Friday, October 27, 2017

Lesson 5-1: Isosceles Triangles (Day 51)

This is what Theoni Pappas writes on page 300 of her Magic of Mathematics:

"In his dissertation, Euler pointed out that the Konigsberg bridge problem seemed geometric in nature, but that Euclidean geometry did not seem to apply since the bridge problem did not deal with 'magnitudes' nor could it be resolved using 'quantitative calculations.'"

This is the third page of the subsection on the Konigberg bridge problem and Euler. As this is first and foremost a Geometry blog, it's interesting to look at what she's saying about Geometry here.

What exactly is Euclidean geometry, anyway? Surely, Euclidean geometry is a geometry based on the five postulates of Euclid. We see Pappas mention "magnitudes" here, and without a doubt, magnitude and measurement are essential to Euclidean geometry. His fourth postulate, after all, is "All right angles are congruent." This postulate requires that we measure something -- in this case, angles. His famous fifth postulate also requires measuring angles (as in "less than two right angles") And his third postulate is "A circle can be drawn with any center and any radius." Now we must measure radius, which is a distance, namely that between the center and a point on the circle.

It's possible to describe Euclidean geometry as the study of certain transformations -- the ones that preserve particular properties of figures. These transformations are the isometries. We've completed Chapter 4 of the U of Chicago text, where we learn that reflections -- our first isometries -- preserve the ABCD properties. Two of these are measurements -- A for Angle Measure, D for Distance. And so once again, we can't have Euclidean geometry without returning to quantitative calculations.

But the Konigsberg bridge problem, on the other hand, isn't a problem of Euclidean geometry, as Euler found out. The lengths of the bridges are irrelevant, as are the angles formed by the bridges. So we discover that this is a problem not of geometry, but of topology. Pappas writes:

"Thus, Euler's solution to the Konigsberg bridge problem acted as a catalyst and introduction to the field of topology."

Why, then, do the bridges of Konigsberg appear in the U of Chicago text -- which is ostensibly a book on Euclidean geometry? Let's return to Lesson 1-4 of the text, where we read:

Fourth description of a point:
A point is a node of a network.

Later on in Lesson 1-7, we learn of the Point-Line-Postulate, which implies that by "point," we no longer mean a node of a network. It's the postulates -- whether the U of Chicago's or Euclid's -- that characterize what exactly a "point" is. Lesson 1-4 is provided as a non-example of a Euclidean point.

Pappas continues by telling us about the ultimate fate of the bridges of Konigsberg:

"Today only three of the seven original bridges remain -- the Honey bridge, the High bridge, and the Wooden bridge."

The lone picture here is captioned, "One of the three original remaining bridges of Konigsberg."

OK, let's stop talking about Chapter 1 and begin Chapter 5 of the U of Chicago text.

Lesson 5-1 of the U of Chicago text is called "Isosceles Triangles." In the modern Third Edition of the text, isosceles triangles appear in Lesson 6-2. But the new Lesson 6-2 expands beyond the old Lesson 5-1 to discuss scalene triangles as well (which don't appear until Lesson 13-7 in the old text).

Two years ago, I made several changes to this lesson. That, I was inspired both by a teacher who commented on this blog and a traditionalist blogger who wrote about how isosceles triangles are taught in modern texts. This year I'm reverting to the U of Chicago text, so we're going back three years to our old worksheet. This first worksheet also includes the Converse to the Isosceles Triangle Theorem, which doesn't appear in Lesson 5-1 (but it does appear in Lesson 6-2, Third Edition).

