circle So that if any angle do not touch the circumference or if the lines that in close that angle doo not ende in the extremities of the corde line but ende other in some other part of the faid corde or in the circumference or that any one of them do so eande then is not that angle accompted to be drawen in the faid cantle of the circle And this promised nowe will I cumme to the meaninge of the theoreme I sette forthe a circle whiche is A. B. C. D and his centre E in this circle I drawe a line D. C whereby there ar made two cantels a more and a lesser The lesser is D. F. C and the geater is D.A.C.C. In this greater cantle I drawe two angles the firste is D. A.C and the second is D. B.C which two angles by reason they are made bothe in one cantle of a circle that is the cantle D.A.B. C therefore are they both equall 〈…〉 Now doth there appere an other triangle whose angle lighteth on the centre of the circle and that triangle is D. E. C whose angle is double to the other angles as is declared in the lxiiij Theoreme whiche maie stande well enough with this Theoreme for it is not made in this cantle of the circle as the other are by reason that his angle doth not light in the circumference of the circle but on the centre of it The .lxvi. theoreme Euerie figure of foure sides drawen in a circle hath his two contrarie angles equall vnto two right angles Example The circle is A. B. C. D and the figure of foure sides in it is made of the sides B. C and C. D and D. A and A.B. Now if you take any two angles that be contrary as the angle by A and the angle by C I saie that those .ij. be equall to .ij. right angles Also if you take the angle by B and the angle by D whiche two are also contrary those two angles are like waies equall to two right angles But if any man will take the angle by A with the angle by B or D they can not be accompted contrary no more is not the angle by C. estemed contrary to the angle by B or yet to the angle by D for they onely be accompted contrary angles whiche haue no one line common to them bothe Suche is the angle by A in respect of the angle by C for there both lynes be distinct where as the angle by A and the angle by D haue one common line A. D and therfore can not be accompted contrary angles So the angle by D and the angle by C haue D. C as a common line and therfore be not contrary angles And this maie you iudge of the residewe by like reason The lxvij Theoreme Vpon one right lyne there can not be made two cantles of circles like and vnequall and drawent towarde one parte Example Cantles of circles be then called like when the angles that are made in them be equall But now for the Theoreme let the right line be A. E.C on whiche I draw a cantle of a circle whiche is A.B.C. Now saieth the Theoreme that it is not possible to draw an other cantle of a circle whiche shall be vnequall vnto this first cantle that is to say other greatter or lesser then it and yet be lyke it also that is to say that the angle in the one shall be equall to the angle in the other For as in this example you see a lesser cantle drawen also that is A. D.C so if an angle were made in it that angle would be greatter then the angle made in the cantle A. B. C and therfore ban not they be called lyke cantess but and if any other cantle were made greatter then the first then would the angle it it be lesser then that in the firste and so nother a lesser nother a greater cantle can be made vpon one line with an other but it will be vnlike to it also The .lxviij. Theoreme Lyke cantelles of circles made on equall righte lynes are equall together Example What is mentby like cantles you haue heard before and it is easie to vnderstand that suche figures are called equall that be of one bygnesse so that the one is nother greater nother lesser then the other And in this kinde of comparison they must so agree that if the one be layed on the other they shall exactly agree in all their boundes so that nother shall excede other Nowe for the example of the Theoreme I haue set forthe diuers varieties of cantles of circles amongest which the first and seconde are made vpō equall lines and ar also both equall and like The third couple ar ioyned in one and be nother equall nother like but expressyng an absurde deformitee whiche would folowe if this Theoreme wer not true And so in the fourth couple you maie see that because they are not equall cantles therfore can not they be like cantles for necessarily it goeth together that all cantles of circles made vpon equall right lines if they be like they must be equall also The lxix Theoreme In equall circles suche angles as be equall are made vpon equall arch lines of the circumference whether the angle light on the circumference or on the centre Example Firste I haue sette for an exaumple twoo equall circles that is A. B. C. D whose centre is K and the second circle E. F. G. H and his centre L and in eche of thē is there made two angles one on the circumference and the other on the centre of eche circle and they be all made on two equall arche lines that is B.C.D. the one and F.G.H. the other Now