Lesson 4-4 of the U of Chicago text introduces formal, two-column proofs. As a first example of a proof, the text gives Proposition 1 from Euclid. Since Euclid was among the first to write formal proofs and this was his first theorem, the students' first proof will be one of the oldest proofs written in the whole world.

I already mentioned David Joyce's website and his Euclid pages. Here's a link to Proposition 1:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI1.html

And here's what Joyce writes about why Euclid chose this one to be first:

"This proposition is a very pleasant choice for the first proposition in the

*Elements.*The construction of the triangle is clear, and the proof that it is an equilateral triangle is evident. Of course, there are two choices for the point

*C,*but either one will do.

"Euclid could have chosen proposition I.4 [SAS Congruence -- dw] to come first, since it doesn’t logically depend on the previous three, but there are some good reasons for putting I.1 first. For one thing, the

*Elements*ends with constructions of the five regular solids in Book XIII, so it is a nice aesthetic touch to begin with the construction of a regular triangle. More important, though, is I.1 is needed in I.2, and that in I.3. Propositions I.2 and I.3 give constructions for moving lines, and I.4, although not logically dependent on I.2 or I.3, does use the concept of superposition which involves, in some sense, moving points and lines."

The U of Chicago text gives a two-column proof of Euclid's Proposition 1. There are a few differences between a two-column proof in this text and those found in most other texts. First, the U of Chicago labels the two columns "Conclusions" and "Justifications" -- whereas most other texts label them "Statements" and "Reasons." Also, most books start their two-column proofs with the "Given" information, but the U of Chicago skips the "Given" lines. On this blog, I plan on doing proofs the way most traditional texts do them, with "Statements," "Reasons," and "Given."

Notice that this theorem is truly a construction -- if available, teachers can have the students use a compass to draw the circles and a straightedge to draw the segments. This sounds like something mentioned in the Common Core Standards:

CCSS.MATH.CONTENT.HSG.CO.D.13

Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle.

The question is, must the equilateral triangle and the square be inscribed in the circle, or must only the regular hexagon be so inscribed? After all, Euclid's equilateral triangle is

*not*inscribed in the circle -- for a triangle to be inscribed, all three vertices must lie on the

*same*circle, but the vertices of Euclid's triangle lie on

*different*circles. Unfortunately, the standards are vague here.

Now Example 1 in the text gives another example of a two-column proof. Many books refer to this as the Alternate

*Exterior*(not Interior) Angle Theorem. But we're saving proofs based on parallel lines until a little later. The midpoint proof in Example 2, of course, can be given.

I decided that this is a good time to include two of the proofs that I gave last year -- the Uniqueness of Perpendiculars Theorem and the Line Perpendicular to Mirror Theorem. The first theorem is actually Euclid's Proposition 12:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI12.html

I chose these two theorems because they fit here the best. After all, Joyce points out that the same double-equilateral-triangle construction is used here as in Proposition 1. Also, Example 2 in Lesson 4-1 hints at this same proof. The difference between our proof and Euclid's is that we explicitly mention reflections in our proof, since Common Core demands that we use reflections in proof. The Line Perpendicular to Mirror Theorem follows directly from the Uniqueness of Perpendiculars Theorem, and tells us that a line perpendicular to the reflecting line (the mirror) is invariant -- that is, its image is identical to the preimage. Finally, these two theorems on perpendiculars provide a great segue into Lesson 4-5, the Perpendicular Bisector Theorem.

Moreover, the Uniqueness of Perpendiculars and the Line Perpendicular to Mirror Theorems can be used to prove some of the properties of reflections on the coordinate plane from yesterday's post. If you recall, we are restrained by the fact that we need dilations and similarity to prove many of the properties of coordinates. Also, recall that so far, the geometry posted so far this school year is neutral geometry that doesn't require a Parallel Postulate. (This restraints are

*not*unrelated -- similarity only works in Euclidean geometry. Technically, I have posted one non-neutral worksheet this year -- it asks the students to find the perimeter and area of a rectangle, but rectangles don't even exist except in Euclidean geometry. In that case, the ability to interact with students learning about area took priority over trying to maintain neutral geometry.)

It's possible to do some work with coordinates while still in neutral geometry. We can follow Dr. Franklin Mason's Lesson 13.1. Here Dr. M defines the

*x-*and

*y*-axes to be any two perpendicular lines, and uses the Ruler Postulate to place coordinates on the axes. Then the coordinates of any point are defined to be the

*projections*of the point onto the

*x*- and

*y*-axes. As Dr. M writes, it's the Uniqueness of Perpendiculars Theorem that guarantees that the coordinates of any point are unique.