This is what I wrote three years ago about today's lesson:

Section 5-1 of the U of Chicago text covers isosceles triangles. Now the most important result of this section is, of course, the Isosceles Triangle Theorem, which states that if a triangle has two equal sides, then the angles opposite them are equal. But this first theorem in this lesson is something called the Isosceles Triangle Symmetry Theorem. Here's a statement of the theorem and its proof as given by the U of Chicago:

Isosceles Triangle Symmetry Theorem:
The line containing the bisector of the vertex of an isosceles triangle is a symmetry line for the triangle

Given: Isosceles triangle ABC with vertex angle A bisected by m
Prove: m is a symmetry line for triangle ABC

There are three things given. Each leads to conclusions that are used later in the proof. First, since m is an angle bisector, because of the Side-Switching Theorem, when ray AB is reflected over m, its image is ray AC. Thus B' is on ray AC. Second, it is given that A is on the reflecting line, so A' = A. Hence, since reflections preserve distance, AB' = AB. Third, it is given that triangle ABC is isosceles with vertex angle A, so AB = AC. Now put all of these conclusions together. By the Transitive Property of Equality, AB' = AC. So B' and C are points on ray AC at the same distance from A, and so B' = C. By the Flip-Flop Theorem, C' = B. So, by the Figure Reflection Theorem, the reflection image of triangle ABC is triangle ACB, which is the sufficient condition for the symmetry of the triangle to line m. QED

Now, when a pre-Common Core teacher first sees this proof, this is not what is expected. Such a teacher expects to give the point where BC and m intersect the name D. Then one tries to prove the triangles ABD and ACD congruent. We have AB = AC by the definition of isosceles, angles BAD and CAD equal by definition of angle bisector, and AD = AD by the Reflexive Property of Equality. And so by SAS, the two triangles and congruent, and so angles B and C are equal by CPCTC.

But we must note that in Chapter 7, we use isometries to prove SAS as a theorem. So we could have the Isosceles Triangle Theorem derived from SAS, which in turn is derived using isometries. Or we can skip the middleman and just derive the Isosceles Triangle Theorem from isometries. This is definitely in the line of Common Core thinking:

Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself.

In this section we found a reflection that maps an isosceles triangle to itself -- namely the reflection over the line containing the angle bisector of the vertex angle. Later on in Chapter 5, we'll find the isometries that map the other figures mentioned in that standard (rectangles, etc.) to themselves.

A few things about the traditional proofs. First of all, these proofs usually involve adding a new point D, where the angle bisector of BAC intersects side BC. But, according to Dr. Hung-Hsi Wu, we can't be sure that these even intersect at all without a Crossbar Axiom. But in the U of Chicago, we don't need it because the point D is never mentioned in the proof.

The most common proof of the Isosceles Triangle Theorem uses an angle bisector and SAS. Notice that we also could have used a median and SSS. But the SAS proof is usually better because some texts use the Isosceles Triangle Theorem to prove SSS. (In fact, Wu uses ITT to prove SSS.) So a proof of ITT using SSS would be circular.

Euclid himself proves the Isosceles Triangle Theorem as his Proposition I.5:

Euclid's proof also uses SAS (which is his Proposition I.4), but his proof is more complicated than any text appearing in a high school text because he was also concerned with the exterior angles at B and C, not just the interior angles of the triangle. According to Euclid, the Isosceles Triangle Theorem was the Pons Asinorum, or bridge of SSA --- well, actually SSA spelled backwards. In a way, we cross a bridge when we enter Chapter 5, as here the somewhat more difficult proofs of geometry begin.

I must mention one of my favorite proofs of ITT here, although it's also based on SAS and so not appropriate for the students right now. According to the above link, it can be attributed to the Greek geometer Pappus, who lived about 500 years after Euclid. Here is the proof given directly from the above link:

The two triangles BAC and CAB have two sides equal to two sides, namely side BA of the first triangle equals side CA of the second triangle, and side AC of the first triangle equal to side AB of the second, and the contained angles are equal, namely angle BAC of the first triangle equals angle CAB of the second, therefore, by I.4, the corresponding parts of the two triangles are equal, in particular, the angle B in the first triangle equals the angle C of the second. QED

In two-column form, the Pappus proof would look like this:

Given: AB = AC
Prove: Angles B and C are equal.

