Lesson 12-9: The AA and SAS Similarity Theorems (Day 129)

Today on her Mathematics Calendar 2019, Theoni Pappas writes:

The dimensions of this rectangular parallelepiped are 3 * 4 * 12. What is the length of AD?

[Here's the additional given info: AD is a diagonal of the box.]

Since we're looking for the diagonal of a box, we can use the Diagonal of a Box Formula that we learned recently in Lesson 11-6:

Diagonal of a Box Formula:
In a box with dimensions a, b, and c, the length of the diagonal is sqrt(a^2 + b^2 + c^2).

So let's just use that formula now:

d = sqrt(a^2 + b^2 + c^2)
d = sqrt(3^2 + 4^2 + 12^2)
d = sqrt(9 + 16 + 144)
d = sqrt(169)
d = 13

Therefore the desired diagonal is 13 units -- and of course, today's date is the thirteenth.

This is what I wrote last year about today's lesson:

Now let's get on with today's lesson. Today is supposed to be Lesson 12-9 of the U of Chicago text, on the AA and SAS Similarity Theorems, since today is Day 129. But there's a problem here.

Chapter 12 is the only chapter of the U of Chicago text with a full ten lessons. This causes a wrinkle in our digit-based pacing plan. Chapter 13, for example, has only eight lessons, so we can cover Lessons 13-1 to 13-8 on Days 131 to 138, then use Day 139 to review for the Chapter 13 Test to be given on Day 140. Then Lesson 14-1 can begin the next day, Day 141. For some shorter chapters, such as Chapter 11, this is even easier. The last lesson of the chapter is 11-6 on Day 116, then this leaves four days to review for and give the Chapter 11 Test.

But with ten lessons in Chapter 12, the pattern would have us cover Lesson 12-9 on Day 129, 12-10 on Day 130, and then 13-1 on Day 131 with no time for the Chapter 12 Test. The best thing to do seems to be to squeeze in the Chapter 12 Test on the same day as Lesson 12-10.

I'll preserve the rest of last year's post, which is a discussion of SAS Similarity -- including a system where SAS~ is assumed as a postulate instead of proved as a theorem:

I've referred to the American mathematician George David Birkhoff several times. Birkhoff was the mathematician who first came up with the Ruler and Protractor Postulates, which are often two of the first postulates to appear in a modern Geometry text. Actually, Birkhoff showed that only four postulates are required to derive all of Euclidean geometry:

• Through any two points, there is exactly one line.
• The Ruler Postulate
• The Protractor Postulate
• SAS~

This is astounding -- only four postulates are required? Even Euclid himself had five postulates, and now Birkhoff claimed that he can derive Euclid's geometry in only four? And as Birkhoff's first postulate is the same as Euclid's, the American's other three postulates should somehow be equivalent to the Greek sage's other four. But how can this be?

Let's think about what theorems can be proved from Birkhoff's postulates. One should immediately jump out at us -- from SAS Similarity, we should be able to prove SAS Congruence. In fact, the proof is almost trivial -- two figures are congruent iff they are similar with scale factor 1. Thus if we assume any similarity statement as a postulate, we can immediately prove the corresponding congruence statement.

We can also see how to derive some of Euclid's other postulates. That a line can be extended indefinitely goes back to the Ruler Postulate -- every point on a line corresponds to a real number, so just as the real numbers go on indefinitely, so do the points on a line. That we may draw a circle with any center and radius also comes from the Ruler Postulate, though this is less obvious. But think about it -- starting from the center, we can imagine placing a ruler in any direction. If the center is marked with the real number 0, we consider the point marked with the real number r, the radius. The locus of all points found in this manner is the desired circle. That all right angles are equal obviously comes from the Protractor Postulate, as all right angles measure 90 degrees.

This leaves us with just one of Euclid's postulates -- but it's the one that caused the most trouble for millennia, the Fifth Postulate. It's possible to derive the Fifth Postulate from Birkhoff's axioms, but this is very complicated. But as we try to work it out, we'll learn much about Birkoff's geometry.

But first, let's think about what we've proved so far. We have Euclid's first four postulates and we also have SAS Congruence. This means that we can prove Euclid's first 28 propositions -- the ones that don't require a Parallel Postulate.

In particular, we can prove his Proposition 26, which is ASA. Recall that Dr. Franklin Mason also reproduces Euclid's proof, where he uses SAS to prove ASA.

Now the we have ASA Congruence, we can derive another similarity theorem. We can do it the same way that it's done in many pre-Core Geometry texts -- one similarity result is assumed as a postulate, and we use that postulate and the corresponding triangle congruence statements to derive each similarity theorem. In such texts, AA~ is usually the postulate, and so these texts use SAS to prove SAS~ and SSS to prove SSS~. We are given two triangles satisfying, say, the SSS~ condition and we wish to prove them similar -- the trick is to come up with a third triangle that is both similar to one of the given triangle via the AA~ Postulate and congruent to the other via SSS. This proves that the two given triangles must also be similar.

