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Current time:0:00Total duration:7:24

Proving the ASA and AAS triangle congruence criteria using transformations

Video transcript

- [Instructor] What we're going to do in this video is show that if we have two different triangles that have one pair of sides that have the same length, so these blue sides in each of these triangles have the same length, and they have two pairs of angles where, for each pair, the corresponding angles have the same measure, so this gray angle here has the same measure as this angle here, and then these double orange arcs show that this angle ACB has the same measure as angle DFE. And so we're gonna show that if you have two of your angles and a side that had the same measure or length, that we can always create a series of rigid transformations that maps one triangle onto the other. Or another way to say it, they must be congruent by the rigid transformation definition of congruency. And the reason why I wrote angle side angle here and angle angle side is to realize that these are equivalent. Because if you have two angles, then you know what the third angle is going to be. So for example, in this case right over here, if we know that we have two pairs of angles that have the same measure, then that means that the third pair must have the same measure as well. So we'll know this as well. So if you really think about it, if you have the side between the two angles, that's equivalent to having an angle, an angle, and a side. Because as long as you have two angles, the third angle is also going to have the same measure as the corresponding third angle on the other triangle. So let's just show a series of rigid transformations that can get us from ABC to DEF. So the first step, you might imagine, we've already shown that if you have two segments of equal length that they are congruent. You can have a series of rigid transformations that maps one onto the other. So what I want to do is map segment AC onto DF. And the way that I could do that is I could translate point A to be on top of point D, so then I'll call this A prime. And then when I do that, this segment AC is going to look something like this. I'm just sketching it right now. It's going to be in that direction. But then, and the whole, the rest of the triangle is going to come with it. So let's see, the rest of that orange side, side AB, is going to look something like that. But then we could do another rigid transformation, which is rotate about point D or point A prime, they're the same point now, so that point C coincides with point F. And so just like that, you would have two rigid transformations that get us, that map AC onto DF. And so A prime, where A is mapped, is now equal to D, and F is now equal to C prime. But the question is where does point B now sit? And the realization here is that angle measures are preserved. And since angle measures are preserved, we are either going to have a situation where this angle, let's see, this angle is angle CAB gets preserved. So then it would be C prime, A prime, and then B prime would have to sit someplace on this ray. Or if we're gonna preserve the measure of angle CAB, B prime is going to sit someplace along that ray. Because an angle is defined by two rays that intersect at the vertex or start at the vertex. And because this angle is preserved, that's the angle that is formed by these two rays. You could say ray CA and ray CB. We know that B prime also has to sit someplace on this ray as well. So B prime also has to sit someplace on this ray, and I think you see where this is going. If B prime, because these two angles are preserved, because this angle and this angle are preserved, have to sit someplace on both of these rays, they intersect at one point, this point right over here that coincides with point E. So this is where B prime would be. So that's one scenario, in which case we've shown that you can get a series of rigid transformations from this triangle to this triangle. But there's another one. There is a circumstance where the angles get preserved. But instead of being on, instead of the angles being on the, I guess you could say the bottom right side of this blue line, you could imagine the angles get preserved such that they are on the other side. So the angles get preserved so that they are on the other side of that blue line. And then the question is, in that situation, where would B prime end up? Well, actually, let me draw this a little bit, let me do this a little bit more exact. Let me replicate these angles. So I'm going to draw an arc like this, an arc like this, and then I'll measure this distance. It's just like this. We've done this in other videos, when we're trying to replicate angles. So it's like that far, and so let me draw that on this point right over here, this far. So if the angles are on that side of line, I guess we could say DF or A prime, C prime, we know that B prime would have to sit someplace on this ray. So let me draw that as neatly as I can, someplace on this ray. And it would have to sit someplace on the ray formed by the other angle. So let me see if I can draw that as neatly as possible. So let me make a arc like this. I probably did that a little bit bigger than I need to, but hopefully it serves our purposes. I measured this distance right over here. If I measure that distance over here, it would get us right over there. So B prime either sits on this ray, or it could sit, or and it has to sit, I should really say, on this ray, that goes through this point and this point. And it has to sit on this ray. And you can see where these two rays intersect is right over there. So the other scenario is if the angles get preserved in a way that they're on the other side of that blue line, well, then B prime is there. And then we could just add one more rigid transformation to our series of rigid transformations, which is essentially or is a reflection across line DF or A prime, C prime. Why will that work, to map B prime onto E? Well, because reflection is also a rigid transformation, so angles are preserved. And so as this angle gets flipped over, it's preserved. As this angle gets flipped over, the measure of it, I should say, is preserved. And so that means we'll go to that first case where then these rays would be flipped onto these rays, and B prime would have to sit on that intersection. And there you have it. If you have two angles, and if you have two angles, you're gonna know the third, if you have two angles and a side that have the same measure or length, if we're talking about angle or a side, well, that means that they are going to be congruent triangles.
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