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Pythagorean Proof

  1. Mar 27, 2015 #1
    Hey,

    This is kind of silly but I think I found a very simple proof for the Pythagorean Theorem. I've been looking around sort of seeing if I could find it somewhere else, but haven't so far. This may not be the forum for this, but I was wondering if anyone had seen something like this before (it sort
    of looks like similar triangles but isn't). Sorry if this post is misplaced X| .
     

    Attached Files:

  2. jcsd
  3. Mar 27, 2015 #2
    How do you know to use a/c = cosine if that's the result you are trying to prove?
    Yes, you will reach the correct conclusion, but you are already using a property of that result, which happens to be correct, but.. how would you know that?
     
  4. Mar 27, 2015 #3
    I think from the second triangle you get there. Cosine is defined as x / r. So looking at the second triangle cos(alpha) = a / c. From the first (upper) triangle it doesn't look that way; you have to flip it over (which I did) and use a as the base rather than c1; this is done in the lower illustration.

    Maybe it would be easier if I wrote that as c * cos(alpha) = a
     
  5. Mar 27, 2015 #4
    Yes, I understand your illustration and everything is correct, but the proof itself is not convincing, because you prove a statement by using the result of the very same statement., namely the trig definition.
     
  6. Mar 27, 2015 #5
    How is the definition of the cosine a result of anything? Using cosines here is only a somewhat indirect way of using similar triangles.

    Instead of noting that cos(alpha) = a/c and cos(alpha) = c1/a you can get a/c = c1/a directly by noting that the triangle with a and c1 as sides is similar to the triangle with a, b and c as sides. (because two angles are the same). The same goes for b/c = c2/b.
    that gives you proof #6 from this set;
    http://www.cut-the-knot.org/pythagoras/
     
  7. Mar 27, 2015 #6

    HallsofIvy

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    It is not the definition of sine and cosine that depend upon the Pythagorean theorem, but the fact that [tex]sin^2(x)+ cos^2(x)= 1[/tex] does.
     
  8. Mar 27, 2015 #7
    But I don't see where this is used in the proof.
     
  9. Mar 28, 2015 #8

    epenguin

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    I think that is OK, congratulations! :approve:

    Though you are not the first to discover it. I for example have discovered it several times! :oldbiggrin: And I was not the first either. My old schoolteacher certainly knew it, as I remembered after discovering it.

    It is essentially "Proof by similar triangles".

    You do not need to mention cosines (which only exist because similar triangles do). Just say a/c = c/a etc.

    I think it is the simplest proof around. Or rather the easiest, psychologically. Because behind it are some theorems about similar triangles (that feel, or have become, obvious) so I don't know it's logically simpler than some other proof if you really take them from the start.

    But you can feel pleased with yourself.
     
  10. Mar 28, 2015 #9
    This is great!
     
  11. Mar 30, 2015 #10
    yes it is essentially proof by similar triangles. For example you name the ratio c1/a as cos(alpha) and later you say its cos(alpha) is equal to a/c. But we can get this equality c1/a=a/c directly by the similarity of triangles without the intermediate naming. It is because orthogonal triangles are similar as long as they have one of their other angles equal and because cosine (as well as sine) of an angle alpha is uniquely defined in all the orthogonal triangles that have this angle alpha and thats because all those orthogonal triangles are similar!

    I have to say though this proof is a smart way to avoid mentioning similarity of triangles and mention cosine of an angle instead, however as i said above , it is the similarity of triangles that constitutes "our hidden basement" upon which we build the definition of cosine (and sine) of an angle.
     
    Last edited: Mar 30, 2015
  12. Mar 30, 2015 #11
    Sorry I'm so late to respond. I see that since I use a cosine of a certain angle twice (either alpha or beta) I end up needing to assume similar triangles for the cosine ratio to be constant in both uses. I'm reading it now but there looks to be a concrete proof of similar triangles rather than raw intuition (postulate) such that the logic of similar triangles is more fundamental than the intuition (what would be a postulate) behind a cosine being constant at all distances.

    Thanks for the responses by the way.
     
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