The elements of geometrie of the most auncient philosopher Euclide of Megara. Faithfully (now first) translated into the Englishe toung, by H. Billingsley, citizen of London. Whereunto are annexed certaine scholies, annotations, and inuentions, of the best mathematiciens, both of time past, and in this our age. With a very fruitfull præface made by M. I. Dee, specifying the chiefe mathematicall scie[n]ces, what they are, and wherunto commodious: where, also, are disclosed certaine new secrets mathematicall and mechanicall, vntill these our daies, greatly missed

¶The 13. Probleme The 13. Proposition. In a Cube geuen, to inscribe a Dodecahedron.

TAke a Cube ADFL.* 1.1 And diuide euery one of the sides therof into two equall partes in the pointes T, H, K, P: G, L, M, F: and pkQs. And drawe these right lines TK, GF, pQ, Hk, Ps, and LM: which lines againe diuide into two equall partes in the pointes N, V, Y, I, Z, X. And draw these right lines NY, VX, and IZ: Now the three lines NY, VX, and IZ, together with the diameter of the cube, shall cut the one the other into two equall partes in the centre of the cube, by the 3••. of the eleuenth: let that centre be the point O. And not to

[illustration]

stand long about the demonstration, vn∣derstand all these right lines to be equall and parallels to the sides of the cube and to cut the one the other right angled wise, by the 29. of the first. Let their halfes, namely, FV, GV, HI, and kI, and the rest such like, be deuided by an ex∣treme and meane proportion, by the 30. of the sixth: whose greater segmēts let be the lines FS, GB, HC, and kE, &c. and drawe these right lines GI, GE, BC, and BE. Now forasmuch as the line GI is equall to the whole line GV, which is the halfe of the side of the cube:* 1.2 and the line IE is equall to the line BV, that is, to the lesse segmēt: therfore, the squares of the lines GI and IE, are triple to the square of the line GB, by the 4. of the thirtenth: But vnto the squares of the lines GI and IE, the square of the line GE is equall, by the 47. of the first•• for the angle GIE is a right angle: Where∣fore the square of the line GE is triple to the square of the line GB. And foras∣much as the line FG is erected perpen∣dicularly to the plaine AGkL, by the 4. of the eleuenth: for it is erected per∣pendicularly to the two lines AG and GI: therefore the angle BGE is a right angle: for the line GE is drawen in the plaine AGkL. Wherefore the line BE, containing in power the two lines BG and GE, by the 47. of the first, is in power quadruple to the line GB (for the line GE was proued to be in power triple to the same line GB): Wherefore the line BE is in length double to the line BG, by the ••0. of the sixth. But (by construction) the line CE is double the line IE: Wherefore the halfes GB and IE, are in propor∣tion the one to the other, as their doubles BE and CE: by the 15. of the fifth. Wherefore the line CE is the greater segment of the line BE diuided by an extreme and meane proportion. And forasmuch as the selfe same thing may be proued touching the line BC: therefore the lines BE and BC, are equall, making an Isosceles triangle. Now let vs proue that three angles of the Pentagon of the Dodecahedron are set at the pointes B, C, E: and the other two angles are set betwene the lines BC and BE.

Forasmuch as the circle which containeth the triangle BCE circumscribeth the Pentagon whose side is the line CE, by the 11. of the fourth:* 1.3 Extend the plaine of the triangle BCE, by the parallel lines dB and HE, cutting the line AD, namely, the diameter of AD the base of the cube in the point I: and let it cut the line Ah the diameter of the cube in the point m. And by the point I drawe in the base AD, a parallel line vnto the line Ad: which let be Il. And forasmuch as from the triangle AHN is, by the parallel line lI, taken away the triangle AlI, like vnto the whole triangle AHN,* 1.4 by the Co∣rollary of the 2. of the sixth: the lines Al, and lI, shall be equall. But as the line HA is to the line Ad, so (by the 2. of the sixth) is the line Hl to the line lI, or to the line lA, which is equall to the line lI. And the greater segment of the line HA. (which is halfe the side of the cube) is, as before hath bene proued, the line Ad, that is, the line GB, which is equall to the line Ad (by the 33. of the first).

