(Of course this is in my new district, where today is only Day 37. In my old district, it's Day 45.)
8:00 -- This is second period. (As usual, "first period" = zero period.) It's a special ed "Academic Improvement" class -- similar to the "Academic Enrichment" that I covered yesterday in the other district.
Students are supposed to show me their weekly planners. (This is similar to yesterday's Enrichment class, except today's planners are weekly, not everyday the class meets.) The girl who didn't complete her planner in September fortunately does her planner today.
8:55 -- Second period ends and third period begins. Third period is Algebra 1B, the second of two years that special ed students spend in Algebra I. The students are now working in Chapter 7 of the Glencoe text, which is on exponents and polynomials. Specifically, they are learning how to use the two division laws of exponents (quotient of powers, power of a quotient).
One question on which the students particularly struggle is -2y^7/(14y^5). Notice that the correct answer is -y^2/7, but it's not obvious why y^2 is in the numerator with 7 in the denominator. I tell them my usual trick -- use the laws on "little numbers" and regular math on "big numbers." But even then it's not obvious why y^2 is on the top and 7 is on the bottom.
Perhaps it's also helpful to look at the starting problem -- for both "big numbers" and "little numbers," if the original problem has the greater number on top then so does the final answer, and if the original problem has the greater number on bottom then so does the final answer. Thus y^2 is on top because 7 > 5 and y^7 starts out on top, and 7 is on the bottom because 14 > 2 and 14 is on the bottom. (This also works for negative exponents, if every positive number is greater than every negative number.)
10:18 -- An earthquake drill is held. Actually, this is the Great California Shakeout, which is held the third Thursday in October. I wrote about this annual earthquake drill two years ago back at the -- oops, ixnay on the arterchay illdray!
Actually, I take that back. I promised that I wouldn't write about the old charter school during our side-along reading of Eugenia Cheng. Right now, our side-along reading is Sue Teele's book, and I said nothing about avoiding the charter during Teele. In fact, during Chapter 7 of Teele I want to compare the author's lessons to the ones I had at the old charter.
Right now I can't help but think about the old charter school, since that year and today are the only times that I've been on a school campus during the annual drill. Anyway, this is wrote I two years ago about the earthquake drill:
You see, on the third Thursday in October, every school in California is supposed to have an earthquake drill. In theory, the drill should be on 10/20 at 10:20, but ours is at 9:00 instead. In part, this is because at most elementary (and K-8) schools, 10:20 is right in the middle of someone's recess. Also, the other problem is the weather -- usually here in California, there's at least one rainy day and one 90+ degree day (Santa Ana winds) in October. We've actually had both such days this week, with today being the scorcher. So if the kids are stuck outside for 45 minutes, we'd all prefer it to be at 9:00 rather than 10:20.
I'll just say it now -- this week was the first Santa Ana winds week in October. And we're not an elementary school, so we don't have to worry about avoiding recesses. Our drill is on 10/18 at 10:18, when it is quite warm outside on the football field where all students must meet.
If you compare today to the September 21st post, you'll notice that on the regular schedule, 10:18 is during tutorial. Having a drill during tutorial is awkward, since attendance must be taken during the drill (and students normally choose which tutorial to attend).
In other words a 10:18 drill doesn't quite fit the regular schedule, so we switch to a different schedule which does fit with a 10:18 drill. This schedule turns out to be the assembly schedule. On assembly days there is no tutorial -- instead there are periods 3A and 3B. If there were an actual assembly, then half the school would attend the assembly during 3A and the rest during 3B -- students not at the assembly would attend third period. Instead, today 3A is the actual third period, which is followed by snack break. After snack, period 3B begins at 10:00. The drill begins 18 minutes later.
11:00 -- In theory, fourth period is supposed to start at 10:50, but in reality, the earthquake drill doesn't end until around 11. This is our favorite math class, Geometry!
The students are learning about conjectures in Lesson 2-1 of the Glencoe text. Notice that conjectures don't really appear in the U of Chicago text until Lesson 5-3 (since that text mainly focuses on conjectures about quadrilaterals). So once again, this slow-moving special ed class that started late is somehow ahead of our U of Chicago text! We observe that Chapter 2 of the Glencoe text is on reasoning and proof. Much Glencoe Chapter 2 material appears in the same-numbered chapter in U of Chicago, but other material doesn't appear until Chapter 13. We should be lucky, then, that conjectures at least appear in Chapter 5.
On the Lesson 2-1 worksheet, some students figure out the first few conjectures, which are of the form "Make a conjecture about the next item in the sequence." The first question has 4 dots, 6 dots, 8 dots, 10 dots, ..., while the next is the sequence 5, -10, 15, -20, ....
Then students are to make conjectures based on Geometry. The first such question on the worksheet directs students to make a conjecture based on the given: "ABC is a right angle." One girl tries to answer "ABC is 90 degrees." The only problem is that this isn't a conjecture, since it can be deduced directly from "ABC is a right angle." I tell them that a better conjecture might be something like "ABCD is a rectangle," but then some students wonder where D is. The truth, of course, is that I conjectured the existence of point D. Students are still confused.
