see textExample.Fyrst before I declare the examples, it shal be mete to shew the true vnderstãdyng of this theorem.Bias lyne.Therfore by theBias line, I meane that lyne, whiche in any square figure dooth runne from corner to corner. And euery square which is diuided by that bias line into equall halues from corner to corner (that is to say, into .ij. equall triangles) those be countedto stande aboute one bias line, and the other squares, whiche touche that bias line, with one of their corners onely, those doo I callFyll squares,Fyll squares.accordyng to the greke name, which isanapleromata,ἀναπληρώματαand called in latinsupplementa, bycause that they make one generall square, includyng and enclosyng the other diuers squares, as in this exãpleH.C.E.N.is one square likeiamme, andL.M.G.C.is an other, whiche bothe are made aboute one bias line, that isN.M,thanK.L.H.C.andC.E.F.G.are .ij. fyll squares, for they doo fyll vp the sydes of the .ij.fyrste square lykeiammes, in suche sorte, that all them foure is made one greate generall squareK.M.F.N.Nowe to the sentence of the theoreme, I say, that the .ij. fill squares,H.K.L.C.andC.E.F.G.are both equall togither, (as it shall bee declared in the booke of proofes) bicause they are the fill squares of two likeiammes made aboute one bias line, as the exaumple sheweth. Conferre the twelfthe conclusion with this theoreme.The xxxiij. Theoreme.In all right anguled triangles, the square of that side whiche lieth against the right angle, is equall to the .ij. squares of both the other sides.Example.see textA.B.C.is a triangle, hauing a ryght angle inB.Wherfore it foloweth, that the square ofA.C,(whiche is the side that lyeth agaynst the right angle) shall be as muche as the two squares ofA.B.andB.C.which are the other .ij. sides.¶By thesquareof any lyne, you muste vnderstande a figure made iuste square, hauyng all his iiij. sydes equall to that line, whereof it is the square, so isA.C.F,the square ofA.C.LykewaisA.B.D.is the square ofA.B.AndB.C.E.is the square ofB.C.Now by the numbre of the diuisions in eche of these squares, may you perceaue not onely what the square of any line is called, but also that the theoreme is true, and expressed playnly bothe by lines and numbre. For as you see, the greatter square (that isA.C.F.) hath fiue diuisions on eche syde, all equall togyther, and those in the whole square are twenty and fiue. Nowe in the left square, whiche isA.B.D.there are but .iij. of those diuisions in one syde, and that yeldeth nyne in the whole. So lykeways you see in the meane squareA.C.E.in euery syde .iiij. partes, whiche in the whole amount vnto sixtene. Nowe adde togyther all the partes of the two lesser squares, that is to saye, sixtene and nyne, and you perceyue that they make twenty and fiue, whyche is an equall numbre to the summe of the greatter square.By this theoreme you may vnderstand a redy way to know the syde of any ryght anguled triangle that is vnknowen, so that you knowe the lengthe of any two sydes of it. For by tournynge the two sydes certayne into theyr squares, and so addynge them togyther, other subtractynge the one from the other (accordyng as in the vse of these theoremes I haue sette foorthe) and then fyndynge the roote of the square that remayneth, which roote (I meane the syde of the square) is the iuste length of the vnknowen syde, whyche is sought for. But this appertaineth to the thyrde booke, and therefore I wyll speake no more of it at this tyme.The xxxiiij. Theoreme.If so be it, that in any triangle, the square of the one syde be equall to the .ij. squares of the other .ij. sides, than must nedes that corner be a right corner, which is conteined betwene those two lesser sydes.Example.As in the figure of the laste Theoreme, bicauseA.C,made in square, is asmuch as the square ofA.B,and also as the square ofB.C.ioyned bothe togyther, therefore the angle that is inclosed betwene those .ij. lesser lynes,A.B.andB.C.(that is to say) the angleB.whiche lieth against the lineA.C,must nedes be a ryght angle. This theoreme dothe so depende of the truthe of the laste, that whan you perceaue the truthe of the one, you can not iustly doubt of the others truthe, for they conteine one sentence, contrary waies pronounced.The .xxxv. theoreme.If there be set forth .ij. right lines, and one of them parted into sundry partes, how manyor few so euer they be, the square that is made of those ij. right lines proposed, is equal to all the squares, that are made of the vndiuided line, and euery parte of the diuided line.Example.see textThe ij. lines proposed arA.B.andC.D,and the lyneA.B.is deuided into thre partes byE.andF.Now saith this theoreme, that the square that is made of those two whole linesA.B.andC.D,so that the lineA.B.stãdeth for the lẽgth of the square, and the other lineC.D.for the bredth of the same. That square (I say) wil be equall to all the squares that be made, of the vndiueded lyne (which isC.D.) and euery portion of the diuided line. And to declare that particularly, Fyrst I make an other lineG.K,equall to the line.C.D,and the lineG.H.to be equal to the lineA.B,and to bee diuided into iij. like partes, so thatG.M.is equall toA.E,andM.N.equal toE.F,and then musteN.H.nedes remaine equall toF.B.Then of those ij. linesG.K,vndeuided, andG.H.which is deuided, I make a square, that isG.H.K.L,Inwhich square if I drawe crosse lines frome one side to the other, according to the diuisions of the lineG.H,then will it appear plaine, that the theoreme doth affirme. For the first squareG.M.O.K,must needes be equal to the square of the lineC.D,and the first portiõ of the diuided line, which isA.E,for bicause their sides are equall. And so the secondesquare that isM.N.P.O,shall be equall to the square ofC.D,and the second part ofA.B,that isE.F.Also the third square which isN.H.L.P,must of necessitee be equal to the square ofC.D,andF.B,bicause those lines be so coupeled that euery couple are equall in the seuerall figures. And so shal you not only in this example, but in all other finde it true, that if one line be deuided into sondry partes, and an other line whole and vndeuided, matched with him in a square, that square which is made of these two whole lines, is as muche iuste and equally, as all the seuerall squares, whiche bee made of the whole line vndiuided, and euery part seuerally of the diuided line.The xxxvi. Theoreme.If a right line be parted into ij. partes, as chaunce may happe, the square that is made of the whole line, is equall to bothe the squares that are made of thesameline, and the twoo partes of it seuerally.Example.see textThe line propounded beyngA.B.and deuided, as chaunce happeneth, inC.into ij. vnequall partes, I say that the square made of the hole lineA.B,is equal to the two squares made of the same line with the twoo partes of itselfe, as withA.C,and withC.B,for the squareD.E.F.G.is equal to the two other partial squares ofD.H.K.GandH.E.F.K,but that the greater square is equall to the square of the whole lineA.B,and thepartiall squares equall to the squares of the second partes of the same line ioyned with the whole line, your eye may iudg without muche declaracion, so that I shall not neede to make more exposition therof, but that you may examine it, as you did in the laste Theoreme.The xxxvij.Theoreme.If a right line be deuided by chaunce, as it maye happen, the squarethatis made of the whole line, and one of the partes of it which soeuer it be, shal be equall to that square that is made of the ij. partes ioyned togither, and to an other square made of that part, which was before ioyned with the whole line.Example.see textThe lineA.B.is deuided inC.into twoo partes, though not equally, of which two partes for an example I take the first, that isA.C,and of it I make one side of a square, as for exampleD.G.accomptinge those two lines to be equall, the other side of the square isD.E,whiche is equall to the whole lineA.B.Nowmay it appeare, to your eye, that the great square made of the whole lineA.B,and of one of his partes that isA.C,(which is equall withD.G.) is equal to two partiall squares, whereof the one is made of the saide greatter portionA.C,in as muche as not onlyD.G,beynge one of his sides, but alsoD.H.beinge the other side, are eche of them equall toA.C.The second square isH.E.F.K,in which the one sideH.E,is equal toC.B,being the lesser parte of the line,A.B,andE.F.is equall toA.C.which is the greater parte of the same line. So that those two squaresD.H.K.GandH.E.F.K,bee bothe of them no more then the greate squareD.E.F.G,accordinge to the wordes of the Theoreme afore saide.The xxxviij. Theoreme.If a righte line be deuided by chaunce, into partes, the square that is made of that whole line, is equall to both the squares that ar made of eche parte of the line, and moreouer to two squares made of the one portion of the diuided line ioyned with the other in square.see textExample.The labels A and B were transposed in the illustration as an alternative to transposing all occurrences of A.C and C.B in the text.Lette the diuided line beeA.B,and parted inC,into twoo partes: Nowe saithe the Theoreme, that the square of the whole lyneA.B,is as mouche iuste as the square ofA.C,and the square ofC.B,eche by it selfe, and more ouer by as muche twise, asA.C.andC.B.ioyned in one square will make. For as you se, the great squareD.E.F.G,conteyneth in hym foure lesser squares, of whiche the first and the greatest isN.M.F.K,and is equall to the square of the lyneA.C.The second square is the lest of them all, that isD.H.L.N,and it is equall to the square of the lineC.B.Then are there two other longe squares both of one bygnes, that isH.E.N.M.andL.N.G.K,eche of them both hauyng .ij. sides equall toA.C,the longer parte of the diuided line, and there other two sides equall toC.B,beeyng he shorter parte of the said lineA.B.So is that greatest square, beeyng made of the hole lyneA.B,equal to the ij. squares of eche of his partes seuerally, and more by as muche iust as .ij. longe squares, made of the longer portion of the diuided lyne ioyned in square with the shorter parte of the same diuided line, as the theoreme wold. And as here I haue put an example of a lyne diuided into .ij. partes, so the theoreme is true of all diuided lines, of what number so euer the partes be, foure, fyue, orsyxe. etc.Thistheoreme hath great vse, not only in geometrie, but also in arithmetike, as herafter I will declare in conuenient place.The .xxxix. theoreme.If a right line be deuided into two equall partes, and one of these .ij. partes diuided agayn into two other partes, as happeneth the longe square that is made of the thyrd or later part of that diuided line, with the residue of the same line, and the square of the mydlemoste parte, are bothe togither equall to the square of halfe the firste line.Example.see textThe lineA.B.is diuided into ij. equal partes inC,and that parteC.B.is diuided agayne as hapneth inD.Wherfore saith the Theorem that the long square made ofD.B.andA.D,with the square ofC.D.(which is the mydle portion) shall bothe be equall to the square of half the lyneA.B,that is to saye, to the square ofA.C,or els ofC.D,which make all one. The long squareF.G.N.O.whiche is the longe square that the theoreme speaketh of, is made of .ij. long squares, wherof the fyrst isF.G.M.K,and the seconde isK.N.O.M.The square of the myddle portion isL.M.O.P.and the square of the halfe of the fyrste lyne isE.K.Q.L.Nowe by the theoreme, that longe squareF.G.N.O,with the iuste squareL.M.O.P,muste bee equall to the greate squareE.K.Q.L,whyche thynge bycause it seemeth somewhat difficult to vnderstande, althoughe I intende not here to make demonstrations of the Theoremes, bycause it is appoynted to be done in the newe edition of Euclide, yet I wyll shew you brefely how the equalitee of the partes doth stande. And fyrst I say, that where the comparyson of equalitee is made betweene the greate square (whiche is made of halfe the lineA.B.) and two other, where of the fyrst is the longe squareF.G.N.O,and the second is the full squareL.M.O.P,which is one portion of the great square all redye, and so is that longe squareK.N.M.O,beynge a parcell also of the longe squareF.G.N.O,Wherforeas those two partes are common to bothe partes compared in equalitee, and therfore beynge bothe abated from eche parte, if the reste of bothe the other partes bee equall, than were those whole partes equall before: Nowe the reste of the great square, thosetwo lesser squares beyng taken away,is that longe squareE.N.P.Q,whyche is equall to the long squareF.G.K.M,beyng the rest of the other parte. And that they two be equall, theyr sydes doo declare. For the longest lynes that isF.KandE.Qare equall, and so are the shorter lynes,F.G,andE.N,and so appereth the truthe of the Theoreme.The .xl. theoreme.If a right line be diuided into .ij. euen partes, and an other right line annexed to one ende of that line, so that it make one righte line with the firste. The longe square that is made of this whole line so augmented, and the portion that is added, with the square of halfe the right line, shall be equall to the square of that line, whiche is compounded of halfe the firste line, and the parte newly added.Example.see textThe fyrst lynepropoundedisA.B,and it is diuided into ij. equall partes inC,and an other ryght lyne, I meaneB.Dannexed to one ende of the fyrste lyne.Nowesay I, that the long squareA.D.M.K,is made of the whole lyne so augmẽted, that isA.D,and the portiõ annexed, ytisD.M,forD.Mis equall toB.D,wherfore ytlong squareA.D.M.K,with thesquare of halfe the first line, that isE.G.H.L,is equall to the great squareE.F.D.C.whiche square is made of the lineC.D.that is to saie, of a line compounded of halfe the first line, beyngC.B,and the portion annexed, that isB.D.And it is easyly perceaued, if you consyder that the longe squareA.C.L.K.(whiche onely is lefte out of the great square) hathanotherlonge square equall to hym, and to supply his steede in the great square, and that isG.F.M.H.For their sydes be of lyke lines in length.The xli. Theoreme.If a right line bee diuided by chaunce, the square of the same whole line, and the square of one of his partes are iuste equall to the lõg square of the whole line, and the sayde parte twise taken, and more ouer to the square of the other parte of the sayd line.Example.see textA.B.is the line diuided inC.AndD.E.F.G,is the square of the whole line,D.H.K.M.is the square of the lesser portion (whyche I take for an example) and therfore must bee twise reckened. Nowe I saye that those ij. squares are equall to two longe squares of the whole lineA.B,and his sayd portionA.C,and also to the square of the other portion of the sayd first line, whiche portion isC.B,and his squareK.N.F.L.In this theoreme there is no difficultie, if you cõsyder that the litle squareD.H.K.M.is .iiij. tymes reckened, that is to say, fyrst of all as a parte of the greatest square, whiche isD.E.F.G.Secondly he is reknedby him selfe. Thirdely he is accompted as parcell of the long squareD.E.N.M,And fourthly he is taken as a part of the other long squareD.H.L.G,so that in as muche as he is twise reckened in one part of the comparisõ of equalitee, and twise also in the second parte, there can rise none occasion of errour or doubtfulnes therby.The xlij. Theoreme.If a right line be deuided as chance happeneth the iiij. long squares, that may be made of that whole line and one of his partes with the square of the other part, shall be equall to the square that is made of the whole line and the saide first portion ioyned to him in lengthe as one whole line.Example.see textThe firste line isA.B,and is deuided byC.into two vnequall partes as happeneth.Thelong square of yt, and his lesser portionA.C,is foure times drawen, the first isE.G.M.K,the seconde isK.M.Q.O,the third isH.K.R.S,and the fourthe isK.L.S.T.And where as it appeareth that one of the little squares (I meaneK.L.P.O) is reckened twise, ones as parcell of the second long square and agayne as parte of thethirde long square, to auoide ambiguite, you may place one insteede of it, an other square of equalitee, withit. thatis to saye,D.E.K.H,which was at no tyme accompting as parcell of any one of them, and then haue you iiij. long squares distinctly made of the whole lineA.B,and his lesser portionA.C.And within them is there a greate full squareP.Q.T.V.whiche is the iust square ofB.C,beynge the greatter portion of the lineA.B.And that those fiue squares doo make iuste as muche as the whole square of that longer lineD.G,(whiche is as longe asA.B,andA.C.ioyned togither) it may be iudged easyly by the eye, sith that one greate square doth comprehẽd in it all the other fiue squares, that is to say, foure long squares (as is before mencioned) and one fullsquare. whichis the intent of the Theoreme.The xliij. Theoreme.If a right line be deuided into ij. equal partes first, and one of those parts againintoother ij. parts, as chaũce hapeneth, the square that is made of the last part of the line so diuided, and the square of the residue of that whole line, are double to the square of halfe that line, and to the square of the middle portion of the same line.Example.see textThe line to be deuided isA.B,and is parted inC.into two equall partes, and thenC.B,is deuided againe into two partes inD,so that the meaninge of the Theoreme, is that thesquare ofD.B.which is the latter parte of the line, and the square ofA.D,which is the residue of the whole line. Those two squares, I say, ar double to the square of one halfe of the line, and to the square ofC.D,which is the middle portion of those thre diuisions. Which thing that you maye more easilye perceaue, I haue drawen foure squares, whereof the greatest being marked withE.is the square ofA.D.The next, which is marked withG,is the square of halfe the line, that is, ofA.C,Andthe other two little squares marked withF.andH,be both of one bignes, by reason that I did diuideC.B.into two equall partes, so that you amy take the squareF,for the square ofD.B,and the squareH,for the square ofC.D.Now I thinke you doubt not, but that the squareE.and the squareF,ar double so much as the squareG.and the squareH,which thing theeasyeris to be vnderstande, bicause that the greate square hath in his side iij. quarters of the firste line, which multiplied by itselfe maketh nyne quarters, and the squareF.containeth but one quarter, so that bothe doo make tenne quarters.ThenG.contayneth iiij. quarters, seynge his side containethtwoo, andH.containeth but one quarter, whiche both makebut fiue quarters, and that is but halfe of tenne.Whereby you may easylye coniecture,that the meanynge of the the-oreme is verified in thefigures of this ex-ample.The xliiij. Theoreme.If a right line be deuided into ij. partes equally, and an other portion of a righte lyne annexed to that firste line, the square of this whole line so compounded, and the square of the portion that is annexed, ar doule as much as the square of the halfe of the firste line, and the square of the other halfe ioyned in one with the annexed portion, as one whole line.Example.see textThe line isA.B,and is diuided firste into twoo equal partes inC,and thẽ is there annexed to it an other portion whiche isB.D.Now saith the Theoreme, that the square ofA.D,and the square ofB.D,ar double to the square ofA.C,and to the square ofC.D.The lineA.B.cõtaining four partes, then must needes his halfe containe ij. partes of such partes I supposeB.D.(which is the ãnexed line) to containe thre, so shal the hole line cõprehend vij. parts, and his square xlix. parts, where vnto if you ad yesquareof the annexed lyne, whiche maketh nyne, than those bothe doo yelde, lviij. whyche must be double to the square of the halfe lyne with the annexed portion. The halfe lyne by it selfe conteyneth but .ij. partes, and therfore his square dooth make foure. The halfe lyne with the annexed portion conteyneth fiue, and the square of it is .xxv, now put foure to .xxv, and it maketh iust .xxix, the euen halfe of fifty and eight, wherby appereth the truthe of the theoreme.The .xlv. theoreme.In all triangles that haue a blunt angle, the square of the side that lieth against the blunt angle, is greater than the two squares of the other twoo sydes, by twise as muche as is comprehended of the one of those .ij. sides (inclosyng the blunt corner) and the portion of the same line, beyng drawen foorth in lengthe, which lieth betwene the said blunt corner and a perpendicular line lightyng on it, and drawen from one of the sharpe angles of the foresayd triangle.Example.For the declaration of this theoreme and the next also, whose vse are wonderfull in the practise of Geometrie, and in measuryng especially, it shall be nedefull to declare that euery triangle that hath no ryght angle as those whyche are called (as in the boke of practise is declared) sharp cornered triangles, and bluntcorneredtriangles, yet may they be brought to haue a ryght angle, eyther by partyng them into two lesser triangles,or els by addyng an other triangle vnto them, whiche may be a great helpe for the ayde of measuryng, as more largely shall be sette foorthe in the boke of measuryng. But for this present place, this forme wyll I vse, (whiche Theon also vseth) to adde one triangle vnto an other, to bryng the blunt cornered triangle into a ryght angled triangle, whereby the proportion of the squares of the sides in suche a blunt cornered triangle may the better bee knowen.see textFyrst therfore I sette foorth the triangleA.B.C,whose corner byC.is a blunt corner as you maye well iudge, than to make an other triangle of yt with a ryght angle, I must drawe forth the sideB.C.vntoD,and frõ the sharp corner byA.I brynge a plumbe lyne or perpẽdicular onD.And so is there nowe a newe triangleA.B.D.whose angle byD.is a right angle. Nowe accordyng to the meanyng of the Theoreme, I saie, that in the first triangleA.B.C,because it hath a blunt corner atC,the square of the lineA.B.whiche lieth against the said blunte corner, is morethen the square of the lineA.C,and also of the lyneB.C,(whiche inclose the blunte corner) by as muche as will amount twise of the lineB.C,and that portionD.C.whiche lieth betwene the blunt angle byC,and the perpendicular lineA.D.The square of the lineA.B,is the great square marked withE.The square ofA.C,is the meane square marked withF.The square ofB.C,is the least square marked withG.And the long square marked withK,is sette in steede of two squares made ofB.C,andC.D.For as the shorter side is the iuste lengthe ofC.D,so the other longer side is iust twise so longe asB.C,WherforeI saie now accordyng to the Theoreme, that the greatte squareE,is more then the other two squaresF.andG,by the quantitee of the longe squareK,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.Labeling of rectangle K is conjectural. Other configurations of B, C and D will also fit, but the printed illustration (B and D on the right, nothing on the left) will not. The text requires a rectangle with BC as one side and CD as the other, doubled as in the next illustration (H.K).