Proving the SAS 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 and we have two sets of corresponding sides that have the same length. For example, this blue side has the same length as this blue side here, and this orange side has the same length as this orange side here.
And the angle that is formed between those sides, so we have two corresponding angles right over here that they also have the equal measure. So we could think about we have a side, an angle, a side, a side, an angle, and a side. If those have the same lengths or measures, then we can deduce that these two triangles must be congruent by the rigid motion definition of congruency.
Or the shorthand is if we have side angle side in common and the angle is between the two sides, then the two triangles will be congruent. So to be able to prove this, in order to make this deduction, we just have to say that there's always a rigid transformation if we have a side angle side in common that will allow us to map one triangle onto the other.
Because if there is a series of rigid transformations that allow us to do it, then by the rigid transformation definition the two triangles are congruent. So the first thing that we could do is we could reference back to where we saw that if we have two segments that have the same length like segment AB and segment DE.
If we have two segments with the same length, then they are congruent. You could always map one segment onto the other with a series of rigid transformations. The way that we could do that in this case is we could map point B onto point E. So this would be now I’ll put B prime right over here.
And if we just did a transformation to do that, so if we just did it, if we just translated like that, then side—oops—then side would that orange side would be something like that. But then we could do another rigid transformation that rotates about point E or B prime that rotates that orange side and the whole triangle with it onto DE.
In which case, once we do that second rigid transformation, point A will not coincide with D, or we could say A prime is equal to D. But the question is, where would C now sit? Well, we can see the distance between A and C. In fact, we can use our compass for it. The distance between A and C is right, is just like that.
And so, since all these rigid transformations preserve a distance, we know that C prime, the point that C gets mapped to after those first two transformations, C prime, its distance is going to stay the same from A prime. So C prime is going to be someplace along this curve right over here.
We also know that the rigid transformations preserve angle measures and so we also know that as we do the mapping, the angle will be preserved. So either side AC will be mapped to this side right over here. And if that's the case, then F would be equal to C prime, and we would have found our rigid transformation based on SAS.
And so, therefore, the two triangles would be congruent. But there's another possibility that the angle gets conserved but side AC is mapped down here. So there's another possibility that side AC, due to our rigid transformations or after our rigid—the first series set of rigid transformations, it looks something like this.
So it looks—it looks something like that, in which case C prime would be mapped right over there. And in that case, we can just do one more rigid transformation. We can just do a reflection about DE or A prime B prime to reflect point C prime over that to get right over there.
How do we know that C prime would then be mapped to F? Well, this angle would be preserved due to the rigid transformation. So as we flip it over, as we do the reflection over DE, the angle would be preserved and A prime C prime will then map to DF.
And then we'd be done. We will—we have to show that there's always a series of rigid transformations as long as you meet this SAS criteria that can map one triangle onto the other, and therefore they are congruent.