Comment by thaumasiotes

> This doesn’t seem super obvious to me, and it’s a bit more than just assuming area scales with the square of hypotenuse length, it indeed needs to be a constant fraction.

The second half of your sentence is not correct; if area scales with the square of any one-dimensional measurement (including hypotenuse length, because the hypotenuse is one-dimensional), that is sufficient to prove the theorem.

The statement you're looking for is: "triangle A is similar to triangle C with a length ratio of a/c, therefore the area of triangle A is equal to the area of triangle C multiplied by the square of that ratio".

It is in fact necessary that the area will scale with the square of hypotenuse length, because the hypotenuse is one-dimensional and area is two-dimensional. If you decided to measure the area of the circle that runs through the three corners of the triangle, the triangle's area would scale linearly with that.

It isn't clear to me what scenario you're thinking might mess with the proof.

> 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.

So, for similar shapes, you can set your own measurements.

1. Say I have two triangles X and Y and they're similar. I take a straightedge, mark off the length of the longest side (x) of triangle X, and say "this length is 1". Then I calculate the area of triangle X. It will be something. Call it k.

2. Now I take a second straightedge, mark off the length of the longest side (y) of triangle Y, and I label that length "1". I can calculate the area of triangle Y and, by definition, it must be k. But it is equal to k using a scale that differs from the scale I used to measure triangle X.

3. We can ask what the area of triangle Y would be if I measured it using the ruler marked in "x"es instead of the one marked in "y"s. This is easier if we have the same area in a shape that's easier to measure. So construct a square, using the "y" ruler, with area equal to k.

4. Now measure that square with the "x" ruler. The side length, measured in y units, is √k. Measured in our new x units, it's (y/x)√k. When we square that, we find that the x-normed area is equal to... k(y/x)².

This is why it's obvious that k must be constant for similar triangles. k is just a name for the scale-free representation of whatever it is that you're measuring. It has to be constant because, when you change the scaling that you use to label a shape, the shape itself doesn't change. And that's what similarity means.