First of all, I showed my student the two triangles that I mentioned in the previous post, and asked him to locate the circumcenter, orthocenter, and centroid of each. This time, he had a much easier time calculating these three centers. I'm actually surprised that our simple formula for the centroid given the coordinates does not appear in Glencoe. Instead, the text only states that the centroid is 2/3 of the way along each median, from the vertex to the opposing side. This makes sense in synthetic geometry, but not in coordinate geometry -- unless one is prepared to use the Distance Formula to figure out what 2/3 of each distance is.

By the way, you may be wondering what happened to the fourth center, the

*incenter*. The reason we omit incenter is that the equations of angle bisectors are difficult to calculate -- in general, we'd need trigonometry to calculate them. So we stick to circumcenter, orthocenter, and centroid.

I will include the lesson that I gave my student -- the one that I didn't have time to write up and post here in advance -- as part of tomorrow's planned Activity Day.

Afterwards we discussed an exercise in Glencoe where the students are to prove that if two triangles are congruent, then so are their medians. Because of Glencoe's high level of formalism, the proof given in the back of the text contains 13 steps, including three "Definition of Congruence" steps, as well as a substitution/simplification step mislabeled as "Addition Property."

Based on how the Glencoe text presents its Chapter 5, I will make a few more changes for next year to the way that I will present my material. In particular, a full proof of the SAS Inequality of Section 5.5 (U of Chicago 7-8) requires the Triangle Inequality, which I've delayed until now. I kept juggling Section 7-8 around back when we did Chapter 7 (as well as make a few other references to the Triangle Inequality when discussing SSS Congruence). Indeed, Section 7-8 will no longer be part of the first semester. My first semester will end with Section 7-7 instead. I haven't decided whether I will keep the Concurrency Theorems squeezed in after Section 7-7, or instead delay them until now, just as Glencoe includes both concurrency (circumcenter, orthocenter, etc.) and inequalities for triangles in the same Chapter 5.

And that takes us to the topic of today's worksheet -- the Triangle Inequality. In the U of Chicago, the Triangle Inequality was given as a postulate, yet it can be proved as a theorem. Many texts, including the Glencoe text, do prove the Triangle Inequality as a theorem, and this is what we will do.

The proof of the Triangle Inequality begins in Glencoe's Section 5-2, where we must prove two theorems, which the U of Chicago calls the Unequal Sides and Unequal Angles Theorems. My student told me that he had no problem understanding these two theorems -- he wanted just a quick review of Indirect Proof in Section 5-3 before moving on to the Triangle Inequality in 5-4. (This is why I'm squeezing in the Triangle Inequality now, rather than prove only the Unequal Sides and Unequal Angles Theorems today and save the Triangle Inequality for next week.)

Dr. Franklin Mason also proves these theorems. In many ways, Dr. M's Chapter 5 is similar to the same numbered chapter in Glencoe, except that Dr. M saves the concurrency results for a separate chapter, Chapter 10. Both Dr. M and Glencoe follow the same sequence of theorems, in which each theorem is built from the previous theorem in the list:

Exterior Angle Theorem (abbreviated TEAE in Dr. M)

Exterior Angle Inequality (TEAI)

Unequal Sides Theorem (TSAI)

Unequal Angles Theorem (TASI)

Triangle Inequality (too important to be abbreviated!)

SAS Inequality (Hinge)

The U of Chicago follows the same pattern, except that the Unequal Angles Theorem is

*not*used to prove the Triangle Inequality. Instead, the Triangle Inequality is merely a postulate. And since Unequal Angles isn't used to prove the Triangle Inequality, the U of Chicago didn't have to wait until

Chapter 13 to state the Triangle Inequality. Instead, the Triangle Inequality Postulate is given in Chapter 1, and the SAS Inequality, which depends on that postulate in its proof, is given in Chapter 7, still well before Chapter 13.

My blog attempted to restore the Dr. M-Glencoe order by delaying the Triangle Inequality. But I screwed up by not delaying the SAS Inequality as well. This is why next year, I plan on delaying SAS Inequality, so that the full logical sequence is given.

But the first four theorems in the list are proved in U of Chicago's Section 13-7. Since I briefly mentioned the Exterior Angle Theorem (TEAE) at the end of the first semester, and the TEAI follows almost trivially from TEAE, my worksheet skips directly to the Unequal Sides Theorem. Its proof is given in the two-column format. Here I reproduce that proof, starting with a Given step:

Unequal Sides Theorem (Triangle Side-Angle Inequality, TSAI):

If two sides of a triangle are not congruent, then the angles opposite them are not congruent, and the larger angle is opposite the longer side.

