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

Let ABCDEF, GHKLMN be two prisms of height,

let one have the parallelogram AF as base, and the ot triangle GHK,

and let the parallelogram AF be double of the triangle I say that the prism ABCDEF is equal to the GHKLMN.

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For let the solids AO, GP be completed.

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

therefore the parallelogram AF is equal to the paralle HK.

But parallelepipedal solids which are on equal bas of the same height are equal to one another; 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; therefore the prism ABCDEF is equal to the GHKLMN.

Therefore etc.

Q. E.

BOOK XII.

HISTORICAL NOTE.

The predominant feature of Book XII. is the use of the m exhaustion, which is applied in Propositions 2, 3-5, 10, 11, 12, a slightly different form) in Propositions 16-18. We conclude theref for the content of this Book Euclid was greatly indebted to Eudoxus, the discovery of the method of exhaustion is attributed. The evid this attribution comes mainly from Archimedes. (1) In the prefac the Sphere and Cylinder 1., after stating the main results obtained by regarding the surface of a sphere or a segment thereof, and the volu surface of a right cylinder with height equal to its diameter as compa those of a sphere with the same diameter, Archimedes adds: "Hav discovered that the properties mentioned are true of these figures, feel any hesitation in setting them side by side both with my former i tions and with those of the theorems of Eudoxus on solids which are he most irrefragably established, namely that any pyramid is one third pa prism which has the same base with the pyramid and equal height [i XII. 7], and that any cone is one third part of the cylinder which has base with the cone and equal height [i.e. Eucl. XII. 10]. For, thou properties also were naturally inherent in the figures all along, yet th in fact unknown to all the many able geometers who lived before and had not been observed by any one.' (2) In the preface to the known as the Quadrature of the Parabola Archimedes states the " assumed by him and known as the "Axiom of Archimedes" (see note above) and proceeds: "Earlier geometers (oi póτepov yewμérpai) h used this lemma; for it is by the use of this same lemma that th shown that circles are to one another in the duplicate ratio of their [Eucl. XII. 2], and that spheres are to one another in the triplicate ratio diameters [Eucl. XII. 18], and further that every pyramid is one third p prism which has the same base with the pyramid and equal height [Euc also, that every cone is one third part of the cylinder which has the s with the cone and equal height [Eucl. XII. 10] they proved by assuming lemma similar to that aforesaid." Thus in the first passage two the Eucl. XII. are definitely attributed to Eudoxus; and, when Archime in the second passage, that "earlier geometers" proved these two 1 by means of the lemma known as the "Axiom of Archimedes": lemma similar to it respectively, we can hardly suppose him to be al

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BOOK XII

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 (fl. about 368—-5 B.C.). Two at least are of earlier date, those of Eucl. XII. 2 and XII. 7.

(a) Simplicius (Comment. in Aristot. Phys. p. 61, ed. Diels) quotes Eudemus as saying, in his History of Geometry, that Hippocrates of Chios (fl. say 430 B.C.) first laid it down (ero) that similar segments of circles are in the ratio of the squares on their bases and that he proved this (edeíkvvev) by proving (ek Toû deîgai) 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.

(b) 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 VII, 1907, pp. 321-363). This is nothing less than a considerable portion of a work under the title ̓Αρχιμήδους περὶ τῶν μηχανικών θεωρημάτων πρὸς Ἐρατοσθένην ἔφοδος, which Method,” addressed to Eratosthenes, is the pódiov 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 l.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 (fl. towards the end of 5th cent. B.C.). The words "without proving it" (xwpis árodeíέews) 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

becomes indefinitely small. To give only one instance, we may 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 meet

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

EB = BD.

Draw AF parallel to ED meeting CE produced in F.

Produc

H so that CK = KH, where K is the point in which CH meets A suppose CH to be a lever.

Let any diameter MNPO be drawn meeting the curve in P and 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 H, the straight line MO with centre of gravity at N, and the straight with centre of gravity at H, will balance about K.

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

Hence Archimedes infers that an area equal to that of the I segment hung at H will balance (about K) the triangle CFA hu centre of gravity, the point X (a point on CK such that CK = 3X therefore that

(area of triangle CFA): (area of segment) = HK: KX

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