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. Probleme. The 17. Proposition. To make a Dodecahedron, and to comprehend it in the sphere geuen, wherin were comprehended the foresayd solides: and to proue that the side of the dodecahedron is an irrationall line of that kind which is called a re∣siduall line.

TAke two playne superficieces or bases of the foresayde cube,* 1.1 which let be the two squares ABCD and CBEF, cutting the one the other in the line BC per∣pendicularly according to the nature of a cube. And (by the 10. of the first) di∣uide euery one of the lines AB, BC, CD, DA, EF, EB, and FC into two equall partes in the poyntes G, H, K, L, M, N, X. And drawe these right lines GK and HL, cutting the one the other in the point P, and likewise draw the right lines MH and NX cutting the one the other in the poynt O. And diuide euery one of these right lines NO, OX, HP, and LP, by an extreme and meane proportion in the

[illustration]

points R, S, T, ct, and let their greater segments be RO, OS, TP, and Pct. And (by the 12. of the eleuenth) frō the poynts R, O, S, rayse vp to the outward part of the playne superficies E∣BCF of the foresayd cube, perpendicular lines RV, & OYSZ: and let eche of those perpēdi∣cular lines be equall to one of these lines RO, OS or TP, which perpendiculars shalbe paral∣lels (by the 6. of the eleuenth), and likewise from the pointes T, P, ct, rayse vp vnto the out∣ward part of the playne superficies ABCD of the sayd cube, these perpendicular lines TW, Pst, and ctl, eche of which perpendicular lines put equall also to the line OS, or OR or TP•• and the sayd perpendiculars shalbe parallels (by the foresayd 6. of the eleuenth)•• And draw these right lines YH, HW, BW, WC, CZ and CB.* 1.2 Now I say that the pentagon figure VBWCZ is equilater and in one and the self ••ame plaine superficies, and moreouer is equi∣angle. Draw these right lines TB, ••B, and ••B. And forasmuch as the right line NO is diuided by an extreme and meane proportion in the poynt R, and his greater segment is the line RO, therefore the square of the lines NO and NR are treble to the square of the line RO (by the 4. of this booke). But the line ON is equall to the line NB, and the line OR to the line RV. Wherefore the squares of the lines BN and RN are treble to the square of the line RV. But vnto the squares of the lines BN and NR is equall the square of the line BR (by the 47. of the first). Wherefore the square of the line BR is treble to the square of the line RV. Wherefore the squares of the lines BR and RV are quadruple to the square of the line RV. But vnto the squares of the lines BR and RV is equall the square of the line

BV (by the 47. of the first): for the angle BRV is a right angle (by the 2. diffinition of the eleuenth). Wherefore the square of the line BV is quadruple to the square of the line VR. Wherefore the line BV is double to the line RV (by the Corollary of the 20. of the sixth). And the line ZV is also double to the line RV (for* 1.3 that the line SR is double to the line O∣R, that is, to the line RV which is equall to the line OS). Wherfore the line BV is equall to the line VZ. And forasmuch as the two lines BN

[illustration]

and NR are equall to the two lines BH and H∣T, namely, the wholes and the lesse segmēts, and they comprehend right angles, namely, of the squares, BO, and BP, therefore (by the 4. of the first) the bases BR and BT are equall. And for∣asmuch as the lines BR, and BT are equall, and the two lines RV, and TW are also by constru∣ction equall, and the angles BRV, and BTW are by supposition right angles: therefore againe (by the 4. of the first) the bases BV and BW are equall: but the line BV, is proued equall to the line VZ. Wherfore the line BW is also equall to the line VZ. In like sort also may we proue that either of these lines WC, CZ is equal to the same line VZ. Wherefore the pentagon figure BVZ∣CW is equilater.

