Lesson 7-2: Triangle Congruence Theorems (Day 72)

Let's look at pages 91-99 of Jim Ottaviani's T-Minus: The Race to the Moon. We left off just as we were preparing to launch Apollo 8 -- a lunar orbit:

Gene Kranz: [My wife] thought we were going to Acapulco. Guess I'd better tell her there's a change in our holiday plans.
Admiral (Hickam Air Force Base -- Honolulu, Hawaii): Christmas?
Gene: Sir, yes, sir.
Admiral: Okay, young man. What have you got to say?
Gene: Well... in this mission... we will learn how to ride a Saturn V. We will learn what it means to leave earth's gravitational field. We will learn how good our radar tracking and onboard guidance computer is. We will learn how to reenter earth's atmosphere from another planet. And we need you to pick up the astronauts when they splash down. Admiral McCain, NASA realizes you've made your plans already... but we're asking you to re-deploy the um... fleet to --

(Ottaviani explains, "Admiral John McCain = father of future Senator John McCain, who was a prisoner of war in Vietnam at the time.)

Adm. John McCain Sr.: Best damn briefing I've ever had. Give Mr. Kraft and NASA what they want.
Gene: Yes, sir! Okay guys...

We now move ahead to the launch. Let's check in to Mission Control:

Gene: How does it feel up there?
Bill: Oof! V-very... good. Very good.

00:00:03:09 GET Launch Escape Tower jettisoned.
00:00:08:45 S-II Rocket separates from the S-IVB
00:00:08:48 S-IVB Rocket burn
...and, much later...
00:03:20:59 GET CSM separates from S-IVB.

It's now time for Gene to inform the crew that it's time to leave earth.

Gene: Go for TLI.

(Ottaviani explains, "TLI = Trans-Lunar Injection = going to the moon!)

Mike (at Houston CAPCOM): Heh.
Gene: What's so funny, Mike?
Mike: There has to be a better way of telling the first three human beings they can leave earth, don't you think?
Gene: No, why?
Mike: Oh. Never mind.
Gene: Apollo 8, Houston. We will have LOS in about 30 seconds, and we'll pick you up at 02:37:30 GET.

(Ottaviani explains, "LOS = Loss of Signal... remember, there's no communications satellite system!)

00:04:48:35 GET
Bill Anders: Boy, it's really hand to describe what this earth looks like. I'm looking out my center window... and the window is bigger than the earth right now. I can see most of South America...

00:11:49:24 GET
Bill: Man, you're looking pretty small down there, Houston.
Gene: We're carrying a big stick, though.

01:07:15:30 GET
Gene: Do you have a picture now?
Bill: Negative. We are having no joy. You don't have a lens cover on there, do you?
Gene: No, we checked that...
Bill: It was a very exciting ride on that big Saturn, but it worked perfectly, and we are looking forward now, of course, to the day after tomorrow... We have to get back to our passive thermal control in the barbeque mode so that we don't get one side of the spacecraft too hot for too long...

Let's skip up to our final scene for today, when they actually approach the moon:

Bill: Look at that... Wow, is that pretty! Hey, don't take that picture. It's not scheduled.
Jim Lovell: What?
Bill: You got color film, Jim? Hand me that roll of color quick, will you...
Jim: Yes, I'm looking..
Bill: Hurry up! Got one? Anything, quick. Well, I think we missed it.
Jim: Hey, I got it right here.
Bill: Let -- let me get it out this window. It's a lot clearer.
Jim: Take several of them! Here, give it to me.
Bill: Wait a minute, let's get the right setting. Here now, just calm down. Yeah...

We'll learn more about this picture of the moon in our next post.


Today on her Mathematics Calendar 2019, Theoni Pappas writes:

If segments BD and AE bisect each other at C, then Triangles ABC = ECD by:
(1) AAA
(2) ASA
(3) SAS
(4) SAA
(5) SSS

This is a straightforward triangle congruence problem. We are given that two segments bisect each other, from which we conclude AC = EC and BC = DC (two S's). What's left is to note that Angles ACB and ECD are vertical angles, hence also congruent (the A). Therefore, we conclude that Triangles ABC and ECD are congruent by SAS -- choice (3), and of course, today's date is the third.

