 Hi, I'm Zor. Welcome to Unizor education. I'm continuing with miniseries related to parallel lines. These miniseries might be just a bit more difficult, but nevertheless they're really not very difficult at all. Now, they're all based on whatever we discussed before in the previous lecture about miniseries one. This is miniseries two. And obviously they're based on the most important fundamental property of the parallel lines. That's the characteristic property of the angles when two parallel lines are intersected by some transversal. So, without further ado, let me just go one by one. In the right triangle, with one acute angle of 30 degrees, the leg opposite to this angle is equal to half of hypotenuse. Okay. So, you have 30 degrees right triangle, and this is obviously 60 degrees. Now, I have to prove that the shorter leg, which is across the 30 degrees angle, is half the hypotenuse. Okay, how can we do it? Actually, it's very simple. Let's continue. On this side, reflect, basically. This point, reflect to this. So, since this is a perpendicular, these two segments are congruent by construction, basically. Now, what do we have right now? Well, we have these two triangles, ABG and GBC are both right triangles with common leg and another leg equal to each other by construction. So, they are completely congruent to each other, which means this is also 60 degrees and this is also 30 degrees. Now, what it means, it means that in the triangle ABC, all three angles are 60 degrees. Equal angle triangles, so to speak. And in the triangle across equal length, equal angles by equal size. So, it's equilateral triangle. Since it's equilateral triangle, BC is congruent to AC. And since DC is half of AC, that's why DC is half of hypotenuse BC in the original triangle BCD. All right. A small additional construction of a triangle on that side was some kind of necessary consideration to prove the theorem. Some of all interior angles of a convex polygon with n sides equal to 180 times n minus 2 if you have convex polygon. Let's assume this is convex polygon. Now, convex polygon has a property that from any point within it, all vertices are visible, which means these lines do not cross any sides of the polygon. That's what convex actually implies. Now, if this is true, and we know that there are n sides of this polygon, well, let's just consider all these small triangles. Each one of them has 180 degrees as some of these angles. So if we will summarize them all together, we will have 180 times n. Now, some of all these angles for all these triangles, what's the difference between these some and only interior angles of the polygon? Obviously, we have 360 degree extra when we have summarized all angles because all these angles, they were included into our calculations, but they are not interior angles. All other, however, do combining together make up all the interior angles of the polygon. So I have to subtract 360 which is equal exactly to what's necessary to do. Okay. Now, some of all exterior angles of a convex polygon count only one exterior angle for each vertex with n sides equal to 360. All right, so if you have, again, some kind of polygon and you have one exterior angle on each side, then some of these, in this case, five angles, but in general, it doesn't really depend on the number of sides of polygon as long as the polygon is convex. Then the sum of these angles is supposed to be equal to 360 degrees. All right, let me think about how to prove it. All right, so let's choose one particular vertex and connect it with all... No, I think it's not necessary. I think there is an easier way to do it. Each interior angle, let's call this angle one, two, three, four, and five. Now, each corresponding interior angle is equal to 180 degrees minus one, 180 degrees minus two, et cetera, 180 degrees minus angle number n, whatever the n is, five in this particular case. Now, these are interior angles. Now, we know that the sum of all interior angles is supposed to be 180 times n minus two. Now, if you will summarize this, we will have 180 times n minus s. s is sum of these angles. Angle number one, number two, et cetera, angle number n. These are our angles. I put this sign of the angle so you don't really confuse angle number one and number one. So, that's what we have here, right? We have this equation, 180 times n minus sum of all exterior angles. Now, these are some, this is sum of all interior angles which by previous many theorems we have proved that this is 180 times n minus two. Well, therefore, from this equation what do we see? 180 times n minus two is times n minus 360 is equal to 180 times n minus s. Well, from which obviously s is equal to 360. That's what it's supposed to be. By the way, as you have noticed I did not really prepare myself for this particular lecture with all the proofs. I'm just trying to solve these problems as we speak in real time. Well, that's why I just decided to do one way then realize that maybe it's not necessary and use another one. All right, if point A is centrally symmetrical with point A prime relative to a center P. So, you have one point A and center P and it's centrally symmetrical which means A T A prime is one line and A P is congruent to B A prime. And point B is centrally symmetrical. Okay, let's say point B is also centrally symmetrical. Let's say this is B and this is B prime. Okay, that's what. Segments A B and A prime B prime are congruent and parallel. So, these two segments are congruent and