Elementa

Euclid

Elementa

Euclid · Elem 13 · perseus-eng2 · More by Euclid

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13.p5697

I have already given, in the note to IV. 10, the evidence upon which the construction of the five regular solids is attributed to the Pythagoreans. Some of them, the cube, the tetrahedron (which is nothing but a pyramid), and the octahedron (which is only a double pyramid with a square base), cannot but have been known to the Egyptians. And it appears that dodecahedra have been found, of bronze or other material, which may belong to periods earlier than Pythagoras' time by some centuries (for references see Cantor's Geschichte der Mathematik I_{3}, pp. 175-6).

13.p5698

It is true that the author of the scholium No. I to Eucl. XIII. says that the Book is about the five so-called Platonic figures, which however do not belong to Plato, three of the aforesaid five figures being due to the Pythagoreans, namely the cube, the pyramid and the dodecahedron, while the octahedron and the icosahedron are due to Theaetetus. This statement (taken probably from Geminus) may perhaps rest on the fact that Theaetetus was the first to write at any length about the two last-mentioned solids. We are told indeed by Suidas (S. V. Θεαίτητος) that Theaetetus first wrote on the ’five solids’ as they are called. This no doubt means that Theaetetus was the first to write a complete and systematic treatise on all the regular solids; it does not exclude the possibility that Hippasus or others had already written on the dodecahedron. The fact that Theaetetus wrote upon the regular solids agrees very well with the evidence which we possess of his contributions to the theory of irrationals, the connexion between which and the investigation of the regular solids is seen in Euclid's Book XIII.

13.p5699

Theaetetus flourished about 380 B.C., and his work on the regular solids was soon followed by another, that of Aristaeus, an elder contemporary of Euclid, who also wrote an important book on Solid Loci, i.e. on conics treated as loci. This Aristaeus (known as the elder ) wrote in the period about 320 B.C. We hear of his Comparison of the five regular solids from Hypsicles (2nd cent. B.C.), the writer of the short book commonly included in the editions of the Elements as Book XIV. Hypsicles gives in this Book some six propositions supplementing Eucl. XIII.; and he introduces the second of the propositions (Heiberg's Euclid, Vol. v. p. 6) as follows:

13.p5700

The same circle circumscribes both the pentagon of the dodecahedron and the triangle of the icosahedron when both are inscribed in the same sphere. This is proved by Aristaeus in the book entitled Comparison of the five figures.

13.p5701

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.

13.p5702

After Euclid, Apollonius wrote on the comparison of the dodecahedron and the icosahedron inscribed in one and the same sphere. This we also 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.

13.prop.1.p5703

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.

13.prop.1.p5704

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.

13.prop.1.p5705

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.

13.prop.1.p5706

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]

13.prop.1.p5707

And CE is the rectangle AB, BC, and FH the square on AC; therefore CE is equal to FH.

13.prop.1.p5708

And, since BA is double of AD, while BA is equal to KA, and AD to AH, therefore KA is also double of AH.

13.prop.1.p5709

But, as KA is to AH, so is CK to CH; [VI. 1] therefore CK is double of CH.

13.prop.1.p5710

But LH, HC are also double of CH.

13.prop.1.p5711

Therefore KC is equal to LH, HC.

13.prop.1.p5712

But CE was also proved equal to HF; therefore the whole square AE is equal to the gnomon MNO.

13.prop.1.p5713

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.

13.prop.1.p5714

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.

13.prop.1.p5715

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.

13.prop.1.p5716

Therefore etc. Q. E. D.

13.prop.2.p5717

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.

13.prop.2.p5718

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.

13.prop.2.p5719

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

13.prop.2.p5720

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

13.prop.2.p5721

Therefore the gnomon MNO is quadruple of AH.

13.prop.2.p5722

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.

13.prop.2.p5723

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

13.prop.2.p5724

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. 1]

13.prop.2.p5725

But LH, HB are also double of HB; therefore KB is equal to LH, HB.

13.prop.2.p5726

But the whole gnomon MNO was also proved equal to the whole CG; therefore the remainder HF is equal to BG.

13.prop.2.p5727

And BG is the rectangle CD, DB, for CD is equal to DG; and HF is the square on CB; therefore the rectangle CD, DB is equal to the square on CB.

13.prop.2.p5728

Therefore, as DC is to CB, so is CB to BD.

13.prop.2.p5729

But DC is greater than CB; therefore CB is also greater than BD.

13.prop.2.p5730

Therefore, when the straight line CD is cut in extreme and mean ratio, CB is the greater segment.

13.prop.2.p5731

Therefore etc. Q. E. D.

13.prop.2.p5732

Lemma. That the double of AC is greater than BC is to be proved thus.

13.prop.2.p5733

If not, let BC be, if possible, double of CA.

13.prop.2.p5734

Therefore the square on BC is quadruple of the square on CA; therefore the squares on BC, CA are five times the square on CA.

13.prop.2.p5735

But, by hypothesis, the square on BA is also five times the square on CA; therefore the square on BA is equal to the squares on BC, CA: which is impossible. [II. 4]

13.prop.2.p5736

Therefore CB is not double of AC.

13.prop.2.p5737

Similarly we can prove that neither is a straight line less than CB double of CA; for the absurdity is much greater.

13.prop.2.p5738

Therefore the double of AC is greater than CB. Q. E. D.

13.prop.3.p5739

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

13.prop.3.p5740

For let any straight line AB be cut in extreme and mean ratio at the point C, let AC be the greater segment, and let AC be bisected at D; I say that the square on BD is five times the square on DC.

13.prop.3.p5741

For let the square AE be described on AB, and let the figure be drawn double.

13.prop.3.p5742

Since AC is double of DC, therefore the square on AC is quadruple of the square on DC, that is, RS is quadruple of FG.

13.prop.3.p5743

And, since the rectangle AB, BC is equal to the square on AC, and CE is the rectangle AB, BC, therefore CE is equal to RS.

13.prop.3.p5744

But RS is quadruple of FG; therefore CE is also quadruple of FG.

13.prop.3.p5745

Again, since AD is equal to DC, HK is also equal to KF.

13.prop.3.p5746

Hence the square GF is also equal to the square HL.

13.prop.3.p5747

Therefore GK is equal to KL, that is, MN to NE; hence MF is also equal to FE.

13.prop.3.p5748

But MF is equal to CG; therefore CG is also equal to FE.

13.prop.3.p5749

Let CN be added to each; therefore the gnomon OPQ is equal to CE.

13.prop.3.p5750

But CE was proved quadruple of GF; therefore the gnomon OPQ is also quadruple of the square FG.

13.prop.3.p5751

Therefore the gnomon OPQ and the square FG are five times FG.

13.prop.3.p5752

But the gnomon OPQ and the square FG are the square DN.

13.prop.3.p5753

And DN is the square on DB, and GF the square on DC.

13.prop.3.p5754

Therefore the square on DB is five times the square on DC. Q. E. D.

13.prop.4.p5755

If a straight line be cut in extreme and mean ratio, the square on the whole and the square on the lesser segment together are triple of the square on the greater segment.

13.prop.4.p5756

Let AB be a straight line, let it be cut in extreme and mean ratio at C, and let AC be the greater segment; I say that the squares on AB, BC are triple of the square on CA.

13.prop.4.p5757

For let the square ADEB be described on AB, and let the figure be drawn.

13.prop.4.p5758

Since then AB has been cut in extreme and mean ratio at C, and AC is the greater segment, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17]

13.prop.4.p5759

And AK is the rectangle AB, BC, and HG the square on AC; therefore AK is equal to HG.

13.prop.4.p5760

And, since AF is equal to FE, let CK be added to each; therefore the whole AK is equal to the whole CE; therefore AK, CE are double of AK.

13.prop.4.p5761

But AK, CE are the gnomon LMN and the square CK; therefore the gnomon LMN and the square CK are double of AK.

13.prop.4.p5762

But, further, AK was also proved equal to HG; therefore the gnomon LMN and the squares CK, HG are triple of the square HG.

13.prop.4.p5763

And the gnomon LMN and the squares CK, HG are the whole square AE and CK, which are the squares on AB, BC, while HG is the square on AC.

