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 5. Proposition. The 5. Probleme. In an Icosahedron geuen, to describe a Dodecahedron.

TAke an Icosahedron, one of whose solide angles let be Z.* 1.1 Now forasmuch as (by those thinges which haue bene proued in the 16. of the thirtenth) the ba∣ses of the triangles which contayne the angle of the Icosahedron doo make a pentagon inscribed in a circle, let that pentagon be ABGDE, which is made of the fiue bases of the triangles, whose playne superficiall angles re∣mayning make the solide angle geuen, namely, Z. And take the centres of the circles which contayne the foresaid triangles, which centers let be the poyntes I, T, K, M, L: and draw these right lines IT, TK, KM, ML, LI.* 1.2 Now then a perpendicular line drawne from the poynt Z to the playne superficies of the pentagon ABGDE, shall fall vpon the centre of the circle which contayneth the pentagon ABGDE (by those thinges which haue bene proued in the selfe same 16. of the thirtenth). Moreouer perpendicular lines drawne from the centre to the sides of the pentagon ABGDE shall in the poyntes C, N, O where they fall cut the right lines AB, BG, GD into two equall partes (by the 3. of the third). Draw these right lines CN and NO. And forasmuch as the angles CBN and NGO are equall, and are contained vn∣der equall sides, therefore the base CN is equall to the base NO (by the 4. of the first). More∣ouer

perpendicular lines dr••••••e from the poynt Z to the b••s•••• of the pentagon ABGDE, shall likewise cutte the bases into two equall partes (by th•••• of the third). For the perpendi∣culars passe by the centre (by the corollary

[illustration]

of the 12. of the thirtēth): Wherfore th••se perpendicular lines shall fal vpō the points C, N, O. And now forasmuch as the right lines ZI, IG are equall to the right lines ZT, TN, & also to the right lines ZK, KO (by reason of the likenes of the equall tri∣angles): therefore the line IT is a parallell to the line CN, and so also is the line TK to the line NO (by the 2. of the sixt). Wher∣fore the angles ITK, and CNO are equal (by the 11. of the eleuenth). Agayne foras∣much as the triangles CBN, and NGO are Isoscels triangles, therefore the angles BCN and BNC are equall (by the 5. of the first). And by the same reason the angles GNO, and GON are equall. And moreouer the angles BCN and BNC are equall to the angles GNO, and GON, for that the triangles CBN and NGO are equall and like. B•••• the three angles BNC, CNO, ONG, are equall to two right angles (by the 13. of the first): for that vpon the right line B•• are set the right lines CN & ON. And the three angles of the triangle CBN, namely, the angles BNC, BCN or GNO (for the angle G∣NO is equall to the angle BCN as it hath bene proued) and NBC are also equall to two right angles (by the 32. of the first). Wherefore taking away the angles BNC & GNO, the angle remayning, namely, CNO is equall to the angle remayning, namely, to CBN. Wher∣fore also the angle ITK (which is proued to be equall to the angle CNO) is equall to the an∣gle CBN. Wherefore ITK is the angle of a pentagon. And by the same reason may be pro∣••ed that the rest of the angles, namely•• the angles TIL, ILM, LMK, MKT, are equall to the rest of the angles, namely to BAE, AED, EDG, DGB. Wherefore ITKML is an e∣quilater and equiangle pentagon (by the 4. of the first) For the equall bases of the pentagon ITKML doo subtend equall angles set at the point Z, and comprehended vnder equall sides. Moreouer it is manifest that the pentagon ITKML is in one and the selfe same playne superficies. For foras••uch as the angles ONC and NCP are in one and the selfe s••me playne superficies, namely in the superficies ABGDE: But vnto the same playne super∣ficies the playne superficieces of the angles KTI and TIL are parallels (by the 15. of the ele∣uenth). And the triangles KTI and TIL concurre: wherefore they are in one and the selfe same playne superficies (by the corollary of the 16. of the eleuēth). And by the same reasō so may we proue that the triangles ILM, LMK, MKT are in the selfe same playne super∣ficies wherein are the triangles KTI and TIL. Wherefore the pentagon ITKML is in one and the selfe same playne superficies. Wherefore the solide angle of the Icosahedron, name¦ly the solide angle at the poynt Z subtendeth an equilater and equiangle pentagon plaine su∣perficies, which pentagon hath his plaine superficiall angles in the centres of the triangles which make the solide angle Z. And in like sort may we proue that the other eleuen solide angles of the Icosahedron, eche of which eleuen solide angles are equall and like to the solide angle Z (by the 16. of the thirtenth) are subtended vnto pentagons equall, and like, and in like sort set to the pentagon ITKML. And forasmuch as in those pētagons the right lines, which ioyne together the centers of the bases, are common sides, it followeth that those 12. pentagons include a solide which solide is therefore a d••dec••hedron (by the 24. diffinition of the eleuenth): and is, by the first diffinition of this booke, described in the Icosahedron, fiue

