Is Pythagoras' theorem about distances or areas? In $\mathbb{R}^2$ with the 1-norm or $\infty$-norm, Pythagoras' theorem is false for lengths of sides of a "right-angled'' triangle, but it is true for areas of shapes on the sides. For example, given a triangle with coordinates $(0,0)$, $(4,0)$, $(0,3)$, the sides have length 3,4,7, or 3,4,4 (depending on the norm); but the areas of squares on the sides are still 9,16,25 because Lebesgue measure is independent of the norm.
Is there a relation between measures and norms that are "compatible'' to them? In this case, Lebesgue measure seems to fit more naturally with Euclidean distance because the area of a rectangle, for example, is proportional to the product of the lengths of its sides, but I don't know if there is a simple formula in terms of the 1-lengths of its sides. If there is a theory on this it doesn't get mentioned in textbooks.
 A: You're right that the Pythagorean theorem is naturally a statement about areas. Euclid's proof involves geometric dissection, and there's the slick proof by dropping a perpendicular from the right angle to the hypotenuse, which divides the original triangle into two smaller similar triangles, whose areas obviously sum to the area of the original triangle, and whose respective hypotenuses are the legs of the original triangle.

There does not exist a formula for the area of a rectangle in terms of the lengths of its sides in the $1$-norm, say, because the lengths of the sides do not uniquely determine the area (i.e., the product of the $2$-norms of the sides).
For example, if $a$ and $b$ are positive, the square with corners
$$
(a, 0),\quad
(a + b, a),\quad
(b, a + b),\quad
(0, b)
$$
has sides of ($1$-norm) length $s = a + b$. Its area,
$$
A = a^{2} + b^{2} = (a + b)^{2} - 2ab = s^{2}  - 2ab,
$$
is not uniquely specified by $s$.
(I suppose one could rotate a rectangle to be axis-oriented before using the side lengths to compute area, but Euclidean rotation doesn't preserve non-Euclidean norms, so the process seems artificial.)
A: 
Answer: Pythagoras' theorem is all about distances and symmetries
  in the Euclidean Plane.

