 Welcome back everyone to our lecture series Math 3130 Modern Geometry for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Misadine. It's good to have you. In part 31 of our series, I want to talk about today the axioms of area. Throughout this course, we've talked about a lot of undefined terms such as incidents, points, lines, between, congruence. I think that basically covers most of them. Maybe I missed one or two there, but it turns out that's just the tip of the iceberg on the type of undefined terms we could have in a geometric axiomatic system. Today, we're going to talk about yet another undefined term, the notion of area. After all, if you had to explain to someone what area is, do so. Could you do it really quickly without using really synonyms to describe it? Areas, how much stuff is in a two-dimensional figure inside it? It's not so easy. In fact, we're going to treat it as a logically undefined term for axiomatic system. I want to talk about this today about area in general congruence geometry before we talk about area of hyperbolic geometry in the next lecture. With this undefined term area, we try to make meaning out of it using our axioms. First of all, by area, we mean a function from triangles in a congruence geometry to a unique real number, which of the triangles denoted ABC, we'll call it the area of the triangle ABC. We'll often use A to denote the area when it's clear from context. In which case then, we have to define axioms so we can understand what this undefined term means. For area, there's essentially three axioms. The first axiom we'll call it the positive axiom, which says that the area of each triangle is a positive real number. I had mentioned earlier the areas of function. That's really putting the cart in front of the horse there. The positivity axiom of areas specifying that associate to each triangle is a positive real number, which we call area. Axiom 1 is a definition of what we mean by this undefined term, but that seems backhanded there. Really, that's what we always do with axioms. We're giving meaning to the meaningless, the undefined terms there. Area will always be a positive real number. No triangle can have zero area. That's not a triangle. No triangle can have negative area, not even situations like hyperbolic geometry. Now, admittedly, one only needs a notion of congruence in order to define area. These area axioms, this area undefined term, will make sense even when we are in places like elliptic geometry and we don't have the full-blown axioms of neutral geometry. The second axiom of area we're going to call it congruence preservation. It gets not the most clever name out there, but it'll serve its purposes. This axiom tells us that if two triangles are congruent, then they will have equal areas. After all, area is a real number, and so we can talk about equality of those. Congruent triangles will have equal areas. And lastly, the third axiom, the additivity axiom, tells us that if we have a point D that sits between A and B, then we can actually dissect the triangle into two pieces. I should have left myself some space here on the board, but let me kind of sketch out what I mean by this. So if we have a triangle whose vertices are A, B, and C, and we pick any point that's between one side, if we pick a point between two vertices of the triangle, so D sits between A and B, if we consider the two triangles, so we cut this in half along that median there, then the triangle A, D, C, if we take its area, which is a number, and we add it to the triangle D, B, C, the sum of the, so we take the area of this one and the area of this one over here, if we add these two areas together, we retain the area of the original triangle. So that's our additivity there. Now in terms, so area is a defined, I should say areas defined for triangles, but we can extend this to the area for polygons in general, because after all a polygon is just going to be a union of triangles, maybe something like this. Just as an example of a polygon, well as this triangle has area, so does this one, so does this one, so does this one, each of these triangles has an area, and as the polygon is the union of all of those triangles, we'll take the area of the polygon, so the area of our polygon is just going to be the sum of the areas of each and every one of the triangles that shows up in this sum. And in fact, this way of defining area for a polygon is going to be a well-defined notion, because after all when one has a polygon, there could be multiple ways, multiple different ways of partitioning the polygon into triangles. Well, as two different partitions by triangles, we'll always have a common refinement. Because of the additivity principle, every refinement will have the same area as the original partition did, and so since partitions have common refinements, we see that area will be well-defined. And this sort of this way of extending area to general polygons and getting that the area as a well-defined function for polygons, this reason is the exact same reasoning we used to extend the notion of defect in hyperbolic geometry from triangles to polygons and showed why defect is