Chapter 7 Test (Day 46)

Today I subbed in a sixth grade math class. Students were working out of the Glencoe California Common Core math text -- the same text that I mentioned for eighth grade earlier on the blog.

Here is the table of contents for the sixth grade text:

1. Ratios and Rates
2. Fractions, Decimals, and Percents
3. Compute with Multi-Digit Numbers
4. Multiply and Divide Fractions
5. Integers and the Coordinate Plane
6. Expressions
7. Equations
8. Functions and Inequalities
9. Area
10. Volume and Surface Area
11. Statistical Measures
12. Statistical Displays

The students appeared to be finishing the test for Chapter 2 that they started yesterday. We just ended the first quarter, so the class ought to be finishing the Chapter 3 Test today, not the Chapter 2 Test. I am currently scheduled to sub in another middle school math class on Monday and Tuesday.

Returning to the U of Chicago text, here's the Chapter 7 Test. As usual, I provide some rationale for the answers.

1. The magnitude, or angle of rotation, is about 90 degrees. Notice that points M, N, O are collinear, as are O, S, T, in order to encourage students to look at the angle at which these two lines intersect.

2. The justification is: rotations preserve distance. Notice that we have a Reflection Postulate stating that reflections preserve distance, and a rotation is merely the composite of two reflections, each of which preserves distance. Therefore the entire rotation preserves distance as well. Indeed, we see that rotations preserve all four of the ABCD properties, as well as orientation, since the first reflection switches orientation and the second switches it back. Wu proves the properties of rotations independently of the properties of reflections, but in U of Chicago, rotations depend on reflections.

3. Angle POP" measures 180 degrees. The angle of rotation is always double the angle at which the two reflecting lines intersect. Since the reflecting lines are perpendicular, the magnitude is 180.

4. Make sure that the students rotate the triangle counterclockwise, since the angle is positive. (And speaking of positive and negative angles, notice that the answer to Question 1 is positive 90 or negative 90, because it is specified that MNP  is the preimage and STU is the image.) We see that the students are to rotate the equilateral triangle 120 degrees -- and notice that the rotation of an equilateral triangle 120 degrees ends up looking like a translation. This might make the rotation easier to perform, or it could trick the students.

5. Similarly, the rotation of an equilateral triangle 180 degrees ends up looking like a glide reflection.

6. No, since there is no AAA Congruence Theorem.

7. Yes, by SAS.

8. Yes, by HL.

9. Yes, by ASA.

10. a. SAS
      b. Angle G
      c. FG
      d. EG

11. a. SSS
      b. A <-> A, B <-> D, C <-> C

12. a. SAS
      b. N <-> O, O <-> Q, P <-> R

13. 1. FG = IG, EG = HG
      2. Vertical Angles Theorem
      3. SAS Congruence Theorem (steps 1, 2, 1)

14. This one is tricky because it appears that nothing is given!
Proof:
Statements                    Reasons
1. Circle O                    1. Given
2. AO = CO, BO = DO   2. Definition of Circle
3. Angle AOB = COD     3. Vertical Angles Theorem
4. Triangle ABO = CDO 4. SAS Congruence Theorem (steps 2, 3, 2)

15, It's Halloween weekend, so of course there must be a pentagram!
Proof:
Statements                           Reasons
1. Regular Pentagon ABCDE 1. Given
2. AB = BC = CD,                2. Definition of Regular Polygon
    Angle ABC = BCD
3. Triangle ABC = BCD        3. SAS Congruence Theorem (steps 2, 2, 2)
4. AC = BD                          4. CPCTC

16. Here's one of my favorites! We are given that the reflection of A is B, so by the definition of reflection, m is the perpendicular bisector of AB, so m is perpendicular to line AB. Similarly, the reflection of C is D, so by the definition of reflection, m is the perpendicular bisector of CD, so m is perpendicular to line CD. Therefore, by the Two Perpendiculars Theorem, lines AB and CD are in fact parallel. QED (Notice that in a way, these last few questions are previews of Chapter 5 -- we see that ABCD is an isosceles trapezoid.)

Today is a test day, so there should be some discussion about traditionalists. Well, what can I say that hasn't already been posted all this week with the Common Core elementary math problems going viral this week.

A second problem has appeared this week. A student was asked whether 103 - 28 = 75 is reasonable, but the student lost a point for finding the exact answer rather than subtracting.

Recall that in both cases, I agree with the traditionalists. I understand why so many problems like this appear -- in order to help students who can't perform the standard algorithms. They are frustrating when student who can perform them are asked to answer the Common Core way.

Enjoy your Halloween weekend!

Review for Chapter 7 Test (Day 45)

Today is Day 45, which marks the end of the first quarter on this blog. Because of this, I'm posting another test.

Last year at this point, I posted what I officially called the "Chapter 4 Test," and it was all about rotations and parallel lines. Now I've changed our Rotation Unit so that it covers congruent triangles instead of parallel lines.

So I must change the test. I could keep the questions about rotations and replace the parallel line questions with questions from the Chapter 7 Test. But now there's another problem -- I never actually posted a Chapter 7 Test last year!

This is because last year, Chapter 7 was the last chapter I covered before the first semester final. For the last chapter of a semester, it's sometimes tough to decide whether to give a test for that last unit or just roll into the final. When I was a student teacher, I taught Algebra I and II -- and we gave our Algebra II students a separate test for the last chapter, but not the Algebra I kids. Last year I decided to follow my Algebra I class and incorporate the Chapter 7 test questions into the final.

But this year, Chapter 7 appears at the end of the quarter, not the semester, so there is no first semester final to give now. So instead, I will post what I will now call the "Chapter 7 Test" -- and it will contain rotation questions from the old test as well as Chapter 7 material. As usual, all questions on today's review worksheet come from the SPUR section of the U of Chicago text.

The quarter mark is a good time for me to discuss my philosophy regarding grades. Last year I wrote about various grading scales. 

A grade represents a percentage, a scale from 0 to 100. One of the most commonly used grading scales in schools is something like this:

90-100 = A
80-90  = B
70-80  = C
60-70 = D
0-60   = F

Now let's return to school grades. For batting averages .300 is excellent, but in school 30% is surely a failing grade. For winning percentages .600 is one of the best teams in the majors, but in school 60% is the lowest passing grade. All of the grade boundaries -- from B to A, C to B, and so on -- occur in the upper half of the scale. The lower half of the scale -- from 0 to 50 percent -- is irrelevant as far as determining letter grades is concerned.

To the extent that C is average, the average student is earning between 70% and 80%. So we expect the average score in the class to be around the midpoint of this range -- 75%. So the average batter gets one hit every four at-bats, the average team wins two out of every four games, and the average student gets three out of four questions right.

