*The Fractal Geometry of Nature*is "Of Men and Ideas," and consists of Chapters 40 through 42. In this section, Mandelbrot gives a brief biographical sketch of many of the mathematicians who contributed to the development of fractals -- though he admits that most mathematicians don't have a bio as interesting as that of Evariste Galois -- whose bio I've discussed earlier on the blog.

Of all the mathematicians whose bio appears in Chapter 40, Paul Levy is the most important to Mandelbrot himself. He writes, "Paul Levy acknowledged no pupil, but came closest to be my [Mandelbrot's] mentor." According to Mandelbrot, Levy once said, "In order to interest children in geometry, one should proceed as quickly as possible to theorems they are not tempted to consider evident," rather than slog through the basics definitions and postulates. Otherwise, they will lose interest and wonder what's the point of learning Geometry. In some ways, this is how Michael Serra's

*Discovering Geometry*is organized -- by interest rather than by provability. Levy became a professor at the French Ecole Polytechnique -- the famous school of mathematics that Galois once dreamt of attending, but never did.

In Chapter 41, Mandelbrot discusses the history of dimension, going all the way back to Euclid. And then in the final chapter of the book, he tells how he first came up with the idea of fractals. As the title of the book implies, he invented fractals because he wanted to describe the shapes of nature.

And so we finally reach Lesson 7-2 of the U of Chicago text, the Triangle Congruence Theorems. I will be able to demonstrate how SSS, SAS, and ASA follow from the definition of congruence in terms of isometries.

Let's start with ASA, since as I said earlier this week, we can use the same proof of ASA directly out of the U of Chicago text. Here is the proof as given in the text:

ASA Congruence Theorem:

If, in two triangles, two angles and the included side of one are congruent to two angles and the included side of the other, then the triangles are congruent.

Proof:

Given

*A*=

*FDE*, and Angle

*B*=

*FED*. Consider the image Triangle

*A'B'C'*of Triangle

*ABC*under an isometry mapping

*A'B'C'*and

*DEF*form a figure with two pairs of congruent angles.

Think of reflecting Triangle

*A'B'C'*over line

*DE*. Applying the Side-Switching Theorem to Angle

*C'DF*, the image of Ray

*A'C'*is Ray

*DF*. Applying the Side-Switching Theorem to Angle

*C'EF*, the image of Ray

*B'C'*is Ray

*EF*. This forces the image of

*C'*to be on both Ray

*DF*and Ray

*EF*, and so the image of

*C'*is

*F*. Therefore the image of Triangle

*A'B'C'*is Triangle

*DEF*.

So if originally two angles and the included side are congruent (

*A*=

*D*, Angle

*B*=

*E*) then Triangle

*ABC*can be mapped onto Triangle

*DEF*by an isometry. (First map

*ABC*over the line

*DE*.) Thus, by the definition of congruence, Triangle

*ABC*is congruent to Triangle

*DEF*. QED

Now as we said earlier, the U of Chicago text uses the Isosceles Triangle Theorem to prove SAS, when we instead want to use SAS to prove the Isosceles Triangle Theorem. But as it turns out, we can write a proof of SAS that's not much different from the ASA proof given above. In this proof, we will discuss more in detail how we perform the first isometry -- the one that maps Triangle

*ABC*to the position

*A'B'C'*, from which we can perform the final reflection.

SAS Congruence Theorem:

If, in two triangles, two sides and the included angle of one are congruent to two sides and the included angle of the other, then the triangles are congruent.

Proof:

Given

*A*=

*FDE*. Our first isometry will be to map

*A*onto

*D*. Now in many Common Core texts, we begin with the translation mapping

*A*to

*D*(and arguably, even Euclid himself, in the proof by "superposition" of his Proposition I.4, began by translating

*A*to

*D*). Our problem is that translations aren't neutral, yet we want our proof to be valid in neutral geometry.

