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How many ways are there to prove Pythagorean Theorem?

  1. Dec 21, 2003 #1
    This thread is intended just for fun.

    My favorite proof of the pythagorean theorem uses algebra, together with a very simple picture. A square is inscribed in another square, and then we use the fact that the whole area is equal to the sum of its parts. The resulting figure consists of an inner square surrounded by four right triangles of equal area.

    The reason I like this proof more than any other, is because of its utter simplicity. The only thing in algebra you need, is how to carry out the following product:

    [tex] (a+b)(a+b) [/tex]

    Using the axioms of algebra, you can conlude that

    [tex] (a+b)(a+b) = a^2 + 2ab + b^2 [/tex]

    If anyone is interested in the picture I am referring to, I can provide a link to a site which already has the picture.
  2. jcsd
  3. Dec 22, 2003 #2


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    Thanks for posting that one StarThrower. I didn't need the picture to work out the details of the proof. very neat. :-)

    Funny thing but I'd never even seen a proof of Pythagorus's Theorum before. I guess you learn that theorum so early at school and at that level you don't usual do proofs, so since then I's never really even thought much about it or looked for a proof.
  4. Dec 22, 2003 #3
    i heard that there is a book dedicated to this subject that contains over 200 proofs, including an original proof by president garfield.
  5. Dec 24, 2003 #4
    in this webpage it says:"Pythagorean theorem (which according to Wells has at least 367 proofs)."-http://www.ics.uci.edu/~eppstein/junkyard/euler/
  6. Dec 27, 2003 #5
    Every demonstration I know of the pythagorean relationship between the sides of a right triangle seems to be a variation of three basic demonstrations:

    (1) (a + b)2

    (2) (a - b)2

    (3) split c in two proportionally



    (1) A large square of sides a + b imbeds a smaller tilted square of sides c. This is the demonstration described by the initiator of this topic. It is a good candidate for being the original demonstration. The drawing was found in an ancient chinese manuscript.

    The demonstration attributed to US President Garfield uses a trapezoid with sides a and b and base a + b. But if you duplicate the trapezoid, invert the copy and fit the fourth sides together you get the same square a + b with the c square embedded. The area of the Garfield trapezoid is (a + b) * (a + b)/2, or (a + b)2/2. So this is clearly just a variation of the (a + b)2 demonstration.

    (2) Split two rectangles of area ab diagonally to form four identical right triangles. Arrange these four triangles into a square on the long side c. This leaves a hole of size (a - b)2. If you prefer, (b - a)2 yields the same thing, in case b is larger than a. in case a = b there is no square in the center. The triangles fit exactly and it is easy to show 2a2 = c2. Otherwise, the area c2 is equal to the area of two rectangles of area ab and the square hole of area (a - b)2. Set one rectangle with a on the bottom and b vertical. Set the other rectangle next to it with b on the bottom and a vertical. The (a - b)2 square can fit on top to form exactly a2 + b2. The nice thing about this demonstration is that there is no piece left over. Bertrand Russell included this demonstration in his Illustrated Story of Philosophy and Carl Sagan used this demonstration in his TV series Cosmos.

    (3) Draw a perpendicular line from the vertex of the right angle of a right triangle to the long side (hypotenuse). This splits its length c into two parts. There are two triangles similar to the original right triangle formed, so the sides are proportional. Suppose x is the part of c on the a side of the perpendicular and y is the part of c on the b side of the perpendicular. then:

    x + y = c
    x/a = a/c
    y/b = b/c

    . The conclusion a2 + b2 can be derived from these. This is the euclidean demonstration.

    It is even more geometrical to split the square c2 instead of the line c. Draw a square on the hypotenuse c opposite the right triangle. Draw the perpendicular all the way down to the far side of the square on c. This produces two rectangles out of the square. It is also easy to copy the right triangle off the far side of this square so its sides are parallel to the original. make two more copies by rotating this copy by a right angle clockwise around the left vertex and rotating the copy by a right angle counterclockwise by a right triangle. This shows that the vertical distance between the horizontal side of the first copy and the horizontal side of the original triangle is a and the horizontal distance between the verticle side of the first copy and the vertical side of the original triangle is b. Now replace the square on c by removing the copy of the right triangle inside the square with the original right triangle itself. This forms two parallelograms. The base of one parallelogram is a and its vertical distance is a; the base of the other parallelogram is b and its horizontal distance is b. So c2 = a2 + b2. To make it more fun, the parallelograms that replace the two rectangles of the square on c can be shifted into the original right triangle and rocked into the bases of squares drawn on the two short sides (legs) of the original right triangle. These have areas a2 and b2. It would make a lovely moving picture, shifting and rocking and shifting again.

    I consider this c2 split as just a variation of the splitting of c. Eric Temple Bell used this to demonstrate the pythagorean relationship in Mathematics the Queen and Servant of Science.

    If anyone knows a demonstration of the pythagorean relation that is essentially different from these three types, I would be pleased to see it. Trigonometric attempts would be nice, but one must avoid using anything that is already derived from the pythagorean relation (for example, sin2 + cos2 = 1).
  7. Dec 30, 2003 #6
    I think I've read somewhere that it can be proved that there are an infinity of proofs to the Pythagorean Theorem. Of course, I might be wrong. So, could anyone please verify that?
  8. Jan 3, 2004 #7
    I have never heard this claim before.
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