> Do you have a good or favorite proof or way to see it intuitively???
> ... But still I dont "Grok" (Heinlein) it.

I thought about it last night, and came up with something
that might help.

Given an a,b,h triangle, we know that h is a function of a and b,
i.e. h=F(a,b).

Take the partial derivative with respect to a. Looking at
a diagram, the triangle with sides "delta h" and "delta a"
is, in the limit, similar to the a,b,h triangle, so that

dh/da = a/h

Multiplying both sides by h, we have

h dh/da =a

Integrating gives us

(1/2) h^2 = (1/2) a^2 + f(b)

When a=0, we have that h=b , so that f(b) must be (1/2) b^2.
Multiplying by 2 gives the Pythagorean identity:
h^2 = a^2 + b^2.

So, Pythagoras' identity can be viewed as the integral version
of a differential relationship involving similar triangles.

Incidentally, after writing the above I did a search to
see if the above proof has been published. Of course it
has! It's essentially number 40 in the list of 43 proofs at
<http://www.cut-the-knot.org/pythagoras/index.shtml>
It was published in 1988 in an issue of The Mathematical
Intelligencer.

All the best,

Walter

--- monkeyEinstein wrote:
> Hi Walter,
> This is Dave C. of the StonyBrook Aquatic team.
> I was looking at you connections on Pythagoras.
> Although I know how to "prove" a^2 + b^2= h^2 for a,b,h=hypoteneuse
> the sides of a rt. triangle, I still have a hard time seeing why in
> an intuitive way.
> Do you have a good or favorite proof or way to see it intuitively???
> I saw a cut and paste proof I liked in long ago in a book The Ascent
> of Man. But still I dont "Grok" (Heinlein) it.
>
> Maybe an algebraic proof would be better. Like...
> "Certaintly a+b=h (n=1) can't be true, so how about a^n + b^n = h^n.
> For n large, it certaintly cant be true.
> Now there is a right angle, which divides the plane into 4 equal
> regions...." I give up.
>
> And what about the generalization to non-right angles triangles?
> (Law of cosines I think)
> Have you "connected" that to anything?
>
> Got to go,
> Take Care,
> Dave