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 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, * 1.1 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.

Notes

• * 1.1

  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.