Most students don't finish this worksheet. The late-running earthquake drill doesn't help, of course.
11:35 -- Fourth period ends. Like all special ed teachers in this district, this regular teacher co-teaches another class. For fifth period, I travel to a junior English class. The main teacher in this class is having her students watch a video on The Crucible. Yes, the class is still watching The Crucible -- she is having the students watch each act as they read it. In September they watched Act I, and today they view Act III. (Actually, there is a sub in this class as well, which factors into the decision to have them watch the video today.)
12:30 -- It is time for lunch. After working through much of the day, I enjoy a well-needed break during the teacher's conference period.
2:10 -- It is now seventh period. As I explained in September, the regular teacher is the tennis coach, and now it's currently the girls' tennis season. (The team competes in a contest today.)
As I reflect upon this class, I wonder how I could have taught the class better. In September, my focus resolution for this class was:
3. Move on from past incidents instead of bringing them up with students.
I'm not sure how well I follow this resolution today. I remember from September that fourth period is the class to worry about. The regular teacher tells me that any names left for her will be assigned to Saturday school. So I tell fourth period that I'm especially keeping an eye on this class because I recall what the class was like last month.
This might count as bringing up a past incident -- and since I want to err on the side of being too hard on myself, I'll say that yes, this definitely counts as bringing up a past incident. I should have announced that the bad students will receive Saturday school without mentioning September at all.
The third period earthquake drill is especially tough. We are supposed to take attendance during the regular third period (3A) and then again at the drill. This is the first time that I ever taken roll both in class and at a drill -- and ended up with more students at the drill than in class! That's because the special aide and I catch two students who are ditching most of third period. One of them shows up at the start of 3B, and the other doesn't appear until outside at the drill itself.
Neither student stays in the correct location during the drill. The later student continues leaving to talk to another class during the drill. I later find out that he spends time at the nurse during the day -- I don't know whether this is before or after the drill. (Then again, that doesn't explain why he's holding a Jack in the Box bag upon his arrival.) The other guy is worse -- he runs around the football field and steals some candy from another student, and the other students cheer him on. (If he really wants to intercept an item, run up and down the field, and have everyone cheer for him, then maybe he should join the football team.) I add both names to the Saturday school list.
In fourth period, students continue to talk loudly. I believe that I've improved from September only in that I don't use "how many questions are answered on the worksheet" as the main criterion for writing names on the bad list (especially since Saturday school is on the line). I ultimately write down one guy's name because he starts to eat some fruit snacks in class. The aide tells him not to eat it, yet he keeps on eating anyway.
But the aide writes down the names of five additional students (plus the girl she works one-on-one with) for various reasons (including the guy who originally passes out the fruit snacks). She leaves it up to the regular teacher whether all of these students get Saturday school or not.
By the way, the other student she worked with in September has indeed been expelled and sent to the continuation school. I wonder whether another guy in this class is next -- he is absent today. Three Saturday school notices arrive for him. I observe that one of them is for missing too many classes, and another is for missing a previously assigned Saturday school.
I wonder whether there's anything I could have done better in this class. I remember telling one girl not to talk, and she claims that she's helping another guy on the worksheet. But she along with the other students keep getting louder. Perhaps I should have told her "No, you can't talk," as apparently letting her talk ultimately costs me control of the class. (The answer I'd give her if she asks "Why?" should be "Because I said so.")
Many students talked throughout the period, only stopping to copy my answer from the board. This doesn't mesh well with conjectures questions that are supposed to be open-ended. This is why other math teachers ask "Which one doesn't belong?" or "What do you notice and wonder?" so that the students make thinking about open-ended questions a habit.
I also might have gained the students' attention if I'd given more interesting conjecture questions beyond those on the worksheet. In past years, I wrote about sequences such as "George, John, Thomas, James," with the next answer being "James" (Monroe, the fifth president).
Finally, I've never played the "Who Am I?" game in this district, since by now I want to focus on classroom management without gimmicky games. But today might have been the one day when this game might actually fits -- recall that another name for this game is "Conjectures." Anyway, this all gives me something to think about the next time I sub in this class.
Today on her Mathematics Calendar 2018, Theoni Pappas writes:
Angles a, b, c, d, and e total 1026 degrees. How many degrees is x?
(Here is the given info from the diagram -- the first five are angles of an octagon. Each of the other three angles of the octagon measures x.)
This clearly isn't a regular octagon, or even a convex octagon. But at any rate, the angles of the octagon sum to (8 - 2)180 = 1080 degrees, by the Polygon-Sum Theorem of Lesson 5-7. Since five angles sum to 1026 degrees, this leaves 54 degrees for the three angles each of measure x. In other words, 3x = 54, and so x = 18. Therefore the desired angle is 18 degrees -- and of course, today's date is the eighteenth.
Chapter 2 of Sue Teele's Rainbows of Intelligence is called "Recent Research on the Brain." Here's how it begins:
"As we examine the relationship between the theory of multiple intelligences and the changes we should make in education, I offer a new model that builds on this theory and considers the complex pathways that individuals utilize to process information.