And so, unlike the batter who gets a hit then makes an out and sees that overall his average has risen, the student who gets 100% on one test and 0% the next can never have a higher letter grade -- only a lower letter grade is possible. Therefore, a grade is much more likely to drop very rapidly than it is to rise very rapidly. Since the average grade is around 75%, a student would have to receive 150% on a test in order for the grade to rise as rapidly as a 0% drops it. (And all of this is assuming that the student isn't at one of those schools that has abolished D grades and makes 70% the lowest passing score -- that 10% difference means that a student would have to earn 160% on a test to have the same impact on the grade as a 0%!)

I've seen students come up to me and ask me why their grade has dropped so quickly -- especially when their grades have dropped from B to D seemingly overnight, while no one's grade is rising from D to B as quickly. They think that I'm a mean teacher and a harsh grader -- when the true reason is the nature of the grading scale, where the F grade takes up three-fifths of the scale.

Now it's the end of the quarter, and grades are about to come out. Let's assume for simplicity that there are equal numbers of points possible in the first and second quarters (in reality, the second quarter may have more points because that quarter has a final exam while the first quarter doesn't).

Let's say a student is earning 10% at the quarter. Then this student will surely fail the class -- even if 100% is earned the second quarter, the average grade is only 55%, an F. This is akin to a baseball team who is 11 games back with only 10 to play -- that team has been mathematically eliminated from winning the division.

A student earning 20% at the quarter can still pass the class -- but only if 100% is earned the second quarter, which isn't very likely. And a student who gets 30% the first quarter would still need to earn 90% (an A) the second quarter to pass the class. The type of student who would earn only 30% in a quarter is not the type of student who would earn an A the second quarter -- which typically covers more difficult material than the first quarter. This is akin to a baseball team who is nine games back with only 10 to play -- although the team hasn't been mathematically eliminated, they are realistically eliminated from the division. It would take a miracle for the team to come back and claim the division, and a similar miracle is needed for the student to get any semester grade other than F.

In fact, in some ways, any quarter grade below 50% would realistically eliminate a student from receiving a grade higher than F at the semester. A 50% quarter grade would require a second quarter of 70%, a C, to pass the class -- and this might be doable, if the student works a little harder during the second quarter.

(And that, of course, assumes that 60% is the lowest passing grade. If the 50% student is at a no-D school, then that student will need 90%, an A, in the second semester to get any letter other than F on the semester report card -- once again, realistic elimination.)

When I am teaching class, what I want is for as few students as possible to be mathematically or realistically eliminated from getting any semester grade other than F, when there are still plenty of weeks separating them from the end of the semester. I have no sympathy for a student who is realistically eliminated when there are only nine or ten days left in the semester, but nine or ten weeks, that's another matter.

A player on a baseball team that is realistically eliminated from the playoffs may be traded to a team that has a shot of winning. And even if the player isn't traded, he will still play out the rest of the season, no matter how little his heart is in it, because he's under contract to do so. Even a college student realistically eliminated from passing a class can drop out and take a W instead. But for high school students, there is no choice but for the student to take the F.

By the way, another complaint about grades is that it seems so difficult to get an A. A student with an 88% late in the semester asks me whether it's possible to get an A in the class. The truth be told, it's virtually impossible to raise that grade the two percentage points necessary to get an A -- one would essentially have to earn 100% on every remaining assignment. This is because ninety percent is just a very, very difficult target to reach if one is currently below it. It's much easier to raise a 68% two points up to 70% than it is to go from 88 to 90, or even 89 to 90. After all, if you have a 68%, you can get an 85% on the next assignment and raise it up to 70% easily. The equivalent for our 88% student would be to get a 105% to raise it up to 90%, but the highest possible score is 100%. In other words, there are plenty of possible scores above 70 to help raise your grade to a C, but there just aren't enough grades between 90 and 100 to help raise your grade to an A. Students think I'm just being mean or stingy with the A's, but in reality, it's the percentage scale that assigns three-fifths of it to the lowest grade that makes it difficult to raise one's grades, especially from F to D or from B to A.

The reason that I'm bringing up grades again is that in the news, there is a school district up in Northern California that is implementing a controversial new grading system:

http://www.pressdemocrat.com/news/4640349-181/rohnert-park-cotati-schools-rethink

I mentioned last year that one way to avoid realistic elimination would be to allow students to retake the tests on which they received a low score. But in this article, there is mentioned of two separate -- and I repeat, separate -- grading scales that help students avoid realistic elimination.

The first is to record scores of zero as equaling 50%. After all, it's the devastating effect of 0% on the grades that eliminate students from passing the course. The average of 0% and 100% is 50%, an F, but the average of 50% and 100% is 75%, a C.

Back when I was an Algebra I student, my teacher did something similar. When she recorded test scores, she only recorded a letter grade, so if one received an F, one can't tell whether the F was really a score of 1% or 59%. Then to determine the average at the quarter or semester, she converted the grades back into percentages by letting each letter represent the midpoint of the interval. So an A was 95%, a B was 85%, a C was 75%, and a D was 65%. But an F was converted into 55%. So this represents a similar scheme to the 0 -> 50 system, since a score of 5% instantly became 55%.

Now this California district had already implemented the 0 -> 50 system, but now it is switching to yet another system. The grading scale will now look like this:

80-100 = A
60-79 = B
40-59 = C
20-39 = D
0-19 = F

Notice that this equal-interval scale has the same effect as the 0 -> 50 scale -- the average of 0% and 100% is still 50%, but that is now a C. Both systems are designed to lessen the effect of a zero so that fewer students are eliminated from passing the class with many weeks left in the semester. So because of this, there is no need to do both. It's makes no sense to change zeros to 50% and then say that 50% is a C.

I was under the impression that in converting to the equal-interval scale, the district would eliminate the 0 -> 50 conversion once and for all. But some comments to this and related articles makes it sound as if the district really is giving out C's for missing assignments. I agree with the commenters who say that doing both is a terrible idea.

A quote from the article:

Lowery, who has tried to retain the traditional grading system, said he has a student who early in the semester had a grade in the 80s range and another in the 25 percent range, but then stopped coming to class. Because his missed assignments all received a 50 percent, the student has a 49 percent average, instead of the 3 percent average he would have under the previous grading system, Lowery said.

Notice that under the traditionalist scale, 49% is still an F, So 0 -> 50 has no effect on this student's letter grade -- it would be F no matter what. If we use the equal-interval scale, then we should not do the 0 -> 50 conversion on top of that. So the student's grade would still be 3%, which is still an F even under the equal-interval scale.