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.see textFyrst I sette foorth the triangleA.B.C,and in yt I draw a plũbe line from the angleC.vnto the lineA.B,and it lighteth inD.Nowe by the theoreme the square ofB.C.is not so muche as the square of the other two sydes, that ofB.A.and ofA.C.by as muche as is twise conteyned in the lõg square made ofA.B,andA.D, A.B.beyng the line or syde on which the perpendicular line falleth, andA.D.beeyng that portion of the same line whiche doth lye betwene the perpendicular line, and the sayd sharpe angle limitted, whiche angle is byA.For declaration of the figures, the square marked withE.is the square ofB.C,whiche is the syde that lieth agaynst the sharpe angle, the square marked withG.is the square ofA.B,and the square marked withF.is the square ofA.C,and the two longe squares marked withH.K,are made of the hole lineA.B,and one of his portionsA.D.And truthe it is that the squareE.is lesser than the other two squaresC.andF.by the quantitee of those two long squaresH.andK.Wherby you may consyder agayn, an other proportion of equalitee,that is to saye, that the squareE.with the twoo longsquaresH.K,are iuste equall to the other twoo squaresC.andF.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 at 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 falleth, is thrise vsed, yet is his square but ones 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, would credite the possibilitie 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 circumferẽce of a circle, and a right line drawen frome the one to the other, that line must needes fal within the circle.Example.see textThe circle isA.B.C.D,the ij. poinctes areA.B,the righteline that is drawenne frome the one to the other, is the lineA.B,which as you see, must needes lyghte within the circle. So if you putte the pointes to beA.D,orD.C,orA.C,otherB.C,orB.D,in any 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. How be it, 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 he 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.see textExample.The circle isA.B.C.D,the line that passeth by the centre, isA.E.C,the line that goeth beside the centre isD.B.Nowesaye I, that the lineA.E.C,dothe cutte that other lineD.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, accordinge as the Theoreme would.The xlix. Theoreme.If twoo right lines drawen in a circle doo crosse one an other, and doo notpasse bythe centre, euery of them dothe not deuide the other into equall partions.Example.see textThe circle isA.B.C.D,and the centre isE,the one lineA.C,and the other isB.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 cuytte the other into equall portions. For as you may easily iudge,A.C.hath one portiõ lõger and an other shorter, and so like wiseB.D.Howbeit, it is not so to be vnderstãd, but one of them may be deuided into ij. euẽ parts,but bothe to bee cutte equally in the middle, is not possible, onles both passe through the cẽtre, therfore much rather whẽ 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.see textExample.This theoreme seemeth of it selfe so manifest, that it neadeth nother demonstration nother declaraciõ. Yet for the plaine vnderstanding of it, I haue sette forthe a figure here, where ij. circles be drawẽ, so that one of them doth crosse the other (as you see) in the pointesB.andG,and their centres appear at the firste sighte to bee diuers. For the centre of the one isF,and the centre of the other isE,which diffre as farre asondre as the edges of the circles, where they bee most distaunte in sonder.The Li. Theoreme.If two circles be so drawen, that one of them do touche the other, then haue they not one centre.see textExample.There are two circles made, as you see, the one isA.B.C,and hath his centre byG,the other isB.D.E,and his centre is byF,so that it is easy enough to perceaue that their centres doe dyffer as muche a sonder, as the halfe diameter of the greater circle is lõger then the half diameter of the lesser circle. And so must it needes be thought and said of all other circles in lyke kinde.The .lij. theoreme.If a certaine pointe be assigned in the diameter of a circle, distant from the centre of the said circle, and from that pointe diuerse lynes drawen to the edge and circumference of the same circle, the longest line is that whiche passeth by the centre, and the shortest is the residew of the same line. And of al the other lines that is euer the greatest, that is nighest to the line, which passeth by the centre. And cõtrary waies, that is the shortest, that is farthest from it. And amongest thẽ all there can be but onely .ij. equall together, and they must nedes be so placed, that the shortest line shall be in the iust middle betwixte them.Example.see textThe circle isA.B.C.D.E.H,and his centre isF,the diameter isA.E,in whiche diameter I haue taken a certain point distaunt from the centre, and that pointe isG,from which I haue drawen .iiij. lines to the circumference, beside the two partes of the diameter, whiche maketh vp vi. lynes in all. Nowe for the diuersitee in quantitie of these lynes, I saie accordyng to the Theoreme, that the line whiche goeth by the centre is the longest line, that is to saie,A.G,and the residewe of the same diameter beeyngG.E,is the shortest lyne. And of all the other that lyne is longest, that is neerest vnto that parte of the diameter whiche gooeth by the centre, and that is shortest, that is farthest distant from it, wherefore I saie, thatG.B,is longer thenG.C,and therfore muche more longer thenG.D,sithG.C,also is longer thenG.D,and by this maie you soone perceiue, that it is not possible to drawe .ij. lynes on any one side of the diameter, whiche might be equall in lengthe together, but on the one side of the diameter maie you easylie make one lyne equall to an other, on the other side of the same diameter, as you see in this exampleG.H,to bee equall toG.D,betweene whiche the lyneG.E,(as the shortest in all the circle) doothe stande euen distaunte from eche of them, and it is the precise knoweledge of their equalitee, if they beequallydistaunt from one halfe of the diameter. Where as contrary waies if the one be neerer to any one halfe of the diameter then the other is, it is not possible that they two may be equall in lengthe, namely if they dooe ende bothe in the circumference of thecircle, and be bothe drawen from one poynte in the diameter, so that the saide poynte be (as the Theoreme doeth suppose) somewhat distaunt from the centre of the said circle. For if they be drawen from the centre, then must they of necessitee be all equall, howe many so euer they bee, as the definition of a circle dooeth importe, withoute any regarde how neere so euer they be to the diameter, or how distante from it. And here is to be noted, that in this Theoreme, by neerenesse and distaunce is vnderstand the nereness and distaunce of the extreeme partes of those lynes where they touche the circumference. For at the other end they do all meete and touche.The .liij. Theoreme.If a pointe bee marked without a circle, and from it diuerse lines drawen crosse the circle, to the circumference on the other side, so that one of them passe by the centre, then that line whiche passeth by the centre shall be the loongest of them all that crosse the circle. And of the other lines those are longest, that be nexte vnto it that passeth by the centre. And those ar shortest, that be farthest distant from it. But among those partes of those lines, whiche ende in the outewarde circumference, that is most shortest, whiche is parte of the line that passeth by the centre, and amongeste theothere eche, of thẽ, the nerer they are vnto it, the shorter they are, and the farther from it, the longer they be. And amongest them all there can not be more then .ij. of any one lẽgth, and they two muste be on the two contrarie sides of the shortest line.Example.see textTake the circle to beA.B.C,and the point assigned without it to beD.Now say I, that if there be drawen sundrie lines fromD,and crosse the circle, endyng in the circumference on the cõtrary side, as here you see,D.A, D.E, D.F,andD.B,then of all these lines the longest must needes beD.A,which goeth by the centre of the circle, and the nexte vnto it, that isD.E,is the longest amongest the rest. And contrarie waies,D.B,is the shorteste, because it is farthest distaunt fromD.A.And so maie you iudge ofD.F,because it is nerer vntoD.A,then isD.B,therefore is it longer thenD.B.And likewaies because it is farther of fromD.A,then isD.E,therfore is it shorter thenD.E.Now for those partes of the lines whiche bee withoute the circle (as you see)D.C,is theshortest. becauseit is the parte of that line which passeth by thecentre, AndD.K,is next to it in distance, and therefore also in shortnes, soD.G,is farthest from it in distance, and therfore is the longest of them. NowD.H,beyng nerer thenD.G,is also shorterthen it,and beynge farther of, thenD.K,is longer then it. So that for this parte of the theoreme (as I think) you do plainly perceaue the truthe thereof, so the residue hathe no difficulte. For seing that the nearer any line is toD.C,(which ioyneth with the diameter) the shorter it is and the farther of from it, the longer it is. And seyng two lynes can not be of like distaunce beinge bothe on one side, therefore if they shal be of one lengthe, and consequently of one distaunce, they must needes bee on contrary sides of the saide lineD.C.And so appeareth the meaning of the whole Theoreme.And of this Theoreme dothe there folowe an otherlyke. whicheyou maye calle other a theoreme by it selfe, or else a Corollary vnto this laste theoreme, I passe not so muche for the name. But his sentence is this:when so euer any lynes be drawen frome any pointe, withoute a circle, whether they crosse the circle, or eande in theutteredge of his circumference, those two lines that bee equally distaunt from the least line are equal togither, and contrary waies, if they be equall togither, they ar also equally distant from that least line.For the declaracion of this proposition, it shall not need to vse any other example, then that which is brought for the explication of this laste theoreme, by whiche you may without any teachinge easyly perceaue both the meanyng and also the truth of this proposition.The Liiij. Theoreme.If a point be set forthe in a circle, and frõ that pointe vnto the circumference many lines drawen, of which more then two are equal togither, then is that point the centre of that circle.Example.see textThe circle isA.B.C,and within it I haue sette fourth for an example three prickes, which areD.E.andF,from euery one of them I haue drawẽ (at the leaste) iiij. lines vnto the circumference of the circle but fromeD,I haue drawen more, yet maye it appear readily vnto your eye, that of all the lines whiche be drawen fromE.andF,vnto the circumference, there are but twoo equall, and more can not bee, forG.E.norE.H.hath none other equal to theim, nor canne not haue any beinge drawen from the same pointE.No more canL.F,orF.K,haue anye line equall to either of theim, beinge drawen from the same pointeF.And yet from either of those two poinctes are there drawen twoo lines equall togither, asA.E,is equall toE.B,andB.F,is equall toF.C,but there can no third line be drawen equall to either of these two couples, and that is by reason that they be drawen from a pointe distaunte from the centre of the circle. But fromD,althoughe there be seuen lines drawen, to the circumference, yet all bee equall, bicause it is the centre of the circle. And therefore if you drawe neuer so mannye more from it vnto the circumference, all shall be equal, so that this is the priuilege (as it were of the centre) and therfore no other point can haue aboue two equal lines drawen from it vnto the circumference. And from allpointesyou maye drawe ij. equall lines to the circumference of the circle, whether that pointe be within the circle or without it.The lv. Theoreme.No circle canne cut an other circle in morepointes then two.see textExample.The first circle isA.B.F.E,the second circle isB.C.D.E,and they crosse one an other inB.and inE,and in no more pointes. Nother is it possible that they should, but other figures ther be, which maye cutte a circle in foure partes, as you se in this exãple. Where I haue set forthe one tunne forme, and one eye forme, and eche of them cutteth euery of their two circles into foure partes. But as they be irregulare formes, that is to saye, suche formes as haue no precise measure nother proportion in their draughte, so can there scarcely be made any certaine theorem of them. But circles are regulare formes, that is to say, such formes as haue in their protracture a iuste and certaine proportion, so that certain and determinate truths may be affirmed of them, sith they ar vniforme and vnchaungable.The lvi. Theoreme.If two circles be so drawen, that the one be within the other, and that they touche one an other: If a line bee drawen by bothe their centres, and so forthe in lengthe, that line shall runne to that pointe, where the circles do touche.see textExample.The one circle, which is the greattest and vttermost isA.B.C,the other circle that is yelesser, and is drawen within the firste, isA.D.E.The cẽtre of the greater circle isF,and the centre of the lesser circle isG,the pointe where they touche isA.And now you may see the truthe of the theoreme so plainely, that it needeth no farther declaracion. For you maye see, that drawinge a line fromF.toG,and so forth in lengthe, vntill it come to the circumference, it wyll lighte in the very poincteA,where the circles touche one an other.The Lvij. Theoreme.If two circles bee drawen so one withouteanother, that their edges doo touche and a right line bee drawnenne frome the centre of the one to the centre of the other, that line shall passe by the place of their touching.Example.see textThe firste circle isA.B.E,and his centre isK,Thesecõd circle isD.B.C,and his cẽtre isH,the point wher they do touch isB.Nowe doo you se that the lineK.H,whiche is drawenfromK,that is centre of the firste circle, vntoH,beyng centre of the second circle, doth passe (as it must nedes by the pointeB,) whiche is the verye poynte wher they do to touche together.The .lviij. theoreme.One circle can not touche an other in more pointes then one, whether they touche within or without.see textExample.For the declaration of this Theoreme, I haue drawen iiij. circles, the first isA.B.C,and his centreH.the second isA.D.G,and his centreF.the third isL.M,and his centreK.the .iiij. isD.G.L.M,and his centreE.Nowe as you perceiue the second circleA.D.G,toucheth the first in the inner side, in so much as it is drawen within the other, and yet it toucheth him but in one point, that is to say inA,so lykewaies the third circleL.M,is drawen without the firste circle and touchethhym, as you maie see, but in one place. And now as for the .iiij. circle, it is drawen to declare the diuersitie betwene touchyng and cuttyng, or crossyng. For one circle maie crosse and cutte a great many other circles, yet can be not cutte any one in more places then two, as the fiue and fiftie Theoreme affirmeth.The .lix. Theoreme.In euerie circle those lines are to be counted equall, whiche are in lyke distaunce from thecentre, Andcontrarie waies they are in lyke distance from the centre, whiche be equall.see textExample.In this figure you see firste the circle drawen, whiche isA.B.C.D,and his centre isE.In this circle also there are drawen two lines equally distaunt from the centre, for the lineA.B,and the lineD.C,are iuste of one distaunce from the centre, whiche isE,and therfore are they of one length. Again thei are of one lengthe (as shall be proued in the boke of profes) and therefore their distaunce from the centre is all one.The lx. Theoreme.In euerie circle the longest line is the diameter, and of all the other lines, thei are still longestthat be nexte vnto the centre, and they be the shortest, that be farthest distaunt from it.Example.see textIn this circleA.B.C.D,I haue drawen first the diameter, whiche isA.D,whiche passeth (as it must) by the centreE,Thenhaue I drawen ij. other lines asM.N,whiche is neerer the centre, andF.G,that is farther from the centre.The fourth line also on the other side of the diameter, that isB.C,is neerer to the centre then the lineF.G,for it is of lyke distance as is the lyneM.N.Nowe saie I, thatA.D,beyng the diameter, is the longest of all those lynes, and also of any other that maie be drawen within that circle, Andthe other lineM.N,is longer thenF.G.Also the lineF.G,is shorter then the lineB.C,for because it is farther from the centre then is the lyneB.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 twothynges, Asfor exaumple, If two squares be sette foorthe, whereof the one containeth in it fiue square feete, 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 measureof 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 circumferẽce 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.see textIn this circleA.B.C,the diameter isA.C,the perpendicular line, which maketh a right angle with the diameter, isC.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 thyngis so plainly seene of the eye, that it needeth no farther declaracion. For euery man wil easily consent, that betwene the croked lineA.F,(whiche is a parte of the circumferẽce of the circle) andA.E(which is the said perpẽdicular line) there can none other line bee drawen in that place where they make the angle. Nowe for the residue of the theoreme. The angleD.A.B,which is made in the semicircle, is greater then anye sharpe angle that may bee made of ryghtelines. andyet is it a sharpe angle also, in as much as it is lesser then a right angle, which is the angleE.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 lineE.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 greater then the other, for they twoo doo make the whole right angle, so that if anye corner coulde be 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 ofthecircle to the pointe where they touche, thatline whiche is drawenne frome the centre, shall be a perpendicular line to the touch line.Example.see textThe circle isA.B.C,and his centre isF.The touche line isD.E,and the point wher they touch isC.Now by reason that a right line is drawen frome the centreF.vntoC,which is the point of the touche, therefore saith the theoreme, that the sayde lineF.C,muste needes bee a perpendicular line vnto the touche lineD.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.see textThe circle isA.B.C,and the centre of it isG.The touche line isD.C.E,and the pointe where it toucheth, isC.Noweit 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 cẽtre in him. For if the saide line shoulde go beside the centre, asF.C.doth, then dothe it not make righte angles with the touche line, which in thetheoremeis 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 circũferẽce.see textExample.ThecircleisA.B.C.D,and his centre is E: the angle on the centre isC.E.D, and the angle on the circumference isC.A.Dttheir commen ground line, isC.F.D.Now say I that the angleC.E.D,whiche is on the centre, is twise so greate as the angleC.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.see textBefore I declare this theoreme by example, it shall bee needefull to declare, what is it to be vnderstande by the wordes in this theoreme. For the sentence canne not be knowen, onles theuerymeaning of the wordes be firste vnderstand. Therefore when it speaketh of angles 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 circle. So that if any angle do not touch the circumference, or if the lines that inclose that angle, doo not ende in the extremities of the corde line, but ende other in some other part of the said 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 said cantle of the circle. And this promised, nowe will I cumme to the meaninge of the theoreme. I sette forthe a circle whiche isA.B.C.D,and his centreE,in this circle I drawe a lineD.C,whereby there ar made two cantels, a more and a lesser. The lesser isD.E.C,and the geater isD.A.B.C.In this greater cantle I drawe two angles, the firste isD.A.C,and the second isD.B.Cwhich two angles by reason they are made bothe in one cantle of a circle (that is the cantleD.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 isD.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.
