Monday, October 5, 2015

Lesson 4-5: The Perpendicular Bisector Theorem (Day 28)

Today I subbed in a high school math class. First period was another statistics class, while second through fifth periods was Integrated Math I. Once again, this is a regular stats class, not an AP class, and just like the last time I subbed for stats, the students were preparing for a test. The test is on histograms, or bar graphs. Once again, this is a very slow-moving class, as the material on the test is still Chapter 1 stuff! (I refer to the same text as last time, Practice of Statistics, published by W.H. Freeman and Company.)

Let's focus on the integrated math classes. The text is California Integrated Mathematics, published by Houghton Mifflin Harcourt. This text isn't divided into chapters -- instead, it consists of 25 different "modules." Here they are:

1. Quantitative Reasoning
2. Algebraic Models
3. Functions and Models
4. Patterns and Sequences
5. Linear Functions
6. Forms of Linear Equations
7. Linear Equations and Inequalities
8. Multi-Variable Categorical Data
9. One-Variable Data Distributions
10. Linear Modeling and Regression
11. Solving Systems of Linear Equations
12. Modeling with Linear Systems
13. Piecewise-Defined Functions
14. Geometric Sequences and Exponential Functions
15. Exponential Equations and Models
16. Tools of Geometry
17. Transformations and Symmetry
18. Congruent Figures
19. Lines and Angles
20. Triangle Congruence Criteria
21. Applications of Triangle Congruence
22. Properties of Triangles
23. Special Segments in Triangles
24. Properties of Quadrilaterals
25. Coordinate Proof Using Slope and Distance

The students were just beginning Module 3. Lesson 3.1 is on graphing relationships. The students had to distinguishing between linear and nonlinear equations, and among linear graphs, between proportional and nonproportional functions.

In this class, I had the freshmen write complete answers to each question, as in, "It's nonproportional because the graph doesn't pass through the origin." Naturally, the students grumbled when I made them write these sentences out, but we know that the Common Core tests, such as the PARCC and SBAC, require students to explain their answers.

I notice that this text has well over a thousand pages. I see the words "Volume 1" and "Volume 2" written inside the book, but the students have hardcovers that aren't divided. After seeing so many large integrated texts, I'm starting to wonder why the books have to be so big. After all, integrated math consists of the same math as the traditionalist pathway only in a different order -- it's not as if an integrated text needs to cover all of Algebra I and all of Geometry in one year. The size of these integrated texts will convince many parents and teachers that it's better to be on the traditionalist pathway with its more reasonably-sized texts.

The division between Volume 1 and Volume 2 is between Modules 13 and 14. This means that the second semester will cover two modules on exponential functions (including so-called "geometric" sequences), and then the rest is on geometry. As usual, this is a geometry blog, so I will focus more on the geometry lessons. Here is an approximate correspondence between Chapters 16 through 25 of the Houghton Mifflin Harcourt Math I text and the U of Chicago Geometry text:

16. Tools of Geometry (U of Chicago Ch. 1 and 2)
17. Transformations and Symmetry (U of Chicago Ch. 4)
18. Congruent Figures (U of Chicago Ch. 6)
19. Lines and Angles (U of Chicago Ch. 3)
20. Triangle Congruence Criteria (U of Chicago Ch. 7)
21. Applications of Triangle Congruence (U of Chicago Ch. 7 continued)
22. Properties of Triangles (parts of U of Chicago Ch. 13)
23. Special Segments in Triangles (parts of various chapters, including today's lesson!)
24. Properties of Quadrilaterals (U of Chicago Ch. 6)
25. Coordinate Proof Using Slope and Distance (U of Chicago Ch. 11)

This is a Common Core text, so of course transformations come early. The transformations of Chapter 17 are used to prove SSS, SAS, and ASA in Chapter 20. In Chapter 22, the Isosceles Triangle Theorem is proved using the "quick and dirty" Pappus proof (i.e., that an isosceles triangle is congruent to itself!) -- one of the few texts that actually uses that proof. Notice that similarity doesn't appear in this text (it's delayed until Math II), yet the slope criteria for parallel and perpendicular lines are nonetheless proved in Chapter 25 using what this text calls "slope triangles" -- only congruence is used, not similarity.