well-defined for all polygons. And the arguments there are essentially the same. What we're going to see, we'll talk about this in the next lecture, but what we're going to see is that the hyperbolic defect function actually satisfies the axioms of area, and therefore bizarre as it might seem, defect is an area function in hyperbolic geometry. What I want to do today, though, in this lecture is focus mostly on area in elliptic geometry, but let's first kind of continue on with this general notion of area in geometry. We say that two polygons are equivalent if there exists a partition of adjacent, non-overlapping triangles for each polygon such that there exists a congruence preserving one-to-one correspondence between the triangles among the two partitions. It's kind of a mouthful there. But the idea is if we can take a polygon and dissect it into, you know, let's start off with say a triangle, right? If we cut that triangle into two right triangles like this, right, and then we copy, we sort of like copy and move pieces around, you know, let's say we don't want it, we don't want this poly, or this triangle on the right, let's say we cut it off, and then we move it, say like down here or something, something like this. This polygon we've created, this was a quadrilateral now, this new polygon is equivalent to the original one. We just dissected it and moved some triangles around. That's what we mean by equivalence, that there's this congruence one-to-one correspondence between the triangles and the two different partitions. Well, from this definition of equivalence, we can actually state a theorem relevant to area that polygons that are equivalent, that is polygons are equivalent if and only if they have equal area. This having equal area gives an equivalence relation on the set of polygons, that makes sense. And it turns out that this notion of equivalence we just defined on polygons is the equivalence relationship defined by area. Now, equivalence implying equal area is fairly straightforward. I mean, there is an argument that has to be made there, but it's a pretty quick application of the area axioms that I'll leave for the student to do here. But the basic idea is by the congruence preservation axiom congruent triangles will be have equal area. And so if two polygons are equivalent, there's this congruence one-to-one correspondence between the two polygons looking at these sub triangles. Well, that that congruence correspondence will those we equal areas for each of the triangles involved, it's the same number of triangles. So all of those are there. Additivity of a polygon, that is the activity area of a polygon, you add those up, that's all equal. That direction is pretty easy. The other direction is much more challenging to show that if two polygons have equal area that they're equivalent, because after all, if they are polygons, we don't necessarily have to have the same number of vertices and such, you have to find some way of constructing this one-to-one congruence preservation between the triangles. It's certainly doable. It's a very lengthy combinatorial argument. It comes down to like intersecting triangles and triangles overlap triangles. You get more triangles and quadrilateral. There's a lot to it. I'm not going to provide the detail here. Intuitively, it kind of feels clear that's not a good enough proof technique, but that'll be sufficient for this lecture series here as well. And with that, we actually can talk about what we've got an area geometry. If we have a congruence geometry that satisfies the three axioms of area, then we call it an area geometry. And honestly, when I say congruence geometry, I even am willing to accept a weaker notion of congruence geometry than we originally defined, because the way we've defined the congruence geometry to remind you is we took Hilbert's axioms of incidence, then we stacked on top of that the axioms of order or betweenness, depending on what you want to call them. That gave us an order to geometry, and then we stacked on top of that the six Hilbert axioms of congruence, and that's what we call a congruence geometry. In order to define area, all one has to do is we have to know what a triangle is. That does come from the order axioms or triangles is the intersection of three angles, angles has to do with betweenness. We need betweenness preservation. So we do not know we need to know what congruence means. So as long as we have a strong idea of what congruence of triangles is, that is what congruence of triangles means, then we have a notion of area. And so that's what we mean by congruence geometry, we just need to know what congruence of triangles are. And so the side angle side axiom is basically what we need to have. And so we need to have a geometry for which side angle side makes sense. And we define triangles using order, but one can define what a triangle is without using the Hilbert axioms of order, like in an elliptic geometry or spherical geometry, we can talk about exactly what we mean by a triangle, even though betweenness doesn't quite work, trichotomy fails, for example, if you're on a sphere. And so in that context, we