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

¶The third booke of Eu∣clides Elementes. (Book 3)

* 1.1THis third booke of Euclide entreateth of the most perfect figure, which is a circle. Where∣fore it is much more to be estemed then the two bookes goyng before, in which he did set forth the most simple proprieties of rightlined figures. For sciences take their dignities of the worthynes of the matter that they entreat of. But of al figures the circle is of most absolute perfection, whose proprieties and passions are here set forth, and most certainely demō∣strated. Here also is entreated of right lines subten∣ded to arkes in circles: also of angles set both at the circumference and at the centre of a circle, and of the varietie and differences of them. Wherfore the readyng of this booke, is very profitable to the attayning to the knowledge of chordes and arkes. It teacheth moreouer which are circles con∣tingēt, and which are cutting the one the other: and also that the angle of contin∣gence is the least of all acute rightlined angles: and that the diameter in a circle is the longest line that can be drawen in a circle. Farther in it may we learne how, three pointes beyng geuen how soeuer (so that they be not set in a right line), may be drawen a circle passing by them all three. Agayne, how in a solide body, as in a Sphere, Cube, or such lyke, may be found the two opposite pointes. Whiche is a thyng very necessary and commodious: chiefly for those that shall make instru∣mentes seruyng to Astronomy, and other artes.

Definitions. * 1.2Equall circles are such, whose diameters are equall, or whose lynes drawen from the centres are equall.

The circles A and B are equal, if theyr diameters, namely, EF and CD be equall: or if their semidiameters, whiche are lynes drawen from the center to the circumference•• namely AF and BD be equall.

* 1.3The reason why circles

[illustration]

take theyr equalitie, of the e∣qualitie of their diameters or semidiameters is, for that a circle is described by one re∣uolution or turnyng about of the semidiameter, hauing one of his endes fixed. As if you imagine the lyne AE to haue his one point namely A fastened, and the other end namely E to moue round till

it come to the place where it begā to moue, it shal fully describ•• the whole circle. Wherefore if the semidiameters bee equall, the circles of necessitye must also be equall: and also the diameters.

By thys also is knowen the definition of vnequall circles.* 1.4

Circles whose diameters or semidiameters are vnequall, are also vnequall. And that circl•• which hath the greater diameter or semidiameter, is the greater circle: and that circle which hath the lesse diameter or semidiameter, is the lesse circle.

As the circle LM is greater

[illustration]

then the circle IK, for that the diameter LM is greater then the diameter IK: or for that the semi∣diameter GL is greater then the semidiameter HI.

A right line is sayd to touch a circle,* 1.5 which touching the cir∣cle and being produced cutteth it not.

As the right lyne EF drawen from the point E, and passyng by a point of the circle, namely, by the point G to the point F on∣ly

[illustration]

toucheth the circle GH, and cutteth it not, nor entreth within it. For a right line entryng within a circle, cutteth and deui∣deth the circle. As the right lyne KL de∣uideth and cutteth the circle KLM, and entreth within it: and therfore toucheth it in two places. But a right lyne tou∣chyng a circle, which is commonly called a cōtingent lyne,* 1.6 toucheth the circle one∣ly in one point.

Circles are sayd to touch the one the other,* 1.7 which touching the one the other, cut not the one the other.

As the two circles AB and BC touch the

[illustration]

one the other. For theyr circumferences touch together in the poynt B. But neither of them cutteth or deuideth, the other. Neither doth any part of the one enter within the o∣ther.* 1.8 And such a touch of circles is euer in one poynt onely: which poynt onely is common to them both. As the poynt B is in the confe∣rence of the circle AB, and also 〈…〉〈…〉 the ••••••••••∣ference of the circle BC.

* 1.9Circles may touch together two maner of wayes, either outwardly the one wholy without the other: or els the one being contayned within the other.

As the circles DE and DF: of which the one DE contay∣neth

[illustration]

the other, namely DF: and touch the one the other in the poynt D: and that onely poynt is common to them both: neither doth the one enter into the other. If any part of the one enter into any part of the other, then the one cutteth and deuideth the other, and toucheth the one the other not in one poynt onely as in the other before, but in two point••s, and haue also a superficies common to them both. As the cir∣cles GHK and HLK cut the one the other in two poyntes H and K: and the one entreth into the other: Al∣so the superficies HK is common to them both: For it is a part of the circle GHK, and also it is a part of the circle HLK.

* 1.10Right lines in a circle are sayd to be equally distant from the cen∣tre, when perpendicular lines drawen from the centre vnto those lines are equall. And that line is sayd to be more di∣stant, vpon whom falleth the greater perpendicular line.

As in the circle ABCD whose centre is E, the two lynes

[illustration]

AB and CD haue equall distance from the centre E: bycause that the lyne EF drawen from the centre E perpendicularly vpon the lyne AB, and the lyne EG drawen likewise perpendi∣larly from the centre E vpon the lyne CD are equall the one to the other. But in the circle HKLM whose centre is N the lyne HK hath greater distance from the centre N then hath the lyne LM: for that the lyne ON drawen from the centre N perpendicularly vppon the lyne HK is greater then the lyne NP which is drawen frō the centre N perpendicularly vpon the lyne LM.

So likewise in the other figure the lynes AB and DC in the circle ABCD are equidist••nt from the centre G•• bycause the lynes OG and GP perpendicularly drawen from the centre G vppon the sayd lynes AB and DC are equall. And the lyne AB hath greater distance from the centre G then hath the the lyne EF, bycause the lyne OG perpendicul••rly dr••wen from the centre G to the lyne AB is gre••ter then the lyne HG whiche is perpendicularly drawen from the c••••tre G to the lyne EF.

* 1.11A section or segment of a circle, is a figure cōprehended vnder a right line and a portion of the circumference of a circle.

As the figure ABC is a section of a circle

[illustration]

bycause it is comprehended vnder the right lyne AC and the circumference of a circle ABC. Likewise the figure DEF is a section of a circle, for that it is comprehended vnder the right lyne DF, and the circūference DEF. And the figure ABC for that it cōtaineth within it the centre of the circle is called the greater section of a circle: and the figure DEF is the lesse section of a circle, bycause it is wholy without the centre of the circle as it was noted in the 16. Definition of the first booke.

An angle of a section or segment, is that angle which is con∣tayned vnder a right line and the circūference of the circle.* 1.12

As the angle ABC in the section ABC is an angle of a sec∣tion,

[illustration]

bycause it is contained of the circumference BAC and the right lyne BC. Likewise the angle CBD is an angle of the section BDC bycause it is contayned vnder the circumference BDC, and the right lyne BC. And these angles are commonly called mixte angles,* 1.13 bycause they are contayned vnder a right lyne and a crooked. And these portions of circumferences are commonly called arkes,* 1.14 and the right lynes are called chordes,* 1.15 or right lynes subtended. And the greater section hath euer the greater angle, and the lesse section the lesse angle.

An angle is sayd to be in a section,* 1.16 whē in the circumference is taken any poynt, and from that poynt are drawen right lines to the endes of the right line which is the base of the segment, the angle which is contayned vnder the right lines drawen from the poynt, is (I say) sayd to be an angle in a section.

As the angle ABC is an angle is the section ABC, bycause

[illustration]

from the poynt B beyng a poynt in the circumference ABC are drawen two right lynes BC and BA to the endes of the lyne AC which is the base of the section ABC. Likewise the angle ADC is an angle in the section ADC, bycause from the poynt D beyng in the circūference ADC are drawen two right lynes, namely, DC & DA to the endes of the right line AC which is also the base to the sayd section ADC. So you see, it is not all one to say,* 1.17 an an∣gle of a section, and an angle in a section. An angle of a section cō∣sisteth of the touch of a right lyne and a crooked. And an angle in a section is placed on the circumference, and is contayned of two right lynes. Also the greater section hath in it the lesse angle, and the lesse section hath in it the greater angle.

But when the right lines which comprehend the angle do re∣ceaue any circumference of a circle,* 1.18 then that angle is sayd to be correspondent, and to pertaine to that circumference.

As the right lynes BA and BC which containe the angle AB

[illustration]

C, and receaue the circumference ADC therfore the angle ABC is sayd to subtend and to pertaine to the circūference ADC. And if the right lynes whiche cause the angle, concurre in the centre of a circle: then the angle is sayd to be in the centre of a circle. As the angle EFD is sayd to be in the centre of a circle, for that it is comprehended of two right lynes FE and FD: whiche concurre and touch in the centre F. And this angle likewise subtendeth the circumference EGD: whiche circumference also, is the measure of the greatnes of the angle EFD.

A Sector of a circle is (an angle being set at the centre of a circle) a figure contayned vnder the right lines which make that angle,* 1.19 and the part of the circumference re∣ceaued of them.

As the figure ABC is a sector of a circle, for that it hath an angle

[illustration]

at the centre, namely the angle BAC, & is cōtained of the two right lynes AB and AC (whiche contayne that angle and the circumfe∣rence receaued by them.

* 1.20Like segmentes or sections of a circle are those, which haue equall angles, or in whom are equall angles.

* 1.21Here are set two definitions of like sections of

[illustration]

a circle. The one pertaineth to the angles whiche are set in the centre of the circle and receaue the circumferēce of the sayd sections:* 1.22 the other per∣taineth to the angle in the section, whiche as be∣fore was sayd is euer in the circumference. As if the angle BAC, beyng in the centre A and re∣ceaued of the circumference BLC be equall to the angle FEG beyng also in the centre E and receaued of the circumference FKG, then are the two sections BCL and FGK lyke by the first definition. By the same definition also are the other two sections like, name∣ly BCD, and FGH, for that the angle BAC is equall to the angle FEG.

* 1.23Also by the second definition if B

[illustration]

AC beyng an angle placed in the cir∣cumference of the section BCA be e-angle EDF beyng an angle in the se∣ction EFD placed in the circumfe∣rence, there are the two sections BCA, and EFD lyke the one to the o∣ther. Likewise also if the angle BGC beyng in the section BCG be equall to the angle EHF beyng in the sectiō EHF the two sections BCG and EFH are lyke. And so is it of angles beyng equall in any poynt of the circumference.

Euclide defineth not equall Sections:* 1.24 for they may infinite wayes be described. For there may vppon vnequall right lynes be set equall Sections (but yet in vne∣quall circles) For from any circle beyng the greater, may be cut of a portion equall to a portion of an other circle beyng the lesse. But when the Sections are equall, and are set vpon equall right lynes, theyr circumferences also shalbe equall. And right lynes beyng deuided into two equall partes, perpendicular lynes drawen from the poyntes of the diuision to the cir∣cumferēces

[illustration]

shalbe equall. As if the two secti∣ons ABC and DEF, beyng set vppon equall ryght lynes AC & DF, be equall: then if ech of the two lynes AC & DF be deuided into two e∣quall partes in the poyntes G and H, & from the sayd poyntes be drawen to the circumferences two perpendicular lynes BG and EH, the sayd perpendicular lynes shalbe equall.

The 1. Probleme. The 1. Proposition. To finde out the centre of a circle geuen.

SVppose that there be a circle geuen ABC. It is requi∣red to finde out the centre of the circle ABC. Draw in it a right line at all aduentures, and let the same be AB.* 1.25 And (by the 10. of the first) deuide the line AB into two equall partes in the poynt D. And (by the 11. of the same) frō the poynt D raise vp vnto AB a perpendicular line DC, & (by the second petition) extend DC vnto y^e point E. And (by the 10. of the first) deuide

[illustration]

the line CE into two equall partes in the poynt F. Then I say that the point F is the centre of the circle ABC.* 1.26 For if it be not, let some other point, namely G, be the centre. And (by the first peti∣tion) draw these right lines GA, GD, and GB. And for asmuch as AD is equall vnto DB, and DG is common vnto thē both, therefore these two lines AD and DG are equall to these two lines GD and DB, the one to the other, and (by the 15. definition of the first) the base GA is equall to the base GB. For they are both drawen from the cen∣tre G to the circumference: therefore (by the 8. of the first) the angle ADG is equall to the angle BDG. But when a right line standing vpon a right line maketh the angles on eche side equall the one to the other, eyther of those angles (by the 10. definition of the first) is a right angle. VVherefore the angle BDG

is a right angle: but y^e angle FDB is also a right angle by construction. VVher∣fore (by the 4. petition) the angle FDB is equall to the angle BDG, the grea∣ter to the lesse, which is impossible. VVherefore the poynt G is not the centre of the circle ABC. In like wise may we proue that no other poynt besides F is the centre of the circle ABC. VVherefore the poynt F is the centre of the circle ABC: which was required to be done.

Correlary.

* 1.27 Hereby it is manifest, that if in a circle a right line do deuide a right line into two equall partes, and make right angles on eche side: in that right line which deuideth the other line into two e∣quall partes is the centre of the circle.

The 1. Theoreme. The 2. Proposition. If in the circūference of a circle be takē two poyntes at all ad∣uentures: a right line drawen from the one poynt to the other shall fall within the circle.

SVppose that there be a circle ABC. And in the circumference ther∣of, let there be takē at all aduentures these two poyntes A & B. Then I say that a right line drawen from A to B shall fall within the circle ABC.* 1.28 For if it do not, let it fall without the circle, as the line AEB doth, which if it be possible imagine to be a right line. And (by the Proposition going before) take the centre of the circle, and let the same be D. And (by the first petition) draw lines from D to A, and from D to B. And extend DF to E. And for asmuch as (by the 15. definition of y^e first)

[illustration]

DA is equall vnto DB. Therefore the an∣gle DAE is equall to the angle DBE. And for asmuch as one of the sides of the triangle DAE, namely the side AEB is produced, therefore (by the 16. of the first) the angle DEB, is greater then the angle DAE. But the angle DAE is equall vnto the angle DBE. VVherfore the angle DEB is grea∣ter then the angle DBE. But (by the 18. of the first) vnto the greater angle is subtended the greater side. VVherefore the side DB is greater then the side DE. But (by the 15. definition of the first) the line DB is

equall vnto the line DF. VVherfore the line DF is greater then the line DE, namely, the lesse greater then the greater: which is impossible. VVherfore a right line drawen from A to B falleth not without the circle. In like sort also may we proue that it falleth not in the circumference. VVherefore it falleth within the circle. If therefore in the circumference of a circle be taken two poyntes at all ad∣uentures: a right line drawen from the one poynt to the other shall fall within the circle: which was required to be proued.

The 2. Theoreme. The 3. Proposition. If in a circle a right line passing by the centre do deuide an o∣ther right line not passing by the cētre into two equall partes: it shall deuide it by right angles. And if it deuide the line by right angles, it shall also deuide the same line into two equall partes.

SVppose that there be a circle ABC,* 1.29 and let there be in it drawen a right line passing by the centre, and let the same be CD, deuiding an other right line AB not passing by y^e centre into two equall partes in the poynt F. Then I say that the angles at the poynt of the deuision are right angles.* 1.30 Take (by y^e first of the third)

[illustration]

the centre of the circle ABC, and let the same be E. And (by the first petition) drawe lines from E to A & from E to B.* 1.31 And for asmuch as the line AF is equall vnto the line FB, and the line FE is common to them both, therfore these two lines EF and FA are equall vnto these two lines EF & FB. And the base EA is equall vnto the base EB (by the 15. defini∣tion of the first). VVherefore (by the 8. of the first) the angle AFE is equall to the angle BFE. But when a right line standing vpon a right line doth make the angles on eche side equall the one to the other, eyther of those angles is (by the 10. defini∣tion of the first) a right angle. VVherfore either of these angles AFE, & BFE is a right angle. VVherefore the line CD passing by the centre, and deuiding the line AB not passing by the centre into two equall partes, maketh at the point of the deuision right angles.

