Segment congruence equivalent to having same length

- [Instructor] So, what I have here are a few definitions that will be useful for a proof we're going to do that connects the worlds of congruence of line segments to the idea of them having the same length. So, first of all, there's this idea of rigid transformations, which we've talked about in other videos, but just as a refresher, these are transformations that preserve distance between points. So, for example, if I have points A and B, a rigid transformation could be something like a translation because after I've translated them, notice the distance between the points is still the same. It could be like that. It includes rotation. Let's say I rotated about point A as the center of rotation. That still would not change, that still would not change my distance between points A and B. It could even be things like taking the mirror image. Once again, that's not going to change the distance between A and B. What's not a rigid transformation? Well, one thing you might imagine is dilating, scaling it up or down. That is going to change the distance, so rigid transformation are any transformations that preserve the distance between points. Now, another idea is congruence, and in the context of this video, we're going to be viewing the definition of congruence as two figures are congruent if and only if there exists a series of rigid transformations which will map one figure onto the other. You might see other definitions of congruence in your life, but this is the rigid transformation definition of congruence that we will use, and we're going to use these two definitions to prove the following, to prove that saying two segments are congruent is equivalent to saying that they have the same length. So, let me get some space here to do that in. So first, let me prove that if segment AB is congruent to segment CD, then the length of segment AB, which we'll just denote as AB without the line over it, is equal to the length of segment CD. How do we do that? Well, the first thing to realize is if AB, if AB is congruent to CD, then AB can be mapped onto CD with rigid transformations, rigid transformations. That comes out of the definition of congruence. And then we could say, "Since "the transformations are rigid, "distance is preserved, "preserved," and so, that would imply that the distance between the points are going to be the same. AB, the distance between points AB, or the length of segment AB, is equal to the length of segment CD. That might almost seem too intuitive for you, but that's all we're talking about. So, now, let's see if we can prove the other way. Let's see if we can prove that if the length of segment AB is equal to the length of segment CD, then segment AB is congruent to segment CD, and let me draw them right over here, just to, so, let's say I have segment AB right over there, and I'll draw another segment that has the same length, so maybe it looks something like this, and this is obviously hand-drawn. So, then, let's call this CD. So, in order to prove this, I have to show, "Hey, if I have two segments with the same length, "that there's always a set of rigid transformations "that will map one segment onto the other, "which means, by definition, they are congruent." So, let me just construct those transformations. So, my first rigid transformation that I could do is to translate, translate, and I'll underline the name of the transformation, segment AB, so that point A is on top of point C, or A is mapped onto C, and you could see that there's always a translation to do that. It would be doing that, and of course we would translate. B would end up like that, and so, after this translation, it's going to be A right over there. A is going to be there, and then B is going to be right over there. Now, the second step I would do is then rotate AB about A, so A is the center of rotation, so I'm gonna rotate it so that point B lies on ray CD. Well, what does this transformation do? Well, since point A is the center of rotation, A is going to stay mapped on top of C from our first translation, but now B is rotated, so it sits on top of the ray that starts at C and goes through D and keeps going, and where will B be on that ray? Well, since the distance between B and A is the same as the distance between D and C, and A and C are the same point, and now B sits on that ray, B will now sit right on top of D because AB is equal to CD. B will be mapped onto, onto D, and just like that, we've shown that if the segment lengths are equal, there is always a set of rigid transformations that will map one segment onto the other. Therefore, since A and B have been mapped onto C and D, we know that A, that segment AB is congruent to segment CD, and we are done. We have proven what we set out to prove both ways.