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Current time:0:00Total duration:5:47

CC Math: HSG.CO.B.8

- [Instructor] 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. And to show that, we just have to show that there's always a series
of rigid transformations that maps triangle ABC onto triangle EDF. 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 onto 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, and then side AB would be moving in this, would be on this direction over here. And 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. And 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, 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 save C prime or C will be mapped to some point on that circle if you take
it 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, and so you could view those
two curves as constraints so we know that C prime has to
sit on both of these curves. So 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've 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 have already coincided on 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 if we think about it, E, 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 defined the radius of this arc and we know that point C Prime in this case the point
C prime in this case is the same distance
from D as F is, as F is. And so one way to think about it, imagine a line between F and, I could 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. 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. 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 they 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 is 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, and 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 perpendicular bisector, when you do the reflection C prime will then coincide with F. And a reflection is a
rigid transformation, so we would be all good.