But the rationale for either 0 -> 50 or equal-interval is so that students can raise their grades with plenty of time left in the semester. So let's say that it's today -- the end of the first quarter -- and the student who has missed so many days in the first quarter has returned to class and wants to work hard during the second quarter in order to get a passing grade at the semester.

Under the 0 -> 50 system, the student has 49% at the quarter. So that student only needs a 71% the second quarter to raise the grade to D. He even has an outside chance -- working extremely hard -- at getting 91% the second quarter to raise the grade to C.

Under the equal-interval system, the student now has a 3% He only needs 37% the second quarter to get a D, and has an outside chance at getting 77% the second quarter to get a C.

On the other hand, under traditionalist grading with neither 0 -> 50 or equal-interval, even getting 100% the second semester only raises the grade to 51.5%. It is mathematically impossible for the student to receive any grade other than F at the semester. So the student, who might have considered working hard the second quarter to raise his grade, instead has no incentive to attend class the second quarter and might as well skip class the second quarter, just as he did the first quarter.

Under either 0 -> 50 or equal-interval, the student can work hard for a D or even a C. Under the traditionalist scale, the student stays at home. So, ironically, more actual learning can occur under the 0 -> 50 or equal-interval scales than under the traditionalist scale.

In a related article, a teacher from this district defends the alternative grading scales:

http://www.pressdemocrat.com/opinion/4665017-181/close-to-home-in-defense?page=2

Now let’s look at student Y. Through no fault of her own and for myriad possible reasons, she comes to my English class with a deficit of skills in reading, writing and even staying organized. Her early assignments earn abysmal grades, and she determines to do better. Through our school’s various interventions — and especially through her own hard work — she masters the skills being taught in my class.

By the end of the semester, she is scoring in the high 80s on assessments similar to the ones she failed earlier.

Under what logic should her early, low grades be counted against her?

Is it not right to give her a grade reflective of her mastery of the class material?

Now here are some responses to her article:

Chum Rum

Why not just grade incrementally. You got an F in the 1st month, a D in the 2nd month, a C in the 3rd month, and a B in the 4th month. See? You're improving and we can see your progress!

Well, let's find out what the problem is. Let's assume that the D, C, and B grades mentioned by the commenter are the highest possible grades in their respective ranges, but the F here is a zero.

Under the 0 -> 50 system:
F: 50%
D: 69%
C: 79%
B: 89%
Semester Average: 71.75%
Grade: C

Under the equal-interval system:
F: 0%
D: 39%
C: 59%
B: 79%
Average: 44.25%
Grade: C

Under the traditionalist system:
F: 0%
D: 69%
C: 79%
B: 89%
Average: 59.25%
Grade: F

So even though this girl has done such amazing things -- going from hardly knowing any English at all to being one of the top students in the class -- she just can't quite make it to sixty percent for the overall semester average, so under the traditionalist system, the only letter that can appear on her report card is F. In other words, she was realistically eliminated from passing with two months left in the semester.

The same commenter later on wrote:

Chum Rum

So if you totally suck at math, but buy the answers to the final off of a dumpster diver and ace the last test you should totally get an A+, because you suddenly 'got it'.

Another person responded to say that no, cheaters get an F no matter what the scale is (provided, of course, that they are caught cheating.) On the other hand, one sometimes reads about students who do no work the entire semester yet still get a perfect score on the final -- of course, they could've earned 100% on the final the very first day of class, because they already knew the material beforehand. In this case, they are commenting on a bureaucracy that won't let students who already know the material challenge or skip the course -- but that's a separate matter.

PD Reader

Cate Woods: "Under what logic should her early, low grades be counted against her?"

Under what logic should that student have been advanced to the next grade in the first place when they clearly didn't have the skills to do so? THAT is where the school system is failing our kids.

This takes us to other issues such as retention -- which is said not tot work -- or using other schemes to divide the students into classes, such as the Path Plan mentioned earlier on the blog. But then one must be careful when dividing students into classes, since then it could turn into tracking, which has several political, economic, and demographic implications.

Indeed, here's another commenter:

SS

I am a Professor, so I am an educator, with over 25 years at this.
The article from a few days ago said, a student that does not take a test or turn in homework will receive the grade of 50, AND that 50 is now a middle C.
So by this grading method, a student could Never attend and pass.
Let me ask the RP/Cotati School Board, are teachers now to be graded the same way......that they don't even have to show up and they get a passing grade?

[First of all, one should use either 0 -> 50 or equal-interval, but not both. So there should be no 0 -> C grades. dw]

When I was a kid you had to learn all the stuff taught (okay, we were supposed to)....and then education began to get dumbed down, and it was called Outcome based, so we have to pick X 'learning outcomes' (was about 10....but remember the table of contents for the text had many more items). A few years ago that list of 10 learning outcomes was deemed Too Difficult, so it got whittled down to maybe 5, and now the whole bend is Competency Based. (Let's jump in Mr. Peabody's WayBack Machine and....oh yeah, years ago when the "outcomes"...what you were required to know, was listed in the table of contents) So now the PC's (professional competencies) are a list of maybe 4 or 5 things like, multicultural understanding, group work.

["Multicultural understanding" -- just as I implied earlier, tracking has several demographic implications. -- dw]

No mention or grading of the actual material in the name of the course.....but we have a really good touch-feely, lets all gather in a circle and sing Kumbahya.
The "bar" of education should always be a reach, and not down where it has been and this insane concept of, we wouldn't want anyone to feel bad.......

[But under the traditionalist scale for our Student Y above, the "bar" is out of reach, because the she has worked very hard but has only a 59.25% F to show for it.]

And are employers going to rate employees by this system?

As for that last question, suppose an employer decided to use the following pay schedule -- instead of paying by the hour, let's say employees have to work at least 60 hours a week to get any money at all (just as students have to score 60% to receive any credit at all, under the traditionalist plan). And if one works fewer than 60 hours, then one has to make up the hours the following week such that the average number of hours worked is 60. So someone who falls behind will have trouble catching up and thus ever making any money at all. And let's say you have to wait 20 weeks before the earliest weeks aren't included in the average. Are employers going to pay employees by this system? No, but that is how the traditionalist grading system appears to many students.

Michael 

There are a few things that need to be addressed.
- If you are grading based on mastery of standards, then
each standard is reported on via the report card. Some students will get good grades for all the standards, some for a few of the standards, and some for none of the standards.
- If you are averaging grades across a subject area, that doesn’t report ‘progress’ in any way, shape or form. Many content areas (standards) are taught over the course of a semester. Just look at Student Y in
the example from the article.
- Homework should not count towards any grade. There is no accounting for who actually did the work, such as parent help or searching for answers on the internet.

