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

M. Iohn Dee, his sundry Inuentions and Annotacions, very necessary, here to be added and considered.

A The••reme.

If fower right lines be in continuall proportion, and vpon the squar•• of the first, as a base, be erec∣ted a rectangle parallelipipedon, whose heith is the fourth line: that rectangle parallelipipedon is equal to the Cube made of the second line. And if vpon the square of the fourth line, as a base, be erected a rectangle parallelipipedon, whose heith is the first line, that parallelipipedon is equall to the Cube made of the third line.

Suppose AB,

[illustration]

CD, E••, and ••H to be fower right lines in cōtinuall proportion:* 1.1 and vpō the square of AB (which let be AI) as a base, l••t be erected a rec∣tāgle parallelipi∣pedō, hauing his heith IK, equall to GH, the fourth line. And let that parallelipipedon be AK. Of the se∣cond line CD, let a Cube be made: whose square base, let be noted wi••h CQ: and let his heith b•• no∣ted by QL: & let the whole Cube be signified by C∣L. I say that AK is equall to CL. Let the like cōstructi∣on be for the cube of the third line: that is, vpon the square of ••H (which suppose to be GN) let a rectangle paralle∣lipipedon be e∣rected,

hauyng his heith NO, equall to A••, the first line: which parallelipipedon let be noted with GO. And suppose the cube of the third line (••F) to be ••M•• whose square base, let be noted by •••••• and hys heith by RM. I say now (secondly) that GO is equal to ••M. For the first part consider,* 1.2 that AI (the square base of AK) is to CQ, the square base of CL, as A•• is to the third line EF, by the ••. Corollary of the 20. of the sixth. But as A••, is to EF, so (by alternate proportion) is CD to GH, to CD. The cubes roote, is QL, the same cubes heith equall: and to GH is IK (by construction) equall: wherefore, as AI is to CQ, so is QL to IK. The bases therefore and heithes of AK and CL, are reciprocally in proporti∣on: wherefore by the second part of this 34. proposition, AK and CL are equall.* 1.3 For proofe of the se∣cond part of my theoreme, I say, that as AB, CD, ••F, and GH, are in continuall proportion forward, so are they backward in continuall proportion, as by the fourth of the fift may be proued. Wherefore now considering GH to be as first, and so A•• to be the fourth: the square base GN, is to the square base ••K, as GH is to CD, by the 2. corollary of the 20. of the sixth: But as GH is to CD, so is •••• to A••, by alternate proportion: to the Cubik roote ••F, is RM (the heith of the same Cube ••M) equall. And to AB, is the heith NO equall, by construction: wherfore as GN is to ••R, so is RM to NO. Therfore by the second part of this 34. proposition, GO is equall to EN. If fowre right lines (therefore) be in continu∣all proportion &c. as in the proposition: which was required to be demonstrated.

Notes

• * 1.1

  Construction.
• * 1.2

  Demonstratiō of the first part.
• * 1.3

  Demonstratiō of the second part.