13.prop.4.p5764

Therefore the squares on AB, BC are triple of the square on AC. Q. E. D.

13.prop.5.p5765

If a straight line be cut in extreme and mean ratio, and there be added to it a straight line equal to the greater segment, the whole straight line has been cut in extreme and mean ratio, and the original straight line is the greater segment.

13.prop.5.p5766

For let the straight line AB be cut in extreme and mean ratio at the point C, let AC be the greater segment, and let AD be equal to AC.

13.prop.5.p5767

I say that the straight line DB has been cut in extreme and mean ratio at A, and the original straight line AB is the greater segment.

13.prop.5.p5768

For let the square AE be described on AB, and let the figure be drawn.

13.prop.5.p5769

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]

13.prop.5.p5770

And CE is the rectangle AB, BC, and CH the square on AC; therefore CE is equal to HC.

13.prop.5.p5771

But HE is equal to CE, and DH is equal to HC; therefore DH is also equal to HE.

13.prop.5.p5772

Therefore the whole DK is equal to the whole AE.

13.prop.5.p5773

And DK is the rectangle BD, DA, for AD is equal to DL; and AE is the square on AB; therefore the rectangle BD, DA is equal to the square on AB.

13.prop.5.p5774

Therefore, as DB is to BA, so is BA to AD. [VI. 17]

13.prop.5.p5775

And DB is greater than BA; therefore BA is also greater than AD. [V. 14]

13.prop.5.p5776

Therefore DB has been cut in extreme and mean ratio at A, and AB is the greater segment. Q. E. D.

13.prop.6.p5777

If a rational straight line be cut in extreme and mean ratio, each of the segments is the irrational straight line called apotome.

13.prop.6.p5778

Let AB be a rational straight line, let it be cut in extreme and mean ratio at C, and let AC be the greater segment; I say that each of the straight lines AC, CB is the irrational straight line called apotome.

13.prop.6.p5779

For let BA be produced, and let AD be made half of BA.

13.prop.6.p5780

Since then the straight line AB has been cut in extreme and mean ratio, and to the greater segment AC is added AD which is half of AB, therefore the square on CD is five times the square on DA. [XIII. 1]

13.prop.6.p5781

Therefore the square on CD has to the square on DA the ratio which a number has to a number; therefore the square on CD is commensurable with the square on DA. [X. 6]

13.prop.6.p5782

But the square on DA is rational, for DA is rational, being half of AB which is rational; therefore the square on CD is also rational; [X. Def. 4] therefore CD is also rational.

13.prop.6.p5783

And, since the square on CD has not to the square on DA the ratio which a square number has to a square number, therefore CD is incommensurable in length with DA; [X. 9] therefore CD, DA are rational straight lines commensurable in square only; therefore AC is an apotome. [X. 73]

13.prop.6.p5784

Again, since AB has been cut in extreme and mean ratio, and AC is the greater segment, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17]

13.prop.6.p5785

Therefore the square on the apotome AC, if applied to the rational straight line AB, produces BC as breadth.

13.prop.6.p5786

But the square on an apotome, if applied to a rational straight line, produces as breadth a first apotome; [X. 97] therefore CB is a first apotome.

13.prop.6.p5787

And CA was also proved to be an apotome.

13.prop.6.p5788

Therefore etc. Q. E. D.

13.prop.7.p5789

If three angles of an equilateral pentagon, taken either in order or not in order, be equal, the pentagon will be equiangular.

13.prop.7.p5790

For in the equilateral pentagon ABCDE let, first, three angles taken in order, those at A, B, C, be equal to one another; I say that the pentagon ABCDE is equiangular.

13.prop.7.p5791

For let AC, BE, FD be joined.

13.prop.7.p5792

Now, since the two sides CB, BA are equal to the two sides BA, AE respectively, and the angle CBA is equal to the angle BAE, therefore the base AC is equal to the base BE, the triangle ABC is equal to the triangle ABE, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend, [I. 4] that is, the angle BCA to the angle BEA, and the angle ABE to the angle CAB; hence the side AF is also equal to the side BF. [I. 6]

13.prop.7.p5793

But the whole AC was also proved equal to the whole BE; therefore the remainder FC is also equal to the remainder FE.

13.prop.7.p5794

But CD is also equal to DE.

13.prop.7.p5795

Therefore the two sides FC, CD are equal to the two sides FE, ED; and the base FD is common to them; therefore the angle FCD is equal to the angle FED. [I. 8]

13.prop.7.p5796

But the angle BCA was also proved equal to the angle AEB; therefore the whole angle BCD is also equal to the whole angle AED.

13.prop.7.p5797

But, by hypothesis, the angle BCD is equal to the angles at A, B; therefore the angle AED is also equal to the angles at A, B.

13.prop.7.p5798

Similarly we can prove that the angle CDE is also equal to the angles at A, B, C; therefore the pentagon ABCDE is equiangular.

13.prop.7.p5799

Next, let the given equal angles not be angles taken in order, but let the angles at the points A, C, D be equal; I say that in this case too the pentagon ABCDE is equiangular.

13.prop.7.p5800

For let BD be joined.

13.prop.7.p5801

Then, since the two sides BA, AE are equal to the two sides BC, CD, and they contain equal angles, therefore the base BE is equal to the base BD, the triangle ABE is equal to the triangle BCD, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4] therefore the angle AEB is equal to the angle CDB.

13.prop.7.p5802

But the angle BED is also equal to the angle BDE, since the side BE is also equal to the side BD. [I. 5]

13.prop.7.p5803

Therefore the whole angle AED is equal to the whole angle CDE.

13.prop.7.p5804

But the angle CDE is, by hypothesis, equal to the angles at A, C; therefore the angle AED is also equal to the angles at A, C.

13.prop.7.p5805

For the same reason the angle ABC is also equal to the angles at A, C, D.

13.prop.7.p5806

Therefore the pentagon ABCDE is equiangular. Q. E. D.

13.prop.8.p5807

If in an equilateral and equiangular pentagon straight lines subtend two angles taken in order, they cut one another in extreme and mean ratio, and their greater segments are equal to the side of the pentagon.

13.prop.8.p5808

For in the equilateral and equiangular pentagon ABCDE let the straight lines AC, BE, cutting one another at the point H, subtend two angles taken in order, the angles at A, B; I say that each of them has been cut in extreme and mean ratio at the point H, and their greater segments are equal to the side of the pentagon.

13.prop.8.p5809

For let the circle ABCDE be circumscribed about the pentagon ABCDE. [IV. 14]

13.prop.8.p5810

Then, since the two straight lines EA, AB are equal to the two AB, BC, and they contain equal angles, therefore the base BE is equal to the base AC, the triangle ABE is equal to the triangle ABC, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend. [I. 4]

13.prop.8.p5811

Therefore the angle BAC is equal to the angle ABE; therefore the angle AHE is double of the angle BAH. [I. 32]

13.prop.8.p5812

But the angle EAC is also double of the angle BAC, inasmuch as the circumference EDC is also double of the circumference CB; [III. 28, VI. 33] therefore the angle HAE is equal to the angle AHE; hence the straight line HE is also equal to EA, that is, to AB. [I. 6]

13.prop.8.p5813

And, since the straight line BA is equal to AE, the angle ABE is also equal to the angle AEB. [I. 5]

13.prop.8.p5814

But the angle ABE was proved equal to the angle BAH; therefore the angle BEA is also equal to the angle BAH.

13.prop.8.p5815

And the angle ABE is common to the two triangles ABE and ABH; therefore the remaining angle BAE is equal to the remaining angle AHB; [I. 32] therefore the triangle ABE is equiangular with the triangle ABH; therefore, proportionally, as EB is to BA, so is AB to BH. [VI. 4]

13.prop.8.p5816

But BA is equal to EH; therefore, as BE is to EH, so is EH to HB.

13.prop.8.p5817

And BE is greater than EH; therefore EH is also greater than HB. [V. 14]

13.prop.8.p5818

Therefore BE has been cut in extreme and mean ratio at H, and the greater segment HE is equal to the side of the pentagon.