But now suppose that the line CD do deuide the line AB in such sort that it maketh right angles.* 1.32 Then I say that it deuideth it into two equall partes, that is, y^t the line AF is equall vnto the line FB.* 1.33 For the same order of construction remayning, for asmuch as the line EA is equall vnto the line EB (by the 15. de∣finition

of the first). Therefore the angle EAF is equall vnto the angle EBF (by the 5. of the first). And the right angle AFE is (by the 4. petition) equall to the right angle BFE. VVherefore there are two triangles EAF, & EBF hauing two angles equall to two angles, & one side equall to one side, namely the side EF which is common to them both, and subtendeth one of the equall angles, wherefore (by the 26. of the first) the sides remayning of the one, are equall vn∣to the sides remayning of the other. VVherefore the line AF is equall vnto the line FB. If therefore in a circle a right line passing by the centre do deuide an other right line not passing by the centre into two equall partes, it shall deuide it by right angles. And if it deuide the line by right angles it shall also deuide the same line into two equall partes: which was required to be demonstrated.

The 3. Theoreme. The 4. Proposition. If in a circle two right lines not passing by the centre, deuide the one the other: they shall not deuide eche one the other into two equall partes.

SVppose that there be a circle ABCD, and let there be in it drawen two right lines not passing by the centre and deuiding the one the other, and let the same be AC and BD, which let deuide the one the other in the poynt E. Then I say that they deuide not eche

[illustration]

the one the other into two equall partes.* 1.34 For if it be possible let them deuide eche the one the other into two equall partes, so that let AE be equall vnto EC, & BE vnto ED. And take the centre of the circle ABCD, which let be F. And (by the first petition) draw a line from F to E. Now for asmuch as a certaine right line FE passing by the centre deuideth an other line AC not passing by the centre into two equall partes, it maketh where the deuisi∣on is right angles (by the 3. of the third). VVherfore the angle FEA is a right angle. Againe for asmuch as the right line FE, passing by the centre, deuideth the right line BD not passing by the centre into two equall partes, therefore (by the same) it maketh where y^e deuision is right angles. VVherfore the angle FEB is a right angle. And it is proued that the angle FEA is a right angle. VVher∣fore (by the 4. petition) the angle FEA is equall vnto the angle FEB, namely the lesse angle vnto the greater: which is impossible. VVherefore the right lines AC and BD deuide not eche one the other into two equall partes. If therfore in a circle two right lines not passing by the centre, deuide the one the other, they

shall not deuide eche one the other into two equall partes: which was required to be demonstrated.

In this Proposition are two cases.* 1.35 For the lines cutting the one the other, do ey∣ther, neyther of them passe by the centre, or the one of them doth passe by the cen∣tre, & the other not. The first is declared by the author. The second is thus proued.

Suppose that in the circle ABCD the line

[illustration]

BD passing by the centre doe cut the line AC not passing by the centre.* 1.36 Then I say that the lines AC and BD do not deuide the one the other in∣to two equall partes. For by the former Proposi∣tion the line BD passing by the centre and deui∣ding the line AC into two equall partes,* 1.37 it shall also deuide it perpendicularly. And for asmuch as the line AC deuideth the line BD into two equall partes & right angled wise: therfore by the Correl∣lary of the first of thys booke, the line AC passeth by the centre of the circle: which is cōtrary to the supposition. Wherfore the lines AC and BD do not deuide the one the other into two equall partes: which was required to be proued.

The 4. Theoreme. The 5. Proposition. If two circles cut the one the other, they haue not one and the same centre.

SVppose that these two circles

[illustration]

ABC,* 1.38 and CBG do cut the one the other in the poyntes C and B. Then I say that they haue not one & the same centre.* 1.39 For if it be possi∣ble let E be centre to them both. And (by the first petition) draw a line from E to C. And draw an other right line EFG at all aduentures. And for asmuch as the poynt E is the centre of the circle ABC, therefore (by the 15. definition of the first) the line EC is equall vnto the line EF. Agayne for asmuch as the poynt E is the centre of the circle CBG, there∣fore (by the same definition) the line EC is equall vnto the line EG. And it is proued that the line EC is equall vnto the line EF: wherefore the line EF also is equall vnto the line EG, namely the lesse vnto the greater: which is impossi∣ble. VVherfore the poynt E is not the centre of both the circles ABC, & CBG. I•• like sort also may we proue that no other poynt is the centre of both the sayd

circles. If therefore two circles cut the one the other, they haue not one and the same centre: which was required to be proued.

The 5. Theoreme. The 6. Proposition. If two circles touch the one the other, they haue not one and the same centre.

SVppose that these two circles ABC, & CDE do touch the one the other in the poynt C.* 1.40 Then I say that they haue not one and the same centre. For if it be possible let the point F be centre vnto them both. And (by the first petition) draw a line from F to C: and

[illustration]

drawe the line FEB at all aduentures. And for asmuch as the poynt F is the cen∣tre of the circle ABC, therfore (by the 15. definition of the first) the line FC is equall vnto the line FB. Agayne forasmuch as the poynt F is the centre of y^e circle CDE, therefore (by the same definition) the line FC is equall vnto the line FE. And it is proued, that the line FC is equall vnto the line FB, wherefore the line FE also is e∣quall vnto the line FB, namely the lesse vnto y^e greater: which is impossible. VVher∣fore the poynt F is not the centre of both the circles ABC and CDE. In like sort also may we proue that no other poynt is the centre of both the sayd circles. If therefore two circles touch the one the other: they haue not one and the same centre: which was required to be demonstrated.

* 1.41In thys Proposition are two cases: for the circles

[illustration]

touchyng the one the other, may touch eyther within or without. If they touch the one the other within, then is it by the former demonstration manifest, that they haue not both one and the selfe same centre. It is also manifest if they touch the one the other without: for that euery cen∣tre is in the middest of hys circle.

The 6. Theoreme. The 7. Proposition. If in the diameter of a circle be taken any poynt, which is not

the centre of the circle, and from that poynt be drawen vnto the circumference certaine right lines: the greatest of those lines shall be that line wherein is the centre, and the lest shall be the residue of the same line. And of all the other lines, that which is nigher to the line which passeth by the centre is greater then that which is more distant. And from that point can fall within the circle on ech side of the least line onely two equall right lines.

SVppose that there be a circle ABCD: and let the diameter thereof be AD. And take in it any poynt besides the centre of the circle, and let the same be F.* 1.42 And let the centre of the circle (by the 1. of y^e third) be the poynt E. And from the poynt F let there be drawen vnto the circumference ABCD these right lines FD, FC, and FG. Then I say that the line FA is the greatest: and the line

[illustration]

FD is the lest. And of the other lines, the line FB is greater then the line FC, and the line FC is greater then the line FG.* 1.43 Drawe (by the first petition) these right lines BE, CE, and GE. And for asmuch as (by the 20. of the first) in eue∣ry triangle two sides are greater then the third,* 1.44 therefore y^e lines EB and EF are greater then the residue, namely then the line FB. But the line AE is e∣quall vnto the line BE (by the 15. defi∣nition of the first). VVherefore the lines BE and EF are equall vnto the line AF. VVherefore the line AF is greater then then the line BF. Agayne for asmuch as the line BE is equall vnto CE (by the 15. definition of the first) and the line FE is common vnto them both, therefore these two lines BE and EF are equall vnto these two CE and EF. But the angle BEF is greater then the angle CEF. VVherefore (by the 24. of the first) the base BF is greater then the base CF: and by the same reason the line CF is greater then the line FG. Agayne for asmuch as the lines GF and FE are greater then the line EG (by the 20. of the first).* 1.45 But (by the 15. definition of the first) the line EG is equall vnto the line ED: VVherefore the lines GF and FE are greater then the line ED, take away EF, which is cōmon to thē both, wherfore y^e residue GF is grea∣ter then the residue FD. VVherefore the line FA is the greatest, and the line FD is the lest, and the line FB is greater then the line FC, and the line FC

is greater then the line FG.* 1.46 Now also I say that from the poynt F there can be drawen onely two equall right lines into the circle ABCD on eche side of the least line, namely FD. For (by the 23. of the first) vpon the right line geuen EF and to the poynt in it, namely E, make vnto the angle GEF an equall an∣gle FEH: and (by the first petition) draw a line from F to H. Now for asmuch as (by the 15. definition of the first) the line EG is equall vnto the line EH, and the line EF is common vnto them both, therefore these two lines GE and EF are equall vnto these two lines HE and EF, and (by construction) the angle GEF is equall vnto the angle HEF. VVherefore (by the 4. of y^e first) the base FG is equall vnto the base FH.* 1.47 I say moreouer that from the poynt F can be drawen into the circle no other right line equall vnto the line FG. For if it possible let the line•• FK be equall vnto the line FG. And for asmuch as FK is equall vnto FG. But the line FH is equall vnto the line FG, therefore the line FK is equall vnto the line FH. VVherfore the line which is nigher to the line which passeth by the centre is equall to that which is farther of, which we haue before proued to be impossible.

* 1.48Or els it may thus be demonstrated.

[illustration]

Draw (by the first petition) a line from E to K: and for asmuch as (by y^e 15. de∣finitiō of y^e first) y^e line GE is equall vnto y^e line EK, and the line FE is common to them both, and the base GF is equall vnto the base FK, therefore (by the 8. of the first) the angle GEF is equall to the angle KEF. But the angle GEF is equall to the angle HEF. VVhere∣fore (by the first common sentence) the angle HEF is equall to the angle KEF the lesse vnto the greater: which is impossible. VVherefore from the poynt F there can be drawen into the circle no other right line equall vnto the line GF. VVherefore but one onely. If therefore in the diameter of a circle be taken any poynt, which is not the centre of the cir∣cle, and from that poynt be drawen vnto the circumference certaine right lines: the greatest of those right lines shall be that wherein is the centre: and the least shall be the residue. And of all the other lines, that which is nigher to the line which passeth by the centre is greater then that which is more distant. And from that poynt can fall within the circle on ech side of the least line onely two equall right lines: which was required to be proued.

¶ A Corollary.

* 1.49Hereby it is manifest, that two right lines being drawen frō any one poynt of the diameter, the one of one side, and the other of the other side, if with the diame∣ter they make equall angles, the sayd two right lines are equall. As in thys place are the two lines FG and FH.

The 7. Theoreme. The 8. Proposition. If without a circle be taken any poynt, and from that poynt be drawen into the circle vnto the circumference certayne right lines, of which let one be drawen by the centre and let the rest be drawen at all aduentures: the greatest of those lines which fall in the concauitie or hollownes of the circumference of the circle, is that which passeth by the centre: and of all the other lines that line which is nigher to the line which passeth by the centre is greater then that which is more distant. But of those right lines which end in the conuexe part of the circumfe∣rence, that is the least which is drawen from the poynt to the diameter: and of the other lines that which is nigher to the least is alwaies lesse then that which is more distant. And from that poynt can be drawen vnto the circumference on ech side of the least onely two equall right lines.

SVppose y^t the circle geuen be ABC,

[illustration]

& without y^e circle ABC, take the point D: and frō y^e same point draw certain right lines into y^e circle vnto the cir∣cumference, & let thē be DA, DE, DF, & DC: & let y^e line DA passe by y^e centre. Then I say, of y^e right lines which fall in the concauitie of y^e circumference AEFC, y^t is, within y^e circle, y^e greatest is y^t which passeth by y^e centre, that is, DA. And of those lines which fall vpon y^e conuex part of y^e circumfe∣rence, y^e lest is y^t which is drawen frō y^e point D vnto y^e end of y^e diameter AG. And of the right lines falling w^tin the circumferēce, the line DE is greater then y^e line DF, & the line DF is greater then y^e line DC. And of the right lines which end in y^e conuex part of the circumference, y^t is, without y^e circle, that (which is nigher vnto DG y^e lest, is alwayes lesse then y^t which is more distāt, that is, the line DK is lesse then the line DL, and the line DL is lesse then the line DH. Take (by the first of the third) the centre of the circle ABC,* 1.50 and let the

same be M:* 1.51 and (by the first petition) drawe these right lines ME, MF, MC, MH, ML, and MK. And for asmuch as (by the 15. definition of the first) the line AM is equall vnto the line EM, put the line MD common to them both. VVherefore the line AD is equall vnto the lines EM and MD. But the lines EM and MD are (by the 20. of the first) greater then the line ED. VVherefore the line AD also is greater then the line ED. Agayne for asmuch as (by the 15. definition of the first) the line ME is equall vnto the line MF, put the line MD common to them both: VVherefore the lines EM and MD are equall to the lines FM and MD, and the angle EMD is greater then the angle FMD: VVherefore (by the 19. of the first) the base ED is greater then the base FD. In like sort also may we proue that the line FD is greater then the line CD. VVherefore the line DA is the greatest, and the line DE is grea∣ter then the line DF, and the line DF is greater then the line DC.

* 1.52And for asmuch as (by the 20. of the

[illustration]

first) ••••e lines MK and KD are greater then the line MD. But (by the 15. defini∣tion of the first) the line MG is equall vn∣to the line MK. VVherefore the residue KD is greater then y^e residue GD. VVher∣fore the line GD is lesse then the line KD. And for asmuch as from the endes of one of the sides of the triangle MLD, namely, MD are drawen two right lines MK and KD meeting within the triangle, therfore (by the 21. of the first) the lines MK and KD are lesse then the lines ML & LD, of which the line MK is equall vnto the line ML. VVherefore the residue DK is lesse then the residue DL. In like sort also may we proue that the line DL is lesse then the line DH. VVherefore the line DG is the lest, and the line DK is lesse then the line DL, and the line DL is lesse then the line DH.

* 1.53Now also I say that from the poynt D can be drawen vnto the circumference on eche side of DG the least onely two equall right lines. Vpon the right line MD, and vnto the poynt in it M make (by the 23. of the first) vnto the an∣gle KMD an equall angle DMB. And (by the first petition) drawe a line from D to B. And for asmuch as (by the 15. definition of the first) the line MB is equall vnto the line MK put the line MD common to the both, wher∣fore these two lines MK and MD are equall to these two lines BM and MD the one to the other, and the angle KMD is (by the 23. of the first) equall to the angle BMD: VVhere••ore (by the 4. of the first) the base DK is equall

to the base DB.

Now I say that from the poynt D on

[illustration]

that side that the line DB is,* 1.54 can not be drawen vnto the circumference any other line besides DB equall vnto the right line DK. For if it be possible let there be drawen an other line besides DB, and let the same be DN. And for asmuch as the line DK is equall vnto the line DN. But vnto the line DK is equall the line DB. Therfore (by the first common sentence) the line DB is equall vnto the line DN. VVherefore that which is nigher vnto DG the least, is equall to y^e which is more distant•• VVhich we haue before proued to be impossible.

