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

¶ Very needefull Problemes and Corollaryes by Master Ihon Dee inuented: whose wonderfull vse also, be partely declareth.

A Probleme. 1.

Two circles being geuē: to finde two right lines, which haue the same proportion, one to the other, that the geuen circles haue, o••e to the other••

Suppose A and B, to be the diameters of two circles geuen: I say that two right lines are to be foūde, hauing that proportiō, that the circle of A hath to the circle of B. Let to A & B (by the 11 of the sixth) a third proportionall line be found, which suppose to be C.* 1.1 I say now that A hath to C, that pro∣portion which the circle of A hath to the cir∣cle

[illustration]

of B. For forasmuch as A, B, and C, are (by construction) three proportionall lines,* 1.2 the square of A is to the square of B, as A is to C, (by the Corollary of the 20. of the sixth)•• but as the square of the line A is to the square of the line B, so is the circle whose diameter is the line A, to the circle whose diameter is the line B, by this second of the eleuēth. Wherfore the circles of the line•• A and B, are in the proportion of the right lines A and C. Therefore two circles be••ng geuen, we haue found two right lines hauing the same proportion betwene thē, that the circles geuen, haue one to the other: which ought to be done.

A Probleme. 2.

Two circles being geuen, and a

[illustration]

right line: to finde an other right line, to which the line geuē shall haue that proportion, which the one circle hath to the other.

Suppose two circles geuē: which let be A & B, & a right line geuē, which let be C: I say that an other right line is to be ••ounde, to which the line C shall haue that proportion that the circle A,* 1.3 hath to the circle B. As the diameter of the circle A, is to the dia∣meter of the circle B, so let the line C be to a fourth line, (by the 12. of the 〈…〉〈…〉 line be D. And, by the 11. of the sixth, let a thirde line proportionall be found, to the lines C & D, which let be E•• I say now, that the line C hath to the line E, that proportion which the circle A, hath to the circle B. For (by construction) the line•• C, D, and E,* 1.4 are proportionall: therefore the square of C•• is to the square of D, as C is to E, by the Corollary of the 10. of the sixth. But by construction, as the diameter of the circle A, is to the diameter of the circle B, so is C, to D: wherefore as the square of the diameter of the circle A, is to the square of the diameter of the circle B, so is the square of the line C to the square of the line D, by the 22. of the sixth. But as the square of the diameter of A, the circle, is to the square of the diameter of the circle D, so is the cir∣cle A, to the circle B, by the second of the twelfth: wherefore by 11. of the fiueth, as the circle A, is to

the circle B, so is the square of the line C, to the square of the li••e D. But it is proued, that ••s the square of the line C, is to the square of the line D, so is the line C to the line E. Wherefore by the 11. of the fiueth, as the circle A, is to the circle B, so is the line C to the line E. Two circles being geuen there∣••ore, and a right line, we haue found a right line, to which the right line geu••n, hath that proportion, which the one circle hath to the other. Which ought to be done.

Note.

The difference betwene this Probleme, and that next before, is this: there, although we had two circles geuen,* 1.5 and two lines were found in that proportion the one to the other, in which the geuen circles were,: and here likewise are two circles geuē, and two line•• also are had in the same proportiō, that the geuē circles are: yet there we tooke at pleasure the first of the two lines, wherunto we framed the second proportionally, to the circles geuen. But here the first of the two lines, is assigned, poynted, a••d determined to vs: and not our choyse to be had therein, as was in the former Probleme.

A Probleme. 3.

A circle being geuen, to finde an other circle, to which the geu••n c••rcle is in any proportion geuē in two right lines.

