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

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Title
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
Author
Euclid.
Publication
Imprinted at London :: By Iohn Daye,
[1570 (3 Feb.]]
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Subject terms
Geometry -- Early works to 1800.
Link to this Item
http://name.umdl.umich.edu/A00429.0001.001
Cite this Item
"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." In the digital collection Early English Books Online. https://name.umdl.umich.edu/A00429.0001.001. University of Michigan Library Digital Collections. Accessed June 7, 2024.

Pages

¶ A Corollary. Hereby it is manifest, that in a semicircle the angle BAD is equall to the angle DBA: but in a section lesse then a semicir∣cle, it is lesse: in a section greater then a semicircle, it is greater.

There is also an other generall wy to finde out the

[illustration]
foresaid centre,* 1.1 which will serue indifferently for any section whatsoeuer: And that is thus. Take in the cir∣cumference geuen or section ABC, three pointes at all auentures which let be A, B, C. Ad draw these lins AB and BC (by the first peticion) And (by the 10. of the first) deuide into two equall partes ithr of the sayd lines. the line AB in the point D, & 〈◊〉〈◊〉 line BC in th point E. And (by the 11. of the first) from th points

Page [unnumbered]

D and E rayse vp vnto the lines AB and BC perpendicular lynes DF and EF. Now forasmuch as either of these angles BDF,* 1.2 and BEF is a right angle, a right line produ∣ced from the point D to the point E, shall deuide either of the said angles: and foras∣much as it falleth vppon the right lines DF and EF, it shall make the inward angles on one and the selfe same side, namely, the angles DEF

[illustration]
and EDF lesse then two right angles. Wherefore (by the fift peticion) the lines DF and EF being produced shall concurre. Let them concurre in the point F. And forasmuch as a certaine right line DF deuideth a cer∣taine right lyne AB into two equall partes and per∣pendicularly, therfore (by the corollary of the first of this booke) in the line DF is the centre of the circle, & by the same reason the centre of the selfe same circle shalbe in the right line EF. Wherfore the centre of the circle wherof ABC is a section, is in the point F, which is commō to either of the lines DF and EF. Wherfore a section of a circle being geuē, namely, the section ABC, there is described the circle of the same section: which was required to be done.

And by this last generall way,* 1.3 if there be geuen three pointes, set howsoeuer, so that they be not all three in one right line, a man may describe a circle which shall passe by all the said three pointes. For as in the example before put, if you suppose onely the 3. pointes A, B, C, to be geuen and not the circumference ABC to be drawen, yet follow∣ing the selfe ame order you did before, that is, draw a right line from A to B and an other from B to C and deuide the said right lines into two equall parts, in the points D and E, and erect the perpendicular lines DF and EF cutting the one the other in the point F, and draw a ight line from F to B: and making the centre the point F, and the space FB describe a circle, and it shall passe by the pointes A & C: which may be pro∣ued by drawing right lines from A to F, and from F to C. For forasmuch as the two sies AD and DF of the triangle ADF are equall to the two sides BD and DF of the triangle BDF (for by supposition the line AD is equall to the line DB, and the lyne DF is common to them bot) and the angle ADF is equall to the angle BDF (for they are both right angles) therfore (by the 4. of the first) the base AF is equall to the base BF. And by the same reason the line FC is equall to the line FB. Wherefore these three lines FA, FB and FC are equall the one to the other. Wherefore makyng the centre the point F and the space FB, it shall also passe by the pointes A and C. Which was required to be done. This proposition is very necessary for many things as you shal afterward see.

Campane putteth an other way, how to describe the

[illustration]
whole circle of a sectiō geuen.* 1.4 Suppose that the section be AB. It is required to describe the whole circle of the same section. Draw in the section two lines at all aduen∣tures AC and BD: which deuide into two equall parts AC in the point E, and BD in the pointe F. Then from the two pointes of the deuisions draw within the secti∣on two perpendicular lines EG and FH which let cutte the one the other in the point K. And the centre of the circle shall be in either of the said perpendicular lines by the corollary of the first of this booke. Wherfore the point K is the centre of the cir∣cle: which was required to be done.

But if the lines EG & FH do not cut the one the o∣ther,

[illustration]
but make one right line as doth GH in the secōd figure: which happeneth when the two lines AC and BD are equidistant. Then the line GH, being applyed to either part of the circumference geuen, shall passe by the centre of the circle, by the selfe same Corollary. For the lines EG and FH cannot be equidistant. For then one and the self same circumference should haue two centres. Wherfore the line HG being deuided in∣to

Page 100

two equall partes in the point K, the said point K shall be the centre of the section.

Pelitarius here addeth a briefe way how to finde out the centre of a circle, which is commonly vsed of Artificers.

Suppose that the circumference be ABCD, whose centre it is required to finde out.* 1.5 Take a point in the circumference geuen which let be A, vppon which describe a circle with what openyng of the compasse you will, which let be EFG. Then take an other point in the circumference geuen which let be B, vpon which describe an other circle with the same opening of the compasse that the cir∣cle

[illustration]
EFG was described, and let the same be EHG, which let cut the circle EFG in the two pointes E and G. (I haue not here drawen the whole circles, but onely those partes of them which cut the one the other for auoyding of confusion) And drawe from those centres these right lines AE, BE, AG, and BG, which foure lines shall be equall, by reason they are semidiameters o equall circles. And draw a right line from A to B, and so shall there be made two Isosceles triangles AEB, and AGB vnto whom the line AB is a common base. Now then deuide the line AB into two equal partes in the point K which must nedes fall betwene the two circumference EFG and EHG, otherwise the part should be greater then his whole. Drawe a line from E to K and pro∣duce it to the point G. Now you see that there are two Isosceles triangles deuided into foure equall triangles EAK, EBK, GAK and GBK. For the two sides AE and AK of the triangle AEK are equall to the two sides BE and BK of the triangle BEK, and the base EK is common to them both. Wherefore the two an∣gles at the point K of the two triangles AEK and BEK are by the 8. of the first equall and therfore are right angles. And by the same reason the other angles at the poynte K are right angles. Wherfore EG is one right lyne by the 14. of the first. Which foras∣much as it deuideth the line AB perpendicularly, therefore it passeth by the center by by the corollary of the first of this booke. And so if you take two other poyntes, name C and D in the circumference geuen, and vpon thē
[illustration]
describe two circles cuttyng the one the other in the pointes L and M, and by the said poyntes pro∣duce a right line, it shall cutte the lyne EG beyng produced in the pointe N, which shall be the cen∣tre of the circle by the same Corollary of the first of this booke, if you imagine the light line CD to be drawen and to be deuided perpendicularly by the lyne LM, which it must needes be as we haue before proued. And here note that to do this me∣chanically not regardyng demonstration, you neede onely to marke the poyntes where the cir∣cles cut the one the other, namely, the poyntes E, G, and L, M, and by those poyntes to produce the lines EG and LM till they cut the one the other, and where they cut the one the other, there is the centre of the circle, as you see here in the seconde figure.

Notes

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