see textExample.Fyrst before I declare the examples, it shal be mete to shew the true vnderstãdyng of this theorem.Bias lyne.Therfore by theBias line, I meane that lyne, whiche in any square figure dooth runne from corner to corner. And euery square which is diuided by that bias line into equall halues from corner to corner (that is to say, into .ij. equall triangles) those be countedto stande aboute one bias line, and the other squares, whiche touche that bias line, with one of their corners onely, those doo I callFyll squares,Fyll squares.accordyng to the greke name, which isanapleromata,ἀναπληρώματαand called in latinsupplementa, bycause that they make one generall square, includyng and enclosyng the other diuers squares, as in this exãpleH.C.E.N.is one square likeiamme, andL.M.G.C.is an other, whiche bothe are made aboute one bias line, that isN.M,thanK.L.H.C.andC.E.F.G.are .ij. fyll squares, for they doo fyll vp the sydes of the .ij.fyrste square lykeiammes, in suche sorte, that all them foure is made one greate generall squareK.M.F.N.Nowe to the sentence of the theoreme, I say, that the .ij. fill squares,H.K.L.C.andC.E.F.G.are both equall togither, (as it shall bee declared in the booke of proofes) bicause they are the fill squares of two likeiammes made aboute one bias line, as the exaumple sheweth. Conferre the twelfthe conclusion with this theoreme.The xxxiij. Theoreme.In all right anguled triangles, the square of that side whiche lieth against the right angle, is equall to the .ij. squares of both the other sides.Example.see textA.B.C.is a triangle, hauing a ryght angle inB.Wherfore it foloweth, that the square ofA.C,(whiche is the side that lyeth agaynst the right angle) shall be as muche as the two squares ofA.B.andB.C.which are the other .ij. sides.¶By thesquareof any lyne, you muste vnderstande a figure made iuste square, hauyng all his iiij. sydes equall to that line, whereof it is the square, so isA.C.F,the square ofA.C.LykewaisA.B.D.is the square ofA.B.AndB.C.E.is the square ofB.C.Now by the numbre of the diuisions in eche of these squares, may you perceaue not onely what the square of any line is called, but also that the theoreme is true, and expressed playnly bothe by lines and numbre. For as you see, the greatter square (that isA.C.F.) hath fiue diuisions on eche syde, all equall togyther, and those in the whole square are twenty and fiue. Nowe in the left square, whiche isA.B.D.there are but .iij. of those diuisions in one syde, and that yeldeth nyne in the whole. So lykeways you see in the meane squareA.C.E.in euery syde .iiij. partes, whiche in the whole amount vnto sixtene. Nowe adde togyther all the partes of the two lesser squares, that is to saye, sixtene and nyne, and you perceyue that they make twenty and fiue, whyche is an equall numbre to the summe of the greatter square.By this theoreme you may vnderstand a redy way to know the syde of any ryght anguled triangle that is vnknowen, so that you knowe the lengthe of any two sydes of it. For by tournynge the two sydes certayne into theyr squares, and so addynge them togyther, other subtractynge the one from the other (accordyng as in the vse of these theoremes I haue sette foorthe) and then fyndynge the roote of the square that remayneth, which roote (I meane the syde of the square) is the iuste length of the vnknowen syde, whyche is sought for. But this appertaineth to the thyrde booke, and therefore I wyll speake no more of it at this tyme.The xxxiiij. Theoreme.If so be it, that in any triangle, the square of the one syde be equall to the .ij. squares of the other .ij. sides, than must nedes that corner be a right corner, which is conteined betwene those two lesser sydes.Example.As in the figure of the laste Theoreme, bicauseA.C,made in square, is asmuch as the square ofA.B,and also as the square ofB.C.ioyned bothe togyther, therefore the angle that is inclosed betwene those .ij. lesser lynes,A.B.andB.C.(that is to say) the angleB.whiche lieth against the lineA.C,must nedes be a ryght angle. This theoreme dothe so depende of the truthe of the laste, that whan you perceaue the truthe of the one, you can not iustly doubt of the others truthe, for they conteine one sentence, contrary waies pronounced.The .xxxv. theoreme.If there be set forth .ij. right lines, and one of them parted into sundry partes, how manyor few so euer they be, the square that is made of those ij. right lines proposed, is equal to all the squares, that are made of the vndiuided line, and euery parte of the diuided line.Example.see textThe ij. lines proposed arA.B.andC.D,and the lyneA.B.is deuided into thre partes byE.andF.Now saith this theoreme, that the square that is made of those two whole linesA.B.andC.D,so that the lineA.B.stãdeth for the lẽgth of the square, and the other lineC.D.for the bredth of the same. That square (I say) wil be equall to all the squares that be made, of the vndiueded lyne (which isC.D.) and euery portion of the diuided line. And to declare that particularly, Fyrst I make an other lineG.K,equall to the line.C.D,and the lineG.H.to be equal to the lineA.B,and to bee diuided into iij. like partes, so thatG.M.is equall toA.E,andM.N.equal toE.F,and then musteN.H.nedes remaine equall toF.B.Then of those ij. linesG.K,vndeuided, andG.H.which is deuided, I make a square, that isG.H.K.L,Inwhich square if I drawe crosse lines frome one side to the other, according to the diuisions of the lineG.H,then will it appear plaine, that the theoreme doth affirme. For the first squareG.M.O.K,must needes be equal to the square of the lineC.D,and the first portiõ of the diuided line, which isA.E,for bicause their sides are equall. And so the secondesquare that isM.N.P.O,shall be equall to the square ofC.D,and the second part ofA.B,that isE.F.Also the third square which isN.H.L.P,must of necessitee be equal to the square ofC.D,andF.B,bicause those lines be so coupeled that euery couple are equall in the seuerall figures. And so shal you not only in this example, but in all other finde it true, that if one line be deuided into sondry partes, and an other line whole and vndeuided, matched with him in a square, that square which is made of these two whole lines, is as muche iuste and equally, as all the seuerall squares, whiche bee made of the whole line vndiuided, and euery part seuerally of the diuided line.The xxxvi. Theoreme.If a right line be parted into ij. partes, as chaunce may happe, the square that is made of the whole line, is equall to bothe the squares that are made of thesameline, and the twoo partes of it seuerally.Example.see textThe line propounded beyngA.B.and deuided, as chaunce happeneth, inC.into ij. vnequall partes, I say that the square made of the hole lineA.B,is equal to the two squares made of the same line with the twoo partes of itselfe, as withA.C,and withC.B,for the squareD.E.F.G.is equal to the two other partial squares ofD.H.K.GandH.E.F.K,but that the greater square is equall to the square of the whole lineA.B,and thepartiall squares equall to the squares of the second partes of the same line ioyned with the whole line, your eye may iudg without muche declaracion, so that I shall not neede to make more exposition therof, but that you may examine it, as you did in the laste Theoreme.The xxxvij.Theoreme.If a right line be deuided by chaunce, as it maye happen, the squarethatis made of the whole line, and one of the partes of it which soeuer it be, shal be equall to that square that is made of the ij. partes ioyned togither, and to an other square made of that part, which was before ioyned with the whole line.Example.see textThe lineA.B.is deuided inC.into twoo partes, though not equally, of which two partes for an example I take the first, that isA.C,and of it I make one side of a square, as for exampleD.G.accomptinge those two lines to be equall, the other side of the square isD.E,whiche is equall to the whole lineA.B.Nowmay it appeare, to your eye, that the great square made of the whole lineA.B,and of one of his partes that isA.C,(which is equall withD.G.) is equal to two partiall squares, whereof the one is made of the saide greatter portionA.C,in as muche as not onlyD.G,beynge one of his sides, but alsoD.H.beinge the other side, are eche of them equall toA.C.The second square isH.E.F.K,in which the one sideH.E,is equal toC.B,being the lesser parte of the line,A.B,andE.F.is equall toA.C.which is the greater parte of the same line. So that those two squaresD.H.K.GandH.E.F.K,bee bothe of them no more then the greate squareD.E.F.G,accordinge to the wordes of the Theoreme afore saide.The xxxviij. Theoreme.If a righte line be deuided by chaunce, into partes, the square that is made of that whole line, is equall to both the squares that ar made of eche parte of the line, and moreouer to two squares made of the one portion of the diuided line ioyned with the other in square.see textExample.The labels A and B were transposed in the illustration as an alternative to transposing all occurrences of A.C and C.B in the text.Lette the diuided line beeA.B,and parted inC,into twoo partes: Nowe saithe the Theoreme, that the square of the whole lyneA.B,is as mouche iuste as the square ofA.C,and the square ofC.B,eche by it selfe, and more ouer by as muche twise, asA.C.andC.B.ioyned in one square will make. For as you se, the great squareD.E.F.G,conteyneth in hym foure lesser squares, of whiche the first and the greatest isN.M.F.K,and is equall to the square of the lyneA.C.The second square is the lest of them all, that isD.H.L.N,and it is equall to the square of the lineC.B.Then are there two other longe squares both of one bygnes, that isH.E.N.M.andL.N.G.K,eche of them both hauyng .ij. sides equall toA.C,the longer parte of the diuided line, and there other two sides equall toC.B,beeyng he shorter parte of the said lineA.B.So is that greatest square, beeyng made of the hole lyneA.B,equal to the ij. squares of eche of his partes seuerally, and more by as muche iust as .ij. longe squares, made of the longer portion of the diuided lyne ioyned in square with the shorter parte of the same diuided line, as the theoreme wold. And as here I haue put an example of a lyne diuided into .ij. partes, so the theoreme is true of all diuided lines, of what number so euer the partes be, foure, fyue, orsyxe. etc.Thistheoreme hath great vse, not only in geometrie, but also in arithmetike, as herafter I will declare in conuenient place.The .xxxix. theoreme.If a right line be deuided into two equall partes, and one of these .ij. partes diuided agayn into two other partes, as happeneth the longe square that is made of the thyrd or later part of that diuided line, with the residue of the same line, and the square of the mydlemoste parte, are bothe togither equall to the square of halfe the firste line.Example.see textThe lineA.B.is diuided into ij. equal partes inC,and that parteC.B.is diuided agayne as hapneth inD.Wherfore saith the Theorem that the long square made ofD.B.andA.D,with the square ofC.D.(which is the mydle portion) shall bothe be equall to the square of half the lyneA.B,that is to saye, to the square ofA.C,or els ofC.D,which make all one. The long squareF.G.N.O.whiche is the longe square that the theoreme speaketh of, is made of .ij. long squares, wherof the fyrst isF.G.M.K,and the seconde isK.N.O.M.The square of the myddle portion isL.M.O.P.and the square of the halfe of the fyrste lyne isE.K.Q.L.Nowe by the theoreme, that longe squareF.G.N.O,with the iuste squareL.M.O.P,muste bee equall to the greate squareE.K.Q.L,whyche thynge bycause it seemeth somewhat difficult to vnderstande, althoughe I intende not here to make demonstrations of the Theoremes, bycause it is appoynted to be done in the newe edition of Euclide, yet I wyll shew you brefely how the equalitee of the partes doth stande. And fyrst I say, that where the comparyson of equalitee is made betweene the greate square (whiche is made of halfe the lineA.B.) and two other, where of the fyrst is the longe squareF.G.N.O,and the second is the full squareL.M.O.P,which is one portion of the great square all redye, and so is that longe squareK.N.M.O,beynge a parcell also of the longe squareF.G.N.O,Wherforeas those two partes are common to bothe partes compared in equalitee, and therfore beynge bothe abated from eche parte, if the reste of bothe the other partes bee equall, than were those whole partes equall before: Nowe the reste of the great square, thosetwo lesser squares beyng taken away,is that longe squareE.N.P.Q,whyche is equall to the long squareF.G.K.M,beyng the rest of the other parte. And that they two be equall, theyr sydes doo declare. For the longest lynes that isF.KandE.Qare equall, and so are the shorter lynes,F.G,andE.N,and so appereth the truthe of the Theoreme.The .xl. theoreme.If a right line be diuided into .ij. euen partes, and an other right line annexed to one ende of that line, so that it make one righte line with the firste. The longe square that is made of this whole line so augmented, and the portion that is added, with the square of halfe the right line, shall be equall to the square of that line, whiche is compounded of halfe the firste line, and the parte newly added.Example.see textThe fyrst lynepropoundedisA.B,and it is diuided into ij. equall partes inC,and an other ryght lyne, I meaneB.Dannexed to one ende of the fyrste lyne.Nowesay I, that the long squareA.D.M.K,is made of the whole lyne so augmẽted, that isA.D,and the portiõ annexed, ytisD.M,forD.Mis equall toB.D,wherfore ytlong squareA.D.M.K,with thesquare of halfe the first line, that isE.G.H.L,is equall to the great squareE.F.D.C.whiche square is made of the lineC.D.that is to saie, of a line compounded of halfe the first line, beyngC.B,and the portion annexed, that isB.D.And it is easyly perceaued, if you consyder that the longe squareA.C.L.K.(whiche onely is lefte out of the great square) hathanotherlonge square equall to hym, and to supply his steede in the great square, and that isG.F.M.H.For their sydes be of lyke lines in length.The xli. Theoreme.If a right line bee diuided by chaunce, the square of the same whole line, and the square of one of his partes are iuste equall to the lõg square of the whole line, and the sayde parte twise taken, and more ouer to the square of the other parte of the sayd line.Example.see textA.B.is the line diuided inC.AndD.E.F.G,is the square of the whole line,D.H.K.M.is the square of the lesser portion (whyche I take for an example) and therfore must bee twise reckened. Nowe I saye that those ij. squares are equall to two longe squares of the whole lineA.B,and his sayd portionA.C,and also to the square of the other portion of the sayd first line, whiche portion isC.B,and his squareK.N.F.L.In this theoreme there is no difficultie, if you cõsyder that the litle squareD.H.K.M.is .iiij. tymes reckened, that is to say, fyrst of all as a parte of the greatest square, whiche isD.E.F.G.Secondly he is reknedby him selfe. Thirdely he is accompted as parcell of the long squareD.E.N.M,And fourthly he is taken as a part of the other long squareD.H.L.G,so that in as muche as he is twise reckened in one part of the comparisõ of equalitee, and twise also in the second parte, there can rise none occasion of errour or doubtfulnes therby.The xlij. Theoreme.If a right line be deuided as chance happeneth the iiij. long squares, that may be made of that whole line and one of his partes with the square of the other part, shall be equall to the square that is made of the whole line and the saide first portion ioyned to him in lengthe as one whole line.Example.see textThe firste line isA.B,and is deuided byC.into two vnequall partes as happeneth.Thelong square of yt, and his lesser portionA.C,is foure times drawen, the first isE.G.M.K,the seconde isK.M.Q.O,the third isH.K.R.S,and the fourthe isK.L.S.T.And where as it appeareth that one of the little squares (I meaneK.L.P.O) is reckened twise, ones as parcell of the second long square and agayne as parte of thethirde long square, to auoide ambiguite, you may place one insteede of it, an other square of equalitee, withit. thatis to saye,D.E.K.H,which was at no tyme accompting as parcell of any one of them, and then haue you iiij. long squares distinctly made of the whole lineA.B,and his lesser portionA.C.And within them is there a greate full squareP.Q.T.V.whiche is the iust square ofB.C,beynge the greatter portion of the lineA.B.And that those fiue squares doo make iuste as muche as the whole square of that longer lineD.G,(whiche is as longe asA.B,andA.C.ioyned togither) it may be iudged easyly by the eye, sith that one greate square doth comprehẽd in it all the other fiue squares, that is to say, foure long squares (as is before mencioned) and one fullsquare. whichis the intent of the Theoreme.The xliij. Theoreme.If a right line be deuided into ij. equal partes first, and one of those parts againintoother ij. parts, as chaũce hapeneth, the square that is made of the last part of the line so diuided, and the square of the residue of that whole line, are double to the square of halfe that line, and to the square of the middle portion of the same line.Example.see textThe line to be deuided isA.B,and is parted inC.into two equall partes, and thenC.B,is deuided againe into two partes inD,so that the meaninge of the Theoreme, is that thesquare ofD.B.which is the latter parte of the line, and the square ofA.D,which is the residue of the whole line. Those two squares, I say, ar double to the square of one halfe of the line, and to the square ofC.D,which is the middle portion of those thre diuisions. Which thing that you maye more easilye perceaue, I haue drawen foure squares, whereof the greatest being marked withE.is the square ofA.D.The next, which is marked withG,is the square of halfe the line, that is, ofA.C,Andthe other two little squares marked withF.andH,be both of one bignes, by reason that I did diuideC.B.into two equall partes, so that you amy take the squareF,for the square ofD.B,and the squareH,for the square ofC.D.Now I thinke you doubt not, but that the squareE.and the squareF,ar double so much as the squareG.and the squareH,which thing theeasyeris to be vnderstande, bicause that the greate square hath in his side iij. quarters of the firste line, which multiplied by itselfe maketh nyne quarters, and the squareF.containeth but one quarter, so that bothe doo make tenne quarters.ThenG.contayneth iiij. quarters, seynge his side containethtwoo, andH.containeth but one quarter, whiche both makebut fiue quarters, and that is but halfe of tenne.Whereby you may easylye coniecture,that the meanynge of the the-oreme is verified in thefigures of this ex-ample.The xliiij. Theoreme.If a right line be deuided into ij. partes equally, and an other portion of a righte lyne annexed to that firste line, the square of this whole line so compounded, and the square of the portion that is annexed, ar doule as much as the square of the halfe of the firste line, and the square of the other halfe ioyned in one with the annexed portion, as one whole line.Example.see textThe line isA.B,and is diuided firste into twoo equal partes inC,and thẽ is there annexed to it an other portion whiche isB.D.Now saith the Theoreme, that the square ofA.D,and the square ofB.D,ar double to the square ofA.C,and to the square ofC.D.The lineA.B.cõtaining four partes, then must needes his halfe containe ij. partes of such partes I supposeB.D.(which is the ãnexed line) to containe thre, so shal the hole line cõprehend vij. parts, and his square xlix. parts, where vnto if you ad yesquareof the annexed lyne, whiche maketh nyne, than those bothe doo yelde, lviij. whyche must be double to the square of the halfe lyne with the annexed portion. The halfe lyne by it selfe conteyneth but .ij. partes, and therfore his square dooth make foure. The halfe lyne with the annexed portion conteyneth fiue, and the square of it is .xxv, now put foure to .xxv, and it maketh iust .xxix, the euen halfe of fifty and eight, wherby appereth the truthe of the theoreme.The .xlv. theoreme.In all triangles that haue a blunt angle, the square of the side that lieth against the blunt angle, is greater than the two squares of the other twoo sydes, by twise as muche as is comprehended of the one of those .ij. sides (inclosyng the blunt corner) and the portion of the same line, beyng drawen foorth in lengthe, which lieth betwene the said blunt corner and a perpendicular line lightyng on it, and drawen from one of the sharpe angles of the foresayd triangle.Example.For the declaration of this theoreme and the next also, whose vse are wonderfull in the practise of Geometrie, and in measuryng especially, it shall be nedefull to declare that euery triangle that hath no ryght angle as those whyche are called (as in the boke of practise is declared) sharp cornered triangles, and bluntcorneredtriangles, yet may they be brought to haue a ryght angle, eyther by partyng them into two lesser triangles,or els by addyng an other triangle vnto them, whiche may be a great helpe for the ayde of measuryng, as more largely shall be sette foorthe in the boke of measuryng. But for this present place, this forme wyll I vse, (whiche Theon also vseth) to adde one triangle vnto an other, to bryng the blunt cornered triangle into a ryght angled triangle, whereby the proportion of the squares of the sides in suche a blunt cornered triangle may the better bee knowen.see textFyrst therfore I sette foorth the triangleA.B.C,whose corner byC.is a blunt corner as you maye well iudge, than to make an other triangle of yt with a ryght angle, I must drawe forth the sideB.C.vntoD,and frõ the sharp corner byA.I brynge a plumbe lyne or perpẽdicular onD.And so is there nowe a newe triangleA.B.D.whose angle byD.is a right angle. Nowe accordyng to the meanyng of the Theoreme, I saie, that in the first triangleA.B.C,because it hath a blunt corner atC,the square of the lineA.B.whiche lieth against the said blunte corner, is morethen the square of the lineA.C,and also of the lyneB.C,(whiche inclose the blunte corner) by as muche as will amount twise of the lineB.C,and that portionD.C.whiche lieth betwene the blunt angle byC,and the perpendicular lineA.D.The square of the lineA.B,is the great square marked withE.The square ofA.C,is the meane square marked withF.The square ofB.C,is the least square marked withG.And the long square marked withK,is sette in steede of two squares made ofB.C,andC.D.For as the shorter side is the iuste lengthe ofC.D,so the other longer side is iust twise so longe asB.C,WherforeI saie now accordyng to the Theoreme, that the greatte squareE,is more then the other two squaresF.andG,by the quantitee of the longe squareK,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.Labeling of rectangle K is conjectural. Other configurations of B, C and D will also fit, but the printed illustration (B and D on the right, nothing on the left) will not. The text requires a rectangle with BC as one side and CD as the other, doubled as in the next illustration (H.K).