Tag Archives: dynamic geometry software

Tangents to a Circle

Is there anything intuitive about tangents drawn to a circle from the same point outside of a circle?

I sent this Quick Poll to my students without telling them anything.

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I watched and listened as students talked with each other.

They drew diagrams, some to scale, and some not, and decided that the tangent segments drawn to the circle from the same point outside the circle were congruent.

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How do you know?

BK and his team knew for sure because they constructed the diagram using dynamic geometry software.

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Why do the tangent segments have to be congruent?

Students practiced look for and make use of structure. What do you see that isn’t pictured?

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Many students drew in some diameters.

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What do you see that isn’t pictured?

Some students recognized that a radius drawn to a point of tangency will be perpendicular to the tangent.

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What do you see that isn’t pictured?

A kite!

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Why is it a kite?

The radii are congruent.

Segment AC is congruent to itself.

The triangles are right, so angles B and D are congruent.

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We got ∆ABC congruent to ∆ADC by HL.

Then the tangent segments are congruent because the triangles are congruent.

And then back to the dynamic geometry software to make more sense of the diagram we had been given.

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What kind of #AskDontTell opportunities are you providing the learners in your care this week?


Posted by on March 15, 2015 in Circles, Geometry


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Inscribed Angles

Circles: CCSS-M G-C.A Understand and apply theorems about circles

  1. Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.
  2. (+) Construct a tangent line from a point outside a given circle to the circle.

We started our unit on circles looking at a diagram with a right triangle both inscribed in a circle and circumscribed about a circle. What do you notice? What do you wonder?

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By the end of the unit, we will be able to generalize the relationship between the sides of the right triangle and the radii of the inscribed and circumscribed circles.

My students don’t come to me knowing all of the vocabulary associated with circles, but the longer we teach with our new standards, the more I am convinced students can learn vocabulary through the modeling of using it properly and by practicing using it properly. Geometry vocabulary doesn’t have to be reduced to copying definitions from the glossary of the textbook onto a notecard (an apology those former students who had me before I figured this out).

For example, we started with a brief look at the Geometry Nspired activity Circles – Angles and Arcs.

Before generalize the relationship between a central angle and its intercepted arc, I sent a Quick Poll. The wording of the Quick Poll added “major arc” to students’ vocabulary.

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For an inscribed angle, I started with a poll just to see how intuitive the relationship is between the angle measure and intercepted arc before any kind of learning episode to explore the relationship.

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About one-third of the students intuited the relationship.

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I didn’t show the results. Instead, we looked at another page in the TNS document. What do you notice as you move point A or C?

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I sent the poll again, and we used their results to generalize the relationship between the measure of an inscribed angle and its intercepted arc.

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And then we thought about why.

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We checked again to be sure that everyone was getting what they needed to about central angles, inscribed angles, and intercepted arcs.

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And then we looked at cyclic quadrilaterals. Without me telling them anything, 12 answered correctly before the bell rang.

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And so the next lesson began with the results from this question. Which answer is correct? And why?

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And so the #AskDontTell journey continues … one lesson at a time.


Posted by on February 23, 2015 in Angles & Triangles, Circles, Geometry


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Pythagorean Theorem Proofs

CCSS-M G-SRT.B.4. Prove theorems about triangles. Theorems include: a line parallel to one side of a triangle divides the other two proportionally, and conversely; the Pythagorean Theorem proved using triangle similarity.

We’ve been teaching our CCSS Geometry course for three years now, and this is the first year that we have been able to spend more than a little class time on proofs of the Pythagorean Theorem. (Our students are coming to us knowing more mathematics than three years ago. Our students are coming to us more willing to take risks and use the Standards for Mathematical Practice than three years ago. We are making progress just in time for our legislators to decide that collaborating with other states to write standards and assessments was a bad idea.)

We started with the Mathematics Assessment Project formative assessment lesson (FAL) on Proofs of the Pythagorean Theorem. This FAL is one that includes student work. Students focus on SMP3: construct a viable argument and critique the reasoning of others.

As students practice look for and make use of structure, I asked them to share what they noticed and wondered.

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Then we looked specifically at a diagram drawn to scale, and students noted what they knew to be true (and why).

