Comment by aaplok

Comment by aaplok 4 days ago

You got a bunch of responses already, here is an intuitive reason.

In similar triangles all distances are scaled by a factor k, by definition. Then, intuitively the areas are scaled by a factor of k^2, since you obtain an area by multiplying two distances.

So the ratio of the area over the hypothenuse is scaled by a factor of k^2/k=k.

It is not hard to confirm the intuition that the areas are scaled by a factor of k^2, since it is precisely the product of the lengths of the two sides adjacent to the right angle.

xeonmc 4 days ago

I think it would be clearer with an explicit function of a rectangle's area with respect to its diagonals:

                   A = x y
                   x = r cos(θ)
                   y = r sin(θ)
             => A(r) = r²cos(θ)sin(θ)
                A(c) = A(a) + A(b)
    = c²cos(θ)sin(θ) = a²cos(θ)sin(θ) + b²cos(θ)sin(θ)
               => c² = a² + b²

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