whereon the head thereof you shall find the Book it belongeth to and the Propositions or uses continued in their order EIGHT BOOKS OF Euclid's Elements With the Vses of each PROPOSITION The FIRST BOOK EUCLID's Design in this Book is to give the first Principles of Geometry and to do the same Methodically he begins with the Definitions and Explication of the most ordinary Terms then he exhibits certain Suppositions and having proposed those Maxims which natural Reason teacheth he pretends to put forward nothing without Demonstration and to convince any one which will consent to nothing but what he shall be obliged to acknowledge in his first Propositions he treateth of Lines and of the several Angles made by their intersecting each other and having occasion to Demonstrate their Proprieties and compare certain Triangles he doth the same in the First Eight Propositions then teacheth the Practical way of dividing an Angle and a Line into two equal parts and to draw a Perpendicular he pursues to the propriety of a Triangle and having shewn those of Parallel Lines he makes an end of the Explication of this First Figure and passeth forwards to Parallelograms giving a way to reduce all sorts of Polygons into a more Regular Figure He endeth this Book with that Celebrated Proposition of Pythagoras and Demonstrates that in a Rectangular Triangle the Square of the Base is equal to the sum of the Squares of the Sides including the Right Angle DEFINITIONS 1. A Point is that which hath no part This Definition is to be understood in this sence The quantity which we conceive without distinguishing its parts or without thinking that it hath any is a Mathematical Point far differing from those of Zeno which were alltogether indivisible since one may doubt with a great deal of Reason if those last be possible which yet we cannot of the first if we conceive them as we ought 2. A Line is a length without breadth The sense of this Definition is the same with that of the foregoing the quantity which we consider having length without making any reflection on its breadth is that we understand by the word Line although one cannot draw a real Line which hath not a determinate breadth it is generally said that a Line is produced by the motion of a Point which we ought well to take notice of seeing that by a motion after that manner may be produced all sorts of quantities imagine then that a Point moveth and that it leaveth a trace in the middle of the way which it passeth the Trace is a Line 3. The Two ends of a Line are Points 4. A streight Line is that whose Points are placed exactly in the midst or if you would rather have it a streight Line is the shortest of all the Lines which may be drawn from one Point to another 5. A Superficies is a quantity to which is given length and breadth without considering the thickness 6. The extremities of a Superficies are Lines 7. A plain or straight Superficies is that whose Lines are placed equally between the extremities or that to which a streight Line may be applyed any manner of way Plate I. Fig. 1. I have already taken notice that motion is capable of producing all sorts of quantity whence we say that when a Line passeth over another it produces a superficies or a Plain and that that motion hath a likeness to Arithmetical Multiplication imagine that the Line AB moveth along the Line BC keeping the same situation without inclining one way or the other the Point A shall describe the Line AD the Point B the Line BC and the other Points between other Parallel Lines which shall compose the Superficies ABCD. I add that this motion corresponds with Arithmetical Multiplication for if I know the number of Points which are in the Lines AB BC Multiplying of them one by the other I shall have the number of Points which Composeth the Superficies ABCD as if AB contains four points and BC six saying Four times Six are Twenty Four the Superficies AB CD should be Composed of Twenty Four Points Now I may take for a Mathematical Point any quantity whatsoever for Example a Foot provided I do not subdivide the same into Parts 8. A plain Angle is the opening of Two Lines which intersect each other and which Compose not one single Line Fig. 2. As the opening D of the Lines AB CB which are not parts of the same Line A Right Lined Angle is the opening of two streight Lines It is principally of this sort of Angles which I intend to treat of at present because experience doth make me perceive that the most part of those who begin do mistake the measuring the quantity of an Angle by the length of the Lines which Composeth the same Fig. 3 4. The most open Angle is the greatest that is to say when the Lines including an Angle are farther asunder than those of another Angle taking them at the same distance from the Points of intersection of their Lines the first is greater than the Second so the Angle A is greater than E because if we take the Points B and D as far distant from the Point A as the Points G and L are from the Points E the Points B and D are farther asunder than the Points G and L from whence I conclude that if EG EL were continued the Angle E would be of the same Measure and less than the Angle A. We make use of Three Letters to express an Angle and the Second Letter denotes the Angular Point as the Angle BAD is the Angle which the Lines BA AD doth form at the Point A the Angle BAC is that which is formed by the Lines BA AC the Angle CAD is comprehended