Proving the SSS triangle congruence criterion using transformations | Geometry | Khan Academy

What we're going to do in this video is see that if we have two different triangles where the corresponding sides have the same measure. So this orange side has the same length as this orange side. This blue side has the same length as this blue side. This gray side has the same length as this gray side. Then we can deduce that these two triangles are congruent to each other based on the rigid transformation definition of congruence.

To show that, we just have to show that there's always a series of rigid transformations that maps triangle ABC onto triangle EFD. So how do we do that? Well, first of all, in other videos, we showed that if we have two line segments that have the same measure, they are congruent. You can map one to the other using rigid transformations.

So let's do a series of rigid transformations that maps AB onto ED. And you could imagine how to do that. You would translate point A. You would translate this entire left triangle so that point A coincides with point E. Then side AB would be moving in this direction over here. Then you would rotate around this point right over here. You could call that A prime, so this is going to be equal to A prime. You rotate around that so that side AB coincides with side ED, and we've talked about that in other videos.

So at that point, D would be equal to B prime, the point to which B is mapped. But the question is, where is C? If we can show that for sure C is either at point F or with another rigid transformation we can get C to point F, then we would have completed our proof. We would have been able to show that with a series of rigid transformations, you can go from this triangle, or you can map this triangle onto that triangle.

To think about where point C is, this is where this compass is going to prove useful. We know that point C is exactly this far away from point A. I will measure that with my compass, so I could do it this way as well. Point C is exactly that far from point A, and so that means that point C needs to be someplace on this curve right over here, on this arc that I'm doing. These are some of the points that are exactly that far away. I can do a complete circle, but you see where this is going.

So point C, I guess we could say C prime, or C will be mapped to some point on that circle from A's perspective because that's how far C is from A. But then we also know that C is this far from B, so let me adjust my compass again. C is that far from B, and so if B is mapped to this point, this is where B prime is, then C prime, where C is mapped, is going to be someplace along this curve.

You could view those two curves as constraints, so we know that C prime has to sit on both of these curves. It's either going to sit right over here where F is. And so if my rigid transformation got us to a point where C sits exactly where F is, well then our proof is complete. We have come up with a rigid transformation.

Now, another possibility is when we do that transformation, C prime ends up right over here. So what could we then do to continue to transform rigidly so that C prime ends up with F? Remember, the other two points, E and D, have already coincided with E and D. So we just have to get C prime to coincide with F.

Well, one way to think about it is if we think about it, point E is equidistant to C prime and F. We see this is going to be equal to—we could put three hashtags there because once again that defines the radius of this arc. We know that point C prime in this case is the same distance from D as F is, and so one way to think about it is to imagine a line between F and—I could get, let me get my straight edge here so it looks a little bit neater—imagine a line that connects F and this C prime.

And once again we're in the case where C prime immediately didn't go to F, where C prime ended up being on this side, so to speak. And you can see that point E, because it is equidistant to C prime and F, it must sit on the perpendicular bisector of the segment FC. The same thing about point D or B prime. This must be the perpendicular bisector because this point is equidistant to F as it is to C prime.

The set of points whose distance is equal to F and C prime will form the perpendicular bisector of FC. So we know that this orange line is a perpendicular bisector of FC. Why is that helpful? Well, that tells us if when we do that first transformation to make AB coincide with EF, if C prime doesn't end up here and it ends up there, we just have to do one more transformation. We just have to do a reflection about ED or about A prime B prime, however you want to view it, about this orange line.

Then C will coincide with F because orange is a perpendicular bisector. So I could do something like this: this length is the same as this length. And since it's a perpendicular bisector, when you do the reflection, C prime will then coincide with F. A reflection is a rigid transformation, so we would be all good.