Tuesday, April 19, 2016

My Most Popular Lesson This Year (Day 143)

Yesterday I wrote about the most popular lesson on the blog. Today I'm writing about the most popular lesson again, but I count only those written during the current 2015-16 school year. Among those, my most popular lesson, from back in October, was on Lesson 4-5 of the U of Chicago text, the Perpendicular Bisector Theorem, with a hit count of 97.

Why is this post so popular? Well, there were several topics in that may have drawn attention to it. I began by describing an Integrated Math I text, so I discussed what Math I is like as well as compared the geometry modules of the text to a traditionalist Geometry course.

Then I moved on to describe some books I purchased at the library book sale. This post was written just as I was about to start a new side-along reading book on Mandelbrot and fractals. (We're just about to start my newest side-along reading book this week, by the way.) I also purchased another old Geometry text that day, published by Prentice-Hall, so I compared that text to the U of Chicago text.

But I suspect that the main reason that post drew many views is my discussion of the Perpendicular Bisector Theorem itself, as well as its converse. I wondered how easy it is to prove the theorem and its converse using only Common Core transformations, not triangle congruence. The U of Chicago proves the forward theorem almost trivially using reflections, but the converse was more difficult.

In fact, the converse was not proved anywhere in the U of Chicago text. The other two texts I mentioned earlier in that post -- HMH Integrated Math I and Prentice-Hall -- give proofs of the converse, but these proofs required the Parallel Postulate. And so I wondered throughout the post whether the Perpendicular Bisector Converse was provable in neutral geometry.

A commenter told me that the Perpendicular Bisector Converse was indeed provable using only Common Core transformations -- and he provided me the proof. I ended up giving his proof in another post two weeks later. So I suspect that our discussion of the proof of the Perpendicular Bisector Converse drew so many readers to that post.

Here I will reproduce that post from back in October. Let's pick it up right in the middle of my discussion of the Prentice-Hall text, when I first brought up Perpendicular Bisector. Then when I start wondering about how to prove the converse, I jump forward two weeks to when I complete that proof:

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.)

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:

Converse of the Perpendicular Bisector Theorem:
If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

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

Let m be the line containing the angle bisector of angle APB. First, since m is an angle bisector, because of the Side-Switching Theorem, when ray PA is reflected over m, its image is PB. Thus A', the reflection image of A, is on ray PB. Second, P is on the reflecting line m, so P' = P. Hence, since reflections preserve distance, PA' = PA. Third, it is given that PA = PB. Now put all of these conclusions together. By the Transitive Property of Equality, PA' = PB. So A' and B are points on ray PB at the same distance from P, and so A' = B. That is, the reflection image of A over m is B.

But, by definition of reflection, that makes m the perpendicular bisector of AB -- and we already know that P is on it. Therefore P is on the perpendicular bisector m of segment AB. QED

I conclude today's post by including a two-sided worksheet. The first side is from early October when we proved the Perpendicular Bisector Theorem, and the other side is from two weeks later when we proved the converse. The two of them together make a complete lesson on the biconditional theorem.

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