how well simple let's just draw a parallel line now let's think AX plus XY plus YB if this is a B prime as you see you see XY is a constant so no matter where point B is if you will shift it by this particular segment to the left so regardless of where this particular maybe it will be in this in this position let's say this position X prime Y prime and it will be this way after you shift it you still have that the sum of these two is different from the total sum by a constant so basically what I'm saying is if you shift B to the left by this length parallel to the line to point C and then you solve this problem between A and C then it actually solves the problem where you have this particular segment inserted in between because all you have to do is just add a constant to already minimal way from A to X to C so A to X to C plus constant which is this length is always A from A to X from X to Y and from Y to B no matter where XY is positioned so we find the point X as the point like in this particular case with a minimal sum of distances AX plus XC so after you shift it to the left point B to a position C then you just have to have C prime symmetrical to C relative to this line so they are on the same perpendicular on the same distance draw a straight line from A to C prime and this is where your X is supposed to be then measure this distance from X and that's would be your Y so this sorry this would be your minimum minimum minimum way minimum distance all right so again you shift B to the left and basically you reduce by this segment which basically reduced the problem to minimize the distance with this constant piece in between to the previous problem where you just have to find one point okay that's it that actually completes my set of simple problems related to straight lines parallelograms trapezoid rhombuses triangles etc everything which precedes the circles let's let's put it this way now again these are simple problems these are your toolbox you have to use them for more complex problems and little creative things which we did in every problem are very useful so you have to really like think about okay if it's a trapezoid maybe you have to shift either your side or your diagonal to the right or to the left whatever the additional construction is required the more elementary problems you solved like these simple ones where you usually need just one simple extra line or something like this to basically reduce the problem to something which you have already done before that's important so the more problems you solve like these the better you're off and that's exactly where your creativity actually is being developed as you just think about okay how can i reduce this problem to that problem which i already know so that's the kind of a philosophical skill which i'm making about creativity and about mathematics which in my personal opinion develops this type of creativity which can be used anywhere okay anyway my next set of theoretical lectures will be about circles and then will be another problems and then i'm planning to have some more difficult problems now meanwhile unizor is your focal point for i would say advanced and deeper kind of knowledge about school level mathematics advanced level school level mathematics whatever and what's important is you can obviously just listen to lectures solve the problems etc without any kind of control or processes if you if you wish but i do encourage parents who are interested in um more controlled environment and studying deeper the mathematics with their children they have the ability to do it with unizor by enrolling their children to this or that program to a specific topic or a bunch of topics and which how they pass exams check the scores and basically control you know pass or fail asking your child to repeat the same course again and to go through the same exam again to get better better score all right that's it thank you very much and good luck