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Therefore (1. 26) the triangles DTU, GTS are equal in all respects, so that

DT = TG,
UT = TS.


If there be two prisms of equal height, and one have a parallelogram as base and ihe other a triangle, and if the parallelogram be double of the triangle, the prisms will be equal.

Let ABCDEF, GHKLMN be two prisms of equal height, let one have the parallelogram AF as base, and the other the triangle GHK, and let the parallelogram AF be double of the triangle GHK; I say that the prism ABCDEF is equal to the prism GHKLMN.

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For let the solids AO, GP be completed.

Since the parallelogram AF is double of the triangle GHK, while the parallelogram HK is also double of the triangle GHK,

[1. 34] therefore the parallelogram AF is equal to the parallelogram HK.

But parallelepipedal solids which are on equal bases and of the same height are equal to one another ;

[xi. 31] therefore the solid AO is equal to the solid GP.

And the prism ABCDEF is half of the solid AO, and the prism GHKLMN is half of the solid GP; [xi. 28] therefore the prism ABCDEF is equal to the prism GHKLMN. Therefore etc.

Q. E. D.

This proposition is made use of in XII. 3, 4. The phraseology is interesting because we find one of the parallelogrammic faces of one of the triangular prisms called its base, and the perpendicular on this plane from that vertex of either triangular face which is not in this plane the height.

The proof is simple because we have only to complete parallelepipeds which are double the prisms respectively and then use xi. 31. It has to be borne in mind, however, that, if the parallelepipeds are not rectangular, the proof in xi. 28 is not sufficient to establish the fact that the parallelepipeds are double of the prisms, but has to be supplemented as shown in the note on that proposition. XII. 4 does, however, require the theorem in its general form.



The predominant feature of Book xii. is the use of the method of exhaustion, which is applied in Propositions 2, 3-5, 10, 11, 12, and (in a slightly different form) in Propositions 16—18. We conclude therefore that for the content of this Book Euclid was greatly indebted to Eudoxus, to whom the discovery of the method of exhaustion is attributed. The evidence for this attribution comes mainly from Archimedes. (1) In the preface to On the Sphere and Cylinder 1., after stating the main results obtained by himself regarding the surface of a sphere or a segment thereof, and the volume and surface of a right cylinder with height equal to its diameter as compared with those of a sphere with the same diameter, Archimedes adds: “Having now discovered that the properties mentioned are true of these figures, I cannot feel any hesitation in setting them side by side both with my former investigations and with those of the theorems of Eudoxus on solids which are held to be most irrefragably established, namely that any pyramid is one third part of the prism which has the same base with the pyramid and equal height (i.e. Eucl. X11. 7), and that any cone is one third part of the cylinder which has the same base with the cone and equal height [i.e. Eucl. xii. 10]. For, though these properties also were naturally inherent in the figures all along, yet they were in fact unknown to all the many able geometers who lived before Eudoxus and had not been observed by any one. (2) In the preface to the treatise known as the Quadrature of the Parabola Archimedes states the "lemma” assumed by him and known as the “Axiom of Archimedes” (see note on x. I above) and proceeds : “Earlier geometers (oi npótepov yeupérpai) have also used this lemma; for it is by the use of this same lemma that they have shown that circles are to one another in the duplicate ratio of their diameters [Eucl. XII. 2], and that spheres are to one another in the triplicate ratio of their diameters [Eucl. XII. 18), and further that every pyramid is one third part of the prism which has the same base with the pyramid and equal height (Eucl. xii. 7]; also, that every cone is one third part of the cylinder which has the same base with the cone and equal height (Eucl

. xii. 10) they proved by assuming a certain lemma similar to that aforesaid." Thus in the first passage two theorems of Eucl. xii. are definitely attributed to Eudoxus ; and, when Archimedes says, in the second passage, that "earlier geometers” proved these two theorems by means of the lemma known as the “Axiom of Archimedes” and of a lemma similar to it respectively, we can hardly suppose him to be alluding to

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any other proof than that given by Eudoxus. As a matter of fact, the lemma used by Euclid to prove both propositions (XII. 3—5 and 7, and xii. 10) is the theorem of Eucl. x. 1. As regards the connexion between the two "lemmas” see note on x. I.

We are not, however, to suppose that none of the results obtained by the method of exhaustion had been discovered before the time of Eudoxus (Al. about 368--5 B.C.). Two at least are of earlier date, those of Eucl. XII. 2

and xil. 7.