Birkhoff can use the same trick, except SAS~ is the postulate this time. We can now use SAS~ and ASA to prove a similarity theorem -- but which one? It's the one that corresponds to ASA. A moment's reflection should convince you that the similarity corresponding to ASA is in fact AA~! It can't be ASA~ -- what does it mean for one pair of sides to be proportional? Today's Lesson 12-9 of the U of Chicago text points out that both ASA and AAS correspond to AA~. (Recall that the text proves all three of AA~, SAS~, SSS~ as theorems by using a dilation to produce similar triangles.)

So now we have proved AA~. The next result is often called the Third Angle Theorem (which Dr. M abbreviates as TAT) -- if two angles of one triangle are congruent to two angles of another triangle, then the third pair of angles is also congruent. This easily follows from AA~ as follows -- since the two triangles have two congruent pairs of angles, they are already similar by AA~. And if two polygons are similar, then all of the corresponding pairs are congruent.

Here's a note of warning -- notice that we have not yet proved a Triangle Sum Theorem. If two angles of a triangle have measure x and y, then all we know is that there exists some function f such that the third angle has measure f (x, y). We don't know that f (x, y) = a - x - y for some constant a, much less that a = 180 degrees.

So the Third Angle Theorem is a much weaker result than the full Triangle Sum. But we can use TAT to prove an interesting result:

-- Every Lambert quadrilateral is a rectangle.

And now you're thinking -- what the...? You thought that we were done with all this Lambert and Saccheri nonsense, and here I go talking about Lambert quadrilaterals again!

But notice that we've already proved the first four Euclid postulates thus far. In other words, so far we're in neutral geometry. And so I'm going to use terms that a neutral geometer would use, such as Lambert quadrilateral. And besides, a Lambert quadrilateral only means a quadrilateral with three right angles -- and of course, a rectangle is a quadrilateral with four right angles. We could have said:

-- If three angles of a quadrilateral are right angles, then so is the fourth.

Also, my claim that we can use TAT to prove this theorem is strange. TAT is all about the third angle of a triangle, and now we'll use it to prove something about the fourth angle of a quadrilateral?

Well, here's the proof. Let's call our Lambert quadrilateral ABCD, and declare that A, B, C are the right angles. So our goal is to prove that angle D is also a right angle.

Let's extend sides AB and CD a little. That is, we choose point E on ray AB beyond point B, and point F on ray CD beyond point D. Notice that rays AB and CD point in opposite directions. So when we connect E and F, we have that EF intersects both BC and AD, and we'll call the points of intersection G and H respectively.

Now we look at triangles AEH and BEG. In these triangles, angles A and EBG are congruent as they are both right angles (with EBG forming a linear pair with the given right angle B), and the triangles have angle E in common. Thus AEH and BEG are similar, and by TAT the third angles, AHE and BGE, must be congruent.

Notice we have vertical angles -- AHE and DHF are congruent, as are BGE and CGF. Since AHE and BGE are congruent, we conclude that CGF and DHF are congruent.

Now we consider triangles GFC and HDF. We just proved that CGF and DHF are congruent, and the triangles have angle F in common. Thus CGF and DHF are similar, and by TAT the third angles, C and FDH, must be congruent.

But we are given that C is a right angle. Thus FDH is also a right angle, and (with FDH forming a linear pair with CDH, the same as the given angle D) D is a right angle. Therefore the quadrilateral ABCD is a rectangle. QED

So we just proved that every Lambert quadrilateral is a rectangle. Now as it turns out, it's known in neutral geometry that the statement "every Lambert quadrilateral is a rectangle" is one of many statements equivalent to Euclid's Fifth Postulate. The proof is not simple -- here's a link to a neutral geometry course where this is proved:

http://personal.bgsu.edu/~warrenb/Courses/09STheorems.pdf

The two relevant proofs are listed as 9->6, which takes us from "every Lambert quadrilateral is a rectangle" to Triangle Sum, and then 6->2, which takes us from Triangle Sum to Playfair. This is sufficient, since the most commonly mentioned Parallel Postulate is Playfair. So this concludes how we get from Birkhoff's four axioms to a Parallel Postulate.

Both David Joyce and David Kung say that there should be as few postulates as possible, and you can count me as a third David who agrees. If the fewness of the postulates were all that mattered, Birkhoff's axiomatization would be the winner.

But that's not all that matters. Imagine a high school Geometry course trying to get from Birkhoff's axioms to Playfair. Some of the steps are reasonable for high school students -- SAS to ASA is not commonly done, but the indirect proof does appear on Dr. M's website. The phrase "Lambert quadrilateral" should never be spoken in a high school course, but if we write it as "if three angles of a quadrilateral are right angles, then so is the fourth," the proof isn't terrible.