Or it may thus be demonstrated.* 1.55 Draw (by the first petition) a line from M to N. And for asmuch as (by the 15. definition of the first) the line KM is equall vnto the line MN, and the line MD is common to them both. And the base KD is e∣quall to the base DN (by supposition) therefore (by the 8. of the first) the an∣gle KMD is equall to the angle DMN. But the angle KMD is equall to the angle BMD. Wherfore the angle BMD is equall to the angle NMD, the lesse vnto the greater: which is impossible. Wherefore from the poynt D can not be drawen vnto the circumference ABC on eche side of DG the lest, more then two equall right lines. If therefore without a circle be taken any poynt and from that poynt be drawen into the circle vnto the circumference certaine right lines, of which let one be drawen by the centre, and let the rest be drawen at all adventures: the greatest of those right lines which fall in y^e concauitie or hollow∣nes of the circumference of the circle is that which passeth by the centre. And of all the other lines, that line which is nigher to the line which passeth by the cen∣tre, is greater then that which is more distant. But of those right lines which end in the conuexe part of the circumference, that line is the lest which is drawen from the poynt to the dimetient: and of the other lines that which is nigher to the least is alwayes lesse then that which is more distant. And from that poynt can be drawen vnto the circumference on ech side of the lest only two equall right lines: which was required to be proued.

Thys Proposition is called commonly in old bookes amongest the barbarous, Ca••d•• Panonis,* 1.56 that is, the Peacockes taile.

¶ A Corollary.

Hereby it is manifest,* 1.57 that the right lines, which being drawen from the poynt

geuen without the circle, and fall within the circle, are equally distant from the least, or from the greatest (which is drawen by the centre) are equall the one to the other: but contrarywyse if they be vnequally distant, whether they light vpon the concaue or conuexe circumference of the circle, they are vnequall.

The 8. Theoreme. The 9. Proposition. If within a circle be taken a poynt, and from that poynt be drawen vnto the circumference moe then two equall right lines, the poynt taken is the centre of the circle.

SVppose that the circle be ABC, and within it let there be taken the poynt D. And from D let there be drawen vnto the circumference ABC moe then two equall right lines,* 1.58 that is, DA, DB, and DC. Then I say that the poynt D is the centre of the circle ABC. Draw (by the first petition) these right lines

[illustration]

AB and BC:* 1.59 and (by the 10. of the first) deuide thē into two equall partes in the poyntes E and F: namely, the line AB in the poynt E, and the line BC in the poynt F. And draw y^e lines ED and FD, and (by the second pe∣tition) extend the lines ED and FD on eche side to the poyntes K, G, and H, L. And for asmuch as the line AE is equall vnto the line EB, and the line ED is common to them both, there∣fore these two sides AE and ED are equall vnto these two sides BE, and ED: and (by supposition) the base DA is equall to the base DB. Wherfore (by the 8. of the first) the angle AED is equall to the angle BED. Wherfore eyther of these angles AED and BED is a right angle. Wherefore the line GK deuideth y^e line AB into two equall partes and maketh right angles. And for asmuch as, if in a circle a right line deuide an other right line into two equall partes in such sort that it maketh also right angles, in y^e line that deuideth is the centre of the circle (by the Correllary of the first of the third). Therfore (by the same Correllary) in the line GK is the centre of the circle ABC. And (by the same reason) may we proue that in y^e line HL is the centre of the circle ABC, and the right lines GK, and HL haue no other poynt common to them both besides the poynt D. Wherefore the poynt D is the centre of the circle ABC. If therefore within a circle be taken a poynt, and from that point be drawen vnto the circumference more then two equall right lines, the poynt taken is the centre of the circle: which was required to be proued.

¶ An other demonstration.

Let there be taken within the circle ABC the poynt D.* 1.60 And from the poynt D let there be drawen vnto the circumference more then two equall right lines, namely, DA, DB, and DC. Then I say that the poynt D is the centre of the circle. For if not, then if it be possible

[illustration]

let the point E be the centre: and draw a line from D to E, and extend DE to the poyntes F and G. Wherefore the line FG is the diameter of the circle ABC. And for asmuch as in FG the diameter of the circle ABC is taken a poynt, namely D, which is not the centre of that circle, therefore (by the 7. of the third) the line DG is y^e grea∣test, and the line DC is greater then the line DB, and the line DB is grea∣te•• then the line DA. But the lines DC, DB, DA, are also equall (by supposi∣tion): which is impossible. Wherefore the poynt E is not the centre of the circle ABC. And in like sort may we proue that no other poynt besides D. Wherefore the poynt D is the centre of the circle ABC: which was required to be proued.

The 9. Theoreme. The 10. Proposition. A circle cutteth not a circle in moe pointes then two.

FOr if it be possible let the circle ABC cut the circle DEF in mo pointes then two,* 1.61 that is, in B, G, H, & F. And drawe lines frō B to G, and from B to H. And (by y^e 10. of the first) deuide either of the lines BG & BH into two equall partes, in y^e pointes

[illustration]

K and L. And (by the 11. of the first) from the poynt K raise vp vnto y^e line BH a perpendicular line KC, and likewise from the poynt L raise vp vnto y^e line BG a perpendicular line LM, and extend the line CK to the poynt A, and LNM to the poyntes X and E. And for asmuch as in the circle ABC, the right line AC deuideth the right line BH into two equall partes and maketh right angles, therfore (by the 3. of the third)

in the line AC is the centre of the circle ABC. Agayne, for asmuch as in the selfe same circle ABC the right line NX, that is, the line ME deuideth the right line BG into two equall partes and maketh right angles, therefore (by the third of the third) in the line NX is the centre of the circle ABC. And it is proued that it is also in the line

[illustration]

AC. And these two right lines AC and NX meete together in no other poynt besides O. Where∣fore the poynt O is the centre of the circle ABC. And in like sort may we proue that the poynt O is the centre of the circle DEF. Wherefore the two circles ABC and DEF deuiding the one the other haue one and the same cen∣tre: which (by the 5. of the third) is impossible. A circle therfore cutteth not a circle in moe poyntes then two: which was required to be proued.

An other demonstration to proue the same.

Suppose that the circle ABC do cut the circle DGF in mo poyntes then two,* 1.62 that is, in B, G, F, and H. And (by the first of the third) take the centre of the circle ABC and let the same be the poynt K. And draw these right lines KB, KG, and KF. Now for asmuch as

[illustration]

within the circle DEF is taken a cer∣taine poynt K, and from that poynt are drawen vnto the circumference moe then two equall right lines, namely, KB, KG, and KF: therefore (by the 9. of the third) K is the centre of the circle DEF. And the poynt K is the centre of the circle ABC. Wherefore two cir∣cles cutting the one the other haue one and the same centre: which (by the 5. of the third) is impossible. A circle therfore cutteth not a circle in moe pointes then two: which was required to be demon∣strated.

The 10. Theoreme. The 11. Proposition. If two circles touch the one the other inwardly, their centres

being geuen: a right line ioyning together their centres and produced, will fall vpon the touch of the circles.

SVppose that these two circles ABC, and ADE do touch the one the other in the poynt A.* 1.63 And (by the first of the third) take the centre of the circle ABC, and let the same be F: and likewise y^e centre of the circle ADE, and let the same be G. Then I say that a right line drawen from F to G and being produced, will fall vpon the poynt A. For if not,* 1.64 then if it be possible let it fall as the line FGDH doth. And draw these right lines AF, & AG. Now for asmuch as the lines AG and

[illustration]

GF are (by the 20. of the first) grea∣ter then the line FA, that is, then the line FH, take away the line GF which is common to them both. Wherefore the residue AG is greater then the residue GH. But the line DG is equall vnto the line GA (by the 15. definition of the first). Wherefore the line GD is greater then y^e line GH: the lesse then the greater: which is impossible. Wher∣fore a right line drawen from the poynt F to the poynt G and produced, falleth not besides the poynt A, which is y^e point of the touch. Wherefore it fallet•• vpon the touch. If therefore two circles touch the one y^e other inwardly, their centres being geuen, a right line ioyning together their centres and produced, will fall vpon the touch of the circles: which was re∣quired to be proued.

An other demonstration to proue the same.

But now let it fall as GFC falleth, and extend y^e line GFC to the poynt H:* 1.65 and drawe these right lines AG and AF. And for asmuch as the lines AG and GF are (by the 20. of the first) greater then the line AF. But the line AF is equall vnto the line CF, that is, vnto the line FH. Take away the line FG common to them both. Wherfore the residue AG is greater then y^e residue GH, that is, the line GD is greater then the line GH: the lesse greater then y^e grea∣ter: which is impossible.

Which thing may also be proued by the 7. Proposition of this booke. For for asmuch as the line HC is the diameter of the circle ABC,* 1.66 & in it is taken a poynt which is not the centre, namely, the poynt G, therefore the line GA is greater then the line GH by the sayd 7. Proposition. But the line GD is equall to the line GA (by the definition of a circle). Wherefore the line GD is greater then the line GH, namely, the part gr••ater then the whole: which is impossible.

The 11. Theoreme. The 12. Proposition. If two circles touch the one the other outwardly, a right line drawen by their centres shall passe by the touch.

SVppose that these two circles ABC and ADE do touch the one the o∣ther outwardly in the poynt A. And (by the third of the third) take the centre of the circle ABC,* 1.67 and let the same be the poynt F: and likewise the centre of the circle ADE, and let the same be the poynt G. Then I say that a right line drawen from the poynt F to the poynt G shall passe by the poynt of the touch, namely, by the poynt A. For if not, then if it be possible, let it passe as the right line FCDG doth. And

[illustration]

draw these right lines AF & AG. And for asmuch as the poynt F is the centre of y^e circle ABC, ther∣fore the line FA is equall vnto the line FC. Againe for asmuch as the poynt G is the centre of the circle ADE, therefore the line GA is equall to the line GD. And And it is proued that the line FA is equall to the line FC. Wherefore the lines FA and AG are equall vnto the lines FC and GD: Wherefore the whole line FG is greater then the lines FA and AG. But it is also lesse (by the 20. of the first): which is impossible. Wherfore a right line drawen from the poynt F to the poynt G shall passe by the poynt of the touch, namely, by the poynt A. If there∣fore two circles touch the one the other outwardly, a right line drawen by their centres, shall passe by the touch: which was required to be demonstrated.

¶ An other demonstration after Pelitarius.

* 1.68Suppose that the two circles ABC and DEF do touch the one the other out∣wardly in the poynt A: And let G be the centre of the circle ABC: From which poynt produce by the touch of the circles the line GA to the poynt F of the circum∣ference DEF. Which for asmuch as it passeth not by the centre of the circle DEF (as the aduersary affirmeth) draw

[illustration]

from the same centre G an other right line GK, which if it be possi∣ble let passe by the centre of the cir∣cle DEF, namely, by the poynt H: cutting the circumference ABC in the poynt B, & the circūference DEF in the poynt D, & let the op∣posite poynt therof be in the point K. And for asmuch as frō the poynt G taken without the circle DEF is drawen the line GK passing by the centre H, and frō the same poynt is•• drawen also an

other line not passing by the contre, namely, the line GF. Therefore (by the 8. of thys booke) the outward part GD. of the line GK shall be lesse then the outward part GA of the line GF•• But the line GA is equall to the line GB. Wherfore the line GD is lesse then the line GB•• namely, the whole lesse then the part•• which is absurde.

The 12. Theoreme. The 13. Proposition. A circle can not touch an other circle in moe poyntes then one, whether they touch within or without.

FOr if it be possible, let the circle ABCD touch y^e circle EBFD first inwardly in moe poyntes then one, that is, in D and B.* 1.69 Take (by the first of the third) the centre of the circle ABCD, and let the same be y^e point G: and likewise y^e centre of the circle EBFD, and let the same be y^e poynt H. Wherefore (by the 11. of the same) a right line drawen from the poynt G to the poynt H and produced, will fall vp∣on the poyntes B and D: let it so fall as the line BGHD doth. And for asmuch as the poynt G is the centre of the circle ABCD, therefore (by the 15. definiti∣on of the first) the line BG is equall to

[illustration]

the line DG. Wherfore the line BG is greater then then the line HD: Wher∣fore the line BH is much greater then the line HD. Agayne for asmuch as the poynt H is the centre of the circle EBFD, therefore (by the same defini∣tion) the line BH is equall to the line HD: and it is proued that it is m••ch greater then it: which is impossible. A circle therefore can not touch a circle in∣wardly in moe poyntes then one.

Now I say that neither outwardly also a circle toucheth a circle in moe poyntes then one.* 1.70 For if it be possible, let the circle ACK touch y^e circle ABCD outwardly in moe poyntes thē one, that is, in A and C: And (by the first petition) draw a line from the poynt A to the poynt C. Now for asmuch as in y^e circumference of either of the circles ABCD, and ACK, are taken two poyntes at all aduentures, namely, A and C, therefore (by the second of the third) a right line ioyning together those poyntes shall fall within both the circles. But it falleth within the circle ABCD, & without the circle ACK: which is absurde. Wherefore a circle shall not touch a circle out∣wardly in moe pointes then one, and it is proued y^t neither also inwardly. Where∣fore a circle can not touch an other circle in moe poyntes then one, whether they

touch within or without: which was required to be demonstrated.

¶ An other demonstration after Pelitarius and Flussates.

* 1.71Suppose that there be two circles ABG and ADG, which if it be possible, let touch the one the other outwardly in moe poyntes then one, namely, in A and G. Let the centre of the circle ABG be the poynt I, and let the centre of the circle ADG be the poynt K. And draw a right line from the poynt I to the poynt K, which (by the 12. of thys booke) shall passe both by the

[illustration]

poynt A and by the poynt G: which is not possible: for then two right lines should include a superficies, contrary to the last common sentence. It may also be thus demonstrated. Draw a line from the centre I to the centre K, which shall passe by one of the touches, as for example by the poynt A. And draw these right lines GK and GI, and so shall be made a tri∣angle, whose two sides GK and GI shall not be greater then the side IK: which i•• contrary to the 20. of the first.

But now if it be possible, let the foresayd circle ADG touch the circle ABC inward∣ly in moe poyntes then one,* 1.72 namely, in the pointes A and G: and let the centre of the circle ABC be the poynt I, as before: and let

[illustration]

the centre of the circle ADG be the poynt K, as also before. And extend a line from the poynt I to the poynt K, which shall fall vpon the touch (by the 11. of thys booke). Draw also these lines KG, and IG. And for asmuch as the line KG is equall to the line KA (by the 15. definition of the first) adde the line KI common to them both. Wherefore the whole line AI is equall to the two lines KG and KI: but vnto the line AI is equall the line IG (by the definition of a circle). Wher∣fore in the triangle IKG the side IG is not lesse then the two sides IK and KG: which is con∣trary to the 20. of the first.

The 13. Theoreme. The 14. Proposition. In a circle, equall right lines, are equally distant from the cē∣tre. And lines equally distant from the centre, are equall the one to the other.

