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## High school geometry

### Course: High school geometry > Unit 3

Lesson 2: Triangle congruence from transformations- Proving the SSS triangle congruence criterion using transformations
- Proving the SAS triangle congruence criterion using transformations
- Proving the ASA and AAS triangle congruence criteria using transformations
- Why SSA isn't a congruence postulate/criterion
- Justify triangle congruence

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# Proving the SSS triangle congruence criterion using transformations

CCSS.Math:

We can prove the side-side-side (SSS) triangle congruence criterion using the rigid transformation definition of congruence. Created by Sal Khan.

## Want to join the conversation?

- I think there is a mistake at5:35.Should Sal have put 4 hash marks on that line? Thanks(18 votes)
- Yeah! you could have read the text box that is on the bottom right corner if you are
**not**on full screen!(7 votes)

- What is even triangle congruence criterion? Before we try to prove it. Did i miss something along the way?(5 votes)
- A Triangle Congruence Criterion is a way of proving that two triangles are congruent. There are four types of criterians. There is SSS (Side, Side, 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. Another example is SAS (Side, Angle, Side). 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. You might be thinking, why can’t I rearrange the letters in the criterion? Well, we can rearrange SAS into SSA. But Sal shows in one of his videos that SSA is not a criterion that shows that two triangles are congruent. But, there is ASA and AAS. The difference between these two, is that in ASA, one of the sides in a triangle has to be shared by two angles. In simple terms, two angles have to be shown on that same side. 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. If you are confused, which is most probably true, then you can ask me questions on this same comment you posted.(13 votes)

- At1:52, what does Sal mean when he said "compass?" Did he mean the compass we use in science?(2 votes)
- The tool in Sal's hand is called a compass, and it's different from a directional compass you would use to navigate.

Given two points on our paper, we use a compass to draw a very accurate circle centered on one point, and with its circumference passing through the other point.

Along with a compass, you usually also have a straightedge, which is a ruler with no markings on it. With these two tools, you can play a sort of game by making challenges. Using only compass and straightedge, can you make an equilateral triangle? Given an angle, can you use a compass and straightedge to bisect it perfectly? Draw a regular pentagon? Construct an angle of exactly 20º? And so on.(16 votes)

- Howdy! I have a big problem. The videos that I am supposed to be watching don't have anything to do with the questions. For example, he is talking about the congruence criterion in all of the videos, however, the questions don't really address that. They ask me three questions, to justify the steps above, whether they are at the same distance from DE or whether they are on the same ray and the like, did you put the wrong questions, or am I just not understanding them? HELP PLEASE(8 votes)
- I understand what he is saying, but none of those relate to the practice questions. Can someone help me with the practice problems?(7 votes)
- The practice problems are meant for you to prove that the triangles are congruent. These videos show the proofs for different scenarios and then talk about the congruency postulate proved. Watch all the videos to learn about the postulates, and then work on the practice problems.(0 votes)

- How did he demonstrate that point E and D must sit on the perpendicular bisector of line FC? I think I am missing some logic. Somebody please help me with this.(6 votes)
- Because since F and C1 are on the same vertical line because it is a reflection, it would be exactly 90 degrees. Imagine DE is perfectly horizontal, so F and Cprime would be on exactly the same vertical line, since they are reflections of eachother.I can't explain it very well but I tried my best.(2 votes)

- I don't understand the perpendicular bisector scene!

What is a perpendicular bisector?(4 votes)- If you break it into two words perpendicular (at a 90 degree angle) and bisector (cut into two equal parts), the for any line segment (including the side of a triangle), we can draw a line such that it goes through the midpoint of the segment and is at a right angle to the line.(4 votes)

- What exactly is
**SSS**?(3 votes)- Let’s say that we have two triangles, triangle A and triangle A’. SSS (Side-Side-Side) is a theorem that says that if all the sides of A is equal to all the sides of A’, triangles A and A’ are congruent. In other words, SSS states that if the measure three sides of one triangle are equal to the measure of three sides of another triangle, the two triangles are congruent.(5 votes)

- so if both the triangles are the same just in different rotations they will still both be the same out come.(5 votes)
- I don't understand the perpendicular bisector scene!

What is a perpendicular bisector?(4 votes)

## Video transcript

- [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.