## Tuesday, September 30, 2014

### More on Rotations (Day 38)

So far, our work on rotations focuses the composite of two reflections in intersecting lines, for that's how the U of Chicago text defines them. But ordinarily, when we think about rotations, we want to think about its center and magnitude -- not the two reflections. Yet the U of Chicago text doesn't spend enough time on rotations do accomplish this. So we must go to another source.

Recall that for Wu, rotations are of primary importance. But the Wu link that I gave yesterday doesn't give exercises or problem sets, since it mainly focuses on definitions and theorems. Luckily, here's a link to the Common Core Geometry curriculum developed for New York State.

https://www.engageny.org/resource/high-school-geometry

I suspect that much of the Empire State's curriculum is based on Wu. We see that transformations are covered in Topic C, Lessons 12-21. Rotations come first (Lesson 13) -- just like Wu. Reflections come next (Lesson 14) and then translations (Lesson 16). But the giveaway that Topic C is based on Wu is Lesson 18 -- which uses 180-degree rotations to construct parallel lines and ultimately prove the Alternate Interior Angles Theorem!

As it turns out, Lesson 15 covers the relationship between a reflection and a rotation -- so it's just yesterday's lesson. And Lesson 17 is essentially the Perpendicular Bisector Theorem, which we've already covered last week. Our focus today will be on Lesson 13, on rotations.

https://www.engageny.org/resource/geometry-module-1-topic-c-lesson-13

Notice that unlike this blog, this New York lesson prescribes what questions are meant to be the Opening Activity, Exercises, Exit Ticket, and Problem Se, as well as how many minutes are to be devoted to each learning task. I don't do that -- but of course, I did sort of hint that the folding tasks at the beginning of my reflection lessons would make good Opening Activities. Actually, yesterday's task of reflecting the letter F in the y-axis and then the x-axis would also make a good Opening Activity or Anticipatory Set -- I considered posting a separate page to perform the reflections, but I want to conserve paper and avoid excessive visits to the Xerox machine. Of course, nothing is stopping the teacher from having the students take a blank sheet of paper, draw a coordinate plane with a large F in the first quadrant, and then perform the reflections, possibly even by folding.

We observe that the New York lesson also uses positive angles to denote counterclockwise and negative angles to denote clockwise. But it also uses the abbreviation CW for clockwise. Also, the New York lesson uses the letter R to denote a rotation, with two subscripts -- the first being the center and the second being the magnitude (that is, the angle). The U of Chicago uses the letter r for reflection, but no letter for translation. This blog doesn't use subscript function notation at all.

Also, I found it interesting that the New York lesson uses the Greek letter theta to denote the measure of an angle. I rarely see the theta symbol used in math texts until Precalculus, or perhaps an Honors Algebra II class with Trigonometry.

For all our talk about how the Wu and New York lessons don't define rotations as a composition of reflections, it does define rotations as a composition of rotations. In particular, a rotation of 180 degrees is the composition of two 90-degree rotations, and rotations with magnitudes greater than 180 are also the composition of rotations whose magnitudes add up to the given magnitude. In the U of Chicago text, 180 degrees is not a special case -- it's included as part of the Two Reflection Theorem for Rotations (where a "non-obtuse" angle could be a right angle, so twice its measure would be a straight angle, or 180 degrees). For larger angles, the text states that we can add or subtract multiples of 360 until the magnitude is between -180 and 180 degrees.

The New York lesson begins with its Opening Task, where the students cut out an angle and then use it to perform the rotation of a given figure. The first exercise asks students to use a protractor to measure the magnitude of a rotation. The second exercise is basically the same task as the first.

But at Exercise 3, things begin to get interesting. This exercise is similar to the first two, where we're rotating the letter M. But there's a problem -- we don't know the center of rotation! So the lesson gives the following algorithm for finding the center of rotation:

a. Draw a segment connecting points A and A'.
b. Using a compass and straightedge, find the perpendicular bisector of this segment.
c. Draw a segment connecting points B and B'.
d. Find the perpendicular bisector of this segment.
e. The point of intersection of the two perpendicular bisectors is the center of rotation. Label this point P.

Can we prove that this algorithm works? It looks a bit familiar -- because it is similar to the construction of an circle through three noncollinear points. But I threw that construction out when we did Section 4-5 -- because the proof that the construction works requires a Parallel Postulate. In this case, even if we had a Parallel Postulate, it's not obvious how to prove that the construction from a-e above works -- we have four points, not just three, and there's nothing saying that three of the points, or even all four of them, aren't collinear. It could be that the perpendicular bisectors of AA' and BB' do not intersect at all, much less intersect at the center of the rotation.

As it turns out, we don't have to worry about that. We already know -- that is, we are given -- that the points A' and B' are rotation images of A and B respectively. That is, we are given that there exists a point P such that P is the center of said rotation. And so we can prove that P lies on both of the perpendicular bisectors -- that is, we can prove that the bisectors must intersect!

And so here is our proof, in paragraph form:

Given: A', B' are the images of A, B under a rotation centered at C
Prove: C lies on the perpendicular bisectors of AA', BB'.

Proof:
The result follows directly from the proof of the Two Reflections Theorem for Rotations. Indeed, here we label the center C rather than P for emphasis -- result (3) of yesterday's proof tells us that AC = A'C (and similarly, BC = B'C). By definition of equidistant, C is equidistant from A and A' (and likewise, C is equidistant from B and B'). Therefore, by the Converse to the Perpendicular Bisector Theorem, C lies on the perpendicular bisector of AA' (and similarly that of BB' as well). QED (The proof sketch given in the New York lesson mentions chords of a circle, which the U of Chicago text doesn't cover until Chapter 15. But we just used that Converse instead.)

Notice that this lesson is clearly a straightedge and compass lesson. But we can find the perpendicular bisector of a segment by folding -- in this case, we fold A over so that it lies on top of A'. The crease of the fold is the perpendicular bisector of AA'. This may seem a bit tricky -- we used folds earlier in order to reflect points, and yes, the reflection image of A over the first fold line is A', yet A' is supposed to be a rotation image of A, not a reflection image. As it turns out, A' can be the image of A via either a reflection or a rotation. The folding is simply to find the perpendicular bisector, not to perform a reflection.

We notice that finding the center rotation is more difficult than finding a line of reflection. If we are given a point A and its reflection image A' -- as long as A' is not A itself, we immediately know the reflecting line to be the perpendicular bisector of AA'. But to find the center of rotation, it's not enough to know a single point and its image A' (unless A' is A itself) -- we saw that a second point B and its image B' were necessary. Because of this, we say that a reflection has one degree of freedom, while a rotation has two degrees of freedom. (A translation, like a reflection, also has only one degree of freedom.)