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.see textFyrst I sette foorth the triangleA.B.C,and in yt I draw a plũbe line from the angleC.vnto the lineA.B,and it lighteth inD.Nowe by the theoreme the square ofB.C.is not so muche as the square of the other two sydes, that ofB.A.and ofA.C.by as muche as is twise conteyned in the lõg square made ofA.B,andA.D, A.B.beyng the line or syde on which the perpendicular line falleth, andA.D.beeyng that portion of the same line whiche doth lye betwene the perpendicular line, and the sayd sharpe angle limitted, whiche angle is byA.For declaration of the figures, the square marked withE.is the square ofB.C,whiche is the syde that lieth agaynst the sharpe angle, the square marked withG.is the square ofA.B,and the square marked withF.is the square ofA.C,and the two longe squares marked withH.K,are made of the hole lineA.B,and one of his portionsA.D.And truthe it is that the squareE.is lesser than the other two squaresC.andF.by the quantitee of those two long squaresH.andK.Wherby you may consyder agayn, an other proportion of equalitee,that is to saye, that the squareE.with the twoo longsquaresH.K,are iuste equall to the other twoo squaresC.andF.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 at 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 falleth, is thrise vsed, yet is his square but ones 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, would credite the possibilitie 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 circumferẽce of a circle, and a right line drawen frome the one to the other, that line must needes fal within the circle.Example.see textThe circle isA.B.C.D,the ij. poinctes areA.B,the righteline that is drawenne frome the one to the other, is the lineA.B,which as you see, must needes lyghte within the circle. So if you putte the pointes to beA.D,orD.C,orA.C,otherB.C,orB.D,in any 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. How be it, 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 he 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.see textExample.The circle isA.B.C.D,the line that passeth by the centre, isA.E.C,the line that goeth beside the centre isD.B.Nowesaye I, that the lineA.E.C,dothe cutte that other lineD.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, accordinge as the Theoreme would.The xlix. Theoreme.If twoo right lines drawen in a circle doo crosse one an other, and doo notpasse bythe centre, euery of them dothe not deuide the other into equall partions.Example.see textThe circle isA.B.C.D,and the centre isE,the one lineA.C,and the other isB.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 cuytte the other into equall portions. For as you may easily iudge,A.C.hath one portiõ lõger and an other shorter, and so like wiseB.D.Howbeit, it is not so to be vnderstãd, but one of them may be deuided into ij. euẽ parts,but bothe to bee cutte equally in the middle, is not possible, onles both passe through the cẽtre, therfore much rather whẽ 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.see textExample.This theoreme seemeth of it selfe so manifest, that it neadeth nother demonstration nother declaraciõ. Yet for the plaine vnderstanding of it, I haue sette forthe a figure here, where ij. circles be drawẽ, so that one of them doth crosse the other (as you see) in the pointesB.andG,and their centres appear at the firste sighte to bee diuers. For the centre of the one isF,and the centre of the other isE,which diffre as farre asondre as the edges of the circles, where they bee most distaunte in sonder.The Li. Theoreme.If two circles be so drawen, that one of them do touche the other, then haue they not one centre.see textExample.There are two circles made, as you see, the one isA.B.C,and hath his centre byG,the other isB.D.E,and his centre is byF,so that it is easy enough to perceaue that their centres doe dyffer as muche a sonder, as the halfe diameter of the greater circle is lõger then the half diameter of the lesser circle. And so must it needes be thought and said of all other circles in lyke kinde.The .lij. theoreme.If a certaine pointe be assigned in the diameter of a circle, distant from the centre of the said circle, and from that pointe diuerse lynes drawen to the edge and circumference of the same circle, the longest line is that whiche passeth by the centre, and the shortest is the residew of the same line. And of al the other lines that is euer the greatest, that is nighest to the line, which passeth by the centre. And cõtrary waies, that is the shortest, that is farthest from it. And amongest thẽ all there can be but onely .ij. equall together, and they must nedes be so placed, that the shortest line shall be in the iust middle betwixte them.Example.see textThe circle isA.B.C.D.E.H,and his centre isF,the diameter isA.E,in whiche diameter I haue taken a certain point distaunt from the centre, and that pointe isG,from which I haue drawen .iiij. lines to the circumference, beside the two partes of the diameter, whiche maketh vp vi. lynes in all. Nowe for the diuersitee in quantitie of these lynes, I saie accordyng to the Theoreme, that the line whiche goeth by the centre is the longest line, that is to saie,A.G,and the residewe of the same diameter beeyngG.E,is the shortest lyne. And of all the other that lyne is longest, that is neerest vnto that parte of the diameter whiche gooeth by the centre, and that is shortest, that is farthest distant from it, wherefore I saie, thatG.B,is longer thenG.C,and therfore muche more longer thenG.D,sithG.C,also is longer thenG.D,and by this maie you soone perceiue, that it is not possible to drawe .ij. lynes on any one side of the diameter, whiche might be equall in lengthe together, but on the one side of the diameter maie you easylie make one lyne equall to an other, on the other side of the same diameter, as you see in this exampleG.H,to bee equall toG.D,betweene whiche the lyneG.E,(as the shortest in all the circle) doothe stande euen distaunte from eche of them, and it is the precise knoweledge of their equalitee, if they beequallydistaunt from one halfe of the diameter. Where as contrary waies if the one be neerer to any one halfe of the diameter then the other is, it is not possible that they two may be equall in lengthe, namely if they dooe ende bothe in the circumference of thecircle, and be bothe drawen from one poynte in the diameter, so that the saide poynte be (as the Theoreme doeth suppose) somewhat distaunt from the centre of the said circle. For if they be drawen from the centre, then must they of necessitee be all equall, howe many so euer they bee, as the definition of a circle dooeth importe, withoute any regarde how neere so euer they be to the diameter, or how distante from it. And here is to be noted, that in this Theoreme, by neerenesse and distaunce is vnderstand the nereness and distaunce of the extreeme partes of those lynes where they touche the circumference. For at the other end they do all meete and touche.The .liij. Theoreme.If a pointe bee marked without a circle, and from it diuerse lines drawen crosse the circle, to the circumference on the other side, so that one of them passe by the centre, then that line whiche passeth by the centre shall be the loongest of them all that crosse the circle. And of the other lines those are longest, that be nexte vnto it that passeth by the centre. And those ar shortest, that be farthest distant from it. But among those partes of those lines, whiche ende in the outewarde circumference, that is most shortest, whiche is parte of the line that passeth by the centre, and amongeste theothere eche, of thẽ, the nerer they are vnto it, the shorter they are, and the farther from it, the longer they be. And amongest them all there can not be more then .ij. of any one lẽgth, and they two muste be on the two contrarie sides of the shortest line.Example.see textTake the circle to beA.B.C,and the point assigned without it to beD.Now say I, that if there be drawen sundrie lines fromD,and crosse the circle, endyng in the circumference on the cõtrary side, as here you see,D.A, D.E, D.F,andD.B,then of all these lines the longest must needes beD.A,which goeth by the centre of the circle, and the nexte vnto it, that isD.E,is the longest amongest the rest. And contrarie waies,D.B,is the shorteste, because it is farthest distaunt fromD.A.And so maie you iudge ofD.F,because it is nerer vntoD.A,then isD.B,therefore is it longer thenD.B.And likewaies because it is farther of fromD.A,then isD.E,therfore is it shorter thenD.E.Now for those partes of the lines whiche bee withoute the circle (as you see)D.C,is theshortest. becauseit is the parte of that line which passeth by thecentre, AndD.K,is next to it in distance, and therefore also in shortnes, soD.G,is farthest from it in distance, and therfore is the longest of them. NowD.H,beyng nerer thenD.G,is also shorterthen it,and beynge farther of, thenD.K,is longer then it. So that for this parte of the theoreme (as I think) you do plainly perceaue the truthe thereof, so the residue hathe no difficulte. For seing that the nearer any line is toD.C,(which ioyneth with the diameter) the shorter it is and the farther of from it, the longer it is. And seyng two lynes can not be of like distaunce beinge bothe on one side, therefore if they shal be of one lengthe, and consequently of one distaunce, they must needes bee on contrary sides of the saide lineD.C.And so appeareth the meaning of the whole Theoreme.And of this Theoreme dothe there folowe an otherlyke. whicheyou maye calle other a theoreme by it selfe, or else a Corollary vnto this laste theoreme, I passe not so muche for the name. But his sentence is this:when so euer any lynes be drawen frome any pointe, withoute a circle, whether they crosse the circle, or eande in theutteredge of his circumference, those two lines that bee equally distaunt from the least line are equal togither, and contrary waies, if they be equall togither, they ar also equally distant from that least line.For the declaracion of this proposition, it shall not need to vse any other example, then that which is brought for the explication of this laste theoreme, by whiche you may without any teachinge easyly perceaue both the meanyng and also the truth of this proposition.The Liiij. Theoreme.If a point be set forthe in a circle, and frõ that pointe vnto the circumference many lines drawen, of which more then two are equal togither, then is that point the centre of that circle.Example.see textThe circle isA.B.C,and within it I haue sette fourth for an example three prickes, which areD.E.andF,from euery one of them I haue drawẽ (at the leaste) iiij. lines vnto the circumference of the circle but fromeD,I haue drawen more, yet maye it appear readily vnto your eye, that of all the lines whiche be drawen fromE.andF,vnto the circumference, there are but twoo equall, and more can not bee, forG.E.norE.H.hath none other equal to theim, nor canne not haue any beinge drawen from the same pointE.No more canL.F,orF.K,haue anye line equall to either of theim, beinge drawen from the same pointeF.And yet from either of those two poinctes are there drawen twoo lines equall togither, asA.E,is equall toE.B,andB.F,is equall toF.C,but there can no third line be drawen equall to either of these two couples, and that is by reason that they be drawen from a pointe distaunte from the centre of the circle. But fromD,althoughe there be seuen lines drawen, to the circumference, yet all bee equall, bicause it is the centre of the circle. And therefore if you drawe neuer so mannye more from it vnto the circumference, all shall be equal, so that this is the priuilege (as it were of the centre) and therfore no other point can haue aboue two equal lines drawen from it vnto the circumference. And from allpointesyou maye drawe ij. equall lines to the circumference of the circle, whether that pointe be within the circle or without it.The lv. Theoreme.No circle canne cut an other circle in morepointes then two.see textExample.The first circle isA.B.F.E,the second circle isB.C.D.E,and they crosse one an other inB.and inE,and in no more pointes. Nother is it possible that they should, but other figures ther be, which maye cutte a circle in foure partes, as you se in this exãple. Where I haue set forthe one tunne forme, and one eye forme, and eche of them cutteth euery of their two circles into foure partes. But as they be irregulare formes, that is to saye, suche formes as haue no precise measure nother proportion in their draughte, so can there scarcely be made any certaine theorem of them. But circles are regulare formes, that is to say, such formes as haue in their protracture a iuste and certaine proportion, so that certain and determinate truths may be affirmed of them, sith they ar vniforme and vnchaungable.The lvi. Theoreme.If two circles be so drawen, that the one be within the other, and that they touche one an other: If a line bee drawen by bothe their centres, and so forthe in lengthe, that line shall runne to that pointe, where the circles do touche.see textExample.The one circle, which is the greattest and vttermost isA.B.C,the other circle that is yelesser, and is drawen within the firste, isA.D.E.The cẽtre of the greater circle isF,and the centre of the lesser circle isG,the pointe where they touche isA.And now you may see the truthe of the theoreme so plainely, that it needeth no farther declaracion. For you maye see, that drawinge a line fromF.toG,and so forth in lengthe, vntill it come to the circumference, it wyll lighte in the very poincteA,where the circles touche one an other.The Lvij. Theoreme.If two circles bee drawen so one withouteanother, that their edges doo touche and a right line bee drawnenne frome the centre of the one to the centre of the other, that line shall passe by the place of their touching.Example.see textThe firste circle isA.B.E,and his centre isK,Thesecõd circle isD.B.C,and his cẽtre isH,the point wher they do touch isB.Nowe doo you se that the lineK.H,whiche is drawenfromK,that is centre of the firste circle, vntoH,beyng centre of the second circle, doth passe (as it must nedes by the pointeB,) whiche is the verye poynte wher they do to touche together.The .lviij. theoreme.One circle can not touche an other in more pointes then one, whether they touche within or without.see textExample.For the declaration of this Theoreme, I haue drawen iiij. circles, the first isA.B.C,and his centreH.the second isA.D.G,and his centreF.the third isL.M,and his centreK.the .iiij. isD.G.L.M,and his centreE.Nowe as you perceiue the second circleA.D.G,toucheth the first in the inner side, in so much as it is drawen within the other, and yet it toucheth him but in one point, that is to say inA,so lykewaies the third circleL.M,is drawen without the firste circle and touchethhym, as you maie see, but in one place. And now as for the .iiij. circle, it is drawen to declare the diuersitie betwene touchyng and cuttyng, or crossyng. For one circle maie crosse and cutte a great many other circles, yet can be not cutte any one in more places then two, as the fiue and fiftie Theoreme affirmeth.The .lix. Theoreme.In euerie circle those lines are to be counted equall, whiche are in lyke distaunce from thecentre, Andcontrarie waies they are in lyke distance from the centre, whiche be equall.see textExample.In this figure you see firste the circle drawen, whiche isA.B.C.D,and his centre isE.In this circle also there are drawen two lines equally distaunt from the centre, for the lineA.B,and the lineD.C,are iuste of one distaunce from the centre, whiche isE,and therfore are they of one length. Again thei are of one lengthe (as shall be proued in the boke of profes) and therefore their distaunce from the centre is all one.The lx. Theoreme.In euerie circle the longest line is the diameter, and of all the other lines, thei are still longestthat be nexte vnto the centre, and they be the shortest, that be farthest distaunt from it.Example.see textIn this circleA.B.C.D,I haue drawen first the diameter, whiche isA.D,whiche passeth (as it must) by the centreE,Thenhaue I drawen ij. other lines asM.N,whiche is neerer the centre, andF.G,that is farther from the centre.The fourth line also on the other side of the diameter, that isB.C,is neerer to the centre then the lineF.G,for it is of lyke distance as is the lyneM.N.Nowe saie I, thatA.D,beyng the diameter, is the longest of all those lynes, and also of any other that maie be drawen within that circle, Andthe other lineM.N,is longer thenF.G.Also the lineF.G,is shorter then the lineB.C,for because it is farther from the centre then is the lyneB.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 twothynges, Asfor exaumple, If two squares be sette foorthe, whereof the one containeth in it fiue square feete, 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 measureof 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 circumferẽce 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.see textIn this circleA.B.C,the diameter isA.C,the perpendicular line, which maketh a right angle with the diameter, isC.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 thyngis so plainly seene of the eye, that it needeth no farther declaracion. For euery man wil easily consent, that betwene the croked lineA.F,(whiche is a parte of the circumferẽce of the circle) andA.E(which is the said perpẽdicular line) there can none other line bee drawen in that place where they make the angle. Nowe for the residue of the theoreme. The angleD.A.B,which is made in the semicircle, is greater then anye sharpe angle that may bee made of ryghtelines. andyet is it a sharpe angle also, in as much as it is lesser then a right angle, which is the angleE.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 lineE.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 greater then the other, for they twoo doo make the whole right angle, so that if anye corner coulde be 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 ofthecircle to the pointe where they touche, thatline whiche is drawenne frome the centre, shall be a perpendicular line to the touch line.Example.see textThe circle isA.B.C,and his centre isF.The touche line isD.E,and the point wher they touch isC.Now by reason that a right line is drawen frome the centreF.vntoC,which is the point of the touche, therefore saith the theoreme, that the sayde lineF.C,muste needes bee a perpendicular line vnto the touche lineD.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.see textThe circle isA.B.C,and the centre of it isG.The touche line isD.C.E,and the pointe where it toucheth, isC.Noweit 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 cẽtre in him. For if the saide line shoulde go beside the centre, asF.C.doth, then dothe it not make righte angles with the touche line, which in thetheoremeis 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 circũferẽce.see textExample.ThecircleisA.B.C.D,and his centre is E: the angle on the centre isC.E.D, and the angle on the circumference isC.A.Dttheir commen ground line, isC.F.D.Now say I that the angleC.E.D,whiche is on the centre, is twise so greate as the angleC.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.see textBefore I declare this theoreme by example, it shall bee needefull to declare, what is it to be vnderstande by the wordes in this theoreme. For the sentence canne not be knowen, onles theuerymeaning of the wordes be firste vnderstand. Therefore when it speaketh of angles 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 circle. So that if any angle do not touch the circumference, or if the lines that inclose that angle, doo not ende in the extremities of the corde line, but ende other in some other part of the said 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 said cantle of the circle. And this promised, nowe will I cumme to the meaninge of the theoreme. I sette forthe a circle whiche isA.B.C.D,and his centreE,in this circle I drawe a lineD.C,whereby there ar made two cantels, a more and a lesser. The lesser isD.E.C,and the geater isD.A.B.C.In this greater cantle I drawe two angles, the firste isD.A.C,and the second isD.B.Cwhich two angles by reason they are made bothe in one cantle of a circle (that is the cantleD.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 isD.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.