Suppose the circle A••C geuen, and therefore his semidiameter is geuen: whereby his diameter also is geuen: which diameter let be AC. Let the proportion geuen, be that which is betwene ••, F, two right lines. I say, a circle is to be found, vnto which A••C hath that proportion that •• hath to ••. As •• is to ••,* 1.6 so let AC the diameter, be to an o∣ther

[illustration]

right line, by the 12. of the sixth. Which line suppose to be 11. Betwene AC and •••• finde a middle proportionall line, by the 13. of the sixth: which let be LN. By the 10. of the first, deuide ••N, into two equall pa••tes: and let that be done in the point ••. Now vpon ••••, (o being made the center) describe a circle: which let be ••N••.* 1.7 I say that ABC, is to LMN, as •• is to F. For seing that AC, LN, and H, are three right lines in con∣tinuall proportion (by construction) therefore (by the Corollary of the 20. of the sixth) as AC is to H, so is the square of A•• to the square of LN. But AC is to H, as •• is to ••, by construction. Wherefore the square of AC is to the square of LN, as E is F: but as the square of the diameter AC, is to the square of the diameter LN, so is the circle ABC to the circle ••MN, by this 2. of the twelueth, wherefore by the 11. of the fiueth, the circle ABC is to the circle LMN, as •• is to F. A circle being geuen (therefore) an other circle is founde, to which the geuen circle is in any proportion geuen betwene two right lines: which ought to be done.

A Probleme. 4.

Two circles being geuen, to finde one circle equall to them both.

Suppose the two circles geuē, haue their diameters A•• & CD. I say that a circle must be ••ound equall to the two circles whose diameters are A•• and CD: vnto the

[illustration]

line A••,* 1.8 at the point A, erect a perpendicular line A••: from which (sufficiently produced) cut a line equall to CD, which let be AF. By the first peticion draw from F to •• a right line: so is FA•• made a tri∣angle rectangle. I say now that a circle whose diameter is F••, is equal to the two circles whose diameters are A•• and ••D.* 1.9 For by the 47. of the first, the square of F•• is equall to the squares of A•• & AF. Which AF is (by construction) equall to CD•• wherefore the square of F•• is equall to the squares of AB and CD. But circles are one to an other, as the squares of their diameters are one to the other, by this second of the twelueth. Therefore the circle whose diameter is •••• is equall to the circles whose diameters are A•• and CD. Therefore two circles being geuen we haue found a circle equall to them both. Which was required to be done.

A Corollary. 1.

Hereby it is made euident, that in all triangles rectangle, the circles, semicircles, quadrants, o•• any other portions of circles described vpon the subtendent line, is equall to the two circles, semicircles, quadrants, or any two other like portions of circles, described on the two lines comprehending the right angle, like to like being compared.

For like partes haue that proportion betwene them selues, that their whole magnitudes haue, of which they are like partes, by the 15. of the fifth. But of the whole circles, in the former probleme it is euident: and therefore in the fornamed like portions of circl••s, it is a true consequent.

A Corollary. 2.

By the former probleme, it is also manifest, vnto circles three, fower, fiue, or to how many so euer one will geue, one circle may be geuen equall.

For if first, to any two, by the former probleme, you finde one equall, and then vnto your found circle and the third of the geuen circles, as two geuen circles, finde one other circle equall, and then to that second found circle, and to the fourth of the first geuen circles•• as two circles, one new circle be found equall, and so proceede till you haue once cuppled orderly, euery one of your propoūded circles (except the first and second already doone) with the new circle thus found for so the last found circle is equall to all the first geuen circles. If ye doubt, or sufficiently vnderstand me not: helpe your selfe by the discourse and demonstration of the last proposition in the second booke, and also of the 31. in the sixth booke.

¶ A Probleme 5.

Two vnequall circles being geuen, to finde a circle equall to the excesse of the greater to the lesse.

Suppose the two vnequal circles geuē, to be ABC & DEF, & let ABC be the greater:* 1.10 whose dia∣meter suppose to be AC: & the diameter of DEF suppose to be DF. I say a circle must be found equal to that excesse in magnitude, by which ABC is greater th•• DEF. By the first of the fourth, in the circle ABC. Apply a right line equall to DF:

[illustration]

whose one end let be at C, and the other let be at B. Frō B to A draw a right line. By the 30. of the third, it may appeare,* 1.11 that ABC is a right angle: and thereby ABC, the triangle is rectangled: wher∣fore by the first of the two corollaries, here before, the circle ABC is equall to the circle DEF, (For BC by constructi∣on is equall to DF) and more ouer to the circle whose diameter is AB. That circle therefore whose diameter is AB, is the circle conteyning the magnitude, by which ABC is greater then DEF. Wherefore two vnequal circles being geuen, we haue found a circle equall to the excesse of the greater to the lesse: which ought to be doone.

