• Archimedes, On the Equilibrium of Planes Props. 6-7
• translated by Henry Mendell (Cal. State U., L.A.)

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Prop. 6: Commensurable magnitudes balance when they have the same ratio for their lengths conversely as their weights.
(general diagram)

(diagram 1) Let there be commensurable magnitudes A, B whose centers are A, B and let there be a length ED, and let it be as A is to B so is length DG to length GE. One must prove that when A, B are put together on each side, the center of weight is G.

For since as A is to B so is DG to GE, but A is commensurable with B, and therefore GD is commensurable with GE, i.e., a straight line with a straight line. (diagram 2) Thus there is a common measure of EG, GD. Let it be N, and (diagram 3) let each of DH, DK lie equal to EG, and EL be equal to DG. (diagram 4) Thus LE is equal to EH. Therefore, LH is double DG, and HK is double GE. Thus N measures each of LH, HK, since also it measures their halves. And since as A is to B so DG is to GE, but as DG is to GE so is LH to HK, since each is respectively double. And therefore as A is to B so is LH to HK. (diagram 5) And as many times as LH is of N, let A be so many times of Z. Therefore, as LH is to N, so is A to Z. And as KH is to LH, so is B to A. (diagram 6) Therefore, ex aequale, as KH is to N so is B to Z. Therefore, KH is an equal multiple of N as B is of Z. But A was also proved as being a multiple of Z. Thus Z is a common measure of A, B. And so, LH having been divided in parts equal to N, and A into parts equal to Z, while the segments in LH which are equal in magnitude to N, will be equal in number to the segments in A, which are equal to Z. (diagram 7) Thus, if a magnitude equal to Z is placed on each of the segments in LH, having the center of weight on the middle of the segment, then the totality of magnitudes are equal to A and E will be the center of the weight composed from all of them. For the totality will be even in number and the ones on each side of E will be equal in number since LE is equal to HE. (diagram 8) Similarly, it will be proved that even if a magnitude equal to Z is placed on each side of the segments in KH, having its center of weight on the middle segment, then the totality of the magnitudes will be equal to B and D will be the center of the weight composed from all of them. (diagram 9) And so A placed on it will be at E, while B placed on it will be at D. In fact, there will be magnitudes equal to one another lying in a straight line whose centers of weight will be equally distant from one another, while being even in multitude. And so it is clear that the bisection of the line having the centers of the magnitudes in between is the center of the magnitude composed from all of them. Since LE is equal to GD, and EG to DK, therefore the whole LG is also equal to GK. Thus, point G is the center of the weight of the magnitude composed from all of them. (diagram 1) Therefore, when A is placed at E and B at D, they balance at G.

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Prop. 7: And then even if the magnitudes are incommensurable, they will similarly balance when they have the same ratio from their lengths conversely as their magnitudes.

Let magnitudes AB, G be incommensurable with lengths DE, EZ, and let AB have to G the same ratio which ED has to length EZ. I say that E is the center of weight of both AB, G. (general diagram)

(diagram 1) For if AB does not balance when placed at Z with G placed at D, either AB is larger than it would be to balance G or not. (diagram 2) Let it be larger, and (diagram 3) let there be subtracted from AB a magnitude less than the excess by which AB is larger than it would be to balance G, so that the remainder A is commensurable with G (proof of this lemma, scholion to Theodosius, Sphaerica iii 9). And so, since magnitudes A, G are commensurable and A has a smaller ratio to G than DE to EZ, A and G do not balance at distances DE, EZ, when A is placed at Z, and G at D. For the same reasons, it won't happen even if G is larger than it would be to balance AB.

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