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Proving the SAS triangle congruence criterion using transformations

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

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  • duskpin ultimate style avatar for user akshaj.vishnubhatla
    So high school geometry is grade 10 math right?
    (11 votes)
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    • aqualine ultimate style avatar for user 592815
      It would be 10th grade if you don't have pre-ap or ap classes. So if you do have gate geometry you will do it in 9th grade and do Algebra 2 next year. If you don't have a gate then you do it in 10th grade and have Algebra 2 next year.
      (2 votes)
  • starky tree style avatar for user vlin
    I wish I had Sal as a teacher in 8th grade for algebra I failed algebra that year and I thought I wasn't doing good enough but when I retook algebra in 9th I got all A's on the test. Maybe I was just my 8th-grade algebra teacher I don't know
    (4 votes)
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  • male robot johnny style avatar for user Emileigh Shaw
    So there's this little apostrophe-looking thing that keeps appearing on certain letters (around the triangles) and I'm a little confused as to what that is. If anyone can tell me what that stands for, I would really appreciate it! :)

    (It's throughout the video when those symbols show up.)
    (0 votes)
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    • mr pink green style avatar for user David Severin
      that is the prime mark (think of transformers and optimus prime) which is an indication that a transformation has been performed on a point or shape. So the pre-image (starting point) would have points such as ABC, but after it is transformed (through translation, rotation, or reflection), the new figure would be A'B'C' which would be congruent, but not necessarily the same orientation. The video talks about A going to A', B going to B', and C going to C'.
      (8 votes)
  • leafers tree style avatar for user ✨Damian✨
    How do you prove triangles congruent with attitude?
    Do it with SAS.
    (3 votes)
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    • hopper cool style avatar for user ‎‎‎ ㅤ ㅤ

      Great Question! If any two angles and the side included between the angles of one triangle are equivalent to the corresponding two angles and side included between the angles of the second triangle, then the two triangles are said to be congruent by ASA rule.

      If you didn't know, there are five ways to prove triangles are congruent.

      1.Side-Side-Side (SSS) theorem.
      2.Side-Angle-Side (SAS) theorem.
      3.Angle-Angle-Side (AAS) theorem.
      4.Angle-Side-Angle (ASA) theorem.
      5.Hypotenuse-Leg (HL) theorem.

      If two pairs of corresponding angles and the pair of included sides are congruent, then the triangles are congruent. If two pairs of corresponding angles and a pair of non-included sides are congruent, then the triangles are congruent.

      To prove that altitudes of a triangle are concurrent, we have to prove that the line segment joining the orthocentre and a vertex considering the altitudes drawn from the other two vertices of triangle meet at the orthocentre.

      Hope this helps.
      (1 vote)
  • starky tree style avatar for user cora
    Wouldn't c' be mapped to F anyway without this extra stuff?
    (2 votes)
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    • primosaur ultimate style avatar for user Conner
      The way that Sal showed it, yes it would be. However it would not have been if a reflection was already used. Say a was translated to d then the triangle was rotated about point a however many degrees it was needed to make b on the opposite side of a' from e, then reflected on a vertical line through point a', c' would be on the opposite side of f on line a', b'. If this was the case then it would need to be reflected.
      (2 votes)
  • piceratops seedling style avatar for user Liam Hanson
    Video won't play. Help?
    (2 votes)
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  • winston baby style avatar for user Amin Sahebi
    Hi, Sal
    Could you tell me which compass you think is the best (most durable) as mine keeps moving (out of the circle) and breaking?
    (2 votes)
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  • blobby green style avatar for user RWGINDY
    Around Sal says that the angle "looks something like this" and then he draws it using a straightedge. Is there a more precise way to construct the angle for the proof using a compass and then the straightedge?

    For example, can something like this be done? We know F=C' and E=B'. We can draw a line between them. So, line FE = line C'B'. Can we use line FE as a radius around point E to create a circle that will intersect with a circle around point D with line DF as a radius and then connect the points and reflect over line CE?
    (2 votes)
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  • piceratops tree style avatar for user unison_research
    also is it just me or does his voice kind of sound like an AI voice
    (1 vote)
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  • scuttlebug green style avatar for user 25burchk
    How does a compass give the exact degree or measure?
    (1 vote)
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    • starky sapling style avatar for user Serena Crowley
      A compass is used in mathematics for drawing and drafting to create arcs, circles, or other geometric figures that can be determined by measuring intersecting line segments. It can be used to bisect lines, bisecting angles, find midpoints, help solve problems in geometry, and for measuring distances or more precise distances on the maps.
      (1 vote)

Video transcript

- [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.