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Current time:0:00Total duration:4:31

CCSS.Math:

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 but they also have the equal measure so we could think about we have a side and angle a side a side and 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 congruence 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 they'll 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 on to the other because if there is a series of rigid transformations that allow us to do it then by the rigid transformation definition the true 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 a B and segment de if we have two segments with the same length that they are congruent you can always map one segment on to the other with a series of rigid transformations the way that we could do that in this case is we could map point B on to 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 a if we just translated like that then side whoops then side B a would that or inside 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 that coincide with D or we could say a prime is equal to D but the question is where would see 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 it's just like that and so since all these rigid transformations preserve 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 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 that 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 our first series set over two 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 could 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 D F 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 si s criteria that can map one triangle on to the other and therefore they are congruent