For the problem set, I decided to omit the part about a straightedge and compass. The reason for this is that the New York lesson assumes that the students can construct 45-, 60-, and 120-degree angles using only a straightedge and compass. Technically, the students in our course can do this -- we know how to draw perpendicular lines with their 90-degree angles and bisect them to get 45 degrees, and the First Theorem of Euclid's Elements lesson gives us equilateral triangles, hence 60 degrees -- but we can't prove the angle measures of an equilateral triangle until Chapter 5. So we expect the students just to use a protractor. Problem Set 3 discusses vertical angles. Interestingly enough, this is how Wu proves the Vertical Angles Theorem -- but we already proved it in Chapter 3 another way. The answers to Problem Set 4 and 5 are "rectangle" and "rhombus" respectively, but we don't define either of these terms until Chapter 5 as well. So I threw these out.

The New York lesson also mentions Geogebra as a possible tool to perform the rotations -- I suggested that earlier this month as well.

## Monday, September 29, 2014

### Section 6-3: Rotations (Day 37)

We have entered the part of our plan that I am devoting to the writings of Dr. Hung-Hsi Wu. As I mentioned last week, I don't like jumping around the U of Chicago text so much, but it is impossible for me to cover the Wu curriculum without such skipping. Let me explain.

First, here's a link to the Wu text to which I am referring -- I've given it before, but I want to give it again because this is what I'm covering right now:

http://math.berkeley.edu/~wu/CCSS-Geometry_1.pdf

Now our focus is on the high school course which begins on page 73 at the above link, not the eighth grade course. But this is a good place to remind everyone that much of geometry taught in the early years, before the eighth grade, aren't that much different before and after Common Core. The eighth grade is when the focus on transformational geometry begins.

Now our goal is to prove Wu's Theorem 12:

"Thoerem [sic] 12. If a pair of alternate interior angles or a pair of corresponding angles
of a transversal with respect to two lines are equal, then the lines are parallel."

This theorem finally gets us into our usual Parallel Tests in terms of alternate interior, same-side interior, and corresponding angles. Now the proof of Theorem 12 depends on Theorem 1:

"Theorem 1. Let L be a line and O be a point not lying on L. Let R be the 180-degree
rotation around O. Then R maps L to a line parallel to L itself."

And Theorem 1 discusses rotations. Therefore, our focus right now will be on the concept of a rotation, our next major transformation after the reflection.

The U of Chicago text already contains a section on rotations -- but not until Section 6-3. (Once again, this refers to my own copy of the text, which is dated 1991. I've heard that newer versions of the text place rotations right in Chapter 4, along with reflections and translations.) Yet I plan on adding on material from Wu himself and other sources, since the U of Chicago doesn't prove anything like Wu's first theorem in Section 6-3, or anywhere else.

Notice that my order of the transformations (reflection-rotation-translation) differs from both that of the U of Chicago (reflection-translation-rotation) and that of Wu (rotation-reflection-translation). I must emphasize that one major problem with any reordering of chapters is circularity -- we may end up placing a theorem whose proof depends on another theorem prior to that theorem. Wu places rotations first because his proof that reflections are even well-defined depend on rotations. The U of Chicago text must put reflections first because it uses reflections to define rotations. This text ensures that reflections are themselves well-defined by including such in the Reflection Postulate.

Since I already started the U of Chicago order by giving reflections first, we must remind ourselves why I don't continue the U of Chicago order and give translations next. The problem is that the U of Chicago's definition of translation is not valid in non-Euclidean geometry and requires a Parallel Postulate to be valid -- so we end up with circularity if we try to derive the properties of parallel lines from translations. On the other hand, the definition of rotation is valid in non-Euclidean geometry, so we can give it without worrying about a Parallel Postulate.

So reflections and rotations, unlike translations, have the common property that they work in all types of geometry. Another common property is that reflections and rotations -- but not translations (unless it is the identity translation) -- have fixed points. The fixed point of a rotation is its center, while a reflection has an entire line (its axis) full of fixed points. In higher math, we find out that one can perform certain reflections and rotations in the coordinate plane using matrix multiplication (which require that the origin be a fixed point), but translations require addition instead.

Also, the fixed points of reflections and rotations allow polygons to have either reflectional or rotational symmetry -- but not translational symmetry. We expect a symmetry of a regular polygon to have at least fixed point -- namely the center of the polygon. As it turns out, only infinite figures (tessellations) can have translational symmetry.

And so I present Section 6-3 of my U of Chicago text, rotations. We start with a definition:

A rotation is the composite of two reflections over intersecting lines.

Some texts define rotations the way you'd expect them to be defined -- in terms of a center and angle of rotation -- and then prove that the composition of two reflections is a rotation. But here, we define rotations to be that composite, then prove that it has the properties you expect. That theorem is called the Two Reflection Theorem for Rotations:

If m intersects l, then the composite of the reflection over m following the reflection over l "turns" figures twice the non-obtuse angle between l and m, measured from l to m, about the point of intersection of the lines.

The text provides the following proof. We are reflecting over lines l and m, and C is the point where they intersect. A is an arbitrary point that we're reflecting. We first reflect A over l to obtain A*, then reflect A* over m to obtain A'. D and E are arbitrary points on lines l and m, respectively, and are used to form angles.

Proof:
(1) C is on both reflecting lines. So C' = C* = C.
(2) By the Figure Reflection Theorem, angle ACD reflected over l is angle A*CD and A*CE reflected over m is angle A'CE. Since reflections preserve angle measure, angles ACD and A*CD have the same measure (call it x) and angles A*CE and A'CE have the same measure (call it y). From the Angle Addition Postulate, the measure of angle DCE is x + y and the measure of angle ACA' is 2(x + y). By substitution, angle ACA' has twice the measure of angle DCE.
(3) Since AC = A*C and A*C = A'C (reflections preserve distance), AC = A'C. QED

In my proof, I prefer to use A' and A" to represent the two images, to emphasize that we are performing two reflections. As in many texts, positive is counterclockwise and negative is clockwise, so we can simply used signed numbers to represent the angle of rotation.

Notice that in preparation for Wu, we especially want to emphasize 180-degree rotations. By the Two Reflection Theorem for Rotations, to obtain a 180-degree rotation, the angle between the two reflecting lines must be half of 180, or 90 degrees. This is why many of my examples have perpendicular reflecting lines.

For these exercises, I include many exercises from Lesson 6-3, but I throw out most of the review Questions, since these are review from mostly Chapter 5 and the first part of Chapter 6, which we obviously haven't covered yet. I only include Question 23 -- but it's a tricky question about the hands of a clock (yet it's relevant here as the clock's hands are actually rotating). I also include the bonus (Exploration) Question 24. Students will have to figure out that if the rotation image of a point A is the point B, then any point equidistant from A and B -- that is, any point on the perpendicular bisector of AB -- can serve as the center of the rotation.