saieth the Theoreme that if the angle B. A. D be equall to the angle F. E. H then are they made in equall circles and on equall arch lines of their circumference Also if the angle B. K.D be equall to the angle F. L.H then be they made on the centres of equall circles and on equall arche lines so that you muste compare those angles together whiche are made both on the centres or both on the circumference and maie not conferre those angles wherof one is drawen on the circumference and the other on the centre For euermore the angle on the centre in suche sorte shall be double to the angle on the circumference as is declared in the three score and foure Theoreme The .lxx. Theoreme In equall circles those angles whiche bee made on equall arche lynes are euer equall together whether they be made on the centre or on the circumference Example This Theoreme doth but conuert the sentence of the last Theoreme before and therfore is to be vnderstande by the same examples for as that saith that equall angles occupie equall archesynes so this saith that equal arche lines causeth equal angles consideringe all other circumstances as was taughte in the laste theoreme before so that this theoreme dooeth affirming speake of the equalitie of those
in that one point F and those iij. angles be equal to the iij. angles of the triangle assigned whiche thinge doth plainely appeare in so muche as they bee equall to ij right angles as you may gesse by the sixt theoreme And the thre angles of euerye triangle are equall also to ij righte angles as the two and twenty theoreme dothe show so that bicause they be equall to one thirde thinge they must needes be equal togither as the cōmon sentence saith Thē do I draw a line frome G. to H and that line maketh a triangle F.G.H. whose angles be equall to the angles of the triangle appointed And this triangle is drawen in a circle as the conclusion didde wyll The proofe of this conclusion doth appeare in the seuenty and iiij Theoreme THE XXX CONCLVSION To make a triangle about a circle assigned whiche shall haue corners equall to the corners of any triangle appointed First draw forth in length the one side of the triangle assigned so that therby you may haue ij vtter angles vnto which two vtter angles you shall make ij other equall on the centre of the circle proposed drawing thre halfe diameters frome the circumference whiche shal enclose those ij angles thē draw iij. touche lines which shall make ij right angles eche of them with one of those semidiameters Those iij. lines will make a triangle equally cornered to the triangle assigned and that triangle is drawē about a circle apointed as the cōclusiō did wil. Example A. B.C is the triangle assigned and G. H.K is the circle appointed about which I muste make a triangle hauing equall angles to the angles of that triangle A.B.C. Fyrst therefore I draw A.C. which is one of the sides of the triangle in length that there may appeare two vtter angles in that triangle as you se B. A. D and B. C E. Then drawe I in the circle appointed a semidiameter whiche is here H. F for F. is the cētre of the circle G. H.K. Then make I on that centre an angle equall to the vtter angle B. A. D and that angle is H.F. K. Likewaies on the same cētre by drawyng an other semidiameter I make an other angle H. F. G equall to the second vtter angle of the triangle whiche is B. C. E. And thus haue I made .iij. semidiameters in the circle appointed Then at the ende of eche semidiameter I draw a touche line whiche shall make righte angles with the semidiameter And those .iij. touch lines mete as you see and make the triangle L. M. N whiche is the triangle that I should make for it is drawen about a circle assigned and hath corners equall to the corners of the triangle appointed for the corner M. is equall to C. Likewaies L. to A and N. to B whiche thyng you shall better perceiue by the vi Theoreme as I will declare in the booke of proofes THE XXXI CONCLVSION To make a portion of a circle on any right line assigned whiche shall conteine an angle equall to a right lined angle appointed The angle appointed maie be a sharpe angle a right angle other a blunte angle so that the worke must be diuersely handeled according to the diuersities of the angles but consideringe the hardenes of those seuerall woorkes I wyll omitte them for a more meter time and at this tyme wyll she we you one light waye which serueth for all kindes of angles and that is this When the line is proposed and the angle assigned you shall ioyne that line proposed so to the other twoo lines contayninge the angle assigned that you shall make a triangle of theym for the easy dooinge whereof you may enlarge or shorten as you see cause nye of the two lynes contayninge the angle appointed And when you haue made a triangle of those iij. lines then accordinge to the doctrine of the seuē and twety coclusiō make acircle about that triangle And so haue you wroughte the request of this conclusion whyche yet you maye woorke by the twenty and eight conclusion also so that of your line appointed you make one side of the triāgle be equal to