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As we started to examine the student work proofs so that students could critique the proofs, SC asked to go back to the previous page. I wonder what will happen if we reflect the outer right triangles about their hypotenuses into the center square.

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What do you think will happen?

The triangles will make a square.

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I think I’ve said before that technology slows me down in the classroom. Students notice and wonder more than they did before, and the technology gives us the chance to see what happens so that we can make sense of why it happens mathematically. I am not the only expert in the room. The student who gets mathematics without technology is not the only expert in the room. Our use of technology increases our confidence and lifts all in the room to experts. And so the journey continues …

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Posted by on February 1, 2015 in Dilations, Geometry, Right Triangles


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The Center of Dilation

CCSS-M 8.G.A.4

Understand that a two-dimensional figure is similar to another if the second can be obtained from the first by a sequence of rotations, reflections, translations, and dilations; given two similar two-dimensional figures, describe a sequence that exhibits the similarity between them.

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Are the figures similar?

Is one triangle a dilation of the other?

If so, where is the center of dilation?

I use the Class Capture feature of TI-Nspire Navigator to watch my students work.

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Who has enough information to show whether one triangle is a dilation of another?

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Whose work would you select for a whole class discussion?

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Are the triangles similar?

How do you know?

(Of course we didn’t get to this one during class … but wouldn’t you always rather have too much to do rather than too little to do?)


Posted by on January 11, 2015 in Dilations, Geometry


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Midpoint Quadrilaterals

Midpoint Quadrilaterals


Prove theorems about parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals.

While I am not exactly certain what “and conversely” modifies in this standard, I do want my students to think about not only the necessary conditions for naming a figure a parallelogram but also the sufficient conditions.

Our learning goals for the unit on Polygons include the following I can statement:

I can determine sufficient conditions for naming special quadrilaterals.

I’ve sent Quick Polls before asking students to determine whether the given information is sufficient for naming the figure a parallelogram.

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Luckily the teachers with whom I work have kindly let me know how pathetic the questions are, and so I no longer send them. So how can we get students to determine the sufficient information for naming a figure a parallelogram without giving them the list from their textbook to use and memorize?

I started this lesson by showing three (pathetically drawn) figures with some given information and sending a poll for them to mark each figure that gives sufficient information for a parallelogram (more than one, if needed). Granted it’s only a bit better than the Yes/No Quick Polls, but it is better, and it did give students more opportunity to construct a viable argument and critique the reasoning of others than the one-at-a-time polls.

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For an item like this, I especially like showing students the results without showing the correct answer, as that leaves room for even more conversation about math.

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Next I asked them to construct a non-special quadrilateral and then its midpoint quadrilateral.

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(Yes, Connor your polygon can be concave.)

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What do you notice?

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It’s a parallelogram.

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How do you know?

I blog to reflect on my practice in the classroom. And so what I know now is that I should have asked students to measure and/or construct auxiliary lines using a sufficient amount of information to show that their midpoint quadrilateral was a parallelogram. Everyone wouldn’t have measured the exact same parts, and I could have used Class Capture to select students to present their information to the class. But I didn’t think of that during the lesson. The students played with their construction, some recognizing that the midpoint quadrilateral is a parallelogram no matter how they arranged their original vertices.

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Others recognizing that every successive midpoint quadrilateral would also be a parallelogram.

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And none connecting what we had done at the beginning of the lesson with what we were doing now.

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And none proving why the figure had to be a parallelogram. I feel like the proof of why should come after we study dilations. But I like students figuring out that the figure is a parallelogram during our unit on polygons.

So maybe, eventually, we will move dilations earlier in the course.

Or maybe we can revisit the why-they-are-parallelograms after or during the dilations unit.

Either way, I’m grateful for a do-over next year as the journey continues …

Or maybe we can revisit the why after or during the dilations unit.

Either way, I’m grateful for a do-over next year as the journey continues …

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Posted by on January 5, 2015 in Geometry, Polygons


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Vertical Angles Are Congruent

CCSS-M.G-CO.C.9. Prove theorems about lines and angles. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment’s endpoints.