Twice a year -- the first Saturdays of April and October -- my local library hosts a book sale. I've bought many of the texts that I mention here on the blog from these sales -- and that even includes my copy of the U of Chicago text.

This time around, I found an old workbook from a Geometry text. As it's a consumable text in poor condition, I was able to purchase it for just a quarter. The publisher of this text is Prentice Hall, but it's not identical to the Prentice Hall text mentioned on Dr. David Joyce's website.

Because this is only the workbook, there is no publishing date listed in the text. Also, the names of the lessons are given, but not the names of the chapters. Here is my best guess as to the titles of each chapter of my Prentice Hall text, along with approximate correspondences with the U of Chicago text (no listed chapter means that the same numbered chapter in each text correspond):

1. Essentials of Geometry
2. Reasoning and Proof
3. Parallel and Perpendicular Lines
4. Congruent Triangles (U of Chicago Ch. 7)
5. Relationships within Triangles (U of Chicago Ch. 13)
6. Quadrilaterals (U of Chicago Ch. 5)
7. Measuring Length and Area (U of Chicago Ch. 8)
8. Similarity (U of Chicago Ch. 12)
9. Right Triangles and Trigonometry (U of Chicago Ch. 14)
10. Surface Area and Volume of Solids
11. Properties of Circles (U of Chicago Ch. 15)
12. Properties of Transformations (U of Chicago Ch. 6)

First of all, we notice that this is definitely not a Common Core text, since transformations don't appear until the final chapter. It will likely be several more years until I can find a Common Core text at a library book sale!

Even though this isn't the Prentice Hall text mentioned by Dr. Joyce, Chapter 1 of each text appears to be nearly identical. Much to Joyce's dismay, constructions appear in Lesson 1-5 -- this includes the perpendicular and angle bisectors (Level 3 of last week's Euclid game). And of course, the Distance Formula has to appear in Lesson 1-6. You know, if I were in charge of purchasing Geometry texts for a district, I would systematically reject any Geometry text which presents the Distance Formula in the first chapter. This is how strongly I agree with Joyce that the Distance Formula should wait until after the Pythagorean Theorem is taught.

(As it turns out, the Houghton Mifflin Harcourt text also places the Distance Formula in its first geometry chapter, Chapter 16. The text states that it's derived from the Pythagorean Theorem, and Pythagoras is even used to prove subsequent theorems in the book, such as HL in Chapter 21, even though neither similarity nor area is used to prove Pythagoras.)

As this is only the workbook, I can't tell what postulates or theorems the text uses. But there are some clues in this workbook as to which statements are postulates. Indeed, here's the last question on the practice worksheet for Lesson 3-1:

Given: a | | b
Prove: Angle 1 = Angle 3

Statements                Reasons
1. a | | b                    1. Given
2. Angle 1 = Angle 2  a. __________
3. Angle 2 = Angle 3  b. __________
3. Angle 1 = Angle 3  c. __________

But when we look at the diagram, we notice that Angles 1 and 3 are corresponding angles, while Angles 1 and 2 are alternate interior angles! This implies that the Prentice Hall text most likely assumes the Alternate Interior Angle Consequence as a postulate, while the Corresponding Angles Consequence is the theorem that this exercise proves.

Now giving priority to alternate interior angles (by which I mean deriving all the other consequences from the Alternate Interior Angles Consequence) fits well with Dr. Hung-Hsi Wu's proofs using rotations of 180 degrees, and my worksheets from last year come were based on Dr. Wu's ideas. But I've since dropped this and want to give priority to corresponding angles instead -- this was after seeing a practice PARCC question which assumes that corresponding angles have priority. Students would be confused by seeing the PARCC question, trying to go back to the workbook to review for the PARCC, and then finding this question on the Lesson 3-1 page. On this blog it's the PARCC that matters more than all other considerations, and so I will give priority to corresponding angles rather than alternate interior angles.