Now let's do the reflections. The key is that the graph of

*x*=

*h*is a line perpendicular to the

*x*-axis for any value of

*h*, and likewise

*y*=

*k*is perpendicular to the

*y*-axis for any value of

*k*. We then reflect the point (

*h*,

*k*) by finding the images of the lines

*x*=

*h*and

*y*=

*k*. For example, if the mirror is the

*y*-axis, then any line of the form

*y*=

*k*must be mapped to itself -- this is because

*y*=

*k*is perpendicular to the

*y*-axis, and so by the Line Perpendicular to Mirror Theorem,

*y*=

*k*is an invariant line. That the image of

*x*=

*h*must be

*x*= -

*h*follows from the Reflection Postulate and Ruler Postulate applied to the

*x*-axis.

We are also able to perform reflections over

*y*=

*x*in neutral geometry. The first step is to prove that the graph of

*y*=

*x*really is a line. The key to the proof is that we let (

*a*,

*a*) be any point on the graph of

*y*=

*x*, and we look at the quadrilateral with vertices (0, 0), (

*a*, 0), (

*a*,

*a*), and (0,

*a*). In Euclidean geometry, this is a square. In neutral geometry, we can show that this is a Lambert quadrilateral (i.e., half of a Saccheri quadrilateral), and also a kite. We end up using the Isosceles Triangle Theorem and its converse, along with the Kite Symmetry Theorem, to show not only that

*y*=

*x*must be a line, but that it's the bisector of the angle formed between the axes. Then we can use the Reflection Postulate and Side-Switching Theorem to show that the reflection image of

*x*=

*a*must be

*y*=

*a*.

Unfortunately, we can't reflect over an arbitrary horizontal or vertical mirror in neutral geometry. The reflections over the axes,

*y*=

*x*, and

*y*= -

*x*work because they map the axes either to themselves or to each other. But let's say we want to reflect over the mirror

*x*= 1. We can't prove in neutral geometry that the line

*y*= 1 is perpendicular to the mirror

*x*= 1, and so we can't prove that

*y*= 1 is invariant. We want to show that the point (

*h*,

*k*) is mapped to (2 -

*h*,

*k*) -- ironically, we can prove that

*x*=

*h*really is mapped to

*x*= 2 -

*h*, but we can't prove that

*y*=

*k*is mapped to itself (though it sounds easier).

I'm not sure whether I want to post these proofs for the students yet -- not because the proofs are hard, but because they have enough to worry about as they first learn about proofs. The cases where the mirror is either of the axes can be proved using today's theorems. The mirrors

*y*=

*x*and

*y*= -

*x*can wait until the first part of Chapter 5, where we prove properties of isosceles triangles and kites, but still before we give a parallel postulate. The mirrors

*x*=

*h*and

*y*=

*k*require a parallel postulate, but we only need to consider the properties of rectangles, not similar triangles. No other mirrors appear on the Common Core test -- thank goodness for that!

Regarding the worksheets, as I mentioned earlier, I like the idea of having the students experiment with each theorem by folding the paper (for reflections) before actually proving the theorem. In some ways, this is the Michael Serra approach, except that the proof is given immediately after -- not near the end of the book. Because of this, I'm including several worksheets today. (Don't say I didn't warn you!) I have the statement of each theorem on one side of the page -- large enough to encourage folding -- and a two-column proof on the other, with a few Reasons left blank for the students to write in.

And as for the Exercises, notice that Question 12 contains a flow proof. Many other geometry texts emphasize flow proof as a third type of proof, after paragraph and two-column proofs. But in the U of Chicago texts, a flow proof appears in this question and then never again in the text. Since it's a rewriting of the Alternate Exterior Angle Theorem from Example 1, I'm throwing it out. Let's not confuse the students with flow proofs just yet -- and I'll probably leave flow proof out altogether.

Question 24 is the Exploration/Bonus Question. It directs students to discover that the altitudes of a triangle are congruent -- that is, that they meet at a point (called the orthocenter). Notice that many texts cover the four concurrency proofs for triangles (centroid, circumcenter, incenter, and orthocenter), but these aren't covered fully in the U of Chicago text. The orthocenter never appears in the text again after this Exploration Question. Actually, one concurrency question appears in the Common Core Standards:

CCSS.MATH.CONTENT.HSG.CO.C.10

Prove theorems about triangles.

*Theorems include ... the medians of a triangle meet at a point*.

The proof that the medians meet at the centroid requires similar triangles, and so the proof must wait until the second semester. But the proof that the perpendicular bisectors meet at the circumcenter is the easiest of the concurrency proofs, and it appears in the very next section, 4-5. Incidentally, the proof that the altitudes meet at a point entails constructing a larger triangle and showing that the altitudes of the smaller triangle extend to the perpendicular bisectors of the larger triangle (so the orthocenter of the smaller triangle is the circumcenter of the larger triangle)! But this proof requires parallel lines and Playfair, so it must wait. Still, the students can still explore this in the Exercises.

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