can still have area, and we'll actually talk about elliptic area a little bit later. So just a little caveat here, when we talk about congruence, we just need to have congruence of triangles. And then we get an area geometry. We'll revisit this issue when we come back to, when we come back to elliptic geometry later this semester, or I should say later in the series. So I want to take, take the rest of this lecture and kind of focus on Euclidean area. And that terminology itself might seem very alien to the student right now. What do you mean by Euclidean area? Is area a Euclidean thing? And the answer is absolutely yes. I mean, because first of all, when I ask you what's the area of a rectangle and you reply base times, base times width there, or base times height, length times width, whatever you want to call these two dimensions there, it's the product of the two dimensions, really your price should have been, what's a rectangle? You know, the fact that we're talking about rectangles means we're talking about Euclidean geometry. Rectangles only exist in Euclidean geometry. So if that kind of makes sense, okay, the area of a rectangle is Euclidean, but what about like a triangle or a right triangle or parallelograms? Those things exist in hyperbolic geometry and other non Euclidean geometries. Can we talk about area for those as well? And we will, but the area function will be fundamentally different in like, say hyperbolic or elliptic geometry compared to Euclidean. And in fact, the classic formulas that we know from geometry class will in fact only be true in Euclidean geometry. So let's kind of first start off with, if we're in Euclidean geometry, why does the area of a rectangle have to be length times width? Is there anything that determines that? Well, if we start off with a rectangle, so we'll take the rectangle ABCD, which I'll draw for you on the screen. So this is a genuine rectangle we're in Euclidean geometry. So we have these four right angles, like so. Whoops. Let me fix that. And so we'll, so we have a specific length of the rectangle. So let's say that the side AB, we'll call that the length of the rectangle. And let's say that the side CB is its width. Now, because rectangles are parallelograms and Euclidean geometry, opposite sides of parallelograms are congruent, it doesn't matter which side we talk about the length of the width. There's a little bit of ambiguity there normally, but Euclidean parallel posture gives that the length is well defined, whether we mean AB or DC doesn't matter. And width as well, we could talk about BC or AD didn't matter. So what we're going to do is we're going to take one nth of the length. So take your pot, your favorite positive integer, and we're going to find points. We can find points say BN between A and B, and CN that sits between D and C, so that the length of the segment ABN is exactly one nth of the segment AB. That is, it's one, it's L over N. So we can divide the length like so. And so because, so we get something like this, and at the moment we see now we have a secondary quadrilateral because this side right here is going to be congruent. So ABN is going to be congruent to DC and how do we know that? Well, because we have still a parallelogram, let's see, do we know what's a parallelogram yet? We do know, well, I guess I might take it back, we don't have to worry about parallelograms all yet. We do know we have two right angles from the original rectangle, and we chose the points BN and CN so that the corresponding side lengths are congruent to L over N. So they'll be congruent to each other. Since we have a secondary quadrilateral in Euclidean geometry, that does force that the summit angles are right, and therefore we have a rectangle. And by the additivity principle of area, because we did this once to get one, we got a rectangle, but then we could kind of tile the original rectangle using these congruent these congruent rectangles. And so by additivity, we're going to get the area, the area of the original rectangle, ABCD, this is going to equal one or N times the area of these smaller ones, ABN, CNDN. And this comes from the additivity of those. Since, well, additivity and congruence preservation, because we can tile using congruent rectangles here, here, here, all along the way, we can tile the original rectangle using congruent smaller rectangles. And so then we get the area, the larger one is one and the smaller one as the smaller one we did. And so this right here is essentially using the notion of similarity that exists in Euclidean geometry. It's sort of a horrible pun right here by a similar argument. I'm talking about the similarities of the polygons, the smaller rectangle is similar to the larger rectangle. Well, I mean, not I should be careful here. I mean, it's not the lengths, the lengths are similar, the widths are actually equal to each other. We didn't shrink those down at all. So it's not exactly proportional in the way I usually mean by similarity here. But but in Euclidean geometry, you can take smaller versions of the things, again, only distorting the length, not the width. But you can tile, you can tile rectangles