13.prop.8.p5819

Similarly we can prove that AC has also been cut in extreme and mean ratio at H, and its greater segment CH is equal to the side of the pentagon. Q. E. D.

13.prop.9.p5820

If the side of the hexagon and that of the decagon inscribed in the same circle be added together, the whole straight line has been cut in extreme and mean ratio, and its greater segment is the side of the hexagon.

13.prop.9.p5821

Let ABC be a circle; of the figures inscribed in the circle ABC let BC be the side of a decagon, CD that of a hexagon, and let them be in a straight line; I say that the whole straight line BD has been cut in extreme and mean ratio, and CD is its greater segment.

13.prop.9.p5822

For let the centre of the circle, the point E, be taken, let EB, EC, ED be joined, and let BE be carried through to A.

13.prop.9.p5823

Since BC is the side of an equilateral decagon, therefore the circumference ACB is five times the circumference BC; therefore the circumference AC is quadruple of CB.

13.prop.9.p5824

But, as the circumference AC is to CB, so is the angle AEC to the angle CEB; [VI. 33] therefore the angle AEC is quadruple of the angle CEB.

13.prop.9.p5825

And, since the angle EBC is equal to the angle ECB, [I. 5] therefore the angle AEC is double of the angle ECB. [I. 32]

13.prop.9.p5826

And, since the straight line EC is equal to CD, for each of them is equal to the side of the hexagon inscribed in the circle ABC, [IV. 15, Por.] the angle CED is also equal to the angle CDE; [I. 5] therefore the angle ECB is double of the angle EDC. [I. 32]

13.prop.9.p5827

But the angle AEC was proved double of the angle ECB; therefore the angle AEC is quadruple of the angle EDC.

13.prop.9.p5828

But the angle AEC was also proved quadruple of the angle BEC; therefore the angle EDC is equal to the angle BEC.

13.prop.9.p5829

But the angle EBD is common to the two triangles BEC and BED; therefore the remaining angle BED is also equal to the remaining angle ECB; [I. 32] therefore the triangle EBD is equiangular with the triangle EBC.

13.prop.9.p5830

Therefore, proportionally, as DB is to BE, so is EB to BC. [VI. 4]

13.prop.9.p5831

But EB is equal to CD.

13.prop.9.p5832

Therefore, as BD is to DC, so is DC to CB.

13.prop.9.p5833

And BD is greater than DC; therefore DC is also greater than CB.

13.prop.9.p5834

Therefore the straight line BD has been cut in extreme and mean ratio, and DC is its greater segment. Q. E. D.

13.prop.10.p5835

If an equilateral pentagon be inscribed in a circle, the square on the side of the pentagon is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the same circle.

13.prop.10.p5836

Let ABCDE be a circle, and let the equilateral pentagon ABCDE be inscribed in the circle ABCDE.

13.prop.10.p5837

I say that the square on the side of the pentagon ABCDE is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the circle ABCDE.

13.prop.10.p5838

For let the centre of the circle, the point F, be taken, let AF be joined and carried through to the point G, let FB be joined, let FH be drawn from F perpendicular to AB and be carried through to K, let AK, KB be joined, let FL be again drawn from F perpendicular to AK, and be carried through to M, and let KN be joined.

13.prop.10.p5839

Since the circumference ABCG is equal to the circumference AEDG, and in them ABC is equal to AED, therefore the remainder, the circumference CG, is equal to the remainder GD.

13.prop.10.p5840

But CD belongs to a pentagon; therefore CG belongs to a decagon.

13.prop.10.p5841

And, since FA is equal to FB, and FH is perpendicular, therefore the angle AFK is also equal to the angle KFB. [I. 5, I. 26]

13.prop.10.p5842

Hence the circumference AK is also equal to KB; [III. 26] therefore the circumference AB is double of the circumference BK; therefore the straight line AK is a side of a decagon.

13.prop.10.p5843

For the same reason AK is also double of KM.

13.prop.10.p5844

Now, since the circumference AB is double of the circumference BK, while the circumference CD is equal to the circumference AB, therefore the circumference CD is also double of the circumference BK.

13.prop.10.p5845

But the circumference CD is also double of CG; therefore the circumference CG is equal to the circumference BK.

13.prop.10.p5846

But BK is double of KM, since KA is so also; therefore CG is also double of KM.

13.prop.10.p5847

But, further, the circumference CB is also double of the circumference BK, for the circumference CB is equal to BA.

13.prop.10.p5848

Therefore the whole circumference GB is also double of BM; hence the angle GFB is also double of the angle BFM. [VI. 33]

13.prop.10.p5849

But the angle GFB is also double of the angle FAB, for the angle FAB is equal to the angle ABF.

13.prop.10.p5850

Therefore the angle BFN is also equal to the angle FAB.

13.prop.10.p5851

But the angle ABF is common to the two triangles ABF and BFN; therefore the remaining angle AFB is equal to the remaining angle BNF; [I. 32] therefore the triangle ABF is equiangular with the triangle BFN.

13.prop.10.p5852

Therefore, proportionally, as the straight line AB is to BF, so is FB to BN; [VI. 4] therefore the rectangle AB, BN is equal to the square on BF. [VI. 17]

13.prop.10.p5853

Again, since AL is equal to LK, while LN is common and at right angles, therefore the base KN is equal to the base AN; [I. 4] therefore the angle LKN is also equal to the angle LAN.

13.prop.10.p5854

But the angle LAN is equal to the angle KBN; therefore the angle LKN is also equal to the angle KBN.

13.prop.10.p5855

And the angle at A is common to the two triangles AKB and AKN.

13.prop.10.p5856

Therefore the remaining angle AKB is equal to the remaining angle KNA; [I. 32] therefore the triangle KBA is equiangular with the triangle KNA.

13.prop.10.p5857

Therefore, proportionally, as the straight line BA is to AK, so is KA to AN; [VI. 4] therefore the rectangle BA, AN is equal to the square on AK. [VI. 17]

13.prop.10.p5858

But the rectangle AB, BN was also proved equal to the square on BF; therefore the rectangle AB, BN together with the rectangle BA, AN, that is, the square on BA [II. 2], is equal to the square on BF together with the square on AK.

13.prop.10.p5859

And BA is a side of the pentagon, BF of the hexagon [IV. 15, Por.], and AK of the decagon.

13.prop.10.p5860

Therefore etc. Q. E. D.

13.prop.11.p5861

If in a circle which has its diameter rational an equilateral pentagon be inscribed, the side of the pentagon is the irrational straight line called minor.

13.prop.11.p5862

For in the circle ABCDE which has its diameter rational let the equilateral pentagon ABCDE be inscribed; I say that the side of the pentagon is the irrational straight line called minor.

13.prop.11.p5863

For let the centre of the circle, the point F, be taken, let AF, FB be joined and carried through to the points, G, H, let AC be joined, and let FK be made a fourth part of AF.

13.prop.11.p5864

Now AF is rational; therefore FK is also rational.

13.prop.11.p5865

But BF is also rational; therefore the whole BK is rational.

13.prop.11.p5866

And, since the circumference ACG is equal to the circumference ADG, and in them ABC is equal to AED, therefore the remainder CG is equal to the remainder GD.

13.prop.11.p5867

And, if we join AD, we conclude that the angles at L are right, and CD is double of CL.

13.prop.11.p5868

For the same reason the angles at M are also right, and AC is double of CM.

13.prop.11.p5869

Since then the angle ALC is equal to the angle AMF, and the angle LAC is common to the two triangles ACL and AMF, therefore the remaining angle ACL is equal to the remaining angle MFA; [I. 32] therefore the triangle ACL is equiangular with the triangle AMF; therefore, proportionally, as LC is to CA, so is MF to FA.

13.prop.11.p5870

And the doubles of the antecedents may be taken; therefore, as the double of LC is to CA, so is the double of MF to FA.

13.prop.11.p5871

But, as the double of MF is to FA, so is MF to the half of FA; therefore also, as the double of LC is to CA, so is MF to the half of FA.

13.prop.11.p5872

And the halves of the consequents may be taken; therefore, as the double of LC is to the half of CA, so is MF to the fourth of FA.