(As it turns out, the Houghton Mifflin Harcourt text mentioned earlier in this post actually gives priority to same-side interior angles. This is the first text I've seen that consistent gives priority to same-side interior angles, for both the Test and the Consequence.)

Meanwhile, today's lesson is on the Perpendicular Bisector Theorem. So we may ask, where does this theorem appear in the Prentice Hall text? Apparently, it doesn't -- most texts don't emphasize the Perpendicular Bisector Theorem as much as the U of Chicago text does. Ironically, Joyce's version of the Prentice Hall text gives Theorem 4-12 as the Perpendicular Bisector Theorem and the next theorem as its converse. But there is no sign of the theorem in Chapter 4 of the workbook.

(As it turns out, the Houghton Mifflin Harcourt text gives both the Perpendicular Bisector Theorem and its converse in Chapter 23. The given proof of the converse is quite complex -- it's an indirect proof that, just like the HL proof, uses the Pythagorean Theorem!)

Actually, here's one lesson in the workbook that does hint at the Perpendicular Bisector Theorem -- Lesson 11-6, which is on loci. After all, the statement of the Perpendicular Bisector Theorem and its converse is that the locus of all points in a plane equidistant from the endpoints of a given segment is the perpendicular bisector of the segment. (Lesson 11-6 is part of the chapter on circles, since a circle is the locus of all points a given distance from a given point.)

The Prentice Hall workbook is one of the two geometry books that I've purchased on Saturday. The other book is called The Fractal Geometry of Nature, by Benoit B. Mandelbrot. Back in July, I mentioned that another text I own, the the HRW text, has a lesson on fractal geometry. This lesson appears in Chapter 11, which consists of fun topics to do after the standardized tests are given. Over the next few weeks, I will read Mandlebrot's book and discuss it here on the blog, so that we can learn more about this mysterious new geometry.

This is what I wrote last year about today's lesson:

Lesson 4-5 of the U of Chicago text covers the Perpendicular Bisector Theorem. This theorem is specifically mentioned in the Common Core Standards, so it's important that we prove it.

There are many ways to prove the Perpendicular Bisector Theorem. The usual methods involve showing that two right triangles are congruent. But that's not how the U of Chicago proves it. Once we have reflections -- and we're expected to use reflections in Common Core, the proof becomes very nearly a triviality.

Here's the proof, based on the U of Chicago text but rewritten so that the column "Conclusions" and "Justifications" become "Statements" and "Reasons," and with "Given" as the first step (as I explained in one of the posts last week):

Given: P is on the perpendicular bisector m of segment AB.
Prove: PA = PB

Statements                        Reasons
1. m is the perp. bis. of AB 1. Given
2. m reflects A to B            2. Definition of reflection
3. P is on m                       3. Given
4. m reflects P to P            4. Definition of reflection
5. PA = PB                       5. Reflections preserve distance.

So the line m becomes our reflecting line -- that is, our mirror. Since m is the perpendicular bisector of AB, the mirror image of A is exactly B. After all, that was exactly our definition of reflection! And since P is on the mirror, its image must be itself. Then the last step is the D of our ABCD properties that are preserved by reflections. The tricky part for teachers is that we're not used to thinking about the definition or properties of reflections as reasons in a proof. Well, it's time for us to start thinking that way!

Now the text writes:

"The Perpendicular Bisector Theorem has a surprising application. It can help locate the center of a circle."

This is the circumcenter of the triangle, one of our concurrency proofs. I mentioned last week that this is the easiest of the concurrency proofs. At first, it appears to be a straightforward application of the Perpendicular Bisector Theorem plus the Transitive Property of Equality. But there's a catch:

"If m and n intersect, it can be proven that this construction works."

Here m and n are the perpendicular bisectors of AB and BC, respectively. But the text doesn't state how to prove that these two lines must intersect. As it turns out, the necessary and sufficient conditions for the lines to intersect is for the three points AB, and C to be noncollinear. Well, that's no problem since right at the top of the page, it's stated that the three points are noncollinear -- and besides, we don't expect there to be a circle through three collinear points anyway. (And if this is part of a concurrency proof, then the three points are the vertices of a triangle, so they are clearly noncollinear.)