Wherefore the 〈◊〉〈◊〉 segment of the line Hl is the line ••A. And as the whole line Hl is to the grea∣ter segment, so shall the same greater segment Hl be to the lesse segment lA, by the 5. of the thir∣tenth. Wherefore the line HA is di∣uided

[illustration]

by an extreme and meane pro∣portion in the point l. But in the tri∣angle AHN, the line NA, which is drawen frō the centre of the base AD, is in the point I cut like vnto the line All, by thu parallel line lI (by the same second of the sixth): for the lines HN and Il, are parallels, by construction. Wherefore the line NA is in the point I diuided by an extreme and meane proportion by the superficies dBEH. And forasmuch as the line YON which coupleth the centres of the opposite bases, is a parallel to the line HE: A plaine superficies extended by the line YON, parallel wise to the plaine dB∣EH: the two plaines shall cut the lines AO and AN (the semidiameter of the cube, and the semidiameter of the base AD) into the selfe same proportions in the pointes m and I, by the 17. of the eleuēth. But the line AN is in the point I diuided by an extreme & meane pro∣portion: Wherefore the semidiameter of the cube is in the point m diuided by an extreme and meane proportion by the plaine of the triangle BCE. And forasmuch as the rest of the triangles de∣scribed in the cube after the like maner, may by the same reasons be proued to be in a plaine which cutteth the semidiameter of the cube by an extreme and means proportion: it is manifest that three plaines of the Dodecahedron shall vnder euery angle of the cube concurre in one & the self same point o•• the semidiameter being cut by an extreme and meane proportion. Now resteth to proue that the right lines which couple that point of the semidiameter with the angles of the triangle BEC, are e∣quall: whereby may be proued that the Pentagons are equilater, and equiangle.

Take the two bases of the cube.

[illustration]

Whereon are set the triangle BCE,* 1.5 namely, the bases AF and Ak, take also the same diameter of the cube that was before, namely, Ah: and let the side set at the poynt n, of the secti∣on of the diameter by an extreame & meane proportion, be the line Cn or Bn: and let the centre of the cube be as before the point O. And extend the line Cn to the line Bd, and let it con∣curre with it in the point a. And foras∣much as the plaine which passeth by the line HCE and the centre O (cut∣ting the cube into two equall partes) is parallel to AF the base of the cube by construction: imagine that by the poynt n, be extended a playne super∣ficies parallel to the former parallel playnes,* 1.6 which shall cutte the semidi∣ameter OA & the line Ca, proporti∣onally in the point n, by the 17. of the eleuenth: For those lines doo touch the extreame parallel plaines exten∣ded by the lines HE and EO, and by the lines Ad and dB. But it is proued that the line OA is diuided by an ex∣••••eame and meane proportion in the