So instead, we simply reflect

*A*onto

*D*-- that is, the mirror is the perpendicular bisector of

*A'*is

*D*, while the images

*B'*and

*C'*can be anywhere on the plane.

Next, we must map the whole segment

*onto our destination,*~~A'B'~~

*A'*is already at

*D*, we can perform (just as Euclid did) a rotation centered at

*D*. How many degrees should this rotation be? The answer is that it's exactly the measure of Angle

*B'DE*. Then this rotation maps Ray

*DB'*to Ray

*DE*, and therefore segment

*DE*that's the correct distance from

*D*, and it has the name

*E*). At this point, after the reflection and rotation, the image

*A"*is still

*D*and the image

*B"*is now

*E*, so all we have to do is figure out where

*C"*is.

Now it could be the case that

*C"*is already

*F*-- in which case, we'd already be done. (Euclid erroneously made the assumption that

*C"*is always

*F*.) This is why the U of Chicago text always draws the case where

*C"*is not

*F*, in hopes that one final reflection, over line

*DE*, will map

*C"*to

*F*.

So far, this part of the proof isn't particular to SAS. All of the congruence theorems will begin with this same isometry -- reflect

*A*to

*D*, then rotate

*DB'*to

*DE*. (By the way, it's possible to dispense with the rotation and use only reflections in the proof. Instead of rotating, use the angle bisector of

*B'DE*as the mirror. Then by the Side-Switching Theorem, Ray

*DB'*maps to

*DE*.)

Notice that I have written

*C"*double-primed. This is because

*C*mapped to

*C'*under the first reflection and then

*C'*maps to

*C"*under the rotation (or second reflection). We would have to write a third prime for the final reflection, to show that

*C'"*is

*F*. (The U of Chicago text shows only a single isometry mapping

*ABC*under the isometry mapping

*ABC*, without any prime symbols. I'm hoping that this will actually be less confusing to the students -- calling the image

*ABC*drives home that the fact that all six parts of both triangles are already known to be congruent, and that it's only the congruence of the parts of

*ABC*and

*DEF*that remains to be proved.

Now here's the part of the proof that's particular to SAS. Instead of using isosceles triangles as in the U of Chicago proof, we notice that just as in the ASA proof, since Angles

*CAB*(which has been moved to

*CDE*) and

*FDE*are given to be congruent, the Side-Switching Theorem once again tells us that the reflection over line

*DE*maps Ray

*DC*to Ray

*DF*, and thus segment

*DC*that's the correct distance from

*D*, and it has the name

*F*), so the image of

*C*is

*F*. Therefore the image of Triangle

*ABC*is Triangle

*DEF*.

So if originally two sides and the included angle are congruent (

*A*=

*D*) then Triangle

*ABC*can be mapped onto Triangle

*DEF*by an isometry. (First map

*ABC*over the line

*DE*.) Thus, by the definition of congruence, Triangle

*ABC*is congruent to Triangle

*DEF*. QED

When presenting this proof in class, we can start with this SAS proof so that students can see how to perform the opening reflection and rotation. Then when we get to ASA and the other proofs, we can just say "there exists an isometry" mapping

*ABC*to the reflection of

*DEF*, so that the proof only needs to discuss the final reflection.

Now let's go for SSS. I mentioned earlier that the proof in the U of Chicago text creates a kite -- and the properties of kites ultimately go back to isosceles triangles. I wrote that we can avoid the Isosceles Triangle Theorem by using the Converse to the Perpendicular Bisector Theorem instead. Here is the new proof:

SSS Congruence Theorem:

If, in two triangles, three sides of one are congruent to three sides of the other, then the triangles are congruent.

Proof:

Given

*A*is mapped to

*D*, and

*B*is mapped to

*E*, and we are now ready to reflect

*C*over line

*DE*.