the āgleassigned as youre selfe mai easily gesse Example First for example of a sharpe āgle let A. stād B.C. shal be that lyne assigned Thē do I make a triangle by adding B. C as a thirde side to those other ij which doo include the āgle assigned and that triāgle is D E. F so that E. F. is the line appointed and D. is the angle assigned Then doe I drawe a portion of a circle about that triangle from the one ende of that line assigned vnto the other that is to saie from E. a long by D. vnto F whiche portion is euermore greatter then the halfe of the circle by reason that the angle is a sharpe angle But if the angle be right as in the second exaumple you see it then shall the portion of the circle that containeth that angle euer more be the iuste halfe of a circle And when the angle is a blunte angle as the thirde exaumpse dooeth propounde then shall the portion of the circle euermore be lesse then the halfe circle So in the seconde example G. is the right angle assigned and H. K. is the lyne appointed and L.M.N. the portion of the circle aunsweryng thereto In the third exaumple O. is the blunte corner assigned P. Q. is the line and R. S. T. is the portion of the circle that containeth that blūt corner and is drawen on R. T. the lyne appointed THE XXXII CONCLVSION To cutte of from any circle appoineed a portion containyng an angle equall to a right lyned angle assigned When the angle and the circle are assigned first draw a touch line vnto that circle and then drawe an other line from the pricke of the touchyng to one side of the circle so that thereby those two lynes do make an angle equall to the angle assigned Then saie I that the portion of the circle of the contrarie side to the angle drawen is the parte that you scke for Example A. is the angle appointed and D. E. F. is the circle assigned frō which I must cut away a portiō that doth contain an angle equall to this angle A. Therfore first I do draw a touche line to the circle assigned and that touch line is B. C the very pricke of the touche is D from whiche D. J. drawe a lyne D. E so that the angle made of those two lines be equall to the angle appointed Then say I that the arch of the circle D. F. E is the arche that I seke after For if I doo deuide that arche in the middle as here it is done in F. and so draw thence two lines one to A and the other to E then will the angle F be equall to the angle assigned THE XXXIII CONCLVSION To make a square quadrate in a circle assigned Draw .ij. diameters in the circle so that they runne a crosse and that they make .iiij.
two squares made of B. C and C. D. For as the shorter side is the iuste lengthe of C. D so the other longer side is iust twise so longe as B. C Wherfore I saie now accordyng to the Theoreme that the greatte square E is more then the other two squares F. and G by the quantitee of the longe square K wherof I reserue the profe to a more conuenient place where I will also teache the reason howe to fynde the lengthe of all suche perpendicular lynes and also of the line that is drawen betweene the blunte angle and the perpendicular line with sundrie other very pleasant conclusions The .xlvi. Theoreme In sharpe cornered triangles the square of anie side that lieth against a sharpe corner is lesser then the two squares of the other two sides by as muche as is comprised twise in the long square of that side on whiche the perpendicular line falleth and the portion of that same line liyng betweene the perpendicular and the foresaid sharpe corner Example Fyrst I sette foorth the triangle A. B. C and in yt I draw a plūbe line from the angle C. vnto the line A. B and it lighteth in D. Nowe by the theoreme the square of B.C. is not so muche as the square of the other two sydes that of B. A. and of A.C. by as muche as is twise conteyned in the lōg square made of A. B and A. D A. B. beyng the line or syde on which the perpendicular line falleth and A.D. beeyng that portion of the same line whiche doth lye betwene the perpendicular line and the sayd sharpe angle limitted whiche angle is by A. For declaration of the figures the square marked with E. is the square of B. C whiche is the syde that lieth agaynst the sharpe angle the square marked with C. is the square of A. B and the square marked with F. is the square of A. C and the two longe squares marked with H. K are made of the hole line A. B and one of his portions A. D. And truthe it is that the square E. is lesser than the other two squares C. and F. by the quantitee of those two long squares H. and K. Wherby you may consyder agayn an other proportion of equalitee that is to saye that the square E. with the twoo long squares H. K are iuste equall to the other twoo squares C. and F. And so maye you make as it were an other theoreme That in al sharpe cornered triangles where a perpendicular line is drawen frome one angle to the side that lyeth againste it the square of anye one side with the ij longesquares made of that hole line whereon the perpendicular line doth lighte and of that portion of it which ioyneth to that side whose square is all ready taken those thre