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:
    enter image description here

  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):
    enter image description here     enter image description here

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

  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.

enter image description here

(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:

enter image description here

  • 1
    $\begingroup$ Yes ${}{}{}{}{}{}{}{}{}{}{}{}{} $ $\endgroup$
    – fleablood
    Sep 26, 2018 at 16:01
  • $\begingroup$ We can prove that the rectangle with side lengths 1 and 2 has the same area as the square with side length $\sqrt{2}$, but we cannot check it by directly measuring the areas (because $\sqrt{2}$ is irrational = immeasurable). That's also the reason why we cannot check (by measurement) that the two rectangles in step 4 have the same area - except when $a,b,c$ are rational numbers. $\endgroup$ Sep 26, 2018 at 16:14
  • $\begingroup$ @fleablood: To which question(s) does your answer relate? (Or did you want to indicate politely that there are too many questions-in-one?) $\endgroup$ Sep 26, 2018 at 16:29
  • $\begingroup$ I was tongue in checkily saying your very long and exhaustive post boils down to is this equivalent to Euclids argument. And the answer is ... "yes". $\endgroup$
    – fleablood
    Sep 26, 2018 at 16:45
  • $\begingroup$ The main difference between Euclid's and my exposition might be: Euclid (in II.14) constructs $\sqrt{ab}$ while I (in step 4) construct just $ab$. The constructions might be comparably intricate, but conceptually $ab$ is easier than $\sqrt{ab}$. $\endgroup$ Sep 26, 2018 at 17:22

2 Answers 2


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

enter image description here

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

  • $\begingroup$ To construct this decomposition for an $a\times b$ rectangle: (1) draw a circle with diameter $a+b$, and the perpendicular chord cutting the diameter into $a$ and $b$ lengths; half the chord is the side length $c$ (2) draw a circle with diameter $b>a$, and a chord of length $c$ from one endpoint of that diameter; the diameter and the chord are two sides of the big right triangle in your diagram, and the rest follows. $\endgroup$ Feb 17, 2019 at 18:27

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.

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.

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.

  • $\begingroup$ Which "particular dissection [I] have in mind" do you have in mind? $\endgroup$ Sep 26, 2018 at 17:37
  • $\begingroup$ I don't know what you had in mind. From the comments, it seemed that incommensurate sides were a problem, so I thought perhaps you had thought of a way to deal with commensurate sides. I have revised the answer to acknowledge the fact that I do not know whether you considered one type of dissection, multiple types, or no specific types. In any case, the way we deal with incommensurate sides is the pieces in the dissection get rotated. $\endgroup$
    – David K
    Sep 26, 2018 at 23:05
  • $\begingroup$ Thanks a lot. My question boils down to: By which specific construction can I dissect the rectangle $a \times b$ such that the pieces cover the rectangle $c \times ab/c$ as constructed in step 4. Sorry, but that's not clear to me, not even after your valuable hint to proposition II.5 which I found difficult to relate to my problem. $\endgroup$ Sep 27, 2018 at 7:36
  • $\begingroup$ Do you believe that if there is a dissection that transforms figure $A$ to figure $B,$ and another that transforms figure $B$ to figure $C,$ there is a dissection from $A$ to $C$? If you do not accept this, or if you insist on a detailed drawing of the explicit dissection from $A$ to $C$ (rather than accepting one of $A$ to $B$ and another from $B$ to $C$), a general proof of the rectangle-to-rectangle dissection will be tedious, difficult, confusing, and (in my opinion) not worth attempting. $\endgroup$
    – David K
    Sep 27, 2018 at 12:50

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