What is not being discussed here is the move to grade students based on each content standard or unit of study. This method produces far more accurate information that can help those at school and at home help a student succeed. A grade averaged across a content area does
not accurately reflect student knowledge. These teachers and administrators should be honest about what is graded, when, and why. In addition, they should be reporting on how those results are communicated to parents. A clearly defined report card that reports
mastery of each content area (standard) in each subject is the clearest ‘snapshot’ of student progress.

Notice that this was also previously mentioned here on the blog -- it takes us to the Raenbo Certificate of Academic Proficiency proposal.

Here's what I say about 0 -> 50 and equal-interval grading. If -- and I repeat, if -- one were to use equal-interval grading, then here's how I would recommend doing it:

-- Don't use 0 -> 50 grading at the same time.
-- Take full advantage of the use of the equal-interval scale. In a higher Algebra I or Algebra II class, add all those topics that are missing from Common Core that help prepare for Calculus. Under the traditionalist grading scale, if there are too many advanced questions on the test, too many students will score below 60% and fail. But if it takes only 20% to pass, then there's room to include more advanced questions to separate the A's from the B's without too many students getting F's. Sure, one only needs 80% to get an A, but they should have to work to get that 80% by answering questions that prepare them for Calculus.

If one were to use a 0 -> 50 type of grading scale, then here's what I recommend:

-- Don't convert zeros or low scores on the final to 50%. The whole point of 0 -> 50 is to protect students when there is plenty of time left in the semester. There's no need to protect them by the time they reach the final.

Well, here's today's review worksheet. You can use whatever grading scale you want when it's time to grade the actual test.

Lesson 7-5: The SSA Condition and HL Congruence (Day 44)

Lesson 7-5 of the U of Chicago text is on SSA and HL. I've already mentioned how I'll be able to prove HL without using AAS, since we have to wait before I can give an AAS proof.

Meanwhile, we know that SSA is invalid, but the U of Chicago text provides us with an SsA Congruence Theorem, where the size of the S's implies that it must be the longer of the congruent sides that is opposite the congruent angle. The text doesn't provide a proof of SsA because the proof is quite difficult -- certainly too difficult for high school Geometry students.

But you know how I am on this blog. I'm still curious as to what a proof of SsA entails, even if we don't ask high school students to prove it. Last year, I mentioned how SsA leads to the ambiguous case of the Law of Sines.

I've noticed that when using the Law of Sines to solve "both" triangles for the SsA case, the "second" triangle ends up having an angle sum of greater than 180 degrees. We can use the Unequal Sides Theorem to see why this always occurs -- that theorem tells us that the angle opposite the "s" must be smaller than the angle opposite the "S" (the known angle). We know that if two distinct angles between 0 and 180 have the same sine, then they are supplementary. So the second triangle would have angle sum of at least 180 minus the smaller angle plus the larger angle -- which always must be greater than 180. (This is often called the Saccheri-Legendre Theorem, named for two mathematicians with whom we're already familiar and associate with non-Euclidean geometry. Of course spherical geometry is not neutral, and Saccheri-Legendre fails in spherical geometry.)

That the sum of the angles of a triangle can never be greater than 180 (but we don't necessarily know that it's exactly 180) is neutral, but the Law of Sines is not neutral. Nonetheless, it is known that SsA is a neutral theorem. So the Law of Sines can't be behind the secret proof that U of Chicago text doesn't print in its text.

But one of the questions from the U of Chicago text hints at how to prove SsA -- and it's one that I included on my HL/SsA worksheet last year. Here is the question:

Follow the steps to make a single drawing of a triangle given the SSA condition:
a. Draw a ray XY.
b. Draw angle ZXY with measure 50 and XZ = 11 cm.
c. Draw circle Z with radius 9 cm. Let W be a point where circle Z and ray XY intersect.
d. Consider triangle XZW. Will everyone else's triangle be congruent to yours?

The answer is that they most likely won't. In step (c) we are to let W be a point -- not the point -- where the circle and the ray intersect. This implies that there could be more than one point where they intersect -- and in fact, there are two such points. But there can never be a third such point, because a circle and a ray (or line) intersect in at most two points. We can prove this indirectly using the Converse of the Perpendicular Bisector Theorem:

Lemma:
A line and a circle intersect in at most two points.

Indirect Proof:
Assume towards a contradiction that there exists a circle O that intersects line l in at least three points A, B, and C, and without loss of generality, let's say that B is between A and C. By the definition of circle, O is equidistant from A, B, and C. From the Converse of the Perpendicular Bisector Theorem, since O is equidistant from A and B, O lies on m, the perpendicular bisector of AB -- and again using the converse, since O is equidistant from B and C, O lies on n, the perpendicular bisector of BC. We know that m and n are distinct lines because m intersects l at the midpoint of AB and n intersects l at the midpoint of BC -- and those midpoints are distinct because B lies between them.

Now both m and n are said to be perpendicular to l, since each is the perpendicular bisector of a segment of l. So by the Two Perpendiculars Theorem, m and n must be parallel -- and yet O is known to lie on both lines, a blatant contradiction. Therefore a line and a circle can't intersect in three points, so the most number of points of intersection is two. QED

So let's prove SsA now using this lemma and the construction from the problem above. Once again, the proof must be indirect. Even though most of our congruence theorem proofs call the two triangles ABC and DEF, I will continue to use the letters XZW so that it matches the above question.

Given: AB = XZ < BC = ZW, Angle A = Angle X
Prove: Triangles ABC and XZW are congruent.

Proof:
We begin by performing the usual isometry that maps AB to XZ. As usual, we wish to show that the final reflection over line XZ must map C to W -- that is, C' must be W.

So assume towards a contradiction that C' is not W. As usual, the given pair of congruent angles allows us to use the Flip-Flop Theorem to map ray AC to ray XY. Just as in the problem from the text, we know that C' must be a point on ray XY (since ray AC maps to ray XY), and it must be the correct distance from Z (since reflections preserve distance). Now the set of all points that are the correct distance from Z is the circle mentioned in the above problem. We know that the intersection of a circle and a ray is at most two points, and so the assumption that C' is not W implies that W must be one of the two points of intersection, and C' must be the other point.

We must show that this leads to a contradiction in the SsA case -- that is, when AB and XZ are longer than the sides BC and ZW. We see that if C' and W are distinct points equidistant from Z, then the triangle ZC'W must be isosceles, and so its base angles ZC'W and ZWC' must be congruent.