using smaller rectangles. Tile ends and hyperbolic geometry are fundamentally different. Tessellations, those type of things, but we can tile a rectangle using smaller rectangles. And so that gives us this relationship that the area of the bigger rectangle will be in times the area of the smaller one. Well, we can do that with the length, but we can also do it for the width. We can tile the width of this thing vertically, like you see before. And so then we get another rectangle, which is going to be one inch of the original one. And so this smaller one will be proportional to however small we want them to be. It's going to be small, small like that. And so because because when we put these things all together, then the area, the what am I trying to say here, the length of a rectangle is directly proportional to the area of the rectangle. That's what I'm trying to get from this first statement right here, that the the length is proportional to the area. And likewise, the width would be proportional to the area as well. And so therefore, we come to this equation right here, the area of a rectangle, it'll be proposed jointly proportional to the length and the width. Right, that's what we see right here. And then we have this proportion constant whenever one has, whenever someone has a proportion, we have a constant that measures how close the proportions relate to each other. All right, and this proportionality constant k is arbitrary. And in fact, it defines a measurement scale, you can think of. For example, if anchor of acres is our desired measurement for area, you know, people who have farms and houses like to measure their property size and acres, we could measure distance and miles, which then would be appropriate to measure area also in square miles. But if we want to measure an area, right, what's the conversion from a square mile to say an anchor, right? Well, the area of a rectangle would actually look like a equals 640. So with this example right here, the area looks like 640 times length and width, where length and width are being measured in miles, but a is being measured in acres. And the idea here is K, your your proportionality constant is just the conversion of 640 acres per square mile. Now, often in practice, though, we will measure area as a square unit of length and the same length that we use to measure the same distance I should say to measure the length and width of the rectangle. And therefore, in that situation, you often take K to equal one. And in that setting, you then regain the usual area formula for a rectangle area's length times width. And so because of the because of the proportionality arguments we can make about rectangles in Euclidean geometry, the area of a rectangle has to be length times width. And so this in some regard could be taken as an equivalent to the Euclidean parallel postulate. From this important formula, that is the area formula of a rectangle, we can recapture in fact all all of the usual area formulas of polygons and other various geometric shapes. So one could find the area of a right triangle of a standard triangle of a parallelogram of a trapezoid of anything you could think of. All of these things come from the area formula of this rectangle. So I want to illustrate this with showing that the area of a parallelogram is equal to the product of its length, the product of its base and its height. So we have to be a little bit careful, we understand what these words even mean. So if we have a parallelogram, like you see in front of us, we'll call this parallelogram ABCD. So by base of the parallelogram, what we're talking about is really just one side of it, the length of the side. And when one draws this parallelogram, we typically refer to the base as the bottom side, assuming there is a vertical side, the base. So we did a similar thing when we talked about security quadrilateral is the base. Now the height of the parallelogram is the distance between opposite sides of the parallelogram. And as the name suggests, opposite sides are parallel and Euclidean geometry, this distance, which we might illustrate like this right here, this distance, the height, it doesn't matter which points you choose, because parallel lines and Euclidean geometry are equidistant. And so the distance between one point of the line is the same as any other point of the line. And so when we talk about if you call say the base little b and the height little h, then the area is going to be base times height, which is essentially the same thing for rectangles. After all, rectangles are special types of parallelograms for which with a rectangle, the height is actually a side length of the rectangle. And the base is always a side length. So this gives us the formula we had before. So with that in mind, we can be clear of what we're trying to say when we say the area of a parallelogram is base times height. So let's, let's let's prove this formula. Now, if we had a rectangle, we would be done just like I said a moment ago. This, this formula for the area of a parallelogram generalizes the notion of rectangles. So we don't have to prove it for rectangles. So let's assume