## Monday, July 28, 2014

### A Preview: Perpendicular and Parallel Lines

In this post I want to preview specific examples of the geometry that I will provide. As I mention earlier, it will be based on my U of Chicago text, rearranged to satisfy Common Core.

Here is a key example: in Chapter 3, the U of Chicago text covers angles, proofs, and then parallel and perpendicular lines. Let's assume that by this point, we've already covered angles (which I repeat are covered awfully late in the text) and we're ready for perpendicular lines. In the text, Section 3-5 begins by defining perpendicular:

U of Chicago:
"Two segments, rays, or lines are called perpendicular if and only if the lines containing them form a 90-degree angle."

The text then proceeds to prove two theorems relating perpendicular and parallel lines:

Two Perpendiculars Theorem:
"If two coplanar lines m and n are each perpendicular to the same line l, then they are parallel to each other."

Perpendicular to Parallels Theorem:
"In a plane, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other."

The text gives an informal proof of these two theorems as being special cases of the Corresponding Angles Postulate and its converse (itself a postulate in this text) in which the corresponding angles happen to be right angles. But this is unacceptable for our purposes. We wish to have only one postulate for parallel lines, the Parallel Postulate of Playfair, and as it turns out, one direction can be proved without any Parallel Postulate. It turns out that the Two Perpendiculars Theorem is provable without Playfair, while the Perpendicular to Parallels Theorem requires that postulate.

Our proof will be very simple. It will depend on reflections -- but once again, Common Core Geometry is all about reflections (and rotations, etc.). This is why I am rearranging the text order so that reflections are covered first. Let's recall our definition of reflection:

"For a point P not on a line m, the reflection image of P over line m is the point Q if and only if m is the perpendicular bisector of segment PQ. For a point P on m, the reflection image of P over line m is P itself."

Of course, we'll have to define "perpendicular" (and "perpendicular bisector") before we can define reflection, but we won't prove any theorems about perpendicular lines until after we've discussed the properties of reflections.

Notice that Section 3-6 covers the construction of perpendicular lines. We refer back to the Common Core Standards:

CCSS.MATH.CONTENT.HSG.CO.D.12
"Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line."

And we see "constructing perpendicular lines" specifically mentioned in the standard. So we definitely want to do exactly that right here on this blog. The problem is that many classrooms lack the materials mentioned on that list -- especially a compass. Teachers often omit the sections of the text requiring straightedge-and-compass constructions.

The text gives the following steps to construct the perpendicular bisector of a segment AB:

Step 1: Circle A containing B (Compass rule)
Step 2: Circle B containing A (Compass rule)
Step 3: Circles A and B intersect at new points C and D (Point rule)
Step 4: Line CD (Straightedge rule)

Notice that while many texts merely require that the compass setting be any length more than half of AB, this text has the setting be exactly the length of AB.

The text states that we cannot prove that this algorithm works -- at least not yet. Notice that this is exactly what Joyce criticizes about Chapter 1 of the Prentice-Hall text. He prefers that no construction be given without a proof that it works. Well, here's a proof that this works:

We observe that this construction (two circles centered at A and B, passing through the other point, and intersecting at C and D) is exactly the proof given in Section 4-4 -- which is stated to be the first theorem in Euclid's Elements. And, as Proposition 1 in Euclid tells us, this makes triangle ABC (as well as ABD) equilateral. Now, as equilateral triangles are isosceles, a theorem in Section 5-1 tells us that the perpendicular bisector of the base of both triangles (AB) are also their medians, so that it passes through the other vertices (C and D). QED

Now the proofs in Section 5-1 about isosceles triangles are proved using reflections. So it helps to have Section 4-1 and its definition of reflection. In that section there is also a construction on how to find the reflection image of a point over a line -- but once again, the proof that it works depends on the isosceles triangles of Section 5-1.

Notice what's going on here. We need to jump all the way to section 5-1 just to be able to prove that the constructions in Chapters 3-4 work -- and that's only to satisfy Joyce's preference that proofs accompany constructions. And those constructions are the ones which require straightedge and compass -- the compass that many students won't have in the first place.

But there's a way out of this. Perhaps we can generate reflection images without either constructing them or proving that the constructions work. Let's look at that Common Core Standard again:

"CCSS.MATH.CONTENT.HSG.CO.D.12
"Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding ..."

Ah ha! We can just use paper folding! So to reflect a point P about a line l, we simply fold the paper over line l, and the point where P lands is the image of P, which we'll call P' (P prime). And to find the perpendicular bisector of AB, we just fold the paper so that point A lands on top of point B, and the crease line is the perpendicular bisector. We don't actually prove any of this -- all of this is just to get the students to generate reflections and perpendicular bisectors without being bogged down with proofs that they work, or even needing compasses to construct them.