* 1.73SVppose that there be a circle

[illustration]

ABCD, and let there be in it drawen these equall right lines AB and CD. Then I say that they are equally distant from the centre.* 1.74 Take (by the first of the third) the centre of the cir∣cle ABCD, and let the same be y^e poynt E. And (by the 12. of the first) from the point E draw vnto the lines AB & CD

perpendicular lines EF and EG. And (by the first petition) draw these right lines AE and CE.* 1.75 Now for asmuch as a certaine right line EF drawen by the centre cutteth a certaine other right line AB not drawen by the centre, in such sort that it maketh right angles, therefore (by the third of the third) it deuideth it into two equall partes. Wherefore the line AF is equall to the line FB. Wher∣fore the line AB is double to the line AF: and by the same reason also the line CD is double to the line CG. But the line AB is equall to the line CD. Wher∣fore the line AF is also equall to the line CG. And for asmuch as (by the 15. de∣finition of the first) the line AE is equall to the line EC, therefore the square of the line EC is equall to the square of the line AE. But vnto the square of the line AE, are equall (by the 47. of the first) the squares of the lines AF & FE: for the angle at the poynt F is a right angle. And (by y^e selfe same) to the square of the line EC are equall the squares of the lines EG and GC: for the angle at the poynt G is a right angle. Wherefore the squares of the lines AF and FE are equall to the squares of the lines CG and GE: of which the square of the line AF is equall to the square of the line

[illustration]

CG: for the line AF is equall to the line CG. Wherefore (by the third common sentence) the square remayning, namely, the square of the line FE, is equall to the square remayning, namely, to the square of the line EG. Wherefore the line EF is equall to the line EG.* 1.76 But right lines are sayd to be equally distant from y^e cen∣tre, when perpendicular lines drawen frō the centre to those lines, are equall (by the 4. definition of the third). Wherfore the lines AB and CD are equally distant from the centre.

But now suppose that the right lines AB and CD be equally distant from the centre,* 1.77 that is, let the perpendicular line EF be equall to the perpendicular line EG. Then I say that the line AB is equall to the line CD. For the same order of construction remayning, we may in like sort proue that the line AB is double to the line AF, and that the line CD is double to the line CG. And for asmuch as the line AE is equall to the line CE, for they are drawen from y^e cen∣tre to the circumference, therfore the square of the line AE is equall to y^e square of the line CE. But (by the 47. of the first) to the square of the line AE are equall the squares of the lines EF and FA. And (by the selfe same) to y^e square of the line CE are equall the squares of the lines EG and GC. Wherfore the squares of the lines EF and FA are equall to the squares of the lines EG and GC. Of which the square of the line EG is equall to the square of the line EF, for the line EF is equall to the line EG. Wherefore (by the third common sen∣tence) the square remayning, namely, the square of the line AF, is equall to the square of the line CG. Wherefore the line AC is equall vnto the line CG. But

the line AB is double to the line AF, and the line CD is double to the line CG. Wherefore the line AB is equall to the line CD. Wherefore in a circle equall right lines are equally distant from the centre. And lines equally distant from the centre, are equall the one to y^e other: which was required to be proued.

¶ An other demonstration for the first part after Campane.

* 1.78Suppose that there be a circle ABDC, whose centre let be the poynt E. And draw in it two equall lines AB and CD. Then I say that they are equally distant from the centre. Draw from the centre vnto the lines AB

[illustration]

and CD, these perpendicular lines EF and EG. And (by the 2. part of the 3. of this booke) the line AB shall be equally deuided in the poynt F. and the line CD shall be equally deuided in the poynt G. And draw these right lines EA, EB, EC, and ED. And for asmuch as in the triangle AEB the two sides AB and AE are equall to the two sides CD and CE of the triangle CED, & the base EB is equall to the base ED. therefore (by the 8. of the first) the angle at the point A shall be equall to the angle at the point C. And for asmuch as in the triangle AEF the two sides AE and AF are equall to the two sides CE and CG of the triangle CEG, and the angle EAF is equall to the angle CEG, therefore (by the 4. of the first) the base EF i•• equall to the base EG: which for asmuch as they are perpendicular lines, therefore the lines AB & CD are equally distant frō the centre, by the 4. definition of this booke.

The 14. Theoreme. The 15. Proposition. In a circle, the greatest line is the diameter, and of all other lines that line which is nigher to the centre is alwayes greater then that line which is more distant.

SVppose that there be a circle

[illustration]

ABCD, and let the diameter thereof be the line AD, and let the centre thereof be the poynt E. And vnto the diameter. AD let the line BC be nigher then the line FG. Then I say that the line AD is the greatest, and the line BC is greater then y^e line FG. Draw (by the 12. of the first) from the centre E to the lines BC and FG per∣pendicular lines EH and EK.* 1.79 And for asmuch as the line BC is nigher vn∣to the centre then the line FG, therfore

(by the 4. definition of the third) the line EK is greater then the line EH. And (by the third of the first) put vnto the line EH an equall line EL. And (by the 11. of the first) from the point L raise vp vnto the line EK a perpen∣dicular line LM: and extend the line LM to the poynt N. And (by the first petition) draw these right lines, EM, EN, EF, and EG. And for asmuch as the line EH is equall to the line EL, therefore (by the 14. of the third,* 1.80 and by the 4. definition of the same) the line BC is equall to the line MN. Againe for asmuch as the line AE is equall to

[illustration]

the line EM, and the line ED to the line EN, therefore the line AD is e∣quall to the lines ME and EN. But the lines ME and EN are (by the 20. of the first) greater then the line MN. Wherefore the line AD is greater then the line MN. And for asmuch as these two lines ME and EN are equall to these two lines FE and EG (by the 15. definition of the first) for they are drawen from the centre to the circumfe∣rence, and the angle MEN is greater then the angle FEG, therefore (by the 24. of the first) the base MN is greater then the base FG. But it is proued that the line MN is equall to the line BC: Wherefore the line BC also is grea∣ter then the line FG. Wherefore the diameter AD is the greatest, and the line BC is greater then the line FG. Wherefore in a circle, the greatest line is the diameter, and of all the other lines, that line which is nigher to y^e centre is alwaies greater then that line which is more distant: which was required to be proued.

¶An other demonstration after Campane.

In the circle ABCD, whose centre let be the poynt E, draw these lines, AB, AC, AD, FG, and HK, of which let the line AD be the diameter of the circle.* 1.81 Then I say that the line AD is the greatest of all the lines.

[illustration]

And the other lines eche of the one is so much greater then ech of the other, how much nigher it is vnto the centre. Ioyne together the endes of all these lines with the centre, by drawing these right lines EB, EC, EG, EK, EH, and EF. And (by the 20. of the first) the two sides EF and EG of the triangle EFG, shall be greater then the third side FG. And for asmuch as the sayd sides EF & EG are equall to the line AD (by the definition of a circle) therefore the line AD is greater then the line FG. And by the same reason it is greater then euery one of the rest of the lines, if they be put to be bases of tri∣angles: for that euery two sides drawen frō the

centre are equall to the line AD. Which is

[illustration]

the first part of the Proposition. Agayne, for as∣much as the two sides EF and EG of the tri∣angle EFG, are equall to the two sides EH and EK of the triangle EHK, and the angle FEG is greater then the angle HEK, therfore (by the 24. of the first) the base FG is greater then the base HK. And by the same reason may it be proued, that the line AC is greater then the line AB. And so is manifest the whole Pro∣position.

The 15. Theoreme. The 16. Proposition. If from the end of the diameter of a circle be drawen a right line making right angles: it shall fall without the circle: and betwene that right line and the circumference can not be drawen an other right line: and the angle of the semicircle is greater then any acute angle made of right lines, but the o∣ther angle is lesse then any acute angle made of right lines.

SVppose that there be a circle ABC: whose centre let be the point D, and let the diameter therof be AB. Then I say y^t a right line drawen from the poynt A, making with the diameter AB right angles, shall fall without the circle.* 1.82 For if it do not, then if it be possible, let it fall within the circle as the line AC doth,

[illustration]

and draw a line from the point D to the point C.* 1.83 And for asmuch as (by the 15. definition of the first) the line DA is equall to the line DC, for they are drawen from the centre to the circum∣ference, therefore the angle DAC is equall to the angle ACD. But the an∣gle DAC is (by supposition) a right angle: Wherfore also the angle ACD is a right angle. Wherefore the angles DAC and ACD, are equall to two right angles: which (by the 17. of the first) is impossible. Wherefore a right line drawen from the poynt A, making with the diameter AB right angles, shall not fall within y^e circle. In like sort also may we proue, that it falleth not in

the circumference. Wherefore it falleth without, as the line AE doth.

I say also,* 1.84 that betwene the right line AE, and the circumference ACB, can not be drawen an other right line. For if it be possible, let the line AF so be drawen. And (by the 12. of the first) from the poynt D draw vnto the line FA a perpendicular line DG: And for asmuch as AGD is a right angle, but DAG is lesse then a right angle, therefore (by the 19. of the first) the side AD is greater then the side DG. But the line DA is equall to the line DH, for they are drawen from the centre to the circumference. Wherefore the line DH is greater then the line DG: namely, the lesse greater then the greater: which is impossible. Wherefore betwene the right line AE and the circumference ACB, can not be drawen an other right line.

I say moreouer,* 1.85 that the angle of

[illustration]

the semicircle contayned vnder y^e right line AB and the circūference CHA, is greater then any acute angle made of right lines. And the angle remayning cōtayned vnder y^e circumference CHA and the right line AE, is lesse then any acute angle made of right lines. For if there be any angle made of right lines greater then that angle which is con∣tayned vnder the right line BA and the circumference CHA, or lesse then that which is contayned vnder the cir∣cumference CHA and the right line AE, then betwene the circumference CHA and the right line AE, there shall fall a right line, which maketh the angle contayned vnder the right lines, grea∣ter then that angle which is contayned vnder the right line BA and the cir∣cumference CHA, and lesse then the angle which is contayned vnder the cir∣cumference CHA and the right line AE. But there can fall no such line, as it hath before bene proued. Wherfore no acute angle contained vnder right lines, is greater then the angle contayned vnder the right line BA and the circumfe∣rence CHA, nor also lesse then the angle contayned vnder the circumference CHA and the line AE.

Correlary.

Hereby it is manifest that a perpendicular line drawen frō the end of the diameter of a circle toucheth the circle: and that a right line toucheth a circle in one poynt onely. For it was proued (by the 2. of the third) that a right line drawen from two pointes taken in

the circumference of a circle, shall fall within the circle. Which was required to be demonstrated.

The 2. Probleme. The 17. Proposition. From a poynt geuen, to draw a right line which shall touch a circle geuen.

* 1.86SVppose that the poynt geuen be A, and let the circle geuen be BCD. It is required from the poynt A to draw a right line which shall touch the circle BCD. Take (by the first of the third) the centre of the circle, and let the same be E. And (by the first petition) draw the right line ADE. And making the centre E, and the space AE, describe (by y^e third petition) a circle AFG. And from the poynt D raise vp (by the 11. of the first) vnto the line EA a perpendicular line DF. And (by the first petition) drawe these lines EBF and AB. Then I

[illustration]

say, that from the point A is drawen to the circle BCD a touch line AB.* 1.87 For for asmuch as the point E is the centre of the circle BCD, and also of the circle AFG, therfore the line EA is equall to the line EF, and the line ED to y^e line EB, for they are drawen from the centre to the circumference. Wherefore to these two lines AE and EB, are equall these two lines EF & ED, and the angle at the poynt E is common to them both: Wherefore (by the 4. of the first) the base DF is equall to the base AB, and y^e triangle DEF is equall to the triangle EBA, and the rest of the angles remayning to the rest of the angles remayning. Wherefore the angle EDF is equall to y^e angle EBA. But the angle EDF is a right angle: Wherfore also the angle EBA is a right angle, and the line EB is drawen from the centre. But a perpendicular line drawen from the end of the diameter of a circle, toucheth the circle (by y^e Corella∣ry of the 16. of the third). Wherefore the line AB toucheth the circle BCD. Wherfore from the point geuen, namely, A, is drawen vnto y^e circle geuē BCD, a touch line AB: which was required to be done.

¶ An addition of Pelitarius.

* 1.88Vnto a right lyne which cutteth a circle, to drawe a parallel line which shall touch the circle.

S••ppose that the right lyne AB do out the circle ABC in the poyntes A and B. It is required to drawe vnto the line AB a parallel lyne

[illustration]

which shall touche the circle. Let the centre of the circle be the point D. And deuide the lyne AB into two equall partes in the point E. And by the point E and by the centre D, draw the diameter CDEF. And from the point F (which is the ende of the dia∣meter) rayse vp (by the 11. of the first) vnto the dia∣meter CF a perpendicular line GFH. Then I say that the lyne GFH (which by the correllary of the 16. of this booke toucheth the circle) is a parallel vnto the line AB.* 1.89 For forasmuch as the right line CF fallyng vpon either of these lines AB & GH ma∣keth all the angles at the point •• right angles (by the 3. of this boke) and the two angles at the point Fare supposed to be right angles: therfore (by the 29. of the first) the lines AB and GH are parallels: which was required to be done. And this Probleme is very commodious for the inscribing or circumscribing of figures in or about circles.

The 16. Theoreme. The 18. Proposition. If a right lyne touch a circle, and from the centre to the touch be drawen a right line, that right line so drawen shalbe a per∣pendicular lyne to the touche lyne.

SVppose that the right line DE do touch the circle ABC in the point C. And take the centre of the circle ABC, and let the same be F.* 1.90 And (by the first petition) from the poynt F to the poynt C drawe a right line FC. Then I say, that CF is a perpendicular line to DE. For if not, draw (by the 12. of the first) from the

[illustration]

poynt F to the line DE a perpendicular line FG. And for asmuch as the angle FGC is a right angle, therefore the angle GCF is an acute angle: Wherefore the angle FGC is greater then y^e angle FCG, but vnto the greater angle is subtended the greater side (by the 19. of the first). Wherefore the line FC is greater then the line FG. But the line FC is equall to the line FB, for they are drawen from the centre to the circumference: Wherfore the line FB also is greater then the line FG, namely, the lesse then the greater: which is impossible. Wherefore the line FG is not a perpendicular line vnto the line DE. And in like sort may we proue, that no other line is a perpendicular line vnto y^e line DE besides the line FC: Wher∣fore the line FC is a perpendicular line to DE. If therefore a right line touch

a circle, & from y^• centre to y^e touch be drawen a right line, y^t right line so drawen shall be a perpendicular line to y^e touch line: which was required to be proued.

¶ An other demonstration after Orontius.

Suppose that the circle geuen be ABC, which let the right lyne DE touch in the point C.* 1.91 And let the centre of the circle be the point F. And draw a right line from F to C. Then

[illustration]

I say that the line FC is perpendicular vnto the line DE. For if the line FC be not a perpēdiculer vnto the line DE, then, by the conuerse of the x. definition of the first boke, the angles DCF & FCE shal be vnequall: & therfore the one is grea∣ter then a right angle, and the other is lesse then a right angle. (For the angles DCF and FCE are by the 13. of the first equall to two right an∣gles) Let the angle FCE, if it be possible, be grea∣ter then a right angle, that is, let it be an obtuse angle. Wherfore the angle DCF ••hal be an acute angle. And forasmuch as by suppositiō the right line DE touche••h the circle ABC, therefore it cutteth not the circle. Wherefore the circumfe∣rence BC falleth betwene the right lines DC & CF: & therfore the acute and rectiline angle DCF shall be greater then the angle of the semicircle BCF which is contayned vnder the circumferēce BC & the right line CF. And so shall there be geuē a rectiline & acute angle greater then the angle of a semicircle: which is contrary to the 16. propo∣sition of this booke. Wherfore the angle DCF is not lesse then a right angle. In like sort also may we proue that it is not greater then a right angle. Wherfore it is a right angle, and therfore also the angle FCE is a right angle. Wherefore the right line FC is a per∣pendicular vnto the right line DE by the 10. definition of the first•• which was required to be proued.

The 17. Theoreme. The 19. Proposition. If a right lyne doo touche a circle, and from the point of the touch be raysed vp vnto the touch lyne a perpendicular lyne, in that lyne so raysed vp is the centre of the circle.

