Prove Pythagoras theorem through dimensional analysis

Here is one formulation of this argument; it is the same as the one suggested by user8268 in the above comments, but perhaps this formulation will make it clearer why this is a proof by dimensional analysis:

• You want to prove that the sum of the squares on each of the non-hypotenuse sides equals the square on the hypotenuse.

• You generalize, and instead prove that for any shape, if you scale it by $a$, and then by $b$, the sum of the resulting areas is the area of the shape scaled by $c$. (We began with the case of the unit square.)

• By thinking about how areas scale, it suffices to check for one particular shape.

• We check it by taking the shape to be the original triangle (to be pedantic: scaled so that its hypotenuse has length one). This case is clear: just drop a perpendicular from the vertex opposite the hypotenuse to the hypotenuse, and see note that the triangle with hypotenuse length $c$ is the sum of two similar triangle of hypotenuse lengths $a$ and $b$.

The dimensional analysis is in the third step. The point is in the final equality that is proved, i.e. in the final proof of $a^2 + b^2 = c^2$, these quantities are not the area of any particular shape, but rather are the scaling factors for the areas of the original triangle after scaling its lengths by $a$, $b$, and $c$. This is why it is a proof by dimensional analysis.

[I originally posted this here, and you can see the comments there for some historical background on this particular argument.]