## Friday, September 26, 2014

### Chapter 3 Test (Day 36)

As usual, let me give some insight on the test:

1. C' is the same as C, but D' goes up diagonally to the left. This is tricky because the line of reflection is not perfectly vertical.

2. I' goes up diagonally to the right.

3. There are two symmetry lines -- the segment joining the two points and its perpendicular bisector.

4. The angle measures 62 degrees.

5. The angle measures 2x degrees.

6. The reflection image over line AD of ray AB is ray AC. This is tricky because it's been a while since we've seen the Side-Switching Theorem.

7. This is officially the Figure Reflection Theorem -- just make the right vertices correspond.

8. Reflections preserve distance.

9. The orientation is clockwise.

10. The orientation is counterclockwise, because reflections switch orientation.

11. There are three pairs: angles B and C, angles BAD and CAD, angles ADB and CDB.

12. There is one line of symmetry -- the line containing the angle bisector. This follows from the Angle Symmetry Theorem.

13. F' = E follows from the Flip-Flop Theorem. FG = EH is because reflections preserve distance.

14. Proof:
Statements          Reasons
1. MO = MN       1. Given
2. M' = N            2. Given
3. MO = NO       3. Reflections preserve distance
4. MNO is equil. 4. Definition of equilateral

It's possible to add more details, such as O' = O, Transitive Property, etc.

15. The rectangle has two lines of symmetry, one horizontal, one vertical.

16. The isosceles triangle has one line of symmetry, and it's horizontal.

17. The images of the vertices are (1,3), (7,1), and (6,-2).

18. The image is (c, -d).

19. The angle measures 140 degrees.

20. The shortest distance is the perpendicular distance.

## Thursday, September 25, 2014

### Review for Chapter 3 Test (Day 35)

I'm calling this the Chapter 3 Test, even though it contains only two problems from that chapter of the U of Chicago text. I want my tests to be numbered in order, even if the section lessons and quizzes aren't.

By the way, after I take a break from the U of Chicago text to entertain Hung-Hsi Wu, I plan on going back to the chapters mostly in order, at least for the rest of the semester. It's easier, when adapting my materials from the U of Chicago text, to stick to the U of Chicago order. Many of the examples and questions from the text draw from lessons earlier in the text. But in the second semester, I plan on shaking the order up a little, to make it resemble Dr. Franklin Mason's order more. In particular, Chapter 12 (Similarity) will precede Chapter 11 (Coordinate Geometry).

Many of these questions on the review come from the SPUR questions in Chapter 4 -- indeed, often the same questions that occurred on last week's quiz. Of course, I will change them to different questions on the actual test.

Also, in deference to those who don't have access to compasses in the classroom, all of these questions are labeled "draw" rather than "construct." Of course, teachers in classrooms with compass and straightedge in the classroom can have the students construct the requested figures.

## Wednesday, September 24, 2014

### Activity: Tic-Tac-Toe (Day 34)

Here are a few more notes on constructions. First of all, we notice that the text writes three rules for using a straightedge and compass. I want to draw attention to the Compass rule:

Compass rule:
A compass can draw a circle with center at a point A and containing a second point B.
Also, a compass can be lifted keeping the same radius.
[Emphasis the U of Chicago's]

Notice the line that the text italicizes. Also, a compass can be lifted keeping the same radius. The point is that back in the days of Euclid, that part was not included! Apparently, for Euclid, a compass was collapsible, so that once lifted, it can't be used to transfer distances. It can only be used to draw a circle with known points at the center and on the circle (that is, whose center and point on the circle adhere to the Point rule).

Interestingly enough, most of the constructions in the text already adhere to this stricter rule. Euclid's Proposition I, of course, adheres to the Collapsible Compass rule. So do the perpendicular bisector and other related perpendicular constructions. My construction of a parallel to a line l through a point  P not on the line can converted to one with a collapsible compass, as follows:

Step 1. Circle P intersects l in points A and B
Step 2. Subroutine: m, the perpendicular bisector of AB
Step 3: Subroutine: k, the perpendicular to m through point P on m

This works because the radius of circle P can be any length, provided it is sufficiently large enough to intersect l in two points, so we aren't transferring distance. As usual, by definition of circle, PA = PB, so P lies on the perpendicular bisector m of AB. The subroutine in step 3 is already partly done since we already have a circle P intersecting m in two points -- call them C and D. Then step 3 merely entails constructing the perpendicular bisector of CD.

But the constructions that do entail transferring distance are the copy constructions. Indeed, the very simple construction for copying a segment requires transferring distance. Yet there is a way to copy a segment with a collapsible compass. We already saw Euclid's First Theorem -- now let's take a look at Euclid's Second Theorem:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI2.html

This construction is quite complex. We don't expect high school students to master these collapsible compass constructions -- especially since they can use compasses that aren't so restricted. The italicized part of the Compass rule tells us that our compasses are not of the collapsible type -- yet interestingly enough, the constructions in the U of Chicago text are collapsible.

David Joyce tells us that he expects proofs for the simple constructions -- which presumably includes the segment (and angle) copy construction. I wonder whether he's expecting a proof similar to Euclid's Proposition II -- especially since the construction where the compass is not of the collapsible type is so trivial to prove.

Now for today's activity -- I want to have the students construct something that involves parallel and perpendicular lines. So let's construct a tic-tac-toe board. Actually, this symbol has more names than any other symbol that I can think of: number sign, pound sign, sharp sign, octothorpe, and hashtag.

Here's an algorithm that can construct this symbol. The four points A, B, C, and D will be the four vertices of the center square:

Step 1: Circle A containing B
Step 2: Circle B containing A
Step 3: Line AB, intersecting Circles A and B at points X and Y respectively
Step 4: Subroutine: k, the perpendicular bisector of XB
Step 5: Subroutine: l, the perpendicular bisector of AY
Step 6: Lines k and l intersect Circles A and B at points C and D respectively
Step 7: Line CD

Of course, students might come up with other ways to construct the symbol of many names. Once again, there's no need for me to post the worksheet.

## Tuesday, September 23, 2014

### Section 3-6: Constructing Perpendiculars (Day 33)

Section 3-6 of the U of Chicago text deals with constructions. So, we're finally here. The students will need a straightedge and compass to complete this lesson.

