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Hypsicles proceeds (pp. 7 sqq.) to give a proof of this theorem. Allman pointed out (Greek Geometry from Thales to Euclid, 1889, pp. 201—2) that this proof depends on eight theorems, six of which appear in Euclid's Book XIII. (in Propositions 8, 10, 12, 15, 16 with Por., 17); two other propositions not mentioned by Allman are also used, namely XIII. 4 and 9. This seems, as Allman says, to confirm the inference of Bretschneider (p. 171) that, as Aristaeus' work was the newest and latest in which, before Euclid's time, this subject was treated, we have in Eucl. XIII. at least a partial recapitulation of the contents of the treatise of Aristaeus.

This we also After Euclid, Apollonius wrote on the comparison of the dodecahedron and the icosahedron inscribed in one and the same sphere. learn from Hypsicles, who says in the next words following those about Aristaeus above quoted: "But it is proved by Apollonius in the second edition of his Comparison of the dodecahedron with the icosahedron that, as the surface of the dodecahedron is to the surface of the icosahedron [inscribed in the same sphere], so is the dodecahedron itself [i.e. its volume] to the icosahedron, because the perpendicular is the same from the centre of the sphere to the pentagon of the dodecahedron and to the triangle of the icosahedron."

1

BOOK XIII. PROPOSITIONS.

PROPOSITION I.

If a straight line be cut in extreme and mean ratio, the square on the greater segment added to the half of the whole is five times the square on the half.

For let the straight line AB be cut in extreme and mean ratio at the point C,

and let AC be the greater segment; let the straight line AD be produced in a straight line with CA, and let AD be made half of AB; I say that the square on CD is five times the square on AD.

For let the squares AE, DF be described on AB, DC, and let the figure in DF be drawn ; let FC be carried through to G.

Now, since AB has been cut in extreme and mean ratio at C, therefore the rectangle AB, BC is equal to the square on AC.

[vi. Def. 3, VI. 17]

P

N

M

H

K

E

B

And CE is the rectangle AB, BC, and FH the square on AC;

therefore CE is equal to FH.

And, since BA is double of AD,

while BA is equal to KA, and AD to AH,

therefore KA is also double of AH.

But, as KA is to AH, so is CK to CH;

therefore CK is double of CH.

But LH, HC are also double of CH.
Therefore KC is equal to LH, HC.

[VI. I]

But CE was also proved equal to HF;

therefore the whole square AE is equal to the gnomon MNO.

And, since BA is double of AD,

the square on BA is quadruple of the square on AD, that is, AE is quadruple of DH.

But AE is equal to the gnomon MNO;

therefore the gnomon MNO is also quadruple of AP; therefore the whole DF is five times AP.

And DF is the square on DC, and AP the square on DA; therefore the square on CD is five times the square on DA. Therefore etc.

Q. E. D.

The first five propositions are in the nature of lemmas, which are required for later propositions but are not in themselves of much importance.

It will be observed that, while the method of the propositions is that of Book II., being strictly geometrical and not algebraical, none of the results of that Book are made use of (except indeed in the Lemma to XIII. 2, which is probably not genuine). It would therefore appear as though these propositions were taken from an earlier treatise without being revised or rewritten in the light of Book II. It will be remembered that, according to Proclus (p. 67, 6), Eudoxus "greatly added to the number of the theorems which originated with Plato regarding the section" (i.e. presumably the "golden section"); and it is therefore probable that the five theorems are due to Eudoxus.

That, if AB is divided at C in extreme and mean ratio, the rectangle AB, BC is equal to the square on AC is inferred from vi. 17. AD is made equal to half AB, and we have to prove that (sq. on CD) = 5 (sq. on AD).

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whence, adding the sq. on AD to each, we have

(sq. on CD) 5 (sq. on AD).

=

The result here, and in the next propositions, is really seen more readily by means of the figure of II. II.

In this figure SR=AC+}AB, by construction; and we have therefore to prove that

(sq. on SR) = 5 (sq. on AR).

R

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This is obvious, for

(sq. on SR) = (sq. on RB)

= sum of sqs. on AB, AR

= 5 (sq. on AR).

The MSS. contain a curious addition to XIII. 1—5 in the shape of analyses and syntheses for each proposition prefaced by the heading :

"What is analysis and what is synthesis.

