 In this video, we're going to state and prove the so-called Hinge theorem, which is a theorem of congruent geometry, so the geometry which we have the notions of congruence between this is incidence with their appropriate axioms as well. So the statement of the theorem is the following. We have two triangles ABC and triangle A prime B prime C prime, for which we suppose the segments AB and A prime B prime are congruent to each other, and we suppose that the segment AC is congruent to A prime C prime. So if we illustrate this just really quickly, we have something that basically looks like the following, and I'm going to try to draw these a little bit so that it looks kind of like the distance is the same here. So here's A, here's A prime, here's the point B, here's the point C, here's the point B prime, and here's the point C prime. And so by assumption, the segment AB is congruent to the segment A prime B prime, and the segment A prime C prime is congruent to the segment AC. You'll notice, of course, I omitted the third side of these triangles. This, of course, was intentional for the reasons that I'll explain in just a second. So the statement of the Hinge theorem is then going to be that if the angle A is larger than the angle A prime, that implies that the side BC is larger than the side B prime C prime and vice versa. So if the side was larger than the angle is going to be larger as well. And so I want you to compare these pictures here. And so I could have done a better example to distort this thing. Maybe I ought to have done so. I'll do that in just a second. But the idea is, when you have a bigger angle, you get a bigger side length, like so. Again, this isn't the best example here, but the larger the angle, the larger the side length. And so to even better further illustrate this and also explain the name behind it, why is it called the Hinge theorem? Well, I want you to think of like a door that's attached by its hinge to a doorway. The dimensions of the door don't change. The dimensions of the doorway do not change as well. But as you allow the door to open, the larger the angle gets, the larger the corresponding segment gets as well. And so for these segments here, here's A, here's B, like so. As we allow, and so each of these segments are approximately the same length. I should at least try to draw them there. So consider these all possible points C. We'll say C, well, I'll call C prime, C double prime, C triple prime, something like this. And so notice that as the angle enlarges, you have the corresponding segment there. And so then again, the angle gets larger and the segment gets larger. And then as the angle gets even longer, we get the segment is longer, like so. And we do this one final time. As the segment gets longer, it must have been that the angle was bigger as well. So the bigger the angle, the larger the segment and vice versa. So as that door is opening, as the angle between the angle of the hinge itself, because the position of the hinge gives you the angle of your doorway, as the hinge gets larger, as its angle gets larger, the gap between the doorway and the door gets bigger, bigger, bigger so that we can walk through it. That's what the hinge theorem is all about. And so the hinge theorem, of course, it's an if and only of statements, we have to go in both directions. The first direction is we're going to assume that the angle A is larger than the angle A prime, or more specifically, the angle BAC is larger than the angle B prime, A prime, C prime. And so with that in mind, we're going to try to prove that the segment BC is larger than the segment B prime, C prime. And so let me scooch the page down a little bit and we're going to draw a picture. And so our picture is going to look something like the following. We're going to have our triangle ABC. I will put it right here. So here is A, here is C, here is B. And for the sake of convenience, I'm not necessarily going to try to draw these triangles to scale, because there are some congruent sides there, but the second triangle is going to be on the screen, mostly just for the sake of reference, but I will intentionally make it look smaller. And again, I'm not trying to make these look like similar triangles or anything like that. So we have these triangles right here, ABC and A prime, B prime, C prime. Remember the assumptions we have here, we're assuming the segment AB is congruent to the segment A prime, B prime. We're also assuming the segment A prime, C prime is congruent to the segment AC. So those are the assumptions here. And then we have the further assumption that angle A is larger than angle A prime here. So this angle is larger than this angle here. And as such, that means there exists some interior point D to the angle BAC. So that when we draw that ray, you get something like this. So we have some point D inside the angle BAC. So if we look at the angle BAD, that's going to congruent to the angle B prime, A prime, C prime. But of course, by segment translation, we can copy the segment A prime, C prime somewhere along this ray. And I'm actually going to do that without velocity of generality. I can assume that the point that forms the congruence will be D itself. And so we might get something like the following. Here is our point D. And so we can then assume that the segment A prime, C prime is congruent to the segment AD. But we also still have the congruence of angles. The angle BAD is congruent to the angle B prime, A prime, C prime, like so. And so then I need to finish up a triangle here. So we have the triangle BAD. We have the triangle B prime, A prime, C prime. And I claim that these two triangles are congruent to each other by side angle side. Because after all, A prime, C prime is congruent to AD. We have that the angles are congruent to each other. And then we have the segment AB is congruent to A prime B prime. So side angle side comes into play here. These two triangles are congruent to each other. And as corresponding parts of congruent triangles are congruent, we get the congruence of the segment BD, which is this segment right here. It has to be congruent to its corresponding side, which would be the segment B prime, C prime right there. So we'll call those ones are going to be congruent. So what we said there was B prime, C prime is congruent to the segment BD. The next thing we're going to do is we're going to introduce the point of intersection that is between BC and AD. Because the ray AD is interior to the angle BAC. And so by the crossbar theorem, there's a point of