Comment by chronial
Draw a square around Einstein's face. Call the side length of the square a and the area of the square A. We have A=a^2. Einstein takes up some portion p < 1 of that area, so Einstein has area E = pA. Now we scale the whole thing by factor f. So the new square has side lengths fa, and thus area A' = (fa)^2 = f^2×a^2 = f^2×A. Since the relative portion the face takes up doesn't change with scaling, the face now has size pA' = p×f^2×A = f^2 × pA = f^2 E.
Does that help or was that not the part you were missing?
No, that part is fine: I'm happy with the fact that it works with arbitrary shapes. What bothers me is that the area on the hypotenuse is equal to the sum of the areas on the other two sides, when the triangle has a right angle.
This somewhat like saying that I'm troubled by the fact that 1+1=2, I know. But that's a potentially distracting sidetrack, let's not get into that one.