* 1.4Now I say that it is in one and the self same playne super••icies. Forasmuch as the line ZV is a parallell to the line SR (as was before proued) but vnto the same line SR, is the line CB a parallell (by the 28. of the first). Wherfore (by the 9. of the eleuenth) the line VZ is a parallell to the line CB. Wherefore, by the seuēth of the e∣leuenth, the right lines which ioyne thē together are in the selfe same playne wherein are the parallell lines. Wherefore the Trapesium BVZ∣C is in one playne. And the triangle BWC is in one playne (by the 2. of the eleuenth). Now to proue that the Trapesium BVZC & the triangle BWC are in one and the self same plaine, we must proue that the right lines YH, and HW are made directly one right line: which thing is thus proued. Forasmuch as the line HP is diuided by an extreme and meane pro∣portion in the point T, and his greater segment is the line PT, therefore as the line HP is to the line PT, so is the line PT to the line TH. But the line HP is equall to the line HO, and the line PT to either of these lines TW and OY. Wherefore as the line HO is to the line O∣Y, so is the line WT to the line TH. But the lines HO and TW being sides of like proportion are parallels (by the 6. of the eleuenth): (For either of them is erected perpendicularly to the plaine superficies BD)•• and the lines TH and OY are parallels, which are also sides of like proportion, by the same 6. of the eleuenth, (For either of them is also erected perpendicu∣larly to the playne superficies BF.) But when there are two triangles, hauing two sides pro∣portionall to two sides, so set vpon one angle, that their sides of like proportiō are also parallels (as the triangles YOH and HTW are)•• whose two sides, OH & HT, being in the two ba∣ses of the cube making an angle at the point H, the sides remayning of those triangles shal (by the 32. of the sixth) be in one right line. Wherfore the lines YH & HW make both one right line. But euery right line is (by the 3. of the eleuenth) in one & the self same plaine superficies. Wherefore if ye draw a right line from B to Y, there shalbe made a triangle BWY, which shalbe in one and the selfe same plaine (by the 2. of the eleuenth). And therefore the whole

pentagon figure VBWCZ is in one and the selfe same playne superficies.

Now also I say that it is equiangle.* 1.5 For forasmuch as the right line NO is diuided by an extreame and meane proportion in the point R, and his greater segment is OR, therefore as both the lines NO and OR added together is to the line ON, so (by the 5. of this booke) is the line ON to the line OR. But the line OR is equall to the line OS. Wherefore as the line SN is to the line NO, so is the line NO to the line OS. Wherfore the line SN is diuided by an extreme and meane proportion in the point O, and his greater segment is the line NO. Wherefore the squares of the lines NS and SO are treble to the square of the line NO (by the 4. of this booke). But the line NO is equall to the NB, and the line SO to the line SZ: wherfore the squares of the lines NS and ZS are treble to the square of the line NB: wher∣fore the squares of the lines ZS, SN and NB, are quadruple to the square of the line NB. But vnto the squares of the lines SN & NB (by the 47. of the first) is equal the square of the line SB: wherefore the squares of the lines BS and SZ, that is, the square of the line BZ, by the 47. of the first, (for the angle ZSB is a right angle by positiō) is quadruple to the square of the line NB. Wherfore the line BZ is double to the line BN (by the Corollary of the 20. of the sixth). But the line BC is also double to the line BN. Wherefore the line BZ is equall to the line BC. Now forasmuch as these two lines BV and VZ are equall to these two lines B∣W and WC, and the base BZ is equall to the base BC, therefore (by the 8. of the first) the an∣gle BVZ is equall to the angle BWC. And in like sort (by the 8. of the first) may we proue that the angle VZC is equall to the angle BWC (prouing first that the lines CB and CV are equal: which are proued equal by this, that the line NS is equal to the line XR, and ther∣fore the line CR is equal to the line BS, by the 47. of the first: wherfore also by the same y^e line CV is equal to the line BZ, that is, to the line BC (for the lines BC & BZ are proued equal.) Wherefore the three angles BWC, BVZ, and VZC are equall the one to the other. But if in an equilater pentagon figure there be thre angles equall the one to the other, the pentagon is (by the 7. of the thirtēth) equiangle: wherfore the pentagon BVZCW is equiangle. And it is also proued that it is equilater. Wherfore the pentagon BVZCW is both equilater & e∣quiangle. And it is made vpon one of the sides of the cube, namely, vpon BC. * 1.6 If therefore vpon euery one of the twelve sides of the cube be vsed the like construction, there shal then be made a dodecahedron contayned vnder twelue pentagons equilater and equiangle.

