 Hi, I'm Zor. Welcome to your new Zor education. This lecture will be about some very trivial theorems about triangles in geometry. I call it mini theorems actually. Notes for this lecture contain all the conditions for these theorems, and I do recommend you to go through these notes and consider proving it yourself, basically, because they're very, very simple. In this lecture, I will prove all these theorems, obviously, and again, I think it should be considered as some kind of a confirmation that what you did was right, that how you proved it is the method which can be really used for proving the theorems. In some cases, maybe I will come up with a method which is different from whatever you came up with. That's okay. I mean, this is my method and this is your method, whatever. In any case, it's very, very useful, and that's probably one of the main purposes of the whole course, like to solve problems. Okay, these are very simple problems. These are very trivial theorems, as I said. So try to do it yourself first, and then, well, just listen to this lecture and check yourself. All right, so I have this list of theorems which, again, is in the notes for this lecture, and I will go one by one. Prove that in there is also a triangle, two regions, two congruent sides are congruent. Okay, so the first mini theorem is you have an isosceles triangle. These two sides are equal. Now, you have two regions, two congruent sides. A, E is the medium, which means these two segments are congruent, and C, D is medium, which means these two segments are congruent. But because these sides, A, B, and B, C are congruent, this is isosceles triangle by condition, the halves of these sides are all congruence, congruent among themselves. So this is given. Now, we have to prove that these two mediums, C, D, and A, E, are congruent among themselves. Well, obviously, in this case, as in many others, if you would like to prove that something is congruent to something else, most likely we have to include them into congruent triangles and prove the congruence of the triangles using one of the characteristic theorems about congruent triangles, like side, side, side, or side, angle, side, et cetera. So what can be done in this particular case? Well, it's quite obvious. Since triangle is isosceles, two angles at the base, as it was proved before, are congruent to each other. So in the triangles, A, E, C, and A, D, C, consider these two triangles. Now, these two sides, A, D, and E, C, are congruent. These are halves of the congruent sides of the isosceles triangle. A, C is common for both of them. And the angle between them, these two angles, between A, D and A, C in this triangle, and between A, C and A, C in this triangle. They're also congruent. So you have side, angle, side in this triangle. Side, angle, side in this triangle. So obviously these triangles are congruent, and from the congruence of these two triangles, you derive the congruence of A, D, and C, D and the theorem. Prove that in an isosceles triangle to angle bisectors to congruent sides are congruent. Okay, so instead of mediums, we have exactly the same type of triangle, but now we have angle bisectors. This is bisector and this is bisector. And we have to prove that they're congruent to each other. Well, same thing again, but in this case we will do it slightly differently. Since we know that these angles at the base of isosceles triangle are congruent to each other, halves of these, and this is the half, because these are bisectors of these triangles. Half of these are also congruent to each other. So in the same two triangles as before, A, B, C, D, E. But now we have a slightly different theorem to use for congruent triangles. Now, what is this? In the triangle A, D, C, you have actually two angles congruent to triangle A, E, C. Now which one? The angle between these two sides, between this and this, of this triangle. It's an angle, the big angle actually at the base. And it's congruent to an angle between these two sides of this triangle. Now, half of this are again angles in the triangles. Now this half is angle in this triangle. This half. And this half is an angle in this triangle. So again, these two triangles are congruent to each other because they have an angle between these two. Then the side, which is common for both. And the second angle, this one. So I'm talking about this triangle over the A, D, C. So you have angle, side, and angle. In the triangle A, E, C, you have also the same angle, same with this one. Side, which is common. And angle, which is again equal to this one. So we have congruence of these two triangles by angle, side, angle, theorem. And that's why A, E, and C, G are congruent to each other. Now, I will not change the same thing again. But now we are talking about altitudes. So first theory was about median. Second was about angle by sectors. The third is about two altitudes. So you have altitude to this, and altitude to this side. And the theory is again the same. These two altitudes, two sides of isosceles triangles are congruent among themselves. Now, we're talking about altitudes to congruent sides. Well, again, this is right triangles because it's altitudes, right? Now, in the right triangles, it's sufficient to have only two elements congruent to each other if you have two right triangles to prove their congruence. And again, we have common hypotenuse in both cases and you have an acute angle. In this triangle, you have a acute angle congruent to, in this triangle, to this angle. So these two right triangles are congruent and that's why two legs of catechic are congruent to each other. So basically to summarize these three first mini theorems, if you have an isosceles triangle and then you have either two medians, two congruent sides or two angle by sectors or two altitudes, they are all congruent to each other. All right. Number four, prove that segments of two perpendicular by sectors to two congruent sides of a isosceles triangle with the midpoint of one side crossing by sector to the other side are congruent. Okay, now the third element for, or the fourth element for a isosceles triangle is perpendicular by sector of this one and this one. So A, B, C. So you have midpoints. So these are congruent segments. These are half of congruent