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The Diagonals of an Isosceles Trapezoid

01 Dec

It was the day before the test on Polygons, and so I thought that writing a proof and then giving feedback on another team’s proof might be helpful.

1a Screen Shot 2014-11-30 at 5.05.51 PM

Students worked alone for a few minutes, thinking about what was given and what could be implied. Then they worked with their team to talk about their ideas and to begin to plan a proof.

Some were off to a good start.

1b Screen Shot 2014-11-30 at 4.25.41 PM

2 Screen Shot 2014-11-30 at 4.28.16 PM

Some were obviously practicing look for and make use of structure.

3 Screen Shot 2014-11-30 at 4.27.20 PM

Some were stuck.

3b Screen Shot 2014-11-30 at 5.13.27 PM

I talked to several groups, listening to their plan, asking a few questions to get them unstuck.

And then I got out colored paper on which to write the team proof.

 

The clock was ticking, but I thought that surely they would be able to trade proofs with another team for feedback within a few minutes.

 

I talked to another group. They were reflecting ∆ABC about line AC.

What will be the image of ∆ABC about line AC?

The answer? ∆ACD.

Of course that is wrong. It seems so obvious that ∆ABC is not congruent to ∆ACD. And I’m also wondering how that helps us prove that AC=BD, since BD isn’t in either of those triangles. But that’s where this team of students is. I now have the opportunity to support their productive struggle, or I can stop productive struggle in its tracks by giving them my explanation.

 

My choice? Scissors. And Paper. And more time.

What happens if you reflect ∆ABD about line AC?

4 Screen Shot 2014-11-30 at 4.31.22 PM

 

Oh! The triangles aren’t congruent.

5 Screen Shot 2014-11-30 at 4.31.34 PM

 

So are there triangles that are congruent that can get us to the diagonals?

∆ABC is congruent to ∆BAD.

6 Screen Shot 2014-11-30 at 4.31.49 PM

 

How do you know?

A reflection.

About what?

This pencil!

7 Screen Shot 2014-11-30 at 4.32.03 PM

 

So what is significant about the line that the pencil is making?

It’s a line of symmetry for the trapezoid.

It goes through the midpoints.

8 Screen Shot 2014-11-30 at 4.32.18 PM 9 Screen Shot 2014-11-30 at 4.32.27 PM

 

(One of the team members was using dynamic geometry software to reflect ∆ABC in the midst of our conversation, but I don’t have pictures of her work.)

 

So the plan was for team to write their proofs on the colored paper and then trade with other teams for feedback. Great idea, right? So how do you proceed with 15 minutes left? Proceed as planned and let them give feedback with no whole class discussion? Or have a whole class discussion to connect student work? Because as it turned out, no two teams proved the diagonals congruent the same way. I chose the latter.

 

I asked the first team to share their work.

 

Their proof needs work. But they have a good idea.

10 Screen Shot 2014-12-01 at 9.07.05 AM

They proved ∆AMD≅∆BMC, which makes the corresponding sides congruent, so with substitution and Segment Addition Postulate, we can show that the diagonals are congruent.

11 Screen Shot 2014-11-30 at 4.37.50 PM

 

Next I asked the team to share who proved ∆ABC≅∆BAD using a reflection about the line that contains the midpoints of the bases. Their written proof needs work, too. But they had a good idea.

12 Screen Shot 2014-11-30 at 4.37.34 PM

 

Another team proved ∆ACD≅∆BDC.

 

Another team constructed the perpendicular bisectors of the bases. Since the bases are parallel, a line perpendicular to one will be perpendicular to the other. I’m not sure they got to a reason that the perpendicular bisectors have to be concurrent. They could have used ∆AZD≅∆BZC to show that. Instead, they used a point Z on both of the perpendicular bisectors (they know that any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment) to reason that ∆AZB and ∆DZC are isosceles & then used Segment Addition Postulate and substitution to show that the diagonals are congruent. Not perfect. But a good start.

13 Screen Shot 2014-11-30 at 4.36.55 PM

 

NCTM’s Principles to Actions discussion on support productive struggle in learning mathematics says, “Teachers sometimes perceive student frustration or lack of immediate success as indicators that they have somehow failed their students. As a result, they jump in to ‘rescue’ students by breaking down the task and guiding students step by step through the difficulties. Although well intentioned, such ‘rescuing’ undermines the efforts of students, lowers the cognitive demand of the task, and deprives students of opportunities to engage fully in making sense of the mathematics.”

 

So while I didn’t rescue my students, we also never made it to an exemplary proof that the diagonals of an isosceles trapezoid are congruent. Did they learn something about make sense of problems and persevere in solve them? Sure. Is that enough?

 

Would it be helpful to lead off next year’s lesson with this student work? Or does that take away the productive struggle?

 

Is it just that we have to find a balance of productive struggle and what exemplary work looks like, which is easier in some lessons than others? If so, I failed at that balance during this lesson. Even so, the journey continues …

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5 Comments

Posted by on December 1, 2014 in Geometry, Polygons, Rigid Motions

 

Tags: , , , ,

5 responses to “The Diagonals of an Isosceles Trapezoid

  1. howardat58

    December 2, 2014 at 2:11 pm

    This really got me going yesterday. Showing that the perp bisector of AB is also the perp bisector of CD turned out to be short but tricky.
    here is the picture:

    G is the mid point of AB and the perp bisector of AB is GH
    Reflect the picture in line GH
    Point C goes to C’ on the line CD (angle GHC is a right angle)
    If H is not the mid point of CD then C’ and D are not the same point,
    and so AD ≠ AC’ (not generally an isosceles triangle)
    Consequently AD ≠ AC
    But they are equal, and so C’ and D are the same point, and H is the mid point of CD
    So GH is the common perp bisector.

    This is an example of the contrapositive, if I follow your definitions in your previous post correctly, in which P implies Q is equivalent to (not Q) implies (not P)

    About your question at the end, who says this? Some sort of balance overall, yes, but surely not in every lesson. Of course, you can give them a nicely laid out version to ponder over, with a firm reminder that THIS is not THE answer !
    And do the same next time, they are going to get more out ot this way than by being led by the nose.

     
    • jwilson828

      December 2, 2014 at 4:34 pm

      I love your proof by contradiction (and yes, contrapositive) … thanks for playing with the idea!

       
      • howardat58

        December 2, 2014 at 5:14 pm

        I had another idea, which involves dropping perpendiculars from A and B onto the line CD. Oh look, we have a rectangle. I’ll leave this with you !

         
  2. jwilson828

    December 2, 2014 at 5:16 pm

    That’s fair enough after the work you did on the last one!

     

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