SVppose that the right line DE do

[illustration]

touch the circle ABC in the point C. And from C raise vp (by y^e 11. of the first) vnto the line DE a perpendicu∣lar line CA. Then I say, that in the line CA is the centre of the circle. For if not, then if it be possible, let y^e centre be with∣out the line CA, as in y^e poynt F. And (by the first petition) draw a right line from C to F.* 1.92 And for asmuch as a certaine right line DE toucheth the circle ABC, and from the centre to the touch is drawen a right line CF, therefore (by the 18. of the third) FC is a perpendicular line to

DE. Wherefore the angle FCE is a right angle. But the angle ACE is also a right angle: Wherefore the angle FCE is equall to the angle ACE, namely, the lesse vnto the greater: which is impossible•• Wherefore the poynt F is not the centre of the circle ABC. And in like sort may we proue, y^t it is no other where but in the line AC. If therefore a right line do touch a circle, and from the point of the touch be raised vp vnto the touch line a perpendicular line, in that line so raised vp is the centre of the circle: which was required to be proued.

The 18. Theoreme. The 20. Proposition. In a circle an angle set at the centre, is double to an angle set at the circumference, so that both the angles haue to their base one and the same circumference.

SVppose that there be a circle ABC, and at the centre thereof,* 1.93 namely, the poynt E, let y^e angle BEC be set, & at the circumference let there be set the angle BAC, and let them both haue one and the same base, namely, the circumference BC. Then I say, that the angle BEC is double to the angle BAC. Draw y^e right

[illustration]

line AE, and (by the second petition) ex∣tend it to the poynt F. Now for asmuch as the line AE is equall to the line EB,* 1.94 for they are drawen from the centre vnto the circumference, the angle EAB is e∣quall to the angle EBA (by the 5. of the first). Wherefore the angles EAB and EBA are double to the angle EAB. But (by the 32. of the same) the angle BEF is equall to the angles EAB and EBA: Wherefore the angle BEF is double to the angle EAB. And by the same reason the angle FEC is double to the angle EAC. Wherefore the whole angle BEC is double to the whole angle BAC.

Againe, suppose that there be set an other angle at the circumference,* 1.95 and let the same be BDC. And (by the ••irst petition) draw a line from D to E. And (by the second petition) extend the line DE vnto the poynt G. And in like sort may we proue, that the angle GEC is double to the angle EDC. Of which the an∣gle GEB is double to the angle EDB. Wherfore the angle remayning BEC is double to the angle remayning BDC. Wherfore in a circle an angle set at the centre, is double to an angle set at the circumference, so that both the angles haue to their base one and the same circumference: which was required to be demon∣strated.

The 19. Theoreme. The 21. Proposition. In a circle the angles which consist in one and the selfe same section or segment, are equall the one to the other.

SVppose y^t there be a circle ABCD, & in the segment therof BAED, let there consist these angles BAD and BED. Then I say, that the angles BAD and BED are equall the one to the other. Take (by the first of the third) the centre of the circle

[illustration]

ABCD,* 1.96 and let the same be the point F. And (by the first petition) draw these lines BF and FD.* 1.97 Now for asmuch as the angle BFD is set at the centre, and the angle BAD at the circumference, and they haue both one and y^e same base, name∣ly, the circumference BCD, therefore the angle BFD is (by the Proposition going before) double to the angle BAD: and by the same reason the angle BFD is al∣so double to the angle BED. Wherefore (by the 7. common sentence) the angle BAD is equall to the angle BED. Wherefore in a circle the angles which consiste in one and the selfe same segment, are equall the one to the other: which was required to be proued.

* 1.98In this proposition are three cases. For the angles consisting in one and the self same segment, the segment may either be greater thē a semicircle, or lesse then a semicircle, or els iust a semicircle. For the first case the demonstration before put serueth.

But now suppose that the angles BAD and BED do consist in the sectiō BAD, which let be lesse then a semicircle. Euen in this case al∣so

[illustration]

I say that the angles BAD and BED are e∣quall. For draw a right line from A to E. And let the lines AD and BE cutte the one the other in the poynt G, wherefore the segment ACE is greater then a semicircle. And therfore by the first part of this proposition the angles whiche are in it, namely, the angles ABE and EDA are equall the one to the other. And forasmuch as in the triangle ABG the inward and oppo∣site angles ABG and GAB are equall to the outwarde angle BGD, and by the same reason the two angles EDG and GED of the triangle DEG are equall to the selfe same outward angle BGD. Wherfore the two angles ABG and GAB are equall to the two angles EDG and GED

by th•• first commō senten••••. From which ••f there

[illustration]

be taken equall angles, namely, ABG, and EDG, the angl•• remainyng BAG shall be equall to the angle remayning DEG, that is, the angle BAD to the angle DEB (by the third common sen∣tence) which was required to be proued.

The selfe same construction and demonstrati∣on will also serue,* 1.99 if the angles were set in a semi∣circle as it is pla••ne to see, in the figure here set.

The 20. Theoreme. The 22. Proposition. If within a circle be described a figure of fower sides, the an∣gles therof which are opposite the one to the other, are equall to two right angles.

SVppose that there be a circle ABCD, and let there be described in it a figure of fower sides, namely, ABCD. Then I say, that the angles thereof which are opposite the one to the other, are equall to two right angles. Draw (by the first petition) these right lines AC and BD.* 1.100 Now for asmuch as (by the 32. of the first) the three angles of euery triangle are equall to two right angles: therfore y^• three angles of the triangle ABC, namely,* 1.101 CAB, ABC, and BCA, are equall to two right angles. But (by the 21. of the third) the angle CAB is equall to the

[illustration]

angle BDC, for they consist in one and the self same segmēt, namely, BADC. And (by the same Proposition) the an∣gle ACB is equall to the angle ADB, for they consist in one and the same seg∣ment ADCB. Wherefore the whol•• angle ADC is equall to y^e angles BAC and ACB: put the angle ABC com∣mon to them both. Wherefore the angles ABC, BAC, and ACB, are equall to the angles ABC and ADC. But the angles ABC, BAC, and ACB, are equall to two right angles. Wherefore the angles ABC and ADC are equall to two right angles. And in like sort also may we proue, that the angles BAD and DCB are equall to two right

angles. If therefore within a circle be described a figure of fower sides, the an∣gles thereof which are opposite the one to y^e other, are equall to two right angles•• which was required to be proued.

The 21. Theoreme. The 23. Proposition. Vpon one and the selfe same right line can not be described two like and vnequall segmentes of circles, falling both on one and the selfe same side of the line.

* 1.102FOr if it be possible, let there be described vpon the right line AB two like & vnequall sections of circles, namely, ACB & ADB, falling both on one and the selfe same side of the line AB. And (by the first petition) drawe the right line ACD, and (by the third petition) drawe right lines from C to B, and from D to B. And for asmuch as the segment ACB is like to the segment ADB: and like

[illustration]

segmētes of circles are they which haue equall angles (by the 10. definition of the third). Where••ore the angle ACB is equall to the angle ADB, namely, the outward angle of y^e triangle CDB to the inward angle: which (by the 16. of the first) is impossible. Wherfore vp∣on one and the self same right line can not be described two like & vnequall seg∣mentes of circles, falling both on one & the selfe same side of the line: which was required to be demonstrated.

* 1.103Here Campane addeth that vpon one and the selfe same right lyne cannot be described two like and vnequall sections neither on one and the selfe same side of the lyne, nor on the opposite side. That they can not be described on one and the selfe same side, hath bene before demonstrated, and that neither also on the oppo∣site side, Pelitarius thus demonstrateth.

Let the section ABC be set vppon the lyne AC, and vpon the other side let be set the section ADC vppon the selfe same lyne AC,

[illustration]

and let the section ADC be lyke vnto the secti∣on ABC. Then I say that the sections ABC and ADC being thus set are not vnequal. For if it be pos∣sible let the section ADC be the greater. And de∣uide the line AC into two equal partes in the point E. And draw the right lyne BED deuiding the lyne AC right angled wise. And draw these right lynes AB, CB, AD and CD.* 1.104 And forasmuch as the section ADC is greater then the section ABC, the perpen¦dicular lyne also ED shall be greater then the per∣pendicular lyne EB: as is before declared in the ende of the definitions of this third booke. Wher∣fore

from the lyne ED ••ut of a ly••e equall to the lyne EB: ••hich 〈◊〉〈◊〉 be EF. And draw these right lynes AF and CF. Now then (by the 4. of the first) the triangle AEB shall be equall to the triangle AEF•• and the angle EBA shall be equall to the angle EFA. And by the same reason the angle EBC shall be equall to the angle EFC. Wherefore the whole angle ABC is equall to the whole angle AFC. But by the 21. of the first, the an∣gle AFC is greater then the angle ADC. Wherfore also the angle ABC is greater then the angle ADC. Wherefore by the definition the sections ABC and ADC are not lyke, which is contrary to the supposition. Wherefore they are not lyke and vnequall: which was required to be proued.

The 22. Theorme. The 24. Proposition. Like segmentes of circles described vppon equall right lines, are equall the one to the other.

SVppose that vpon these equall right lines AB and CD be described these like segmentes of circles, namely, AEB and CFD. Then I say, that the segment AEB is equall to the segment CFD. For putting the segment AEB vpon the segment CFD, and the poynt A vpon y^e poynt C, and the right line AB vpon the right line CD, the poynt B also shall fall vpon the poynt D, for y^e line AB is equall to the line CD. And the right line AB ex∣actly agreing with

[illustration]

the right line CD, y^• segment also AEB shall exactly agree with the segment CFD.* 1.105 For if the right line AB do exactly agree with the right line CD, and the segment AEB do not exactly agree with the segment CFD, but dif∣fereth as the segment CGD doth: Now (by the 13. of the third) a circle cutteth not a circle in more pointes then two, but the circle CGD cutteth y^e circle CFD in more pointes thē two, that is, in the points C, G, and D: which is (by the same) impossible. Wherefore the right line AB exactly agreing with the right line CD, the segment AEB shall not but exactly agree with the segment CFD: Wherefore it exactly agreeth with it, and is equall vnto it. Wherefore like seg∣mentes of circles described vpon equall right lines, are equall the one to the other: which was required to be proued.

This Proposition may also be demonstrated by the former proposition For if the se∣ctions AEB and CFD being like and set vpon equall right lines AB and CD,* 1.106 should be vnequall, then the one beyng put vpon the other, the great••r shall exceede the lesse: but the line AB is one line with the line CD: so that therby shal follow the contrary of the former Proposition.

Suppose that there be two right lines AB & CD which let be equall: and vpon thē let there be set these like sections ABK, and CDE.* 1.107 Then I say that the said sections are equall. For if not, then let CED be the greater section. And deuide the two lines AB and CD into two equall partes, the line AB in the pointe F, and the line CD in the

[illustration]

point G. And erect two perpendicular lines FK and GE. And draw these right lines AK & KB: EC, ED. And forasmuch as the section CED is the greater, therefore the perpendicular line GE is greater then the perpendicular FK: From the lyne GE cut of a line equall to the line FK, which let be GH: and draw these right lines CH and HD. And forasmuch as in the triangle AKF the two sides AF and FK are equall to the two sides CG & GH of the triangle CH∣G, and the angles at the pointes F and G are equal (for that they are right angles) ther∣fore (by the 4. of the first) the base A K is equall to the base CH, and the angle AKF to the angle CHG. And by the same reason the angle BKF is equall to the angle DHG. Wherfore the whole angle AKB is equall to the whole angle CHD. But the angle CHD is greater then the angle CED by the 21. of the first. Wherfore also the angle AKD is greater then the angle CED. Wherfore the sections are not lyke, which is contrary to the supposition.

The 3. Probleme. The 25. Proposition. A segment of a circle beyng geuen to describe the whole cir∣cle of the same segment.

SVppose that y^e segment geuen be

[illustration]

ABC. It is required to describe the whole circle of the same seg∣ment ABC.* 1.108 Deuide (by the 10. of the first) the line AC into two equall partes in the poynt D. And (by the 11. of the same) from the poynt D raise vp vnto the line AC a perpendicular line BD. And (by the first petition) draw a right line from A to B. Now then the angle ABD being compared to y^e angle BAD, is either greater then it, or equall vnto it, or lesse then it.

* 1.109First let it be greater. And (by the 23. of the same) vpon the right line BA, and vnto the poynt in it A, make vnto the angle ABD an equall angle BAE. And (by the second petition) extend the line BD vnto the poynt E. And (by the first petition) draw a line from E to C. Now for asmuch as the angle ABE is e∣quall to the angle BAE,* 1.110 therefore (by the 6. of the first) the right line EB is equall to the right line AE. And for asmuch as the line AD is equall to y^e line DC, and the line DE is common to them both: therefore these two lines AD

••nd DE, are equall to these t••o lines CD and DE the one to the other. And the angle ADE •••• (by the 4. petition) equall to the angle CDE, for either of them is a right angle•• Wherefor•• (by the 4. of the first) the base AE is equall to the base CE. But it is pro••ed, that the line AE is ••quall to y^e line BE. Wher∣fore the line BE also is equall to the line CE. Wherefore these three lines AE, EB, and EC, are equall the one to the other. Wherefore making the centre E, and the space either AE, or EB, or EC, describe (by the third petition) a cir∣cle, and it shall passe by the poyntes A, B, C. Wherefore there is described the whole circle of the segment geuen, And it is manifest, that the segment ABC is les••e then a semicircle, for the centre E falleth witho••t it.

The like demo••stration also will

[illustration]

serue if the angle ABD be equall to the angle BAD.* 1.111 For the line AD being equall to either of these lines, BD, and DC, there are three lines•• DA, DB, and DC, equall the one to the other. So that the point D shall be the centre of the circle being complete, and ABC shall be a semicircle.

But if the angle ABD be lesse then the angle

[illustration]

BAD,* 1.112 then (by the 23. of the first) vpon the right line BA, and vnto the point in it A, make vnto the angle ABD an equall angle within y^• segment ABC. And so the ce••tre of the circle shall fall in y^e line DB, and it shall be the point E: and the segment ABC shall be greater then a semicircle, Wherefore a segment being geuen, there is described the whole circle of the same segment: ••hich was required to be done.

¶ 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 w••y to finde out the

[illustration]

foresaid centre,* 1.113 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. A••d draw these lin••s AB and BC (by the first peticion) And (by the 10. of the first) deuide into two equall partes ••ith••r 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•• point••s

D and E rayse vp vnto the lines AB and BC perpendicular lynes DF and EF. Now forasmuch as either of these angles BDF,* 1.114 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,* 1.115 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 si••es 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.* 1.116 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

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.* 1.117 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.

The 23. Theoreme. The 26. Proposition. Equall angles in equall circles consist in equall circūferences, whether the angles be drawen from the centres, or from the circumferences.