The problem is that the proof is not neutral. To prove that if AB, and C are noncollinear, then the perpendicular bisectors must intersect, requires the Parallel Postulate (Playfair)! The proof is very similar to that of the Two Reflection Theorem for Translations in Lesson 6-2, except this one is an indirect proof. Assume that the two lines m and n don't intersect -- that is, that they are parallel (not skew, since everything is happening in a plane containing AB, and C). We are given that AB is perpendicular to m (since the latter is the perpendicular bisector of the former) and that BC is perpendicular to n. So, by the Perpendicular to Parallels Theorem and the Two Perpendiculars Theorem (as the proof in Lesson 6-2, as I mentioned last year, leaves out the fact that the Perpendicular to Parallels Theorem is required to get both AB and BC to be perpendicular to the same line -- m or n -- before we can apply the Two Perpendiculars Theorem), AB and BC are the same line and the three points are collinear. But this contradicts the assumption that the three points are noncollinear! Therefore m and n must intersect. QED

The Two Perpendiculars Theorem doesn't require Playfair, but the Perpendicular to Parallels Theorem does, so the above proof requires Playfair. As it turns out, in hyperbolic geometry, it's possible for three points to be noncollinear and yet no circle passes through them -- and so there's a triangle with neither a circumcenter nor a circumcircle!

And, if one has any lingering doubts that the existence of a circle through the three noncollinear points requires a Parallel Postulate, here's a link to Cut the Knot, one of the oldest mathematical websites still in existence. It was first created the year after I passed high school geometry as a student, and it has recently been redesigned:

Of the statements that require Euclid's Fifth Postulate to prove, we see listed at the above link:

"3. For any three noncollinear points, there exists a circle passing through them."

Now I didn't plan on giving Playfair's Postulate until Chapter 5. But here's something I noticed -- Playfair is used to prove the Parallel Consequences -- that is, the theorems that if parallel lines are cut by a transversal, then corresponding (or alternate interior) angles are congruent -- of which the Perpendicular to Parallels Theorem is a special case. But only that special case is needed to prove our circumcenter theorem. Indeed, I saw that the proof that the orthocenters are concurrent needs only that special case, and so does the Two Reflection Theorem for Translations.

But the Perpendicular to Parallels Theorem is tricky to prove. And we've already seen what other texts do when a theorem is tricky to prove, yet useful to prove medium- or higher-level theorems later on. We just simply declare the theorem to be a postulate! So instead of giving Playfair in Chapter 5, I state the Perpendicular to Parallels Postulate. (Note -- this is what I did last year. I have yet to decide whether I will change this for this year or not.)

Notice that the Perpendicular to Parallels Postulate can then be used to prove full Playfair. A proof of Playfair is given in the text at Lesson 13-6. The only changes we need to make to that proof is making sure that angles 1, 2, and 3 are all right angles. The new Step 2 can read:

2. The blue line is the line passing through P perpendicular to line l, which uniquely exists by the Uniqueness of Perpendiculars Theorem (which we proved on this blog last week). So angle 1 measures 90 degrees. So, by the new Perpendicular to Parallels Postulate, since the blue line is perpendicular to l, it must be perpendicular to both x and y, as both are parallel to l. So angles 2 and 3 both measure 90 degrees.

Then Playfair is used to prove the full Parallel Consequences. Notice that like Dr. Franklin Mason, we plan on adding a new postulate. But unlike his Triangle Exterior Angle Inequality Postulate, my postulate can replace Playfair, while Dr. M's TEAI Postulate must be used in addition to Playfair.

The final thing I want to say about a possible Perpendicular to Parallels Postulate is that this postulate is worth adding if it will make things easier for the students. I believe that it will. Notice that the full Parallel Consequences require identifying corresponding angles, alternate interior angles, same-side interior angles, and so on, and students may have trouble remembering which is which. But the Perpendicular to Parallels Postulate simply states that in a plane, if m and n are parallel and l is perpendicular to m, then l is perpendicular to n -- no need to remember what alternate or same-side interior angles are! So, in the name of making things easier for the students, I just might include this postulate in Chapter 5.