poynt n: wherefore the line Ca, is also 〈◊〉〈◊〉 by an extr••ame and meane proportion in the poynt n. Agayne forasmuch as BCE is an Isoscels triangle, and it is proued that the line BI cutteth the base CE into two equall partes in the poynt I, the angles BIC and BIE•• shall be right angles. Imagine by the line BI and the centre O a plaine to passe (cutting the cube into two equall partes) parallel to the base AD. And vnto those plaines let there be imagined an other parallel plaine passing by the poynt n: which let be ne: which shall cutte the semidiameter AO and the halfe side of the cube, namely, the line lH, like, in the pointes n and c by the 17. of the eleuenth. Wherefore the line IH is in the poynt e diuided by an extreame 〈◊〉〈◊〉 meane proportiō. Wherfore the line He is equall to the line CI or IE: namely, ech are lesse segmēts. And forasmuch as the line Ie is to the line IC (which is equall ••o the ••ine EH) as the whole is to the greater segment, take away from the whole line Ie th•• greater segm•••• IC: there shall remayne the lesse segment Ce by the 5. of the thirtenth: Wherefore the line I•• is diui∣ded by an extreame & meane proportion in the point C. Againe vnto the same playnes imagine an o∣ther playne to passe by the point a, parallel wise, and let the same be ag. Now then (by the same 17. of the eleuenth) the lines Ca and Cg are in like sort cut in the pointes n and e. But the line Ca was in the point n cutte by an extreame and meane proportion, wherefore the line Cg shall be cutte in the poynt e, by an extreame & meane proportion. But the line IC is to the line Ce, as the greater segment is to the lesse: wherfore the line Ce, is to the line eg, as the greater segment to the lesse: and therefore their proportion is as the whole line IC is to the greater segment Ce, and as the greater segment Ce is to the lesse segment eg: wherefore the whole line Ceg which, maketh the greater segment and the lesse, is equall to the whole line IC or IE. And forasmuch as two parallel plaine superficieces (namely, that which is extended by IOB and that which is extended by the line ag) are cutte by the playne of the triangle BCE, which passeth by the lines ag and IB, their common sections ag and IB shall be parallels (by the 16. of the eleuenth). But the angle BIE or BIC is a right angle, wherefore the angle agC is also a right angle (by the 29. of the first) and those right angles are contayned vnder equall sides, namely, the line gC is equall to the line CI, and the line ag to the line BI, by the 33. of the first: wher∣fore the bases Ca and CB are equall, by the 4. of the first. But of the line CB the line CE was proued to be the greater segment: wherefore the same line CE is also the greater segment of the line Ca: but cn was also the greater segment of the same line Ca. Wherefore vnto the line CE, the line cn which is the side of the dodecahedron, and is set at the diameter, is equall. And by the same reason the rest of the sides, which are set at the diameter may be proued e••uall to lines equall to the line CE. Wherfore the pentagon inscribed in the circle where in is contained the triangle BCE is, by the 11. of the fourth equiangle, and equilater. And forasmch as two pentagons, set vpon euery one of the bases of the cube doo make a dodecahedron, and sixe bases of the cube doo receaue twelue angles of the dodecahedron: and the 8. semidiameters doo in the pointes where they are cutte by an extreame and meane proporti∣on receaue the rest: therefore the 12. pentagon bases contayning 20. solide angles doo inscribe the do∣decahedron in the cube: by the 1. diffinition of this booke. Wherefore in a cube geuen is inscribed a dodecahedron: which was required to be done.

First Corollary.

The diameter of the sphere, which containeth the dodecahedron, containeth in power these two sides, namely, the side of the Dodecahedron, and the side of the cube wherein the Dodecahedron is inscribed. For in the first figure a line drawne from the centre O, to the poynt B the angle of the Dodecahedron, namely the line OB, containeth in pow∣er these two lines OV the halfe side of the cube, and VB the halfe side of the dodecahedron, by the 47. of the first. Wherefore by the 15. of the fiueth, the double of the line OB, which is the diameter of the sphere containing the Dodecahedron, containeth in power the double of the other lines OV and VB, which are the sides of the cube, and of the dodecahedron.

¶ Second Corollary.

The side of a cube diuided by an extreme and meane proportion, maketh the lesse segment the side of the dodecahedron inscribed in it: and the greater seg∣ment the side of the cube inscribed in the same Dodecahedron: For it was before pro∣ued, that the side of the dodecahedron is the greater segment of BE the side of the triangle BEC•• but the side BE (which is equall to the line•• GB and SF) is the greater segmēt of GF the side of the cube: which line ••E (subtending th•• angle of the pentagon) was (by the ••. of this booke) the side of the cube inscribed in the dodecahedron.

Third Corollary.

The side of a cube, is equal to the sides of a Dodecahedron inscribed in it, and circumscribed about it. For it was manifest by this proposition, that the side of a cube ma∣keth the lesse segment, the side of a Dodecahedron inscribed in it, namely, as in the first figure the line BS the side of the Dodecahedron inscribed, is the lesse segmēt of the line GF the side of the cube. And it was proued in the 17. of the thirtenth, that the same side of the cube subtēdeth the angle of the penta∣gon of the Dodecahedron circumscribed: and therefore it maketh the greater segment the side of the Dodecahedron or of the pentagon, by the first corollary of the same. Wherefore it is equal to bothe those segments.

Notes

• * 1.1

  First part of the construc∣tion.
• * 1.2

  First part of the demon∣s••ration.
• * 1.3

  Second part of the con∣struction.
• * 1.4

  Second part of the De∣monstration.
• * 1.5

  Third part of the construc∣tion.
• * 1.6

  Third part of the demon∣stration.