Comment by layer8

Comment by layer8 4 days ago

2 replies

> This proof assumes that the area a triangle is some function k c^2 of the hypotenuse c where k is constant for similar triangles.

I would state it differently: Given the specific triangle, the ratio between the area of the square on the hypotenuse and the area of the triangle is some constant k that is invariant under scaling of the triangle.

This should be intuitively obvious: When you have a picture with some shapes in it, scaling the picture won’t change the relative proportions of the shapes in the picture. You don’t have to know the absolute dimensions of the picture to determine the area ratio of two shapes within the picture. The constant k above will be different for differently shaped triangles, but will be the same for triangles of the same shape (same angles).

So, for a given triangle T1 with hypotenuse c we have, for some k: area(T1) = k x c²

Now, we subdivide T1 into two smaller triangles T2 and T3 that have the property of both being a scaled version of T1, and their hypotenuses being the a and b of T1. Hence we have:

area(T2) = k x a²

area(T3) = k x b²

all with the same k, since they have the same shape (only differing by scaling).

Because we have

area(T1) = area(T2) + area(T3),

it follows that

k x c² = k x a² + k x b²

and since k ≠ 0, we get

c² = a² + b².

In pictures (as far as possible with ASCII art):

   _______
  |       |\   
  |   a²  | \  
  |       |T2\ 
  |       |.‘ 
   ¯¯¯¯¯¯¯

        +  

             .
           .T3\
          |¯¯¯¯|
          | b² |
          |____|
           

        =

                 / \
              /     \
           /         \
          |\    c²    \
          | \         /
          |T1\     /
          |   \ /
           ¯¯¯¯
(The three are supposed to be the same exact picture, just rotated and scaled, and T1 can be split into T2 and T3, hence c² must also be the sum of a² and b².)

The underlying “miracle” is that you can subdivide any right triangle into two smaller copies of itself. The Pythagorean theorem then follows immediately from that. This subdivision capability is something that might be amenable to some further underlying explanation.

layer8 4 days ago

P.S.: T2/T3 are also mirrored, not just scaled, from T1, but obviously that doesn’t change anything about their area.