see textExample.Fyrst before I declare the examples, it shal be mete to shew the true vnderstãdyng of this theorem.Bias lyne.Therfore by theBias line, I meane that lyne, whiche in any square figure dooth runne from corner to corner. And euery square which is diuided by that bias line into equall halues from corner to corner (that is to say, into .ij. equall triangles) those be countedto stande aboute one bias line, and the other squares, whiche touche that bias line, with one of their corners onely, those doo I callFyll squares,Fyll squares.accordyng to the greke name, which isanapleromata,ἀναπληρώματαand called in latinsupplementa, bycause that they make one generall square, includyng and enclosyng the other diuers squares, as in this exãpleH.C.E.N.is one square likeiamme, andL.M.G.C.is an other, whiche bothe are made aboute one bias line, that isN.M,thanK.L.H.C.andC.E.F.G.are .ij. fyll squares, for they doo fyll vp the sydes of the .ij.fyrste square lykeiammes, in suche sorte, that all them foure is made one greate generall squareK.M.F.N.Nowe to the sentence of the theoreme, I say, that the .ij. fill squares,H.K.L.C.andC.E.F.G.are both equall togither, (as it shall bee declared in the booke of proofes) bicause they are the fill squares of two likeiammes made aboute one bias line, as the exaumple sheweth. Conferre the twelfthe conclusion with this theoreme.The xxxiij. Theoreme.In all right anguled triangles, the square of that side whiche lieth against the right angle, is equall to the .ij. squares of both the other sides.Example.see textA.B.C.is a triangle, hauing a ryght angle inB.Wherfore it foloweth, that the square ofA.C,(whiche is the side that lyeth agaynst the right angle) shall be as muche as the two squares ofA.B.andB.C.which are the other .ij. sides.¶By thesquareof any lyne, you muste vnderstande a figure made iuste square, hauyng all his iiij. sydes equall to that line, whereof it is the square, so isA.C.F,the square ofA.C.LykewaisA.B.D.is the square ofA.B.AndB.C.E.is the square ofB.C.Now by the numbre of the diuisions in eche of these squares, may you perceaue not onely what the square of any line is called, but also that the theoreme is true, and expressed playnly bothe by lines and numbre. For as you see, the greatter square (that isA.C.F.) hath fiue diuisions on eche syde, all equall togyther, and those in the whole square are twenty and fiue. Nowe in the left square, whiche isA.B.D.there are but .iij. of those diuisions in one syde, and that yeldeth nyne in the whole. So lykeways you see in the meane squareA.C.E.in euery syde .iiij. partes, whiche in the whole amount vnto sixtene. Nowe adde togyther all the partes of the two lesser squares, that is to saye, sixtene and nyne, and you perceyue that they make twenty and fiue, whyche is an equall numbre to the summe of the greatter square.By this theoreme you may vnderstand a redy way to know the syde of any ryght anguled triangle that is vnknowen, so that you knowe the lengthe of any two sydes of it. For by tournynge the two sydes certayne into theyr squares, and so addynge them togyther, other subtractynge the one from the other (accordyng as in the vse of these theoremes I haue sette foorthe) and then fyndynge the roote of the square that remayneth, which roote (I meane the syde of the square) is the iuste length of the vnknowen syde, whyche is sought for. But this appertaineth to the thyrde booke, and therefore I wyll speake no more of it at this tyme.The xxxiiij. Theoreme.If so be it, that in any triangle, the square of the one syde be equall to the .ij. squares of the other .ij. sides, than must nedes that corner be a right corner, which is conteined betwene those two lesser sydes.Example.As in the figure of the laste Theoreme, bicauseA.C,made in square, is asmuch as the square ofA.B,and also as the square ofB.C.ioyned bothe togyther, therefore the angle that is inclosed betwene those .ij. lesser lynes,A.B.andB.C.(that is to say) the angleB.whiche lieth against the lineA.C,must nedes be a ryght angle. This theoreme dothe so depende of the truthe of the laste, that whan you perceaue the truthe of the one, you can not iustly doubt of the others truthe, for they conteine one sentence, contrary waies pronounced.The .xxxv. theoreme.If there be set forth .ij. right lines, and one of them parted into sundry partes, how manyor few so euer they be, the square that is made of those ij. right lines proposed, is equal to all the squares, that are made of the vndiuided line, and euery parte of the diuided line.Example.see textThe ij. lines proposed arA.B.andC.D,and the lyneA.B.is deuided into thre partes byE.andF.Now saith this theoreme, that the square that is made of those two whole linesA.B.andC.D,so that the lineA.B.stãdeth for the lẽgth of the square, and the other lineC.D.for the bredth of the same. That square (I say) wil be equall to all the squares that be made, of the vndiueded lyne (which isC.D.) and euery portion of the diuided line. And to declare that particularly, Fyrst I make an other lineG.K,equall to the line.C.D,and the lineG.H.to be equal to the lineA.B,and to bee diuided into iij. like partes, so thatG.M.is equall toA.E,andM.N.equal toE.F,and then musteN.H.nedes remaine equall toF.B.Then of those ij. linesG.K,vndeuided, andG.H.which is deuided, I make a square, that isG.H.K.L,Inwhich square if I drawe crosse lines frome one side to the other, according to the diuisions of the lineG.H,then will it appear plaine, that the theoreme doth affirme. For the first squareG.M.O.K,must needes be equal to the square of the lineC.D,and the first portiõ of the diuided line, which isA.E,for bicause their sides are equall. And so the secondesquare that isM.N.P.O,shall be equall to the square ofC.D,and the second part ofA.B,that isE.F.Also the third square which isN.H.L.P,must of necessitee be equal to the square ofC.D,andF.B,bicause those lines be so coupeled that euery couple are equall in the seuerall figures. And so shal you not only in this example, but in all other finde it true, that if one line be deuided into sondry partes, and an other line whole and vndeuided, matched with him in a square, that square which is made of these two whole lines, is as muche iuste and equally, as all the seuerall squares, whiche bee made of the whole line vndiuided, and euery part seuerally of the diuided line.The xxxvi. Theoreme.If a right line be parted into ij. partes, as chaunce may happe, the square that is made of the whole line, is equall to bothe the squares that are made of thesameline, and the twoo partes of it seuerally.Example.see textThe line propounded beyngA.B.and deuided, as chaunce happeneth, inC.into ij. vnequall partes, I say that the square made of the hole lineA.B,is equal to the two squares made of the same line with the twoo partes of itselfe, as withA.C,and withC.B,for the squareD.E.F.G.is equal to the two other partial squares ofD.H.K.GandH.E.F.K,but that the greater square is equall to the square of the whole lineA.B,and thepartiall squares equall to the squares of the second partes of the same line ioyned with the whole line, your eye may iudg without muche declaracion, so that I shall not neede to make more exposition therof, but that you may examine it, as you did in the laste Theoreme.The xxxvij.Theoreme.If a right line be deuided by chaunce, as it maye happen, the squarethatis made of the whole line, and one of the partes of it which soeuer it be, shal be equall to that square that is made of the ij. partes ioyned togither, and to an other square made of that part, which was before ioyned with the whole line.Example.see textThe lineA.B.is deuided inC.into twoo partes, though not equally, of which two partes for an example I take the first, that isA.C,and of it I make one side of a square, as for exampleD.G.accomptinge those two lines to be equall, the other side of the square isD.E,whiche is equall to the whole lineA.B.Nowmay it appeare, to your eye, that the great square made of the whole lineA.B,and of one of his partes that isA.C,(which is equall withD.G.) is equal to two partiall squares, whereof the one is made of the saide greatter portionA.C,in as muche as not onlyD.G,beynge one of his sides, but alsoD.H.beinge the other side, are eche of them equall toA.C.The second square isH.E.F.K,in which the one sideH.E,is equal toC.B,being the lesser parte of the line,A.B,andE.F.is equall toA.C.which is the greater parte of the same line. So that those two squaresD.H.K.GandH.E.F.K,bee bothe of them no more then the greate squareD.E.F.G,accordinge to the wordes of the Theoreme afore saide.The xxxviij. Theoreme.If a righte line be deuided by chaunce, into partes, the square that is made of that whole line, is equall to both the squares that ar made of eche parte of the line, and moreouer to two squares made of the one portion of the diuided line ioyned with the other in square.see textExample.The labels A and B were transposed in the illustration as an alternative to transposing all occurrences of A.C and C.B in the text.Lette the diuided line beeA.B,and parted inC,into twoo partes: Nowe saithe the Theoreme, that the square of the whole lyneA.B,is as mouche iuste as the square ofA.C,and the square ofC.B,eche by it selfe, and more ouer by as muche twise, asA.C.andC.B.ioyned in one square will make. For as you se, the great squareD.E.F.G,conteyneth in hym foure lesser squares, of whiche the first and the greatest isN.M.F.K,and is equall to the square of the lyneA.C.The second square is the lest of them all, that isD.H.L.N,and it is equall to the square of the lineC.B.Then are there two other longe squares both of one bygnes, that isH.E.N.M.andL.N.G.K,eche of them both hauyng .ij. sides equall toA.C,the longer parte of the diuided line, and there other two sides equall toC.B,beeyng he shorter parte of the said lineA.B.So is that greatest square, beeyng made of the hole lyneA.B,equal to the ij. squares of eche of his partes seuerally, and more by as muche iust as .ij. longe squares, made of the longer portion of the diuided lyne ioyned in square with the shorter parte of the same diuided line, as the theoreme wold. And as here I haue put an example of a lyne diuided into .ij. partes, so the theoreme is true of all diuided lines, of what number so euer the partes be, foure, fyue, orsyxe. etc.Thistheoreme hath great vse, not only in geometrie, but also in arithmetike, as herafter I will declare in conuenient place.The .xxxix. theoreme.If a right line be deuided into two equall partes, and one of these .ij. partes diuided agayn into two other partes, as happeneth the longe square that is made of the thyrd or later part of that diuided line, with the residue of the same line, and the square of the mydlemoste parte, are bothe togither equall to the square of halfe the firste line.Example.see textThe lineA.B.is diuided into ij. equal partes inC,and that parteC.B.is diuided agayne as hapneth inD.Wherfore saith the Theorem that the long square made ofD.B.andA.D,with the square ofC.D.(which is the mydle portion) shall bothe be equall to the square of half the lyneA.B,that is to saye, to the square ofA.C,or els ofC.D,which make all one. The long squareF.G.N.O.whiche is the longe square that the theoreme speaketh of, is made of .ij. long squares, wherof the fyrst isF.G.M.K,and the seconde isK.N.O.M.The square of the myddle portion isL.M.O.P.and the square of the halfe of the fyrste lyne isE.K.Q.L.Nowe by the theoreme, that longe squareF.G.N.O,with the iuste squareL.M.O.P,muste bee equall to the greate squareE.K.Q.L,whyche thynge bycause it seemeth somewhat difficult to vnderstande, althoughe I intende not here to make demonstrations of the Theoremes, bycause it is appoynted to be done in the newe edition of Euclide, yet I wyll shew you brefely how the equalitee of the partes doth stande. And fyrst I say, that where the comparyson of equalitee is made betweene the greate square (whiche is made of halfe the lineA.B.) and two other, where of the fyrst is the longe squareF.G.N.O,and the second is the full squareL.M.O.P,which is one portion of the great square all redye, and so is that longe squareK.N.M.O,beynge a parcell also of the longe squareF.G.N.O,Wherforeas those two partes are common to bothe partes compared in equalitee, and therfore beynge bothe abated from eche parte, if the reste of bothe the other partes bee equall, than were those whole partes equall before: Nowe the reste of the great square, thosetwo lesser squares beyng taken away,is that longe squareE.N.P.Q,whyche is equall to the long squareF.G.K.M,beyng the rest of the other parte. And that they two be equall, theyr sydes doo declare. For the longest lynes that isF.KandE.Qare equall, and so are the shorter lynes,F.G,andE.N,and so appereth the truthe of the Theoreme.The .xl. theoreme.If a right line be diuided into .ij. euen partes, and an other right line annexed to one ende of that line, so that it make one righte line with the firste. The longe square that is made of this whole line so augmented, and the portion that is added, with the square of halfe the right line, shall be equall to the square of that line, whiche is compounded of halfe the firste line, and the parte newly added.Example.see textThe fyrst lynepropoundedisA.B,and it is diuided into ij. equall partes inC,and an other ryght lyne, I meaneB.Dannexed to one ende of the fyrste lyne.Nowesay I, that the long squareA.D.M.K,is made of the whole lyne so augmẽted, that isA.D,and the portiõ annexed, ytisD.M,forD.Mis equall toB.D,wherfore ytlong squareA.D.M.K,with thesquare of halfe the first line, that isE.G.H.L,is equall to the great squareE.F.D.C.whiche square is made of the lineC.D.that is to saie, of a line compounded of halfe the first line, beyngC.B,and the portion annexed, that isB.D.And it is easyly perceaued, if you consyder that the longe squareA.C.L.K.(whiche onely is lefte out of the great square) hathanotherlonge square equall to hym, and to supply his steede in the great square, and that isG.F.M.H.For their sydes be of lyke lines in length.The xli. Theoreme.If a right line bee diuided by chaunce, the square of the same whole line, and the square of one of his partes are iuste equall to the lõg square of the whole line, and the sayde parte twise taken, and more ouer to the square of the other parte of the sayd line.Example.see textA.B.is the line diuided inC.AndD.E.F.G,is the square of the whole line,D.H.K.M.is the square of the lesser portion (whyche I take for an example) and therfore must bee twise reckened. Nowe I saye that those ij. squares are equall to two longe squares of the whole lineA.B,and his sayd portionA.C,and also to the square of the other portion of the sayd first line, whiche portion isC.B,and his squareK.N.F.L.In this theoreme there is no difficultie, if you cõsyder that the litle squareD.H.K.M.is .iiij. tymes reckened, that is to say, fyrst of all as a parte of the greatest square, whiche isD.E.F.G.Secondly he is reknedby him selfe. Thirdely he is accompted as parcell of the long squareD.E.N.M,And fourthly he is taken as a part of the other long squareD.H.L.G,so that in as muche as he is twise reckened in one part of the comparisõ of equalitee, and twise also in the second parte, there can rise none occasion of errour or doubtfulnes therby.The xlij. Theoreme.If a right line be deuided as chance happeneth the iiij. long squares, that may be made of that whole line and one of his partes with the square of the other part, shall be equall to the square that is made of the whole line and the saide first portion ioyned to him in lengthe as one whole line.Example.see textThe firste line isA.B,and is deuided byC.into two vnequall partes as happeneth.Thelong square of yt, and his lesser portionA.C,is foure times drawen, the first isE.G.M.K,the seconde isK.M.Q.O,the third isH.K.R.S,and the fourthe isK.L.S.T.And where as it appeareth that one of the little squares (I meaneK.L.P.O) is reckened twise, ones as parcell of the second long square and agayne as parte of thethirde long square, to auoide ambiguite, you may place one insteede of it, an other square of equalitee, withit. thatis to saye,D.E.K.H,which was at no tyme accompting as parcell of any one of them, and then haue you iiij. long squares distinctly made of the whole lineA.B,and his lesser portionA.C.And within them is there a greate full squareP.Q.T.V.whiche is the iust square ofB.C,beynge the greatter portion of the lineA.B.And that those fiue squares doo make iuste as muche as the whole square of that longer lineD.G,(whiche is as longe asA.B,andA.C.ioyned togither) it may be iudged easyly by the eye, sith that one greate square doth comprehẽd in it all the other fiue squares, that is to say, foure long squares (as is before mencioned) and one fullsquare. whichis the intent of the Theoreme.The xliij. Theoreme.If a right line be deuided into ij. equal partes first, and one of those parts againintoother ij. parts, as chaũce hapeneth, the square that is made of the last part of the line so diuided, and the square of the residue of that whole line, are double to the square of halfe that line, and to the square of the middle portion of the same line.Example.see textThe line to be deuided isA.B,and is parted inC.into two equall partes, and thenC.B,is deuided againe into two partes inD,so that the meaninge of the Theoreme, is that thesquare ofD.B.which is the latter parte of the line, and the square ofA.D,which is the residue of the whole line. Those two squares, I say, ar double to the square of one halfe of the line, and to the square ofC.D,which is the middle portion of those thre diuisions. Which thing that you maye more easilye perceaue, I haue drawen foure squares, whereof the greatest being marked withE.is the square ofA.D.The next, which is marked withG,is the square of halfe the line, that is, ofA.C,Andthe other two little squares marked withF.andH,be both of one bignes, by reason that I did diuideC.B.into two equall partes, so that you amy take the squareF,for the square ofD.B,and the squareH,for the square ofC.D.Now I thinke you doubt not, but that the squareE.and the squareF,ar double so much as the squareG.and the squareH,which thing theeasyeris to be vnderstande, bicause that the greate square hath in his side iij. quarters of the firste line, which multiplied by itselfe maketh nyne quarters, and the squareF.containeth but one quarter, so that bothe doo make tenne quarters.ThenG.contayneth iiij. quarters, seynge his side containethtwoo, andH.containeth but one quarter, whiche both makebut fiue quarters, and that is but halfe of tenne.Whereby you may easylye coniecture,that the meanynge of the the-oreme is verified in thefigures of this ex-ample.The xliiij. Theoreme.If a right line be deuided into ij. partes equally, and an other portion of a righte lyne annexed to that firste line, the square of this whole line so compounded, and the square of the portion that is annexed, ar doule as much as the square of the halfe of the firste line, and the square of the other halfe ioyned in one with the annexed portion, as one whole line.Example.see textThe line isA.B,and is diuided firste into twoo equal partes inC,and thẽ is there annexed to it an other portion whiche isB.D.Now saith the Theoreme, that the square ofA.D,and the square ofB.D,ar double to the square ofA.C,and to the square ofC.D.The lineA.B.cõtaining four partes, then must needes his halfe containe ij. partes of such partes I supposeB.D.(which is the ãnexed line) to containe thre, so shal the hole line cõprehend vij. parts, and his square xlix. parts, where vnto if you ad yesquareof the annexed lyne, whiche maketh nyne, than those bothe doo yelde, lviij. whyche must be double to the square of the halfe lyne with the annexed portion. The halfe lyne by it selfe conteyneth but .ij. partes, and therfore his square dooth make foure. The halfe lyne with the annexed portion conteyneth fiue, and the square of it is .xxv, now put foure to .xxv, and it maketh iust .xxix, the euen halfe of fifty and eight, wherby appereth the truthe of the theoreme.The .xlv. theoreme.In all triangles that haue a blunt angle, the square of the side that lieth against the blunt angle, is greater than the two squares of the other twoo sydes, by twise as muche as is comprehended of the one of those .ij. sides (inclosyng the blunt corner) and the portion of the same line, beyng drawen foorth in lengthe, which lieth betwene the said blunt corner and a perpendicular line lightyng on it, and drawen from one of the sharpe angles of the foresayd triangle.Example.For the declaration of this theoreme and the next also, whose vse are wonderfull in the practise of Geometrie, and in measuryng especially, it shall be nedefull to declare that euery triangle that hath no ryght angle as those whyche are called (as in the boke of practise is declared) sharp cornered triangles, and bluntcorneredtriangles, yet may they be brought to haue a ryght angle, eyther by partyng them into two lesser triangles,or els by addyng an other triangle vnto them, whiche may be a great helpe for the ayde of measuryng, as more largely shall be sette foorthe in the boke of measuryng. But for this present place, this forme wyll I vse, (whiche Theon also vseth) to adde one triangle vnto an other, to bryng the blunt cornered triangle into a ryght angled triangle, whereby the proportion of the squares of the sides in suche a blunt cornered triangle may the better bee knowen.see textFyrst therfore I sette foorth the triangleA.B.C,whose corner byC.is a blunt corner as you maye well iudge, than to make an other triangle of yt with a ryght angle, I must drawe forth the sideB.C.vntoD,and frõ the sharp corner byA.I brynge a plumbe lyne or perpẽdicular onD.And so is there nowe a newe triangleA.B.D.whose angle byD.is a right angle. Nowe accordyng to the meanyng of the Theoreme, I saie, that in the first triangleA.B.C,because it hath a blunt corner atC,the square of the lineA.B.whiche lieth against the said blunte corner, is morethen the square of the lineA.C,and also of the lyneB.C,(whiche inclose the blunte corner) by as muche as will amount twise of the lineB.C,and that portionD.C.whiche lieth betwene the blunt angle byC,and the perpendicular lineA.D.The square of the lineA.B,is the great square marked withE.The square ofA.C,is the meane square marked withF.The square ofB.C,is the least square marked withG.And the long square marked withK,is sette in steede of two squares made ofB.C,andC.D.For as the shorter side is the iuste lengthe ofC.D,so the other longer side is iust twise so longe asB.C,WherforeI saie now accordyng to the Theoreme, that the greatte squareE,is more then the other two squaresF.andG,by the quantitee of the longe squareK,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.Labeling of rectangle K is conjectural. Other configurations of B, C and D will also fit, but the printed illustration (B and D on the right, nothing on the left) will not. The text requires a rectangle with BC as one side and CD as the other, doubled as in the next illustration (H.K).