Prove Pythagoras theorem through dimensional analysis I've recently become acquainted with Buckingham's Pi theorem for the first time . Then I've found an excercise that says:
Use dimensional analysis to prove the Pythagoras theorem. [Hint: Drop a perpendicular to the hypotenuse of a right-angle triangle and consider the resulting similar triangles.]
Any ideas? Thanks.
 A: 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.]
A: I recently came across a proof of Pythagoras Theorem via dimensional analysis in a book by Paul J Nahin called Mrs. Perkins's Electric Quilt, which goes as follows.
Let there be a right triangle with sides $a, b, c$, with $c$ the hypotenuse and let $\phi$ be one acute angle of this triangle. Since area has dimensions of length squared, and given $c, \phi$ we can uniquely determine the triangle, the area must be $$\Delta = c^2 f(\phi)$$
Now, drop the altitude on $c$. We get two right triangles with hypotenuses $a, b$ and one acute angle $\phi$. Hence, there areas are $$\Delta_1 = a^2 f(\phi) \qquad \Delta_2 = b^2 f(\phi)$$
But, $$\Delta = \Delta_1 + \Delta_2 \implies c^2 f(\phi) = a^2 f(\phi) + b^2 f(\phi)$$
$$\implies \boxed{c^2 = a^2 + b^2}$$
