Proving the ASA and AAS triangle congruence criteria using transformations | Geometry | Khan Academy

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. 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. We're going to show that if you have two of your angles and a side that have 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. 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. 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.

Let's just show a series of rigid transformations that can get us from triangle ABC to triangle 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. 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 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 can do another rigid transformation, which is to 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. 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?

The realization here is that angle measures are preserved. 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.

If we're going to preserve the measure of angle CAB, B prime is going to sit someplace along that ray. An angle is defined by two rays that intersect at the vertex or start at the vertex. 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. 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 the—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. 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. Let me draw that on this point right over here, this far. If the angles are on that side of line 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. It would have to sit someplace on the ray formed by the other angle. Let me see if I can draw that as neatly as possible.

Let me make an arc like this; I probably did that a little bit bigger than I need to, but hopefully it serves our purposes. I measure 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 it has to sit, I should really say—on this ray that goes through this point and this point. It has to sit on this ray. 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. 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. As this angle gets flipped over, it's preserved. As this angle gets flipped over, the measure of it, I should say, is preserved.

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. There you have it. If you have two angles, and if you have two angles, you're going to know the third.

If you have two angles and a side that have the same measure or length—if we're talking about an angle or a side—well, that means that they are going to be congruent triangles.