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.see textFyrst I sette foorth the triangleA.B.C,and in yt I draw a plũbe line from the angleC.vnto the lineA.B,and it lighteth inD.Nowe by the theoreme the square ofB.C.is not so muche as the square of the other two sydes, that ofB.A.and ofA.C.by as muche as is twise conteyned in the lõg square made ofA.B,andA.D, A.B.beyng the line or syde on which the perpendicular line falleth, andA.D.beeyng that portion of the same line whiche doth lye betwene the perpendicular line, and the sayd sharpe angle limitted, whiche angle is byA.For declaration of the figures, the square marked withE.is the square ofB.C,whiche is the syde that lieth agaynst the sharpe angle, the square marked withG.is the square ofA.B,and the square marked withF.is the square ofA.C,and the two longe squares marked withH.K,are made of the hole lineA.B,and one of his portionsA.D.And truthe it is that the squareE.is lesser than the other two squaresC.andF.by the quantitee of those two long squaresH.andK.Wherby you may consyder agayn, an other proportion of equalitee,that is to saye, that the squareE.with the twoo longsquaresH.K,are iuste equall to the other twoo squaresC.andF.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 at 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 falleth, is thrise vsed, yet is his square but ones 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, would credite the possibilitie 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 circumferẽce of a circle, and a right line drawen frome the one to the other, that line must needes fal within the circle.Example.see textThe circle isA.B.C.D,the ij. poinctes areA.B,the righteline that is drawenne frome the one to the other, is the lineA.B,which as you see, must needes lyghte within the circle. So if you putte the pointes to beA.D,orD.C,orA.C,otherB.C,orB.D,in any 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. How be it, 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 he 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.see textExample.The circle isA.B.C.D,the line that passeth by the centre, isA.E.C,the line that goeth beside the centre isD.B.Nowesaye I, that the lineA.E.C,dothe cutte that other lineD.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, accordinge as the Theoreme would.The xlix. Theoreme.If twoo right lines drawen in a circle doo crosse one an other, and doo notpasse bythe centre, euery of them dothe not deuide the other into equall partions.Example.see textThe circle isA.B.C.D,and the centre isE,the one lineA.C,and the other isB.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 cuytte the other into equall portions. For as you may easily iudge,A.C.hath one portiõ lõger and an other shorter, and so like wiseB.D.Howbeit, it is not so to be vnderstãd, but one of them may be deuided into ij. euẽ parts,but bothe to bee cutte equally in the middle, is not possible, onles both passe through the cẽtre, therfore much rather whẽ 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.see textExample.This theoreme seemeth of it selfe so manifest, that it neadeth nother demonstration nother declaraciõ. Yet for the plaine vnderstanding of it, I haue sette forthe a figure here, where ij. circles be drawẽ, so that one of them doth crosse the other (as you see) in the pointesB.andG,and their centres appear at the firste sighte to bee diuers. For the centre of the one isF,and the centre of the other isE,which diffre as farre asondre as the edges of the circles, where they bee most distaunte in sonder.The Li. Theoreme.If two circles be so drawen, that one of them do touche the other, then haue they not one centre.see textExample.There are two circles made, as you see, the one isA.B.C,and hath his centre byG,the other isB.D.E,and his centre is byF,so that it is easy enough to perceaue that their centres doe dyffer as muche a sonder, as the halfe diameter of the greater circle is lõger then the half diameter of the lesser circle. And so must it needes be thought and said of all other circles in lyke kinde.The .lij. theoreme.If a certaine pointe be assigned in the diameter of a circle, distant from the centre of the said circle, and from that pointe diuerse lynes drawen to the edge and circumference of the same circle, the longest line is that whiche passeth by the centre, and the shortest is the residew of the same line. And of al the other lines that is euer the greatest, that is nighest to the line, which passeth by the centre. And cõtrary waies, that is the shortest, that is farthest from it. And amongest thẽ all there can be but onely .ij. equall together, and they must nedes be so placed, that the shortest line shall be in the iust middle betwixte them.Example.see textThe circle isA.B.C.D.E.H,and his centre isF,the diameter isA.E,in whiche diameter I haue taken a certain point distaunt from the centre, and that pointe isG,from which I haue drawen .iiij. lines to the circumference, beside the two partes of the diameter, whiche maketh vp vi. lynes in all. Nowe for the diuersitee in quantitie of these lynes, I saie accordyng to the Theoreme, that the line whiche goeth by the centre is the longest line, that is to saie,A.G,and the residewe of the same diameter beeyngG.E,is the shortest lyne. And of all the other that lyne is longest, that is neerest vnto that parte of the diameter whiche gooeth by the centre, and that is shortest, that is farthest distant from it, wherefore I saie, thatG.B,is longer thenG.C,and therfore muche more longer thenG.D,sithG.C,also is longer thenG.D,and by this maie you soone perceiue, that it is not possible to drawe .ij. lynes on any one side of the diameter, whiche might be equall in lengthe together, but on the one side of the diameter maie you easylie make one lyne equall to an other, on the other side of the same diameter, as you see in this exampleG.H,to bee equall toG.D,betweene whiche the lyneG.E,(as the shortest in all the circle) doothe stande euen distaunte from eche of them, and it is the precise knoweledge of their equalitee, if they beequallydistaunt from one halfe of the diameter. Where as contrary waies if the one be neerer to any one halfe of the diameter then the other is, it is not possible that they two may be equall in lengthe, namely if they dooe ende bothe in the circumference of thecircle, and be bothe drawen from one poynte in the diameter, so that the saide poynte be (as the Theoreme doeth suppose) somewhat distaunt from the centre of the said circle. For if they be drawen from the centre, then must they of necessitee be all equall, howe many so euer they bee, as the definition of a circle dooeth importe, withoute any regarde how neere so euer they be to the diameter, or how distante from it. And here is to be noted, that in this Theoreme, by neerenesse and distaunce is vnderstand the nereness and distaunce of the extreeme partes of those lynes where they touche the circumference. For at the other end they do all meete and touche.The .liij. Theoreme.If a pointe bee marked without a circle, and from it diuerse lines drawen crosse the circle, to the circumference on the other side, so that one of them passe by the centre, then that line whiche passeth by the centre shall be the loongest of them all that crosse the circle. And of the other lines those are longest, that be nexte vnto it that passeth by the centre. And those ar shortest, that be farthest distant from it. But among those partes of those lines, whiche ende in the outewarde circumference, that is most shortest, whiche is parte of the line that passeth by the centre, and amongeste theothere eche, of thẽ, the nerer they are vnto it, the shorter they are, and the farther from it, the longer they be. And amongest them all there can not be more then .ij. of any one lẽgth, and they two muste be on the two contrarie sides of the shortest line.Example.see textTake the circle to beA.B.C,and the point assigned without it to beD.Now say I, that if there be drawen sundrie lines fromD,and crosse the circle, endyng in the circumference on the cõtrary side, as here you see,D.A, D.E, D.F,andD.B,then of all these lines the longest must needes beD.A,which goeth by the centre of the circle, and the nexte vnto it, that isD.E,is the longest amongest the rest. And contrarie waies,D.B,is the shorteste, because it is farthest distaunt fromD.A.And so maie you iudge ofD.F,because it is nerer vntoD.A,then isD.B,therefore is it longer thenD.B.And likewaies because it is farther of fromD.A,then isD.E,therfore is it shorter thenD.E.Now for those partes of the lines whiche bee withoute the circle (as you see)D.C,is theshortest. becauseit is the parte of that line which passeth by thecentre, AndD.K,is next to it in distance, and therefore also in shortnes, soD.G,is farthest from it in distance, and therfore is the longest of them. NowD.H,beyng nerer thenD.G,is also shorterthen it,and beynge farther of, thenD.K,is longer then it. So that for this parte of the theoreme (as I think) you do plainly perceaue the truthe thereof, so the residue hathe no difficulte. For seing that the nearer any line is toD.C,(which ioyneth with the diameter) the shorter it is and the farther of from it, the longer it is. And seyng two lynes can not be of like distaunce beinge bothe on one side, therefore if they shal be of one lengthe, and consequently of one distaunce, they must needes bee on contrary sides of the saide lineD.C.And so appeareth the meaning of the whole Theoreme.And of this Theoreme dothe there folowe an otherlyke. whicheyou maye calle other a theoreme by it selfe, or else a Corollary vnto this laste theoreme, I passe not so muche for the name. But his sentence is this:when so euer any lynes be drawen frome any pointe, withoute a circle, whether they crosse the circle, or eande in theutteredge of his circumference, those two lines that bee equally distaunt from the least line are equal togither, and contrary waies, if they be equall togither, they ar also equally distant from that least line.For the declaracion of this proposition, it shall not need to vse any other example, then that which is brought for the explication of this laste theoreme, by whiche you may without any teachinge easyly perceaue both the meanyng and also the truth of this proposition.The Liiij. Theoreme.If a point be set forthe in a circle, and frõ that pointe vnto the circumference many lines drawen, of which more then two are equal togither, then is that point the centre of that circle.Example.see textThe circle isA.B.C,and within it I haue sette fourth for an example three prickes, which areD.E.andF,from euery one of them I haue drawẽ (at the leaste) iiij. lines vnto the circumference of the circle but fromeD,I haue drawen more, yet maye it appear readily vnto your eye, that of all the lines whiche be drawen fromE.andF,vnto the circumference, there are but twoo equall, and more can not bee, forG.E.norE.H.hath none other equal to theim, nor canne not haue any beinge drawen from the same pointE.No more canL.F,orF.K,haue anye line equall to either of theim, beinge drawen from the same pointeF.And yet from either of those two poinctes are there drawen twoo lines equall togither, asA.E,is equall toE.B,andB.F,is equall toF.C,but there can no third line be drawen equall to either of these two couples, and that is by reason that they be drawen from a pointe distaunte from the centre of the circle. But fromD,althoughe there be seuen lines drawen, to the circumference, yet all bee equall, bicause it is the centre of the circle. And therefore if you drawe neuer so mannye more from it vnto the circumference, all shall be equal, so that this is the priuilege (as it were of the centre) and therfore no other point can haue aboue two equal lines drawen from it vnto the circumference. And from allpointesyou maye drawe ij. equall lines to the circumference of the circle, whether that pointe be within the circle or without it.The lv. Theoreme.No circle canne cut an other circle in morepointes then two.see textExample.The first circle isA.B.F.E,the second circle isB.C.D.E,and they crosse one an other inB.and inE,and in no more pointes. Nother is it possible that they should, but other figures ther be, which maye cutte a circle in foure partes, as you se in this exãple. Where I haue set forthe one tunne forme, and one eye forme, and eche of them cutteth euery of their two circles into foure partes. But as they be irregulare formes, that is to saye, suche formes as haue no precise measure nother proportion in their draughte, so can there scarcely be made any certaine theorem of them. But circles are regulare formes, that is to say, such formes as haue in their protracture a iuste and certaine proportion, so that certain and determinate truths may be affirmed of them, sith they ar vniforme and vnchaungable.The lvi. Theoreme.If two circles be so drawen, that the one be within the other, and that they touche one an other: If a line bee drawen by bothe their centres, and so forthe in lengthe, that line shall runne to that pointe, where the circles do touche.see textExample.The one circle, which is the greattest and vttermost isA.B.C,the other circle that is yelesser, and is drawen within the firste, isA.D.E.The cẽtre of the greater circle isF,and the centre of the lesser circle isG,the pointe where they touche isA.And now you may see the truthe of the theoreme so plainely, that it needeth no farther declaracion. For you maye see, that drawinge a line fromF.toG,and so forth in lengthe, vntill it come to the circumference, it wyll lighte in the very poincteA,where the circles touche one an other.The Lvij. Theoreme.If two circles bee drawen so one withouteanother, that their edges doo touche and a right line bee drawnenne frome the centre of the one to the centre of the other, that line shall passe by the place of their touching.Example.see textThe firste circle isA.B.E,and his centre isK,Thesecõd circle isD.B.C,and his cẽtre isH,the point wher they do touch isB.Nowe doo you se that the lineK.H,whiche is drawenfromK,that is centre of the firste circle, vntoH,beyng centre of the second circle, doth passe (as it must nedes by the pointeB,) whiche is the verye poynte wher they do to touche together.The .lviij. theoreme.One circle can not touche an other in more pointes then one, whether they touche within or without.see textExample.For the declaration of this Theoreme, I haue drawen iiij. circles, the first isA.B.C,and his centreH.the second isA.D.G,and his centreF.the third isL.M,and his centreK.the .iiij. isD.G.L.M,and his centreE.Nowe as you perceiue the second circleA.D.G,toucheth the first in the inner side, in so much as it is drawen within the other, and yet it toucheth him but in one point, that is to say inA,so lykewaies the third circleL.M,is drawen without the firste circle and touchethhym, as you maie see, but in one place. And now as for the .iiij. circle, it is drawen to declare the diuersitie betwene touchyng and cuttyng, or crossyng. For one circle maie crosse and cutte a great many other circles, yet can be not cutte any one in more places then two, as the fiue and fiftie Theoreme affirmeth.The .lix. Theoreme.In euerie circle those lines are to be counted equall, whiche are in lyke distaunce from thecentre, Andcontrarie waies they are in lyke distance from the centre, whiche be equall.see textExample.In this figure you see firste the circle drawen, whiche isA.B.C.D,and his centre isE.In this circle also there are drawen two lines equally distaunt from the centre, for the lineA.B,and the lineD.C,are iuste of one distaunce from the centre, whiche isE,and therfore are they of one length. Again thei are of one lengthe (as shall be proued in the boke of profes) and therefore their distaunce from the centre is all one.The lx. Theoreme.In euerie circle the longest line is the diameter, and of all the other lines, thei are still longestthat be nexte vnto the centre, and they be the shortest, that be farthest distaunt from it.Example.see textIn this circleA.B.C.D,I haue drawen first the diameter, whiche isA.D,whiche passeth (as it must) by the centreE,Thenhaue I drawen ij. other lines asM.N,whiche is neerer the centre, andF.G,that is farther from the centre.The fourth line also on the other side of the diameter, that isB.C,is neerer to the centre then the lineF.G,for it is of lyke distance as is the lyneM.N.Nowe saie I, thatA.D,beyng the diameter, is the longest of all those lynes, and also of any other that maie be drawen within that circle, Andthe other lineM.N,is longer thenF.G.Also the lineF.G,is shorter then the lineB.C,for because it is farther from the centre then is the lyneB.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 twothynges, Asfor exaumple, If two squares be sette foorthe, whereof the one containeth in it fiue square feete, 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 measureof 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 circumferẽce 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.see textIn this circleA.B.C,the diameter isA.C,the perpendicular line, which maketh a right angle with the diameter, isC.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 thyngis so plainly seene of the eye, that it needeth no farther declaracion. For euery man wil easily consent, that betwene the croked lineA.F,(whiche is a parte of the circumferẽce of the circle) andA.E(which is the said perpẽdicular line) there can none other line bee drawen in that place where they make the angle. Nowe for the residue of the theoreme. The angleD.A.B,which is made in the semicircle, is greater then anye sharpe angle that may bee made of ryghtelines. andyet is it a sharpe angle also, in as much as it is lesser then a right angle, which is the angleE.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 lineE.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 greater then the other, for they twoo doo make the whole right angle, so that if anye corner coulde be 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 ofthecircle to the pointe where they touche, thatline whiche is drawenne frome the centre, shall be a perpendicular line to the touch line.Example.see textThe circle isA.B.C,and his centre isF.The touche line isD.E,and the point wher they touch isC.Now by reason that a right line is drawen frome the centreF.vntoC,which is the point of the touche, therefore saith the theoreme, that the sayde lineF.C,muste needes bee a perpendicular line vnto the touche lineD.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.see textThe circle isA.B.C,and the centre of it isG.The touche line isD.C.E,and the pointe where it toucheth, isC.Noweit 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 cẽtre in him. For if the saide line shoulde go beside the centre, asF.C.doth, then dothe it not make righte angles with the touche line, which in thetheoremeis 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 circũferẽce.see textExample.ThecircleisA.B.C.D,and his centre is E: the angle on the centre isC.E.D, and the angle on the circumference isC.A.Dttheir commen ground line, isC.F.D.Now say I that the angleC.E.D,whiche is on the centre, is twise so greate as the angleC.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.see textBefore I declare this theoreme by example, it shall bee needefull to declare, what is it to be vnderstande by the wordes in this theoreme. For the sentence canne not be knowen, onles theuerymeaning of the wordes be firste vnderstand. Therefore when it speaketh of angles 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 circle. So that if any angle do not touch the circumference, or if the lines that inclose that angle, doo not ende in the extremities of the corde line, but ende other in some other part of the said 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 said cantle of the circle. And this promised, nowe will I cumme to the meaninge of the theoreme. I sette forthe a circle whiche isA.B.C.D,and his centreE,in this circle I drawe a lineD.C,whereby there ar made two cantels, a more and a lesser. The lesser isD.E.C,and the geater isD.A.B.C.In this greater cantle I drawe two angles, the firste isD.A.C,and the second isD.B.Cwhich two angles by reason they are made bothe in one cantle of a circle (that is the cantleD.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 isD.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.