Now we look at triangle ZC'X. It contains an angle, ZC'X, which forms a linear pair with ZC'W, so its measure must be 180 - m/ZC'W -- that is, m/ZC'X = 180 - m/ZWC' by substitution. It also contains an angle ZXC' -- which (renamed as ZXW), we see must be larger than ZWC' (renamed as ZWX) by the Unequal Sides Theorem -- ZXW is opposite the longer side ZW (in triangle ZXW), so it must be the bigger angle.

So now we add up the measures of two of the three angles in triangle ZC'W -- the angle ZC'X has measure 180 - m/ZWC', and angle ZXC' is known to be greater than ZWC'. So the sum of the two angles is greater than 180 - m/ZWC' + m/ZWC' -- that is, it is greater than 180. That is, the sum of the angles of triangle ZC'W is greater than 180 -- which is a contradiction, since by the Triangle-Sum (actually Saccheri-Legendre) Theorem, the sum must be at most 180.

Therefore the assumption that C' and W are different points is false. So C' must be exactly W. QED

Of course we wouldn't want to torture our students with this proof. It depends on three theorems -- Isosceles Triangle, Unequal Sides, and Triangle-Sum -- that we have yet to prove. On my worksheet, I added an extra note to this problem explain how it leads to SsA.

Lesson 7-4: Overlapping Triangles (Day 43)

Today I subbed at a middle school. It was mostly for sixth and seventh grade English, but during the conference period, I also covered a special education math class for seventh graders.

They were working on adding rational numbers. This time, they had to add numbers that already had like denominators -- but since this is seventh grade, these were signed rational numbers. The trickiest problems for students were ones where they had to keep track of the sign, reduce the fraction, and convert it to a mixed number all in the same problem. There were also a few questions where the students had to explain their answers for Common Core. We already mentioned last week that special ed students often struggle with these.

I know that I shouldn't still be discussing this from yesterday, but I was thinking about what to do when faced with a student who refuses to put squared and cubed units for areas and volumes -- especially considering some of the silly reasons that teachers deduct points for. We want him to see that putting squared and cubed is crucial, not silly. Well, I was thinking about the old trick of turning a dollar into a penny due to carelessness with units:

$1 = 100c = (10c)^2 = ($.10)^2 = $0.01 = 1c

If we want to emphasize the connection to measurement, use meters and centimeters, or even yards and inches, instead of dollars and cents.

Here's another trick that might work to convince students to do extra things to help them in math -- in this case Algebra I -- without making it seem that we're taking off points for correct answers. In Algebra I, we often want students to draw a line through the equal sign to make sure that they're not getting the two sides of the equation confused. But some students can already solve equations and get correct answers without drawing the line. Now here's a possible solution that I may implement if I ever teach Algebra I -- when solving equations in class, I make sure that I always draw the line. Then if I ever forget to draw the line, I give the student who catches my error an extra point. That way, the only person I'm requiring to draw the line is myself -- yet the weaker students see the importance of drawing the line.

Lesson 7-4 of the U of Chicago text covers more proofs. These proofs are trickier, since they involve overlapping triangles.

Last year, I tried to post an activity about overlapping triangles, but this was still in the middle of my scanner not working, and the activity didn't come out too well. So now I'm making this into a regular lesson and rewriting it from scratch.

Once again, I had to be careful not to include any questions that require theorems that we have yet to prove, such as the Isosceles Triangle Theorem, that we're still waiting until next week for. Because the triangles overlap, it may appear that, in Question 4, we need to show that Triangle SUA is isosceles. But as it turns out, we actually don't need to show this to complete the proof.

The bonus question is somewhat interesting here. It asks whether there is a valid congruence theorem for quadrilaterals, SSASS. Last year I tried to solve it, but got confused, so I want to take the time to set the record straight.

As it turns out, SSASS is not a valid congruence theorem for quadrilaterals. A counterexample for SSASS is closely related to a counterexample to SSA for triangles -- we start with two triangles that satisfy SSA yet aren't congruent -- one of these will be acute, the other obtuse. Then we reflect each triangle over the congruent side that is adjacent to the congruent angle. Each triangle becomes a kite -- as the original triangles aren't congruent, the kites can't be congruent either, yet they satisfy SSASS (with the A twice as large as the A of the original triangles).

I tried to prove SSASS by dividing each quadrilaterals into two triangles, then using SAS on the first pair and SSS on the second. The problem with this is that that division doesn't produce two triangles unless the quadrilateral is known to be convex. With our two kites, notice that the acute triangle becomes a convex kite, while the obtuse triangle becomes a nonconvex (or concave) kite -- which is also known as a dart. If both quadrilaterals are already known to be convex, then my proof of SSASS is valid.

One congruence theorem that actually is valid for quadrilaterals is SASAS. We can prove it the same way that we proved SAS for triangles. We put one of the sides -- in this case the congruent side that's between the other two congruent sides -- on the reflecting line. Then we can prove that the two far vertices are on the correct ray, the correct distance from the two vertices on the reflecting line -- this works whether the quadrilateral is convex or concave. We can also prove SASAS by dividing the quadrilateral into triangles. There are separate cases for convex and concave quadrilaterals, but all of them work out.

Other congruence theorems for quadrilaterals are ASASA and AASAS. Another congruence theorem, AAASS, is also valid, but it's similar to AAS in that there's a trivial proof based on the angle-sum that reduces it to ASASA (just as AAS reduces to ASA), only in Euclidean geometry. A neutral proof of AAASS exists, but it's more complicated.

Lesson 7-3: Triangle Congruence Proofs (Day 42)

Lesson 7-3 of the U of Chicago text discusses triangle congruence proofs. Finally, this is what most Geometry students and teachers think of when they hear about "proofs."

There's actually not much that I have to say about this lesson. We already know much about how the proofs in this section go -- generally we are given two triangles with some of the corresponding sides and angles given as congruent. In the easiest examples, the given pairs are already enough to conclude that the triangles are congruent. The U of Chicago text points out that often we must work to get one of the needed pairs -- such as the Reflexive Property of Equality when the two triangles have a side in common, or perhaps the Vertical Angles Theorem to find a pair of congruent angles -- then we use SAS, SSS, or ASA to conclude that the two triangles are congruent. Finally, the students are usually asked to prove one more pair of parts to be congruent, which requires the use of CPCTC.

Now technically speaking, CPCTC was covered in Lesson 6-7, which we skipped over. The U of Chicago text uses the abbreviation CPCF, which stands for corresponding parts of congruent figures, for after all, the property applies to all polygons, not just triangles. But the abbreviation CPCTC is so well-known that I prefer to use CPCTC rather than CPCF. Even though we skipped Lesson 6-7, I believe that the students can figure out CPCTC quickly, so I incorporate it into today's lesson.

