@LostInParadise—My question about the Einstein one would be this: Is there not a way to split any triangle into two similar triangles, and so be able to describe any triangle in that way? I’ve been very roughly sketching them and it seems like I can (but I’m just eyeballing sketches, so maybe not). The line wouldn’t be perpendicular (it would be an angle particular to the original triangle), and there isn’t technically a hypotenuse (just the longest side), but if the three triangles are similar, then isn’t there still some k that relates a2, b2, and c^2 each to the area of their respective triangle? (I’m presuming this isn’t actually correct, so why not? How is it different for a right triangle compared to others?)
I have a related question with yours, @LuckyGuy—I’ll address it to whoever wants to answer: I was trying to play around with using parallelograms instead of rectangles. Basically, I seem to get negative space in the first version with 2 rhombuses, one with side lengths “a” and one with side lengths “b.” When I split the parallelograms and rearrange, I get a parallelogram with side lengths “c” and “d” (because the parallelograms are split along different diagonals). It seems I still get a relationship between the squares of the sides of the triangles. That is, that: a*(height of rhombus a) + b*(height of rhombus b) = c*(height of parallelogram with c as base) = d*(height of parallelogram with d as base).... but in the case of a right triangle, the height is the same as the side length and so things simplify down. Is that true? And if so, is there a more complex/complete version of the idea behind the Pythagorean equation that describes the relationship between the sides of any triangle, provided you know the relevant angle? (Or again, is this not actually correct? And why not?... if it is true, it does seem less immediately useful—would have to know more about the triangle—but would make the Pythagorean equation seem less like it’s out of nowhere).