We are given that

*) =*~~DC~~

*D*is equidistant from

*C*and

*F*. Therefore by Converse Perpendicular Bisector,

*D*lies on the perpendicular bisector of

~~B~~

*) =*~~EC~~

~~E~~

*E*is equidistant from

*C*and

*F*. Therefore by Converse Perpendicular Bisector,

*E*lies on the perpendicular bisector of

*D*and

*E*. Since two points determine a line, this tells us that the perpendicular bisector of

*is exactly line*~~CF~~

*DE*-- and that's convenient, since

*DE*is exactly the mirror over which we wish to reflect!

So line

*DE*is the perpendicular bisector of

*. Therefore, by the definition of reflection (meaning),*~~CF~~

*C*reflected over line

*DE*must be

*F*-- which is exactly what we want to prove. QED

But Lesson 7-2 contains another congruence theorem -- AAS. The U of Chicago, like most texts, prove AAS using ASA plus the Triangle Sum Theorem. But as we already know, the Triangle Sum Theorem isn't neutral -- even though AAS is in fact valid in neutral geometry. Dr. David Joyce says the following about AAS:

There are only two theorems in this very important chapter. There's a trivial proof of AAS (by now the internal angle sum of a triangle has been demonstrated).

The way that Dr. Joyce writes this -- "a trivial proof of AAS" -- implies that there can be another, non-trivial proof of AAS, one that is valid in neutral geometry.

I like the following proof of AAS:

AAS Congruence Theorem:

If, in two triangles, two angles and a non-included side of one are congruent respectively to two angles and the corresponding non-included side of the other, then the triangles are congruent.

Proof:

Given

*A*=

*FDE*, Angle

*C*=

*DFE*. We begin, as in the other proofs, by performing the isometry that maps

*DE*as a mirror, Ray

*DC*is mapped to Ray

*DF*. So we know that the image of

*C*is somewhere on Ray

*DF*, but we don't know yet that

*C'*is exactly

*F*.

We know that Angle

*ACB*(same as

*DCE*) is mapped to Angle

*DC'E*, and as reflections preserve angle measure, these angles are congruent. And we are given that Angles

*ACB*and

*DFE*are congruent, so this tells us that

*DC'E*and

*DFE*are congruent. But this isn't sufficient to identify the images of any rays, since we don't know whether the vertices of the angles,

*C'*and

*F*, are the same point yet.

So let's try an indirect proof -- assume that

*C'*and

*F*are

*not*the same point. This may be tough to visualize, so try drawing a picture. We already know that

*C'*lies on Ray

*DF*, so we can draw

*C'*to be any point on Ray

*DF*other than

*F*. It doesn't matter whether

*C'*is between

*D*and

*F*or on the opposite side of

*F*from

*D*-- both will lead to the same contradiction.

After we label the two angles known to be congruent,

*DC'E*and

*DFE*, we notice something about the diagram we've drawn. We see that lines

*C'E*and

*FE*are in fact two lines cut by the transversal

*DF*, and the two angles

*DC'E*and

*DFE*turn out to be corresponding angles that are congruent. Thus, by the Corresponding Angles Test, lines

*C'E*and

*FE*are parallel! (Recall that the Parallel

*Tests*are neutral, unlike the Parallel

*Consequences*.) And so we have two parallel lines that intersect at

*E*, a blatant contradiction. So the assumption that

*C'*is not

*F*must be false, and so

*C'*is exactly

*F*. QED

Like previous indirect proofs involving parallel lines, this can be converted into a direct proof if we use the U of Chicago definition of parallel. Then

*C'E*and

*FE*are parallel lines with

*E*in common, so they are identical line -- that is,

*C'*lies on

*FE*. Then just as in the ASA proof,

*C'*lies on both

*DF*and

*FE*, so

*C'*is exactly

*F*.

I like this proof as it has the same flavor as the SAS, ASA, and SSS proofs. The main problem with it is that, even though the Parallel Tests are neutral, they won't appear on the blog until November, so students won't be able to complete the proof as of yet.

