Thursday, January 22, 2015

Section 11-4: The Midpoint Formula (Day 97)

Last night I tutored my geometry student. Section 4-5 of the Glencoe text is on ASA and AAS Congruence, so this clearly corresponds to Chapter 7 of the U of Chicago text.

I wanted to show him my worksheet from this blog -- the one where I proved the ASA Congruence Theorem using Common Core principles. At first he only took one glance at my proof outline and was intimidated when he saw that it was a ten-step proof. But then he had to finish his homework assignment, which required what turned out to be a nine-step proof.

Notice that the congruence chapters in the two texts are organized differently. In the U of Chicago, SSS, SAS, ASA, and AAS are all theorems that need to be proved. And so Section 7-2 contains these proofs, so that Section 7-3 has students prove the higher-level theorems that use these medium-level Congruence Theorems, and then Section 7-4 contains more complex proofs, the ones with overlapping triangles.

But Glencoe, like most pre-Common Core texts, gives SSS, SAS, and ASA as postulates. Because of this, the lessons go straight into having the students make higher-level proofs. And some of these proofs had overlapping triangles. So unlike the U of Chicago, with separate sections for the simpler and harder proofs, Sections 4-4 (SSS, SAS) and 4-5 (ASA, AAS) contained both simpler and harder proofs.

We look at Section 7-4 of the U of Chicago, and see Example 1 of a proof with overlapping triangles. Here, B lies on AD and C lies on AE.

Given: AC = ABAD = AE
Prove: Angle D = E

Statements                    Reasons
1. AC = AB, AD = AE  1. Given
2. Angle A = A              2. Reflexive Property of Congruence
3. ABE, ACD cong. tri.  3. SAS Congruence (steps 1, 2, 1)
4. Angle D = E              4. CPCTC

Now Glencoe contains a similar problem, except that the given is different. F is where BE and CD meet.

Given: AB, AC cong. seg., BDF, CDE cong. tri.
Prove: D, E cong. angles

Statements                      Reasons
1. BDF, CDE cong. tri.   1. Given
2. BD cong. CE              2. CPCTC
3. AB cong. AC              3. Given
4. BD = CE, AB = AC    4. Segment Congruence Theorem
5. AB + BD = AC + CE  5. Addition Property of Equality
6. AD = AE                    6. Betweenness Theorem
7. Angle A cong. A          7. Reflexive Property of Congruence
8. ABE, ACD cong. tri.    8. SAS Congruence (steps 3, 7, 6)
9. Angle D cong. E          9. CPCTC

The reason that this proof is so long is that in many texts, including Glencoe, the statements "segment AB is congruent to AC" and "length AB is equal to AC," while equivalent, are not identical. In the U of Chicago, the Segment Congruence Theorem of Chapter 6 tells us that segments of equal lengths are congruent -- that is, that there is an isometry mapping one segment to the other. But after introducing the theorem, the book freely interchanges the congruence of the segments and the equality of the lengths, without mentioning it as a separate step in the proof.

Here the problem is that we need the Betweenness Theorem -- what Glencoe calls the Segment Addition Postulate -- and that only numbers, not figures, can be added. A length is a number, but a segment is a figure -- so we must convert the given segment congruence statement into an equal length statement before we can add anything, and then convert it back to segment congruence before we can invoke SAS. Most geometry teachers just handwave over the distinction between segment congruence and length equality -- as does the U of Chicago text, as do I on this blog, since the congruence symbol is hard to type in ASCII.

But the Glencoe text painfully includes every distinction between segment congruence and length equality -- as became apparent when I looked at one of the answers (a flow proof) in the back of the text. Moreover, my student's math teacher has been known to deduct half-points for leaving out such steps!

Even in the proof I posted above, notice that Step 6 is AD = AE, but I failed to convert this back to the congruence of segments AD and AE. Not only that, but technically speaking, the Segment Addition Postulate tells us that AB + BD = AD and AC + CE = AE. This should be the step that immediately follows step 5, and then the next step should derive AD = AE by substituting the above two equations into step 5.

So a fully complete proof would actually have eleven steps, not nine. The student and I came to the agreement that these nine steps should be sufficient for the teacher not to deduct any points -- but we won't know until today.

To me, having to make the painful distinction between segment congruence and length equality, as well as adding the substitution step, changes this problem to an exercise in formality, rather than help the student to see how Segment Addition can be used in an overlapping triangles proof. Even Dr. Wu admits that he skips steps in proofs when they don't enhance student understanding -- which is the primary goal of teaching proofs.

