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I’ve been going through Euclid Book I and I’ve got to proposition 45. This is a construction that allows the area of a rectilineal figure to be converted into a parallelogram with given angle and side. Reading Heath’s Thirteen books of Euclid’s elements I came across this paragraph in relation to proposition 45 (page 347):

“…geometrical algebra of the Greeks. We have now learnt how to represent any rectilineal area, which can of course be resolved into triangles, by a single parallelogram having one side equal to any given straight line and one angle equal to any given rectilineal angle. Most important of all such parallelograms is the rectangle, which is one of the simplest forms in which an area can be shown. Since a rectangle corresponds to the product of two magnitudes in algebra, we see that application to a given straight line of a rectangle equal to a given area is the geometrical equivalent of algebraical division of the product of two quantities by a third.”

I’m not sure what he means by the last sentence “algebraical division of the product of two quantities by a third.” and how this is related to proposition 45. I can see that the final parallelogram, if it’s a rectangle, has two sides but where does the division come in?

rhody
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    He means that as $Area = h \times w$ then $w = \frac {Area}h$. If we have a rectangle of any area $A$ we can create a parallelogram with area $A$ and any side length $w$. Therefore the altitude of the parallelogram will be $\frac Aw$. We just did division. (I think that is what he means. – fleablood Nov 20 '22 at 08:37
  • I think you are right but the area will be the original rectilineal area we want to turn into a parallelogram with the same area. The width of that as you say is $area/h$ both of which we are given. Hence the construction effectively does division. – rhody Nov 20 '22 at 16:37
  • This theorem converts any polygon into a rectangle - any rectangle, say. So if you make or have a rectangle with area $A$ and sides $a,b$, and then you apply that figure to a line of length $f$, since you preserve the area of the figure with the construction, whatever other side $g$ emerges from the new rectangle with sides $f,g$, you now have the equivalent of $A=ab=fg$, hence $g=\frac{ab}{f}$. – RobinSparrow Jan 31 '24 at 02:46

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