figures I say are equall to the ij squares of the other ij sides of the triangle In whiche you muste vnderstand that the side on which the perpendiculare falseth is thrise vsed yet is his square but one 's mencioned for twise he is taken for one side of the two long squares And as I haue thus made as it were an other theoreme out of this fourty and sixe theoreme so mighte I out of it and the other that goeth nexte before make as manny as woulde suffice for a whole booke so that when they shall bee applyed to practise and consequently to expresse their benefite no manne that hathe not well wayde their wonderfull commoditee woulde credite the posibilitie of their wonderfull vse and large ayde in knowledge But all this wyll I remitte to a place conuenient The xlvij Theoreme If ij points be marked in the circumferēce of a circle and a right line drawen frome the one to the other that line must needes fal with in the circle Example The circle is A. B.C.D the ij poinctes are A. B the righte line that is drawenne frome the one to the other is the line A. B which as you see must needes lyghte within the circle So if you putte the pointes to be A. D or D. C or A. C other B. C or B. D inany of these cases you see that the line that is drawen from the one pricke to the other dothe euermore run within the edge of the circle els canne it be no right line Howbeit that a croked line especially being more croked then the portion of the circumference maye bee drawen from pointe to pointe withoute the circle But the theoreme speaketh only of right lines and not of croked lines The xlviij Theoreme If a righte line passinge by the centre of a circle doo crosse an other right line within the same circle passinge beside the centre if be deuide the saide line into twoo equal partes then doo they make all their angles righte And contrarie waies if they make all their angles righte then doth the longer line cutte the shorter in twoo partes Example The circle is A. B. C. D the line that passeth by the centre is A. E. C the line that goeth beside the centre is D. B. Nowe saye I that the line A. E. C dothe cutte that other line D. B. into twoo iuste partes and therefore all their four angles ar righte angles And contrarye wayes bicause all their angles are righte angles therfore it muste be true that the greater cutteth the lesser into two equal partes acordinge as the Theoreme would The xlix Theoreme If twoo right lines drawen in a circle doo crosse one an other and doo not passe by the centre euery of them dothe not deuide the other into two equall partions Example The circle is A. B. C. D and the centre is E the one line A. C and the other is B. D which two lines crosse one an other but yet they go not by the centre wherefore accordinge to the woordes of the theoreme eche of theim doth cutte the other into equall portions For as you may easily iudge A C. hath one portiō lōger and an other shorter and so like wise B. D. Howbeit it is not so to be vnderstād but one of them may be diuided into ij euē parts but bothe to bee cutte equally in the middle is not possible onles both passe through the cētre therfore much rather whē bothe go beside the centre it can not be that eche of theym shoulde be iustely parted into ij euen partes The L. Theoreme If two circles crosse and cut one an other then haue not they both one centre Example This theoreme seemeth of it selfe so manifest that it neadeth nother demonstration nother declaraciō Yet for the plaine vnderstanding of it I haue sette forthe a figure here where ij circles be drawē so that one of them doth crosse the other as you see in the pointes B. and G and their centres appear at the firste sighte to bee diuers For the centre of the one is F and the centre of the other is E which diffre as farre a sondre as the edges of the circles where they
for it is of lyke distance as is the line M.N. Nowe saie I that A. D beyng the diameter is the longest of all those lynes and also of any other that maie be drawen within that circle And the other line M. N is longer then F. G because it is nerer to the centre of the circle then F. G. Also the line F. G is shorter then the line B. C. for because it is farther from the centre then is the lyne B. C. And thus maie you iudge of al lines drawen in any circle how to know the proportion of their length by the proportion of their distance and contrary waies howe to discerne the proportion of their distance by their lengthes if you knowe the proportion of their length And to speake of it by the waie it is a maruaylouse thyng to consider that a man maie knowe an exacte proportion betwene two thynges and yet can not name nor attayne the precise quantitee of those two thynges As for exaunple If two squares be sette foorthe whereof the one containeth in it fiue square seete and the other contayneth fiue and fortie foote of like square feete I am not able to tell no nor yet anye manne liuyng what is the precyse measure of the sides of any of those .ij. squares and yet I can proue by vnfallible reason that their sides be in a triple proportion that is to saie that the side