Multiple choice questions like this one are common during the first week of the month, in order to keep the number of choices manageable. The Triangle Congruence Theorems are covered in Lesson 7-2 of the U of Chicago text. And guess what -- that's today's lesson! This marks the second perfectly timed Pappas question this calendar year -- in my February 15th post, Pappas had a Distance Formula problem on the day we covered Lesson 11-2. (Oh, and while Pappas uses the abbreviation SAA, the U of Chicago text uses AAS, as we'll see later in today's post.)

The fact that this problem mentions "bisect each other" suggests that it might lead to a proof of a certain parallelogram test, which will appear in Lesson 7-7.

Also, after I complained that there were so few Geometry problems in November, we've already had Geometry problems all three days in December thus far. (This includes a Triangle Inequality problem on December 1st, a non-posting day.)

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

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 AB = DE, Angle A = FDE, and Angle B = FED. Consider the image Triangle A'B'C' of Triangle ABC under an isometry mapping AB onto DE. Triangle 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 (AB = DE, Angle A = D, Angle B = E) then Triangle ABC can be mapped onto Triangle DEF by an isometry. (First map AB onto DE, then reflect the image of Triangle 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 AB = DE, AC = DF, and Angle A = FDE. Our first isometry will be to map A onto D. Now we simply reflect A onto D -- that is, the mirror is the perpendicular bisector of AD. This may be a bit trickier to visualize that the translation, but it works. The image A' is D, while the images B' and C' can be anywhere on the plane.

Next, we must map the whole segment A'B' onto our destination, DE. Since 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 DB' to DE as they both have the same length as the original segment AB (that is, we already have a point on Ray 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 AB to DE, so it uses one fewer prime symbol.) To avoid having to write multiple prime symbols, we will abuse notation and simply refer to the image of Triangle ABC under the isometry mapping AB to DE as Triangle 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 to DF as it's given that they both have the same length as the original segment AC (that is, we already have a point on Ray 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 (AB = DE,  AC = DF, Angle A = D) then Triangle ABC can be mapped onto Triangle DEF by an isometry. (First map AB onto DE, then reflect the image of Triangle 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 AB = DE, AC = DF, and BC = EF. We begin, as in the other proofs, by performing the isometry that maps AB to DE. So 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 AC (mapped to DC) = DF -- that is, D is equidistant from C and F. Therefore by Converse Perpendicular Bisector, D lies on the perpendicular bisector of CF. We are given that BC (mapped to EC) = EF -- that is, E is equidistant from C and F. Therefore by Converse Perpendicular Bisector, E lies on the perpendicular bisector of CF. So we already know two points on the perpendicular bisector of CF, namely D and E. Since two points determine a line, this tells us that the perpendicular bisector of CF is exactly line 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 CF. Therefore, by the definition of reflection (meaning), 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. 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 AB = DE, Angle A = FDE, Angle C = DFE. We begin, as in the other proofs, by performing the isometry that maps AB to DE. And as in the proofs of ASA and SAS, the Side-Switching Theorem implies that using line 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! 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. 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.

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 AB = DE, BC = EF, A and D are right angles. We begin, as in the other proofs, by performing the isometry that maps AB to DE. Now since BC (same as EC) = EF, just as in the proof of SSS, we see that E is equidistant from C and F, so that E lies on the perpendicular bisector of CF.

Now since CAB (same as CDE) and FDE are right angles, line DE is perpendicular to CF. By the Uniqueness of Perpendiculars Theorem, DE is the only line through E perpendicular to CF, so line 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

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). 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. Six 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 DF or on the other side of 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 -- four
 years 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?

Here are the worksheets for today. This year, I couldn't resist adding today's Pappas question to the Exercises (which on a whim I decided to change to red). In my February 15th post, I added the Pappas problem and stated that it might be years until our next perfectly timed calendar problem. (I changed "SAA" to "AAS," of course.)

Well, "years" turned out to be less than ten months. Of course, there might never be another perfectly timed Pappas problem again -- because I don't know whether there will ever be another Pappas calendar again.