Here's a good point to ask ourselves, which constructions do we want to include here? The text itself focuses on the constructions involving perpendicular lines. Well, let's check the Common Core Standards, our ultimate source for what to include:

CCSS.MATH.CONTENT.HSG.CO.D.12
Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line.
CCSS.MATH.CONTENT.HSG.CO.D.13
Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle.
But let's also go back to what David Joyce writes about constructions:
The book [Prentice-Hall 1998 -- dw] does not properly treat constructions. Constructions can be either postulates or theorems, depending on whether they're assumed or proved. For instance, postulate 1-1 above is actually a construction. On pages 40 through 42 four constructions are given: 1) to cut a line segment equal to a given line segment, 2) to construct an angle equal to a given angle, 3) to construct a perpendicular bisector of a line segment, and 4) to bisect an angle. Later in the book, these constructions are used to prove theorems, yet they are not proved here, nor are they proved later in the book. There is no indication whether they are to be taken as postulates (they should not, since they can be proved), or as theorems. At the very least, it should be stated that they are theorems which will be proved later.
David Joyce, after all, emphasizes that at the very least, constructions should be proved. He writes here that they can be proved later -- but of course, he prefers that theorems not be stated until they can be proved.
So which of the theorems in the Common Core list can be proved so far? Let's look back at that list one by one:
Copying a segment: This should be trivial to construct and prove. The student simply uses the straightedge to draw a line, marks a point O on it, and opens up the compass to the length of the given line segment AB to mark the second point P. The proof that these segments AB and OP have the same length simply follows from the definition of straightedge and compass. It's a bit surprising that the U of Chicago text doesn't begin with this as the first construction, as this one should be easy for the students.
Copying an angle: This is the one that we can't prove yet. The usual construction requires SSS to prove. This is one reason why Joyce would prefer that Chapter 8 of his text occur before these constructions in his Chapter 1.
Bisecting a segment; constructing perpendicular lines, including the perpendicular bisector of a line segment: This is the focus of Section 3-6 of the U of Chicago text. Notice that as soon as we've constructed the perpendicular bisector, we've already done the other two constructions (bisecting the segment and drawing its perpendicular). And so, as soon as we prove the perpendicular bisector construction, we are done.
Given: Circle A contains B, Circle B contains A, Circles A and B intersect at C and D.
Prove: Line CD is the perpendicular bisector of AB
Proof (in paragraph form -- can be converted to two columns later):
Just as in the proof of Euclid's first theorem (Section 4-4, two weeks ago), since both B and C lie on circle A, AB = AC by the definition of circle, and since both A and C lie on circle B, AB = BC. Then by the Transitive Property of Equality, AC = BC -- that is, C is equidistant from A and B. So, by the Converse of the Perpendicular Bisector Theorem, C lies on the perpendicular bisector of AB. In the same way, we can prove that D also lies on the perpendicular bisector of AB. And through the two points C and D there is exactly one line -- and that line is the perpendicular bisector of AB. QED
In many texts, it's pointed out that the compass opening for the two circles need not be exactly the same as AB. All that's necessary is for the opening to be greater than half of AB -- that guarantees that the two circles intersect in two points.
The text states that the midpoint of AB has been constructed as a "bonus" -- so we've bisected the segment, as requested. All we need now is to construct perpendicular lines -- and that's exactly what the text does in the next example, construct a perpendicular to line AP through point P on the line, using our perpendicular bisector algorithm as a subroutine.
Bisecting an angle: This is an interesting one. As I mentioned last week, Questions 16 and 17 from Section 4-7 of the text show us how to perform an angle "bisectomy." Notice that this construction is slightly different from the one usually given in texts, since the usual construction requires SSS to prove while this one depends only on reflections:
Given: Circle O contains A, Circle O intersects ray OB at C, Line PQ is the perpendicular bisector of AC
Prove: Ray OP is the angle bisector of Angle AOB
Proof:
Since both A and C lie on circle O, AO = CO by the definition of circle. So, by the Converse of the Perpendicular Bisector Theorem, O lies on the perpendicular bisector of AC. Now let line OP be a reflecting line. The image of A is C, and points O and P, both lying on the mirror, are each their own image. So the image of angle AOP is angle COP. Since reflections preserve angle measure, AOP and COP have the same measure. Therefore, by the definition of angle bisector, ray OP is the angle bisector of AOB. QED
Since we are jumping around the book right now, we can perform a few more constructions that we've skipped over. In Lesson 4-1, we construct the reflection image of point B over the line m:
Given: Circle B intersects m in Q and S, Circles Q and S contain B and B'
Prove: B' is the reflection image of B over m
Proof:
Once again, using the definition of circle and the Transitive Property of Equality, we have both BQ = B'Q and BS = B'S, so once again, by the Converse of the Perpendicular Bisector Theorem, both Q and S lie on the perpendicular bisector of BB'. Since two points determine a line, the line containing Q and S -- line m -- is the perpendicular bisector of BB'. So by the definition of reflection, B' is the mirror image of B over m. QED
The book states that there is another "bonus" -- line BB' is the perpendicular to m through B. So we have another construction and the existence of a line perpendicular to a line through a point not on the given line. (This line is unique by the Uniqueness of Perpendiculars Theorem.) But Common Core asks for the similar construction:
Constructing a line parallel to a given line through a point not on the line: Believe it or not, we can complete this construction right now. Usually, the construction involves copying in angle in a way so that the angle and its copy are corresponding or alternate interior angles, so that the lines are parallel. We have neither given the angle copy construction nor the parallel test for corresponding angles. But we do have the theorem from yesterday -- the Two Perpendiculars Theorem.
So given a line l and a point P not on the line l, we construct the line parallel to l through point P as follows:
Step 1. Subroutine m, a line perpendicular to l.
Step 2. Subroutine k, a line perpendicular to m through P.
then by the Two Perpendiculars Theorem, k and l are parallel. Notice that line m need not contain point P. If it does, then Step 1 is a "construct perpendicular through point not on line" step and Step 2 is "construct perpendicular through point on line." Or one could choose a point Q on l, and then do the "construct perpendicular through point on line" step first instead of second.
I was not able to create a good worksheet for constructions. So I leave you on your own to find a good worksheet.

## Monday, September 22, 2014

### Section 3-5: Perpendicular Lines (Day 32)

This week, I'm returning to Chapter 3 of the U of Chicago text, to pick up some of the lessons we skipped. But even then, I'm not simply doing the second half of Chapter 3 in order. First, I'm still bypassing Section 3-4, since I want to save the Corresponding Angles Postulate until after I'm ready to show Hung-Hsi Wu's method for deriving this property -- which I plan on doing soon.

Officially, I'm doing Section 3-5 now, but then again, not really. Let's consider the contents of this particular section:

-- The definition of perpendicular has already been covered. I'm moved it to Section 3-2 when I defined right angles, because I wanted to get it in before jumping to Chapter 4 on reflections, since reflections are defined in terms of the perpendicular bisector.

-- The Perpendicular to Parallels Theorem is an interesting case. Last week, I mentioned that there are certain interesting and important theorems that are derived from Perpendicular to Parallels -- and these include the properties of translations as well as some of the concurrency theorems. I said that it might be better just to include this as a postulate and use it to prove those other theorems. And as I think about it more and more, I like the idea of including this as a postulate, then using it to prove those other theorems as well as Playfair's Parallel Postulate, and then finally using Playfair to prove the other Parallel Consequences following Dr. Franklin Mason.

