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

¶An Assumpt.

Is a right line be deuided into two partes how soeuer: the rectangle paral∣lelogramme contayned vnder both the partes, is the meane proportionall betwene the squares of the same parts. And the rectangle parallelogramme contained vnder the whole line and one of the partes, is the meane pro∣portionall betwene the square of the whole line and the square of the sayd part.

* 1.1Suppose that there be two squares AB and BC, and let the lines DB and BE so be put that they both make one right line. Wherefore (by the 14. of the first) the lines FB and BG make also both one right line. And make perfect the parallelogramme AC. Then I say, that the rectangle parallelogramme DG is the meane proportionall betwene the squares AB and BC: and moreouer, that the parallelogramme DC is the meane proportionall betwene the squares AC and CB. First the parallelogramme AG is a square. For forasmuch as the line DB is equall to the line BF, and the line BE vnto the line BG, therfore the whole line DE is equall to the whole line FG. But the line DE is equall to either of these lines AH & KC, and the line FG is equall to either of these lines AK and HC (by the 34. of the first).* 1.2 Wher∣fore the parallelogrāme AC is equilater, it is also rectan∣gle

[illustration]

(by the 29. of the first). Wherefore (by the 46. of the first) the parallelogrāme AC is a square. Now for that as the line FB is to the line BG, so is the line DB to the line BE. But as y^e line FB is to the line BG, so (by the 1. of the sixt) is the parallelogrāme AB, which is the square of the line DB, to the parallelogramme DG, and as the line DB is to the line BE, so is the same parallelogrāme DG to the parallelogramme BC, which is the square of the line BE. Wherefore as the square AB is to the pa∣rallelogramme DG, so is the same parallelogramme DG to the square BC. Wherefore the parallelogramme DG is the meane proportionall betwene the squares AB and BC. I say moreouer, that the paral∣lelogramme DC is the meane proportionall betwene the squares AC and CB. For for that as the line AD is to the line DK, so is the line KG to the line GC (for they are ech equall to eche). Wherefore by composition (by the 18. of the fift) as the line AK is to the line KD, so is the line KC to the line CG. But as the line AK is to the line KD, so is the square of the line AK, which is the square AC, to the parallelogramme cōtayned vnder the lines AK and KD, which is the parallelogramme CD: and as the line KC is to the line CG, so also is the parallelogramme DC to the square of the line GC, which is the square BC. Wherefore as the square AC is to the parallelogramme DC, so is the parallelogrāme DC to the square BC. Wherefore the parallelogramme DC is the meane proportionall betwene the squares AC and BC: which was required to be demonstrated.

Notes

• * 1.1

  This Assumpt as was before noted, f••ll••w∣eth most ••ri••f∣ly without farther demon¦stration of the 25. of this booke.
• * 1.2

  Demonstra∣tion.