Given: Triangle

*ABC*with

*BA*>

*BC*

Prove: angle

*C*> angle

*A*

Proof:

Statements Reasons

1. Triangle

*ABC*with

*BA*>

*BC*1. Given

2. Identify point

*C'*on ray

*BA*2. On a ray, there is exactly one point at a given distance from

with

*BC'*=

*BC*an endpoint.

3. angle 1 = angle 2 3. Isosceles Triangle Theorem

4. angle 2 > angle

*A*4. Exterior Angle Inequality (with triangle

*CC'A*)

5. angle 1 > angle

*A*5. Substitution (step 3 into step 4)

6. angle 1 + angle 3 = angle

*BCA*6. Angle Addition Postulate

7. angle

*BCA*> angle 1 7. Equation to Inequality Property

8. angle

*BCA*> angle

*A*8. Transitive Property of Inequality (steps 5 and 7)

The next theorem is proved only informally in the U of Chicago. The informal discussion leads to an indirect proof.

Unequal Angles Theorem (Triangle Angle-Side Inequality, TASI):

If two angles of a triangle are not congruent, then the sides opposite them are not congruent, and the longer side is opposite the larger angle.

Indirect Proof:

The contrapositive of the Isosceles Triangle Theorem is: If two angles in a triangle are not congruent, then sides opposite them are not congruent. But which side is opposite the larger angle? Because of the Unequal Sides Theorem, the larger side cannot be opposite the smaller angle. All possibilities but one have been ruled out. The larger side must be opposite the larger angle. QED

My student told me that he wanted to see one more indirect proof before showing him the Triangle Inequality, so why not show him this one? The initial assumption is, assume that the longer side is

*not*opposite the larger angle. Since the angle opposite the longer side is not

*greater*than the angle opposite the shorter side, the former must be

*less than or equal to*the latter. And these are the two cases that lead to contradictions of Isosceles Triangle Contrapositive and Unequal Sides as listed in the above paragraph proof.

Now finally we can prove the big one, the Triangle Inequality. This proof comes from Dr. M -- but Dr. M writes that his proof goes all the way back to Euclid. Here is the proof from Euclid, where he gives it as his Proposition I.20:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI20.html

Here is the two-column proof as given by Dr. M. His proof has eight steps, but I decided to add two more steps near the beginning. Step 1 is the Given, and Step 2 involves extending a line segment, so that it's similar to Step 2 of the Unequal Sides proof. Indeed, the proofs of Unequal Sides and the Triangle Inequality are similar in several aspects:

Triangle Inequality Theorem:

The sum of the lengths of two sides of any triangle is greater than the length of the third side.

Given: Triangle

*ABC*

Prove:

*AC*+

*BC*>

*AB*

*Proof:*

Statements Reasons

1. Triangle

*ABC*1. Given

2. Identify point

*D*on ray

*BC*2. On a ray, there is exactly one point at a given distance from

with

*CD*= A

*C*an endpoint.

3. angle

*CAD*= angle

*CDA*3. Isosceles Triangle Theorem

4. angle

*BAD*=

*BAC*+

*CAD*4. Angle Addition Postulate

5. angle

*BAD*> angle

*CAD*5. Equation to Inequality Property

6. angle

*BAD*> angle

*CDA*6. Substitution (step 3 into step 5)

7.

*BD*>

*AB*7. Unequal Angles Theorem

8.

*BD*=

*BC*+

*CD*8. Betweenness Theorem (Segment Addition)

9.

*BD*=

*BC*+

*AC*9. Substitution (step 2 into step 8)

10.

*BC*+

*AC*>

*AB*10. Substitution (step 9 into step 7)

To help my student out, I also included another indirect proof in the exercises. We are given a triangle with two sides of lengths 9 cm and 20 cm, and we are asked whether the 9 cm side must be the shortest side. So we assume that it isn't the shortest side -- that is, that the third side must be even shorter than 9 cm. This would mean that the sum of the two shortest sides must be less than 9 + 9, or 18 cm, and so by the Triangle Inequality, the longest side must be shorter than 18 cm. But this contradicts the fact that it is 20 cm longer. Therefore the shortest side must be the 9 cm side. QED

Notice that the U of Chicago text probably expects an informal reason from the students. A full indirect proof can't be given because this question comes from Section 1-9, while indirect proofs aren't given until Chapter 13.

My plan is to give the activity tomorrow, and yet another lesson to help my student out on Tuesday, and then finally finish Chapter 13 with Section 13-8 on Wednesday, since this has nothing to do with Glencoe's Chapter 5. I hope all these worksheets will actually help my student!

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