of the greater square whiche containeth .xlv. foote is three tymes so long iuste as the side of the lesser square that includeth but fiue foote But this seemeth to be spoken out of ceason in this place therfore I will omitte it now reseruyng the exacter declaration therof to a more conuenient place and time and will procede with the residew of the Theoremes appointed for this boke The .lxi. Theoreme If a right line be drawen at any end of a diameter in perpendicular forme and do make a right angle with the diameter that right line shall light without the circle and yet so iointly knitte to it that it is not possible to draw any other right line betwene that saide line and the circumfereÌ„ce of the circle And the angle that is made in the semicircle is greater then any sharpe angle that may be made of right lines but the other angle without is lesser then any that can be made of right lines Example In this circle A. B.C the diameter is A. C the perpendicular line which maketh a right angle with the diameter is E. A whiche line falleth without the circle and yet ioyneth so exactly vnto it that it is not possible to draw an other right line betwene the circumference of the circle and it whiche thyng is so plainly seene of the eye that it needeth no farther declaracion For euery man wil easily consent that betwene the croked line A. F whiche is a parte of the circumfereÌ„ce of the circle and A. E which is the said perpeÌ„dicular line there can none other line bee drawen in that place where they make the angle Nowe for the residue of the theoreme The angle D. A. B which is made in the semicircle is greater then anye sharpe angle that maye bee made of ryghte lines and yet is it a sharpe angle also in as much as it is lesser then a right angle which is the angle E. A.D and the residue of that right angle which lieth without the circle that is to saye E. A.B is lesser then any sharpe angle that can be made of right lines also For as it was before rehersed there canne no right line be drawen to the angle betwene the circumference and the right line E.A. Then must it needes folow that there can be made no lesser angle of righte lines And againe if ther canne be no lesser then the one then doth it sone appear that there canne be no greatter then the other for they twoo doo make the whole right angle so that if anye corner coulde bee made greater then the one parte then shoulde the residue bee lesser then the other parte so that other bothe partes muste be false or els bothe graunted to be true The lxij Theoreme If a right line doo touche a circle and an other right line drawen frome the centre of tge circle to the point where they touch that line whiche is drawenne frome the centre shall be a perpendicular line to the touch line Example The circle is A. B. C and his centre is F. The touche line is D. E and the point wher they touch is C. Now by reason that a right line is drawen frome the centre F. vnto C which is the point of the touche therefore saith the theoreme that the sayde line F. C muste needes bee a perpendicular line vnto the touche line D.E. The lxiij Theoreme If a righte line doo touche a circle and an other right line be drawen from the pointe of their touchinge so that it doo make righte corners with the touche line then shal the centre of the circle bee in that same line so drawen Example The circle is A. B. C and the centre of it is G. The touche line is D. C.E and the pointe where it toucheth is C. Nowe it appeareth manifest that if a righte line be drawen from the pointe where the touch line doth ioine with the circle and that the said lyne doo make righte corners with the touche line then muste it needes go by the centre of the circle and then consequently it must haue the sayde ceÌ„tre in him For if the saide line shoulde go beside the centre as F. C. doth then dothe it not make righte angles with the touche line which in the â—heoreme is supposed The lxiiij Theoreme If an angle be made on the centre of a circle and an other angle made on the circumference of the same circle and their grounde line be one common portion of the circumference then is the angle on the centre twise so great as the other angle on the circuÌ„fereÌ„ce Example The cirle is A. B. C. D and his centre is E the angle on the centre is C. E.D and the angle on the circumference is C. A. D t their commen ground line is C. F.D Now say I that the angle C. E. D whiche is one the centre is twise so greate as the angle C. A.D which is on the circumference The lxv Theoreme Those angles whiche be made in one cantle of a circle must needes be equal togither Example Before I declare this theoreme by an example it shall bee needefull to declare what is to be vnderstande by the wordes in this theoreme For the sentence canne not be knowen onles the uery meaning of the wordes be firste vnderstand Therefore when it speaketh of angâ—es made in one cantle of a circle it is this to be vnderstand that the angle muste touch the circumference and the lines that doo inclose that angle muste be drawen to the extremities of that line which maketh the cantle of the