under the Line CA AD. Fig. 3. The Arch of a Circle is the measure of an Angle thus designing to measure the quantity of the Angle BAD I put one Foot of the Compasses on the Point A and with the other I describe an Arch of a Circle BCD the Angle shall be the greater by how much the Arch BCD which is the measure thereof shall contain a greater portion of a Circle and because that commonly an Arch of a Circle is divided into Three Hundred and Sixty equal Parts called Degrees It is said that an Angle containeth Twenty Thirty Forty Degrees when the Arch included betwixt its Lines contains Twenty Thirty Forty Degrees so the Angle is greatest which containeth the greatest number of Degrees As the Angle BAD is greater than GEL the Line CA divideth the Angle BAD in the middle because the Arches BC CD are equal and the Angle BAC is a part of BAD because the Arch BC is part of the Arch BD. 10. When a Line falling on another Line maketh the Angle on each side thereof equal Those Angles are Right Angles and the Line so falling is a Perpendicular Fig. 5. As if the Line AB falling on CD
the Base BC Let there be imagined another Triangle DEF having an Angle D equal to the Angle A and the Sides DE DF equal to AB AC Now since the Sides AB AC are equal the Four Lines AB AC DE DF shall be equal Demonstration because the Sides AB DE AC DF are equal as also the Angles A and D if we put the Triangle DEF on the Triangle ABC the Line DE shall fall upon AB and DF on AC and FE on BC by the Fourth therefore the Angle DEF shall be equal to ABC and since one part of the Line DE falls on AB the whole Line DI shall fall on AG otherwise Two streight Lines would contain a space therefore the Angle IEF shall be equal to the Angle GBC Now if you should turn the Triangle DEF and make the Line DF to fall on AB and DE on AC because the Four Lines AB DF AC DE are equal as also the Angles A and D the Two Triangles ABC DEF shall ly exactly on each other and the Angles ACB DEF HCB IEF shall be equal Now according to our first Comparison the Angle ABC was equal to DEF and GBC to IEF therefore the Angles ABC ACB which are equal to the same DEF and GBC HCB which are also equal to the same IEF are also equal among themselves I thought fit not to make use of Euclid's Demonstration because it being too difficult for young Learners to Apprehend they are often discouraged to proceed PROPOSITION VI. THEOREM IF Two Angles of a Triangle are equal that Triangle shall be an Isosceles Triangle Let the Angles ABC ACB of the Triangle ABC be equal I affirm that their opposite Sides AB AC are equal to each other let the Triangle DEF have the Base DF equal to BC and the Angle DEF equal to ABC as also DEF equal to ACB since then that we suppose that the Angles ABC ACB are equal the Four Angles ABC ACB DEF DFE are equal Now let us imagine the Base EF be so put on BC that the Point E fall on B it is evident that since they be equal the one shall not exceed the other anywise moreover the Angle E being equal to the Angle B and the Angle F to the Angle C the Line EB shall fall on BA and FD on CA therefore the Lines ED FB shall meet each other in the Point A from whence I conclude that the Line ED is equal to BA Again let us turn over the Triangle DEF placing the Point E on C and F on B which must so happen because BC is supposed equal to EF and because the Angles F and B E and C are supposed equal the Side FB shall fall on BA and ED on CA and the Point D on A wherefore the Lines AC DE shall be equal from whence I conclude that the Sides AB AC are equal to each other since they are equal to the same Side DE. USE Use 6. THis Proposition is made use of in measuring all inaccessible Lines it is said that Thales was the first that measured the Height of the Egyptian Pyramids by their Shadows it may easily be done by this Proposition for if you would measure the Height of the Pyramid AB you must wait till the Sun be elevated 45 degrees above the Horizon that is to say untill the Angle AGB be 45 degrees now by the Proposition the shadow BC is equal to the Pyramid AB for seeing the Angle ABC is a Right Angle and that the Angle ACB is half Right or 45 degrees the Angle CAB shall also be half a Right Angle as shall be proved hereafter Therefore the Angles BCA BAC are equal and by the Sixth the Sides AB BC are also equal I might measure the same without making use of the Shadow by going backwards from B untill the Angle ACB be half a Right Angle which I may know by a Quadrant Those Propositions are made use of in Trigonometry and several other Treatises I Omit the Seventh whose use is only to Demonstrate the Eighth PROPOSITION VIII THEOREM IF Two Triangles be equally Sided they shall also have their opposite Angles equal Let the Sides GI LT HI VT GH LV be equal I say that the Angle GIH shall be equal to the Angle LTV IGH to the Angle L IHG to the Angle V. From the Center H at the distance HI let the Circle IG be described and from the Center G with the Extent GI let the Circle HI be described Demonstration If the Base LV be laid on the Base HG they will agree because they are equal I further add that the Point T shall fall precisely at the Circumference of the Circle IG since we suppose the Lines HL VT are equal it must likewise fall at the Circumference IH seeing GI is equal to LT therefore it shall fall on the Point I which is the Point were those Circles cut each other but if you deny it and say it should fall on some