13.prop.11.p5873

And DC is double of LC, CM is half of CA, and FK a fourth part of FA; therefore, as DC is to CM, so is MF to FK.

13.prop.11.p5874

Componendo also, as the sum of DC, CM is to CM, so is MK to KF; [V. 18] therefore also, as the square on the sum of DC, CM is to the square on CM, so is the square on MK to the square on KF.

13.prop.11.p5875

And since, when the straight line subtending two sides of the pentagon, as AC, is cut in extreme and mean ratio, the greater segment is equal to the side of the pentagon, that is, to DC, [XIII. 8] while the square on the greater segment added to the half of the whole is five times the square on the half of the whole, [XIII. 1] and CM is half of the whole AC, therefore the square on DC, CM taken as one straight line is five times the square on CM.

13.prop.11.p5876

But it was proved that, as the square on DC, CM taken as one straight line is to the square on CM, so is the square on MK to the square on KF; therefore the square on MK is five times the square on KF.

13.prop.11.p5877

But the square on KF is rational, for the diameter is rational; therefore the square on MK is also rational; therefore MK is rational

13.prop.11.p5878

And, since BF is quadruple of FK, therefore BK is five times KF; therefore the square on BK is twenty-five times the square on KF.

13.prop.11.p5879

But the square on MK is five times the square on KF; therefore the square on BK is five times the square on KM; therefore the square on BK has not to the square on KM the ratio which a square number has to a square number; therefore BK is incommensurable in length with KM. [X. 9]

13.prop.11.p5880

And each of them is rational.

13.prop.11.p5881

Therefore BK, KM are rational straight lines commensurable in square only.

13.prop.11.p5882

But, if from a rational straight line there be subtracted a rational straight line which is commensurable with the whole in square only, the remainder is irrational, namely an apotome; therefore MB is an apotome and MK the annex to it. [X. 73]

13.prop.11.p5883

I say next that MB is also a fourth apotome.

13.prop.11.p5884

Let the square on N be equal to that by which the square on BK is greater than the square on KM; therefore the square on BK is greater than the square on KM by the square on N.

13.prop.11.p5885

And, since KF is commensurable with FB, componendo also, KB is commensurable with FB. [X. 15]

13.prop.11.p5886

But BF is commensurable with BH; therefore BK is also commensurable with BH. [X. 12]

13.prop.11.p5887

And, since the square on BK is five times the square on KM, therefore the square on BK has to the square on KM the ratio which 5 has to 1.

13.prop.11.p5888

Therefore, convertendo, the square on BK has to the square on N the ratio which 5 has to 4 [V. 19, Por.], and this is not the ratio which a square number has to a square number; therefore BK is incommensurable with N; [X. 9] therefore the square on BK is greater than the square on KM by the square on a straight line incommensurable with BK.

13.prop.11.p5889

Since then the square on the whole BK is greater than the square on the annex KM by the square on a straight line incommensurable with BK, and the whole BK is commensurable with the rational straight line, BH, set out, therefore MB is a fourth apotome. [X. Deff. III. 4]

13.prop.11.p5890

But the rectangle contained by a rational straight line and a fourth apotome is irrational, and its square root is irrational, and is called minor. [X. 94]

13.prop.11.p5891

But the square on AB is equal to the rectangle HB, BM, because, when AH is joined, the triangle ABH is equiangular with the triangle ABM, and, as HB is to BA, so is AB to BM.

13.prop.11.p5892

Therefore the side AB of the pentagon is the irrational straight line called minor. Q. E. D.

13.prop.12.p5893

If an equilateral triangle be inscribed in a circle, the square on the side of the triangle is triple of the square on the radius of the circle.

13.prop.12.p5894

Let ABC be a circle, and let the equilateral triangle ABC be inscribed in it; I say that the square on one side of the triangle ABC is triple of the square on the radius of the circle.

13.prop.12.p5895

For let the centre D of the circle ABC be taken, let AD be joined and carried through to E, and let BE be joined.

13.prop.12.p5896

Then, since the triangle ABC is equilateral, therefore the circumference BEC is a third part of the circumference of the circle ABC.

13.prop.12.p5897

Therefore the circumference BE is a sixth part of the circumference of the circle; therefore the straight line BE belongs to a hexagon; therefore it is equal to the radius DE. [IV. 15, Por.]

13.prop.12.p5898

And, since AE is double of DE, the square on AE is quadruple of the square on ED, that is, of the square on BE.

13.prop.12.p5899

But the square on AE is equal to the squares on AB, BE; [III. 31, I. 47] therefore the squares on AB, BE are quadruple of the square on BE.

13.prop.12.p5900

Therefore, separando, the square on AB is triple of the square on BE.

13.prop.12.p5901

But BE is equal to DE; therefore the square on AB is triple of the square on DE.

13.prop.12.p5902

Therefore the square on the side of the triangle is triple of the square on the radius. Q. E. D.

13.prop.13.p5903

To construct a pyramid, to comprehend it in a given sphere, and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.

13.prop.13.p5904

Let the diameter AB of the given sphere be set out, and let it be cut at the point C so that AC is double of CB; let the semicircle ADB be described on AB, let CD be drawn from the point C at right angles to AB, and let DA be joined; let the circle EFG which has its radius equal to DC be set out, let the equilateral triangle EFG be inscribed in the circle EFG, [IV. 2] let the centre of the circle, the point H, be taken, [III. 1] let EH, HF, HG be joined; from the point H let HK be set up at right angles to the plane of the circle EFG, [XI. 12] let HK equal to the straight line AC be cut off from HK, and let KE, KF, KG be joined.

13.prop.13.p5905

Now, since KH is at right angles to the plane of the circle EFG, therefore it will also make right angles with all the straight lines which meet it and are in the plane of the circle EFG. [XI. Def. 3]

13.prop.13.p5906

But each of the straight lines HE, HF, HG meets it: therefore HK is at right angles to each of the straight lines HE, HF, HG.

13.prop.13.p5907

And, since AC is equal to HK, and CD to HE, and they contain right angles, therefore the base DA is equal to the base KE. [I. 4]

13.prop.13.p5908

For the same reason each of the straight lines KF, KG is also equal to DA; therefore the three straight lines KE, KF, KG are equal to one another.

13.prop.13.p5909

And, since AC is double of CB, therefore AB is triple of BC.

13.prop.13.p5910

But, as AB is to BC, so is the square on AD to the square on DC, as will be proved afterwards.

13.prop.13.p5911

Therefore the square on AD is triple of the square on DC.

13.prop.13.p5912

But the square on FE is also triple of the square on EH, [XIII. 12] and DC is equal to EH; therefore DA is also equal to EF.

13.prop.13.p5913

But DA was proved equal to each of the straight lines KE, KF, KG; therefore each of the straight lines EF, FG, GE is also equal to each of the straight lines KE, KF, KG; therefore the four triangles EFG, KEF, KFG, KEG are equilateral.

13.prop.13.p5914

Therefore a pyramid has been constructed out of four equilateral triangles, the triangle EFG being its base and the point K its vertex.

13.prop.13.p5915

It is next required to comprehend it in the given sphere and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.

13.prop.13.p5916

For let the straight line HL be produced in a straight line with KH, and let HL be made equal to CB.

13.prop.13.p5917

Now, since, as AC is to CD, so is CD to CB, [VI. 8, Por.] while AC is equal to KH, CD to HE, and CB to HL, therefore, as KH is to HE, so is EH to HL; therefore the rectangle KH, HL is equal to the square on EH. [VI. 17]

13.prop.13.p5918

And each of the angles KHE. EHL is right; therefore the semicircle described on KL will pass through E also. [cf. VI. 8, III. 31.]

13.prop.13.p5919

If then, KL remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points F, G, since, if FL, LG be joined, the angles at F, G similarly become right angles; and the pyramid will be comprehended in the given sphere.

13.prop.13.p5920

For KL, the diameter of the sphere, is equal to the diameter AB of the given sphere, inasmuch as KH was made equal to AC, and HL to CB.

13.prop.13.p5921

I say next that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid

13.prop.13.p5922

For, since AC is double of CB, therefore AB is triple of BC; and, convertendo, BA is one and a half times AC.