sides wh••reof 〈◊〉〈◊〉 set vpon the pentagon ABGDE. Wherefore in an Icosahedron geuen i•• inscribed a dodecahedron: which was required to be done.

An annotation of Hypsi••les.

This is to be noted, that if a man should demaund 〈◊〉〈◊〉 many sides an Icosahedron hath, we may thus answere: It is manifest that an Icosah••r••n is contayned vnder 20. triangles, and that euery triangle consisteth of three right lin••s. Now then multiply the 20. trian∣gles into the sides of one of the triangles, and so shall there be produced 6••. ••he halfe of which is 30. And so many sides hath an Icosahedron. And in like sort in a dodecahedron, foras∣much as 12. pentagons make a dodecahedron, and euery pentagon contayneth ••. right lines•• multiply •••• into 12. and there shall be produced 60. the halfe of which is 30. And so many are the sides of a dodecahedron. And the reason why we take the halfe, i••, for that euery side whe∣ther it be of a triangle or of a pentagon, or of a square as in a cube, ••s taken twise. And by the same reason may you finde out how many sides are in a cube, and in a pyramis, and in an octohedron.

But now agayne if ye will finde out the number of the angles of euery one of the solide figures, when ye haue done the same multiplication that ye did before, di••id•• the same sides, by the number of the plaine superficieces which comprehend one of the angles of the solides As for example, forasmuch as 5. triangles contayne the solide angle of an Icosahedron, diuide 60. by 5. and there will come forth 12. and so many solide angles hath an Icosahed••on. In a dodecahedron, forasmuch as three pentagons comprehend an angle, diuide 60. by 3. and there will come forth 20: and so many are the angles of a dodecahedron. And by the same reason may you finde out how many angles are in eche of the rest of the solide figures.

* 1.3If it be required to be knowne, how one of the plaines of any of the fiue solides being ge∣uen, there may be found out the inclination of the sayd plaines the one to the other, which con¦tayne eche of the solides. This (as sayth Isidorus our greate master) is fo••••d out after this maner. It is manifest that in a cube, the plaines which contayne i••, do•• 〈◊〉〈◊〉 the one the other by a right angle. But in a Tetrahedron, one of the triangles being geuen, let the endes of one of the sides of the sayd triangle be the centers, and let the space be the perpendicular line drawne from the toppe of the triangle to the base, and describe circumfer••nces of a circle, which shall cutte the one the other: and from the intersection to the centers draw right lines, which shall containe the inclination of the plaines cōtayning the Tetrahedron. In an Octo••e∣dron, take one of the sides of the triangle ther••of, and vpon it describe a square, and draw the diagonall line, and making the centres, the endes of the diagonall line, and the space likewise the perpendicular line drawne from the toppe of the triangle to the base, describe circumfe∣rences: and agayne from the common section to the centres draw right lines, and they shall contayne the inclination sought for. In an Icosahedron, vpon the side of one of the tri∣angles thereof, describe a pentagon, and draw the line which subtendeth one of the angles of the sayd pentagon, and making the centres the endes of that line, and the space the perpendi∣cular line of the triangle, describe circumferences: and draw from the common intersectio•• of the circumferences, vnto the centres right lines: and they shall contayne likewise the incli∣nation of the plaines of the icosahedron. In a dodecahedron, take one of the pentagons, and draw likewise the line which subtendeth one of the angles of the pentagon, and making the centres the endes of that line, and the space, the perpendicular line drawne from the section into two equall partes of that line to the side of the pentagon, which is parallel vnto it, de∣scribe circumferences: and from the point of the intersection of the circumferences draw vn∣to the centres right lines: and they shall also containe the inclination of the plaines of the do∣decahedron. Thus did this most singular learned man reason, thinking the de••onstration in euery one of them to be plaine and cleare. But to make the demonstration of them mani∣fest,