Setting: You never studied Euclidean Plane Geometry. Instead, you are looking at $\mathbb{R}^2$ as a vector space, and you want it to correspond to what happens when points, lines, triangles and circles are constructed on a piece of graph paper.
You have linear equations but your problem is finding the $\mathbb{R}^2$ equation for circles.
Response: You want to create a vector length formula that aligns to measuring lengths with a ruler on graph paper. The tick mark of $1$ on the ruler matches up to coordinates $(1,0)$ on the $x\text{-axis}$ on your graph paper. You also purchase a compass at a local art store.
False Start: You come up with the idea of the $1\text{-norm}$, but as soon as you graph the vectors of distance $1$ from the $0$ vector you throw that idea away (the $1\text{-norm}$ 'circle' doesn't match up with the compass drawing).
You are certain that the triangle inequality must hold:
$\|\mathbf {v} +\mathbf {w} \|\leq \|\mathbf {v} \|+\|\mathbf {w} \|$
Or, setting,  $\mathbf {u} = \mathbf {v} +\mathbf {w}$, and applying the binomial theorem,
$\tag 1 \|\mathbf {u}\|^2 \leq \|\mathbf {v} \|^2+\|\mathbf {w} \|^2 + 2 \|\mathbf {v} \| \|\mathbf {w} \| $
You draw some figures of vector tip-to-tail addition,  
imagining the $\mathbf w$ vector moving continuously from the 'same direction' of angle $0$ with $\mathbf v$ to the negative opposite angle of 180 degrees. This gets you excited - progress might be made by adapting properties of the absolute value (in a $\text{ 1-dim }$ space) over to your plane geometry, so you write:
$\tag 2 \|\mathbf {v} \|^2+\|\mathbf {w} \|^2 - 2 \|\mathbf {v} \| \|\mathbf {w} \| \le \|\mathbf {u}\|^2 \leq \|\mathbf {v} \|^2+\|\mathbf {w} \|^2 + 2 \|\mathbf {v} \| \|\mathbf {w} \| $
But you can write (2) this way:
$\tag 3 \|\mathbf {u}\|^2 = \|\mathbf {v} \|^2+\|\mathbf {w} \|^2 -  2 \|\mathbf {v} \| \|\mathbf {w} \| \, \tau $
where $\tau$ varies continuously between $-1$ and $+1$ as the angle that $\mathbf {w}$ makes with $\mathbf {v}$ varies.
Your favorite angle is 90 degrees, and this is the 'midpoint' of vector $\mathbf {w}$'s traversal journey. Is it possible that the unknown decreasing function $\tau$  'should' hit the midpoint of its range, $0$, at this angle?
Now using the norm formula 
$\|\mathbf {u}\| = \sqrt {x^2 + y^2}$ where $\mathbf {u} = (x,y)$,
the graph the circle $x^2 + y^2 = 1$ is exactly what you wanted to see; you've found the Euclidean norm for $\mathbb{R}^2$!
As your theory develops without any hiccups, you can only smile when you find the work of John Molokach on the Law of Cosines and learn about the Polarization identity and its relation to the Law of Cosines.
Another way to get the Euclidean Plane modeled with $\mathbb{R}^2$ coordinates is possible with some help from Euclid - see True Calculus Proof of the Pythagorean Theorem /John Molokach.
A: The Pythagoren theorem doesn't state that for any right angle triangle, the area of a square with one of its edges being the hypotenuse is the sum of the squares of the lengths of the legs. The Pythagorean theorem states that for any right angle triangle, the square of the length of the hypotenuse is the sum of the squares of the lengths of its legs. Some people make the assumption that the area of any square is the square of the length of its edges so they accept a proof like the one in this answer as a proof of the Pythagorean theorem.
The distance formula can be considered a binary function from $\mathbb{R}^2$ to $\mathbb{R}$, in otherwords, a function from $(\mathbb{R}^2)^2$. Most of those who accept that proof of the Pythagorean theorem are probably actually making an additional assumption that $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$ Without the additional assumption, that just shows that the distance formula is $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$ going from any point to any different point and therefore that the Pythagoren theorem holds for all right angle triangles whose axes are parallel to the axes. That's because area can be defined using Calculus. With the additional assumption, you can then show that the Pythagorean theorem holds for all right angle triangles. To show that the distance formula is that even going from a point to itself, we have to make the additional assumption that the distance from any point to itself is 0.
How do we know a function satisfying those properties exists? Because if you define the distance formula to be $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$, you can in fact show that it does satisfies all 3 properties. It's trivial to show that using that definition, the distance from any point to itself is 0. That so called proof of the Pythagorean theorem shows that using that definition of distance, the area of any square is in fact the square of the length of its edges. That definition can also be shown to satisfy $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$ as follows.
$\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R} d((0, 0), (xz - yw, xw + yz)) = \sqrt{(xz - yw)^2 + (xw + yz)^2} = \sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = \sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = \sqrt{(x^2 + y^2)(z^2 + w^2)} = \sqrt{x^2 + y^2}\sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$
You may instead be thinking that you cannot assume that the area of any square is the square of the length of its edges and can only assume that the distance formula satisfies the following properties:


*

*$\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$

*$\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((x, y), (z, w))$ is nonnegative

*$\forall \text{ nonnegative } x \in \mathbb{R}d((0, 0), (x, 0)) = x$

*$\forall x \in \mathbb{R}\forall y \in \mathbb{R}d((0, 0), (x, -y)) = d((0, 0), (x, y))$

*$\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$
but you still saw a proof of the Pythagoren theorem like the one in this answer and that's why you wondered if the Pythagorean theorem was just a theorem about area and not a theorem about distance. It really is a theorem about distance and some people accept that proof because they make the assumption that the area of any square is the square of the length of its edges.
However if you use just those assumptions about distance, you can still show that the distance formula is $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$ and therefore that the area of any square is the square of the length of its edges. How do we know there actually exists a way of defining distance that satisfies all 5 of those properties? Because it's trivial to show that the function that satisfies the first 4 properties it has been shown earlier that that function also satisfies property 5. Now we know that since it satisfies those properties, it also satisfies the property that the area of any square is the square of the length of its edges.
There's still one more property of distance that some people find so intuitive that I never mentioned earlier. That's that $\forall x \in \mathbb{R}d((0, 0), (\cos(x), \sin(x))) = 1$. How do we know that function also satisfies that additional property? Because it can be proven as follows. $\cos$ and $\sin$ can be defined by the following differential equations:


*

*$\cos(0) = 1$

*$\sin(0) = 0$

*$\cos' = -\sin$

*$\sin' = \cos$
$\frac{d}{dx}(\cos^2(x) + \sin^2(x)) = \frac{d}{dx}(\cos^2(x)) + \frac{d}{dx}(\sin^2(x)) = 2\cos(x)(-\sin(x)) + 2\sin(x)\cos(x) = 0$ This shows that $\cos^2(x) + \sin^2(x)$ is constant. Also $\cos^2(x) + \sin^2(x) = 1$ so $\forall x \in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$. Now using the distance formula, we can show that $\forall x \in \mathbb{R}d((0, 0), (\cos(x), \sin(x))) = 1$.