Last year I didn't create a worksheet, but instead just wrote down ten problems. This year, I will create a worksheet and include some of the problems from last year, but others I had to throw out. In particular, I had to drop the proof of the Converse of the Isosceles Triangle Theorem -- we certainly want to avoid the U of Chicago proof that uses AAS, since we haven't taught AAS yet. I want to cover it at the same time as the forward Isosceles Triangle Theorem -- I'm saving both for next week. I also had to throw out a few "review" questions that once again review lessons that we've skipped.

Notice that Question 5 mentions isosceles triangles, but it doesn't actually require either the Isosceles Triangle Theorem or its converse. We only need to use the definition of "isosceles" to get one pair of congruent sides -- the first S. The definition of "midpoint" gives us the second S, and the Reflexive Property of Congruence gives us the third S, so the triangles are congruent by SSS.

Indeed, this is a good time for me to bring up a point about two-column proofs. I can easily see a student being stuck on this question because none of the three S's are given -- they have to work to find all of them. One thing I like to point out is if the students see a long word like "isosceles" -- or even "midpoint" -- in a question, they will usually need the definition of that word. In particular, a big word in the "Given" usually leads to the students using the meaning half of the definition, and a big word in the "Prove" often needs the sufficient condition half. The meanings of the words "isosceles" and "midpoint" lead to two of the three S's, and then they'll probably see that the Reflexive Property gives the third S.

Of course, soon they'll learn about the Isosceles Triangle Theorem and its converse -- and once those appear, nearly every proof involving isosceles triangles will use one or the other. But before students learn about the theorems pertaining to a key term, it's the definition that appears in the proof. We've already seen a few proofs about parallel lines where it's the definition of "parallel" that matters -- but once we've proved the Parallel Tests and Consequences, we hardly ever use the definition of parallel.

Last thing -- in the U of Chicago text, the "Given" is never included as a separate step. I always include it as this is more common -- so our proofs will have one more step than those in the text.

OK, my very last post was about traditionalists vs. Common Core, so I shouldn't be posting another such post right away. But over the weekend, another elementary Common Core horror worksheet went viral, and I want to address it.

There was a third-grade assignment where students are supposed to multiply 5 times 3 using some repeated addition. A student wrote 5 + 5 + 5 = 15, but was marked wrong, because the answer intended by the teacher was 3 + 3 + 3 + 3 + 3 = 15. The next question was to multiply 4 times 6 using an array. The student drew six rows of four dots, but was marked wrong -- the answer intended by the teacher was four rows of six dots.

Just as with the infamous Common Core check, the mathematician Hemant Mehta came in to explain what is going on with these two problems. (Warning: Mehta's blog is mostly about religion and politics -- yet once again, it's the math post that draws the most comments.)

http://www.patheos.com/blogs/friendlyatheist/2015/10/21/why-would-a-math-teacher-punish-a-child-for-saying-5-x-3-15/

Mehta explains that even though multiplication is commutative, drawing 4 times 6 as four rows of six rather than six rows of four is more consistent with the matrices that one learns in Algebra II:

What about the array problem? This one’s even more straightforward.

It actually matters which way you draw the picture. But it’s not something kids will understand until they start using matrices in algebra class.

Even if you’re unfamiliar with matrices, here’s what you need to know: There’s a difference between a 2 x 3 matrix and a 3 x 2 matrix.

So here Mehta argues that a third grade teacher who has her students distinguish between four rows of six and six rows of four will help prepare the students for higher math.

So now let's look at the first problem. Mehta tries to explain here why 5 * 3 should be written only as 3 + 3 + 3 + 3 + 3 and not as 5 + 5 + 5. But I disagree with Mehta here. For just as we appealed to higher math to explain how to draw the array, I can appeal to higher math to explain why, in fact, we should say that 5 + 5 + 5 is correct and 3 + 3 + 3 + 3 + 3 is wrong.

To find out why, the higher math we seek out is set theory -- the same theory that mathematicians all the way from Georg Cantor to Randall Holmes studied. In particular, we consider something called an "ordinal." Instead of 5 * 3, let's consider omega * 3, where omega -- the last letter of the Greek alphabet -- is a particular ordinal. Just as with matrix multiplication, ordinal multiplication is not commutative -- and in fact, omega * 3 is exactly omega + omega + omega. It is definitely not the same as 3 * omega. So using ordinal arithmetic, 5 * 3 = 5 + 5 + 5 is exactly right.

I've decided that it would detract from this Geometry blog to get into a deep discussion of what exactly ordinals are -- except that they have something to do with order (which helps to explain why the order matters). Here's a link that describes ordinal multiplication in more detail:

http://mathworld.wolfram.com/OrdinalMultiplication.html

I can see why students would be frustrated when they lose points for multiply in the wrong order. As Geometry teachers, we can compare this to the student who complains about losing points for not putting square units for area or cubic units for volume. Of course, there are actual reasons that students must put square or cubic units. But juxtaposing these two situations in which students lose points even after doing all of the calculations correctly, we can see why from the students' perspective they are just losing points for silly reasons in both cases. This is indeed, exactly what happened today when I was subbing -- remember that today was my last day at a continuation school. A student -- who had been upset at something that had happened earlier in the day -- didn't want to listen to me when I told him to write down squared or cubed when calculating area or volume. (To believe that
students at a continuation school would listen to me just because I was "the sage on the stage"!)

Again, Mehta's post has drawn more than a thousand replies. In that comment thread, I noticed that one poster actually mentioned ordinal multiplication as an example! A poster named MNb wrote:

MNb:

As a math teacher myself I think this case is very straightforward. The task explicitely was not about the answer, but about the method:

"Use the repeated addition strategy"

I do this all the time myself. If I give a quadratic equation and I ask them to use factorization then using the abc-formula is simply wrong. And I tell my pupils in advance.

[emphasis MNb's]

And it was a poster with the single letter b who responded with ordinals:

b:

You are not teaching students a useful skill. You are teaching them an arbitrary rule that has no applicability outside passing a test that you wrote. No mathematician would say that pq can only mean p groups of q, never q groups of p. It can be either, or neither.

After thinking for a while, I've come up with two cases where a product has an asymmetrical grouping interpretation in real mathematics.

One is that matrices are more commonly divided into column vectors than row vectors, so if you think of a 5-component vector as a group of 5 things, then a 5×3 matrix is more likely to be 3 groups of 5 than 5 groups of 3.

The other is ordinal multiplication, where ω·3 means ω+ω+ω, while 3·ω means 3+3+3+... (ω times), which is less than ω+ω+ω.

Note that in both cases the order is the opposite of what you teach.