I think it good to declare and make open his wordes•• and first in a T••trahedron••

Suppose that there be a Pyramis or Te••rahedrō ABCD

[illustration]

cōtained vnder 4. equilater triangles: & let the toppe ther∣of be the point D. And (by the 10•• of the first) diuide the side AD into two equall parts in the poi••t E•• & draw the lines BE and EG. And forasmuch as ADB and ADC are e∣quilater triangles, and the line AD is diuided into two e∣quall partes, therefore the lines BE and EC fall perpendi∣cularly vpon the line AD, by the 8. of the first. Now I say that the angle BEC is an acute angle. For forasmuch as the line AC is double to the line AE (for by construction the line AD, which is equal to the line AC, is diuided into two equall partes in the point E): therefore the square of the line AC is quadruple to the square of the line AE (by the c••rollary of the 20. of the sixt). But the square of the line AC is equall to the squares of the lines AE and EC (by the 47. of the first): and the square of the line AC is to the square of the line CE (sesquitertia) as 4. to 3: (for the square of the line AC is proued quadruple to the square of the line AE): wherefore the square of the line BC (which is equall to the square of the line AC) is lesse then the squares of the two perpendicu∣lars BE & EC (for it is vnto them in subsesquialter proportiō, namely, as 4. to 6. or 2. to 3.) Wherefore (by the 13. of the second) the angle •• EC is an acute angle. Now forasmuch as the line AD is the common intersection of the two plaines ABD, and ADC, and in either of those plaines to one point of the common section are drawne perpendicular lines ••E and EC which containe an acute angle BEC•• therefore (by the 5. diffinition of the eleuenth) the an∣gle BEC is the inclination of the plaines, and it is geuen. For the line BC, which is the side of the triangle, being geuen, and any one of the lines BE or EC, which is the perpendicular of the equilater triangle, being also ge••e••: make the centres the poynte•• B and C, that is, the endes of one of the sides, and the space the perpendicular of the triangle, and describe circum∣ferences, and they shall cutte the one the other in the poynt ••. And from the poynt B draw to the centres B and C right lines, and they shall containe the inclination of the plain••••: and this is it which Isidorus before sayd. And now that making the centres the poynts B and C, and the space the perpendicular of the triangle, the circl••s described shall cutte the one the other, it is manifest, for either of the lines BE and EC i•• greater then half•• of the line BC.* 1.4 Now if the centers were the poynts B and C, and the space the halfe of the line BC, the circles described shall touch the one the other. But if the space ••e lesse then the halfe, they shal neither touch nor cut the one the other: but if it be greater, they shall vndoubtedly cut.

Againe suppose that vpon the square ABCD be set a pyramis, hauing his altitude the poynt E, and let the triangles which containe it, be

[illustration]

equilater: wherfore the pyramis ABCDE shalbe the halfe of the Octohedron (by the 2 corollary of the 14. of the thirtenth.) Deuide by the 10•• of the first) one side of one of the triangles, namely, the line AE, into two equal partes in the poynt F: and draw the lines BF and DF: wherefore the li••es BF and DF are equal and fal perpendicularly vpon the line AE (by the 4. and 8. of the first) Then I say that the angle BFD, is an ••••tuse an∣gle. For draw the line BD. And forasmuch as AC is a square, and the diameter is ••D: therefo•••• th•• s••uare of the line BD is double to the square of the line DA) by the 47. of the first.) But the square of the li•••• DA is to the square of the line