SVppose that these circles ABC and DEF, be equall. And from their centres, namely, the pointes G and H, let there be drawen these equall angles BGC and EHF: and likewise from their circumferences these equall angles BAC and EDF. Then I say, that the circumference BKC is equall to the circumference ELF. Draw (by the first petition) right lines from B to C,* 1.118 and from E to F. And for asmuch as the circles ABC and DEF are equall, the right lines also drawen from their centres to their circumferences,* 1.119 are (by the first definition of the third) equall the one to the other. Wherefore these two lines BG and

[illustration]

GC, are equall to these two lines EH and HF. And the angle at the poynt G is equall to the angle at the point H: Wherfore (by the 4. of the first) the base BC is equall to y^e base EF. And for asmuch as the angle at the poynt A is equall to the angle at the point D, therefore the segment BAC is like to the segment EDF. And they are described vpon equall right lines BC and EF. But like segmentes of circles described vpon equall right lines, are (by the 24. of the third) equall the one to the other. Wherefore the segment BAC is equall to the segment EDF. And the whole circle ABC is equall to y^e whole circle DEF. Wherefore (by the third common sentence) the circumference re∣mayning BKC is equall to the circumference remayning ELF. Wherefore equall angles in equall circles consist in equall circumferences, whether the angles be drawen from the centres or from the circumferences: which was required to be demonstrated.

The 24. Theoreme. The 27. Proposition. In equall circles the angles which consist in equall circumfe∣rences, are equall the one to the other, whether the angles be drawen from the centres, or from the circumferences.

SVppose y^t these circles ABC, and DEF, be equall. And vpon these equall circumferences of the same circles, namely, vpon BC and EF, let there consist these angles BGC and EHF drawen from the cen∣tres, and also these angles BAC and EDF drawen from the cir∣cumferences. Then I say, that the angle BGC is equall to the angle EHF, and the angle BAC to the angle EDF. If the angle BGC be equall to the an∣gle EHF, then it is manifest, that the angle BAC is equall to y^e angle EDF (by the 20. of the third). But if the angle BGC be not equall to y^e angle EHF,* 1.120 then is the one of them greater then the other. Let the angle BGC be greater And (by the

[illustration]

23. of the first) vpon the right line BG, and vnto the point geuen in it G, make vnto the angle EHF an equall an∣gle BGK. But (by the 26. of y^e third) equall angles in equall circles consist vpō equall circumferences, whether they be drawen from the centres or from the circumferences. Wherefore the circumference BK is equall to the circumference EF. But the circumference EF is equall to the circumference BC: Wherefore the circumference BK also is equall to the cir∣cumference BC, the lesse to the greater: which is impossible. Wherfore the an∣gle BGC is not vnequall to the angle EHF: Wherefore it is equall. And (by the 20. of the third) the angle at the point A is the halfe of the angle BGC: and (by the same) the angle at the point D is the halfe of the angle EHF. Where∣fore the angle at the point A is equall to the angle at the point D. Wherefore in equall circles, the angles which consist in equall circumferences, are equall the one to the other, whether the angles be drawen from the centres or from the circumfe∣rences: which was required to be proued.

The 25. Theoreme. The 28. Proposition. In equall circles, equall right lines do cut away equall cir∣cumferences, the greater equall to the greater, and the lesse e∣quall to the lesse.

SVppose that these circles ABC, and DEF, be equall. And in them let there be drawen these equall right lines BC and EF, which let cut away these circumferences BAC and DEF being the greater, & also these circumferences BGC and EHF being the lesse. Then I say, that the greater circumference BAC is equall to the greater circumference EDF: and the lesse circumference BGC is equall to the lesse circumference EHF.* 1.121 Take (by the first of the third) the centres of the circles, and let the same be the pointes K and L. And draw these right lines, KB, KC, LE, and LF. And for asmuch as the circles are equall,* 1.122 therfore (by the first definition of the third) the lines which are drawen frō the cen∣tres

[illustration]

are equall. Where∣fore these two lines BK and KC, are equall to these two lines EL and LF. And (by suppositi∣on) the base BC is equall to the base EF. Where∣fore (by the 8. of y^e first) the angle BKC is equall to the angle ELF. But (by the 26. of the third) equall angles drawen from the centres, consist vpon equall circumferences. Wher∣fore the circumference BGC is equall to the circumference EHF: and y^e whole circle ABC is equall to the whole circle DEF. Wherefore the circumference remayning BAC, is (by the third common sentence) equall to the circumference remayning EDF. Wherefore in circles, equall right lines do cut away equall circumferences, the greater equall to the greater, and the lesse equall to the lesse: which was required to be proued.

The 26. Theoreme. The 29. Proposition. In equall circles, vnder equall circumferences are subtended equall right lines.

* 1.123 SVppose that these circles ABC and DEF, be equall. And in them let there be taken these equall circumferences, BGC and EHF: and drawe these right lines BC and EF. Then I say, that the right line BC is equall to the right line EF. Take (by the first of y^e third) the centres of the circles, and let them be the pointes K and L,* 1.124 and draw these right lines KB, KC, LE, LF. And for asmuch as the circumference BGC is equall to the circumference EHF,* 1.125 the angle BKC is equall to y^e angle ELF (by the 27. of the third). And for asmuch as the circles ABC and DEF are equall the one to the other, therefore (by the first definition of the third) the lines

whiche are

[illustration]

drawē from the centres are equall. Wherefore these y. lines BK and KC, are e∣quall to these y. lines LE and LF, & they comprehend equall angles. Wherefore (by the 4. of the first) the base BC is e∣quall to the base EF. Wherefore in equall circles vnder equall circumferences, are subtended equall right lines: which was required to be demonstrated.

The 4. Probleme. The 30. Proposition. To deuide a circumference geuen into two equall partes.

SVppose that the circumference geuen be ADB. It is required to de∣uide the circumference ADB into two equall partes.* 1.126 Draw a right line from A to B. And (by the 10. of the first) deuide the line AB into two equall partes in the point C. And (by the 11. of the first) from the point C rayse vp vnto AB a perpendicular line CD. And draw these right lines AD and DB. And forasmuch as the line AC is equall to the line CB,* 1.127 & the line CD is common to them both, there∣fore

[illustration]

these two lines AC and CD are equall to these two lines BC and CD. And (by the 4. peti••ion) the angle ACD is equall to the angle BCD, for either of them is a right right angle. Wherfore (by the 4. of the first) the base AD is equall to the base DB. But equall right lines do cut away equall circum∣ferences, the greater equall to the greater, & the lesse equall to the lesse (by the 28. of the third) And either of these circum∣ferences AD and DB is lesse then a semicircle. Wherfore the circumference AD is equall to the circumference DB. Wherfore the circumference geuen is deui∣ded into two equall partes: Which was required to be done.

The 27. Theoreme. The 31. Proposition. In a circle an angle made in the semicircle is a right angle:

but an angle made in the segment greater then the semicircle is lesse then a right angle, and an angle made in the segment lesse then the semicircle, is greater then a right angle. And moreouer the angle of the greater segment is greater then a right angle: and the angle of the lesse segment is lesse then a right angle.

SVppose that the circle be ABCD, and let the dimetient of the circle be the right line BC, and the cētre therof the point E. And take in the se∣micircle a point at all auentures, and let the same be D. And draw these right lines BA, AC, AD, and DC.* 1.128 Then I say that the angle in the semicircle BAC,

[illustration]

namely, the angle BAC is a right angle. And the angle ABC which is in the segment ABC being greater then the semicircle, is lesse then a right angle. And the angle ADC which is in the segment ADC being lesse then the semicircle is greater thē a right an∣gle. Draw a line from the point A to the point E, and extend the line BA vnto the point F. And forasmuch as the line BE is equall to the line EA, (for they are drawen from the centre to the circumference) therfore the an∣gle EAB is equall to the angle EBA (by the 5. of the first). Againe forasmuch as the line AE is equall to the line EC, the angle ACE is (by the same) equall to the angle CAE. Wherfore the whole angle BAC is equall to these two angles ABC and ACB. But the angle FAC which is an outward angle of the triangle ABC is (by the 32. of the first) equall to the two angles ABC & ACB. Wherfore the angle BAC is equall to the angle F••AC. Wherfore either of them is a right angle. Wherfore the angle BAC which is in the semicircle BAC is a right angle.

* 1.129And forasmuch as (by the 17. of the first) the two angles of the triangle ABC, namely, ABC and BAC are lesse then two right angles, and the angle BAC is a right angle. Therfore the angle ABC is lesse then a right angle, and it is in the segment ABC which is greater then the semicircle.

* 1.130And forasmuch as in the circle there is a figure of foure sides, namely, ABCD. But if within a circle be described a figure of foure sides, the angles therof which are opposite the one to the other are equall to two right angles (by the 22. of the third) Wherfore (by the same) the angles ABC and ADC are equall

to two right angles. But the angle ABC is lesse then a right angle. Wherfore the angle remayning ADC is greater then a right angle, and it is in a segment which is lesse then the semicircle.

Now also I say that the angle of the greater segment, namely,* 1.131 the angle which is comprehended vnder the circumference ABC and the right line AC is greater then a right angle, and the angle of the lesse segment comprehended vnder the circumference ADC, and the right line AC is lesse thē a right angle, which

[illustration]

may thus be proued. Forasmuch as the angle comprehended vnder the right lines BA and AC is a right angle, therfore the angle com∣prehended vnder the circumference ABC and the right line AC is greater then a right angle: for the whole is euer greater then his part (by the 9. common sentence.

Againe forasmuch as the angle compre∣hended vnder the right lines AC and AF is a right angle,* 1.132 therfore the angle com∣prehended vnder the right line CA and the circumference ADC is lesse then a right an∣gle. Wherfore in a circle an angle made in the semicircle is a right angle, but an angle made in the segment greater then the semicircle is lesse then a right angle, and an an∣gle made in the segment lesse then the semicircle, is greater then a right angle. And moreouer the angle of the greater segment is greater then a right angle: & the angle of the lesse segment is lesse then a right angle: which was required to be demonstrated.

An other demonstration to proue that the angle BAC is a right angle.* 1.133 For∣asmuch as the angle AEC is double to the angle BAE (by the 32. of the first) for it is equall to the two inward angles which are opposite. But the inwarde an∣gles are (by the 5. of the first) equall the one to the other, and the angle AEB is double to the angle EAC. Wherfore the angles AEB and AEC are double to the angle BAC. But the angles AEB and AEC are equall to two right an∣gles: Wh••rfore the angle BAC is a right angle. Which was required to be de∣monstrated.

Correlary.

Hereby it is manifest,* 1.134 that if in a triangle one angle be equall to the two other angles remayning the same angle is a right

angle: for that the side angle to that one angle (namely, the angle which is made of the side produced without the trian∣gle) is equall to the same angles, but when the side angles are equall the one to the other, they are also right angles.

¶ An addition of Pelitarius.

If in a circle be inscribed a rectangle triangle, the side opposite vnto the right angle shall be the diameter of the circle.

* 1.135Suppose that in the circle ABC be inscribed a

[illustration]

rectangle triangle ABC, whose angle at the point B let be a right angle. Then I say, that the side AC is the diameter of the circle. For if not, then shall the centre be without the line AC, as in the point E.* 1.136 And draw a line from the poynt A to the point E, & produce it to the circumference to the point D: and let AED be the diameter: and draw a line from the point B to the point D. Now (by this 31. Propositiō) the angle ABD shall be a right angle, and therefore shall be equall to the right angle ABC, namely, the part to the whole: which is ab∣surde. Euen so may we proue, that the centre is in no other where but in the line AC. Wherfore AC is the diameter of the circle: which was required to be proued.

¶ An addition of Campane.

* 1.137By thys 31. Proposition, and by the 16. Proposition of thys booke, it is mani∣fest, that although in mixt angles, which are contayned vnder a right line and the circumference of a circle, there may be geuen an angle lesse & greater then a right angle, yet can there neuer be geuē an angle equall to a right angle. For euery secti∣on of a circle is eyther a semicircle, or greater then a semicircle, or lesse, but the an∣gle of a semicircle is by the 16. of thys booke, lesse then a right angle, and so also is the angle of a lesse section by thys 31. Proposition: Likewise the angle of a greater section, is greater then a right angle, as it hath in thys Proposition bene proued.

The 28. Theoreme. The 32. Proposition. If a right line touch a circle, and from the touch be drawen a right line cutting the circle: the angles which that line and the touch line make, are equall to the angles which consist in the alternate segmentes of the circle.

SVppose that the right line EF do touch the circle ABCD in the point B: and from the point B let there be drawen into the circle ABCD a right line cutting the circle, and let the same be BD. Then I say, that the angles which the line BD together with the

touch line EF do make, are equall to the angles which are in the alternate seg∣mentes of the circle, that is, the angle FBD is equall to the angle which consi∣steth in the segment BAD, and the angle EBD is equall to the angle which consisteth in the segment BCD.* 1.138 Raise vp (by the 11. of the first) from y^e point B vnto the right line EF a perpendicular line BA. And in the circumference BD take a point at all aduentures, and let the same be C. And draw these right lines AD, DC, and CB.* 1.139 And for asmuch as a certaine right line EF tou∣cheth the circle ABC in the point B, and from the point B where the touch is, is raysed vp vnto the touch line a perpen∣dicular

[illustration]

BA. Therfore (by the 19. of the third) in the line BA is the centre of the circle ABCD. Wherfore y^e angle ADB being in the semicircle, is (by the 31. of the third) a right angle. Wherefore the an∣gles remayning BAD and ABD, are equall to one right angle. But the angle ABF is a right angle. Wherefore the an∣gle ABF is equall to the angles BAD and ABD. Take away y^e angle ABD which is common to them both. Wherefore the angle remayning DBF, is equall to the angle remayning BAD, which is in the alternate segment of the circle. And for asmuch as in the circle is a figure of fower sides, namely, ABCD, therfore (by the 22. of the third) the angles which are opposite the one to the other, are equall to two right angles. Wherfore the an∣gles BAD and BCD, are equall to two right angles. But the angles DBF and DBE, are also equall to two right angles. Wherefore the angles DBF and DBE, are equall to the angles BAD and BCD. Of which we haue proued that the angle BAD is equall to the angle DBF. Wherefore the an∣gle remayning DBE, is equall to the angle remayning DCB, which is in the alternate segment of the circle, namely, in the segment DCB. If therfore a right line touch a circle, and from the touch be drawen a right line cutting the circle: the angles which that line and the touch line make, are equall to y^e angles which consist in y^e alternate segmentes of the circle: which was required to be proued.

In thys Proposition may be two cases.* 1.140 For the line drawen from the touch and cutting the circle, may eyther passe by the centre or not. If it passe by the centre, then is it manifest (by the 18. of thys booke) that it falleth perpendicularly vpon the touch line, and deuideth the circle into two equall partes, so that all the angles in eche semicircle, are by the former Proposition, right angles, and therfore equall to the alternate angles made by the sayd perpendicular line and the touch line. If it passe not by the centre, then followe the construction and demonstration be∣fore put.

The 5. Probleme. The 33. Proposition. Vppon a right lyne geuen to describe a segment of a circle, which shall contayne an angle equall to a rectiline angle geuē.

SVppose that the right line geuen be AB, and let the rectiline angle ge∣••••n be C. It is required vpon the right line geuē AB to describe a seg∣ment of a circle which shall contayne an angle equall to the angle C. Now the angle C is either an acute angle, or a right angle, or an ob∣tuse angle.