sides and you have perpendicular by sectors and you have to prove that these segments between two sides, D, E, F, G. The segment D, F and segment G, E are congruent to each other. Well, again, there is nothing difficult in this particular meaning theory either. Again, we have to consider some triangles and try to prove their congruence. But in this case, triangles are on the top side of this drawing. Consider G, B, E. It's the right triangle because these are perpendicular by sectors, right? These are right angles. So G, B, E is a right triangle and B, E is one of its legs, the cataclysm. Triangle B, G, F is also a right triangle and B, D is cataclysm. So we have one common angle and two congruent cataclysm legs of these two right triangles. That's why triangles are congruent and that's why the second set of cataclysm, second legs are congruent to each other. That's it. So now all four elements, which we know of median, bisector, angle bisector, altitude, and perpendicular bisector of the side. So all four elements are congruent to each other in isosceles triangles. Prove that the line perpendicular to an angle bisector cuts from two rays forming this angle congruent segments, assuming that the angle is less than 100 degrees. Okay, so we have an angle and we have bisector. And now we have a line perpendicular to this angle bisector. So we have to prove that these two segments, which this line cuts from these two rays. So you have M, A, B. You have to prove that A, M, and M, B are congruent. What you have is that this is angle bisector and you have that this is a perpendicular. Well, again, very simple, these are two right triangles. M and A and M and B, right? Because this is the right angle. Common cataclysm and two congruent angles because it's an angle bisector. That's why these two right triangles are congruent. And that's why these two legs, A, M, and B, are congruent to each other. Very simple. All right, next. Prove that median A, M over triangle A, B, C from vertex A to opposite side B, C is equidistant from vertices B and C. Okay, so you have a triangle of A, B, C. And you have the median, which means it's crossing the opposite side in its middle. So these two segments are congruent to each other. What you have to prove is that this line, the whole line, A, D, is equidistant from vertices B and C. Now, the distance from the point to a line is measured by its perpendicular, right? So you have perpendicular to line A, G to a median from B, and this is falling to the point M, and from C, which falls to point M. What you have to prove is that these two distances are the same. That's what it means that the whole line, median, is equidistant from two vertices. Okay, as usual, we have to prove that some triangles which include these two segments are congruent to each other. Now, obvious triangles are D, M, D, and C, M, D. And why are they congruent? Well, number one, they are right triangles since these are perpendicular, and number two, they have two elements congruent to each other. These are hypotenoses of both triangles, and the Q-tangle is also, one Q-tangle is also congruent to each other, to another as vertical. So obviously, triangles are congruent, and that's why the M is congruent to C, M. That's it. You see how simple these many theorems are? All you have to do is to find the right triangle to include your element in question and prove the congruence of the triangles. We have to just find what's the right triangle and apply the proper theorem about triangles. Okay. Prove that the lengths of the union A, M, or triangle A, BC from vertex A to BC is less than half some of the lengths of AB and AC lines in between. Okay. Again, we have a triangle, and again, we have medium. So these two segments are congruent, because D is the midpoint of BC. So AB is medium. What you have to prove is, let's use letters C, D, this is A by half and A by half, right? And this is M. So what you have to prove is that medium is smaller than half the sum of two sides it goes in between. So you have to prove that M is less than B plus C divided by two. That's the theory that you have to prove. By the way, speaking about letters, it's usually, it's customary to use capital legend letters for vertices and lowercase legend letters for sides if you want to have some letter for a side. And what's important is that lowercase letter of the side is used for the opposite, the same letter as opposite vertices. Same step here, E of A, and this side is A. C and this side is C. Medium is again, M for medium, H is usually for altitude because it's height. And for angle bisector, sometimes it's L or sometimes whatever. So that's the theory in which we have to prove that the lengths of the medium is less than half of sum of two sides, it lies in between. How to prove this theory, all right. Now, in many cases, to prove certain theorem in geometry, it's not sufficient just to look at the drawing given to you by the condition of the theorem. Sometimes it's necessary to construct something else which would help to prove the theorem. Now, and that's exactly this creativity part which solving the problems really kind of develops in the person, what exactly should we do? What kind of additional drawing, conditional construction, additional elements we have to include into the given theorem and the given condition of the theorem to prove the point? Well, I think the only way to actually learn how to do it is to solve many, many problems and as you solve these problems, this skill will just develop itself basically in you. So again, that's the purpose of the whole course to solve problems and to develop certain degree of creativity, that's what mathematics actually is all about, you develop your creativity. Well, in this case, actually it's a very simple additional drawing which you need, but still somebody, you know, you have to really come up with this. Very simple to do it is the following way. Let's just continue this line AG by the same distance and connect these points, let's call it E. So AG is equal to GE by construction, that's how we constructed it, all right? Obviously this angle and this angle, they are congruent because they are vertical. So what we have is, this is also D because DEG is congruent to ADC, side angle, side, same single