An algorithm to check if two polygons have the same area

Question

With Euclid's propositions I.45 (constructing a rectangle equal to a given polygon) and II.14 (constructing a square equal to a given rectangle) one can reduce the comparison of areas of polygons (which is not a basic operation) to the comparison of lengths (which is a basic operation):

Two polygons have the same area if the squares constructed by I.45 and II.14 have the same side length.

But the constructions described in I.45 and II.14 are quite intricate as are the corresponding proofs that the resulting polygons (the rectangle and the square) have the same area as the polygons from which they are constructed.

I wonder if the following recipe to check if the areas of two arbitrary polygons are the same might be more intuitive, reducing to a minimum the "miracle" that two polygons (one constructed from the other) have the same area (which we have to believe by proof but cannot check in general).

The recipe goes like this:

1. Decompose the polygons to be compared into arbitrary many arbitrary triangles.

2. For each triangle build the parallelogram of twice the area of the triangle:
   

3. For each parallelogram build a rectangle of the same area by decomposing and rearranging it (which can always be done in a systematic way and is probably a special case of proposition I.45):
       

4. For each rectangle with sides $a,b$ build another rectangle of the same area with common height $c$ by this construction:
   

5. Join the rectangles created from all triangles, thus building a "long" rectangle of height $c$.

6. The initial polygons have the same area, when their summed up rectangles of height $c$ have the same length.

Some "miracle" occurs in step 4: It cannot be checked in general - by decomposing and rearranging the rectangles in finitely many steps - that the two rectangles have the same area. But it can be proved from first principles.

(I don't know if it's astonishing that the result doesn't depend on the initial triangulization (step 1). We take it for granted, but I wonder if it needs a proof on its own.)

My question is:

Is this recipe essentially the same as Euclid's constructions I.45 and II.14 (in disguise)? Has it drawbacks? Is it of any educational value? Has it been proposed before (or is even standard)? Are there even simpler ones?

Furthermore: Did I oversee how it can be checked that the two rectangles in step 4 do have the same area? "Checked" opposed to "proved", i.e. by making them commensurable by the same set of covering pieces, tangram-like:

Answer 1

I've found this "proof without words" that any rectangle is equicomposable with a square:

(The question remains open to find this decomposition for a given rectangle. It's about constructing one specific angle - but how?)

Answer 2

It may be that the dissections (if any) that you have considered will not work for irrational ratios, but that does not mean there is no dissection that works.

The dissection implicit in Euclid II.5, plus a dissection that proves the Pythagorean Theorem, together give you enough tools to dissect any rectangle into a square of the same area.

Having done that, you can use the same method to dissect the square into any other rectangle of the same area. So it is possible to go from an arbitrary rectangle to another arbitrary rectangle of the same area.

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Some cases are easier than others. In the example of a rectangle of sides $1$ and $2,$ take a square of side $\sqrt 2$ and cut it along both diagonals. You now have four isosceles right triangles of leg length $1.$ Join two triangles along their hypotenuses, and you now have a square of side $1.$ Do this again with the other two triangles and place the squares next to each other to make the desired rectangle.

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The more general question of comparing polygons seems more complicated, since you need to get triangles of equal area before you can apply the rectangle dissection.