 In this video, we're going to conclude lecture 13, and we're going to do that by finally defining what a triangle is. So, a triangle is defined using three non-colinear points in an ordered geometry called those points ABC. Then we define the triangle denoted the triangle ABC as the intersection of three angles, hence the name tri-angle, there's three angles here. So, the triangle ABC is the intersection of the angles ABC, BCA, and CAB. The interior of a triangle is defined similar to the idea of the interior of an angle. So, the interior of the triangle will denote that as triangle with a little circle here for interior, it represents an open set. The interior of the triangle ABC will then be the intersection of the three interiors of the three angles in play here. So, the interior of triangle ABC is the intersection of the interiors of the angle ABC, BCA, and CAB. I've often used this word boundary as we've talked about these things. This is a very topological term, but it's appropriate to define here. We will define the boundary of a triangle. You have this little symbol that represents the boundary of it. So, it's very natural to explain what the boundary of a triangle would be. It's the union of the three line segments, AB, BC, and CAB. We could also talk about the boundary of an angle. We've never defined that officially, but if you're talking about the boundary of the angle ABC, this would then be the union of the two rays, BA and BC. We could talk about the boundary of a half plane. So, we have a half plane determined by the line AB on the side of C. The boundary of this thing is just going to be the line itself, which is why I often refer to as the boundary line. So, this list language is very much consistent with how we've talked about angles and half planes and such. So, a triangle is then the intersection of three angles. We can talk about the interior of the triangle. We can talk about the boundary of the triangle. We can also talk about the exterior of the triangle. This exterior would be all those points which are not on or in the triangle or anything like that. So, we have all that vocabulary here. And so, as we finish our discussion of order geometry, I wanted to provide a proof of a very simple proposition about triangles. There's two parts here. Suppose that triangle ABC is given and P is an interior point to the triangle. Consider the lines determined by the boundary of the triangle. So, line L is AB, line M is AC, and line N is BC. So, if I were to sketch the picture here, we get something like the following. So, we have these three lines here. Let's label this one A, this one B, this one C. And so, therefore, AB is this line L, then line M is AC. So, we have this one and then N is going to be the line BC. So, we get something like this. So, the triangle of course in question is the triangle ABC, which I'm going to highlight this here with a slightly different color. And then suppose that P is some point interior to the triangle. So, that's just the setup, that's what we're talking about right here. So, the first thing I want to say here, part A of this proposition is that we defined a triangle to be an intersection of three angles, but we could also define the triangle to be the intersection of three half planes. The half planes associated to the lines L, M, and N, and we can use the interior point P for each and every one of those. Now, the proof of part A is going to be pretty simple. It basically comes from the observation that a triangle, by definition, is the intersection of three angles and an angle of itself, the intersection of two half planes. So, take for example the angle ABC, which would be illustrated right here in our diagram. By definition, it is the intersection of the half plane bounded by the line L containing the point C. So, we're looking at this half plane right here and then I do apologize the typo there, that the half plane of the line N containing the point A. So, you get this over here. So, when you look at the intersection, you're going to get this portion of the graph. So, that's the angle ABC. But as a half plane is an equivalence relation, you could represent the points with a different one. So, in particular, you could replace the point C with the point P. That's in the same half plane as L. C and P are on the same side as L. But you could also replace the point A with the point P. Again, A and P are on the same half plane associated to the line N right here. And so, you can make that substitution. So, if we do that for all three angles, you could replace each of the angles. So, we did ABC, you could replace BAC and you could replace, what's the other one, BCA with the half planes. You're going to have the half plane bounded by L with P, the half plane bounded by M and P and the half plane bounded by N and P. You'll get both of those half planes two times. But that doesn't make any bit of a difference for intersections. So, you're going to get that the triangle, which is the intersection of three angles that look like this, will look like the intersection of the half planes with L, M, and N using this interior point P. So, it falls very quickly from the definitions of these things. Now, the second proposition, or I should say the second part of the proposition, Part B, I'm going to leave this as an exercise to the viewer. So, what this tells us is that any ray, PQ, is going to intersect the boundary of the triangle at a unique point. So, P is interior to the triangle. If you take any other point, it could be inside, it could be on the triangle, it could be outside the triangle, it doesn't matter. If you take any other point, Q, and you look at the ray that emanates from P and goes off towards Q, then that ray will necessarily intersect the triangle's boundary at a unique point. And so, like I said, I'm going to leave it as an exercise to the viewer to prove exactly that. Now, before in this video, I do want to offer us a little bit of a hint on this one. So, first, construct the ray AP, like so. And so, when you look at the angle CAB, then because P is interior by the crossbar theorem, you're going to have to have a point of intersection where the ray AP intersects the line segment BC, call that point D, like so. And so, what you need to do is sort of consider two cases. It could be the case that actually Q was a point on this ray and which case then the point of intersection you're looking for is D, right? It's going to intersect the boundary there. That's exactly what we wanted, that's great. So then, consider the possibility that these two rays are different. So, the ray AP is different from the ray PQ. Well, in that situation, then this ray AP is going to dissect my triangle into two triangles, okay? In particular, P is going to be on the boundary of this one. So, what you have going on here is that this line, if I extend the line, if you extend the line by posh's axiom, we're going to get that you look at the triangle ADC. This line, since it crosses at P, it's going to have to cross somewhere else. But then when you look at the triangle ABD, again, because the line intersects at P, it's going to have to intersect at some other point. And these two points of intersection between the two sub triangles, they're going to be on opposite rays. And so, in particular, the ray PQ is only going to hit one of those. That's the gist of it. I'll let you provide the rest of the details. But that ends our discussion for here in lecture 13. That also ends our conversation about order geometries in this lecture series. So, thanks for watching. We'll talk about the congruence axioms next time. If you learned anything about order geometry in these videos, please like them. Subscribe to the channel to learn more about geometry and other mathematical things. And then, if ever you have any questions, please post them in the comments below. Thanks, everyone. Bye.