A Probleme. 6.

A Circle being geuen to finde two Circles equall to the same: which found Circles, shall haue the one to the other, any proportion geuen in two right lines.

Suppose ABC, a circle geuen: and the proportion

[illustration]

geuen, let it be that, which is betwene the two right lines D and E. I say, that two circles are to be found equall to ABC: and with al, one to the other,* 1.12 in the proportiō of D to E. Let the diameter of ABC be AC. As D is to E, so let AC be deuided, by the 10. of the sixth, in the poynt F. At F, to the line AC let a perpēdicular be drawne FB, and let it mete the circūferēce at the poynt B. From the poynt B to the points A and C, let right lines be drawne: BA and BC. I say that the circles whose diamete•• are the lines BA and BC are equall to the circle ABC: and that those circles hauing to their diameters the lines BA and BC are one to the other in the proportion of the line D to the line E. For, first that they are equal, it is e∣uident:

by reason that ABC is a triangle rectangle: wher¦fore

[illustration]

by the 47. of the first the squares of BA and BC are equall to the square of AC: And so by this second it is mani••est, the two circles to be equall to the circle ABC. Secondly as D is to ••, so is AF to FC: by construction.

And as the line AF is to the line FC, so is the square of the line ••A to the square of the line BC. [Which thing, we will briefely p••oue thus. The parallelogramme contayned vnder AC and AF,* 1.13 is equall to the square of BA: by the Lemma after the 32. of the tenth booke: and by the same Lemma or Assumpt, the parallelogramme contayned vn∣der AC and ••C, is equall to the square of the line BC. Wherfore as the first parallelogramme hath it selfe to the second•• so hath the square of BA (equall to the first pa∣rallelogramme) it selfe, to the square of BC, equall to the second parallelogramme. But both the parallelogrāmes haue one heigth, namely, the line AC: and bases, the lines AF and FC: wherefore as AF is to FC, so is the parallelog••amme contayned vnder AC, AF, to the parallelogramme contayned vnder AC, FC, by the fi••st of the sixth.

And therefore as AF is to FC, so is the square of BA to the square of BC.] And as the square of BA is to the square of BC: so is the circle whose diameter is BA, to the circle whose diameter is BC, by this second of the twelfth. Wherefore the circle whose diameter is BA, is to the circle whose diameter is BC, as D is to E. And before we proued them equall to the circle ABC. Wher••fore a circle being geuen, we haue found two circles equall to the same: which haue the one to the other any proportion geuen in two right lines. Which ought to be done.

Note.

* 1.14He••e may you per••eiue an other way how to execute my first probleme, for if you make a right an∣gle conteyned of the diameters geuē, as in this figure suppose them BA and BC: and then subtend the right angle with the line AC: and from the right angle, let fall a line perpendicular to the base AC: that perp••ndicular at the point of his fall, deuideth AC into AF and FC, of the proportion required.

A Corollary.

It followeth of thinges manifestly proued in the demonstration of this probleme, that in a trian∣gle rectangle, if from the right angle to the base, a perpendicular be let fall: the same perpendicular cutteth the base into two partes,* 1.15 in that proportion, one to the other, that the squares of the righ•• lines, conteyning the right angle, are in, one to the other: those on the one side the perpendicular, being compared to those on the other, both square and segment.

A Probleme 7.

Betwene two circles geuen, to finde a circle middell proportionall.

