1. If two areas are enclosed by a straight-line and the section of a right-angled cone, which we are able to apply to the given straight-line, and they do not have the same center of weight, the center of weight of the magnitude composed from both of them will be on the straight-line that joins the center of their weight which divides the mentioned straight-line in a way that the segments of the have the same ratio inversely with the areas.

(diagram 2) Let there be two areas, AB, GD, as stated,(*1e*) points E, Z the centers of their weight, and the ratio which AB has to GD, let ZQ have this to QE. One must show that point Q is the center of the weight of the magnitude composed from both areas AB, GD.

(diagram 3) Let each of ZH, ZK be, in fact, equal to EQ, (diagram 4) and EL equal to ZQ, that is HE. Therefore, LQ will also be equal to KQ, and furthermore as LH is to HK so is AB to GD. For each is double each. (diagram 5) Let the area of AB be applied along LH on each side of LH, so that MN is equal to AB. In fact, point E is the center of the weight of MN (I prop. 9).(*2*) (diagram 6) Let NX be, in fact, filled out, but MN will have a ratio to NX that LH has to HK. But AB also has to GD the ratio of LH to HK. Therefore, as AB is to GD, so too is MN to NX. And alternando. But AB is equal to MN. (diagram 7) Therefore GD is also equal to NX, and point Z is the center of its weight (and the center of NX because of I prop. 9). (diagram 8) And since LQ is equal to QK, and a whole, LK, bisects the opposite sides, point Q is the center of the weight of a whole, PM (I prop. 9). But MP is equal to that from both MN, NX. (diagram 9) Thus too point Q is the center of the weight of that from AB, GD. Assumption 6

Commentary of Eutocius, pp. 278.2-16 (the wording of the text follows more closely the wording of the statement of the theorem, but note the other differences):

On the 2nd book. Having pursued precisely the first [book] and having made clear the what was difficult to understand in it, we consider it also necessary to set out appropriately what was difficultly stated in the second book. He says then in the first sentence of the first theorem, “Let there be supposed areas AB, GD enclosed by a straight-line and the section of a right-angled cone which we are able to apply to the given straight-line.” It is not possible to find this at once from the things proved here. Since it has been proved by him, as also he said in On the Sphere and Cylinder (in the introduction), that such a figure is a third again the triangle having the same base as it and an equal height, but we are able to apply the rectilinear plane that is a third again of a triangle along the given straight line, it is obvious that [we can also do it] to figures of this sort. What was said in the construction are clear through the tenth theorem of the first of these books. (back)

(*1*)This theorem is surrounded with controversy. Why does Archimedes prove it instead of just using Propositions 6-7 of Book I? Moreover, a careful perusal of the theorem would show that it is crucial to the proof that one can apply the area of parabola AB to line LH, as Eutocius notes. The commensurable case, theorem I 6, does not apply the area to the line but supposes that equal commensurable weights can be divided up into units of the common measure. Of course, theorem 7 does neither by reducing the incommensurable case to the commensurable one. In a sense, II 1 and I 6 are different special cases of the general principle of the balance where the use of some sort of distribution of weights is possible.

Is there any guarantee that one can always divide up the weight into its measures? It is plausible that if one can prove two weights commensurable one should be able to do this. For example, given two commensurable circles, one can readily construct a circle that is the common measure by using the diagonal of the square common measure of the squares of the diagonals. Of course, a similar problem exists with the proofs of Eucid, Elements X.

Does Archimedes need the general principle of the balance, I 6-7, in Book II or only the case of the parabola, II 1? In fact, in the logical structure of both books, I 6-7 is only used for I 8. However, I 8 is a fundamental theorem for both books. This issue depends on how one interprets I 6-7, however (more below).

In fact, Archimedes does not seem to need or use II 1 at all in Book II. If so, II 1 stands alone in the logical structure of Book II. The issue here is that sometimes Archimedes infers that two parabolas equal in area have their centers of weight equidistant from some point. This follows equally from II Prop. 1 and the much more trivial I Prop. 4. However, at the same time he will make the exact same claim about polygons. Here, II Prop. 1 will be useless, but I Prop. 4 is required. Hence, in my comments, I ignore applications that may be this sort of trivial application of II Prop. 1 and cite I Prop. 4 instead. The reader should take note of this, however.

We can speculate why Archimedes proves II 1. Perphaps, he wants to show use this application of areas arguments involves a simpler argument. Perhaps, he is concerned about more philosophical issues. Perhaps, he regards Book II as a complete treatise on parabolas and so proves every relevant, important theorem on centers of weight. He doesn't tell us.