It makes sense that the proof of AAS would require a Parallel Test. We know that AAS is valid in neutral -- that is, Euclidean or hyperbolic -- geometry. But it's not valid in spherical geometry. Recall the counterexample to AAS that I gave over the summer, where we have two triangles, each of which has one vertex at the North Pole and a second vertex where the Equator and Prime Meridian meet. But the third vertices, while both are on the Equator, have different longitudes. Then both triangles satisfy AAS (two right angles and side equal to a quadrant), yet they aren't congruent (indeed, the angles at the Pole equal the longitudes of their respective third vertices).

Since there are no parallel lines in spherical geometry, the appearance of parallel lines in our proof blocks AAS from being valid in that geometry. But since AAS is valid in both Euclidean and hyperbolic geometry, the proof should depend on the neutral Parallel Tests -- and it does.

We notice that our remaining congruence theorem, HL, doesn't have the same problem as AAS. The proof of HL doesn't appear until Lesson 7-5, where the U of Chicago derives it from AAS. We know that HL is valid in neutral, but not spherical, geometry -- indeed, our same counterexample to AAS serves as a counterexample to HL (it has two right angles and hence two hypotenuses). So it makes sense that AAS would be used in a proof of HL, as the use of AAS prevents the proof from being valid in spherical geometry.

Now I have an alternate proof of HL that avoids AAS and uses only theorems that have been proved previously on the blog so far. As it turns out, it's not AAS that blocks this proof in spherical geometry, but the Uniqueness of Perpendiculars Theorem (as there exist infinitely many lines perpendicular to the Equator passing through the North Pole).

HL Congruence Theorem:

If, in two right triangles, the hypotenuse and a leg of one are congruent to the hypotenuse and a leg of the other, then the two triangles are congruent.

Proof:

Given

*A*and

*D*are right angles. We begin, as in the other proofs, by performing the isometry that maps

*E*is equidistant from

*C*and

*F*, so that

*E*lies on the perpendicular bisector of

Now since

*CAB*(same as

*CDE*) and

*FDE*are right angles, line

*DE*is perpendicular to

*DE*is the only line through

*E*perpendicular to

*DE*must be that perpendicular bisector that we were discussing earlier. So just as in the proof of SSS, we have by the definition of reflection (meaning),

*C'*is exactly

*F*. QED

(At this point, one might be wondering, if the use of Uniqueness of Perpendiculars invalidates HL in spherical geometry, why isn't the proof of SSS also invalid in spherical geometry, as we used the fact that there is a unique line through

*D*and

*E*in the proof? Well, if there were infinitely many lines through

*D*and

*E*, then

*D*and

*E*would be antipodal points. Then as every line that passes through

*D*also passes through

*E*, the points

*D*,

*E*, and

*F*would be

*collinear*, and so it wouldn't even make sense to refer to "triangle"

*DEF*anymore. It would be just like trying to apply SSS to triangle

*DEF*in Euclidean geometry when

*D*,

*E*, and

*F*are collinear. SSS only applies to

*triangles*.)

So we have a proof of HL that I can post, but I don't have a proof of AAS that can be posted yet, since our AAS proof requires the Corresponding Angles Test that I've yet to cover. I've seen other proofs of AAS before, but many of them use the Triangle Exterior Angle Inequality -- yes, that TEAI that Dr. Franklin Mason formerly used to make some of his proofs neutral. Like Dr. M, I don't really want to use TEAI any more to neutralize proofs either.

I was wondering whether there's a proof of AAS that avoids TEAI (or the Corresponding Angles Test, which can also be proved as a result of TEAI). We know that the proof can't be too easy, or otherwise we could accidentally prove AAS in spherical geometry. As the U of Chicago uses AAS to prove HL and we've already been reversing many of the proofs in the text, I'm wondering whether we might possibly use HL to prove AAS (drawing in altitudes in order to generate right triangles) -- but I wasn't able to find such a proof.