Now it is required to comprehend it in the sphere geuen, and to proue that the side of the dodecahedron is an irrationall line of that kinde which is called a residuall line. Extend the line YO, and let the line extended be YQ: now then the line YQ shall light vppon the diameter of the cube, and shall diuide the one the other into two equall parts.* 1.7 For this is manifest to se by the 39. of the eleuenth. (For if by the two lines NX and MH be drawen two playnes perpendicularly to the bases, and cutting the cube, the common section of those playnes shalbe the line YO produced: for their common section is from the poynt O erected perpendicularly to the plaine EBCF, by the 19. of the eleuenth). Let them cut the one the o∣ther in the point Q: wherefore Q is the centre of the sphere which comprehendeth the cube, and YQ is the halfe diameter of the sphere by that which was demōstrated in the 15. of this booke: wherefore the right lines drawen from the centre Q to all the angles of the cube shalbe equall. And draw a right line from the point V to the point Q. Now forasmuch as the right line NS i•• diuided by an extreme and meane proportion in the point O, and his greater seg∣ment is the line NO, as hath before ben proued, therefore the squares of the lines NS and SO are treble to the square of the line NO, by the 4. of this booke. But the line NS is equal to the line YQ (for the line NO is equal to the line OQ as hath before ben proued, & the line YO to the line OS) being both lesse segmentes: but the line OS is equall to the line TV, for the line RO is equall thereunto: wherefore the squares of the lines QY and YV are treble to the square of the line NO. But vnto the squares of the lines QY & YV the square of the line VQ is equall (by the 47. of the first): wherefore the square of the line VQ is treble to the

square of the line NO. But the semidiameter

[illustration]

of the sphere cōprehēding the said cube is in po∣wer treble to the half of the side of the cube. For we haue before (in the 15. of this booke) taught how to make a cube, and to comprehende it in a sphere, and haue proued that the diameter of the sphere is in power treble to the side of the cube. Now in what proportiō y^e whole is to the whole, in the same is the halfe to the halfe (by the 15. of the fifth). But the line NO is the half of the side of the cube. Wherefore the line VQ is equall to the semidiameter of the sphere cōprehēding the cube. But the point Q is the centre of the sphere cōprehending the cube. Wherefore the point V, which is one of the angles of the dodecahedron, toucheth the superficies of the sphere geuen. In like sort also may we proue, that euery one of the rest of the angles of the dodecahedron toucheth the superficies of the sphere. Wherefore the dodecahedron is comprehended in the sphere geuen.

Now I say, that the side of the dodecahedron is an irrationall line of that kinde which is called a residuall line.* 1.8 For forasmuch as the line NO is diuided by an extreme and meane proportion in the point R, and his greater segment is the line OR, and the line OX is also di∣uided by an extreme and meane proportion in the point S, and his greater segment is the line OS. Wherefore the whole line NX is diuided by an extreme and meane proportion, and his greater segment is the line RS. (For for that as the line ON is to the line OR, so is the line OR to the line NR, and in the same proportion also are their doubles (for the partes of eque∣multiplices haue one and the selfe same proportion with the whole, by the 15. of the fifth). Wherefore as the line NX is to the line RS, so is the line RS to both the lines NR and SX added together. But the line NX is greater then the line RS, by both the lines NR and SX added together. Wherefore the line NX is diuided by an extreme and meane proportion, and his greater segment is the line RS. But the line RS is equall to the line VZ, as hath before bene proued. Wherefore the line NX is diuided by an extreme and meane proportion, and his greater segment is the line VZ. And forasmuch as the diameter of the Sphere is ratio∣nall, and is in power treble to the side of the cube, by the 15. of this booke, therefore the line NX, being the side of the cube, is rationall. But if a rationall line be diuided by an extreme and meane proportion, either of the segmentes is (by the 6. of this booke) an irrationall line of that kinde which is called a residuall line. Wherefore the line VZ being the side of the do∣decahedron, is an irrational line of that kinde which is called a residuall line. Wherfore there is made a dodecahedron, and it is cōprehended in the Sphere geuen, wherein the other solides were contained, and it is proued that the side of the dodecahedron is a residuall line: which was required to be done, and also to be proued.