But I won't include it right now. And so I'll skip that part of the lesson -- and throw out the questions like 4 and 5 that require it.

That makes this lesson rather thin. So we ask, is there anything else that can be included? Let's look at the Common Core Standard that requires the Perpendicular Bisector Theorem again:

Prove theorems about lines and angles. Theorems include [...] points on a perpendicular bisector of a line segment are exactly those equidistant from the segment's endpoints.

We notice a key word there -- exactly. It means that if we rewrote this statement as an if-then statement, then it would have to be written as a biconditional:

A point is on the perpendicular bisector of a segment if and only if it is equidistant from its endpoints.

That is -- we need the converse of the Perpendicular Bisector Theorem. For some strange reason, the U of Chicago text makes zero mention of the converse! As it turns out, we can prove the converse quite easily, but it requires a theorem that's still two sections away. Once we reach Lesson 4-7, then we can finally prove the Converse of the Perpendicular Bisector Theorem.

(Actually, now after having seen the Houghton Mifflin Harcourt text, now I'm wondering whether my own proof is valid. Do I really need the Pythagorean Theorem -- which isn't neutral -- to prove the Converse of Perpendicular Bisector? Is the converse even neutral? I'd better double check this by the time we reach Lesson 4-7 on the blog!)

And so there's not much left in this section -- but one could say that the Perpendicular Bisector Theorem is so important that it nonetheless merits its own section. Notice that I kept Question 6, which is similar to the construction of the circumcircle, except it's given that lines e and f intersect at point C. So we don't need a Parallel Postulate to prove that they intersect.


  1. "(Actually, now after having seen the Houghton Mifflin Harcourt text, now I'm wondering whether my own proof is valid. Do I really need the Pythagorean Theorem -- which isn't neutral -- to prove the Converse of Perpendicular Bisector? Is the converse even neutral? I'd better double check this by the time we reach Lesson 4-7 on the blog!)"

    Hi, David. How about this approach to proving the converse:

    1. Show that angles are reflection-symmetric.

    2. Show that a figure consisting of equal segments that share an endpoint is reflection-symmetric using #1 above.

    3. Prove the converse of the PB Theorem using #2 above.

    What do you think?

    1. James,

      Thanks for your input. That Pythagorean Theorem proof came from a different book, but I usually use my 1991 Geometry text published by the University of Chicago School Mathematics Project. Both the U of Chicago text and the Common Core Standards comes from NCTM ideas, which is why a book published in 1991 contains Common Core transformations.

      The U of Chicago text doesn't prove the converse, but it does give a proof of the Isosceles Triangle Theorem -- which is related because the figure that you form in #2 is two sides of an isosceles triangle. In my October 20th post, I wrote it as a proof of the Perpendicular Converse -- and the proof ends up being essentially identical to yours. So yes, our proof of the converse should work.

      I know that having to do everything using Common Core transformations is tough for all of us Geometry teachers. I notice that your profile links to your own blog on WordPress. I'd like to respond to your most recent post on vertical angles -- and since I don't have a WordPress account, I'll reply to it right here.

      In my U of Chicago text, the Vertical Angle Theorem is proved using algebra in Lesson 3-2, just as in your Approach #1. But you're right -- this only proves that the angles have equal measure, not that they're congruent. Then later on, in Lesson 6-5, we prove the Angle Congruence Theorem, which states that two angles are congruent iff they have the same measure.

      But in that same post from October 20th, I give a worksheet with questions on rotations. One of them asks what type of angles you get when you rotate an angle 180 degrees about its vertex -- and the answer is vertical angles.

      To me, when considering whether it's better to teach Approach #1 or #2, what matters the most is, which one will students actually understand? We won't know until we teach both approaches to our students.

      Thanks for your comment. I will address some of the issues that we discussed here in my next post, to be dated November 2nd.