After this proof, I hardly had time to show my student the full proof of the ASA Congruence Theorem, as planned. What I decided to do was cut out two triangles satisfying the ASA condition, and then showed how I could slide, flip, and turn one triangle so that it coincided with the other. In particular, I showed him exactly how the Side-Switching Theorem was used to show that the reflection image of C' is F.

Afterward, my student noticed another proof in the Glencoe text and wondered whether he could use reflections, as I had to in the proof of the ASA Congruence Theorem, to prove the triangles in that problem congruent. Actually, he had a point there. Notice that in general, in a congruence proof, the first given triangle isn't necessarily a mere reflection image of the second. But in overlapping triangles, the triangles actually are mirror images of each other. Indeed, we look at the diagrams in U of Chicago Lesson 7-4, and notice that every single diagram in the lesson has reflectional symmetry!

In the diagram for this problem, point E lies on AB.

Given: Angles BCE, DCE congruent, Angles BEC, DEC congruent
Prove: AB, AD congruent

Let's compare the traditional Glencoe proof with a Common Core/U of Chicago proof.

Glencoe Proof:
Statements                        Reasons
1. Angles BCE, bla, bla     1. Given
2. CE cong. CE                 2. Reflexive Property of Congruence
3. BCE, DCE cong. tri.     3. ASA Congruence (steps 1, 2, 1)
4. BE cong. DE                 4. CPCTC
5. AEB, CEB linear pair,   5. Definition of linear pair
    AED, CED linear pair
6. Angle AEB cong. AED  6. Congruent Supplements Theorem
7. AE cong. AE                  7. Reflexive Property of Congruence
8. AEB, AED cong. tri.      8. SAS Congruence (steps 4, 5, 6)
9. AB cong. AD                 9. CPCTC

For the Common Core/U of Chicago proof, we'll start out just like the ASA Congruence Theorem proof and proceed from there:

Common Core/U of Chicago Proof:
Statements                                       Reasons
1. Angles BCE, bla, bla                    1. Given
2. Ray EB refl. over line AC is ED  2. Side-Switching Theorem
3. Ray CB refl. over line AC is CD  3. Side-Switching Theorem
4. The image of B lies on ED, CD   4. Figure Reflection Theorem
5. The image of B is D                     5. Line Intersection Theorem
6. The image of A is A                      6. Definition of reflection (point on mirror refl. to itself)
7. AB refl. over AC is AD                 7. Figure Reflection Theorem
8. AB cong. AD                                 8. Definition of congruence (since the refl. is the isometry)

The Common Core proof is slightly shorter -- mainly because of the wasted step in the Glencoe about showing that the angles form a linear pair. And I forget whether Glencoe would make the student add yet another step, that the angles in a linear pair are supplementary! It's because of these extra formalistic steps that make the reflectional proof shorter.

Also, this week I've been subbing at the same continuation school for the second week in a row. One student is preparing to take the national test known as the Armed Services Vocational National Battery -- or ASVAB, for short. He is having trouble passing the ASVAB because it covers much material from the second semester of Algebra I, whereas he has just barely finished the first semester of algebra.

Notice that the ASVAB has two test formats -- computerized and written -- and it is given by the federal government -- specifically the military.

Section 11-4 of the U of Chicago text covers the other important formula of coordinate geometry -- the Midpoint Formula. As the text states, this is one of the more difficult theorems to prove.

In fact, the way we prove the Midpoint Formula is to use the Distance Formula to prove that, if M is the proposed midpoint of PQ, then both PM and MQ are equal to half of PQ. The rest of the proof is just messy algebra to find the three distances. The U of Chicago proof uses slope to prove that M actually lies on PQ. Since we don't cover slope until next week, instead I just use the Distance Formula again, to show that PM + MQ = PQ, so that M is between P and Q. The algebraic manipulation here is one that's not usually used -- notice that instead of taking out the four in the square root of 4x^2 to get 2x (as is done in the last exercise, the review question), but instead we take the 2 backwards inside the radical to get 4, and then distribute that 4 so that it cancels the 2 squared in the denominator.

I don't have nearly as much to say about the Midpoint Formula as the Pythagorean Theorem and its corollary, the Distance Formula. To me, it's a shame that I had to bury the Pythagorean Theorem in the middle of this Coordinate Geometry unit. The main theorem named for a mathematician really deserves its own lesson, but due to time constraints I had to combine it with the Distance Formula the way I just did it in yesterday's lesson.

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