Dr. Randall Holmes, a math professor at Boise State, also tried to find a proof of AAS that avoids the TEAI, but he could not find such a proof. Two years ago, he wrote:

http://math.boisestate.edu/~holmes/math311/M311S13announcements.html

**"AAS proof note:**I'm convinced that there is no way to prove AAS without using the exterior angle theorem, which makes it less attractive as a test proof (because of the need for cases – but see that I actually handle the cases quite compactly below). By the way, the ASA proof does not need cases, because the application of the Angle Construction Postulate in it does not depend on the position of the new point in the same way the application of the Exterior Angle theorem in the AAS proof does."

We can easily why we need two cases in our proof of AAS using TEAI. Recall that in our above proof of AAS, the image

*C'*could either be on the segment

*F*from

*D*. Well in the former case, Angle

*DC'E*is exterior to Triangle

*C'FE*, so by TEAI, Angle

*DC'E*>

*DFE*. And in the latter case, Angle

*DFE*is exterior to that same triangle, so by TEAI, Angle

*DFE*>

*DC'E*. In either case, we have a contradiction since angles

*DC'E*and

*DFE*are known to be congruent.

Now Dr. Holmes isn't merely a geometer -- he's also a set theorist. Set theory ultimately goes back to the mathematician Georg Cantor. (Yes, the same Cantor for whom the Cantor dust is named. But no, he is not one of the mathematicians whose biography is given in Mandelbrot's book.) Now as it turned out, Cantor's original theory led to contradictions (for example, a set containing all sets was problematic for Cantor). Ever since then, set theorists have been trying to find new theories that avoid the contradiction of the Cantor's theory.

Now here's where Holmes comes in -- two months ago, he completed a proof that an alternate set theory, called New Foundations, allows for a set of all sets without contradictions. This theory is complicated -- recall that I once called the Axiom of Choice the set theorists' Parallel Postulate. Well, the Axiom of Choice is not even compatible with New Foundations.

Thus Holmes is undoubtedly an expert at proofs and determining which theorems can be proved using which axioms or postulates. So if someone like Holmes is unable to come up with a proof of AAS without using TEAI (or a theorem derived from TEAI), then who am I even to try?

And so today I post worksheets for SAS, ASA, and SSS, but not AAS yet. We'll get to HL next week since it's not until Lesson 7-5 of the U of Chicago text. I'm still holding out hope that I can find a proof of AAS from HL by the time we get to Lesson 7-5 (but considering what Holmes wrote above, don't count on it).

This means that my next unit in November is starting to get crowded. As I just wrote, we will spend the rest of this month getting up to Lesson 7-5 on HL. Now here is what I plan on covering during the next unit in November:

-- Lessons 5-1 (isosceles triangles) through 5-5 (isosceles trapezoids). The proofs will be rewritten to resemble the standard proofs using SAS, ASA, SSS, and HL. The isosceles trapezoid lesson will be written to fit a neutral geometry.

-- Translations. They will be defined, and their properties given, using the isosceles trapezoids that we just covered in Lesson 5-5.

-- Parallel Line Tests. They will be proved using the properties of translations.

-- AAS Theorem. It will be proved using the Corresponding Angles Test.

-- Parallel Postulate. Now we can prove the Parallel Line Consequences.

-- Finish Chapter 5. The most important theorem in this part of Chapter 5 is Triangle Sum.

Originally our plan carried over from last year devoted two weeks to this unit. It actually makes more sense to spend three weeks on it, so that we can finish it by Thanksgiving. And besides, much of Chapter 6 will end up just being a review of the transformations that we will have already covered, so it's not really a new unit.

Here are the worksheets for today. I've actually written a worksheet for ASA last year, but I couldn't post it due to the computer problems I had last Thanksgiving. I'd written it using the prime-notation from the U of Chicago text, where we begin with some isometry mapping Triangle

*ABC*to

*A'B'C'*as we prepare for the final reflection. The new worksheets for SAS and SSS that I created this year do not refer to Triangle

*A'B'C'*-- instead we abuse the name

*ABC*again.

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