13.prop.13.p5923

But, as BA is to AC, so is the square on BA to the square on AD.

13.prop.13.p5924

Therefore the square on BA is also one and a half times the square on AD.

13.prop.13.p5925

And BA is the diameter of the given sphere, and AD is equal to the side of the pyramid.

13.prop.13.p5926

Therefore the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. Q. E. D.

13.prop.13.p5927

Lemma. It is to be proved that, as AB is to BC, so is the square on AD to the square on DC.

13.prop.13.p5928

For let the figure of the semicircle be set out, let DB be joined, let the square EC be described on AC, and let the parallelogram FB be completed.

13.prop.13.p5929

Since then, because the triangle DAB is equiangular with the triangle DAC, as BA is to AD, so is DA to AC, [VI. 8, VI. 4] therefore the rectangle BA, AC is equal to the square on AD. [VI. 17]

13.prop.13.p5930

And since, as AB is to BC, so is EB to BF, [VI. 1] and EB is the rectangle BA, AC, for EA is equal to AC, and BF is the rectangle AC, CB, therefore, as AB is to BC, so is the rectangle BA, AC to the rectangle AC, CB.

13.prop.13.p5931

And the rectangle BA, AC is equal to the square on AD, and the rectangle AC, CB to the square on DC, for the perpendicular DC is a mean proportional between the segments AC, CB of the base, because the angle ADB is right. [VI. 8, Por.]

13.prop.13.p5932

Therefore, as AB is to BC, so is the square on AD to the square on DC. Q. E. D.

13.prop.14.p5933

To construct an octahedron and comprehend it in a sphere, as in the preceding case; and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron.

13.prop.14.p5934

Let the diameter AB of the given sphere be set out, and let it be bisected at C; let the semicircle ADB be described on AB, let CD be drawn from C at right angles to AB, let DB be joined; let the square EFGH, having each of its sides equal to DB, be set out, let HF, EG be joined, from the point K let the straight line KL be set up at right angles to the plane of the square EFGH [XI. 12], and let it be carried through to the other side of the plane, as KM; from the straight lines KL, KM let KL, KM be respectively cut off equal to one of the straight lines EK, FK, GK, HK, and let LE, LF, LG, LH, ME, MF, MG, MH be joined.

13.prop.14.p5935

Then, since KE is equal to KH, and the angle EKH is right, therefore the square on HE is double of the square on EK. [I. 47]

13.prop.14.p5936

Again, since LK is equal to KE, and the angle LKE is right, therefore the square on EL is double of the square on EK. [id.]

13.prop.14.p5937

But the square on HE was also proved double of the square on EK; therefore the square on LE is equal to the square on EH; therefore LE is equal to EH.

13.prop.14.p5938

For the same reason LH is also equal to HE; therefore the triangle LEH is equilateral.

13.prop.14.p5939

Similarly we can prove that each of the remaining triangles of which the sides of the square EFGH are the bases, and the points L, M the vertices, is equilateral; therefore an octahedron has been constructed which is contained by eight equilateral triangles.

13.prop.14.p5940

It is next required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron.

13.prop.14.p5941

For, since the three straight lines LK, KM, KE are equal to one another, therefore the semicircle described on LM will also pass through E.

13.prop.14.p5942

And for the same reason, if, LM remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points F, G, H, and the octahedron will have been comprehended in a sphere.

13.prop.14.p5943

I say next that it is also comprehended in the given sphere.

13.prop.14.p5944

For, since LK is equal to KM, while KE is common, and they contain right angles, therefore the base LE is equal to the base EM. [I. 4]

13.prop.14.p5945

And, since the angle LEM is right, for it is in a semicircle, [III. 31] therefore the square on LM is double of the square on LE. [I. 47]

13.prop.14.p5946

Again, since AC is equal to CB, AB is double of BC.

13.prop.14.p5947

But, as AB is to BC, so is the square on AB to the square on BD; therefore the square on AB is double of the square on BD.

13.prop.14.p5948

But the square on LM was also proved double of the square on LE.

13.prop.14.p5949

And the square on DB is equal to the square on LE, for EH was made equal to DB.

13.prop.14.p5950

Therefore the square on AB is also equal to the square on LM; therefore AB is equal to LM.

13.prop.14.p5951

And AB is the diameter of the given sphere; therefore LM is equal to the diameter of the given sphere.

13.prop.14.p5952

Therefore the octahedron has been comprehended in the given sphere, and it has been demonstrated at the same time that the square on the diameter of the sphere is double of the square on the side of the octahedron. Q. E. D.

13.prop.15.p5953

To construct a cube and comprehend it in a sphere, like the pyramid; and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube.

13.prop.15.p5954

Let the diameter AB of the given sphere be set out, and let it be cut at C so that AC is double of CB; let the semicircle ADB be described on AB, let CD be drawn from C at right angles to AB, and let DB be joined; let the square EFGH having its side equal to DB be set out, from E, F, G, H let EK, FL, GM, HN be drawn at right angles to the plane of the square EFGH, from EK, FL, GM, HN let EK, FL, GM, HN respectively be cut off equal to one of the straight lines EF, FG, GH, HE, and let KL, LM, MN, NK be joined; therefore the cube FN has been constructed which is contained by six equal squares.

13.prop.15.p5955

It is then required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube.

13.prop.15.p5956

For let KG, EG be joined.

13.prop.15.p5957

Then, since the angle KEG is right, because KE is also at right angles to the plane EG and of course to the straight line EG also, [XI. Def. 3] therefore the semicircle described on KG will also pass through the point E.

13.prop.15.p5958

Again, since GF is at right angles to each of the straight lines FL, FE, GF is also at right angles to the plane FK; hence also, if we join FK, GF will be at right angles to FK; and for this reason again the semicircle described on GK will also pass through F.

13.prop.15.p5959

Similarly it will also pass through the remaining angular points of the cube.

13.prop.15.p5960

If then, KG remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, the cube will be comprehended in a sphere.

13.prop.15.p5961

I say next that it is also comprehended in the given sphere.

13.prop.15.p5962

For, since GF is equal to FE, and the angle at F is right, therefore the square on EG is double of the square on EF.

13.prop.15.p5963

But EF is equal to EK; therefore the square on EG is double of the square on EK; hence the squares on GE, EK, that is the square on GK [I. 47], is triple of the square on EK.

13.prop.15.p5964

And, since AB is triple of BC, while, as AB is to BC, so is the square on AB to the square on BD, therefore the square on AB is triple of the square on BD.

13.prop.15.p5965

But the square on GK was also proved triple of the square on KE.

13.prop.15.p5966

And KE was made equal to DB; therefore KG is also equal to AB.

13.prop.15.p5967

And AB is the diameter of the given sphere; therefore KG is also equal to the diameter of the given sphere.

13.prop.15.p5968

Therefore the cube has been comprehended in the given sphere; and it has been demonstrated at the same time that the square on the diameter of the sphere is triple of the square on the side of the cube. Q. E. D.

13.prop.16.p5969

To construct an icosahedron and comprehend it in a sphere, like the aforesaid figures; and to prove that the side of the icosahedron is the irrational straight line called minor.

13.prop.16.p5970

Let the diameter AB of the given sphere be set out, and let it be cut at C so that AC is quadruple of CB, let the semicircle ADB be described on AB, let the straight line CD be drawn from C at right angles to AB, and let DB be joined; let the circle EFGHK be set out and let its radius be equal to DB, let the equilateral and equiangular pentagon EFGHK be inscribed in the circle EFGHK, let the circumferences EF, FG, GH, HK, KE be bisected at the points L, M, N, O, P, and let LM, MN, NO, OP, PL, EP be joined.

13.prop.16.p5971

Therefore the pentagon LMNOP is also equilateral, and the straight line EP belongs to a decagon.

13.prop.16.p5972

Now from the points E, F, G, H, K let the straight lines EQ, FR, GS, HT, KU be set up at right angles to the plane of the circle, and let them be equal to the radius of the circle EFGHK, let QR, RS, ST, TU, UQ, QL, LR, RM, MS, SN, NT, TO, OU, UP, PQ be joined.