Statements                Reasons
1. AB = AC                1. Given
2. Angle A = Angle A 2. Reflexive Property of Equality
3. AC = AB                 3. Symmetric Property of Equality
4. ^ABC = ^ACB        4. SAS
5. Angle B = Angle C 5. CPCTC

Notice that the closest I could get to a triangle symbol in ASCII is the caret with a strikethrough. I was considering writing an angle symbol as an underlined slash, but I decided that simply writing out the word "Angle" was less confusing.

We see that in the Pappus proof, triangle ABC is congruent to itself -- but under a new name ACB. In a way, this is exactly what symmetry is -- a figure is symmetrical if there is a nontrivial congruence between the figure and itself (under a different name). Triangle ABC is always congruent to triangle ABC, but it's only congruent to ACB if the triangle is isosceles. So we know that there is an isometry from ABC to ACB -- but we don't know that it's a reflection yet, much less a reflection over the line m containing an angle bisector of angle A, although this can be deduced. Notice that the line m never appears in the Pappus proof -- the only proof that doesn't require an auxiliary line.

Now let's return to the U of Chicago proof. This proof is on the longish side -- indeed, many of the heavy-duty theorems have long proofs. It's instructive to cut out an isosceles triangle with the bisector of A drawn in, and ask the students where B would land if the triangle were folded along that angle bisector. Chances are that the students will say that it would land on C. But this doesn't work unless the triangle is isosceles. For the scalene triangle DEF drawn on the U of Chicago page right below the Isosceles Triangle Symmetry Theorem, E will not land on F if the triangle is folded along the bisector of angle D. So we need to prove that it works for isosceles triangles.

This proof contains a common trick in isometry-based proofs -- to prove that the image of B is C, we first show that the image of B lies on ray AC, then show that the image is a point that's the correct distance from A. And the point at just the right distance from A is exactly C.

But there's something missing from this lesson in the U of Chicago. We have the Isosceles Triangle Theorem, but there's no mention of its converse. The text even states, "The Isosceles Triangle Theorem is useful in proofs in which you must go from equal sides to equal angles," but what if one wants to go from equal angles to equal sides?

The most common proof is based on AAS. Once again, we draw the angle bisector of angle A (which intersects BC at D). We are given that angles B and C are equal, and angle BAD = angle CAD by definition of angle bisector and AD = AD by the Reflexive Property. So this gives us ABD and ACD, congruent by AAS, and AB = AC by CPCTC.

In Euclid, the converse is Proposition I.6. His proof is an indirect proof that is not usually given in high school classes:

There is also a Pappus-style proof of this converse. We are given that angles B and C are equal, we have BC = CB by the Reflexive Property, and we have that angles C and B are equal. And so once again, triangles ABC and ACB are congruent, this time by ASA, and so AB = AC by CPCTC.

But is there a proof based on reflections, similar in style to the U of Chicago's proof of the forward Isosceles Triangle Theorem? Here's what such a proof may look like. Instead of letting m be the bisector of angle A, let it be the perpendicular bisector of BC:

Given: Triangle ABC with equal angles B and Cm perpendicular bisector of BC
Prove: m is a symmetry line for triangle ABC

We obtain that the reflection image of B is C immediately, as the reflecting line m is the perpendicular bisector of BC (definition of reflection). Now what is the reflection image of angle B? It must be an angle with vertex C and ray CB as one of its sides, and of the same measure as angle B -- and we're already given such an angle, angle C. So not only is the point C the image of point B, but angle C is the image of angle B. This means that the image of ray BA is ray CA, so that any point on ray BA has an image on ray CA, and vice versa (Flip-Flop Theorem). Now A is clearly on ray BA, and so A' must lie on ray CA. Likewise A is on ray CA, and so A' must like on BA. So A' is a point that lies on both rays BA and CA, but by the Line Intersection Theorem, rays BA and CA have only one point in common -- and that is A itself. So A' must be exactly A itself -- that is, A is a fixed point of the reflection -- that is, A must lie on the reflecting line m, the perpendicular bisector of BC. And since A lies on the perpendicular bisector of BC, we have, from the Perpendicular Bisector Theorem, that A is equidistant from B and C -- that is, AB = AC. QED

But there is a gap in this proof here. How do we know that the image of angle B is not an angle of the same vertex and measure as angle C, but on the opposite side of BC from where A is? (Recall the Two sides of line assumption from the Angle Measure Postulate.) If the image of angle B is on the wrong (alternate) side of BC, then angle B and its image would form alternate interior angles, and so their other sides wouldn't intersect at all, much less at A.