How do we know that vertical angles are congruent, other than “my teacher told me”, or “the dynamic geometry software convinces me”. (Even though we did let our dynamic geometry software convince us, as most students had not before seen measured vertical angles move.)

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Students worked individually first. I monitored their work.

How many times have you heard a student say that they don’t know where to start when writing a proof?

Can the leveled learning progression that Jill Gough (@jgough) and I have written for construct a viable argument and critique the reasoning of others help?


What information is given (or implied) in the diagram?

One student marked given information on the diagram so that she could understand it.

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Another student is on her way to establishing given information and is working on communicating why her conjecture must be true.

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Another student uses his given information and can get to m∠2=m∠4 but should probably show that m∠2+m∠3=m∠3+m∠4 more directly.

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Without realizing it, another student is on her way to establishing the Congruent Supplements Theorem. We can see from her work that she used some angle measures to make sense of why vertical angles have to be congruent.

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And another student with a “congruent supplements” argument but not written exactly the same way.

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So 1 of the 31 students suggested that vertical angles are congruent because of a reflection.

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What information do we need to know to define a reflection?

An object and a line.

So about what line are you reflecting ∠2 or ∠4 to show that the figures are congruent?

By the time I had made it around the room again, TL had decided that the angles should be reflected about the angle bisector of ∠3 and ∠1.

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When we were ready for the whole class discussion, we started with the progression of traditional Euclidean proofs – letting each student I called on adding a bit more to the argument. Then we considered TL’s proof with rigid motions.

His argument makes sense to the class – and in fact if we test the conjecture using technology we can see that it is true:

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But I wonder how we can prove the angle bisector of ∠1 is collinear with the angle bisector of ∠3 without technology. Maybe an indirect proof would work?

So is there another rigid motion that would let us show the congruence of vertical angles?

A rotation?

A rotation of what object about what point using how many degrees?

And so, together, we came up with the following argument to show that vertical angles are congruent using a rotation.

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And so the journey continues … learning more about transformational geometry every day from my students, who see geometry unfold differently than I, because their study of geometry started with rigid motions.


Posted by on November 18, 2014 in Angles & Triangles, Geometry, Rigid Motions


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Proving Triangles Congruent – SAS

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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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Can we show that the two triangles are congruent?

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

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

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

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Can we recover showing congruence from the initial reflection?

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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


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Carrying a Parallelogram Onto Itself

Carrying a Parallelogram Onto Itself

CCSS-M G-CO 5: Given a geometric figure and a rotation, reflection, or translation, draw the transformed figure using, e.g., graph paper, tracing paper, or geometry software. Specify a sequence of transformations that will carry a given figure onto another.

We define a parallelogram as a trapezoid with both pairs of opposite sides parallel. Students constructed a parallelogram based on this definition, and then two teams explored the angles, two teams explored the sides, and two teams explored the diagonals. I monitored while they worked.

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We discussed their results and measurements for the angles and sides, and then proved the results and measurements (mostly through congruent triangles).

I asked what they predicted about the diagonals of the parallelogram before we heard from those teams.

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We saw an interesting diagram from SJ. She explained that she had reflected the parallelogram about the segment that joined midpoints of one pair of opposite sides, which didn’t carry the parallelogram onto itself.

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So how many ways can you carry a parallelogram onto itself?

Or can you?

What if you reflect the parallelogram about one of its diagonals?

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It doesn’t always work for a parallelogram, as seen from the images above. But we can also tell that it sometimes works. For what type of special parallelogram does reflecting about a diagonal always carry the figure onto itself?

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It’s obvious to most of my students that we can rotate a rectangle 180˚ about the point of intersection of its diagonals to map the rectangle onto itself.

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It’s not as obvious whether that will work for a parallelogram. We need help seeing whether it will work.

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One of the Standards for Mathematical Practice is to look for and make use of structure.

Drawing an auxiliary line helps us to see.

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The diagonals of a parallelogram bisect each other. Since X is the midpoint of segment AB, rotating ADBC about X will map A to B and B to A.

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Since X is the midpoint of segment CD, rotating ADBC about X will map C to D and D to C.

We can verify with technology what we think we’ve made sense of mathematically using the properties of a rotation.