DF, as ••. is to 3.) as was in the former proued.) Wherefore the square of the line DB is to the square of the line FD, as 8. is to 3. (namely, as 2. to 1. and 4. to 3. added together), but the line DF is equal to the line FB. Wherefore the square of the line DB, is greater then the squares of the lines DF and FB (for it is to them, as 8. is to 6.) Wherfore the angle BFD, is an obtuse angle (by the 12. of the second.) And forasmuche as the line AE is the common section of the two plaines ABE and ADE cutting the one the other, and in either of those plaines to a poynt in the common section are drawne perpendicular lines, BF and DF, con∣taining an obtuse angle BFD: wherfore the angle BFD (contained of the right lines BF and DF) is the angle of the inclination of the plaines ABE and ADE. If therefore the angle BFD be geuen, the saide inclination also is g••uen. For forasmuch as the triangle of the Octohedron is geuen, and one of the sides of the Octohedron is the line AD, and vpon it is described the square AC, and BD the diameter of that square being geuen, and the lines BF and FD are the perpendiculars of that triangle: wherefore also the angle BFD is ge∣uen. Now then if vpon the side of the triangle be described a square: as the square AC, and the diameter BD be drawne, if also making the centres the poyntes B and D, and the space, the said perpendicular of the triangle, we describe circles, they shall cut the one the other in the poynt F. And the right lines which are drawne from the centres to the poynt F shal con∣taine that inclination, which is comprehended vnder the angle BFD, which is the angle of the inclination of those plaines. And it is manifest that either of the lines BF and FD, is greater then the halfe of the line. For for that, by the demonstration, it was proued that the square of the line BD is to the square of the line FD, as 8. is to 3: therfore the square of half the line BD, is to the square of the line FD, as ••. is to thre (for the square of halfe the line BD is the fourth part of the square of the whole line BD, by the 4. of the second). Wherefore either of the lines BF and FD, is greater then the line BD: wherfore the circles which are described by those lines BF and FD, and hauing their centres the poynts B and D shall cut the one the other. And thus much touching the octohedron.

As touching the Icosahedron, suppose an equilater

[illustration]

pētagon ABCDE, & vpon it let there be set a pyra∣mis hauing his toppe the poynt F: and let the trian∣gles which cōtaine it, be equilater. Now thē the pyra∣mis ABCDEF, shal be a part of the Icosahedrō. Let FC one side of one of the triangles be deuided into two equal partes in the poynt G. And draw the lines BG & GD, which shal be equal & fal perpēdicular∣ly vpon the line FC (as it is easie to se by the demō∣stratiō of the former). Thē I say that y^e angle BGD is an obtuse angle: which thing is manifest. For drawing the line BD, it shall subtend the obtuse an∣gle BCD of the pentagon (which is obtuse, by that which was demonstrated in the ende of the first co∣rollary of the 18. of the 13. booke:) But the angle BGD is greater then the angle BCD, for the lines BG and GD, are lesse then the lines BC and CD: wherefore likewise as in the for∣mer the angle BGD is the inclination of the triangles BFC, and CFD. Wherfore the an∣gle BGD being geuen, the inclination also of the plaines of the Icosahedron shall be geuen. For if vpon the side of the triangle of the Icosahedron be described an equilater pentagon, and then be drawne the line which subtendeth two sides of the pentagon, as in this figure the line BD, if also the perpendiculars BG and GD of the triangles be drawne, the angle BGD shalbe geuen. For if ye make the centres the endes of the line which subtendeth two sides of the pentagon, as the poynts B and D, and the space the perpendicular of the triangle, and so describe circles, they shall cut the one the other in the poynt G, and from the poynt of the in∣ters••ction

G, dra••e vnto the centres B and D right 〈…〉〈…〉 they shal containe the angle of the inclination BGD. And it is manifest, by the description of the figure, that either of the lines BG and GD is greater then 〈◊〉〈◊〉 line BD. Wh••ch 〈◊〉〈◊〉 may also thus be proued. Sup∣pose an equilater HKL, and vpon K•• (one of the sides 〈◊〉〈◊〉) describe an equilater pen∣tagon KMNXL, and draw the line ML. And diuide (••y the 10. of the first) the side KL into two equal parts in the poynt O•• & draw the line H••, which shall be the perpendicular of the triangle HKL (by the 8. of the first.) Then I say that the line HO is greater then half of the line ML, which subtendeth the inclination of the plaines. For from the poynt K draw (by the 1••. of the first) vnto the line ML a perpendicular line KP: and forasmuche as the angle KLP is greater then the third part of a right angle, that is, then the angle KHO•• (For the angle KLM is two fi••••th partes of a righ•• angle,

