368

from which it follows that

BOOK XII

area of parabolic segment = AABC.

The same sort of argument is used for solids, plane sections taking the place of straight lines.

Archimedes is careful to state once more that this method of argument does not constitute a proof. Thus, at the end of the above proposition about the parabolic segment, he adds: "This property is of course not proved by what has just been said; but it has furnished a sort of indication (ëμþaσív Tiva) that the conclusion is true."

Let us now turn to the passage of Plutarch (De Comm. Not. adv. Stoicos XXXIX. 3) about Democritus above referred to. Plutarch speaks of Democritus as having raised the question in natural philosophy (pvσuk@s): "if a cone were cut by a plane parallel to the base [by which is clearly meant a plane indefinitely near to the base], what must we think of the surfaces of the sections, that they are equal or unequal? For, if they are unequal, they will make the cone irregular, as having many indentations, like steps, and unevennesses; but, if they are equal, the sections will be equal, and the cone will appear to have the property of the cylinder and to be made up of equal, not unequal circles, which is very absurd." The phrase "made up of equal...circles" (ἐξ ἴσων συγκείμενος...κύκλων) shows that Democritus already had the idea of a solid being the sum of an infinite number of parallel planes, or indefinitely thin laminae, indefinitely near together: a most important anticipation of the same thought which led to such fruitful results in Archimedes. If then one may hazard a conjecture as to Democritus' argument with regard to a pyramid, it seems probable that he would notice that, if two pyramids of the same height and equal triangular bases are respectively cut by planes parallel to the base and dividing the heights in the same ratio, the corresponding sections of the two pyramids are equal, whence he would infer that the pyramids are equal as being the sum of the same infinite number of equal plane sections or indefinitely thin laminae. (This would be a particular anticipation of Cavalieri's proposition that the areal or solid content of two figures are equal if two sections of them taken at the same height, whatever the height may be, always give equal straight lines or equal surfaces_respectively.) And Democritus would of course see that the three pyramids into which a prism on the same base and of equal height with the original pyramid is divided (as in Eucl. XII. 7) satisfy this test of equality, so that the pyramid would be one third part of the prism. The extension to a pyramid with a polygonal base would be easy. And Democritus may have stated the proposition for the cone (of course without an absolute proof) as a natural inference from the result of increasing indefinitely the number of sides in a regular polygon forming the base of a pyramid.

BOOK XII. PROPOSITIONS.

PROPOSITION I.

Similar polygons inscribed in circles are to one anoth the squares on the diameters.

Let ABC, FGH be circles,

let ABCDE, FGHKL be similar polygons inscribed in and let BM, GN be diameters of the circles;

I say that, as the square on BM is to the square on G is the polygon ABCDE to the polygon FGHKL.

For let BE, AM, GL, FN be joined.

Now, since the polygon ABCDE is similar to the po FGHKL,

the angle BAE is equal to the angle GFL,

and, as BA is to AE, so is GF to FL.

[VI.

Thus BAE, GFL are two triangles which have one equal to one angle, namely the angle BAE to the GFL, and the sides about the equal angles proportional therefore the triangle ABE is equiangular with the tr FGL.

Therefore the angle AEB is equal to the angle FL

H. E. III.

But the right angle BAM is also equal to the right GFN;

therefore the remaining angle is equal to the remaining

Therefore the triangle ABM is equiangular wit triangle FGN.

Therefore, proportionally, as BM is to GN, so is to GF

But the ratio of the square on BM to the square o is duplicate of the ratio of BM to GN,

and the ratio of the polygon ABCDE to the polygon FG. is duplicate of the ratio of BA to GF;

therefore also, as the square on BM is to the square on so is the polygon ABCDE to the polygon FGHKL.

Therefore etc.

Q. E. I

As, from this point onward, the text of each proposition usually o considerable space, I shall generally give in the notes a summary argument, to enable it to be followed more easily.

Here we have to prove that a pair of corresponding sides are in t of the corresponding diameters.

Since s BAE, GFL are equal, and the sides about those proportional,

so that

AS ABE, FGL are equiangular,

LAEB=L FLG.

s AMB, FNG, are ec

Hence their equals in the same segments,
And the right angles BAM, GFN are equal.
Therefore As ABM, FGN are equiangular, so that

B

I

say that, as the circle ABCD is to the circle EFGH the square on BD to the square on FH.

For, if the square on BD is not to the square on the circle ABCD is to the circle EFGH

then, as the square on BD is to the square on FH, the circle ABCD be either to some less area than the EFGH, or to a greater.

First, let it be in that ratio to a less area S.

Let the square EFGH be inscribed in the circle El then the inscribed square is greater than the half of the EFGH, inasmuch as, if through the points E, F, G, draw tangents to the circle, the square EFGH is ha square circumscribed about the circle, and the circle i than the circumscribed square;

hence the inscribed square EFGH is greater than the h the circle EFGH.

Let the circumferences EF, FG, GH, HE be bisec the points K, L, M, N,

and let EK, KF, FL, LG, GM, MH, HN, NE be joi therefore each of the triangles EKF, FLG, GMH, H also greater than the half of the segment of the circle it, inasmuch as, if through the points K, L, M, N we tangents to the circle and complete the parallelograms straight lines EF, FG, GH, HE, each of the triangles

(

372

BOOK XII

[XII. 2

FLG, GMH, HNE will be half of the parallelogram about it,

while the segment about it is less than the parallelogram; hence each of the triangles EKF, FLG, GMH, HNE is greater than the half of the segment of the circle

about it.

Thus, by bisecting the remaining circumferences and joining straight lines, and by doing this continually, we shall leave some segments of the circle which will be less than the excess by which the circle EFGH exceeds the area S.

For it was proved in the first theorem of the tenth book that, if two unequal magnitudes be set out, and if from the greater there be subtracted a magnitude greater than the half, and from that which is left a greater than the half, and if this be done continually, there will be left some magnitude which will be less than the lesser magnitude set out.

Let segments be left such as described, and let the segments of the circle EFGH on EK, KF, FL, LG, GM, MH, HN, NE be less than the excess by which the circle EFGH exceeds the area S.

Therefore the remainder, the polygon EKFLGMHN, is greater than the area S.

Let there be inscribed, also, in the circle ABCD the polygon AOBPCQDR similar to the polygon EKFLGMHŃ; therefore, as the square on BD is to the square on FH, so is the polygon AOBPCQDR to the polygon EKFLGMHN.

[XII. 1]

But, as the square on BD is to the square on FH, so also is the circle ABCD to the area S; therefore also, as the circle ABCD is to the area S, so is the polygon AOBPCQDR to the polygon EKFLGMHN ;

[v. II] therefore, alternately, as the circle ABCD is to the polygon inscribed in it, so is the area S to the polygon EKFLGMÄN.

[v. 16]

But the circle ABCD is greater than the polygon inscribed in it; therefore the area S is also greater than the polygon EKFLGMHN.