And Teele highlights the following statement:
"We each learn in a highly personal, individual way. No two students have the same profile of intelligences."
So this chapter is all about the brain and how it functions. Teele tells us that, while the "left brain" and "right brain" exist, the story is more complicated than that:
"The brain is not simply right or left hemispheres, but a whole entity with constant interaction between both sides. We can define our intelligence as the capacity to carry out a multitude of tasks that range from breathing, crying, registering pain, seeing, and hearing, to making highly complex decisions."
Compare this to Eugenia Cheng's definition of intelligence -- the ability to benefit both the self as well as others.
Teele shows a chart comparing the relative size of a child's brain compared to an adult's:
6 months: 1/2
3 years: 3/4
5 years: 9/10
She writes:
"The brain regulates many body systems such as breathing, pumping blood, and digesting food. Language and use of symbols are functions of the brain."
Teele shows a chart with the main parts of the brain:
- Brain stem
- Limbic system
- Cerebral cortex
The most important part of the brain, where learning occurs, is the cerebral cortex:
"The cellular part of the cortex is called the gray matter. Beneath the gray matter is the white matter, which makes up most of this part of the brain."
As mentioned earlier, the brain is divided into two hemispheres:
Left Hemisphere:
- Right hand touch & movement
- Speech
- Language and writing
- Linear thinking
- Analysis
Right Hemisphere:
- Left hand touch & movement
- Spatial construction
- Face recognition
- Music processing
- Nonverbal matters
One traditionalist, Katharine Beals, named her (now defunct) website "Out in Left Field." Here "left" actually refers to the left hemisphere of the brain. The main idea of her blog is that traditional math appeals to the left hemisphere. But according to Teele, true geniuses such as Da Vinci and Einstein used both hemispheres of their brains together. She writes:
"I have found that when I use a shelled walnut to show what the brain looks like, individuals can visualize the two hemispheres of the brain when they actually see a representation of it. The center of the walnut depicts the corpus callosum, the connection between the two hemispheres."
Teele draws many diagrams of parts of the brain in her book, but I don't draw those pictures. I apologize to all right-brained individuals who might prefer the pictures -- instead I'll just have to use those left-brained words. She writes:
"Neurons look like trees and have branches that are called dendrites. Dendrites are tubular extensions that receive and carry messages to and from other neurons and will grow when stimulated."
Teele explains that as the dendrites branch, abstract thinking increases. But if the brain is not uses, the dendrites do not branch out. She continues:
"The corpus callosum provides pathways for continuous exchanges between the two hemispheres. These pathways allow two to three different intelligences to operate simultaneously from both sides of the brain."
Teele tells us that the cortex is divided into four lobes -- temporal, parietal, occipital, and frontal. The first of these is for speech, but the second is of particular interest to us:
"The parietal lobe is located at the top of the head. It responds to sensations of touch and motor control. Spatial functions such as geometry, map reading, and mathematical reasoning operate in this lobe."
The occipital lobe processes visual information. As for the frontal lobe:
"It sends messages to muscles and glands and is responsible for voluntary movement. It is the area of goal-directed behavior, long-term planning, problem solving, critical thinking, and decision making."
Teele highlights the following statement:
"Humans are the only known species that have prefrontal lobes."
And she explains that synthesis and evaluation can take place in this part of the brain -- in other words, what we know as "higher-level thinking."
Teele tells us that we can ask hypothetical questions to encourage neural branching:
"An example of hypothetical thinking is asking questions like, 'What would have happened to slavery if Abraham Lincoln had not been elected president of the United States when he was?' or 'What would our country be like now if slavery still existed?'"
And she explains how drawing can stimulate certain areas of the brain:
"Advanced drawing ability in gifted young artists is not usually identified until they are of school age. Some advanced young drawers are severely language disabled."
The author moves on to discuss the importance of time to learning. She highlights the following:
"It takes 5 to 6 hours to permanently store a skill in the brain. Do not try to teach too many skills at once that rely on the same part of the brain because the information is not retained in long-term memory."
Finally, Teele writes about the emotional part of the brain. If you recall, distinguishing logic from emotion was important to Eugenia Cheng as well. Teele highlights the following:
"The brain has two memory systems, one for ordinary facts and one for emotionally charged ones."
And she warns us:
"When our emotions increase, our rational thinking decreases."
But teachers with strong emotional intelligence are able to perceive whether they are engaging individual students. Again, logic and emotions can work together.
Teele concludes the chapter by discussing the information we've learned about the brain:
"This new information needs to be closely monitored and integrated into educational theories of how individuals learn and process information."
Lesson 4-5 of the U of Chicago text is "The Perpendicular Bisector Theorem." (This theorem is part of Lesson 5-5 of the modern edition.) 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 segmentAB.
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 ofAB, 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 ofAB 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 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. 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:
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 Lesson 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.
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
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
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
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. 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:
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 Lesson 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.
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