Let the two circles geuen, be

[illustration]

ACD and BEF: I say, that a cir∣cle is to be foūd, which betwene ACD and BEF is middell pro∣portionall.* 1.16 Let the diameter of ACD, be AD, and of BEF, let B•• be the diameter: betwene AD and BF, finde a line middell proportionall, by the 13. of the sixth: which let be HK: I say that a circle, whose diameter is HK is middell proportionall betwene ACD and BEF. To AD, HK, and BF, (three right lines in con¦tinuall proportion, by construc∣tion) let a fourth line be found: to which BF shal haue that pro∣portion, that AD hath to HK:

by the 12. of the sixth, & let that line be ••.* 1.17 It is manifest that the ••ower lines AD, HK, BF, and L, are in continuall proportion. [For by cōstruction, as AD is to HK, so is B•• to L. And by constructi∣on,

on, as AD is to HK, so is HK to BF: wherefore HK is to BF, as BF is to L: by the 11. of the fifth, wher∣fore the 4. lines are in continuall proportion.] Wherefore as the first is to the third, that is AD to BF, so is the square of the first to the square of the second:

that is, the square of AD, to the square of HK: by the corollary of the 20. of the sixth. And by the same corollary, as HK is to L, so is the square of HK to the square of BF. But by alternate proportion, the line AD is to BF, as HK is to L: wherefore the square of AD is to the square of HK, as the square of HK is to the square of BF. Wherefore the square of HK, is middell proportionall, betwene the square of AD and the square of BF. But as the squares are one to the other, so are the circles (whose diameters produce the same squares) one to the other, by this secōd of the twelfth: wherfore the circle whose diameter is the line HK, is middel proportional, betwene the circles whose diameters are the line a AD and BF. Wherefore betwene two circles geuen, we haue found a circle middell proportionall: which was requisite to be doone.

¶A Corollary.

Hereby it is manifest, three lines or more, being in continuall proportion, that the circles hauing those lines to their diameters, are also in continuall proportion.

As of three, our demonstration hath already proued: so of fower, will the proufe go forward•• if you adde a fifth line in continuall proportion to the fower geuen: as we did to the three, adde the fourth: namely, the line L. And so, if you haue 6, by putting to one more, the demonstration will be ••asie and plaine. And so of as many as you will.

A Probleme. 8.

To a circle being geuen, to finde three circles equall: Which three circles shall be in continuall proportion, in any proportion geuen betwene two right lines.

Suppose the circle geuen to be ABC: and the proportion geuen to be that which is betwene the lines X and Y. I say, that three circles are to be geuen, which three, together, shall be equall to the circle ABC: and withall in continuall proportion, in the same proportion which is betwene the right lines X and Y. Let the diameter of ABC, be AC. Of AC, make a square: by the 46. of the first: which which let be ACDE. From the point D

[illustration]

drawe a line, sufficiently long (any way, * 1.18 without the square): which let be DO. At the point D, and from the line DO, cut a line equall to X: which let be DM. At the point M, and from the line MO, cut a line equall to •••• which let be MN. At the point N, to the two lines DM and MN, set a third line proportionall, by the 12. of the sixt: which let be NO. From E (one of the angles of the square ACDE, next to D) draw a right line to O: making per∣fecte the triangle DEO. Now from the pointes M and N, drawe lines, to the ••ide DE, parallel to the side EO: by the 31. of the first: which let be MF and NG. Wher∣fore, by the 2. of the sixt, the side DE, is proportionally cut in the pointes F and G, as DO is cut in the pointes M and N: ther∣fore, as DM is to MN, so is DF, to FG: and as MN is to NO, so is FG to GE. Wherefore, seing DM, MN, and NO, are, by construction, continually proportioned, in the proportion of X to Y: So likewise, are DF, FG, and GE, in continuall propor∣tion, in the proportion of X to Y, by the 11. of the fift. From the pointes F and G, to the opposite ••ide AC, let right lines be drawen parallel to the other sides: which lines, suppose to be FI, and GK: making thereby, three parallelogrammes DI, ••K, and GC, equall to the whole square ACDE. Which three parallelogrammes, by the first of the sixt, are one to an other, as their bases, DF, FG, and GE, are. But DF, FG, and GE, were proued to be in continuall propor∣tion, in the proportion of X to Y: Wherefore, the three parallelogrammes DI, FK, and GC, by the 11. of the fifth, are also in continuall proportion, and in the same, which X is in, to Y. Let three squares be made, equall to the three parallelogrammes DI, FK, and GC: by the last of the second: Let the sides of those squares be, orderly, S, T, and V. Forasmuch as, it was last concluded that the three paral∣lelogrammes,