13.prop.16.p5973

Now, since each of the straight lines EQ, KU is at right angles to the same plane, therefore EQ is parallel to KU. [XI. 6]

13.prop.16.p5974

But it is also equal to it; and the straight lines joining those extremities of equal and parallel straight lines which are in the same direction are equal and parallel. [I. 33]

13.prop.16.p5975

Therefore QU is equal and parallel to EK.

13.prop.16.p5976

But EK belongs to an equilateral pentagon; therefore QU also belongs to the equilateral pentagon inscribed in the circle EFGHK.

13.prop.16.p5977

For the same reason each of the straight lines QR, RS, ST, TU also belongs to the equilateral pentagon inscribed in the circle EFGHK; therefore the pentagon QRSTU is equilateral.

13.prop.16.p5978

And, since QE belongs to a hexagon, and EP to a decagon, and the angle QEP is right, therefore QP belongs to a pentagon; for the square on the side of the pentagon is equal to the square on the side of the hexagon and the square on the side of the decagon inscribed in the same circle. [XIII. 10]

13.prop.16.p5979

For the same reason PU is also a side of a pentagon.

13.prop.16.p5980

But QU also belongs to a pentagon; therefore the triangle QPU is equilateral.

13.prop.16.p5981

For the same reason each of the triangles QLR, RMS, SNT, TOU is also equilateral.

13.prop.16.p5982

And, since each of the straight lines QL, QP was proved to belong to a pentagon, and LP also belongs to a pentagon, therefore the triangle QLP is equilateral.

13.prop.16.p5983

For the same reason each of the triangles LRM, MSN, NTO, OUP is also equilateral.

13.prop.16.p5984

Let the centre of the circle EFGHK the point V, be taken; from V let VZ be set up at right angles to the plane of the circle, let it be produced in the other direction, as VX, let there be cut off VW, the side of a hexagon, and each of the straight lines VX, WZ, being sides of a decagon, and let QZ, QW, UZ, EV, LV, LX, XM be joined.

13.prop.16.p5985

Now, since each of the straight lines VW, QE is at right angles to the plane of the circle, therefore VW is parallel to QE. [XI. 6]

13.prop.16.p5986

But they are also equal; therefore EV, QW are also equal and parallel. [I. 33]

13.prop.16.p5987

But EV belongs to a hexagon; therefore QW also belongs to a hexagon.

13.prop.16.p5988

And, since QW belongs to a hexagon, and WZ to a decagon, and the angle QWZ is right, therefore QZ belongs to a pentagon. [XIII. 10]

13.prop.16.p5989

For the same reason UZ also belongs to a pentagon, inasmuch as, if we join VK, WU, they will be equal and opposite, and VK, being a radius, belongs to a hexagon; [IV. 15, Por.] therefore WU also belongs to a hexagon.

13.prop.16.p5990

But WZ belongs to a decagon, and the angle UWZ is right; therefore UZ belongs to a pentagon. [XIII. 10]

13.prop.16.p5991

But QU also belongs to a pentagon; therefore the triangle QUZ is equilateral.

13.prop.16.p5992

For the same reason each of the remaining triangles of which the straight lines QR, RS, ST, TU are the bases, and the point Z the vertex, is also equilateral.

13.prop.16.p5993

Again, since VL belongs to a hexagon, and VX to a decagon, and the angle LVX is right, therefore LX belongs to a pentagon. [XIII. 10]

13.prop.16.p5994

For the same reason, if we join MV, which belongs to a hexagon, MX is also inferred to belong to a pentagon.

13.prop.16.p5995

But LM also belongs to a pentagon; therefore the triangle LMX is equilateral.

13.prop.16.p5996

Similarly it can be proved that each of the remaining triangles of which MN, NO, OP, PL are the bases, and the point X the vertex, is also equilateral.

13.prop.16.p5997

Therefore an icosahedron has been constructed which is contained by twenty equilateral triangles.

13.prop.16.p5998

It is next required to comprehend it in the given sphere, and to prove that the side of the icosahedron is the irrational straight line called minor.

13.prop.16.p5999

For, since VW belongs to a hexagon, and WZ to a decagon, therefore VZ has been cut in extreme and mean ratio at W, and VW is its greater segment; [XIII. 9] therefore, as ZV is to VW, so is VW to WZ.

13.prop.16.p6000

But VW is equal to VE, and WZ to VX; therefore, as ZV is to VE, so is EV to VX.

13.prop.16.p6001

And the angles ZVE, EVX are right; therefore, if we join the straight line EZ, the angle XEZ will be right because of the similarity of the triangles XEZ, VEZ.

13.prop.16.p6002

For the same reason, since, as ZV is to VW, so is VW to WZ, and ZV is equal to XW, and VW to WQ, therefore, as XW is to WQ, so is QW to WZ.

13.prop.16.p6003

And for this reason again, if we join QX, the angle at Q will be right; [VI. 8] therefore the semicircle described on XZ will also pass through Q. [III. 31]

13.prop.16.p6004

And if, XZ remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through Q and the remaining angular points of the icosahedron, and the icosahedron will have been comprehended in a sphere.

13.prop.16.p6005

I say next that it is also comprehended in the given sphere.

13.prop.16.p6006

For let VW be bisected at A'.

13.prop.16.p6007

Then, since the straight line VZ has been cut in extreme and mean ratio at W, and ZW is its lesser segment, therefore the square on ZW added to the half of the greater segment, that is WA', is five times the square on the half of the greater segment; [XIII. 3] therefore the square on ZA' is five times the square on .

13.prop.16.p6008

And ZX is double of ZA', and VW double of ; therefore the square on ZX is five times the square on WV.

13.prop.16.p6009

And, since AC is quadruple of CB, therefore AB is five times BC.

13.prop.16.p6010

But, as AB is to BC, so is the square on AB to the square on BD; [VI. 8, V. Def. 9] therefore the square on AB is five times the square on BD.

13.prop.16.p6011

But the square on ZX was also proved to be five times the square on VW.

13.prop.16.p6012

And DB is equal to VW, for each of them is equal to the radius of the circle EFGHK; therefore AB is also equal to XZ.

13.prop.16.p6013

And AB is the diameter of the given sphere; therefore XZ is also equal to the diameter of the given sphere.

13.prop.16.p6014

Therefore the icosahedron has been comprehended in the given sphere

13.prop.16.p6015

I say next that the side of the icosahedron is the irrational straight line called minor.

13.prop.16.p6016

For, since the diameter of the sphere is rational, and the square on it is five times the square on the radius of the circle EFGHK, therefore the radius of the circle EFGHK is also rational; hence its diameter is also rational.

13.prop.16.p6017

But, if an equilateral pentagon be inscribed in a circle which has its diameter rational, the side of the pentagon is the irrational straight line called minor. [XIII. 11]

13.prop.16.p6018

And the side of the pentagon EFGHK is the side of the icosahedron.

13.prop.16.p6019

Therefore the side of the icosahedron is the irrational straight line called minor.

13.prop.16.p6020

Porism. From this it is manifest that the square on the diameter of the sphere is five times the square on the radius of the circle from which the icosahedron has been described, and that the diameter of the sphere is composed of the side of the hexagon and two of the sides of the decagon inscribed in the same circle. Q. E. D.

13.prop.17.p6021

To construct a dodecahedron and comprehend it in a sphere, like the aforesaid figures, and to prove that the side of the dodecahedron is the irrational straight line called apotome.

13.prop.17.p6022

Let ABCD, CBEF, two planes of the aforesaid cube at right angles to one another, be set out, let the sides AB, BC, CD, DA, EF, EB, FC be bisected at G, H, K, L, M, N, O respectively, let GK, HL, MH, NO be joined, let the straight lines NP, PO, HQ be cut in extreme and mean ratio at the points R, S, T respectively, and let RP, PS, TQ be their greater segments; from the points R, S, T let RU, SV, TW be set up at right angles to the planes of the cube towards the outside of the cube, let them be made equal to RP, PS, TQ, and let UB, BW, WC, CV, VU be joined.

13.prop.17.p6023

I say that the pentagon UBWCV is equilateral, and in one plane, and is further equiangular.

