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Old May 20, 2003, 05:57   #1
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Geeeeometry (Help!)
Most of you are probably familiar with the Pythagorean Theorem. In a right triangle, a² + b² = c²

I need the most simple way to prove it. Does anybody know?
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Old May 20, 2003, 06:11   #2
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I know but i'm too lazy to write it in English.
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Old May 20, 2003, 06:12   #3
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But...

http://www.usna.edu/MathDept/mdm/pyth.html
http://www.emsl.pnl.gov:2080/docs/mathexpl/ptprove.html
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Old May 20, 2003, 06:20   #4
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Thanks Eli, I think the second link will be most helpful, I'll go have a look at it.

I just have to figure out what "perpendicular" means.
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Old May 20, 2003, 06:30   #5
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At right-angle to.
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Old May 20, 2003, 06:35   #6
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I think you can start bt the Cosine Rule and work it as a special case.

Cosine Rule: a2 = b2 + c2 - 2bc*cos A

A = 90°
cos(A) = 0

Q.E.D.

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Old May 20, 2003, 06:38   #7
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Thanks guys. Learning, learning.
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Old May 20, 2003, 06:42   #8
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if you can get your hands on the book 'Fermats Last Theorem' theres a (proper) proof in the appendics of that (my copy is at work)
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Old May 20, 2003, 07:19   #9
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I remember that once an Applet proving the Pythagorean Theorem won a Sun Contest, i was unable to find it, anyway the following seems to me well done too.

http://www.utc.edu/~cpmawata/geom/geom7.htm

Obviously you need a Java-enabled browser, just try the link, if it starts you have it

P.S. No, i found it, it's here:

http://www.math.ubc.ca/~morey/java/pyth/
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Old May 20, 2003, 07:36   #10
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Quote:
Originally posted by Urban Ranger
I think you can start bt the Cosine Rule and work it as a special case.
Yeah, but then you have to prove the Cosine Rule.

Although there is a trigonometrical proof somewhere...

I like Angelo's first one the best. I think that is the one Pythag himself used.
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Old May 20, 2003, 12:08   #11
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Stephen Hawkings did a good proof of it in the Brief History of Time.
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Old May 20, 2003, 12:15   #12
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Quote:
Originally posted by Immortal Wombat
Yeah, but then you have to prove the Cosine Rule.
Well, you could, but you're not asked to do it.
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Old May 20, 2003, 12:42   #13
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The area of the large square can be found in two ways:

1) Area of large square = (x+y)^2


2) Measure each element in the large square. The area of each triangle is 0.5xy. The area of the tilted square is z^2

Area of large square = 4*(area of each triangle)+ area of tilted square

= 4 (0.5xy)+z^2

1) and 2) give different expressions but must be equivilent because the represent the same area

so (x+y)^2 = 4(0.5xy)+z^2

= x^2+y^2+2xy = 2xy+z^2

= x^2+y^2+z^2 which is the theorem.
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Old May 20, 2003, 13:22   #14
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or you could just make a right triangle that's 3x4x5
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Old May 20, 2003, 13:27   #15
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something a humanities major would say.
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Old May 20, 2003, 14:52   #16
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Thats called an example orange, not a proof.
With that triangle, you could claim all right triangle have sides a,b,c such that a+b=c+2... which of course is not true
 
 

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