¶ Corollary.

Hereby it is manifest, that the side of a cube being diuided by an extreme

and meane proportion, the greater segment thereof is the side of the dode∣cahedron. As it was manifest by the line VZ which was proued to be the greater segment of the right line NX, namely, of the side of the cube.

A further construction of the dodecahedron after Flussas.

Forasmuch as it hath bene proued that the pentagon BVZCW is equilater and equiangle and toucheth one of the sides of the cube.* 1.9 Let vs show also by what meanes vpon eche of the 12. sides of the cube may in like sort be applyed pentagons ioyning one to the other, and composing the 12. bases of the dodecahedron. Draw in the former figure these right lines AI, ID, IL, ctK. Now forasmuch as the line PL was in the point ct diuided like vnto the lines PH, ON. or OX, and vpon the pointes T, P, ct, were erected perpendicular lines equall vnto the line OY, and the rest: namely, vnto the greater segmēt: and the lines T W and ct I were proued parallels, therefore the lines WI and Tct are parallels, by the 7. of the eleuenth, and 33. of the first. Wherefore also, by the 9. of the eleuenth, the lines WI and DC are parallels. Wherefore by the 7. of the eleuenth CWID is a playne superficies. And the triangle AID is a playne superficies, by the 2. of the eleuenth. Now it is manifest that the right lines ID, & IA are equall to the right line WC. For the right lines AL & ••ct (which are equall to the right lines BH, & HT) do make the subtēded lines Act and BT equall by the 4. of the first. And agayne forasmuch as the lines BT and TW contayne a right angle BTW, as also doo the right lines Act and ctI contayne the right an∣gle ActI (for the right lines WT, and Ict are erected perpendicularly vnto one and the selfe same playne ABCD by supposition). And the squares of the lines BT and TW are equall to the squares of the lines Act, and ctI (for it is proued that the line BT is equall to the line Act, and the line TW to the line ctI). And vnto the squares of the lines BT and TW is equall the square of the line BW, by the 47. of the first: likewise by the same vnto the squares of the lines Act and ctI is equall the square of the line AI. Wherefore the square of the line BW is equall to the square of the line AI, wherefore also the line BT is equall to the line AI. And by the same reason are the lines ID and WC equall to the same lines. Now forasmuch as the lines AI and ID, and the lines AL and LD are equall, and the base IL is common to them both, the angles ALI and DLI shalbe equall, by the 8. of the first: and therefore they are right angles, by the 10. diffinition of the first. And by the same reason are the angles WHB, and W∣HC right angles. And forasmuch as the two lines HT and TW are equall to the two lines Lct and ct∣I, and they contayne equall angles, that is, right angles by supposition, therefore the angles WH∣T, and ILct, are equall by the 4. of the first. Wherefore the playne superficies AID is in like sort incli∣ned to the playne superficies ABCD, as the playne superficies BWC is inclined to the same playne A∣BCD, by the 4. diffinition of the eleuenth. In like sort may we proue that the playne WCDI is in like sort inclined to the playne ABCD, as the playne BVZC is to the playne EBCF. For that in the trian∣gles YOH and ctPK which consist of equall sides (eche to his correspondent side), the angles YHO, and ctKP, which are the angles of the inclination, are equall. And now if the right line ctK be extended to the point a, and the pentagon CWIDa be made perfect, we may, by the same reason, proue that that playne is equiangle and equilater, that we proued the pentagon BVZCW to be equaliter and e∣quiangle. And likewise if the other playnes BWIA and AID be made perfect, they may be proued to be equall and like pentagons and in like sort situate, and they are set vpon these common right lines B∣W, WC, WI, AI, and ID. And obseruing this methode, there shall vpon euery one of the 12. ••id••s of the cube be set euery one of the 12. pentagons which compose the dodecahedron.