parallel. Well, look, I mean it's obvious two triangles, this is equal this is equal and the vertical angle. So, triangles are congruent since they are congruent it means these two angles are equal and these triangles are equal and this is congruent to this. So, what do we have? We have congruence of these segments and because this and this angle are alternate interior with these two lines and transversal and these angles are congruent to each other that's why the lines are parallel. So, we have proven the parallelism and congruence of these two segments. That was easy. Quadrangle with both pairs of opposite sides parallel to each other have both pairs of opposite sides and opposite angles. Okay, so we have quadrangle with opposite sides parallel to each other. It's called parallelogram by the way. So, this is parallel to this and this is parallel to this. Well, from parallelism it follows that these two angles are congruent to each other which means this and this are supplemental. Here, this angle is congruent to this one with these parallel and transversal. So, we have this angle congruent to this as corresponding angles with these parallel and the transversal. This is congruent to this because of these parallel and this as a transversal. That's why these two are congruent among themselves since they are separate congruent to the third one. Let's go with this guy. Same type of proof. So, now we have angles opposite to each other being congruent to each other. Okay, now we have to prove that the sides opposite to each other are also congruent to each other. All right, I draw a diagonal. I don't need this anymore. So, I draw a diagonal. Now, since this is it can be considered as too parallel and diagonal being a transversal. Now, what do we have here? These two triangles have come inside. Now, this angle is congruent to this because of these parallel. These are transversal and these are alternate interior. And this angle is congruent to this because of these parallel and the same transversal. And these are alternate interior. So, we have these two triangles, this one and this, congruent to each other because we have a common side of these two triangles which are on both sides of this side. They're congruent to each other and that's why opposite sides are congruent because they lie across the congruent angles. The same thing with these guys. So, if you have a parallelogram which means two parallel lines crossing two other parallel lines, you have opposite angles congruent to each other, this and this and opposite sides congruent to each other. Next, even triangle ABC, point N is a midpoint of side BC that requires some drawing. Triangle ABC point N is a midpoint of side BC. This is N, midpoint. Point M is a point on the side AB side should segment MN is parallel to AC. So, we took the midpoint on the BC and draw a parallel line to the base AC. Then, we have to prove that M is also a midpoint of this segment and the segment MN is half of segment AC. By the way, this is called mid-segment. The line connecting two midpoints is called mid-segment. So, basically this theorem is that you can draw the mid-segment by having only one side divided in half and drawing a parallel line and the second that the mid-segment is half the base which it parallels to. Okay, that's right. Alright, let's think how it can be done. Okay, let's draw from N another line parallel to another side. So, MN is drawn from the N parallel to AC and P is drawn parallel to AB. Now, so what do we see? Well, obviously these are triangles which you have to consider M, B, N and P and C. Now, let's think about what can we have from this. Well, obviously since these two are parallel then these angles are corresponding angles with parallel lines and transversal which means they are congruent. Okay, now these are parallel lines and this is transversal so these are corresponding angles. Right? So, we have angle, angle, angle, angle. Now, how about these two sides? Obviously they're congruent because if you remember M was chosen as a midpoint of segment BC. So, these are two halves. That's why they're all equal. That's why triangles are equal and that's why that's why this is equal to this. Alright, let's consider M and PA. This is a quadrangle, a polygon with four vertices with two sides parallel to each other and two other sides parallel to each other. That's the previous theorem and we know that opposite sides are congruent to each other. So, what we have proven here is that this piece is congruent to this and this is congruent to this. That's why they're congruent to each other. That's why M is midpoint of AB. So, if we draw a parallel line from midpoint on this side, we get the midpoint on this side as well. Now, obviously, P is exactly in the same position because how did we obtain P by drawing a line from a midpoint parallel to this? So, P is also a midpoint because of the same property of parallel line drawn through the midpoint of one side. So, these two are also equal to each other. So, what we have to prove right now is that this line Mn is half of AC. Well, but we have already proven that these two triangles, m, b, n and P and C, are congruent to each other, which means the congruence of these two segments. And again, since this is equal to this and this to this, it makes Mn half of AC. So, what we have proven here is that the midsegment connects two midpoints parallel to the base and its length is half the base. It's parallel to. Okay? Next, given an angle, m, a, n, m, a, n, and