see textExample.Fyrst before I declare the examples, it shal be mete to shew the true vnderstãdyng of this theorem.Bias lyne.Therfore by theBias line, I meane that lyne, whiche in any square figure dooth runne from corner to corner. And euery square which is diuided by that bias line into equall halues from corner to corner (that is to say, into .ij. equall triangles) those be countedto stande aboute one bias line, and the other squares, whiche touche that bias line, with one of their corners onely, those doo I callFyll squares,Fyll squares.accordyng to the greke name, which isanapleromata,ἀναπληρώματαand called in latinsupplementa, bycause that they make one generall square, includyng and enclosyng the other diuers squares, as in this exãpleH.C.E.N.is one square likeiamme, andL.M.G.C.is an other, whiche bothe are made aboute one bias line, that isN.M,thanK.L.H.C.andC.E.F.G.are .ij. fyll squares, for they doo fyll vp the sydes of the .ij.fyrste square lykeiammes, in suche sorte, that all them foure is made one greate generall squareK.M.F.N.
see text
Fyrst before I declare the examples, it shal be mete to shew the true vnderstãdyng of this theorem.Bias lyne.Therfore by theBias line, I meane that lyne, whiche in any square figure dooth runne from corner to corner. And euery square which is diuided by that bias line into equall halues from corner to corner (that is to say, into .ij. equall triangles) those be countedto stande aboute one bias line, and the other squares, whiche touche that bias line, with one of their corners onely, those doo I callFyll squares,Fyll squares.accordyng to the greke name, which isanapleromata,ἀναπληρώματαand called in latinsupplementa, bycause that they make one generall square, includyng and enclosyng the other diuers squares, as in this exãpleH.C.E.N.is one square likeiamme, andL.M.G.C.is an other, whiche bothe are made aboute one bias line, that isN.M,thanK.L.H.C.andC.E.F.G.are .ij. fyll squares, for they doo fyll vp the sydes of the .ij.fyrste square lykeiammes, in suche sorte, that all them foure is made one greate generall squareK.M.F.N.
Nowe to the sentence of the theoreme, I say, that the .ij. fill squares,H.K.L.C.andC.E.F.G.are both equall togither, (as it shall bee declared in the booke of proofes) bicause they are the fill squares of two likeiammes made aboute one bias line, as the exaumple sheweth. Conferre the twelfthe conclusion with this theoreme.
see text
A.B.C.is a triangle, hauing a ryght angle inB.Wherfore it foloweth, that the square ofA.C,(whiche is the side that lyeth agaynst the right angle) shall be as muche as the two squares ofA.B.andB.C.which are the other .ij. sides.
¶By thesquareof any lyne, you muste vnderstande a figure made iuste square, hauyng all his iiij. sydes equall to that line, whereof it is the square, so isA.C.F,the square ofA.C.LykewaisA.B.D.is the square ofA.B.AndB.C.E.is the square ofB.C.Now by the numbre of the diuisions in eche of these squares, may you perceaue not onely what the square of any line is called, but also that the theoreme is true, and expressed playnly bothe by lines and numbre. For as you see, the greatter square (that isA.C.F.) hath fiue diuisions on eche syde, all equall togyther, and those in the whole square are twenty and fiue. Nowe in the left square, whiche isA.B.D.there are but .iij. of those diuisions in one syde, and that yeldeth nyne in the whole. So lykeways you see in the meane squareA.C.E.in euery syde .iiij. partes, whiche in the whole amount vnto sixtene. Nowe adde togyther all the partes of the two lesser squares, that is to saye, sixtene and nyne, and you perceyue that they make twenty and fiue, whyche is an equall numbre to the summe of the greatter square.
By this theoreme you may vnderstand a redy way to know the syde of any ryght anguled triangle that is vnknowen, so that you knowe the lengthe of any two sydes of it. For by tournynge the two sydes certayne into theyr squares, and so addynge them togyther, other subtractynge the one from the other (accordyng as in the vse of these theoremes I haue sette foorthe) and then fyndynge the roote of the square that remayneth, which roote (I meane the syde of the square) is the iuste length of the vnknowen syde, whyche is sought for. But this appertaineth to the thyrde booke, and therefore I wyll speake no more of it at this tyme.
As in the figure of the laste Theoreme, bicauseA.C,made in square, is asmuch as the square ofA.B,and also as the square ofB.C.ioyned bothe togyther, therefore the angle that is inclosed betwene those .ij. lesser lynes,A.B.andB.C.(that is to say) the angleB.whiche lieth against the lineA.C,must nedes be a ryght angle. This theoreme dothe so depende of the truthe of the laste, that whan you perceaue the truthe of the one, you can not iustly doubt of the others truthe, for they conteine one sentence, contrary waies pronounced.
see text
The ij. lines proposed arA.B.andC.D,and the lyneA.B.is deuided into thre partes byE.andF.Now saith this theoreme, that the square that is made of those two whole linesA.B.andC.D,so that the lineA.B.stãdeth for the lẽgth of the square, and the other lineC.D.for the bredth of the same. That square (I say) wil be equall to all the squares that be made, of the vndiueded lyne (which isC.D.) and euery portion of the diuided line. And to declare that particularly, Fyrst I make an other lineG.K,equall to the line.C.D,and the lineG.H.to be equal to the lineA.B,and to bee diuided into iij. like partes, so thatG.M.is equall toA.E,andM.N.equal toE.F,and then musteN.H.nedes remaine equall toF.B.Then of those ij. linesG.K,vndeuided, andG.H.which is deuided, I make a square, that isG.H.K.L,Inwhich square if I drawe crosse lines frome one side to the other, according to the diuisions of the lineG.H,then will it appear plaine, that the theoreme doth affirme. For the first squareG.M.O.K,must needes be equal to the square of the lineC.D,and the first portiõ of the diuided line, which isA.E,for bicause their sides are equall. And so the secondesquare that isM.N.P.O,shall be equall to the square ofC.D,and the second part ofA.B,that isE.F.Also the third square which isN.H.L.P,must of necessitee be equal to the square ofC.D,andF.B,bicause those lines be so coupeled that euery couple are equall in the seuerall figures. And so shal you not only in this example, but in all other finde it true, that if one line be deuided into sondry partes, and an other line whole and vndeuided, matched with him in a square, that square which is made of these two whole lines, is as muche iuste and equally, as all the seuerall squares, whiche bee made of the whole line vndiuided, and euery part seuerally of the diuided line.
see text
The line propounded beyngA.B.and deuided, as chaunce happeneth, inC.into ij. vnequall partes, I say that the square made of the hole lineA.B,is equal to the two squares made of the same line with the twoo partes of itselfe, as withA.C,and withC.B,for the squareD.E.F.G.is equal to the two other partial squares ofD.H.K.GandH.E.F.K,but that the greater square is equall to the square of the whole lineA.B,and thepartiall squares equall to the squares of the second partes of the same line ioyned with the whole line, your eye may iudg without muche declaracion, so that I shall not neede to make more exposition therof, but that you may examine it, as you did in the laste Theoreme.
see text
The lineA.B.is deuided inC.into twoo partes, though not equally, of which two partes for an example I take the first, that isA.C,and of it I make one side of a square, as for exampleD.G.accomptinge those two lines to be equall, the other side of the square isD.E,whiche is equall to the whole lineA.B.
Nowmay it appeare, to your eye, that the great square made of the whole lineA.B,and of one of his partes that isA.C,(which is equall withD.G.) is equal to two partiall squares, whereof the one is made of the saide greatter portionA.C,in as muche as not onlyD.G,beynge one of his sides, but alsoD.H.beinge the other side, are eche of them equall toA.C.The second square isH.E.F.K,in which the one sideH.E,is equal toC.B,being the lesser parte of the line,A.B,andE.F.is equall toA.C.which is the greater parte of the same line. So that those two squaresD.H.K.GandH.E.F.K,bee bothe of them no more then the greate squareD.E.F.G,accordinge to the wordes of the Theoreme afore saide.
see text
The labels A and B were transposed in the illustration as an alternative to transposing all occurrences of A.C and C.B in the text.
Lette the diuided line beeA.B,and parted inC,into twoo partes: Nowe saithe the Theoreme, that the square of the whole lyneA.B,is as mouche iuste as the square ofA.C,and the square ofC.B,eche by it selfe, and more ouer by as muche twise, asA.C.andC.B.ioyned in one square will make. For as you se, the great squareD.E.F.G,conteyneth in hym foure lesser squares, of whiche the first and the greatest isN.M.F.K,and is equall to the square of the lyneA.C.The second square is the lest of them all, that isD.H.L.N,and it is equall to the square of the lineC.B.Then are there two other longe squares both of one bygnes, that isH.E.N.M.andL.N.G.K,eche of them both hauyng .ij. sides equall toA.C,the longer parte of the diuided line, and there other two sides equall toC.B,beeyng he shorter parte of the said lineA.B.
So is that greatest square, beeyng made of the hole lyneA.B,equal to the ij. squares of eche of his partes seuerally, and more by as muche iust as .ij. longe squares, made of the longer portion of the diuided lyne ioyned in square with the shorter parte of the same diuided line, as the theoreme wold. And as here I haue put an example of a lyne diuided into .ij. partes, so the theoreme is true of all diuided lines, of what number so euer the partes be, foure, fyue, orsyxe. etc.
Thistheoreme hath great vse, not only in geometrie, but also in arithmetike, as herafter I will declare in conuenient place.
see text
The lineA.B.is diuided into ij. equal partes inC,and that parteC.B.is diuided agayne as hapneth inD.Wherfore saith the Theorem that the long square made ofD.B.andA.D,with the square ofC.D.(which is the mydle portion) shall bothe be equall to the square of half the lyneA.B,that is to saye, to the square ofA.C,or els ofC.D,which make all one. The long squareF.G.N.O.whiche is the longe square that the theoreme speaketh of, is made of .ij. long squares, wherof the fyrst isF.G.M.K,and the seconde isK.N.O.M.The square of the myddle portion isL.M.O.P.and the square of the halfe of the fyrste lyne isE.K.Q.L.Nowe by the theoreme, that longe squareF.G.N.O,with the iuste squareL.M.O.P,muste bee equall to the greate squareE.K.Q.L,whyche thynge bycause it seemeth somewhat difficult to vnderstande, althoughe I intende not here to make demonstrations of the Theoremes, bycause it is appoynted to be done in the newe edition of Euclide, yet I wyll shew you brefely how the equalitee of the partes doth stande. And fyrst I say, that where the comparyson of equalitee is made betweene the greate square (whiche is made of halfe the lineA.B.) and two other, where of the fyrst is the longe squareF.G.N.O,and the second is the full squareL.M.O.P,which is one portion of the great square all redye, and so is that longe squareK.N.M.O,beynge a parcell also of the longe squareF.G.N.O,Wherforeas those two partes are common to bothe partes compared in equalitee, and therfore beynge bothe abated from eche parte, if the reste of bothe the other partes bee equall, than were those whole partes equall before: Nowe the reste of the great square, thosetwo lesser squares beyng taken away,is that longe squareE.N.P.Q,whyche is equall to the long squareF.G.K.M,beyng the rest of the other parte. And that they two be equall, theyr sydes doo declare. For the longest lynes that isF.KandE.Qare equall, and so are the shorter lynes,F.G,andE.N,and so appereth the truthe of the Theoreme.
see text
The fyrst lynepropoundedisA.B,and it is diuided into ij. equall partes inC,and an other ryght lyne, I meaneB.Dannexed to one ende of the fyrste lyne.
Nowesay I, that the long squareA.D.M.K,is made of the whole lyne so augmẽted, that isA.D,and the portiõ annexed, ytisD.M,forD.Mis equall toB.D,wherfore ytlong squareA.D.M.K,with thesquare of halfe the first line, that isE.G.H.L,is equall to the great squareE.F.D.C.whiche square is made of the lineC.D.that is to saie, of a line compounded of halfe the first line, beyngC.B,and the portion annexed, that isB.D.And it is easyly perceaued, if you consyder that the longe squareA.C.L.K.(whiche onely is lefte out of the great square) hathanotherlonge square equall to hym, and to supply his steede in the great square, and that isG.F.M.H.For their sydes be of lyke lines in length.
see text
A.B.is the line diuided inC.AndD.E.F.G,is the square of the whole line,D.H.K.M.is the square of the lesser portion (whyche I take for an example) and therfore must bee twise reckened. Nowe I saye that those ij. squares are equall to two longe squares of the whole lineA.B,and his sayd portionA.C,and also to the square of the other portion of the sayd first line, whiche portion isC.B,and his squareK.N.F.L.In this theoreme there is no difficultie, if you cõsyder that the litle squareD.H.K.M.is .iiij. tymes reckened, that is to say, fyrst of all as a parte of the greatest square, whiche isD.E.F.G.Secondly he is reknedby him selfe. Thirdely he is accompted as parcell of the long squareD.E.N.M,And fourthly he is taken as a part of the other long squareD.H.L.G,so that in as muche as he is twise reckened in one part of the comparisõ of equalitee, and twise also in the second parte, there can rise none occasion of errour or doubtfulnes therby.
see text
The firste line isA.B,and is deuided byC.into two vnequall partes as happeneth.Thelong square of yt, and his lesser portionA.C,is foure times drawen, the first isE.G.M.K,the seconde isK.M.Q.O,the third isH.K.R.S,and the fourthe isK.L.S.T.And where as it appeareth that one of the little squares (I meaneK.L.P.O) is reckened twise, ones as parcell of the second long square and agayne as parte of thethirde long square, to auoide ambiguite, you may place one insteede of it, an other square of equalitee, withit. thatis to saye,D.E.K.H,which was at no tyme accompting as parcell of any one of them, and then haue you iiij. long squares distinctly made of the whole lineA.B,and his lesser portionA.C.And within them is there a greate full squareP.Q.T.V.whiche is the iust square ofB.C,beynge the greatter portion of the lineA.B.And that those fiue squares doo make iuste as muche as the whole square of that longer lineD.G,(whiche is as longe asA.B,andA.C.ioyned togither) it may be iudged easyly by the eye, sith that one greate square doth comprehẽd in it all the other fiue squares, that is to say, foure long squares (as is before mencioned) and one fullsquare. whichis the intent of the Theoreme.
see text
The line to be deuided isA.B,and is parted inC.into two equall partes, and thenC.B,is deuided againe into two partes inD,so that the meaninge of the Theoreme, is that thesquare ofD.B.which is the latter parte of the line, and the square ofA.D,which is the residue of the whole line. Those two squares, I say, ar double to the square of one halfe of the line, and to the square ofC.D,which is the middle portion of those thre diuisions. Which thing that you maye more easilye perceaue, I haue drawen foure squares, whereof the greatest being marked withE.is the square ofA.D.The next, which is marked withG,is the square of halfe the line, that is, ofA.C,Andthe other two little squares marked withF.andH,be both of one bignes, by reason that I did diuideC.B.into two equall partes, so that you amy take the squareF,for the square ofD.B,and the squareH,for the square ofC.D.Now I thinke you doubt not, but that the squareE.and the squareF,ar double so much as the squareG.and the squareH,which thing theeasyeris to be vnderstande, bicause that the greate square hath in his side iij. quarters of the firste line, which multiplied by itselfe maketh nyne quarters, and the squareF.containeth but one quarter, so that bothe doo make tenne quarters.