we have a non rectangular parallelogram, much like I drew before. So and so if it's not we are in Euclidean geometry, the angle sum of this parallelogram has to add up to be 360 degrees. That's true because we are Euclidean. And for parallelograms, opposite angles are congruent. So like angle C is congruent to angle B, right? And angle D is congruent to angle B right there. And so if one of the if we have a parallelogram and one of the angles was right, let's say for example that angle A was a right angle. Well, that would mean that angle C is also a right angle. And then between B and D that still remains 180 degrees to be accounted for. But since B and D are themselves congruent, you have to have two angles that have to be 180 that are congruent, they'd have to be right angles as well. So if a parallelogram has one right angle, then all angles are going to be right. And that's that's equivalent to Euclidean parallel postulate. But we're in Euclidean geometry, so we're using that fact right now. All right. And so what I'm trying to say here is if we're non rectangular parallelogram, we can assume that some of these angles are acute and some of them are obtuse. The way that I've drawn this picture, we can assume that angle B and angle D, which they're congruent, we can assume those are the obtuse angles. And then therefore A and C would have to be acute. All right. And so with these, with this situation in mind, let E and F be the feet of the perpendicular lines dropped from D and C onto the line AB. So the foot of the perpendicular drop from D, we're going to call F. And then the foot of the perpendicular drop from C, we're going to call E right here. And so as the picture seems to suggest, F is between A and B, and B is between F and E. This is actually why we cared about the obtuse angles before. By assuming that D is an obtuse angle, this will guarantee that F is between A and B. And this also will guarantee that B is between F and E. There is a between this argument there. But if this weren't true, then it would force the quadrilateral to have an angle some greater than 360 degrees. Or basically, we could construct a triangle, which has an angle some greater than 180. So really, the fact that D is obtuse will guarantee this picture even in hyperbolic geometry, even, even though we only care about Euclidean geometry right now. So because this parallelogram is Euclidean, opposite sides of the parallelogram will be congruent with each other. So in particular, the side AD is congruent to BC. Because the, because parallel lines are equidistant, the line DF is congruent to CE. And as these are right triangles, the hypotenuse leg condition applies and guarantees that the triangle AFD is congruent to the triangle BCE. So we got this triangle here, and this triangle right here, they're congruent to each other. And so if we, if we focus our attention on now the quadrilateral FECD, so Foxtrot Echo Charlie Delta, which I'm coloring here in yellow, this is going to be a secondary quadrilateral, because angles F and E are right. And we have that's the the equidistant lines DF and CE are the same. That's what we meant by equidistance between the parallel lines there. So we have a we have a secondary quadrilateral, it's Euclidean, so it has to itself be a rectangle, like so. You'll notice that once the top side of the rectangle is none other than I mean, it's equal to DC, which is congruent to AB, which was the base of the parallelogram. And then the side lengths of this rectangle is just the height. The other the other side of the rectangle is the height of the parallelogram. So you'll notice we've now constructed a rectangle whose area is base times height, right? But this rectangle we've constructed is equivalent to the original parallelogram, because basically we just took this triangle ADF, and we move it over here, because BC is congruent to it. And so the rectangle we formed is equivalent to the parallelogram we started with. And therefore the area of the rectangle was base times height, which will give us the area of the parallelogram. And so this kind of can show you the importance of equivalence equivalency in terms of area. If you have an area problem, that's hard to do. You can always dissect it, move some triangles around our other polygons around to something equivalent. And go go from there. A common argument that we can use in hyperbolic geometry is to take a triangle and to replace it with its equivalent secondary quadrilateral. That's a topic we'll talk about another time here. So the last thing I want to talk about here in this lecture is the idea of a shearing map. Because I mentioned before that all of the area formulas we know from Euclidian-Jomchi can be derived from what we already know about rectangles. We just proved something about parallelograms. You can prove the area formula of a triangle is one half base times height, basically from the argument that a triangle is half of a parallelogram. Again, there's some details to provide there, but I'll leave it for the student to finish that up. Area formulas for everything else will come from those. Now that'll take care of the area