Wednesday, January 21, 2015

The Pythagorean Theorem and the Distance Formula (Day 96)

Section 8-7 of the U of Chicago text is on the Pythagorean Theorem, and Section 11-2 of the same text is on the Distance Formula. I explained yesterday that I will cover these two related theorems in this lesson.

The Pythagorean Theorem is, of course, one of the most famous mathematical theorems. It is usually the first theorem that a student learns that is named for a person -- the famous Greek mathematician Pythagoras, who lived about 2500 years ago -- a few centuries before Euclid. I believe that the only other named theorem in the text is the Cavalieri Principle -- named after an Italian mathematician from 400 years ago. Perhaps the best known named theorem is Fermat's Last Theorem -- named after the same mathematician Fermat mentioned in yesterday's worksheet. (We discuss some other mathematicians such as Euclid and Descartes, but not their theorems.)

It's known that Pythagoras was not the only person who knew of his named theorem. The ancient Babylonians and Chinese knew of the theorem, and it's possible that the Egyptians at least knew about the 3-4-5 case.

We begin with the proof of the Pythagorean Theorem -- but which one? One of my favorite math websites, Cut the Knot (previously mentioned on this blog), gives over a hundred proofs of Pythagoras:

The only other theorem with many known proofs is Gauss's Law of Quadratic Reciprocity. Here is a discussion of some of the first few proofs:

Proof #1 is Euclid's own proof, his Proposition I.47. Proof #2 is simple enough, but rarely seen. Proofs #3 and #4 both appear in the U of Chicago, Section 8-7 -- one is given as the main proof and the other appears in the exercises. Proof #5 is the presidential proof -- it was first proposed by James Garfield, the twentieth President of the United States. I've once seen a text where the high school students were expected to reproduce Garfield's proof.

So far, the first five proofs all involve area. My favorite area-based proof is actually Proof #9. I've tutored students where I've shown them this version of the proof. Just as the Cut the Knot page points out, Proof #9 "makes the algebraic part of proof #4 completely redundant" -- and because it doesn't require the students to know any area formulas at all (save that of the square), I could give this proof right now. In fact, I was considering including Proof #9 on today's worksheet. Instead, I will wait until our next activity day on Friday to post it.

But it's the proof by similarity, Proof #6, that's endorsed by Common Core. This proof has its own page:

Here is Proof #6 below. The only difference between my proof and #6 from the Cut the Knot webpage is that I switched points A and C, so that the right angle is at C. This fits the usual notation that c, the side opposite C, is the hypotenuse.

Given: ACB and ADC are right angles.
Prove: BC * BC + AC * AC = ABAB (that is, a^2 + b^2 = c^2)

Statements                                Reasons
1. ADC, ACBCDB rt. angles   1. Given
2. Angle A = A, Angle B = B     2. Reflexive Property of Congruence
3. ADC, ACB, CDB sim. tri.     3. AA Similarity Theorem
4. AC/AB = AD/AC,                 4. Corresponding sides are in proportion.
    BC/AB = BD/BC
5. AC * AC = AB * AD,           5. Multiplication Property of Equality
    BC * BC = AB * BD
6. BC * BC + AC * AC =         6. Addition Property of Equality
    AB * BD + AB * AD
7. BC * BC + AC * AC =         7. Distributive Property
    AB * (BD + AD)
8. BC * BC + AC * AC =         8. Betweenness Theorem (Segment Addition)
    AB * AB

I mentioned before that, like many converses, the Converse of the Pythagorean Theorem is proved using the forward theorem plus a uniqueness theorem -- and the correct uniqueness theorem happens to be the SSS Congruence Theorem (i.e., up to isometry, there is at most one triangle given three sidelengths). To prove this, given a triangle with lengths a^2 + b^2 = c^2 we take another triangle with legs a and b, and we're given a right angle between a and b. By the forward Pythagorean Theorem, if the hypotenuse of the new triangle is z, then a^2 + b^2 = z^2. (I chose z following the U of Chicago proof.) Then z^2 = c^2 by transitivity -- that is, z = c. So all three pairs of both triangles are congruent -- SSS. Then by CPCTC, the original triangle has an angle congruent to the given right angle -- so it's a right triangle. QED

Interestingly enough, there's yet another link at Proof #6 at Cut the Knot, "Lipogrammatic Proof of the Pythagorean Theorem." At that link, not only is Proof #6 remodified so that it's also an area proof (just like Proofs #1-5), but, as its author points out, slope is well-defined without referring to similar triangles!

The Common Core Standards only require that the Pythagorean Theorem be proved using similarity -- not the concept of slope (which we'll cover here on the blog next week). So now I'm wondering whether it might be easier for the students to understand this derivation of slope.

It was David Joyce who pointed out that slope requires similarity to prove. He also criticized the area-based proof of the Pythagorean Theorem given in the Prentice-Hall text -- but this is because he wanted the Parallel Tests and Consequences to be taught first. (According to Cut the Knot, the Pythagorean Theorem is equivalent to the Parallel Postulate.)

For now, I think I'll stick to my plan to use similarity to prove slope -- but my proof may still be based on the one given at the above link.

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