Now as I imply in this post, most of what I write on this blog is derived from mathematicians like Dr. M and Dr. Wu, who have written extensively about Common Core Geometry. But my plan to include Perpendicular to Parallels as a postulate appears to be original to me. I've searched and I have yet to see any text or website who will derive all the results of parallel lines from a Perpendicular to Parallels Postulate. Then again, what I'm doing here is, in some ways, as old as Euclid.

Let's look at Playfair's Parallel Postulate again:

Through a point not on a line, there is at most one line parallel to the given line.

This is a straightforward, easy to understand rendering of a Parallel Postulate, and Dr. M uses this postulate to derive his Parallel Consequences. But let's look at Euclid's Fifth Postulate, as written on David Joyce's website:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/post5.html

That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

No modern geometry text would word its Parallel Postulate in this manner. For one thing, even though the use of degrees to measure angles dates back to the ancient Babylonians, Euclid never uses degrees in his Elements. So the phrase "less than two right angles" is really just Euclid's way of writing "less than 180 degrees." Indeed, in Section 13-6, the U of Chicago text phrases Euclid's Fifth Postulate as:

If two lines are cut by a transversal, and the interior angles on the same side of the transversal have a total measure of less than 180, then the lines will intersect on that side of the transversal.

But let's go back to the Perpendicular to Parallels Theorem as stated in Section 3-5:

In a plane, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other.

Now count the number of right angles mentioned in this theorem. The transversal being perpendicular to the first line gives us our first right angle, and the conclusion that the transversal is perpendicular to the second line gives us our second right angle. So we have two right angles -- just like Euclid! So in some ways, making the Perpendicular to Parallels Theorem into our Parallel Postulate is making our postulate more like Euclid's Fifth Postulate, not less.

Of course, if we wanted to make our postulate even more like Euclid's, we could write:

If a plane, if a transversal is perpendicular to one line and form an acute angle (that is, less than right) with another, then those two lines intersect on the same side of the transversal as the acute angle.

But this would set us up for many indirect proofs, which I want to avoid. So our Perpendicular to Parallels Postulate is the closest we can get to Euclid without confusing students with indirect proofs.

So this is exactly what I plan on doing. Since we're in Section 3-5, the section that has Perpendicular to Parallels given, I could include it here -- we don't need to worry about how to prove it since I want to make it a postulate. But I already said that I want to wait until Chapter 5 before including any sort of Parallel Postulate. And so the new postulate will be given in that chapter.

Returning to Lesson 3-5:

-- The Perpendicular Lines and Slopes Theorem must also wait. Common Core Geometry gives an interesting way to prove this theorem, but the proof depends on similar triangles, which I don't plan on covering until second semester.

So that leaves us with only one result to be covered in 3-5: the Two Perpendiculars Theorem:

If two coplanar lines l and m are each perpendicular to the same line, then they are perpendicular to each other.

This theorem doesn't require any Parallel Postulate to prove. Indeed, even though I just wrote that I don't want to use indirect proof, in some ways this theorem is just begging for an indirect proof:

Indirect Proof:
Assume that lines l and m are both perpendicular to line n, yet aren't parallel. Then the lines must intersect (as they can't be skew, since we said "coplanar") at some point P. So l and m are two lines passing through point P perpendicular to n. But the Uniqueness of Perpendiculars Theorem (stated on this about two weeks ago) states that there is only one line passing through point P perpendicular to n, a blatant contradiction. Therefore l and m must be parallel. QED

But this would be a very light lesson indeed if all I included is this one theorem. Because of this, I decided to include a theorem that I mentioned back in July -- the Line Parallel to Mirror Theorem (companion to the Line Perpendicular to Mirror Theorem mentioned a few weeks ago):

Line Parallel to Mirror Theorem:
If a line l is reflected over a parallel line m, then l is parallel to its image l'.

As I mentioned in July, I proved this theorem in the same way that Wu proves for a similar theorem involving 180-degree rotations. In some ways, both are special cases of the following theorem:

Lemma:
Suppose T is a transformation with the following properties:
-- The image of a line is a line.
-- Through every point P in the plane, there exists a line L passing through P such that L is invariant with respect to T -- that is, T maps L to itself.
Then any line that doesn't contain a fixed point of T must be parallel to its image.

Notice that there are three confusing concepts in this theorem -- fixed point, invariant line, and reflection-symmetric figure. On the surface, all three mean the same thing: image of the point, line, or figure is itself. But there is a key difference -- an invariant line simply means that the image of any point on the line is also somewhere on the line -- but it need not be that point itself. In other words, not every point on an invariant line is a fixed point. Similarly, not every point on a symmetric figure needs to be a fixed point, nor every line on a symmetric figure an invariant line.

Now all Common Core transformations satisfy the first property -- that the image of a line is itself some line. (Forget about that Geogebra "circle reflection" where the image of a line can be a circle, since that's not a Common Core transformation.) As it turns out, there are five types of Common Core transformations that satisfy the second property:

-- Any reflection
-- Any translation
-- Any glide reflection
-- Any dilation
-- A rotation of 180 degrees

So notice that the only Common Core transformations not satisfying the second property are rotations of angles other than 180 degrees.

Here is a proof of the lemma. (By the way, "lemma" means a short theorem that is mainly used to prove another theorem.) Let l be the original line and l' its image, and let P be any point on l. Since l doesn't contain any fixed points, the image of P can't be P itself -- so instead, it must be some point distinct from P, which we'll call P'. So of course P' lies on l'. Now the point P lies on some invariant line L -- and by invariant, we mean that P' lies on it. Now through the two points P and P', there is exactly one line, and that line is L, not l. Since P lies on l, it means that P' can't lie on l. But this is true for every point P on l. For every point P on l, P' is not on l. So l can't intersect its image l', since every point on l fails to have an image on l. In other words, l and its image l' are parallel. QED

And so to prove the Line Parallel to Mirror Theorem, it suffices to show that any reflection satisfies the hypotheses of the lemma. We know that the reflection image of a line is a line (part of the Reflection Postulate), and we know that through any point P not on the mirror m, there is a line through P perpendicular to m (Uniqueness of Perpendiculars Theorem) -- which matters because the lines perpendicular to the mirror are invariant lines (Line Perpendicular to Mirror Theorem). So the hypotheses of the lemma are satisfied. Any line parallel to the mirror is parallel to its image. QED

This is a valid proof, but it's inappropriate for a high school geometry class. Providing a lemma that works for a wide variety of transformations and then showing that reflections are the correct type of transformation requires a level of abstraction that we don't expect high school students to have. So instead, we must prove the theorem for each of the types of transformations. But doing it in the case of reflections will prepare the students for the proof from Wu, who does it with 180-degree rotations.

