The other day a friend of mine who needed help studying for a math placement test for college asked me why a^2+b^2=c^2 for a right triangle. I said pythagorean theorem. But then I realized I don't actaully know why this is true. Is there actaully a proof for this that I either totally forgot about or never heard?

i will tell you the proof that i learned in 8th grade, although as i understand it, there are hundreds of proofs.

drop an altitude from the right angle to the hypotenuse. let D be the point where the altitude meets the hypotenuse, h the height of the altitude, and x the length of DB. triangle ABC is similar to CBD is similar to ACD.

from ABC similar to ACD, we get AB/AC=CA/DA or c/b=b/x or b²=cx. from ABC similar to CBD, we get AB/CB=BC/BD or c/a=a/c-x, or a²=c²-cx.

put the two equations together, and you have a² + b² = c²

Try this one:

You have a square... cut it into 4 equal triangles with a smaller square in the center.

The triangles are along the edges. The legs are length "a" and "b" with the length of the hypont. being "c."

The total area of the big square is (a + b)^2.
The area of the smaller square is c^2
The area of one triangle is 1/2*a*b

The total area of the big square is the addition of the smaller square plus the combined areas of the triangles

(a + b)^2 = c^2 + 4(1/2*a*b)
a^2 + b^2 + 2ab = c^2 + 2ab
a^2 + b^2 = c^2
QED

Originally posted by lethe
i will tell you the proof that i learned in 8th grade, although as i understand it, there are hundreds of proofs.

drop an altitude from the right angle to the hypotenuse. let D be the point where the altitude meets the hypotenuse, h the height of the altitude, and x the length of DB. triangle ABC is similar to CBD is similar to ACD.

from ABC similar to ACD, we get AB/AC=CA/DA or c/b=b/x or b²=cx. from ABC similar to CBD, we get AB/CB=BC/BD or c/a=a/c-x, or a²=c²-cx.

put the two equations together, and you have a² + b² = c²

haha, in 8th grade and hundreds of proofs...thats funny--makes me feel like an idiot, oh well. I probably was told one in 8th grade too, but i probably wasnt listening...anyways thank you for answering my simple question, i really probably should have tried to figure out a proof before i asked the question, but i was being lazy...

since it seems it was an easy question it would be interesting to see how many proofs we could all come up with, and whether or not many of them are the same thing in a different form...

No thanks to me? Well... :rolleyes:

Originally posted by cephas1012
since it seems it was an easy question it would be interesting to see how many proofs we could all come up with, and whether or not many of them are the same thing in a different form...

I am sure much of what we can think of is already done.
http://www.cut-the-knot.org/pythagoras/index.shtml
Maybe we have some very creative people here on SciForums...

In our Maths class we were shown how to derive the Pythagorean Theorem using algebra. It goes something like this...

Take any two vectors in Rn, u and v.

The length of u (also called the norm of u) is defined as ||u|| = sqrt(u . u)

Given u and v in Rn we define their length to be...

||u|| = sqrt(u . u)

If u = v then the distance apart is obviously 0.

We can take this one step further and find the length of any vector. We generally don't like using square roots so lets square the whole thing...
Remember ||u||^2 = (sqrt(u . u))^2 = (u . u)

||u + v||^2 = ||u + v||^2 = (u + v).(u + v)
||u + v||^2 = u . (u + v) + v . (u + v)
||u + v||^2 = u . u + u . v + v . u + v . v expanding the equation (which we can by the properties of the inner product)
||u + v||^2 = ||u||^2 + ||v||^2 + 2(u . v) (Cauchy-Schwartz Inequality)

Now what we have here is the length of the vector u + v which is constructed using the parallelogram law from u and v separately.

Now let u = a, v = b, and u + v = c

c^2 = a^2 + b^2 + 2(a.b)
c^2 = a^2 + b^2 + 2abcos(q)
This is how the cosine rule came about!

Notice this is useful for non-right angle triangles (ie. where u and v are not perpendicular). So what would happen if u and v were perpendicular??? Well the cos(q) = cos(90) = 0

So c^2 = a^2 + b^2 where c =a (the hypotenuse) and b = v and a = u (the other two sides).

Here is my favorite proof (Devised by Martin Gardner, found at Mathworld (http://mathworld.wolfram.com/PythagoreanTheorem.html)):
http://mathworld.wolfram.com/p3img2448.gif

The pythagorean theorem is equivalent to saying that the area of the large square is equal to the sum of the area of the smaller squares. This is easily proven by shearing the large square area a few times. In the diagram, the grey area is always constant (a shear operation maintains constant area).

(the translation between diagrams 2 & 3 isn't immediately obvious, until you rotate the triangle 90° clockwise around its topmost vertex to see how the lengths add up)

Wow.. nice proof. Thank you Pete.

There are probably more different proofs of the Pythagorean Theorem than any other theorem!

Here is a link to a number of them, including one originally found by President James Garfield (It was his assassination that made Theodore Roosevelt president).