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The following statement seems too obvious to prove or even to mention:

That the area of a rectangle with side lengths $a$ and $b$ equals the area of a rectangle with side lengths $(ab)/c$ and $c$ for arbitrary $c>0$, i.e.

$$ a \times b = ((ab)/c) \times c$$

(Note that $(ab)/c$ is a shorthand for $a/(c/b)$.)

I wonder how this statement would have been formulated by Euclid and how he would have proved it (as a geometrical statement with geometrical methods).

I guess there is no proposition in Euclid's Elements that says exactly this, but maybe there is one which says essentially the same - or there is an easy proof starting from theorems already proved by Euclid.

This is how how I would formulate the statement:

Consider two figures constructed like this:

enter image description here

and like that:

enter image description here

Then they are equal (in size).

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  • $\begingroup$ Basically, in Euclid's geometry a rectangle is defined by two sides : $a$ and $b$ (following your example). Thus (IMO) the "algebraic" expression $((a \times b) / c) \times c$ does not makes sense because the "division" of a rectangle $(a \times b)$ by a side $c$ has no "geomteric interpretation". In conclusion, I agree with your assertion that in Euclid's Elements we cannot find a proposition "that says essentially the same". $\endgroup$
    – Mauro ALLEGRANZA
    Commented Aug 29, 2018 at 13:07
  • $\begingroup$ I doubt that Eucid proved exactly that as you formulate it, nor is that a likely construction for him. See mathcs.clarku.edu/~djoyce/java/elements/bookII/propII6.html for the flavor of Euclid's treatment of multiplication. $\endgroup$
    – Ethan Bolker
    Commented Aug 29, 2018 at 13:09
  • $\begingroup$ I consciously wrote $((ab)/c) \times c$ instead of $((a \times b)/c) \times c$ and declared that $(ab)/c$ is a shorthand for another number created by the second construction (two divisions). $\endgroup$
    – Hans-Peter Stricker
    Commented Aug 29, 2018 at 13:10
  • $\begingroup$ @EthanBolker: It may not be a likely construction for Euclid, but the statement comes quite natural. It's simple and seems provable. $\endgroup$
    – Hans-Peter Stricker
    Commented Aug 29, 2018 at 13:12
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    $\begingroup$ Not sure if this can help, but consider a line segment $AB$ of length $a+b$ with a point $P$ inside such that $AP=a$ and $PB=b$. Let $CP$ be an arbitrary line segment such that $C$ is not on the same line as $A,P,B$ and $CP=c$. Draw a circle $\Gamma$ passing through $A$, $B$, and $C$. Extend $CP$ to meet $\Gamma$ again at $D$. We then have by the Power-of-Point Theorem that $$AP\cdot PB=CP\cdot PD\text{ or }a\cdot b=c\cdot PD\,.$$ $\endgroup$
    – Batominovski
    Commented Aug 29, 2018 at 15:03

1 Answer 1

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Two thoughts:

Euclid's Proposition II.14 is "to construct a square equal to a given rectilinear figure."

Most of the proof consists of showing that any rectangle is equal in area to a square. This follows from the Pythagorean Theorem and an earlier result in the chapter (Proposition II.5) which decomposes a rectangle as the difference of two squares.

It immediately follows that given any rectangle $R$ and any segment $AB$, you can construct another rectangle with side $AB$ and area equal to $R$. I'm not sure how easy it is to see that this rectangle must be the same one as in your construction, though.

Alternately, you could check out the treatment of similarity and area in Hilbert (chapters III and IV). Euclid would have been comfortable with all his constructions (if perhaps not with their interpretations), and he makes extensive use of exactly the sort of parallel-line construction you are using to define multiplication of line segments. I don't think he proves exactly your theorem, but he proves things which I'm sure are equivalent to it with a little work. (In particular, he rigorously proves that area is well-defined, which is sufficient to fill the gap I mentioned in the proof following Euclid — though possibly overkill.)

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  • $\begingroup$ Wonderful answer, thanks a lot! (Maybe Proposition II.14 is what I was looking for.) $\endgroup$
    – Hans-Peter Stricker
    Commented Aug 29, 2018 at 14:51

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