two points P and Q on the AM side, such that segment AP and PQ are congruent. Okay, so the lengths of these segments are the same. Two parallel lines, two parallel lines, one line and two lines. Two parallel lines and they're crossing another in P prime and Q prime. So, if these segments are equal, the theorem states that these two are supposed to be equal as well. Regardless of what kind of parallel lines, this way or this way, as long as the parallel lines cross another side of this angle in some points, then it cuts equal segments from this side if these segments are equal as well. They're not equal among themselves, but if these are equal to each other, then these will be also equal to each other. Okay, how can we prove that? I think it's very similar to the previous problem. I think what we should do is, we can try, from P prime, if I will do parallel to AM, what happens? Let's say it's point R. Well, since these are two parallel lines and this is transversal, then these angles are congruent. Since these are two parallel lines by construction and the transversal, then these two angles are equal to each other. Okay, I think I'm almost right. I think I should start from another point. Let's see. I think instead of parallel from P prime, I have to put parallel from P this way. And this is point R. Yes, I think that would be better. So from the P, I draw a line parallel to AM until it crosses the Q, Q prime in point R. Now what do we have? These two lines are parallel and this is transversal, then these two angles are congruent. Now, since these two, one second, which one do we have to consider now? Yes, since these two lines are parallel and this is transversal, these angles are corresponding to each other, two parallel and the transversal, so these angles correspond to each other. And we have given that these two segments are congruent to each other. That's why triangle A, P prime, P and PRQ are congruent to each other. So both sides, and two angles on both sides. Now if that's true, it means that these two sides are equal to each other. AP prime and PR are congruent because they are corresponding sides in congruent triangles. But at the same time, PR and P prime, Q prime, they are parallel to each other because these are parallel lines. This is the same quadrangle with opposite sides parallel to each other. So the opposite sides are supposed to be congruent. And that's why we have these two pieces, these two segments, both independently congruent to one and that's why their lengths is the same. End of story. There are a certain number of equal segments on one side of the angle. By the way, it's not necessarily two. You can count three, etc. The proof is exactly the same. And then you draw parallel lines to another side of the angle. Then on another side you will also have equal segments. And as an interesting way to divide a segment into any number of parts, equal parts, you can consider the following. For instance, you would like to divide this segment into whatever, seven equal parts. How to do it? Well, have an angle and have any segment you want. Doesn't matter. And put it seven times here. One, two, three, four, five, etc. Then next you connect the end with this and draw parallel lines from corresponding points down. So if these were equal to each other in lengths by construction, then these will be also equal to each other because of the previous theorem. And that's how you divide the particular segment into one, two, three, four, five, five different parts equally. Okay. And one more thing. A theorem about medians of a triangle. So let's say you have three medians. What's interesting is that this particular point, let's say P, D, D, E, F. So the point where medians are crossing each other and we did actually talk about why they're crossing in the same point, all three medians. What's interesting is this point divides each median into two to one ratio. So this piece is twice as big as this one. And the same is this one, these two pieces, and these two pieces. Each piece which is closer to the vertex is twice as big as the one which is close to the opposite side. How can we prove it? All right. Actually, we did something very similar. Let's draw two lines, F, F prime and E, E prime, parallel to A, D. Okay. Now, since this is equal to this, because F is the midpoint, now these are two parallel lines, that's why these two are equal to each other as well. So you have an angle. You have two equal segments here. You draw two parallel lines to another angle and that's why you have these two congruent to each other as well. Same thing here. Since this is congruent to this, you have these two pieces, these two segments, this and this. These are parallel. This is an angle. These segments are equal. That's why these two will be equal. Now, why did I use the same double strike here as here? Well, these two are halves of CD. These two are halves of BD. But D is the midpoint of BC, which means each one is equal to a quarter of BC. That's why they're all equal to each other in lengths. And since we have established that, let's now consider angle E, B, C. So this side, BE, is the median. Here we have one, two, three equal segments on this side and we connect it on this side. That's why these three will be equal as well. So now the angle is this and these are three parallel lines. And that's why we have this situation. Since these three are equal to each other, we have this is one-third and this is two-thirds. That's it. Thank you very much.