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

A Theoreme. 2.

If a right line, deuided by an extreme and meane proportion, be geuen, and to the great segment ••herof, he directly adioyned a line equal to the whole line geuen, that adioyned line, and the said greater segment, do make a line diuided by extreme and meane proportion, whose greater segment is the line ••dioyned.

Suppose the line geuen, deuided by extreame and meane proportion, to be AB deuided in the point C, and his greater segment, let be AC: vnto AC directly adioyne a line equall to AB: let that be AD: I say, that AD, together with AC, (that is DC) is a deuided by extreme and middel proportion, whose greater segment is AD, the line adioyned. Deuide AD, equally in the point E. Now, foras∣much as AE, is the halfe of AD, (by construction,) it is also, the halfe of AB (equall, to AD, by con∣struction): Wherfore by the 1. of the thirtenth, the square of the line composed of AC and AE (which ••ne is EC) is quintuple to the square of the line AE. Wherefore the double of AE, and the line AC

composed, (as in one right line) is a line de∣uided

[illustration]

by extreme and meane proportion, by the conuerse of this third (by me demonstra∣ted): and the double of AE, is the greater segment. But DC is the line composed of the double of AE, & the line AC: and with all, AD is the double of AE. Wherfore, DC, is a line deuided by extreme and meane pro∣portion, and AD, i•• hi•• greater segment. If a right line, therefore, deuided by extreme and meane pro∣portion, be geuen, and to the greater segment thereof, be directly adioyned a line equall to the whole line geuen, that adioyned line, and the sayd greater segment, do make a line diuided by extreame and meane proportion, whose greater segment, is the line adioyned: Which was required to be demon∣strated.

Two other briefe demonstrations of the same.

Forasmuch as, AD is to AC: as AB, is to AC (because AD is equall to AB, by construction): but as AB is to AC, so is AC to CB: by supposition. Therefore by the 11. of the fifth, as AC, is to CB, so is AD to AC. * 1.1 Wherefore, as AC, and CB, (which is AB) is to CB: so is AD, and AC (which is DC) to AC. Therefore, euersedly, as AB, is to AC: so is DC to AD. And it is proued, AD, to be to AC: as AC is to CB. Wherefore as AB is to AC, and AC, to CB: so is DC, to AD, and AD, to AC. But AB, AC, and CB are in continuall proportion, by supposition: Wherfore DC, AD, and AC, are in continuall proportion. Wherefore, by the 3. definition of the sixth booke, DC, is deuided by extreme and middell proportion, and his greatest segment, is AD. Which was to be demonstrated. Note from the marke * 1.2, how this hath two demonstrations. One I haue set in the margent by.

¶A Corollary. 1.

Vpon Euclides third proposition demonstrated, it is made euident: that, of a line deuided by ex∣treame and meane proportion, if you produce the lesse segment, equally to the length of the greater: the line therby adioyned, together with the sayd lesse segment, make a new line deuided by extreame and middle proportion: Whose lesse segment, is the line adioyned.

For, if AB, be deuided by extreme and middell proportion in the point C, AC, being the greater segment, and CB be produced, from the poynt B, making a line, with CB, equall to AC, which let be CQ: and the line thereby adioyned, let be BQ: I say that CQ, is a line also deuided by an extreame and meane proportion, in the point B: and that BQ (the line adioyned) is the lesse segment. For by the thirde, it is proued, that halfe AC, (which, let be, CD) with CB, as one line, composed, hath his powre or square, quintuple to the powre of the

[illustration]

segment CD: Wherfore, by the second of this booke, the double of CD, is de∣uided by extreme and middell propor∣tion•• and the greater segment thereof, shalbe CB. But, by construction, CQ, is the double of CD, for it is equall to AC. Wherefore CQ is deuided by extreme and middle propor∣tion, in the point B: and the greater segment thereof shalbe, CB. Wherefore BQ, is the lesse segment, which is the line adioyned. Therefore, a line being deuided, by extreme and middell proportion, if the lesse segment, be produced equally to the length of the greater segment, the line thereby adioyned to∣gether with the sayd lesse segment, make a new line deuided, by extreme & meane proportion, who••e lesse segment, is the line adioyned. Which was to be demonstrated.

¶A Corollary. 2.

If•• from the greater segment, of a line diuided, by extreme and middle proportion, a line, equall to the lesse segment be cut of: the greater segment, thereby, is also deuided by extreme and meane propor∣tion, whose greater segment•• shall be 〈◊〉〈◊〉 that part of it, which is cut of.

For, taking from AC, a line equall to CB: let AR remayne. I say, that AC, is deuided by an ex∣treme and meane proportion in the point R: and that CR, the line cut of, is the greater segment. For it is proued in the former Corollary that CQ is deuided by extreme and meane proportion in the point B. But AC, is equall to CQ, by construction: and CR is equall to CB by construction: Wherefore the

re••idue, AR is equall to BQ the residue. Seing therfore the whole AC is equall to the whole CQ: and the greater part of AC, which is C∣R

[illustration]

is equal to CB the greater part of CQ and the lesse segmēt also equall to the lesse: and withall seing CQ is proued to be diuided by extreme & meane proportion in the point B, it foloweth of necessity, that, AC, is diuided by extreme and meane proportion in the point R. And seing CB, is the greater segment of CQ: CR shall be the greater segment of AC. Which was to be de∣monstrated.

A Corollary. 3.

It is euident thereby, a line being diuided by extreme and meane proportion, that the line whe••••∣by the greater segment excedeth the lesse, together with the lesse segment, do make a line diuided by extreme and meane proportion: whose lesse segment, is the sayd line of exceesse, or difference betwene the segments.