Once we have this, we can now create some simple proofs. All proofs involving reflections ultimately go back to the Reflection Postulate, which we'll now state in full:

Reflection Postulate (U of Chicago):
Under a reflection:
a. There is a 1-1 correspondence between points and their images. This means that each preimage has exactly one image and each image comes from exactly one preimage.
b. If three points are collinear, then their images are collinear. Reflections preserve collinearity. The image of a line is a line.
c. If B is between A and C, then the image of B is between the images of A and C. Reflections preserve betweenness. The image of a line segment is a line segment.
d. The distance between two preimages equals the distance between their images. Reflections preserve distance.
e. The image of an angle is an angle of the same measure. Reflections preserve angle measure.

Wu gives many of these same properties, except for rotations rather than reflections. The U of Chicago uses the mnemonic A-B-C-D: Reflections preserve Angle measure, Betweenness, Collinearity, and Distance.

Using this, let's prove a theorem. Section 3-6 gives a construction for the following, but we shall prove this using not the construction, but using reflections:

Theorem:
Through a point P not on a line l, there is exactly one line through P perpendicular to l. (Notice that this says perpendicular, not parallel.)

Proof:
We are to prove this using reflections, but how? Well, we have a point P and a line l, so why don't we try reflecting the point P over the line l? Then there exists a unique point P' (by part a of the Reflection Postulate) such that l is the perpendicular bisector of PP'. Then voila! Line PP' is exactly the line that we seek -- a line through P perpendicular to l. Its existence and uniqueness are guaranteed because through two points (P and P') there is exactly one line. QED

And believe it or not, this gives us a quick and dirty proof of the Two Perpendiculars Theorem!

Two Perpendiculars Theorem:
If two coplanar lines m and n are each perpendicular to the same line l, then they are parallel to each other.

Proof:
Suppose that m and n weren't parallel -- that is, suppose that they intersect at some point P. Then through P there are two lines perpendicular to the line l (namely m and n) when by the previous theorem, there's only supposed to be one such line, a contradiction. Therefore the lines m and n must be parallel. QED

The only problem with this proof is that it's an indirect proof, which many books seek to avoid until much later -- the U of Chicago text doesn't cover indirect proofs until Chapter 13. I notice that the U of Chicago text gives an interesting definition of parallel:

Definition:
"Two coplanar lines m and n are parallel lines if and only if they have no points in common, or they are identical" (emphasis mine).

So this definition actually allows a line to be parallel to itself! Although not common in most texts, this alternate definition allows the above indirect proof to become a direct proof:

Proof:
We have that either m and n intersect at some point P, or they don't intersect. In the latter case, we are already done. In the former case, we have both m and n as lines through P perpendicular to l, and by the previous theorem, there is only one such line. So m and n are the same line. Therefore m and n have either no point in common, or every point in common -- and by our definition, both of these implies that m and n are parallel. QED

For completion, let me give the proof of the Perpendicular to Parallels Theorem. As we noted earlier, this proof will require Playfair's Parallel Postulate. It also contains an indirect part:

Perpendicular to Parallels Theorem:
In a plane, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other.

Proof:
Let's say we're given lines k, l, and m, with k parallel to m and l perpendicular to m. Also, the point where l and m intersect, let's call it A, and the intersection of k and l we'll call B. So our goal is to prove that k and l are perpendicular.

Let n be the perpendicular bisector of segment AB. So both m and n are perpendicular to l, and so by the Two Perpendiculars Theorem, m and n are parallel. Now we want to reflect m about line n, to produce a new line, m'. But where is m'? We already know where the image of one point on m is -- namely A, of course. The image of A is exactly B since n is the perpendicular bisector of segment AB (definition of reflection). Since A lies on m, its reflection image B lies on m'. This also implies that the reflection image of l is exactly l itself, since A and B lie on l.

Now can m and m' have any points in common? Suppose they have a point C in common -- that is, C lies on both m and m'. Since C lies on m, its reflection image C'  lies on m', and since C' also lies on m, its own reflection image must lie on m'. But the reflection image of C' is just C itself -- the U of Chicago text calls the Flip-Flop Theorem (since n is the perpendicular bisector of segment CC' and trivially C'C as well). So there are two lines passing through both C and C' (namely m and m') when there is only supposed to be one such line, a contradiction. So m and m' have no points in common -- that is, they are parallel.

Recall that B lies on m'. So m' is a line parallel to m passing through B. By Playfair, there is exactly one such line -- and we already know of such a line, namely k. So the reflection image of m over line n is exactly k. By the Reflection Postulate part e, reflections preserve angles, and since l and m are perpendicular (a 90-degree angle), their reflection images l and k meet at the same angle, so they must be perpendicular. QED

This proof is admittedly long. It may be shortened if we've already proved the following Lemma:

Lemma:
If a line is perpendicular to its reflecting line, its image is identical to itself.
If a line is parallel to its reflecting line, its image is parallel to itself.

Notice that Woo proved a similar lemma for 180-degree rotations -- a line passing through the center of rotation is mapped to itself, and a line not passing to the center is mapped to a line parallel to itself -- but he didn't prove this lemma for reflections. I'd say that the Lemma is worth proving, but it still requires an indirect proof unless we use the trick of defining every line as parallel to itself.