Of course, the proof that takes us from the rectangle theorem to Triangle Sum is very inappropriate for a high school class. And besides -- this proof sequence is the opposite from what we normally want to do in high school Geometry. We want to use Playfair to prove Triangle Sum, not vice versa, and we want to use Triangle Sum in our proof about the angles of a rectangle, not vice versa.

But actually, this isn't even where the problems with Birkhoff's axioms begin. They actually start with the very first proof -- using SAS Similarity to prove SAS Congruence. Yet of all the proofs, this is by far the simplest, almost trivial: two triangles are congruent iff they are similar with scale factor 1.

An argument can be made that similarity is much more important than congruence. Think about it:

• What can we use to set up scale models and maps? (similarity, not congruence)
• What can we use to prove the Pythagorean Theorem? (similarity, not congruence)
• What can we use to derive the trig ratios? (similarity, not congruence)
• What can we use to derive the slope formula? (similarity, not congruence)

...and so on. Congruence matters very little outside of Geometry class, but similarity matters throughout subsequent high school class, college, and careers.

If we were to introduce an SAS~ Postulate, we instantly have a proof of SAS Congruence. Of course, we already proved SAS Congruence, but we can imagine a class where similarity is taught before congruence is. (I actually once saw a text that teaches similarity before congruence.) Then we can get to the Pythagorean Theorem and the slope formula, and we can still teach congruence as a special case of similarity with scale factor 1. And we can just throw in Playfair as an extra postulate, so we don't run into the same problems as Birkhoff's axioms did.

So what is the problem with this approach? Similarity is a more difficult concept than congruence -- and students don't fully understand similarity until they know what congruence is. This is why I tried so hard to avoid teaching similarity until the second semester, in hopes that students can get a good first semester grade without being confused by similarity. So minimizing the number of axioms is secondary to maximizing student understanding. And so in today's lesson, the first of the second semester, we introduce SAS~, from Lesson 12-9 of the U of Chicago text

There's one more thing I want to say about SAS~. Last week, I implied that I would have to assume SAS~ as a postulate first, then use SAS~ to prove the properties of dilations, and finally use dilations to define similar and prove the other similarity theorems.

But this is hardly logical. To see why, let's look at a statement of SAS~:

SAS Similarity Theorem:
If, in two triangles, the ratios of two pairs of corresponding sides are equal and the included angles are congruent, then the triangles are ______________.

That blank there is intentional, to remind you that we haven't defined "similar" yet! And it's circular to use SAS~ to prove that triangles are "similar," use it to prove the properties of dilations, and only then use dilations to define "similar."

All of this stems from that PARCC question that I mentioned last week. The Common Core Standards tell us that we should use dilations to define similarity:

CCSS.MATH.CONTENT.HSG.SRT.A.2
Given two figures, use the definition of similarity in terms of similarity transformations to decide if they are similar; explain using similarity transformations the meaning of similarity for triangles as the equality of all corresponding pairs of angles and the proportionality of all corresponding pairs of sides.

yet that PARCC question does the opposite -- we have to use SAS~ (and therefore know what "similar" means) in order to prove the properties of dilations.

If all I had was the Common Core standard above, I would have introduced a Dilation Postulate that assumes the properties of dilations, then used dilations to derive SAS~ and the other postulates. If all I had was the PARCC question, I would have followed the classical pre-Core definition of similarity, proved SAS~, and then used it to prove the properties of dilations at the very end of the unit.

I decided to go back to Dr. M's website to see how he teaches similarity, and I see that his method is a compromise of the Common Core method and the PARCC method. He teaches similarity in his Chapter 7, and in his Lesson 7.3 he introduces a postulate:

The Polygon Similarity Postulate:
Given a polygon P and a positive quantity k, we may construct a second polygon Q such that P ~ Q, with scale factor k.

In this case, similarity has already been defined classically, while dilation isn't defined yet. Yet this postulate, while it doesn't say specifically that a dilation (with scale factor k, of course) maps P to Q but only that such a Q exists, serves the same purpose as my Dilation Postulate.

We can then prove all three similarity statements -- SAS~, SSS~, and AA~ -- as theorems by using the corresponding congruence theorem plus the Polygon Similarity Postulate. In particular, we use the postulate to produce a triangle similar to the first given triangle, with the correct scale factor that makes it congruent to the second triangle via the corresponding theorem.

2019 Update: It's come to my attention that last year's worksheets are based on this above method of teaching similarity, rather than the U of Chicago method. Despite my describing the old method here on the blog, I have replaced the instruction worksheet for both yesterday's and today's lessons.