(a) Simplicius (Comment. in Aristot. Phys. p. 61, ed. Diels) quotes Eudemus as saying, in his History of Geometry, that Hippocrates of Chios (A. say 430 B.C.) first laid it down (čdeto) that similar segments of circles are in the ratio of the squares on their bases and that he proved this (idelkvcev) by proving (ex toù deigai) that the squares on the diameters have the same ratio as the (whole) circles. We know nothing of the method by which Hippocrates proved this proposition; but, having regard to the evidence from Archimedes quoted above; it is not permissible to suppose that the method was the fully developed method of exhaustion as we know it.

(») As regards the two theorems about the volume of a pyramid and of a cone respectively, which Eudoxus was the first to prove, a new piece of evidence is now forthcoming in the fragment of Archimedes recently brought to light at Constantinople and published by Heiberg (for the Greek text see Hermes XLII., 1907, pp. 235—303; for Heiberg's translation and Zeuthen's notes see Bibliotheca Mathematica vilz, 1907, pp. 321–363). This is nothing less than a considerable portion of a work under the title 'Αρχιμήδους περί των unxavikwv Oewpnuátwv mpòs 'Epatogévnu podos, which “Method," addressed to Eratosthenes, is the podcov on which, according to Suidas, Theodosius wrote a commentary, and which is several times cited by Heron in his Metrica; and it adds a new and important chapter to the history of the integral calculus. In the preface to this work (Hermes l.c. p. 245, Bibliotheca Mathematica 1.c. p. 323) Archimedes alludes to the theorems which he first discovered by means of mechanical considerations, but proved afterwards by geometry because the investigation by means of mechanics did not constitute a rigid proof; he observes, however, that the mechanical method is of great use for the discovery of theorems, and it is much easier to provide the rigid proof when the fact to be proved has once been discovered than it would be if nothing were known to begin with. He goes on : “Hence too, in the case of those theorems the proof of which was first discovered by Eudoxus, namely those relating to the cone and the pyramid, that the cone is one third part of the cylinder, and the pyramid one third part of the prism, having the same base and equal height, no small part of the credit will naturally be assigned to Democritus, who was the first to make the statement (of the fact) regarding the said figure [i.e. property], though without proving it.” Hence the discovery of the two theorems must now be attributed to Democritus (A. towards the end of 5th cent. B.c.). The words without proving it" (xwpis útodeitews) do not mean that Democritus gave no sort of proof, but only that he did not give a proof on the rigorous lines required later; for the same words are used by Archimedes of his own investigations by means of mechanics, which, however, do constitute a reasoned argument. The character of Archimedes' mechanical arguments combined with a passage of Plutarch about a particular question in infinitesimals said to have been raised by Democritus may perhaps give a clue to the line of Democritus argument as regards the prism. The essential

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feature of Archimedes' mechanical arguments in this tract is that he regards an area as the sum of an infinite number of straight lines parallel to one another and terminated by the boundary or boundaries of the closed figure the area of which is to be found, and a volume as the sum of an infinite number of plane sections parallel to one another : which is of course the same thing as taking (as we do in the integral calculus) the sum of an infinite number of strips of breadth dx (say), when dx becomes indefinitely small, or the sum of an infinite number of parallel laminae of depth dz (say), when da becomes indefinitely small. To give only one instance, we may take the case of the area of a segment of a parabola cut off by a chord.

Let CBA be the parabolic segment, CE the tangent at C meeting the

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diameter EBD through the middle point of the chord CA in E, so that

EB = BD. Draw AF parallel to ED meeting CE produced in F. Produce CB to H so that CK = KH, where K is the point in which CH meets AF; and suppose CH to be a lever.

Let any diameter MNPO be drawn meeting the curve in P and CF, CK, CA in M, N, O respectively. Archimedes then observes that

CA: AO = MO:OP (" for this is proved in a lemma”), whence

HK: KN= MO:OP, so that, if a straight line TG equal to PO be placed with its middle point at H, the straight line MO with centre of gravity at N, and the straight line TG with centre of gravity at H, will balance about K.

Taking all other parts of diameters like PO intercepted between the curve and CA, and placing equal straight lines with their centres of gravity at H, these straight lines collected at H will balance (about K) all the lines like MO parallel to FA intercepted within the triangle CFA in the positions in which they severally lie in the figure.

Hence Archimedes infers that an area equal to that of the parabolic segment hung at H will balance (about K) the triangle CFA hung at its centre of gravity, the point X (a point on CK such that CK' = 3XX), and therefore that (area of triangle CFA): (area of segment) = HK: KX

= 3:1,

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