theorem, side, angle, side. That's why these two are congruent as well and this is D. And incidentally, exactly the same thing you could say about this one, because triangle EGC is congruent to BGA, again, side, angle, and side, side, angle, and side. So angles are congruent as vertical and sides are congruent by basically either condition of the theorem because this is a median, which means it's a midpoint of this segment or by construction, how we constructed it. I said we will continue the line and use exactly the same length, so these two are congruent. Now it is actually obvious, this, because AE is equal to two M, obviously, because that's how we constructed it. And AD plus DE is B plus C. And as we know, the straight line between two points is always shorter than any kind of a line which contains different segments. That's why we have this inequality. And from this inequality, this one follows by dividing back to both sides of inequality and the theorem. Now, the last problem which I would like to solve for the last minute theorem is very, very similar to this one. And in this case, it says, proof that sum of lengths of all three medians of a triangle is less than the perimeter, but greater than half the perimeter. So now we have three medians, one, two. Again, what it says, the sum of lengths of all three medians is less than the perimeter, but greater than half the perimeter. Okay, this is the last minute theorem. A, B, C, A divided by two and A divided by two, B divided by two and B divided by two, C divided by two and C divided by two. Now, this is median and I will use the index B because it falls from the vertex B to the side B. And this will be median with index A and median with index C. So what's necessary to prove that sum of three medians, M A plus M B plus M C is less than the perimeter, A plus B plus C, but greater than half the perimeter. So it's two different problems actually in one and we will have to prove them one by one. Okay, let's start from this one. This one is actually obvious, why? Because of the previous problem, which I have just proven. I have proven that M A is less than B plus C divided by two and M B is less than A plus C divided by two, right? Because M B is like in between A and C. That's why by the previous theorem, it's less than half of their sum. And similar with A's, M C lies between A and B. So it's less than A plus B divided by two. And if you will add all these three together, you will have on the left side M A plus M B plus M C. On the right side, you will have B plus C plus A plus C plus A plus B divided by two. And this is two B, two A and two C. So that's why it's A plus B plus C. So this one is proven by using our previous theorem three times for each median and then adding them together. By the way, I think basically used one more thing which I didn't really mention about inequalities and properties of inequalities. This is algebra actually. So that's where the algebra means geometry. And one of the properties of inequalities is if you have two inequalities, one thing is less than another and then the third thing less than the fourth thing. You can add the first and the third and some would be less than the second and the fourth element. Okay, so this one is proven. Now we have to talk about this guy. All right, so how can we prove that? Let me think. I don't know, I have to think about it. Well, probably we can do it this way. I'm not sure, but I think that would work. Let's consider all these small triangles. I would call them, I should use some other letters, I think. We have six, okay, let's just number them. One, two, three, four, five and six. Okay, so one, two, three, four, five and six. Four, five and six, right? Now what can we say using the triangle in ecology about these triangles? Well, in every case, this is smaller than some of this plus this, right? This is a straight line between two points and this is the combination of two segments which connect the same two points. So that's why I have six plus three is greater than C2. One plus six is greater than C2. One plus four is greater than B2, et cetera. So let me just do it in some sequence. I will start, let's say, from the side A and have these two triangles first. So it's two plus three is greater than A2. So A2 is less than two plus three. Another A2 is less than two plus five. A by two is less than two plus five. Now B, this B divided by two is smaller than four plus five and this B divided by two is smaller than one but plus four. Finally C divided by two is smaller than one plus six and the last segment C divided by two is smaller than three plus six. If you will add them together, what happens? You have A divided by two once, twice and B divided by two twice and C divided by two twice. So if you will add all the left parts of these equations, in the equation, sorry, sorry, on the left part you will have basically A plus B plus C, right, which is a perimeter. A plus B plus C. Now, what will be on the right part? You have two plus three, but then you have another two. So two occurs twice, three occurs twice, five occurs twice, four occurs twice and one occurs twice and six occurs twice. So each segment, one, two, three, four, five, five, six occurs twice, which means that all together some of them will be twice the total sum of all the medians. So this will be smaller than two M A plus M B plus M C from which this follows by dividing by two and the proof and the end of this lecture. Yes, as usual, let me remind you that the website Unisor.com is an excellent source for information about mathematics and maybe some other subjects. And what's important, it can help parents and supervisors to basically control the educational process by enrolling the students into this or that particular course and then following the results of the exams which students should take, basically making a decision, parents making a decision about whether to consider this particular course as completed or not if exam is not really satisfactory, if the score is not high enough then the parents can say, okay, do it again, basically. So for homeschooling, it's really an invaluable asset. So I suggest you to go to the Unisor.com and consider the whole set of courses and that's it for today. Thank you very much and good luck.