* 1.141First, let it be an acute angle as

[illustration]

in the first description. And (by the 23 of the first) vpon the right line AB and to the point in it A describe an angle equal to the angle C,* 1.142 and let the same be DAB. Wherfore the angle DAB is an acute angle. From the point A raise vp (by the 11. of y^e first) vnto the line AD a perpendiculer line AF. And (by the 10. of the first) deuide the line AB into two equall partes in the point F. And (by the 11. of the same) from the point F raise vp vnto the line AB a perpendicular lyne FG, and draw a line from G to B.* 1.143 And forasmuch as the line AF is equall to the line FB and the line FG is common to them both, therfore these two lines AF and FG are equall to these two lines FB and FG: and the angle AFG is (by the 4. peticion) equall to the angle GFB. Wherfore (by the 4. of the same) the base AG is equall to the base GB. Wherfore making the centre G and the space GA describe (by the 3. peticion) a circle and it shall passe by the point B: de∣scribe such a circle & let the same be ABE: And draw a line from E to B. Now forasmuch as from the ende of the diameter AE, namely, from the point A is 〈◊〉〈◊〉 a right line AD making together with the right line AE a right an∣gle, therfore (by the correllary of the 16. of the third) the line AD toucheth the circle ABE. And forasmuch as a certaine right line AD toucheth the circle ABE, & from the point A where the touch is, is drawen into y^e circle a certaine right line AB: therfore (by the 32. of the third) the angle DAB is equall to the angle AEB which is in the alternate segment of the circle. But the angle DAB is equall to the angle C, wherfore the angle C is equall to the angle AEB. Wher∣fore vpon the right line geuen AB is described a segment of a circle which con∣tayneth the angle AEB, which is equall to the angle geuen, namely, to C.

* 1.144But now suppose that the angle C be a right angle. It is againe required v∣pon

the right line AB to describe a seg∣ment

[illustration]

of a circle, which shall contayne an angle equal to the right angle C.* 1.145 Describe againe vpon the right line AB and to the point in it A an angle BAD equal to the rectiline angle geuen C (by the 23. of the first) as it is set forth in the second de∣scription. And (by the 10. of the first) de∣uide the line AB into two equall partes in the point F. And making the centre the point F and the space FA or FB describe (by the 3. peticion) y^e circle AEB.* 1.146 Wher∣fore the right line AD toucheth the cir∣cle AEB: for that the angle BAD is a right angle. Wherfore y^e angle BAD is equall to the angle which is in the segment AEB, for the angle which is in a semicircle is a right angle (by the 31. of the third) But the angle BAD is equal to the angle C. Wherfore there is againe described vpon the line AB a segment of a circle, namely, AEB, which containeth an angle equall to the angle geuen namely, to C.

But now suppose that the angle C be an obtuse angle.* 1.147 Vpon the right line AB and to the point in it A describe (by the 23. of the first) an angle BAD equall to the angle C: as it is in the third description.* 1.148 And from the point A rayse vp vnto the line AD a perpendiculer line AE

[illustration]

(by the 11. of the first) And agayne by the 10. of the first) deuide the line AB into two equall partes in the point F. And from the point F. ra••se vp vnto the line AB a perpē∣dicular line FG (by the 11. of the same) & drawe a line from G to B.* 1.149 And now foras∣much as the line AF is equal to the line FB, and the line FG is common to them both, therfore these two lines AF and FG are e∣quall to these two lines BF and FG: and the angle AFG is (by the 4. peticion) equall to the angle BFG: wherfore (by the 4. of the same) the base AG is equall to the base GB. Wherfore making the centre G, and the space GA describe (by the 3. peticion) a circle and it shall passe by the point B: let it be described as the circle AEB is. And forasmuch as from the ende of the diameter AE is drawen a perpendiculer line AD, therefore (by the correllary of the 16. of the third) the line AD toucheth the circle AEB & from the point of the touche, namely; A, is extended the line AB. Wherfore (by the 32. of the third) the angle BAD is equall to the angle AHB which is in the alternate segment of the circle. But the angle BAD is equall to the angle C.

Wherefore the angle which is in the segment AHB is equall to the angle C. Wherfore vpon the right line geuen AB, is described a segment of a circle AHB, which contayneth an angle equall to the angle geuen, namely, C: which was required to be done.

The 6. Probleme. The 34. Proposition. From a circle geuen to cut away a section which shal containe an angle equall to a rectiline angle geuen.

SVppose that the circle geuen be AC and let the rectiline angle geuen be D. It is required frō the circle ABC to cut away a segment which shall contayne an angle equall to the angle D.* 1.150 Draw (by the 17. of the third) a line touching the circle, and let the same be EF: and let it touche in the point B. And (by the 23. of

[illustration]

the first) vpon the right line EF and to the point in it B describe the angle FBC equall to the angle D.* 1.151 Now forasmuch as a certayne right line EF toucheth the cir∣cle ABC in the point B: and frora y^e point of the touche, namely, B, is drawn into the circle a certaine right line BC, there••ore (by the 32. of the third) the angle FBC is equall to the angle BAC which is in the alternate segment. But the angle FBC is equall to the angle D. Wherfore the angle BAC which consisteth in the segment BAC is equall to the angle D. Where∣fore from the circle geuen ABC is cut away a segment BAC, which containeth an angle equall to the rectiline angle geuen: which was required to be done.

The 29. Theoreme. The 35. Proposition. If in a circle two right lines do cut the one the other, the rect∣angle parallelograme comprehended vnder the segmentes or parts of the one line is equall to the rectangle parallelograme comprehended vnder the segment or partes of the other line.

LEt the circle be ABCD, and in it let these two right lines AC and BD c••t the one the other in the point E. Then I say that the rectangle parallelogramme contayned vnder the partes AE and EC is equall to

the rectangle parallelogramme contained vnder the

[illustration]

partes DE and EB.* 1.152 For if the line AC and BD be drawen by the centre, then is it manifest, that for as much as the lines AE and EC are equall to the lines DE and EB by the definition of a circle,* 1.153 the rectangle parallelograme also contayned vnder the lines AE and EC is equall to y^e rectangle paralle∣lograme contained vnder the lines DE and EB.

But now suppose that the lines AC and DB be not extended by the centre,* 1.154 and take (by the 1. of the third) the centre of the cir∣cle ABCD, and let the same be the point F,* 1.155 and from the point F draw to the right lines AC and DB perpendicular lines FG and FH (by the 12. of the first) and draw these right lines FB, FC, and FE.

[illustration]

And forasmuch as a certaine right line FG drawen by the centre,* 1.156 cutteth a certaine right line AC not drawen by the centre in such sorte that it maketh right angles, it therfore deuideth the line AC into two equall partes (by the 3. of the third). Wherfore the line AG is equall to the line GC. and foras∣much as the right line AC is deuided into two e∣quall partes in the point G, and into two vnequall partes in the point E: therfore (by the 5. of the second) the rectangle parallelo∣gramme contained vnder the lines. AE and EC together with the square of the line EG is equall to the square of the line GC. Put the square of the line GF common to them both, wherfore that which is contained vnder the lines AE & EC together with the squares of the lines EG and GF is equall to the squares of the lines GF & GC. But vnto y^e squares of y^e lines EG & GF is equall y^e square of y^e line FE (by the 47. of the first): and to the squares of the lines GC and GF is equall the square of the line FC (by the same) Wherfore that which is contai∣ned vnder the lines AE and EC, together with the square of the line FE is e∣quall to the square of the line FC. But the line FC is equall to the line FB. For they are drawen from the centre to the circumference. Wherfore that which is contained vnder the lines AE and E•• together with the square of the lyne FE is equal to the square of the line FB. And by the same demonstration that which is contained vnder the lines DE and EB together with the square of the line FE is equall to the square of the line FB. Wherfore that which is contained vn∣der the lines AE and EC together with the square of the line EF is equall to that which is contayned vnder the lines DE and EB together with the square of the line EF. Take away the square of the line FE which is common to them both. Wherfore the rectangle parallelogramme remayning which is contayned vnder the lines AE and EC is equall to the rectangle parallelogramme remay∣ning, which is contayned vnder the lines DE and EB. If therefore in a circle two right lines do cut the one the other: the rectangle parallelogramme compre∣hended

vnder the segmentes or parts of the one line is equall to the rectangle pa∣rallelograme comprehended vnder the segmentes or parts of the other line: which was required to be demonstrated.

In thys Proposition are three cases:* 1.157 For eyther both the lines passe by the cen∣tre, or ••eyther of them passeth by the centre: or the one passeth by the centre and the other not. The two first cases are before demonstrated.

But now let one of the lines onely, namely, the line AC passe by the centre, which let be the poynt F,* 1.158 and let it cut the other line, namely, BD, in the poynt E. Now then the line AC deuideth the line BD eyther into two equall partes, or into two vn∣equall partes. Fyrst let it deuide it into two equall partes: Wherefore also it deuideth it ••ight angled wyse by the 3. of thys booke. Drawe a right line from B to F. Where∣fore BEF is a right angled triangle. And for asmuch as the right line AC is deuided into two equall partes in the poynt F, & into two vnequall partes in the poynt E. Ther∣fore the rectangle figure contayned vnder the

[illustration]

lines AE and EC together with the square of the line EF, is equall to the square of the line FC (by the 5. of the second). But vnto the square of the line FC is equall the square of the line BF (for that the lines FB and FC are equall). Ther∣fore that which is cōtayned vnder the lines AE and EC together with the square of the line EF, is ••••uall to the square of the line BF. But vnto the square of the line BF, are equall the squares of the lines BE and EF (by the 47. of the first). Wherefore that which is contayned vnder the lines AE and EC together with the square of the line EF, is equall to the squares of the lines BE and EF. Take away the square of the line EF which is common to them both: Wherefore that which remayneth, namely, that which is contayned vnder the lines AE and EC, is equall to the residue, namely, to the square of the line BE. But the square o•• the line BE is that which is contained vn∣der the lines BE and ED: for (by supposition) the line BE is equall to the line ED. Wher••fore that which is contayned vnder the lines AE & EC, is equall to that which is contayned vnder the lines BE and ED: which was required to be proued.

But now let the line AC passing by the centre,

[illustration]

deuide, the line BD not passing by the centre, vn∣equally in the poynt E. And frō the poynt E raise vp vnto the line AC a perpendicular line EH, which produce on the other side to the poynt G. Where••ore (by the 3. of this booke) the line EH is equall to the line EG. Wherfore as we haue be∣fore proued, that which is contayned vnder the lines AE and EC, is equall to that which is con∣tayned vnder the lines GE & EH: but that which is contayned vnder the lines BE and ED, is also equall to that which is contayned vnder the lines GE and EH, by the second case of thys Propositi∣on: Wherfore that which is contayned vnder the lines AE and EC, is equall to that which is con∣tayned vnder the lines BE and ED: which was agayne required to be proued.

Amongest all the Propositions in this third booke, doubtles thys is one of the chiefest. For it setteth forth vnto vs the wonderfull nature of a circle. So that by

it may be done many goodly conclusions in Geometry, as shall afterward be de∣clared when occasion shall serue.

The 30. Theoreme. The 36. Proposition. If without a circle be taken a certaine point, and from that point be drawen to the circle two right lines, so that the one of them do cut the circle, and the other do touch the circle: the rectangle parallelogramme which is comprehended vnder the whole right line which cutteth the circle, and that portion of the same line that lieth betwene the point and the vtter cir∣cūference of the circle, is equall to the square made of the line that toucheth the circle.

SVppose that the circle be ABC: and without the same circle take a∣ny point at all aduentures, and let the same be D.* 1.159 And from the point D let there be drawen to the circle two right lines DCA and DB, and let the right line DCA cut the circle ACB in the point C, and let the right line BD touch the same. Then I say, that the rectangle parallelo∣gramme contayned vnder the lines AD and DC,* 1.160 is equall to the square of the line BD. Now the line DCA is either drawen by the centre, or not.

First let it be drawen by the centre.* 1.161 And (by

[illustration]

the first of the third) let the poynt F be y^e centre of the circle ABC, and drawe a line from F to B. Wherefore the angle FBD is a right angle.* 1.162 And for asmuch as y^e right line AC is deuided into two equall partes in the poynt F, and vnto it is added directly a right line CD, therfore (by the 6. of the second) that which is contayned vnder the lines AD and DC together with the square of y^e line CF, is equall to the square of the line FD. But the line FC is equall to the line FB, for they are drawen from the centre to y^e circumference: Wher∣fore that which is contayned vnder the lines AD and DC together with the square of the line FB, is equall to the square of the line FD. But y^e square of the line FD, is (by the 47. of the first) equall to the squares of the lines FB and BD (for the angle FBD is a right angle). Wherefore that which is contayned vnder the lines AD and DC together with the square of the line FB, is equall to the

squares of the lines FB and BD. Take away the square of the line FB which is common to them both. Wherefore that which remayneth, namely, that which is contayned vnder the lines AD and DC, is equall to the square made of the line DB which toucheth the circle.

But now suppose that the right line DCA be

[illustration]

not drawen by the centre of the circle ABC.* 1.163 And (by the first of the third) let the point E be y^e cen∣tre of the circle ABC.* 1.164 And from y^e poynt E, draw (by the 12. of the first) vnto the line AC a per∣pendicular line EF, and draw these right lines EB, EC, and ED.* 1.165 Now the angle EFD is a right angle. And ••or asmuch as a certaine right line EF drawen by the centre, cutteth a certayne other right line AC not drawen by the centre, in such sort that it maketh right angles, it deuideth it (by y^e third of the third) into two equall partes. Wherefore the line AF is equall to the line FC. And for asmuch as the right line AC is deuided into two equall partes in the poynt F, & vnto it is added directly an other right line making both one right line, therefore (by the 6. of the second) that which is contayned vnder the lines DA and DC together with the square of the line FC, is equall to the square of the line FD: put the square of the line FE common to them both. Wherefore that which is contayned vnder the lines DA and DC together with the squares of the lines CF and FE, is equall to the squares of the lines FD and FE. But to the squares of the lines FD and FE, is equall the square of the line DE (by the 47. of the first) for the angle EFD is a right angle. And to the squares of the lines CF and FE, is equall the square of the line CE (by the same). Wherfore that which is contayned vn∣der the lines AD and DC together with the square of the line EC, is equall to the square of the line ED. But the line EC is equall to the line EB: for they are drawen from the centre to the circumference. Wherefore that which is con∣tayned vnder the lines AD and DC together with the square of the line EB, is equall to the square of the line ED. But to the square of the line ED, are e∣quall the squares of the lines EB and BD (by the 47. of the first) for the an∣gle EBD is a right angle: Wherefore that which is contayned vnder the lines AD and DC together with the square of the line EB, is equall to the squares of the lines EB and BD. Take away the square of the line EB which is com∣mon to them both: Wherefore the residue, namely, that which is contayned vn∣der the lines AD and DC, is equall to the square of the line DB. If therfore without a circle be taken a certaine point, and from that poynt be drawen to the circle two right lines, so that the one of them do cut the circle, and the other do

••ouch the circle: the rectangle parallelogramme which is comprehended vnder the whole right line which cutteth the circle and that portion of the same line that lieth betwene the poynt and the vtter circumference of the circle, is equall to the square made of the line that toucheth the circle: which was required to be demonstrated.

¶Two Corollaries out of Campane.

If from 〈◊〉〈◊〉 and the selfe same poynt taken without a circle be drawen into the circle lines how many soeuer:* 1.166 the rectangle Parallelogrammes contayned vnder euery one of them and hys outward par••, are equall the one to the other.

And thys is hereby manifest, for that euery one of those rectangle Parallelo∣grammes are equall to the square of the line which is drawen from that poynt and toucheth the circle by thys 36. Proposition. Hereunto he addeth.

If two lines drawen from one and the selfe same point do touch a circle,* 1.167 they are equall the one to the other.

Which although it neede no demonstration, for that the square of eyther of them is equall to that which is contayned vnder the line drawen from the same poynt and hys outward part: yet he thus proueth it.