13.prop.17.p6024

For let RB, SB, VB be joined.

13.prop.17.p6025

Then, since the straight line NP has been cut in extreme and mean ratio at R, and RP is the greater segment, therefore the squares on PN, NR are triple of the square on RP. [XIII. 4]

13.prop.17.p6026

But PN is equal to NB, and PR to RU; therefore the squares on BN, NR are triple of the square on RU.

13.prop.17.p6027

But the square on BR is equal to the squares on BN, NR; [I. 47] therefore the square on BR is triple of the square on RU; hence the squares on BR, RU are quadruple of the square on RU.

13.prop.17.p6028

But the square on BU is equal to the squares on BR, RU; therefore the square on BU is quadruple of the square on RU; therefore BU is double of RU.

13.prop.17.p6029

But VU is also double of UR, inasmuch as SR is also double of PR, that is, of RU; therefore BU is equal to UV.

13.prop.17.p6030

Similarly it can be proved that each of the straight lines BW, WC, CV is also equal to each of the straight lines BU, UV.

13.prop.17.p6031

Therefore the pentagon BUVCW is equilateral.

13.prop.17.p6032

I say next that it is also in one plane.

13.prop.17.p6033

For let PX be drawn from P parallel to each of the straight lines RU, SV and towards the outside of the cube, and let XH, HW be joined; I say that XHW is a straight line.

13.prop.17.p6034

For, since HQ has been cut in extreme and mean ratio at T, and QT is its greater segment, therefore, as HQ is to QT, so is QT to TH.

13.prop.17.p6035

But HQ is equal to HP, and QT to each of the straight lines TW, PX; therefore, as HP is to PX, so is WT to TH.

13.prop.17.p6036

And HP is parallel to TW, for each of them is at right angles to the plane BD; [XI. 6] and TH is parallel to PX, for each of them is at right angles to the plane BF. [id.]

13.prop.17.p6037

But if two triangles, as XPH, HTW, which have two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining straight lines will be in a straight line; [VI. 32] therefore XH is in a straight line with HW.

13.prop.17.p6038

But every straight line is in one plane; [XI. 1] therefore the pentagon UBWCV is in one plane.

13.prop.17.p6039

I say next that it is also equiangular.

13.prop.17.p6040

For, since the straight line NP has been cut in extreme and mean ratio at R, and PR is the greater segment, while PR is equal to PS, therefore NS has also been cut in extreme and mean ratio at P, and NP is the greater segment; [XIII. 5] therefore the squares on NS, SP are triple of the square on NP. [XIII. 4]

13.prop.17.p6041

But NP is equal to NB, and PS to SV; therefore the squares on NS, SV are triple of the square on NB; hence the squares on VS, SN, NB are quadruple of the square on NB.

13.prop.17.p6042

But the square on SB is equal to the squares on SN, NB; therefore the squares on BS, SV, that is, the square on BV —for the angle VSB is right—is quadruple of the square on NB; therefore VB is double of BN.

13.prop.17.p6043

But BC is also double of BN; therefore BV is equal to BC.

13.prop.17.p6044

And, since the two sides BU, UV are equal to the two sides BW, WC, and the base BV is equal to the base BC, therefore the angle BUV is equal to the angle BWC. [I. 8]

13.prop.17.p6045

Similarly we can prove that the angle UVC is also equal to the angle BWC; therefore the three angles BWC, BUV, UVC are equal to one another.

13.prop.17.p6046

But if in an equilateral pentagon three angles are equal to one another, the pentagon will be equiangular, [XIII. 7] therefore the pentagon BUVCW is equiangular.

13.prop.17.p6047

And it was also proved equilateral; therefore the pentagon BUVCW is equilateral and equiangular, and it is on one side BC of the cube.

13.prop.17.p6048

Therefore, if we make the same construction in the case of each of the twelve sides of the cube, a solid figure will have been constructed which is contained by twelve equilateral and equiangular pentagons, and which is called a dodecahedron.

13.prop.17.p6049

It is then required to comprehend it in the given sphere, and to prove that the side of the dodecahedron is the irrational straight line called apotome.

13.prop.17.p6050

For let XP be produced, and let the produced straight line be XZ; therefore PZ meets the diameter of the cube, and they bisect one another, for this has been proved in the last theorem but one of the eleventh book. [XI. 38]

13.prop.17.p6051

Let them cut at Z; therefore Z is the centre of the sphere which comprehends the cube, and ZP is half of the side of the cube.

13.prop.17.p6052

Let UZ be joined.

13.prop.17.p6053

Now, since the straight line NS has been cut in extreme and mean ratio at P, and NP is its greater segment, therefore the squares on NS, SP are triple of the square on NP. [XIII. 4]

13.prop.17.p6054

But NS is equal to XZ, inasmuch as NP is also equal to PZ, and XP to PS.

13.prop.17.p6055

But further PS is also equal to XU, since it is also equal to RP; therefore the squares on ZX, XU are triple of the square on NP.

13.prop.17.p6056

But the square on UZ is equal to the squares on ZX, XU; therefore the square on UZ is triple of the square on NP.

13.prop.17.p6057

But the square on the radius of the sphere which comprehends the cube is also triple of the square on the half of the side of the cube, for it has previously been shown how to construct a cube and comprehend it in a sphere, and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube. [XIII. 15]

13.prop.17.p6058

But, if whole is so related to whole, so is half to half also; and NP is half of the side of the cube; therefore UZ is equal to the radius of the sphere which comprehends the cube.

13.prop.17.p6059

And Z is the centre of the sphere which comprehends the cube; therefore the point U is on the surface of the sphere.

13.prop.17.p6060

Similarly we can prove that each of the remaining angles of the dodecahedron is also on the surface of the sphere; therefore the dodecahedron has been comprehended in the given sphere.

13.prop.17.p6061

I say next that the side of the dodecahedron is the irrational straight line called apotome.

13.prop.17.p6062

For since, when NP has been cut in extreme and mean ratio, RP is the greater segment, and, when PO has been cut in extreme and mean ratio, PS is the greater segment, therefore, when the whole NO is cut in extreme and mean ratio, RS is the greater segment.

13.prop.17.p6063

[Thus, since, as NP is to PR, so is PR to RN, the same is true of the doubles also, for parts have the same ratio as their equimultiples; [V. 15] therefore as NO is to RS, so is RS to the sum of NR, SO.

13.prop.17.p6064

But NO is greater than RS; therefore RS is also greater than the sum of NR, SO; therefore NO has been cut in extreme and mean ratio, and RS is its greater segment.]

13.prop.17.p6065

But RS is equal to UV; therefore, when NO is cut in extreme and mean ratio, UV is the greater segment.

13.prop.17.p6066

And, since the diameter of the sphere is rational, and the square on it is triple of the square on the side of the cube, therefore NO, being a side of the cube, is rational.

13.prop.17.p6067

[But if a rational line be cut in extreme and mean ratio, each of the segments is an irrational apotome.]

13.prop.17.p6068

Therefore UV, being a side of the dodecahedron, is an irrational apotome. [XIII. 6]

13.prop.17.p6069

Porism. From this it is manifest that, when the side of the cube is cut in extreme and mean ratio, the greater segment is the side of the dodecahedron. Q. E. D.

13.prop.18.p6070

To set out the sides of the five figures and to compare them with one another.

13.prop.18.p6071

Let AB, the diameter of the given sphere, be set out, and let it be cut at C so that AC is equal to CB, and at D so that AD is double of DB; let the semicircle AEB be described on AB, from C, D let CE, DF be drawn at right angles to AB, and let AF, FB, EB be joined.

13.prop.18.p6072

Then, since AD is double of DB, therefore AB is triple of BD.

13.prop.18.p6073

Convertendo, therefore, BA is one and a half times AD.

13.prop.18.p6074

But, as BA is to AD, so is the square on BA to the square on AF, [V. Def. 9, VI. 8] for the triangle AFB is equiangular with the triangle AFD; therefore the square on BA is one and a half times the square on AF.

13.prop.18.p6075

But the square on the diameter of the sphere is also one and a half times the square on the side of the pyramid. [XIII. 13]

13.prop.18.p6076

And AB is the diameter of the sphere; therefore AF is equal to the side of the pyramid.

