# How would Euclid have proved that $a \times b = ((ab)/c) \times c$?

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: and like that: Then they are equal (in size).

• 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". Aug 29, 2018 at 13:07
• 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. Aug 29, 2018 at 13:09
• 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). Aug 29, 2018 at 13:10
• @EthanBolker: It may not be a likely construction for Euclid, but the statement comes quite natural. It's simple and seems provable. Aug 29, 2018 at 13:12
• 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 $$AP\cdot PB=CP\cdot PD\text{ or }a\cdot b=c\cdot PD\,.$$ Aug 29, 2018 at 15:03

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.