ThenG.contayneth iiij. quarters, seynge his side containethtwoo, andH.containeth but one quarter, whiche both makebut fiue quarters, and that is but halfe of tenne.Whereby you may easylye coniecture,that the meanynge of the the-oreme is verified in thefigures of this ex-ample.
see text
The line isA.B,and is diuided firste into twoo equal partes inC,and thẽ is there annexed to it an other portion whiche isB.D.Now saith the Theoreme, that the square ofA.D,and the square ofB.D,ar double to the square ofA.C,and to the square ofC.D.The lineA.B.cõtaining four partes, then must needes his halfe containe ij. partes of such partes I supposeB.D.(which is the ãnexed line) to containe thre, so shal the hole line cõprehend vij. parts, and his square xlix. parts, where vnto if you ad yesquareof the annexed lyne, whiche maketh nyne, than those bothe doo yelde, lviij. whyche must be double to the square of the halfe lyne with the annexed portion. The halfe lyne by it selfe conteyneth but .ij. partes, and therfore his square dooth make foure. The halfe lyne with the annexed portion conteyneth fiue, and the square of it is .xxv, now put foure to .xxv, and it maketh iust .xxix, the euen halfe of fifty and eight, wherby appereth the truthe of the theoreme.
For the declaration of this theoreme and the next also, whose vse are wonderfull in the practise of Geometrie, and in measuryng especially, it shall be nedefull to declare that euery triangle that hath no ryght angle as those whyche are called (as in the boke of practise is declared) sharp cornered triangles, and bluntcorneredtriangles, yet may they be brought to haue a ryght angle, eyther by partyng them into two lesser triangles,or els by addyng an other triangle vnto them, whiche may be a great helpe for the ayde of measuryng, as more largely shall be sette foorthe in the boke of measuryng. But for this present place, this forme wyll I vse, (whiche Theon also vseth) to adde one triangle vnto an other, to bryng the blunt cornered triangle into a ryght angled triangle, whereby the proportion of the squares of the sides in suche a blunt cornered triangle may the better bee knowen.
see text
Fyrst therfore I sette foorth the triangleA.B.C,whose corner byC.is a blunt corner as you maye well iudge, than to make an other triangle of yt with a ryght angle, I must drawe forth the sideB.C.vntoD,and frõ the sharp corner byA.I brynge a plumbe lyne or perpẽdicular onD.And so is there nowe a newe triangleA.B.D.whose angle byD.is a right angle. Nowe accordyng to the meanyng of the Theoreme, I saie, that in the first triangleA.B.C,because it hath a blunt corner atC,the square of the lineA.B.whiche lieth against the said blunte corner, is morethen the square of the lineA.C,and also of the lyneB.C,(whiche inclose the blunte corner) by as muche as will amount twise of the lineB.C,and that portionD.C.whiche lieth betwene the blunt angle byC,and the perpendicular lineA.D.
The square of the lineA.B,is the great square marked withE.The square ofA.C,is the meane square marked withF.The square ofB.C,is the least square marked withG.And the long square marked withK,is sette in steede of two squares made ofB.C,andC.D.For as the shorter side is the iuste lengthe ofC.D,so the other longer side is iust twise so longe asB.C,WherforeI saie now accordyng to the Theoreme, that the greatte squareE,is more then the other two squaresF.andG,by the quantitee of the longe squareK,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.
Labeling of rectangle K is conjectural. Other configurations of B, C and D will also fit, but the printed illustration (B and D on the right, nothing on the left) will not. The text requires a rectangle with BC as one side and CD as the other, doubled as in the next illustration (H.K).
see text
Fyrst I sette foorth the triangleA.B.C,and in yt I draw a plũbe line from the angleC.vnto the lineA.B,and it lighteth inD.Nowe by the theoreme the square ofB.C.is not so muche as the square of the other two sydes, that ofB.A.and ofA.C.by as muche as is twise conteyned in the lõg square made ofA.B,andA.D, A.B.beyng the line or syde on which the perpendicular line falleth, andA.D.beeyng that portion of the same line whiche doth lye betwene the perpendicular line, and the sayd sharpe angle limitted, whiche angle is byA.
For declaration of the figures, the square marked withE.is the square ofB.C,whiche is the syde that lieth agaynst the sharpe angle, the square marked withG.is the square ofA.B,and the square marked withF.is the square ofA.C,and the two longe squares marked withH.K,are made of the hole lineA.B,and one of his portionsA.D.And truthe it is that the squareE.is lesser than the other two squaresC.andF.by the quantitee of those two long squaresH.andK.Wherby you may consyder agayn, an other proportion of equalitee,that is to saye, that the squareE.with the twoo longsquaresH.K,are iuste equall to the other twoo squaresC.andF.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 at 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 falleth, is thrise vsed, yet is his square but ones 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, would credite the possibilitie of their wonderfull vse, and large ayde in knowledge. But all this wyll I remitte to a place conuenient.
see text
The circle isA.B.C.D,the ij. poinctes areA.B,the righteline that is drawenne frome the one to the other, is the lineA.B,which as you see, must needes lyghte within the circle. So if you putte the pointes to beA.D,orD.C,orA.C,otherB.C,orB.D,in any 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. How be it, 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.
see text
The circle isA.B.C.D,the line that passeth by the centre, isA.E.C,the line that goeth beside the centre isD.B.Nowesaye I, that the lineA.E.C,dothe cutte that other lineD.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, accordinge as the Theoreme would.
see text
The circle isA.B.C.D,and the centre isE,the one lineA.C,and the other isB.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 cuytte the other into equall portions. For as you may easily iudge,A.C.hath one portiõ lõger and an other shorter, and so like wiseB.D.Howbeit, it is not so to be vnderstãd, but one of them may be deuided into ij. euẽ parts,but bothe to bee cutte equally in the middle, is not possible, onles both passe through the cẽtre, therfore much rather whẽ bothe go beside the centre, it can not be that eche of theym shoulde be iustely parted into ij. euen partes.
see text
This theoreme seemeth of it selfe so manifest, that it neadeth nother demonstration nother declaraciõ. Yet for the plaine vnderstanding of it, I haue sette forthe a figure here, where ij. circles be drawẽ, so that one of them doth crosse the other (as you see) in the pointesB.andG,and their centres appear at the firste sighte to bee diuers. For the centre of the one isF,and the centre of the other isE,which diffre as farre asondre as the edges of the circles, where they bee most distaunte in sonder.
see text
There are two circles made, as you see, the one isA.B.C,and hath his centre byG,the other isB.D.E,and his centre is byF,so that it is easy enough to perceaue that their centres doe dyffer as muche a sonder, as the halfe diameter of the greater circle is lõger then the half diameter of the lesser circle. And so must it needes be thought and said of all other circles in lyke kinde.
see text
The circle isA.B.C.D.E.H,and his centre isF,the diameter isA.E,in whiche diameter I haue taken a certain point distaunt from the centre, and that pointe isG,from which I haue drawen .iiij. lines to the circumference, beside the two partes of the diameter, whiche maketh vp vi. lynes in all. Nowe for the diuersitee in quantitie of these lynes, I saie accordyng to the Theoreme, that the line whiche goeth by the centre is the longest line, that is to saie,A.G,and the residewe of the same diameter beeyngG.E,is the shortest lyne. And of all the other that lyne is longest, that is neerest vnto that parte of the diameter whiche gooeth by the centre, and that is shortest, that is farthest distant from it, wherefore I saie, thatG.B,is longer thenG.C,and therfore muche more longer thenG.D,sithG.C,also is longer thenG.D,and by this maie you soone perceiue, that it is not possible to drawe .ij. lynes on any one side of the diameter, whiche might be equall in lengthe together, but on the one side of the diameter maie you easylie make one lyne equall to an other, on the other side of the same diameter, as you see in this exampleG.H,to bee equall toG.D,betweene whiche the lyneG.E,(as the shortest in all the circle) doothe stande euen distaunte from eche of them, and it is the precise knoweledge of their equalitee, if they beequallydistaunt from one halfe of the diameter. Where as contrary waies if the one be neerer to any one halfe of the diameter then the other is, it is not possible that they two may be equall in lengthe, namely if they dooe ende bothe in the circumference of thecircle, and be bothe drawen from one poynte in the diameter, so that the saide poynte be (as the Theoreme doeth suppose) somewhat distaunt from the centre of the said circle. For if they be drawen from the centre, then must they of necessitee be all equall, howe many so euer they bee, as the definition of a circle dooeth importe, withoute any regarde how neere so euer they be to the diameter, or how distante from it. And here is to be noted, that in this Theoreme, by neerenesse and distaunce is vnderstand the nereness and distaunce of the extreeme partes of those lynes where they touche the circumference. For at the other end they do all meete and touche.
see text
Take the circle to beA.B.C,and the point assigned without it to beD.Now say I, that if there be drawen sundrie lines fromD,and crosse the circle, endyng in the circumference on the cõtrary side, as here you see,D.A, D.E, D.F,andD.B,then of all these lines the longest must needes beD.A,which goeth by the centre of the circle, and the nexte vnto it, that isD.E,is the longest amongest the rest. And contrarie waies,D.B,is the shorteste, because it is farthest distaunt fromD.A.And so maie you iudge ofD.F,because it is nerer vntoD.A,then isD.B,therefore is it longer thenD.B.And likewaies because it is farther of fromD.A,then isD.E,therfore is it shorter thenD.E.Now for those partes of the lines whiche bee withoute the circle (as you see)D.C,is theshortest. becauseit is the parte of that line which passeth by thecentre, AndD.K,is next to it in distance, and therefore also in shortnes, soD.G,is farthest from it in distance, and therfore is the longest of them. NowD.H,beyng nerer thenD.G,is also shorterthen it,and beynge farther of, thenD.K,is longer then it. So that for this parte of the theoreme (as I think) you do plainly perceaue the truthe thereof, so the residue hathe no difficulte. For seing that the nearer any line is toD.C,(which ioyneth with the diameter) the shorter it is and the farther of from it, the longer it is. And seyng two lynes can not be of like distaunce beinge bothe on one side, therefore if they shal be of one lengthe, and consequently of one distaunce, they must needes bee on contrary sides of the saide lineD.C.And so appeareth the meaning of the whole Theoreme.
And of this Theoreme dothe there folowe an otherlyke. whicheyou maye calle other a theoreme by it selfe, or else a Corollary vnto this laste theoreme, I passe not so muche for the name. But his sentence is this:when so euer any lynes be drawen frome any pointe, withoute a circle, whether they crosse the circle, or eande in theutteredge of his circumference, those two lines that bee equally distaunt from the least line are equal togither, and contrary waies, if they be equall togither, they ar also equally distant from that least line.
For the declaracion of this proposition, it shall not need to vse any other example, then that which is brought for the explication of this laste theoreme, by whiche you may without any teachinge easyly perceaue both the meanyng and also the truth of this proposition.
see text
The circle isA.B.C,and within it I haue sette fourth for an example three prickes, which areD.E.andF,from euery one of them I haue drawẽ (at the leaste) iiij. lines vnto the circumference of the circle but fromeD,I haue drawen more, yet maye it appear readily vnto your eye, that of all the lines whiche be drawen fromE.andF,vnto the circumference, there are but twoo equall, and more can not bee, forG.E.norE.H.hath none other equal to theim, nor canne not haue any beinge drawen from the same pointE.No more canL.F,orF.K,haue anye line equall to either of theim, beinge drawen from the same pointeF.And yet from either of those two poinctes are there drawen twoo lines equall togither, asA.E,is equall toE.B,andB.F,is equall toF.C,but there can no third line be drawen equall to either of these two couples, and that is by reason that they be drawen from a pointe distaunte from the centre of the circle. But fromD,althoughe there be seuen lines drawen, to the circumference, yet all bee equall, bicause it is the centre of the circle. And therefore if you drawe neuer so mannye more from it vnto the circumference, all shall be equal, so that this is the priuilege (as it were of the centre) and therfore no other point can haue aboue two equal lines drawen from it vnto the circumference. And from allpointesyou maye drawe ij. equall lines to the circumference of the circle, whether that pointe be within the circle or without it.
see text
The first circle isA.B.F.E,the second circle isB.C.D.E,and they crosse one an other inB.and inE,and in no more pointes. Nother is it possible that they should, but other figures ther be, which maye cutte a circle in foure partes, as you se in this exãple. Where I haue set forthe one tunne forme, and one eye forme, and eche of them cutteth euery of their two circles into foure partes. But as they be irregulare formes, that is to saye, suche formes as haue no precise measure nother proportion in their draughte, so can there scarcely be made any certaine theorem of them. But circles are regulare formes, that is to say, such formes as haue in their protracture a iuste and certaine proportion, so that certain and determinate truths may be affirmed of them, sith they ar vniforme and vnchaungable.
see text
The one circle, which is the greattest and vttermost isA.B.C,the other circle that is yelesser, and is drawen within the firste, isA.D.E.The cẽtre of the greater circle isF,and the centre of the lesser circle isG,the pointe where they touche isA.And now you may see the truthe of the theoreme so plainely, that it needeth no farther declaracion. For you maye see, that drawinge a line fromF.toG,and so forth in lengthe, vntill it come to the circumference, it wyll lighte in the very poincteA,where the circles touche one an other.
see text
The firste circle isA.B.E,and his centre isK,Thesecõd circle isD.B.C,and his cẽtre isH,the point wher they do touch isB.Nowe doo you se that the lineK.H,whiche is drawenfromK,that is centre of the firste circle, vntoH,beyng centre of the second circle, doth passe (as it must nedes by the pointeB,) whiche is the verye poynte wher they do to touche together.
see text
For the declaration of this Theoreme, I haue drawen iiij. circles, the first isA.B.C,and his centreH.the second isA.D.G,and his centreF.the third isL.M,and his centreK.the .iiij. isD.G.L.M,and his centreE.Nowe as you perceiue the second circleA.D.G,toucheth the first in the inner side, in so much as it is drawen within the other, and yet it toucheth him but in one point, that is to say inA,so lykewaies the third circleL.M,is drawen without the firste circle and touchethhym, as you maie see, but in one place. And now as for the .iiij. circle, it is drawen to declare the diuersitie betwene touchyng and cuttyng, or crossyng. For one circle maie crosse and cutte a great many other circles, yet can be not cutte any one in more places then two, as the fiue and fiftie Theoreme affirmeth.
see text
In this figure you see firste the circle drawen, whiche isA.B.C.D,and his centre isE.In this circle also there are drawen two lines equally distaunt from the centre, for the lineA.B,and the lineD.C,are iuste of one distaunce from the centre, whiche isE,and therfore are they of one length. Again thei are of one lengthe (as shall be proued in the boke of profes) and therefore their distaunce from the centre is all one.
see text
In this circleA.B.C.D,I haue drawen first the diameter, whiche isA.D,whiche passeth (as it must) by the centreE,Thenhaue I drawen ij. other lines asM.N,whiche is neerer the centre, andF.G,that is farther from the centre.The fourth line also on the other side of the diameter, that isB.C,is neerer to the centre then the lineF.G,for it is of lyke distance as is the lyneM.N.Nowe saie I, thatA.D,beyng the diameter, is the longest of all those lynes, and also of any other that maie be drawen within that circle, Andthe other lineM.N,is longer thenF.G.Also the lineF.G,is shorter then the lineB.C,for because it is farther from the centre then is the lyneB.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 twothynges, Asfor exaumple, If two squares be sette foorthe, whereof the one containeth in it fiue square feete, 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 measureof 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.
see text
In this circleA.B.C,the diameter isA.C,the perpendicular line, which maketh a right angle with the diameter, isC.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 thyngis so plainly seene of the eye, that it needeth no farther declaracion. For euery man wil easily consent, that betwene the croked lineA.F,(whiche is a parte of the circumferẽce of the circle) andA.E(which is the said perpẽdicular line) there can none other line bee drawen in that place where they make the angle. Nowe for the residue of the theoreme. The angleD.A.B,which is made in the semicircle, is greater then anye sharpe angle that may bee made of ryghtelines. andyet is it a sharpe angle also, in as much as it is lesser then a right angle, which is the angleE.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 lineE.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 greater then the other, for they twoo doo make the whole right angle, so that if anye corner coulde be 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.
see text
The circle isA.B.C,and his centre isF.The touche line isD.E,and the point wher they touch isC.Now by reason that a right line is drawen frome the centreF.vntoC,which is the point of the touche, therefore saith the theoreme, that the sayde lineF.C,muste needes bee a perpendicular line vnto the touche lineD.E.
see text
The circle isA.B.C,and the centre of it isG.The touche line isD.C.E,and the pointe where it toucheth, isC.Noweit 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 cẽtre in him. For if the saide line shoulde go beside the centre, asF.C.doth, then dothe it not make righte angles with the touche line, which in thetheoremeis supposed.
see text
ThecircleisA.B.C.D,and his centre is E: the angle on the centre isC.E.D, and the angle on the circumference isC.A.Dttheir commen ground line, isC.F.D.Now say I that the angleC.E.D,whiche is on the centre, is twise so greate as the angleC.A.D,which is on the circumference.
see text
Before I declare this theoreme by example, it shall bee needefull to declare, what is it to be vnderstande by the wordes in this theoreme. For the sentence canne not be knowen, onles theuerymeaning of the wordes be firste vnderstand. Therefore when it speaketh of angles 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 circle. So that if any angle do not touch the circumference, or if the lines that inclose that angle, doo not ende in the extremities of the corde line, but ende other in some other part of the said 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 said cantle of the circle. And this promised, nowe will I cumme to the meaninge of the theoreme. I sette forthe a circle whiche isA.B.C.D,and his centreE,in this circle I drawe a lineD.C,whereby there ar made two cantels, a more and a lesser. The lesser isD.E.C,and the geater isD.A.B.C.In this greater cantle I drawe two angles, the firste isD.A.C,and the second isD.B.Cwhich two angles by reason they are made bothe in one cantle of a circle (that is the cantleD.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 isD.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.