intersection. We will call that point of intersection E, which would be this point right here. And then of course, by the crossbar theorem, we know that the point of intersection, since it's in the segment BC, E is going to be between BC like we can see illustrated on our graph here. Now we're going to introduce a new point, which we're going to call that point F, for which this point F, really we're introducing a ray in this situation. The ray AF is going to be the bisector of the angle CAD. So we want to cut that angle in half. So we get something like this, so that this angle, the angle CAF will be congruent to the angle right here of DAF. And so we have some point F right here. Now the bisector is the ray, and I'm using the point F to distinguish this. But honestly, I could use any point along the ray other than F to define the ray. And so can I choose F to be a convenient point there? Well, again, by the crossbar theorem, since the bisector is necessarily interior to the angle CAD, we know that this ray intersects the crossbar CE. And so there's some point of intersection right here. We'll call that point F. Again, that way we don't have to introduce too many extra points into this diagram here. And so now I'm ready to do another triangle congruence. I want to compare the triangles ACF with ADF, for which I need to insert that other segment right there. So we have our triangle ACF right there and the triangle ADF like so, for which why are those triangles congruent? Well, AC is congruent to AD because AF with the bisector, we get that these two angles are congruent to each other. And then the third side is AF itself, it's equal to itself, so particularly it's congruent to itself. So we do get that these two triangles are now congruent to each other. And why do we like triangle congruences? Because corresponding parts of congruent triangles are congruent. And therefore we can then get the congruence of the segment FC is congruent to FD. So we get this is congruent to this right there. CF is congruent to FD. Now we're ready to consider an inequality here, a pretty important one. This is the main argument for this direction. So consider the segment B prime C prime to prove this direction of the hinge there. And we have to show that B prime C prime is less than the segment BC, which is exactly what we're going to do right now. So B prime C prime, which is this segment right here, we previously showed that it was congruent to the segment BD, which you can see highlighted right there. Those are congruent. If you look at the triangle BDF, so BDF is this triangle right here. The segment BD is less than, strictly less than the sum of the other two sides of the triangle BF plus FD. This is a consequence of the triangle inequality, which we proved in the previous video of this lecture here. And so this is the triangle inequality in its full glory here. Now, when it comes to segment addition here, if you replace any of the operands with a congruent segment, you'll get a congruent segment sum there. So in particular, we're going to leave BF alone. But if you swap out FD with FC, which we know those segments are congruent, then their sums will still be congruent. But notice what we have here. You have the segment BF and the segment FC. These points are collinear with F sitting in between. So BF plus FC is equal to BC as a segment. And this then proves that the segment B prime C prime is less than the segment BC. Assuming, of course, that the original angle A was larger than the angle A prime. And so that proves the first direction that the larger angle produces the longer side. What about the other direction? So conversely, this time around, let's assume that the segment BC is greater than the segment B prime C prime. Now, in that situation, if the two angles, we're trying to show that angle A is larger than angle A prime. So I'm actually going to use trichotomy here because when we compare the sizes of these two angles, one of three things will happen. We're going to have that angle A is greater than angle A prime. That's what we want. There could be that angle A is congruent to angle A prime. We also could have that angle A is less than angle A prime. So we're going to knock off the other two possibilities. Now, if angle BAC was congruent to angle B prime A prime C prime, be aware that because we know the two segments, like AB, we have still the assumption that AB is congruent to A prime B prime. And we also have the assumption that AC is congruent to the segment A prime C prime. So if angles A and A prime were congruent to each other, we would have a side angle side situation for which the triangles would be congruent and therefore their corresponding sides would be congruent. In particular, the segment BC would correspond to B prime C prime. That would give us a contradiction to this assumption. Therefore, that cannot be the possibility. So angles A and angle A prime are not congruent to each other. Which, like I said, that would show that BC is congruent to A prime, excuse me, BC is congruent to B prime C prime, which is a contradiction. So the other possibility is what if angle A was less than angle A prime. This is a beautiful argument because we already did the other direction. We could actually use it right now. If angle A was less than angle A prime, then the previous argument would imply that BC is less than B prime C prime, which again is a contradiction to our assumption right there. So therefore that possibility can't hold as well. And so by trichotomy here, there's only one other option. Angle A has to be larger than angle A prime. It's a beautiful argument where you have an if and only a statement, but you use the one direction to prove the other direction. It's wonderful. And we can do that because there's these three exhaustive cases and we can rule out one of the cases by the other direction. And then this situation where they're congruent, that's of course ruled out because of the side angle side situation. So the proof of the hinge theorem is now complete. And that also completes lecture 20 about the triangle inequality. You'll notice that in the proof of the hinge theorem, we did use the triangle inequality. So the hinge theorem is an application of the triangle inequality, which is why we included here in lecture 20. Thanks for watching. If you learned anything in these videos, please like them, subscribe to the channel to see more videos like this in the future and post any questions in the comments below and I'll be glad to answer them as soon as I can. Bye everyone.