Let's start with the postulate itself. Earlier, I stated that the most convenient postulate for proofs is the Perpendicular to Parallels Postulate (a theorem in the U of Chicago and most other texts). I've decided that I will give this postulate a new name:

The Fifth Postulate:

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

The name "Fifth Postulate" has several levels of significance. First of all, it obviously refers to Euclid's Fifth Postulate, and we are using this postulate just as Euclid used his -- namely, as the main postulate for parallels. As I mentioned earlier, both my Fifth Postulate and Euclid's actually refer to two right angles. By calling my postulate the "Fifth Postulate," I can still use the phrase "the proof of this theorem depends on the Fifth Postulate" and any reader will know that the theorem is provable in Euclidean geometry as opposed to non-Euclidean geometry.

Here's a link to the page on David Joyce's website regarding Euclid's Fifth Postulate:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/post5.html

Furthermore, one significance of this postulate is that it can be directly applied in a few other theorems, including our translation theorems (for Chapter 6) and the concurrency theorems. These proofs involve a line that are perpendicular to one of two parallel lines and showing that it is perpendicular to the other. So our Fifth Postulate is convenient for these proofs.

Finally, notice that this is actually the

*fifth*named postulate on this blog. (The other four postulates are the Point-Line-Plane Postulate, the Angle Measure Postulate, the Reflection Postulate, and the Plane Separation Postulate.) Technically speaking, we've used more than five postulates as the properties of arithmetic are actually postulates -- but I think of these as being more like Euclid's "Common Notions" than his postulates. Notice that there is more connection to the number five -- we are now in Chapter 5, and in that chapter, the first section requiring a Parallel Postulate is 5-5. I was actually considering waiting until Day 55 before posting this postulate, but ran out of time (due to last week being skipped), but at least I posted this in the 50's.

Indeed, my skipping last week has major consequences for the pacing. I want to be out of Chapter 5 by the end of October, as I want to devote November and December to Chapters 6 and 7 -- these cover translations and triangle congruence which are vital for Common Core. Originally, I wanted to slow down to prove the many consequences of our Fifth Postulate last week and this week, but instead we must finish Chapter 5 by this week. Then again, my pacing rush resembles the problems faced in many actual classrooms -- we must rush Chapter 5 if we want to finish Chapter 7 by the end of the semester.

And so today's lesson will cover more topics than I originally wanted to in one lesson. Notice that even though this lesson is numbered 5-5, much of the material comes from other sections.

First, we stated the Perpendicular to Parallels Theorem, found in Section 3-5 of the U of Chicago text, as our Fifth Postulate.

Second, we prove the Uniqueness of Parallels Theorem, found in Section 13-6. Notice that this is actually Playfair's Parallel Postulate -- John Playfair being a 19th century Scottish mathematician. It's best for us, just as the U of Chicago does, to call it the Uniqueness of Parallels Theorem, since both of us actually prove it (so it's not a postulate). But sometimes on this blog, I just call this statement "Playfair" for short.

Our proof is similar to the U of Chicago's. The first difference is that since our Fifth Postulate requires perpendicular lines, we just draw the blue line in the picture so that it's perpendicular to line

*l*-- that is, make angle 1 90 degrees. Then by the Fifth Postulate, angles 2 and 3 are also 90. The other difference is that we want to avoid indirect proof. Notice that step 3 in the text concludes that if

*x*and

*y*are both lines through

*P*parallel to

*l*, then they are identical -- so there is really only

*one*line through

*P*parallel to

*l*. This isn't actually an indirect proof at all -- many proofs in higher math do this. To show that a number

*x*with a certain property is unique, we simply let

*x*and

*y*be two numbers with that property and then prove that

*x*=

*y*.

The resulting proof looks something like this:

Given:

*x*contains

*P*,

*y*contains

*P*,

*x*||

*l*,

*y*||

*l*

Prove:

*x*and

*y*are identical

Proof:

Statements Reasons

1.

*x*,

*y*contain

*P*1. Given

2. Let

*b*be line through

*P*perp. to

*l*2. Uniqueness of

*Perpendiculars*Theorem

3.

*b*perp. to

*x*,

*b*perp. to

*y*3. Fifth Postulate

4.