13.prop.18.p6077

Again, since AD is double of DB, therefore AB is triple of BD.

13.prop.18.p6078

But, as AB is to BD, so is the square on AB to the square on BF; [VI. 8, V. Def. 9] therefore the square on AB is triple of the square on BF.

13.prop.18.p6079

But the square on the diameter of the sphere is also triple of the square on the side of the cube. [XIII. 15]

13.prop.18.p6080

And AB is the diameter of the sphere; therefore BF is the side of the cube.

13.prop.18.p6081

And, since AC is equal to CB, therefore AB is double of BC.

13.prop.18.p6082

But, as AB is to BC, so is the square on AB to the square on BE; therefore the square on AB is double of the square on BE.

13.prop.18.p6083

But the square on the diameter of the sphere is also double of the square on the side of the octahedron. [XIII. 14]

13.prop.18.p6084

And AB is the diameter of the given sphere; therefore BE is the side of the octahedron.

13.prop.18.p6085

Next, let AG be drawn from the point A at right angles to the straight line AB, let AG be made equal to AB, let GC be joined, and from H let HK be drawn perpendicular to AB.

13.prop.18.p6086

Then, since GA is double of AC, for GA is equal to AB, and, as GA is to AC, so is HK to KC, therefore HK is also double of KC.

13.prop.18.p6087

Therefore the square on HK is quadruple of the square on KC; therefore the squares on HK, KC, that is, the square on HC, is five times the square on KC.

13.prop.18.p6088

But HC is equal to CB; therefore the square on BC is five times the square on CK.

13.prop.18.p6089

And, since AB is double of CB, and, in them, AD is double of DB, therefore the remainder BD is double of the remainder DC.

13.prop.18.p6090

Therefore BC is triple of CD; therefore the square on BC is nine times the square on CD.

13.prop.18.p6091

But the square on BC is five times the square on CK; therefore the square on CK is greater than the square on CD; therefore CK is greater than CD.

13.prop.18.p6092

Let CL be made equal to CK, from L let LM be drawn at right angles to AB, and let MB be joined.

13.prop.18.p6093

Now, since the square on BC is five times the square on CK, and AB is double of BC, and KL double of CK, therefore the square on AB is five times the square on KL.

13.prop.18.p6094

But the square on the diameter of the sphere is also five times the square on the radius of the circle from which the icosahedron has been described. [XIII. 16, Por.]

13.prop.18.p6095

And AB is the diameter of the sphere; therefore KL is the radius of the circle from which the icosahedron has been described; therefore KL is a side of the hexagon in the said circle. [IV. 15, Por.]

13.prop.18.p6096

And, since the diameter of the sphere is made up of the side of the hexagon and two of the sides of the decagon inscribed in the same circle, [XIII. 16, Por.] and AB is the diameter of the sphere, while KL is a side of the hexagon, and AK is equal to LB, therefore each of the straight lines AK, LB is a side of the decagon inscribed in the circle from which the icosahedron has been described.

13.prop.18.p6097

And, since LB belongs to a decagon, and ML to a hexagon, for ML is equal to KL, since it is also equal to HK, being the same distance from the centre, and each of the straight lines HK, KL is double of KC, therefore MB belongs to a pentagon. [XIII. 10]

13.prop.18.p6098

But the side of the pentagon is the side of the icosahedron; [XIII. 16] therefore MB belongs to the icosahedron.

13.prop.18.p6099

Now, since FB is a side of the cube, let it be cut in extreme and mean ratio at N, and let NB be the greater segment; therefore NB is a side of the dodecahedron. [XIII. 17, Por.]

13.prop.18.p6100

And, since the square on the diameter of the sphere was proved to be one and a half times the square on the side AF of the pyramid, double of the square on the side BE of the octahedron and triple of the side FB of the cube, therefore, of parts of which the square on the diameter of the sphere contains six, the square on the side of the pyramid contains four, the square on the side of the octahedron three, and the square on the side of the cube two.

13.prop.18.p6101

Therefore the square on the side of the pyramid is fourthirds of the square on the side of the octahedron, and double of the square on the side of the cube; and the square on the side of the octahedron is one and a half times the square on the side of the cube.

13.prop.18.p6102

The said sides, therefore, of the three figures, I mean the pyramid, the octahedron and the cube, are to one another in rational ratios.

13.prop.18.p6103

But the remaining two, I mean the side of the icosahedron and the side of the dodecahedron, are not in rational ratios either to one another or to the aforesaid sides; for they are irrational, the one being minor [XIII. 16] and the other an apotome [XIII. 17].

13.prop.18.p6104

That the side MB of the icosahedron is greater than the side NB of the dodecahedron we can prove thus.

13.prop.18.p6105

For, since the triangle FDB is equiangular with the triangle FAB, [VI. 8] proportionally, as DB is to BF, so is BF to BA. [VI. 4]

13.prop.18.p6106

And, since the three straight lines are proportional, as the first is to the third, so is the square on the first to the square on the second; [V. Def. 9, VI. 20, Por.] therefore, as DB is to BA, so is the square on DB to the square on BF; therefore, inversely, as AB is to BD, so is the square on FB to the square on BD.

13.prop.18.p6107

But AB is triple of BD; therefore the square on FB is triple of the square on BD.

13.prop.18.p6108

But the square on AD is also quadruple of the square on DB, for AD is double of DB; therefore the square on AD is greater than the square on FB; therefore AD is greater than FB; therefore AL is by far greater than FB.

13.prop.18.p6109

And, when AL is cut in extreme and mean ratio, KL is the greater segment, inasmuch as LK belongs to a hexagon, and KA to a decagon; [XIII. 9] and, when FB is cut in extreme and mean ratio, NB is the greater segment; therefore KL is greater than NB.

13.prop.18.p6110

But KL is equal to LM; therefore LM is greater than NB.

13.prop.18.p6111

Therefore MB, which is a side of the icosahedron, is by far greater than NB which is a side of the dodecahedron. Q. E. D.

13.prop.18.p6112

I say next that no other figure, besides the said five figures, can be constructed which is contained by equilateral and equiangular figures equal to one another.

13.prop.18.p6113

For a solid angle cannot be constructed with two triangles, or indeed planes.

13.prop.18.p6114

With three triangles the angle of the pyramid is constructed, with four the angle of the octahedron, and with five the angle of the icosahedron; but a solid angle cannot be formed by six equilateral and equiangular triangles placed together at one point, for, the angle of the equilateral triangle being two-thirds of a right angle, the six will be equal to four right angles: which is impossible, for any solid angle is contained by angles less than four right angles. [XI. 21]

13.prop.18.p6115

For the same reason, neither can a solid angle be constructed by more than six plane angles.

13.prop.18.p6116

By three squares the angle of the cube is contained, but by four it is impossible for a solid angle to be contained, for they will again be four right angles.

13.prop.18.p6117

By three equilateral and equiangular pentagons the angle of the dodecahedron is contained; but by four such it is impossible for any solid angle to be contained, for, the angle of the equilateral pentagon being a right angle and a fifth, the four angles will be greater than four right angles: which is impossible.

13.prop.18.p6118

Neither again will a solid angle be contained by other polygonal figures by reason of the same absurdity.

13.prop.18.p6119

Therefore etc. Q. E. D.

13.prop.18.p6120

Lemma. But that the angle of the equilateral and equiangular pentagon is a right angle and a fifth we must prove thus.

13.prop.18.p6121

Let ABCDE be an equilateral and equiangular pentagon, let the circle ABCDE be circumscribed about it, let its centre F be taken, and let FA, FB, FC, FD, FE be joined.

13.prop.18.p6122

Therefore they bisect the angles of the pentagon at A, B, C, D, E.

13.prop.18.p6123

And, since the angles at F are equal to four right angles and are equal, therefore one of them, as the angle AFB, is one right angle less a fifth; therefore the remaining angles FAB, ABF consist of one right angle and a fifth.

13.prop.18.p6124

But the angle FAB is equal to the angle FBC; therefore the whole angle ABC of the pentagon consists of one right angle and a fifth. Q. E. D.

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