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Sum of Angles in a Triangle

26 Oct

I’ve talked before about providing students an opportunity to look for and make use of structure while trying to prove the Triangle Sum Theorem. This summer I ran across a task on Illustrative Mathematics with a proof of the Triangle Sum Theorem using transformational geometry. The task is scaffolded quite a bit, and so while I didn’t give my students the task as-is with the scaffolded instructions, those questions ended up playing a big role in our whole class conversation. I asked the question the same way I have before – I just knew because of reading through the task and learning from the task that I wanted my students to recognize another way to prove the Triangle Sum Theorem.

We talked for a moment about our Learning Progression for look for and make use of structure. In geometry, we often have to ask, “What do you see that’s not pictured?” We often have to draw auxiliary lines to help make sense of a figure.

I gave students a diagram of a triangle. And I gave them the following words:

Triangle Sum

Given: ∆ABC

Prove: m∠A+m∠B+m∠C=180°

I then asked the class to think back through our progression in building our deductive system. What do we know? What have we proved? What have we allowed into our systems as postulates? They thought back through our units of study:

Triangles – medians, altitudes, angle bisectors, perpendicular bisectors

Vertical angles are congruent; Angle & Segment Addition Postulates

Parallel Line postulate & theorems – corresponding angles, alternate interior angles, …

Transformations – reflections, rotations, translations

I set the timer for 3 minutes, moved the Learning Mode clip to “Individual”, and watched (monitored) as they thought. Some sat for three minutes thinking without drawing anything. Some drew in an altitude for the triangle. Some composed the triangle into a rectangle or a non-special parallelogram… not all the same way.

2014-10-14 09.45.45     2014-10-14 10.02.16

Students continued to work alone for several more minutes before they even noticed that the timer had finished. I couldn’t believe what I saw besides the traditional responses. On SC’s paper, I saw three triangles: the original, and two images of the original triangles that had been rotated about the sides. I’ve had students proving the Triangle Sum Theorem for years now and never once has someone thought to transform the triangle by rotating it. I asked students to share their work with their team. I listened. And I asked a few probing questions, especially to SC. SC needed to cut out a triangle congruent to the image so that she could describe the resulting rotations. Her first thought was that the triangle had been rotated around one of its vertices.

2014-10-14 09.45.28     2014-10-14 09.44.50

I drew a few of the diagrams that students had created on the board & asked students to take a few more minutes to see if they could justify the Triangle Sum Theorem using one of the diagrams.

Then we talked all together. Several students had used a rectangle to show why the sum of the measures of the angles of the triangle has to equal 180˚. (A few tried to use the interior angles of a quadrilateral summing to 360˚ in their reasoning, so we talked about being unable to use that, since it is really a result of what we are trying to prove.)

Screen Shot 2014-10-14 at 10.06.49 AM

Others used the side of the rectangle parallel to the base of the triangle showing that alternate interior angles congruent and then used the Angle Addition Postulate to finalize their result.

Screen Shot 2014-10-14 at 10.06.27 AM

Next we moved to the diagram of the rotated triangles.

How can we describe the rotation that resulted in the triangle on the left?

Screen Shot 2014-10-14 at 10.06.43 AM

Students suggested rotating 180˚ about the midpoint of segment AB. Point A = Point B’, and Point B = Point A’. We loosely used 1’, 2’, and 3’ to name the angles in the image. We know that ∠1’ is congruent to ∠1 because a rotation preserves congruence. And so then we know that segment A’B is parallel to segment AC since alternate interior angles are congruent.

Similarity, we can rotate the triangle 180˚ about the midpoint of segment BC. Point B = Point C’’ and Point C = Point B’’. We loosely used 1’’, 2’’, and 3’’ to name the angles in the image. We know that ∠3’’ is congruent to ∠3 because a rotation preserves congruence. And so then we know that segment A’’B is parallel to segment AC since alternate interior angles are congruent.

Screen Shot 2014-10-16 at 11.17.20 AM

So how do we know that A’’, B, and A’ are collinear?

Because if they are, then A’A’’ is parallel to AC, and m∠1’+m∠2+m∠3’’=180, which means m∠1+m∠2+m∠3=180.

For one of the first times in class, we actually used the parallel postulate to explain why A’’, B, and A’ are collinear (through a point not on a line, there is exactly one line through the point parallel to the given line). We are still studying Euclidean geometry, after all.

We are always running out of time, and so I was just using the rotation tools on my Promethean Board ActivInspire software in our conversation.

Next year, we will add our dynamic geometry software to help verify and make sense of our results.

10-26-2014 Image001 10-26-2014 Image002 10-26-2014 Image003

In my last CMC-S session yesterday I gave participants just a few minutes to come up with a way to prove the Triangle Sum Theorem using transformations. Of course I was expecting the rotation solution. I’m not sure when I’ll ever quit being surprised by solutions I don’t expect. One participant suggested that we translate ∆ABC using vector AB. We labeled the resulting image ∆A’B’C’. We know that ∠1 ≅∠1’ because a translation preserves angle congruence. Then BC’ is parallel to segment AC because corresponding angles are congruent.

10-26-2014 Image012

Similarly, we translated ∆ABC using vector CB. We labeled the resulting image ∆A’’B’’C’’. We know that ∠2 ≅∠2’’ because a translation preserves angle congruence. Then BA’’ is parallel to segment AC because corresponding angles are congruent.

10-26-2014 Image011

A’’, B, and C’ are collinear by the Parallel Postulate since there can be only one line through B parallel to segment AC. ∠3 ≅∠B’’BB’ because vertical angles are congruent. The Angle Addition Postulate gets us m∠2’’+m∠B’’BB’+m∠1’=180, and then with substitution and the definition of congruent angles, we can conclude that the sum of the measures of the angles of the triangle is 180˚.

And so the journey continues … ever grateful for resources like Illustrative Mathematics, that push me to keep learning – and keep me pushing my students to make connections that we haven’t previously been making, and ever grateful for the educators who attend conferences like CMC-South eager and willing to learn alongside other educators.

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1 Comment

Posted by on October 26, 2014 in Angles & Triangles, Geometry, Rigid Motions

 

Tags: , , , , ,

One response to “Sum of Angles in a Triangle

  1. howardat58

    October 26, 2014 at 2:51 pm

    Here’s a really good one, using transformations:
    Draw a triangle
    Take a pencil and place its center on one of the vertices, pointing along one of the sides.
    Rotate it through the angle so that it is pointing along the other side.
    Translate it in the direction of that side til its center is on the next vertex.
    Repeat all this for the new vertex.
    And again for the third vertex.
    OMG, the pencil is pointing in the opposite direction …. total rotation = sum of the angles.

    This is not doing transformations on the original figure, but on the pencil, which of course is the representation of a straight line.

     

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