But now we ask, is this a good time to provide the Line Parallel to Mirror Theorem? It makes sense that I'm giving it before the Wu proofs, but does it make sense to put it in the same lesson as the Two Perpendiculars Theorem -- especially since neither theorem is used to prove the other? In some ways, these two theorems do have something in common. These two theorems are our first Parallel Tests -- that is, they are used to prove that lines are parallel. They are both of the form "if two lines have some property (such as each being perpendicular to a third line, or one line being the mirror image of another over a parallel third line), then the lines are parallel."

For these proofs, I use the U of Chicago definition of parallel -- that is, two coplanar lines are parallel if and only if they have no points in common, or they are identical. As I mentioned in July, this U of Chicago definition of parallel allows me to avoid indirect proofs. Notice that the following definitions of parallel are equivalent to "Two lines have no points in common or are identical":

-- If two lines have one point in common, then they have every point in common.
-- If two lines have one point not in common, then they have no point in common.

These last two are written in if-then form, and so are convenient to write as the hypothesis (or Given) and conclusion parts of a two-column proof. The first is used in our proof of the Two Perpendiculars Theorem -- we show that if two lines k and l, both perpendicular to line m, have one point P in common, then they have every point in common. The second is used in our proof of the Line Parallel to Mirror Theorem -- if there is a point P that lies on l but not on m, then the two lines have no point in common.

These proofs will still likely confuse students. So I begin with a reminder of what it means for two lines to be parallel, using our inclusive definition. Usually, I try to include the Questions from the U of Chicago text, but I changed Lesson 3-5 so radically that I could only include a few of them.

## Friday, September 19, 2014

### Chapter 4 Quiz (Day 31)

Here is my version of the Chapter 4 Quiz. In this post I explain the answers.

1. Students should draw a point P' on the other side of line m, the same distance as P, but in the opposite direction. In other words, P' should be drawn so that m is the perpendicular bisector of PP'.

2. Students should draw the perpendicular bisector of PP'. But notice that we haven't covered constructions yet, so students should just estimate its position.

3. The measure of the angle is 48 degrees -- exactly double that of the given angle.

4. The symmetry line is the line containing the angle bisector of the given angle. This follows directly from the Angle Symmetry Theorem. The hard part about writing this quiz is that I couldn't include every single property or theorem on the Quiz Review. Students should be familiar with all of the properties of reflections (in other words, the Reflection Postulate) as well as of the results derived from these properties (the theorems).

5. The orientation is counterclockwise, of course, since reflections switch orientation.

6. Reflections preserve distance.

7. There are two pairs of angles -- angles B and C, and angles BAD and CAD.

8. The conclusion follows from the definition of symmetry line and the fact that reflections preserve angle measure. Technically, we need the definition of symmetry line -- we can't just use the Reflection Postulate directly because nowhere in the picture is any mention of a reflection. So we need the definition of symmetry line to get from "line AD is a symmetry line" to "the reflection image of triangle ABC is triangle ACB." But I'd accept it as a correct answer if a student only mentioned that reflections preserve angle measure, since a full two-column proof isn't required.

9. A square has four symmetry lines -- one horizontal, one vertical, two diagonal. This is actually not proved until later on -- but once again, a full proof isn't required.

10. The reflection image is (-a, b). There were actually examples of this in the Exercises, but only where the coordinates were numbers, not variables. If the students are confused, a teacher can change this question to something like the reflection image of (1, 2), and even encourage the students to draw in a quick graph to demonstrate it.

Hopefully I didn't make this quiz too difficult for the students.

## Thursday, September 18, 2014

### Review for Chapter 4 Quiz (Day 30)

We have reached the end of Chapter 4. I am posting a quiz on Chapter 4 for now, because I want to go back and do the second half of Chapter 3 before posting a test.

Today is the thirtieth day of school on this blog. It completes the first half of a 60-day trimester, for schools that divide the year into trimesters. Unlike quaver -- the name that I gave to half of a quarter and appears to be my own original name for this period -- half of a trimester already has an established name. It's called a "hexter."

The name hexter is interesting indeed. But to discover the origin of this name, we must first consider the origins of the words semester and trimester.

Where does the word semester come from? Some people might recognize a prefix semi- meaning "half" -- for example, in geometry a semicircle is half of a circle. Since a semester is half of an academic year, this seems logical -- but it's wrong. As it turns out, the word semester actually means "six months" -- it comes from Latin sex-, "six," plus mes- or mens-, "month." (Notice that in Spanish, the word mes still means "month.") But a semester can't possibly last six months, since then two semesters would be twelve months, the entire year, with no time for summer vacation. As it turns out, the word semester doesn't come directly from Latin, but passed through German. In German universities, the two semesters actually are six months long -- the winter semester lasting from October to March, and the summer semester lasting from April to September. There actually are breaks corresponding to our summer break, but they're actually included as part of the semesters! So semester means "six months," sex- plus mes-, but Latin speakers often drop the letter x when it appears right before the letter m, just as emigrate is really ex- (out of) plus migrate.

Therefore, a trimester actually means "three months" -- since it comes from tri-, "three," plus mes-, which we already identified as "month." It does not mean "one-third of a year." But since the school year is approximately nine, or three-squared, months long, one-third of the year just happens to be around three months. The term of a woman's pregnancy is also around three-squared months, and so some might believe that trimester means one-third of a pregnancy, but it still means "three months."

And so what about hexter? Now hex- is Greek for six (think hexagon), but is a hexter six of something, or one-sixth of something else? This word doesn't contain mes-, so it has nothing to do with six months or one-sixth of a month. On one hand, there are six hexters in a year, so this word, hexter, appears to be one-sixth of an academic year. But a hexter is also six of something -- it is close to six weeks in length, since there are approximately 36 or six-squared weeks in a school year! The answer is that we can't be sure, since the academic term hexter, while used at some schools, doesn't appear in a dictionary where we can discover its etymology.

Finally, notice that hex- is Greek while all the other numerical prefixes for academic terms are derived from Latin. Recall what I wrote about this lack of linguistic purity in geometry, where we have both hexagons (hex-, Greek) and nonagons (non-, Latin). To be linguistically consistent, we ought to use the Latin prefix sex- and call it a "sexter." The problem is that most schools using hexters are middle schools, and students at that age will assume that this has something to do with sexuality, even though the Latin sex, "six," has nothing to do with the Latin sexus, "sex." In order to avoid trying to explain to middle schoolers how "sexter" and "sexual" come from two completely unrelated Latin roots, the schools just throw linguistic purity out the window and use Greek-based "hexter" instead.

Notice that trimesters, and therefore hexters, appear mainly at the middle school level. High schools almost always use semesters instead, as this is what the colleges expect on the transcripts. But I have seen a few high schools give report cards three times per semester -- in other words, the progress report occurs at the end of every hexter. But I've never seen the word "hexter" used to refer to these thrice-a-semester progress reports.