DI, FK, and GC (which are equall to the square ACDE) are also in continuall pro∣portion, in the proportion of X to Y, therefore their equalls, namely, the three squares of S, T, & V, are also equall to the whole square ACDE, and in continuall proportion, in the proportion of X to Y. Wherefore the three circles, whose diameters are S, T, and V, are equall to the circle, whose diameter is AC, the side of the square ACDE, and also in continuall proportion, in the proportion of X to Y: by this second of the twelfth. But, by construction, AC is the diameter of the circle ABC. Wherefore we haue found three circles, equall to ABC: namely, the circle, whose diameter is S: and the circle, whose diameter is T: and the circle, whose diameter is V: which three circles, also, are in continuall propo••tion, in the proportion of X to Y. Wherefore to a circle being geuen, we haue found three cir∣••les equall in any proportion, geuen, betwene two right lines: which was requisite to be done.

¶A Corollary.

Hereby, it is euident, that a circle geuen, we may finde circles 4, 5, 6, 10, 20, 100, 1000, or how many soeuer shall be appointed, being in continuall proportion, in any proportion, geuen betwene two right lines: which circles, all together, shall be equall to the circle geuen.

For, euermore deuiding the one side of the chiefe square (which is made of the diameter of the circle geuen) into so many partes, as circles are to be made: so that betwene those partes be continued the proportion geuen betwene two right lines•• and from the pointes of those diuisions, drawe paral∣lels, perpendiculars to the other side of the said chiefe square: making so many parallelogrammes of the chiefe square, as are circles to be made: and to those parallelogrammes (orderly) making equall squares: it is manifest that the sides of those squares, are the diameters of the circles required to be made.

A Probleme. 9.

Three circles being geuen, to finde three equall to them: which three found circles shall be in con∣tinuall proportion, in any proportion geuen betwene two right lines.

Suppose the three circles

[illustration]

geuen to be A, B, and C, and let the proportion geuen, be that which is betwene the right lines X & Y. I say, three other circles are to be found, equall to A, B, and C, & with all, in continuall proportion, in the proportion of X to Y. By the 2. Corollary of my 4. Probleme,* 1.19 make one circle e∣quall to the three circles A, B, and C. Which one circle sup∣pose to be D: And by the pro∣bleme next before, let three circles be ••ound, equall to D, and with all, in cōtinuall pro∣portion, in the same propor∣tion which is betwene X and Y. Which three circles, sup∣pose to be E, F, and G. I say, that E, F, G, are equall to A, B, C: and with all, in continuall proportion, in the propo••tiō of X to Y.* 1.20 For, by cōstruction, the circle D is equall to the circles A, B, & C: and by con∣struction likewise, the circles E, F, and G, are equall to the same circle D: Wherfore the three circles E, F, & G, are e∣quall to the three circles A, B, & C: and by construction E, F,

and G, are in conti••uall proportion, in the proportion of the line •• to the line Y. Wherefore E, F, and G, are equall to A, B, and C: and in continuall proportion, in the proportion of X to Y. Three circles, therefore, being geuen, we haue found three circles, equall to them, and also in continuall proportion, in any proportion geuen, betwene two right lines. Which was requisite to be done.

¶A Corollary. 1.

It is hereby very manifest, that vnto 4.5.6.10.20. 100. or how many circles soeuer, shall be geuen, we may finde 3.4.5.8.10. or how many soeuer, shall be appointed: which, all together, shall be equall to the circles geuen, how many soeuer they are: and with all, our found circles, to be in continu∣all proportion, in any proportion assigned betwene two right lines geuen.

For, euermore, by the Corollary of the 4. Probleme, reduce all your circles to one: and by the Co∣rollary of my 8. Probleme, make as many circles as you are appointed, equall to the circles geuen, and continuall in proportion, in the same, wherein, the two right lines geuen, are•• And so haue you perfor∣med, the thing required.

Note.