[end b's response]

I was actually looking for a simple traditionalist reply, but didn't find it quickly enough. Most traditionalists would say that the goal is for students to multiply quickly -- to be able to say that 3 times 5 is 15 in one second or so. All this about 3 groups of 5 vs. 5 groups of 3 should only be the means to that end. And I agree with traditionalism at that age (third grade).

Activity: Jeopardy! (Day 41)

I have finished reading Benoit B. Mandelbrot's The Fractal Geometry of Nature. Let me tell you a joke about Mandelbrot:

Q: What does the B. in Benoit B. Mandelbrot stand for?
A: Benoit B. Mandelbrot.

Notice that this makes Mandelbrot's name self-similar, just like the fractals that he studied. Actually, as it turns out, this no joke. Mandelbrot had no middle name, and so he intentionally show the initial so that his name could be self-similar.

The readers of this blog may wonder, why did I read and discuss Mandelbrot's book anyway? It's because in learning something new like fractal geometry (or category theory, like that book that I read over the summer), I can gain a better perspective of my students when they learn new things.

I admit that Mandelbrot's book was a tough read -- especially the second half. There were so many equations that I couldn't understand -- for example, Calculus students learn about differentiation and integration, but Mandelbrot describes something called "integro-differentiation."

But as I have trouble with fractal geometry, my students have trouble with regular Geometry. As I struggle with 2.5-dimensional objects, my students struggle with 2- and 3-dimensional objects. I don't know everything there is to know about geometry, and so I can't expect my students to know everything there is to know about geometry either. And so I can better sympathize with my students when don't understand a Geometry topic the first time I teach it. All in all, I highly recommend Mandelbrot's book to those who want to expand their horizons.

Today is an activity day. Now last year I didn't plan on having any activities between Lessons 7-2 and 7-3, but with the changes to my pacing plan, this year I want an activity at this point.

So what activity should I post? Well, all this week (from last Monday, Day 37, up to and including this Monday, Day 42), I've been subbing in the same continuation school that I was in last month (back on Day 18). And as I wrote in September, every Friday that class plays a game of Jeopardy.

And so I decided to devote today activity to yet another game of Jeopardy -- except that instead of an afterthought to the already scheduled Daffynition Game that I posted in September, this time Jeopardy will be the main activity.

In the actual classroom today, math was one of the four categories. This time, the topic for math is percents and discounts. Unfortunately, the Jeopardy game was canceled because I couldn't come up with good questions for the other categories.

Now for the Jeopardy game that I'm posting for this Geometry blog -- which of course won't be cancelled -- the four categories refer to the four lessons that I posted this week -- 6-3, the extra rotations lesson, 7-1, and 7-2. Of course, the first two lessons are so closely related, as are the last two lessons, that it may be difficult to distinguish the lessons among the categories.

Here's what I came up with for the names of the categories:
1. Rotations = 2 Reflections
2. Properties of Rotations
3. Triangle Congruence (general theorems like SAS, SSS, etc.)
4. Can We Prove Congruence? (specific numbers and shapes are given)

I've decided that I won't post a worksheet for this, because it's best done not on worksheets, but rather an index card for each question. So let me type in the questions right here in this post:

Rotations = 2 Reflections
100. A rotation centered at O maps P to P'. Why is OP = OP'? (Rotations preserve distance.)
200. A rotation maps Angle N to Angle T. Why are they equal? (Rotations preserve angle measure.)
300. The angle between two mirrors is 37. What is the angle of the clockwise rotation? (74)
400. The angle of a rotation is 172. What is the angle between the mirrors? (86)
500. The angle of a rotation is -160. What is the angle between the mirrors? (80)

Properties of Rotations
100. What is the sign of a clockwise rotation? (negative)
200. What is the sign of a counterclockwise rotation? (positive)
300. Rotate a 30-60-90 triangle about the midpoint of its longest side. What do you get? (rectangle)
400. Rotate a 45-45-90 triangle about the midpoint of its longest side. What do you get? (square)
500. Rotate a 45-degree angle about its vertex. What type of angles are 45 degrees? (vertical angles)

Is This a Valid Triangle Congruence Theorem?
100. SSS (yes)
200. AAA (no)
300. ASA (yes)
400. SAS (yes)
500. SSA (no)

Can We Prove Congruence?
100. Triangle ABC has sides 2, 8, 7 cm. Triangle DEF has sides 7, 2, 8 cm. (yes, SSS)
200. Triangle ABC has angles 30-60-90. Triangle DEF has angles 30-60-90. (no)
300. AB = DE = 20, Angle A = D = 60, Angle B = E = 30. (yes, ASA)
400. AB = DE = 20, Angle A = D = 60, AC = DF = 10. (yes, SAS)
500. AB = DE = 20, Angle A = D = 60, BC = DF = 10. (no)

Final Jeopardy
(Draw a kaleidoscope image with hexagonal symmetry. Label the center O, then label three points A, B, and C equally spaced around the center at 12 o'clock, 4 o'clock, and 8 o'clock, respectively.)
What is the angle of the rotation centered at O mapping B to C? (-120)

Now since today was an activity day rather than a test day, I really don't need to have another topic about traditionalists today. But the traditionalist Dr. Katharine Beals is still posting her Math Problem of the Week series comparing Common Core to pre-Core Geometry, and so I will address it.

Last week, Dr. Beals posted one of the two proofs that appear on the PARCC Practice Performance Based Assessment (PBA) exam. Today she posted the other one:

Marcella drew each step of a construction of an angle bisector.
9. Part A
Angle Z is given in Step 1. Describe the instructions for Steps 2 to 5 of the construction. Enter your description in the space provided.
Part B
Marcella wants to explain why the construction produces an angle bisector. She makes a new step with line segments AC and BC added to the proof, as shown.
Using the figure, prove that ray ZC bisects angle AZB. Be sure to justify each statement of your proof.

Now in each post in this series, Beals asks a few "Extra Credit Questions":

1. In how many steps can this proof be completed?
2. Does practice constructing lengthy proofs play any role in preparing for college and career in the 21st century?

We can tell by the way Beals asks the questions what answers she's expecting. Indeed, the second question contains the intended answer to the first -- she wrote "lengthy proofs," so the answer to the first is something like "too many." And we can easily figure out the intended answer to her second question as well -- "No, the long proof doesn't prepare anyone for college or career, so completing it is a complete waste of time." In other words, Beals, who laments that there aren't enough proofs in Common Core Geometry, is now implying that a Geometry proof is a waste of time!

So what's going on here? The only logical explanation is that Beals considers some proofs to be a waste of time and others to be essential to a rigorous Geometry class. So how can we distinguish between a useless proof and an essential proof?