Here is the quiz review:

## Wednesday, September 17, 2014

### Section 4-7: Reflection-Symmetric Figures (Day 29)

Section 4-7 of the U of Chicago text deals with reflection-symmetric figures. A definition is in order:

A plane figure F is a reflection-symmetric figure if and only if there is a line m such that r(F)=F. The line m is a symmetry line for the figure.

In other words, it's what one usually means when one uses the word "symmetry." Some geometry texts use the term "line-symmetric" instead of "reflection-symmetric." Some geometry and algebra texts use the term "axis of symmetry" instead of "symmetry line" -- especially Algebra I texts referring to the axis of symmetry of a parabola. Some biology texts use the term "bilateral symmetry" instead of "reflection (or line) symmetry" - in particular, when referring to symmetry in animals. As animals are three-dimensional, instead of a symmetry line there's a sagittal plane.

Indeed, it is this last topic that makes symmetry most relevant and interesting. Most animals -- including humans -- have bilateral symmetry. I once read of a teacher who came up with an activity where the students look for the most symmetrical human face. The teacher blogged about how students who are normally indifferent to geometry suddenly came fascinated and engaged to learn about the relationship between symmetry and human beauty. Unfortunately, this was more than a year ago, and I can't remember or find what teacher did this activity -- otherwise I'd be posting a link to that teacher's blog right here!

In the Common Core Standards, symmetry is first introduced as a fourth grade topic:

CCSS.MATH.CONTENT.4.G.A.3
Recognize a line of symmetry for a two-dimensional figure as a line across the figure such that the figure can be folded along the line into matching parts. Identify line-symmetric figures and draw lines of symmetry.

Later on, symmetry appears in the high school geometry standards:

CCSS.MATH.CONTENT.HSG.CO.A.3
Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself.

Notice that if a reflection over a line carries a polygon to itself, then that line is a symmetry line. But symmetry lines for polygons formally appears in Chapter 5 of the U of Chicago text. Right here in Chapter 4, we only cover symmetry lines for simpler shapes -- segments and angles. The text reads:

"In the next chapter, certain polygons are examined for symmetry. All of their symmetries can be traced back to symmetries of angles or segments."

For segments, the text presents the Segment Symmetry Theorem:

A segment has exactly two symmetry lines:
1. its perpendicular bisector, and
2. the line containing the segment.

The text gives an informal proof of this -- as the mirror image of an endpoint, there can only be two possible reflections mapping a segment AB to itself. One of them maps A to B and B to A -- and that mirror must be the perpendicular bisector of AB, by the definition of reflection. The other reflection maps A to A and B to B -- which means that both A and B must lie on the mirror, since the image of each is itself. No other symmetry is possible. QED

But we also want to work with angles. The first theorem given is the Side-Switching Theorem:

If one side of an angle is reflected over the line containing the angle bisector, its image is the other side of the angle.

An informal proof: the angle bisector divides an angle into two angles of equal measure. The picture in the U of Chicago text divides angle ABC into smaller angles 1 and 2. Now the reflection must map ray AB onto a ray that's on the other side of the angle bisector BD, but forms the same angle with BD that AB does with BD. And there's already such a ray in the correct place -- ray BC. Notice that part b of the Angle Measure Postulate from Chapter 3 already hints at this -- the "Two sides of line assumption" gives two angles of the same measure, one on each side of a given ray. QED

The other theorem, the Angle Symmetry Theorem, follows from the Side-Switching Theorem:

Angle Symmetry Theorem:
The line containing the bisector of an angle is a symmetry line of the angle.

Earlier this week, I wrote that we'd be able to prove the Converse of the Perpendicular Bisector Theorem after this section. As it turns out, the Side-Switching Theorem is the theorem we need.

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.

Now I was considering giving a two-column proof of this, but it ended up being a bit harder than I would like for the students. But as it turns out, even though the U of Chicago text doesn't prove this converse, in Section 5-1 it gives a paragraph proof of what it calls the "Isosceles Triangle Symmetry Theorem," and the proof of this one and the Converse of the Perpendicular Bisector Theorem are extremely similar. After all, we're given that PA = PB -- so PAB is in fact an isosceles triangle!

Proof:
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

Let's think about what we're trying to prove here. We want the Converse of the Perpendicular Bisector Theorem -- and consider what I wrote earlier about the proof of converses. The proof of the converse of a statement often involves the forward direction of the theorem and a uniqueness statement -- and even though we didn't use the forward direction of the theorem here, we did use a uniqueness statement here. As it turns out, given two distinct points A and B, there exists only one line m such that the mirror image of A over m is B -- and that line is the perpendicular bisector of the segment AB. And so if we can somehow find out another way that the mirror image of A over m is B, we'll have proved that m is the perpendicular bisector of AB. So that's exactly what we did above -- we proved that a certain line (the angle bisector of APB) is the perpendicular bisector of AB.

In this section, we found symmetry lines for simple figures such as segments and angles. But can we find symmetry lines for the simplest figures? As it turns out, a point has infinitely many lines of symmetry -- any line passing through the point is a symmetry line. But a ray has only one line of symmetry -- the line containing the ray.

Finally, does a line have a line of symmetry? This is exactly the answer to Question 25 of this section, in the Exploration/Bonus Section. A line -- considered as a straight angle -- contains more than one symmetry line. This is because any point on the line can be taken as the vertex of that straight angle. Since straight angles measure 180, their angle bisectors must divide them into pairs of 90-degree angles. Therefore, any line perpendicular to a line (straight angle) is a symmetry line of the given line. This is what I called the Line Perpendicular to Mirror Theorem. It implies that a line (straight angle) has infinitely many symmetry lines. (Of course, the line has one more symmetry line that I didn't mention -- namely the line itself.)

I included Question 24, even though it appears to mention corresponding and same-side interior angles formed by two lines and a transversal. But nowhere in the question does it mention anything about the two lines being parallel.

I left out Questions 16 and 17, which give the construction of an angle bisector. I finally plan on going to constructions sometime next week. But here's another video from Square One TV, where doctors have to perform a "bisectomy" on an angle. (Unfortunately, only the entire 30-minute show is available on YouTube -- the "bisectomy" doesn't begin until the 11-and-a-half-minute mark.)

## Tuesday, September 16, 2014

### Section 4-6: Reflecting Polygons (Day 28)

Section 4-6 of the U of Chicago text considers what happens when we reflect an entire polygon -- not just individual points or even a segment or angle.