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Analysis is the assumption of that which is sought as if it were admitted <and the arrival > by means of its consequences at something admitted to be true.

"Synthesis is an assumption of that which is admitted <and the arrival > by means of its consequences at something admitted to be true.”

There must apparently be some corruption in the text; it does not, in the case of synthesis, give what is wanted. B and V have, instead of "something admitted to be true," the words "the end or attainment of what is sought."

The whole of this addition is evidently interpolated. To begin with, the analyses and syntheses of the five propositions are placed all together in four MSS.; in P, q they come after an alternative proof of XIII. 5 (which alternative proof P gives after XIII. 6, while q gives it instead of XIII. 6), in B (which has not the alternative proof of XIII. 5) after XIII. 6, and in b (in which XIII. 6 is wanting, and the alternative proof of XIII. 5 is in the margin, in the first hand) after XIII. 5, while V has the analyses of 1 -3 in the text after XIII. 6 and those of 4-5 in the same place in the margin, by the second hand. Further, the addition is altogether alien from the plan and manner of the Elements. The interpolation took place before Theon's time, and the probability is that it was originally in the margin, whence it crept into the text of P after XIII. 5. Heiberg (after Bretschneider) suggested in his edition (Vol. v. p. lxxxiv.) that it might be a relic of analytical investigations by Theaetetus or Eudoxus, and he cited the remark of Pappus (v. p. 410) at the beginning of his "comparisons of the five [regular solid] figures which have an equal surface," to the effect that he will not use "the so-called analytical investigation by means of which some of the ancients effected their demonstrations." More recently (Paralipomena zu Euklid in Hermes XXXVIII., 1903) Heiberg conjectures that the author is Heron, on the ground that the sort of analysis and synthesis recalls Heron's remarks on analysis and synthesis in his commentary on the beginning of Book II. (quoted by an-Nairizī, ed. Curtze, p. 89) and his quasi-algebraical alternative proofs of propositions in that Book.

To show the character of the interpolated matter I need only give the analysis and synthesis of one proposition. In the case of XIII. I it is in substance as follows. The figure is a mere

straight line.

Let AB be divided in extreme and mean ratio at C, AC being the greater segment;

and let

I say that

(Analysis.)

"For, since

AD = 1 AB.

(sq. on CD) = 5 (sq. on AD).

(sq. on CD) = 5 (sq. on AD),"

A

and (sq. on CD) = (sq. on CA) + (sq. on AD) + 2 (rect. CA, AD), therefore (sq. on CA) + 2 (rect. CA, AD) = 4 (sq. on AD).

B

But

and

rect. BA. AC= 2 (rect. CA. AD),

(sq. on CA) = (rect. AB, BC).

or

Therefore

(rect. BA, AC) + (rect. AB, BC) = 4 (sq. on AD),
(sq. on AB) = 4 (sq. on AD):

and this is true, since
(Synthesis.)
Since

and

therefore

AD=AB.

(sq. on AB) = 4 (sq. on AD),

(sq. on AB)="(rect. BA, AC) + (rect. AB, BC),
4 (sq. on AD) = 2 (rect. DA, AC) + sq. on AC.

Adding to each the square on AD, we have

(sq. on CD) 5 (sq. on AD).

=

PROPOSITION 2.

If the square on a straight line be five times the square on a segment of it, then, when the double of the said segment is cut in extreme and mean ratio, the greater segment is the remaining part of the original straight line.

For let the square on the straight line AB be five times the square on the segment AC of it,

and let CD be double of AC;

I say that, when CD is cut in extreme and mean ratio, the greater segment is CB.

Let the squares AF, CG be described on AB, CD respectively, let the figure in AF be drawn, and let BE be drawn through.

Now, since the square on BA is five times the square on AC, AF is five times AH.

Therefore the gnomon MNO is quadruple of AH.

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And, since DC is double of CA, therefore the square on DC is quadruple of the square on CA, that is, CG is quadruple of AH.

But the gnomon MNO was also proved quadruple of AH; therefore the gnomon MNO is equal to CG.

And, since DC is double of CA,

while DC is equal to CK, and AC to CH, therefore KB is also double of BH.

[VI. I]

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