The obvious guess -- that Beals views proofs based on reflections and other transformations to be useless and those based on triangle congruence as useful -- is invalid because we know that PARCC is expecting a proof based on triangle congruence for Part B. This is because a similar problem appears as Question 25 of the Practice End-of-Year (EOY) Exam. That question asks about the angle bisector construction, and it specifically mentions congruent triangles. Nonetheless, Beals doesn't like this proof appearing on the standardized test.

We think back to Dr. David Joyce, who also emphasizes a strong proof-based course. Here's what he writes in particular about the proof of constructions such as the angle bisector:

The book 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.

So we see that Joyce -- unlike Beals -- believes that constructions such as 4) to bisect an angle should indeed be proved in class.

Beals asks her readers to count the number of steps required in the proof. Let's try writing a proof to find out:

Given: Angle Z
Prove: Ray ZC, as constructed, bisects Angle AZB.

Proof:
Statements                                         Reasons
1. Angle Z                                          1. Given
2. Construct points A, B on Angle Z 2. Compass rule
such that ZA = ZB
3. Construct point C inside Angle Z 3. Compass rule
such that AC = BC
4. ZC = ZC                                         4. Reflexive Property of Equality
5. Triangles AZC, BZC congruent     5. SSS Congruence Theorem (steps 2, 3, 4)
6. Angle AZC = BZC                         6. CPCTC
7. Ray ZC bisects Angle AZB            7. Definition of angle bisector

Now one might argue that the "compass rule" is intended to refer to the collapsible compass, which the compass that appears in the problem is not. Well, if we insist on a collapsible compass, then we can just use Euclid's proof of his Proposition I.9.

Let the angle BAC be the given rectilinear angle.

It is required to bisect it.

I.3
I.Post.1, I.1

Take an arbitrary point D on AB. Cut off AE from AC equal to AD, and join DE. Construct the equilateral triangle DEF on DE, and join AF.

I say that the angle BAC is bisected by the straight line AF.

Since AD equals AE, and AF is common, therefore the two sides AD and AF equal the two sides EA and AF respectively.

I.Def.20, I.8

And the base DF equals the base EF, therefore the angle DAF equals the angle EAF.

Therefore the given rectilinear angle BAC is bisected by the straight line AF.

Q.E.F.

Now Euclid's proof clearly also uses seven steps -- perhaps only six, since "it is required to bisect it" isn't really a step at all.

We can shorten this proof even more if we just put the steps ZA = ZB and AC = BC right into the Given information, since it's given by the use of the compass that these two statements are true.

Given: ZA = ZB and AC = BC
Prove: Ray ZC, as constructed, bisects Angle AZB.

Proof:
Statements                                         Reasons
1. ZA = ZB and AC = BC                   1. Given
2. ZC = ZC                                         2. Reflexive Property of Equality
3. Triangles AZC, BZC congruent     3. SSS Congruence Theorem (steps 1, 1, 2)
4. Angle AZC = BZC                         4. CPCTC
5. Ray ZC bisects Angle AZB            5. Definition of angle bisector

By this point, the proof looks more like a classic Lesson 7-3 proof -- which we'd think is the type of proof of which Beals approves.

Of course, we know that Beals holds the Weeks & Adkins Geometry text in high regard. I can't help but wonder how Weeks & Adkins presents the angle bisector construction -- since unfortunately, Beals doesn't provide a Weeks & Adkins text today for her usual comparison. (I'm also curious about what sort of proof Joyce expects to see for the angle bisector construction.)

So instead, let's look at the Weeks & Adkins page that Beals posted last week, and find out how many steps it takes to solve that problem:

Given: P is a point on the side AB of Triangle ABC such that AP : AB = 1 : 3. Q is the point on BC such that CQ : CB = 1 : 3. AQ and CP intersect at X.
Prove: AX : AQ = 3 : 5.

Proof:
Statements                                         Reasons
1. AP : AB = 1 : 3, CQ : CB = 1 : 3   1. Given
2. BP : AB = 2 : 3, BQ : CB = 2 : 3    2. Subtraction Property of Proportions
3. Angle ABC = ABC                         3. Reflexive Property of Equality
4. Triangle PBQ ~ ABC                     4. SAS Similarity Theorem (steps 2, 3, 2)
5. PQ : AC = 2 : 3                              5. Corresponding sides of similar triangles are proportional.
6. Angle BPQ = BAC                         6. Corresponding angles of similar triangles are congruent.
7. Line PQ | | AC                                7. Corresponding Angles Test
8. Angle QPX = ACX, PQX = CAX   8. Alternate Interior Angles Consequence
9. Triangle QPX ~ ACX                     9. AA Similarity Theorem (steps 8, 8)
10. QX : AX = 2 : 3                            10. Corresponding sides of similar triangles are proportional.
11. AX : AQ = 3 : 5                            11. Addition Property of Proportions

It might be possible to shorten this proof by using the Side-Splitting Theorem (found in Lesson 12-10 in the U of Chicago text). In particular, the converse of that theorem would allow us to jump directly from Step 2 directly to Step 7, a savings of four steps. But the problem is that we need Step 5 in order to find ratios in Steps 10 and 11 -- 2 : 3 is the bridge between the given 1 : 3 and the desired 3 : 5. So I had to stick to this 11-step proof, with both sets of similar triangles.

Beals complains about the seven-step proof of the angle bisector construction, yet has no problem with this 11-step similarity proof. Therefore, I must disagree with Beals on this one.

Now Beals might have had a good point here. One difference between the derivation of SAS from transformations (which she opposes) and the derivation of the Isosceles Triangle Theorem from SAS (which she approves of) is that the latter proof will be much shorter than the former. Our proof of the latter will contain only five steps (much shorter than in many texts), but just look at our proofs of SAS, SSS, and ASA yesterday -- seven, eight, and ten steps respectively. (I do point out that I gave different levels in detail in each proof -- SSS ought to have the longest proof, not ASA.) So we can let students avoid those long proofs by making SAS, SSS, and ASA all postulates. But then she weakened her argument by endorsing the similarity question with its 11-step proof.

It's all about what will engage the students to learn. In another post, Beals writes about how it's students with disabilities who prove that the progressive "guide on the side" classroom is ineffective, and that students will thrive in a traditionalist "sage on the stage" environment. Well, this week, I see how it's the continuation school students who prove that the traditionalist classroom isn't always the best -- the students are the least likely to listen to a teacher just because he/she's a sage on a stage.

As I said about Levy yesterday, the trick is to hook the students with an interesting result. Then we can move on to the formalism of proving the basic theorems of Geometry.