[illustration]

by the 4. of the first, and by the assumpt put after the first co∣rollary of the 18. of the thirtenth booke, and the angle K••O is one third part of one right angle; for the whole angle ••∣HL, wherof the angle KHO is the half, by the 4. of the first, is one third part of two right angles, by the ••••. of the first••) vpon the line ••L, and at the poynt L put vnto the angle KHO an equal angle PL•• (by the 23. of the first.) Wher∣fore the triangles PLR & OHK, shalbe equiangle, by the corollary of the 32. of the first. Wherefore also the line PL shalbe the perpendicular of the 〈◊〉〈◊〉 triangle described vpon the line RL. Wherefore (by the 〈◊〉〈◊〉 added by Flussas after the 12. proposition of the thirtenth booke) the line RL is in power sesquitercia, that is, as 4. is to 3. to the perpendicular LP. But the 〈…〉〈…〉 LR (by the 19. of the first. Wherfore y^e square of the li••e K∣L hath to the square of the line LP a greater proportion thē hath 4. to 3: but it hath to the square of the line HO that proportion that 4. hath to 3. Wher∣fore the line KL hath to the line LP a greater proportion then it hath to the line HO. Wherfore the line HO is greater then the line LP, by the 10. of the fifth.

As concerning a Dodecahedron. Take one of the squares of the cube wheron the Dode∣cahedron is described (by the 17. of the thirtenth): and let the same be ABCD: and let the two plaines of the Dodecahedron set vpon it be AEBFG, and GFDHC. Then I say that here also is geuen the inclination of the two

[illustration]

pentagons. Diuide (by the 10. of the first) the side FG into two equal partes in the poynt K. And from the poynt K draw vnto the line FG in either of the plaines AEB∣FG and GFDHC perpendicular lines KL and KM (by the 11. of the first.) And draw the line ML. First I say that the an∣gle MKL is an obtuse angle. For, by the discourse of the demonstration of the 17. proposition of the 13. boke, where is taught the description of the Dodecahedron, it is mani∣fest, that the line drawne perpendicularly from the poynt K to the square ABCD, is equal to halfe the side of the pentagon. Wherefore it is lesse then halfe of the line ML. Wherefore the angle MKL is an obtuse angle. Moreouer by the former discourse of the 17. proposition of the 13. booke, it was manifest that the square of the line KL is equal to the square of half the side of the cube, and to the square of halfe the side of the pentagon. And forasmuche as

the lines KL and KM are equal, and a••e ••••he greater then halfe of the line ML•• wherfore the angle MKL being geuen, there shall also be 〈◊〉〈◊〉 the inclination of the two plaines of the Dodecahedron. For forasmuch as the side of the square ABCD subtendeth two sides of the pentagon geuen, the pentagon also is geuen, and therefore also is geuen the line ML. But there is also geuen either of the lines MK and KL: for they are drawne perpendicularly frō the section into two equal partes of the line AB, which subtendeth two sides of the pentagon vnto the side of the pentagon, which is a parallel vnto the line AB: namely, to the side FG: Wherfore there is geuen the angle LKM, which is the angle of the inclination sought for. And now touching Isidorus wordes, he sayeth, that the pentagon being geuen, we must draw the line which subtendeth two sides of the pentagon, which line 15 equal to the side of the cube: and making the centres the endes of that line, and the space the perpendicular line, which is drawne from the section of the same line into two equal parts to the side of the pen∣tagon which is parellel to the said line, as in the former description the line KL, or the line KM, describe circumferences, and from the poynt of the intersection of the circumferences draw vnto the centres right lines which shall containe the angle of the inclination. For by that which was sayd before, namely; touching the Icosahedron, it is manifest that the perpendicular KL, is greater then halfe of the 〈◊〉〈◊〉 ML or CD, which is equal vnto it. And therefore the cir∣cles described by those perpendiculars and hauing to their centres the end of the line CD, shall cut the one the other, as was be∣fore proued.

Notes

• * 1.1

  Construction.
• * 1.2

  Demonstra∣tion.
• * 1.3

  That which here followeth concerning the inclination of the plaines of the fiue solides, was before tought (••hough not altogether after the same maner) out of Flussas in the latter ••nde of the 13 booke.
• * 1.4

  The reason of this ••ou shall most plainely see in that which is ad∣ded out of ••roclus a••ter the 22. propo∣sition of the first booke.