¶ Certayne Corollaryes added by Flussas.

First Corollary.

The side of a cube, is equall to the right line which subtendeth the angle of the pentagon of a dode∣cahedron contayned in one and the selfe same sphere with the cube.

For the angles BWC and AID, are subtended of the lines BC and AD. Which are sides of the Cube••

¶Second Corollary.

In a dodecahedron there are sixe sides euery two of which are parallels and opposite, whose secti∣ons into two equall partes, are coupled by three right lines, which in the center of the sphere which contayneth the dodecahedron, deuide into two equall partes and perpendicularly both them selues and also the sides.

For vpon the sixe bases of the cube are set sixe sides of the dodecahedron, as it hath bene pro∣ued (by the lines ZV, WI &c.) which are cutte into two equall partes by right lines, which ioyne together the centers of the bases of the cube, as the line YO produced, and the other like. Which lines coupling together the centers of the bases are three in number, cutting the one the other perpendicu∣larly (for they are parallels to the sides of the cube) and they cutte the one the other into two equall partes in the center of the sphere which contayneth the cube (by that which was demonstrated in the 15. of this booke). And vnto these equall lines, ioyning together the centers of the bases of the cube, are without the bases added equall partes OY, Pct, and the other like, which by supposition are equall to halfe of the side of the dodecahedron. Wherefore the whole lines, which ioyne together the sectoins of the opposite sides of the dodecahedron, are equall, and they cut those sides into two equall partes and perpendicularly.

Third Corollary.

A right line ioyning together the poynts of the sections of the opposite sides of the dodecahedron in∣to two equall partes, being diuided by an extreame and meane proportion: the greater segment thereof shalbe the side of the cube, and the lesse segment the side of the dodecahedron contayned in the selfe same sphere.

For it was proued that the right line YQ is diuided by an extreame and meane proportion in the poynt O, and that his greater segment OQ is halfe the side of the cube, and his lesse segment OY is halfe of the side VZ (which is the side of the dodecahedron). Wherefore it followeth (by the 15. of the fifth) that their doubles are in the same proportion. Wherefore the double of the line YQ which ioy∣neth the poynt opposite vnto the line Y, is the whole: and the greater segment is the double of the line OQ which is the side of the cube: & the lesse segment is the double of the line YO, which is equall to the side of the dodecahedron, namely, to the side VZ.

Notes

• * 1.1

  First part of the construc∣tion.
• * 1.2

  The pentagon VBWCZ•• proued equi∣later.
• * 1.3

  The lin•• SR is equal & pa∣rallel to the line ZV (by the 33. of the first) for they ioyne toge∣ther equall and parallel lines, and are in the same su••e••ficies with them, by the 7. of the ele∣uenth.
• * 1.4

  The pentagon VBWCR, proued to be in one and the selfe same playne su∣perficies.
• * 1.5

  The pentagon VBWCZ, it proued equi∣angle.
• * 1.6

  Looke for a farther con∣struction after Flussas at the ende of the de∣monstration.
• * 1.7

  That the do∣decahedron is contayned in the sphere geuen.
• * 1.8

  That the side of the dodeca∣hedron is a residuall line.
• * 1.9

  Draw in the former figure these lines, ctA, ctL, c••D.