Suppose that there be a circle BCD, whose

[illustration]

centre let be E, and without it take the point A. And from the poynt A drawe two lines AB and AD, which let touch the circle in the poyntes B and D. Then I say, that they are equall. Draw these right lines EB, ED, and AE. And by the 18. of thys booke, eyther of the angles at the poyntes B and D is a right angle. Wherefore (by the 47. of the first) the square of the line AE, is equall to the two squares of the lines AB ••nd EB: and by the same reason, to the two squares of the lines AD and ED. Wherefore the two squares of the lines AB and EB, are equall to the two squares of the lines AD and ED. And for asmuch as the squares of the lines EB and ED are equall, therefore the two other squares of the lines AB and AD are also equall. Wher∣fore the line AB is equall to the line AD•• which was required to be proued.

The same may be proued an other way: Draw a line from B to D. And (by the 5. of the first) the angle EBD is equall to the angle EDB. And for asmuch as the two an∣gles ABE and ADE are equall, namely, for that they are right angles: if you take from them the equall angles EBD & EDB, the two other angles remayning, namely, the angles ABD and ADB shall be equall. Wherefore (by the 6. of the first) the line AB is equall to the line AD.

¶ Hereunto also Pelitarius addeth this Corollary.

From a poynt geuen without •• circle, can be drawen vnto a circle onely two touch lines.* 1.168

The former description remayning, I say that from the poynt A can be drawen vnto

the circle BCD no more touch lines, but the two lines AB and AD. For if it be pos∣sible, let AF also be in the former figure a touch line, touching the circle in the poynt F. And prawe a line from E to F. And the angle at the point F shall be a right angle, by the 18. of this booke: Wherefore it is equall to the angle EBA, which is contrary to the 20. of the first.

This may also be thus proued. For asmuch as all the lines drawen from one and the selfe same poynt & touching a circle are equall, as we haue before proued, but the lines AB and AF can not be equall, by the 8. Proposition of this booke, therefore the line AF can not touch the circle BCD.

The 31. Theoreme. The 37. Proposition. If without a circle be taken a certaine point, and from that point be drawen to the circle two right lines, of which, the one doth cut the circle and the other falleth vpon the circle, and that in such sort, that the rectangle parallelogramme which is cōtayned vnder the whole right line which cutteth the circle, and that portion of the same line that lieth betwene the point and the vtter circumferēce of the circle, is equall to the square made of the line that falleth vpon the circle: then that line that so falleth vpon the circle shall touch the circle.

LEt the circle be ABC: and with∣out

[illustration]

the same circle take a point, and let the same be D,* 1.169 & from the point D let there be drawen to the circle ABC two right lines DCA and DB: and let DCA cut the circle, and DB fall vpon the circle. And that in such sort, that that which is contayned vnder the lines AD and DC, be equall to the square of the line DB. Then I say, that y^e line DB toucheth the circle ABC. Drawe (by the 17. of the third) from the poynt D a right line touching the circle ABC, and let the same be DE.* 1.170 And (by the first of the same) let the point F be the centre of the circle ABC: and draw these right lines FE, FB, and FD. Wherfore the angle FED is a right angle.* 1.171 And for asmuch as the right line DE toucheth the circle ABC, and the right line DCA cutteth the same, therfore (by the Proposition going before) that which is contayned vnder the lines AD and DC, is equall to the square of the line DE. But that which is contayned

vnder the lines AD and DC, is supposed to be equall to the square of the line DB. Wherefore the square of the line DE is equall to the square of the line DB. Wherefore also the line DE is equall to the line DB. And the line FE is equall to the line FB, for they are drawen from the centre to y^e circumference. Now therefore these two lines DE and EF are equall to these two lines DB and BF, and FD is a common base to them both. Wherefore (by the 8. of the first) the angle DEF is equall to the angle DBF. But the angle DEF is a right angle. Wherefore also the angle DBF is a right angle. And y^e line FB being produced, shall be the diameter of the circle. But if from the end of the di∣ameter of a circle be drawen a right line making right angles, the right line so drawen toucheth the circle (by the Correllary of the 16. of the third). Wherfore the right line DB toucheth the circle ABC. And the like demonstration will serue if the centre be in the line AC. If therefore without a circle be taken a cer∣taine point, and from that poynt be drawen to the circle two right lines, of which the one doth cut the circle, and the other falleth vpon the circle, and that in such sort, that the rectangle parallelogramme which is contayned vnder the whole right line which cutteth the circle, and that portion of the same line that lieth be∣twene the poynt and the vtter circumference of the circle, is equall to the square made of the line that falleth vpon the circle: then the line that so falleth vpon the circle shall touch the circle: which was required to be proued.

¶ An other demonstration after Pelitarius.

Suppose that there be a circle BCD, whose

[illustration]

centre let be E:* 1.172 and take a point without it, name∣ly, A: And frō the poynt A drawe two right lines ABD, and AC: of which let ABD cut the circle in the poynt B, & let the other fall vpon it. And let that which is contained vnder the lines AD and AB, be equall to the square of the line AC. Then I say, that the line AC toucheth the circle. For first if the line ABD do passe by the centre, draw the right line CE. And (by the 6. of the second) that which is contayned vnder the lines AD and AB together with the square of the line EB, that is, with the square of the line EC (for the lines EB and EC are equall) is equall to the square of the line AE. But that which is contained vnder the lines AD and AB, is supposed to be equall to the square of the line AC: Wherefore the square of the line AC together with the square of the line CE, is equall to the square of the line AE. Wherefore (by the last of the first) the angle at the point C is a right angle. Wherfore (by the 18. of this boke) the line AC toucheth the circle.

But if the line ABD doo not passe by the centre, drawe from the point A the line AD, in which let be the centre E. And forasmuch as that which is contained vnder this whole line and his outward part, is equall to that which is contained vnder the lines AD and AB by the first Corollary before put, therefore the same is equall to the square of the line AC, wherefore the angle ECA is a right angle as hath before bene proued in the first part of this Proposition. And therfore the line AC toucheth the circle: Which was required to be proued.

The ende of the third booke of Euclides Elementes.

Notes

• * 1.1

  The argument of this booke.
• * 1.2

  The first defi∣nition.
• * 1.3

  Why circles take their equality of their diame∣ters or semi∣diameters.
• * 1.4

  Definition of vnequall cir∣cles.
• * 1.5

  Second defi∣nition.
• * 1.6

  A contigent line.
• * 1.7

  Third defini∣••ion.
• * 1.8

  The touch of circles is 〈◊〉〈◊〉 in one po•••••• onely.
• * 1.9

  Circles may touch toge∣••her two ma∣•••••• of wayes.
• * 1.10

  Fourth defi∣nition.
• * 1.11

  Fift defini∣tion.
• * 1.12

  Sixt defini∣tion.
• * 1.13

  Mixt angles.
• * 1.14

  Arkes.
• * 1.15

  Chordes.
• * 1.16

  Seuenth de∣finition.
• * 1.17

  Difference of an angle of a Section, and of an angle in a Section.
• * 1.18

  Eight defi∣nition.
• * 1.19

  Ninth defi∣nition.
• * 1.20

  Tenth defini∣tion.
• * 1.21

  Two defini∣tions.
• * 1.22

  First.
• * 1.23

  Second.
• * 1.24

  Why Euclide defineth not equall Secti∣ons.
• * 1.25

  Constuction.
• * 1.26

  Demonstra∣tion leading to an impos∣sibilitie.
• * 1.27

  Correlary.
• * 1.28

  Demonstra∣tiō leading to an impossibi∣litie.
• * 1.29

  The first para of this Propo∣sition.
• * 1.30

  Construction.
• * 1.31

  Demonstra∣tion.
• * 1.32

  The second part conuerst of the first.
• * 1.33

  Demonstra∣tion.
• * 1.34

  Demonstra∣tion leading to an impos∣sibilitie.
• * 1.35

  Two cases in this Propo∣sition.
• * 1.36

  Construction for the second case.
• * 1.37

  Demonstra∣tion.
• * 1.38

  Construction.
• * 1.39

  Demonstra∣tiō leading to an impossibi∣litie.
• * 1.40

  Demonstra∣t••on leading to an impos∣sibilitie.
• * 1.41

  Two case•• in thys Pro∣position.
• * 1.42

  Construction.
• * 1.43

  The first part of this Propo∣sition.
• * 1.44

  Demonstra∣tion.
• * 1.45

  Second part.
• * 1.46

  Third part.
• * 1.47

  This demon∣••••rated by an argument lea∣ding to an im∣possibilie.
• * 1.48

  An other de∣monstration of the latter part of the Proposition leading also to an impossi∣bilitie.
• * 1.49

  A Corollary.
• * 1.50

  Construction.
• * 1.51

  The first part of this Propo∣sition.
• * 1.52

  Second part.
• * 1.53

  Third part.
• * 1.54

  This is demō∣strated by an argument lea∣ding to an ab∣surdity.
• * 1.55

  An other de∣monstration of the latter part, leading also to an im∣possibility.
• * 1.56

  This Proposi∣on is commōly called Ca••d•• Panonis.
• * 1.57

  A Corollary.
• * 1.58

  Construction.
• * 1.59

  Demonstra∣tion.
• * 1.60

  An other de∣monstration of the same leading also to an impossi∣bilitie.
• * 1.61

  Demonstra∣tion leading to an impos∣sibilitie.
• * 1.62

  An other de∣monstration of the same leading also to an impossi∣bilitie.
• * 1.63

  Construction.
• * 1.64

  Demonstra∣tion leading to an impos∣sibilitie.
• * 1.65

  An other de∣monstration of the same leading also to an impossi∣bilitie.
• * 1.66

  The same ••∣gaine demon∣strated by an ••rgument lea∣ding to an ab∣surdititie.
• * 1.67

  Demonstra∣ti•• leading to an impossibi∣li••ie.
• * 1.68

  An other de∣monstration after Pelita∣rius leading also to an ab∣surditie.
• * 1.69

  Of circles which touch the one the other inward∣ly.
• * 1.70

  Of circles which touch the one the other out∣wardly.
• * 1.71

  An other de∣monstration after Pelita∣rius & Flus∣sates, of circles which tooch the one the other out∣wardly.
• * 1.72

  Of circles which tooch the one the other in∣wardly.
• * 1.73

  The first part of this Theo∣reme.
• * 1.74

  Construction.
• * 1.75

  Demonstra∣tion.
• * 1.76

  Demonstra∣tion.
• * 1.77

  The second part which is the conuerse of the first.
• * 1.78

  An other de∣monstration of the first part after Campane.
• * 1.79

  Construction.
• * 1.80

  Demonstra∣tion.
• * 1.81

  An other de∣monstration after Cam∣pane.
• * 1.82

  The first part of this Theo∣reme.
• * 1.83

  Demonstra∣tion leading to an absurdi∣tie.
• * 1.84

  Second part.
• * 1.85

  Third part.
• * 1.86

  Construction.
• * 1.87

  Demonstra∣tion.
• * 1.88

  An addition of Pelitarius.
• * 1.89

  This Pro∣bleme commo∣dious for the inscribing and circumscri∣bing of figures in or abou•• circles.
• * 1.90

  Demonstra∣tion leading to an impossi∣bilitie.
• * 1.91

  An other de∣••onstration after Oron∣tius.
• * 1.92

  Demonstra∣tion leading to an impos∣sibilitie.
• * 1.93

  Two cases in thys Pro∣position the one when the angle set at the circumfe∣rence inclu∣deth the cen∣ter.
• * 1.94

  Demonstra∣tion.
• * 1.95

  The other whē the same angle set at the circumfe∣rence inclu∣deth not the center.
• * 1.96

  Construction.
• * 1.97

  Demonstra∣tion.
• * 1.98

  Three cases in this Proposi∣tion.

  The first case.

  The second case.
• * 1.99

  The third case.
• * 1.100

  Construction.
• * 1.101

  Demonstra∣tion.
• * 1.102

  Demons••ra∣tion leading to an impos∣sibilitie.
• * 1.103

  An ad••ition of Campane d••mo••strated by Pelitari••s.
• * 1.104

  Demonstra¦tion leading to an impossi∣bilitie.
• * 1.105

  Demonstra∣tion leading to an impos∣sibilitie.
• * 1.106

  An other de∣monstration.
• * 1.107

  An other de∣mo••stratiō af∣ter Pelitarius.
• * 1.108

  Construction.
• * 1.109

  Three cases in this Proposi∣tion.

  The first case.
• * 1.110

  Demonstra∣tio••.
• * 1.111

  The second case.
• * 1.112

  The third case.
• * 1.113

  An other more ready way.
• * 1.114

  Demonstra∣tion.
• * 1.115

  An addition.
• * 1.116

  An other con∣struction and demonstration of this Propo∣sition after, Campane
• * 1.117

  A ready way to finde out the center of a circle commō∣ly vsed a∣mongst a••ti∣ficers.
• * 1.118

  Construction.
• * 1.119

  Demonstra∣tion.
• * 1.120

  Demonstra∣tion leading to an impos∣sibilitie.
• * 1.121

  Construction.
• * 1.122

  Demonstra∣tion.
• * 1.123

  The conuerse of the former Proposition.
• * 1.124

  Construction.
• * 1.125

  Demonstra∣tion.
• * 1.126

  Construction.
• * 1.127

  Demons••ra∣tio••.
• * 1.128

  The first ••art of this Theo∣reme.
• * 1.129

  Second part.
• * 1.130

  Thir•• part.
• * 1.131

  The fourth part.
• * 1.132

  The fift and last part.
• * 1.133

  An other De∣monstration to proue that the ang••e in a semicircle is a right angle.
• * 1.134

  A Corollary.
• * 1.135

  An addition of P••litarius.
• * 1.136

  Demonstra∣tion lea••ing to an absurdi∣t••••.
• * 1.137

  An addition of Campane.
• * 1.138

  Construction.
• * 1.139

  Demonstra∣tion.
• * 1.140

  Two cases in this Proposi∣tion.
• * 1.141

  Three cases in this Proposi∣tion.

  The first case.
• * 1.142

  Construction.
• * 1.143

  Demonstra∣tio••.
• * 1.144

  The second case.
• * 1.145

  Construction.
• * 1.146

  Demonstra∣tion.
• * 1.147

  The third case.
• * 1.148

  Construction.
• * 1.149

  Demonstra∣tion.
• * 1.150

  Cons••••uction.
• * 1.151

  Demonstra∣tion.
• * 1.152

  Two cases in this Proposi∣tion.

  First case.
• * 1.153

  Demonstra∣tion.
• * 1.154

  The second c••se.
• * 1.155

  Construction.
• * 1.156

  Demonstra∣tion.
• * 1.157

  Three cases in this ••roposi∣tion.
• * 1.158

  The third case.
• * 1.159

  Construction.
• * 1.160

  Two cases in this Proposi∣tion.
• * 1.161

  The first case.
• * 1.162

  Demonstra∣tion.
• * 1.163

  The second case.
• * 1.164

  Construction.
• * 1.165

  Demonstra∣tion.
• * 1.166

  First Corolla∣ry.
• * 1.167

  Second Co∣rollary.
• * 1.168

  Third Corol∣lary.
• * 1.169

  This proposi∣tion is the cō∣uerse of the former.
• * 1.170

  Construction.
• * 1.171

  Demonstra∣tion.
• * 1.172

  An other de∣monstration after Pelita∣rius.