formulas for polygons. If you're talking about areas of general regions like circles and other sort of non-polygonal regions in the plane, you might have to use calculus to talk about the areas of those things. And if you're doing calculus, you need some type of continuity axioms such as like the Dedekins axiom, for example. So we won't delve into all of that, but if you think of like in calculus one, calculus two, the Riemann sum, the area under a curve, you can approximate the area under the curve using rectangles. And then you take the limit as the number of rectangles goes to infinity. That gives you the area of the region. Those integral problems you do in calculus come from areas of rectangles. And so all of those even non-polygonal regions, the area is derived from this Euclidean area formula. So let's talk about the idea of a transvection or a shearing map. What do we mean by that? Well, a shearing map is one that distorts things in the plane. I should say displaces things in the plane according by some type of angle here. So let's say we have an axis, which we could think of as like the x axis. And we're going to take some to shear, say horizontally by a certain direction. What this does is you're going to proportional to how far above the x axis you are is how far you'll move. So if you're one unit above the x axis, you'll move one unit to the right. If you're two units above the x axis, you'll move two units to the right. If you're three units above the x axis, you'll move three units to the right. And if you kind of connect these dots here, people who live on the x axis don't get moved whatsoever. But people who are below the x axis, they are going to move to the left proportional to how far below they are. So that's what this definition means by the signed distance. It occurred to me I never actually read the definition here. In plane geometry, a shearing map is a map that displaces each point in a fixed direction. What I have illustrated here is a horizontal displacement by amount proportional to its signed distance from a line that is parallel to that direction. So if we're like, oh, we're going to shear things horizontally, we take the x axis as this directional map, it's if we're horizontal, that means we're parallel to the x axis. And based upon how far above or below the x axis will move things to the left or right. You can share, of course, in any direction with any axis, we'll think of the x axis right here. If we try to think of how Euclid thought of these things in the elements, if we could think of this in terms of like parallelograms, we could say that a parallelogram has been sheared is if you take the original parallelogram A, B, C and D something like this. And so if it's a parallelogram, that means opposite sides live on side of parallel lines. Well, if we keep one of the sides fixed, so we keep the base fixed, and we take this segment right here DC, and we translate it somewhere else down the line, keeping the same length that we did before. And we connect the dots. We will make a new parallelogram. And this new parallelogram is a shear of the first one. And of course, we can do this for general polygons as well. But this is how one would could share a parallelogram. And one thing that that Euclid noticed and he proved this in the elements is that when you share a parallelogram, the area doesn't change. Just so you're aware that in linear algebra, if you want to shear, this would be multiplication by a map of the following form 1m01. This is a horizontal shear in the m direction. And these things don't change the area of the parallelogram. In linear algebra, this is realized by saying that the determinant of this matrix is one and in determinants is a way of measuring area in linear algebra. But I'm not going to say too much more about that. So let's talk about how the area is invariant when one shears it. What's a geometric argument for that argument I was making earlier about linear algebra. So it suffices to look at the unit square, because if we can show that the unit square, its image is going to be one squared unit, then this will be true for all regions in the plane here. And so let's say we shear the unit square, it is a square after all. And so there's basically two pictures you get if you shear the square up to like reflections and symmetry and things like that. Let's try this again. So let's start off with the unit square, which you see illustrated right here. So we're going to label it A, B, C and D. And let's shear it. Let's shear it in the following manner. And so we get that we have some point E right here, some point F right here. And so we have this parallelogram right here, that's a shear of the unit square. So we want to show that these two things have equal area. That's one possible picture you could have. Another possible picture, another possible picture you could have is something like the following. So let me sketch that real quick. So we get our unit square again, trying to be conscious of how I label things. So if you compare the two pictures, you get something similar. So you get this, it could also be that the shear, and I should have done this