The change in color after performing the rotation verifies my result.

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The dynamic ability of the technology helps us verify our result for more than one parallelogram.

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Is rotating the parallelogram 180˚ about the midpoint of its diagonals the only way to carry the parallelogram onto itself?

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Before I could remind my students to give everyone a little time to think, the team in the back waved their hands madly. 729,000,000˚ works!

Did you try 729 million degrees? Why does it work?

And yes, of course, they tried it. Because they can.

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And they even understand that it works because 729 million is a multiple of 180. (I’ll even assume that SD generated 729 million as a multiple of 180 instead of just randomly trying it.)

We did eventually get back to the properties of the diagonals that are always true for a parallelogram, as we could see there were a few misconceptions from the QP with the student conjectures: the diagonals aren’t always congruent, and the diagonals don’t always bisect opposite angles.

@jgough tells a story about delivering PD on using technology to deepen student understanding of mathematics to a room full of educators years ago. A college professor in the room was unconvinced that any student should need technology to help her understand mathematics.

Jill looked at the professor and said, “Sir, I need you to remove your glasses for the rest of our session.”

He looked up, “Excuse me?”

Jill answered, “I need you to remove your glasses.”

He replied, “I can’t see without my glasses.”

Jill said, “You have a piece of technology (glasses) that others in the room don’t have. You need to remove your glasses.”

The college professor answered, “But others in the room don’t need glasses to see. I do.”

Jill’s point had been made. As the teacher of mathematics, I might not need dynamic action technology to see the mathematics unfold. But we all have students sitting in our classrooms who need help seeing.

What opportunities are you giving your students to enhance their mathematical vision and deepen their understanding of mathematics?


Posted by on October 31, 2014 in Geometry, Polygons, Rigid Motions


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Sum of Angles in a Triangle

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.

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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.

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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.)

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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.

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Next we moved to the diagram of the rotated triangles.

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

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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.

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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.

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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.

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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.

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


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Locating a Warehouse

We changed the Learning Mode to individual. Where would you place a warehouse that needed to be equidistant from all three roads? (From Illustrative Mathematics.)

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Students started sketching on paper, and I set up a Quick Poll so that we could see everyone’s conjecture at the same time.

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We changed the Learning Mode to whole class. With whom do you agree?

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I didn’t intend for us to talk in detail at this point. I wanted students to be able to test their conjecture using their dynamic geometry software. But we had done that the day before for Placing a Fire Hydrant (post to come), and class was cut short during this lesson because of lock-down and evacuation drills. So we did talk in more detail than I had planned. Is the point outside of the triangle equidistant from the three roads? One student vehemently defended her point: I drew a circle with that point as center that touched all three roads. (We have been talking about the distance from a point to a line.) How do you know that the roads are the same distance from the center? They are all radii of the circle. They are perpendicular to the road from the center.

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Could a point inside the triangle of roads be correct? If so, which? We started drawing distances from the points to the lines. Some points were about the same distance from two of the roads but obviously to close to the third road. What’s significant about the point that will be the same distance from all three sides of a triangle? Several students wondered about drawing perpendicular bisectors. Another student vehemently insisted that the point needed to lie on an angle bisector. Would that always work?

Are you going to let us try it ourselves? Well of course! So with about 12 minutes left, students began to construct.

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With about 3 minutes left, I made a student the Live Presenter who showed us that the angle bisectors are concurrent, and used the length measurement tool to show us that the point is equidistant to the sides of the triangle.

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With about 2 minutes left I made another student the Live Presenter who had made a circle inside the triangle. How did you get that circle? What is significant about the circle? It’s inscribed. The center is the where the angle bisectors intersect. So we call that point the incenter. It’s the center of the inscribed circle of a triangle, and the point of concurrency for the angle bisectors. How is this point different from the circumcenter?

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And with 1 minute left: Do you understand what we mean when we say that every point on an angle bisector is equidistant from the two sides?

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And so while I have some record of what every student did during class through Quick Polls and Class Capture and collecting their TNS document once the bell rang, my efforts at closure are foiled again. Maybe one day I’ll actually send one of the Exit Quick Polls that I have made for every lesson.


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