Still, the section begins with a theorem on what happens when we reflect a single point twice. Suppose we have two points, F and G and a reflecting line m. Now suppose I told you that the mirror image of F is G. So where do you think the mirror image of G is? If we drew this out and showed it to a student, chances are the student will say that the mirror image of G is F. The book gives a proof of this fact -- by the definition of reflection, G as the mirror image of F means that m is the perpendicular bisector of FG. But FG is the same segment as GF, so its perpendicular bisector is still m. And so, by the definition of reflection again, this would make F the mirror image of G. QED

The text calls this the Flip-Flop Theorem:

If F and F' are points or figures and r(F) = F', then r(F') = F.

Recall that the text often uses the function notation r(F) to denote the reflection image of F. So the theorem can be written as:

If F and F' are points or figures and the mirror image of F is F', then the mirror image of F' is F.

And one can use even more function notation than the text and write the theorem as:

If F is a point or figure, then r(r(F)) = F.

So here's a two-column proof of the Flip-Flop Theorem:

Given: r(F) = F'
Prove: r(F') = F

Proof:
Statements                        Reasons
1. r(F) = F'                       1. Given
2. m is the perp. bis. of FF' 2. Definition of reflection (meaning)
3. FF' = F'F                      3. Reflexive Property of Equality
4. m is the perp. bis. of F'F 4. Substitution Property of Equality
5. r(F') = F                       5. Definition of reflection (sufficient condition)

Notice that this proof uses both the meaning and the sufficient condition parts of the definition of reflection -- this occurs in other proofs as well. For example, a proof of the theorem "all right angles are congruent" (Euclid's Fourth Postulate) uses both the meaning and the sufficient condition parts of the definition of right angle.

But the above proof is a little strange. We explained earlier the significance of Statement 3 in the above proof -- but the problem is that we need a reason as to why FF' and F'F are the same segment. There is no actual definition, postulate, or theorem that states this directly. The reason I wrote "Reflexive Property" above is that this often occurs in other proofs -- especially triangle congruence proofs that are used to prove that certain quadrilaterals are parallelograms. For example, in Section 7-7, we wish to prove that quadrilaterals with opposite sides congruent are parallelograms. The proof at the beginning of that lesson divides quadrilateral ABCD into two triangles, ABD and CDB, which the text then proves are congruent by SSS. But Step 2 of that proof reads:

2. BD is congruent to DB   2. Reflexive Property of Congruence

And so I did the same in the above proof. Of course, it's awkward to follow a statement that uses the "Reflexive Property" (that some object equals itself) with one that uses the "Substitution Property." (So we're substituting an object for itself?)

Some people may point out that now we're being overly formalistic here. The Flip-Flop Theorem is obviously true -- the two-column proof only serves to confuse the students. Perhaps if even I, as a teacher, have trouble filling in all the steps in the "Reasons" column (like Step 3 above), it means that the proof is so simple that it's better written as a paragraph proof (as the U of Chicago text has done) and not as a two-column proof.

Here's one final way to state the Flip-Flop Theorem:

A reflection is an involution.

An involution is simply a function or translation such that performing it twice on a point or figure gives the original point or figure. Therefore composition of an involution with itself is the identity. In function notation, f(f(x)) = x.

Now the other concept introduced in this chapter is orientation. The important concept, added to the Reflection Postulate as part f, is that reflections switch orientation.

But what exactly is the "orientation" of a polygon? The text explains that, in naming the vertices of a polygon, we can move either clockwise or counterclockwise around the polygon. The important idea here is that if pentagon ABCDE is clockwise and we reflect it, then A'B'C'D'E' is counterclockwise.

Then the book proceeds to tell us that "orientation" is undefined -- just like point, line, and plane. As we mentioned earlier, we only discover what an undefined term is by using postulate. So we have the Point-Line-Plane Postulate to tell us what points, lines, and planes are, and we have part f of the Reflection Postulate to tell us what orientation is. We may not know what orientation actually is, but we do know that whatever it is, reflections switch it.

The idea that reflections switch orientation shows up later on. In particular, translations and rotations preserve orientation, because they are the compositions of two reflections -- so the first reflection switches it, and the second switches it back.

Also, a question that often comes up is, if translations and rotations are the compositions of two reflections, maybe reflections are the composition of two rotations, or two of something else. As it turns out, this is impossible. Reflections can't be the composition of two of the same type of transformation, because of orientation. Either the orientation is switched and switched back, or it isn't switched at all. (If you want a reflection to be some transformation composed with itself, you must do something complicated, such as cut the plane into strips, then translate some of the strips and reflect the others.)

Is it possible to define "orientation"? We think back to Chapter 1, where the term "point," although undefined, can be modeled with an ordered pair. If we know all of the x- and y-coordinates of the vertices of the polygon, then we can plug it into a complicated formula such that if the answer is positive, then the orientation must be counterclockwise, and if the answer is negative, then the orientation must be clockwise. (If it's zero, then the points are collinear, which means that they don't form a polygon at all.) What's cool about the formula is that the number -- not just the sign -- actually means something. In particular, if we divide the number by two, we get the area of the polygon! But I won't give the formula here.

There's also a simpler version of the formula, but it only works if the polygon is convex. Notice the picture of octagon FGHIJKLM in the text. The book points out that determining its orientation is more difficult because it's nonconvex.

A much more intuitive way of thinking about orientation is if the preimage and image aren't figures, but words. If we hold up words to a mirror, then unless we're lucky and choose a word like MOM, the image will be illegible, since reflections reverse orientation. But if we translated the words instead, then we can still read the words (unless by "translation" we mean translation into another language).

One final note about orientation: A well-known math teacher blogger named Kate Nowak -- she calls her blog "Function of Time" or f(t) in function notation -- recently gave an Opening Task to her geometry classes:

Now Nowak gave her classes pairs of figures, and the students had to identify whether the two figures are "the same" or "not the same." As it turned out, the students easily reached a consensus if the two figures have the same orientation, but they disagreed if the orientations were different:

One group: "We said set C is not the same because you have to flip it."
Me [Nowak -- dw]: "Great."
Other group: "Wait a minute, we said set C is the same because we thought flipping was okay."
Me: "Also great."
Yet another group: "So which is it? We said they are the same."
Me: "...   ...   ...  because... ?"

## Monday, September 15, 2014

### Section 4-5: The Perpendicular Bisector Theorem (Day 27)

Section 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

Proof:
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 A, B, 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, that if A, B, 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 A, B, 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, AB is parallel to BC. (The proof in Lesson 6-2, as I mentioned in July, 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 -- be it m or n -- before we can apply the Two Perpendiculars Theorem.) But B is on each line. So AB and BC are the same line (recall that in the text, a line is parallel to itself) 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:

http://www.cut-the-knot.org/triangle/pythpar/Fifth.shtml

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 -- I gave a proof back in July. 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.

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, which we already plan on doing using the rotation trick of Dr. Hung-Hsieh Wu.

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:

CCSS.MATH.CONTENT.HSG.CO.C.9
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 Section 4-7, then we can finally prove the Converse of the Perpendicular Bisector Theorem.

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.