side up here in red. It could be that when you shear, you get something like the following. And so the issue is basically the following. In turn, how are the points D and E related to each other? When you look at the top line, do we have A, D, E, F as a between this relationship, or we have a dash E dash D dash F, both of these pictures are possible, and we have to consider them and up to without the loss of generality, these are the only two pictures that are available to us. Now, because we do have parallelograms and opposite sides are congruent, we do have that A, B is going to be congruent to DC. This is true for both situations. We have that be will be congruent to CF. Again, this is true, whichever the two situations you're in. And then the last part is that BC will be congruent to AD, which is its opposite side. But BC is also the opposite side of the parallelogram EF. So we get that EF is congruent to AD, which will be useful in our observations here, because they're both congruent to BC. So AD was congruent to BC. We mentioned that one. And what else? Yeah, let's go on from there. So if you look at the first picture where E comes after D, what's going to happen is these lines are going to cross at some point, we'll call it G. And so if you use that point G introduced right there, consider, it's going to be the intersection between AF, I'm sorry, not AF. It's going to be the intersection between BC. I might have labeled my points a little incorrectly here. Sorry about that. Not BC. We want BE and we want DC. I have to double check those later. And so I hope my other things aren't labeled so incorrectly here. But if we look at the triangle AFD, yep, that's not right. Sorry about that. What triangle we mean would be the following. Let's take the triangle. Well, there's a couple triangles we want to look at here. But if you take the triangle ABD, this one right here, it is congruent to the triangle DCF. So we have a congruent to triangles there, DEF. And those are congruent by side, side, side. Right. And I think I fixed the typo before I started the lecture here. I think it should be correct what I'm trying to say right here. And so trying to figure out which triangles I'm interested in here. The triangle ABE should be congruent to DCF. So let's go back in time a little bit and see what I meant. So ABE, ABE is congruent to DCF. So ABE is congruent to DCF. Okay, that was the picture I was trying to draw earlier. Is that even correct? Well, I'm perplexing myself right now. I'm sorry about this. I'm going to fix these. I'm going to fix these typos in the lecture notes. Because there's some issues going on here. I apologize for that confusion. But basically, what we're trying to get here is that when you look at the triangle, or look at triangles here, you get a triangle here, you get a triangle here, those are congruent triangles, whatever their labels are, you get this triangle right here is congruent to itself. Notice how this yellow triangle is part of both parallelograms. And then you have this little guy as well. It's kind of part of that's involved in both of them. So you're taking a difference. So you have that big yellow triangle plus that little triangle, the small triangle minus that one. And so you can argue that this square we started off with is equivalence to the rectangle by some dissections. And like I said, I will fix these in the notes to make sure all of these vertices are correctly. That's if you're in the situation where E comes after D. If you have the situation where E comes before D, it's basically the same idea where you take this triangle right here, which is equivalent to this or it's congruent to this one. And so by moving those around, you get equivalent things again. And so the shearing of a of the shear of any map will be equivalent to the original one, despite my little typos in the proof there. And so shearing will be shearing maps will always be will always preserve area there. And so it's really kind of interesting to see that the existence of arbitrary rectangles, which is an equivalent statement to the Euclidean parallel Bachelet, in conjunctions with the area axioms is sufficient to prove the traditional area formula. And then all the other area formulas can be derived from the area of a rectangle. So our usual notion of area can thus be described as Euclidean area. And the conference is also true that in an area geometry, it's Euclidean if and only if rectangles are proportional to its length and width. Next time I want to talk about area of hyperbolic triangles and hyperbolic area just in general. And we're going to see it's very, very different from from our notions of Euclidean area. And so that concludes our video today. I want to thank everyone for listening. If you liked what you what you what you saw here, feel free to subscribe. Make some comments. If you have any questions, please comment below. I have a link in the description of this video to give you a script of what we saw here, which will be updated since this video had some errors in this very last proof. So take a look at that sometime if you're interested. And I'll see you next time. Bye.