Proving the SAS triangle congruence criterion using transformations

- [Instructor] 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 side 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 short hand 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 that they are congruent. You can 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, if we just translated like that, then side, woops, then side B A 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 now 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 just like that. And so since all of these rigid transformations preserve distance, we know that C prime, the point that C gets mapped to after those first two transformations. C prime it's distance is going to stay the same from A prime. So C prime is going to be some place, some place 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 first set of rigid transformations, looks something like this. 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 will be preserved. And A prime C prime will then map to DF. And then we'd be done. We have just shown 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.