This problem appeared in our Alternate Interior Angles Test proof as well -- and it occurs in any proof where we have equal angles and want to map one to the other with an isometry. The solution is the Plane Separation Postulate, which tells us that line BC divides the plane into two half-planes, and we want to prove that angle B and its image each has a side in the same half-plane.

To prove this, suppose we have a point P in the coordinate plane and wish to reflect it over, let's say, the y-axis. So of course P and P' are on opposite sides of the y-axis. But notice that P and P' must be on the same side of the x- (not y-, but x-) axis. This is because line PP', by definition of reflection, is perpendicular to the reflecting line, the y-axis, and the x-axis is also perpendicular to the y-axis. So by the Two Perpendiculars Theorem, line PP' must be parallel to the x-axis, and so the line can hardly intersect x-axis if it is parallel to the x-axis! (Of course if P lies on the x-axis then so does P'.) This tells us that the while the reflection image of a half-plane whose boundary is the mirror must be the opposite half-plane, its reflection image in any mirror perpendicular to the boundary is itself. Any line perpendicular to the boundary line of a half-plane is a symmetry line of that half-plane.

Notice that the Isosceles Triangle and Perpendicular Bisector Theorems are ultimately related, in that the latter is used to prove the converse of the former. Moreover, the converse of the latter can be used to prove the former as well! In fact, my proof of the Converse of the Perpendicular Bisector Theorem actually borrowed portions of the proof of the Isosceles Triangle Theorem. This means that some steps of the proof are redundant. We can shorten the Isosceles Triangle Theorem proof by using the Converse of the Perpendicular Bisector Theorem, as follows. Notice that now m must be the perpendicular bisector of BC, since that's the line we want:

Given: Isosceles triangle ABC with m the perpendicular bisector of BC
Prove: m is a symmetry line for triangle ABC

First, it is given that A is on the reflecting line, so A' = A. Second, it is given that triangle ABC is isosceles with vertex angle A, so AB = AC. Since A is equidistant from B and C, by the Converse of the Perpendicular Bisector Theorem, A lies on the perpendicular bisector of BC, which is m. Third, by the definition of reflection, we have B' = C and C' = B. So, by the Figure Reflection Theorem, the reflection image of triangle ABC is triangle ACB, which is the sufficient condition for the symmetry of the triangle to line m. QED

Returning to 2017, today is an activity day. When this school year began, I planned on finding activities from two particular sources:

-- old activities from two or three years ago on the blog
-- the Exploration questions from the current U of Chicago lesson

So far, I've only used the former, since there are so many activities already posted here. But this time, I'm going to do the latter, for two reasons:

-- I never posted any activities for Lesson 5-1. Three years ago, I posted Lesson 5-1 on a Monday and then posted no activity until after Lesson 5-4 on a Friday. Two years ago, I made more changes and ended up dropping that activity altogether.
-- The Exploration questions for Lesson 5-1 are very important.

This is the relevant Common Core Standard:

Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point.
[emphasis mine]

Well, today we cover base angles of isosceles triangles in the main lesson. But the key here is on the medians of triangles and how they are concurrent. That doesn't appear in the U of Chicago text at all, except in this Exploration question.

Three years ago, I squeezed in the Exploration question as a bonus, but now I'm creating an activity worksheet for this question. Still, this is what I wrote three years ago about medians:

In the exercises, I just had to include Exploration Question 21 as a bonus question, since it refers to the centroid or point where the medians intersect, which is explicitly mentioned in Common Core. I know that this is an exploration question, so nothing is actually proved yet. Earlier I wrote that the actual proof can wait until second semester, but after comparing the U of Chicago with Wu, I see that the proof is better given during first semester. In fact we may be able to give a proof very soon.

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