Proving Triangles Congruent – SAS

15 Nov

CCSS-M. G-CO.B.8. Explain how the criteria for triangle congruence (ASA, SAS, and SSS) follow from the definition of congruence in terms of rigid motions.

This standard made me realize that the textbooks I had used for a long time allowed the ASA, SAS, and SSS Triangle Congruence Theorems into our deductive system as postulates. We’ve always proved SAA and HL, but for some reason I thought the others were in the back of the book in a section of more challenging proofs of theorems. (I at least knew that the proofs weren’t left as an “exercise” for students at the end of the section on congruent triangles.)


How can we use rigid motions to show that SAS always works?

Here is one student’s suggestion.

Screen Shot 2014-11-09 at 5.16.28 PM

We’ve mapped ∆ABC to ∆DEF with C to F using vector CF, and rotating ∆A’B’C’ about F using angle C’A’D will map one triangle on top of the other. But have we used the given SAS? We know that ∠B≅∠E, not that ∠C≅∠F.

Here is another student’s suggestion.

Screen Shot 2014-11-09 at 7.14.27 PM

11-10-2014 Image001

Once you’ve mapped C to F using vector CF, the student suggests rotating the new triangle 180˚ about C.

We know that using dynamic geometry software doesn’t prove our results for us.

11-10-2014 Image005

But using dynamic geometry software does help convince us that we are proving the right thing. I cannot remember where I recently read (a Tweet? a blog post?) that students need to be convinced a statement is true before they will expend effort proving it. It takes a lot of Math Practice 3 for us to make it through explanations for why SSS, SAS, and ASA provide sufficient information for proving triangles congruent.

We can use a translation and a rotation, but we need to map ∆ABC to ∆DEF with B to E using vector BE. We know that ∆A’B’C’ is congruent to ∆ABC because a translation preserves congruence.

11-10-2014 Image003

Then what rotation will ensure that ∠B’ maps onto ∠E?

Rotating ∆A’B’C’ about E using angle C’EF will leave E=B’=B’’. B’’C’’=EF because a rotation preserves congruence. A”B”=DE because a rotation preserves congruence, and ∠B≅∠E because a rotation preserves congruence.

11-10-2014 Image004

If the given triangles do not have the same orientation, a reflection will be necessary, which could then be followed by a translation and/or a rotation as needed. Note: I’ve recently seen different interpretations of “orientation”. We say two figures have the same orientation if the clockwise order of the vertices is the same.

Even if the triangles do have the same orientation, a reflection or sequence of reflections can be used.

Since EF=EC’, E is on the perpendicular bisector of C’F. Reflecting ∆A’B’C’ about the perpendicular bisector of segment C’F will leave E=B’=B’’ since a point on the line of reflection will be its own image.

11-11-2014 Image007

Since ∠DEF≅∠C’EA’ and EA’=ED, A’ and D will also have to coincide after the reflection about the perpendicular bisector of C’F.

11-11-2014 Image008

Thus, ∆ABC≅∆A’B’C’≅∆DEF.

Thinking through the proofs of SSS and SAS make our traditional congruent triangle problems look like a waste of time.

Screen Shot 2014-11-15 at 9.13.40 PM

Can we show that the two triangles are congruent?

Screen Shot 2014-08-26 at 9.23.48 AM

Students look at this and immediately see that one triangle is a rotation of the other 180˚ about the midpoint of segment AC.

Screen Shot 2014-08-26 at 9.24.07 AM


Students look at this problem and see the same rigid motion to prove congruence.

Screen Shot 2014-08-26 at 9.26.12 AM


And another, except that this time, someone initially suggested a reflection about segment AC.

Screen Shot 2014-08-26 at 9.19.15 AMScreen Shot 2014-08-26 at 9.19.35 AM


Can we recover showing congruence from the initial reflection?

Screen Shot 2014-08-26 at 9.22.28 AM

Of course … another reflection about the perpendicular bisector of segment AC shows the given triangles congruent.

And then we think about why that works.

As the journey continues, I am grateful for standards that push my students and me to think outside our comfort zone, giving all of us the opportunity to make sense of problems and persevere in solving them.


From Illustrative Mathematics, Why does SAS work?

Usiskin, Peressini, Marchisotto, Stanley. Mathematics for High School Teachers: An Advanced Perspective, Pearson 2003.


Posted by on November 15, 2014 in Angles & Triangles, Geometry, Rigid Motions


Tags: , , , , ,

8 responses to “Proving Triangles Congruent – SAS

  1. howardat58

    November 15, 2014 at 9:50 pm

    Your last example:
    “Of course … another reflection about DD’ shows the given triangles congruent.”
    Not quite, since D and D’ both stay put and A does not end up on C.
    The second reflection needs to be about the perpendicular bisector of AC

    Regarding the main topic, showing that ASA is a sufficient condition for congruence, How does it follow from the rigid motion definition ?
    Well, consider two triangles, T1 and T2
    Suppose there is a sequence of rigid motions carrying T1 to T2
    Then since angle sizes and line segment lengths are preserved, T1 and T2 have the same angles and the same side lengths.
    So we are left with the problem: Does the knowledge of two angles and the included side enable us to draw the triangle (that is, does it fix the triangle) ?
    The old problem.

    The other way of looking at it is to say “Here are two triangles with matching ASA. Is there a sequence of rigid motions ….?” but you cannot call them “congruent by ASA” as this is what you are trying to show.

    I hope I have put this intelligibly ! It’s ten to midnight here.

    • jwilson828

      November 16, 2014 at 4:21 pm

      Pretty impressive corrections for ten to midnight. This is what happens when I try to write about a lesson that was 3 months ago. Thank you!

  2. Michele Torres

    November 15, 2014 at 9:53 pm

    I think this is what CCSS has in mind for this standard.

    Sent from my iPad

  3. howardat58

    November 16, 2014 at 6:02 am

    It’s morning now !
    If T1 and T2 have matching ASA then to find a transformation of T1 the first thing is to find the rigid motion taking the S of T1 to the S of T2. There always is one. So do it, and there are only four possible results for the new position of T1. A reflection about the now common S, or a reflection about the perpendicular bisector of the S, or both, or neither, will finish the match. This is a proof, but a little informal. So what! I am convinced!

    • jwilson828

      November 16, 2014 at 4:24 pm

      Yes – this is good. The first year, we didn’t formalize SSS or SAS because of how long it took just to realize the transformations needed. Last year, we made it through SAS. This year, we made it through SAS and part of SSS. Maybe we can get through ASA in the next year or two. We’ve just been letting that one in by observation so far.

  4. howardat58

    November 16, 2014 at 6:06 am

    How can anyone confuse SAS with ASA ?
    It doesn’t matter, the same approach described above should work.

  5. howardat58

    November 16, 2014 at 4:32 pm

    Another thought: SAS and the others are dropped in seemingly arbitrarily. I see them not as a way of establishing congruence but a way of specifying what is needed to be able to draw a unique triangle. Have you got your students to construct a triangle, given SAS data (and the others. it’s quite instructive.
    If the criterion (SAS .. ) does fix the triangle then we know all the angles and all the sides, which gets over one of the difficulties with the rigid motion stuff.


    January 21, 2016 at 1:35 am



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