Iohn Dee.

¶ Two new wayes, to deuide any right line geuen by an extreme and meane proportion: demonstrated and added by M. Dee.

A Probleme.

To deuide by an extreme and meane proportion, any right line geuen, in length and position.

Suppose a line geuen in length and position•• to be AB. I say that AB is to be deuided by an extreme, and meane proportion.* 1.3 Deuide AB into two equall parts as in the point C. Produce AB directly, from the point B, to the point D: making BD, equal to BC. To the line AD, and at the point D, let a line be drawen * 1.4perpendicular: by the 11. of the first, which let be DF: (of what length you will). From DF and at the point D, cut of the sixth parte of

[illustration]

DF: by the 9. of the sixth. And let that sixth part, be the line DG. Vppon DF, as a diame∣ter, describe a semicircle: which let be DH∣F. From the point G, rere a line perpendicu∣lar to DF, which suppose to be GH: and let it come to the circumference of DHF, in the point H. Draw right lines, HD, and HF. Pro∣duce DH, from the point H, so long, till a line adioyned with DH, be equall to HF, which let be DI, equall to HF. From the point H, to the point B, (the one ende of our line geuen) let a right line be drawen: as H∣B. From the point I, let a line be drawen, to the line AB: so that it be also parallel to the line HB. Which parallel line suppose to be I∣K: cutting the line AB, at the point K. I say that AB, is deuided by an extreme & meane proportion, in the point K. For the triangle DKI, hauing HB, parallel to IK, hath his sides DK and DI,* 1.5 cut proportionally, by the 2. of the sixth. Wherefore as IH is to HD: so is KB, to BD. And therfore compoundingly, (by the 18. of the fiueth) as DI, is to DH: so is DK to DB. But by construction DI is equall to HF: wherefore by the 7. of the fifth, DI is to DH, as HF is to DH. Wherefore by the 11. of the fifth, DK is to DB, as HF is to DH. Wherefore the square of DK is to the square of DB, as the square of HF, is to the square of DH: by the 22. of the sixth. But the square of HF, is to the square of DH: as the line GF is to the line GD•• by my corrollary vpon the 5, probleme of my additions to the second proposition of the twelfth. Wherefore by the 11. of the fifth, the square of DK is to the square of DB, as the line GF is to the line GD. But by construction, GF is quintuple to GD. Wherefore the square of DK is quintuple to the square of DB: and therefore, the double of DB, is deuided by an ex∣treme and meane proportiō, and BK is the greater segment therof, by the 2. of this thirtenth. Where∣fore seing AB is the double of DB by construction: the line AB is deuided by an extreme and meane

proportion: and his greater segment, is the line BK. Wherefore AB is deuided by an extreme and meane proportion, in the point K. We haue therefore deuided by extreme and meane proportion any line geuen in length and position. Which was requisite to be done.

The second way to execute this probleme.

Suppose the line geuen to be AB. Deuide A•• into two equall parts: as suppose it to be done in the point C. Produce AB from the point B: adioyning a line equall to BC, which let be BD. To the right line AD, and at the point D, erect a perpendicular line equall to BD, let that be DE. Produce ED frō the point D to the point F: making DF to contayne fiue such equall partes, as DE is one. Now vpon EF as a diameter, describe a semicircle which let 〈◊〉〈◊〉 EKF, and let the

[illustration]

point where the circumference of EKF, doth cut the line AB, be the point K. I say that AB, is deuided in the point K, by an extreme and meane proportion. For by the 13. of the sixth ED, DK, & DF, are three lines in continuall proportion, (DK being the middle proportionall)•• Wherefore by the corollary of the 20. of the sixth, as ED is to DF, so is the square of ED, to the square of DK, but by construction, ED, is sub∣quintuple to DF. Wherefore the square of ED, is subquintuple to the square of DK. And therefore the square of DK, is quintuple to the square of ED. And ED is equall to ED, by construction, therefore the square of DK, is quintuple to the square of E D. Wherefore the dou∣ble of BD, is deuided by an extreme and meane proportion: whose greater segment is BK•• by the second of this thirte••th. But by con∣struction, AB, is the double of ••D•• Wherefore AB, is diuided by ex∣treme and meane proportion, and his greater segment, is BK: and thereby, K•• the point of the diuision. We haue therefore deuided by extreme and meane proportion, any right line geuen, in length and po∣sition. Which was to be done.

Note••

Ech of these wayes, may well be executed: But in the first, you haue this auantage: that the diame∣ter is taken at pleasure. Which ••n the second way, is euer iust thrise so long, as the line geuen to be de∣uided.

Iohn Dee.

Notes

• * 1.1

  Therefore, by my second The∣oreme added v∣pon the second proposition, DC is deuided by ex¦treame and meane proporti∣on in the point A. And because AC is bigger then CB: ther∣fore DA is greater then A∣C: wherefore if a right line &c. as in the propo∣sition. Which was to be de∣monstrated.
• * 1.2

  Therefore, by my second The∣oreme added v∣pon the second proposition, DC is deuided by ex¦treame and meane proporti∣on in the point A. And because AC is bigger then CB: ther∣fore DA is greater then A∣C: wherefore if a right line &c. as in the propo∣sition. Which was to be de∣monstrated.
• * 1.3

  Construction.
• * 1.4

  Though I say, perpēdicular: yes you may perceue how in∣finite other p••∣s••••iōs will serue: so that DI and AD make an angle for a tri∣angle to haue his sides propor∣tionally cut. &c.
• * 1.5

  Demonstra∣tion.