*b*intersects

*x*,

*y*at 90 4. Definition of perpendicular

5.

*x*intersects

*y*at 0 degrees 5. Angle Addition Property

6.

*x*and

*y*are identical 6. Zero Angle Property

Another, more interesting proof of Playfair depends on a theorem given in Section 3-4:

Transitivity of Parallelism Theorem:

In a plane, if

*l*||

*m*and

*m*||

*n*, then

*l*||

*n*.

When we covered Lesson 3-4, we skipped this theorem because it was proved using slope, and we don't want to discuss slope until second semester. But we can use our Fifth Postulate to prove the Transitivity of Parallelism Theorem, as follows:

Proof:

Draw any line

*b*perpendicular to

*m*. Then by our Fifth Postulate,

*b*must be perpendicular to both

*l*and

*n*as well. Therefore by the Two Perpendiculars Theorem,

*l*and

*n*are parallel. QED

And now notice that Playfair immediately follows from this. To show that there can be only one line parallel to

*m*through a point

*P*, we let

*l*and

*n*be two lines parallel to

*m*through a point

*P*. By the Transitivity of Parallelism Theorem,

*l*and

*n*are parallel to each other -- yet the lines both contain the same point

*P*. Then, by

*our*definition of parallel, the only way for two parallel lines to have a point in common is for them to be identical, so

*l*and

*n*are the same line. I like this proof better.

Once we have Playfair, we can now derive the Parallel

*Consequences*-- that is, statements of the form "if two parallel lines are cut by a transversal, then ...." Now I was considering deriving Corresponding Angles Consequence the same way that Dr. Franklin Mason does this -- but unfortunately the proof that he gives is indirect. So again we must go back to the trick used for Lesson 13-6 earlier:

Given:

*l*||

*m*with transversal

*t*, intersecting

*m*and

*P*

Prove: Corresponding angles 1 and 2 are equal in measure.

Proof:

Statements Reasons

1.

*l*||

*m*1. Given

2. Let

*n*intersect

*t*at

*P*, 2. Angle Measure Postulate

at corr. angle 3 = angle 1

3.

*l*||

*n*3. Corresponding Angles

*Test*

4.

*m*and

*n*are identical, 4. Uniqueness of Parallels Test (Playfair)

so angle 2 = angle 3

5. angle 1 = angle 2 5. Transitive Property of Equality

This may be easier for students to understand if we think of angle 1 having a specific measure, something like

*x*. We want to prove that angle 2 also has measure

*x*. To prove it, we let

*n*be the line through

*P*so that the angle between

*n*and

*t*already has the measure we want --

*x*. Then by the Corresponding Angles Test,

*n*is a line through

*P*parallel to

*l*. But we already have a line through

*P*parallel to

*l*-- namely

*m*. So by Playfair

*n*is just another name for

*m*. And since

*n*intersects

*t*at

*x*degrees and

*n*really means

*m*,

*m*intersects

*t*at

*x*degrees, which is what we want to prove.

As I mentioned before, this is a trick for proving converses. The Corresponding Angles Consequence is just the converse of the Corresponding Angles Test. The trick to proving converses is to combine the forward statement with a uniqueness criterion. And this is exactly what we did here -- we proved the Corresponding Angles Consequence by combining the Corresponding Angles Test with the Uniqueness of Parallels Theorem (Playfair).

The remaining statements are proved quickly. We prove the Alternate Interior Angles Consequence using the U of Chicago proof from Section 5-6 -- I included this in the exercises.

Then I prove the Same-Side Interior Angles Consequence -- this isn't proved in the U of Chicago text

*per se*, except as part of the Trapezoid Angle Theorem in Section 5-5. I decided to include both this theorem and the theorem about both pairs of base angles of an isosceles trapezoid being equal. But the next theorem, on the symmetry of an isosceles trapezoid, is saved for the next section.

The proofs in this theorem occur in Euclid: Proposition 29 being the Parallel Consequences and Proposition 30 being the Transitivity of Parallels Theorem. These are the first two theorems of Euclid requiring his Fifth Postulate.

With this new image, the worksheet I posted back in July where the Perpendicular to Parallels is given as a theorem is now obsolete.

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