What incredible fruite in the Science of proportions may hereby grow, no mans tounge can suffi∣ciently expresse. And sory I am, that vtterly l••ysure is taken from me, somewhat to specifie in particu∣lar hereof.

¶The key of one of the chiefe treasure houses, belonging to the State Mathematicall.

THat•• which in thes•• 9. Problemes, is said of circles•• is much more sayd of squares, by whose meanes, circles, are thus handled. And therefore seing to all Polygonon right lined figures, equall squares [ 1] may be made, by the falt of the second: and contrariwise, to any square, a right lined figure may be made [ 2] equall, and withall, like to any right lined figure geuen, by the 25 of the sixt. And fourthly, seing vpon [ 3] the said plaine figures•• as vpō base•• may Pri••mes, Parallelipipedons, Pyramids, sided Columnes, Cones, [ 4] and Cylinders, be reared: which being * 1.21 all of one height, shall haue that proportion, one to the o∣ther, that their bases haue, one to the other. And fiftly, seing Spheres, Cones, and Cylinders are one to other in 〈◊〉〈◊〉 knowen proportions: and so may be made, one to the other in any proportion as∣signed. [ 5] And 〈◊〉〈◊〉, seing vnder euery one of the kindes of figures, both plaine, and solide, infinite cases may chaunce, by the ayde of these Problemes, to be soluted and executed: How infinite (then) vpon [ 6] infinite, is the number of practises, either Mathematicall, o•• Mechanicall, to be performed, of compa∣risons betwene diuers kindes, of plaines to plaines, and solides to solides••

Fa••••h••••••ore, to speake of playne superficiall figures, in respect of the con••••r, or Area of the circle, [ 7] sundry mixt line figures, Anular and Lunular figures: and also of circles to be geuen equall to the sayd ••••••sed figures•• and in all proportions els: and euermore thinking of solides, (like high) set vpon a∣ny of those vnused figures, (O Lord) in cōsideration of al the premisses, how infinite, how straunge and [ 8] ••••credible ••••••••••••ation•• and practise••, may (by the ayde and direction of these few problemes) 〈◊〉〈◊〉 redi∣ly into the imagination•• and handes of them, that will bring their minde and intent wholy and fixedly to such mathematicall discourses? In these Elementes, I entend but to geue to young beginners some 〈◊〉〈◊〉, and courage to exercise ••heir owne witts, and talent, in this most pleasant and profitable sci∣••nc••. All ••hinge•• 〈◊〉〈◊〉 not, neither y•••• ••an, in euery place be sayd. Opportunitie, and S••ff••ci••••ty, best are to be allowed.

Notes

• * 1.1

  Construction.
• * 1.2

  Demonstra∣tion.
• * 1.3

  Construction
• * 1.4

  Demonstra∣tion.
• * 1.5

  Di••••reence betwene the first probleme and the se∣cond.
• * 1.6

  Construction.
• * 1.7

  Demonstra∣tion.
• * 1.8

  Cons••ruction.
• * 1.9

  Demonstra∣tion.
• * 1.10

  Cons••ruction.
• * 1.11

  Demonstra∣tion.
• * 1.12

  Cons••••uction.
• * 1.13

  Note this well: for it i•• of great vse.
• * 1.14

  An other way of demonst••at••ō of the f••••st ••ro∣bleme of th•••• addition.
• * 1.15

  Note this pro∣per••ie of a tri∣angle rectangle.
• * 1.16

  Construction.
• * 1.17

  Demonstra∣tion.
• * 1.18

  Though I say, without the square, yet you must thinke, that it may be also within the square: & that diuersly. Wher∣fore this Pro∣bleme may haue diuerse cases, so, but briefly, to a∣••••yde all, may thus be said: cut any side of that square into 3. parts•• in the pro¦portion of X to Y.

  Note the maner of the drift in this demonstra∣tion and con∣struction, mixt∣ly and with no determination to the construc∣tiō•• as commōly i•• in probleme•••• which is here of me so vsed•• for an example to young studētes of variety in art.
• * 1.19

  Construction.
• * 1.20

  Demonstra∣tion.
• * 1.21

  Note and remember one ••e••th in these so∣lids.