Sas theorem12/2/2023 So we know that this orange line is a perpendicular bisector of FC. The set of points whose distance is equal to F and C prime they will form the perpendicular bisector of FC. This must be the perpendicular bisector, because this point is equidistant to F as it is to C prime. To C Prime and F, it must sit on the perpendicular bisector of the segment FC. On this side so to speak, and you can see that point E, because it is equidistant 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 thatĬonnects 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 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 pointĬ prime in this case is the same distanceįrom D as F is, as F is. Well one way to think about it is, if we if we think about it, E, point E is equidistant 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 Now another possibility is when we do that transformation C prime ends up right over here. We've come up with a rigid transformation. Transformation got us to a point where C sits exactly where F is, well then our proof is complete. Right over here where F is and so if my rigid Two curves as constraints so we know that C prime has to C is that far from B, and so if B is mapped to this point, this is where B prime is thenĬ prime where C is mapped is going to be someplace along this curve, and so you could view those It from A's perspective, because that's how far C is from A, but then we also know So point C, I guess weĬould save C prime or C will be mapped to some point on that circle if you take And so that means that pointĬ needs to be someplace, someplace on this curve right over here, on this arc that I'm doing. Point C is exactly that far from point A. And to think about where point C is, this is where this compass Rigid transformations, you can go from this triangle or you can map this We would have been able to show that with a series of 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. But the question is where is C? If we can show that for So at that point, D wouldīe equal to B prime, the point to which B is mapped. You rotate around that so that side AB coincides with side ED. You would translate thisĮntire 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. Rigid transformations that maps AB onto ED. Have two line segments that have the same measure, So how do we do that? Well, first of all, in other videos, we showed that if we Of rigid transformations that maps triangle ABC onto triangle EDF. And to show that, we just have to show that there's always a series These two triangles are congruent to each other based on the rigid transformationĭefinition of congruence. Length as this gray side, then we can deduce that Length as this blue side, this gray side has the same Same length as this orange side, this blue side has the same Sides have the same measure, so this orange side has the Two different triangles where the corresponding Going to do in this video is see that if we have If you are confused, which is most probably true, then you can ask me questions on this same comment you posted. So in SAS and ASA, if A or S is in the middle of two same letters, then that angle or side has to have two sides on the same angle (for SAS) and for ASA, one side has to show two angles on it. In simple terms, two angles have to be shown on that same side. The difference between these two, is that in ASA, one of the sides in a triangle has to be shared by two angles. But Sal shows in one of his videos that SSA is not a criterion that shows that two triangles are congruent. You might be thinking, why can’t I rearrange the letters in the criterion? Well, we can rearrange SAS into SSA. This means that if one angle (it has to be shared by the two other sides, I’ll explain what this means) is equal on both triangles, and two sides on that angle are equal on both triangles, it proves that they are congruent. Another example is SAS (Side, Angle, Side). This means if each of the 3 sides of one of the triangles are equivalent to the other 3 sides on the other one, then they are both congruent. A Triangle Congruence Criterion is a way of proving that two triangles are congruent.
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