other Point as at O then I say the Line HO that is to say VT would be greater then HI and the Line GO that is to say LT would be less than GI which is contrary to the Hypo. from whence I conclude the Triangles do agree in all their parts and that therefore the Angles GIH is equal to the Angle LTV This Proposition is necessary for the establishing the next following moreover when we cannot measure an Angle which is made in a Solid Body by reason we cannot place our Instruments we take the Three Sides of a Triangle and make another on Paper whose Angles we easily measure this is very useful in Dialling and in cutting of Stone and Bevelling of Timber PROPOSITION IX PROBLEM TO Bisect or Divide into two equal parts a Right-Lined Angle given SAT. Take AS equal to AT and draw the Line ST upon which make an equilateral Triangle SVT draw the Right Line VA it shall Bisect the Angle Demonstration AS is equal to AT and the Side AV is common and the Base SV equal to VT therefore the Angle SAV is equal to TAV which was to be done USE THis Proposition is very useful to Divide a Quadrant into Degrees it being the same thing to Divide an Angle or to Divide an Arch into two equal parts for the Line AV Divideth as well the Arch ST at the Angle SAT for if you put the Semi-Diameter on a Quadrant you cut off an Arch of 60 Degrees and Dividing that Arch into Two Equal parts you have 30 Degrees which being again Divided into two equal parts you have 15 Degrees It is true to make an end of this Division you must Divide an Arch in Three which is not yet known to Geometricians Our Sea Compasses are Divided into 32 Points by this Proposition PROPOSITION X. PROBLEM TO Divide a Line given into two equal parts Let the Line AB be proposed to be Divided into Two equal parts erect an Equilateral Triangle ABC
under AB and AC shall be Three times 8 or 24 the square of AC 3 is 9 the Rectangle comprehended under AC 3 and CB 5 is 3 times 5 or 15. It is evident that 15 and 9 are 24. USE A Â 43 C 40. 3 B Â 3 120. Â 9. 129 Â Â THis Proposition serveth likewise to Demonstrate the ordinary practice of Multiplication For Example if one would Multiply the Number 43 by 3 having separated the Number of 43 into two parts in 40 and 3 three times 43 shall be as much as three times 3 which is Nine the Square of Three and Three times Forty which is 120 for 129 is Three Times 43. Those which are young beginners ought not to be discouraged if they do not conceive immediately these Propositions for they are not difficult but because they do imagine they contain some great Mystery PROPOSITION IV. THEOREM IF a Line be Divided into Two Parts the Square of the whole Line shall be equal to the Two Squares made of its parts and to Two Rectangles comprehended under the same parts Let the Line AB be Divided in C and let the Square thereof ABDE be made let the Diagonal EB be drawn and the Perpendicular CF cutting the same and through that Point let there be drawn GL Parallel to AB It is evident that the Square ABDE is equal to the Four Rectangles GF CL CG LF The Two first are the Square of AC and of CB the Two Complements are comprehended under AC CB. Demonstration The Sides AE AB are equal thence the Angles AEB ABE are half Right and because of the Parallels GL AB the Angles of the Triangles of the Square GE by the 29th shall be equal as also the Sides by the 6th of the 1. Thence GF is the Square of AC In like manner the Rectangle CL is the Square of CB the Rectangle GC is comprehended under AC and AG equal to BL or BC the Rectangle LF is comprehended under LD equal to AC and under FD equal to BC. Coroll If a Diagonal be drawn in a Square the Rectangles through which it passeth are Squares USE A 144 B 22 C 12 THis Proposition giveth us the practical way of finding or extracting the Square Root of a Number propounded Let the same be the number A 144 represented by the Square AD and its Root by the Line AB Moreover I know that the Line required AB must have Two Figures I therefore imagine that the Line AB is Divided in C and that AC representeth the first Figure and BC the Second I seek the Root of the First Figure of the Number 144 which is 100 and I find that it is 10 and making its Square 100 represented by the Square GF I Subtract the same from 144 and there remains 44 for the Rectangles GC FL and the Square CL. But because this gnomonicall Figure is not proper I transport the Rectangle FL in KG and so I have the Rectangle KL containing 44. I know also almost all the Length of the Side KB for AC is 10 therefore KC is 20 I must then Divide 44 by 20 that is to say to find the Divisor I double the Root found and I say how many times 20 in 44 I find it 2 times for the Side BL but because 20 was not the whole Side KB but only KC this 2 which cometh in the Quotient is to be added to the Divisor which then will be 22. So I find the same 2 times precisely in 44 the Square Root then shall be 12. You see that the Square of 144 is equal to the Square of 10 to the Square of 2 which is 4 and to twice 20 which are Two Rectangles comprehended under 2 and under 10. PROPOSITION V. THEOREM IF a Right Line be cut into equal parts and into unequal parts the Rectangle comprehended under the unequal parts together with the Square which is of the middle part or difference of the parts is equal to the Square of half the Line If the Line AB is Divided equally in C and unequally in D the Rectangle AH comprehended under the Segments AD DB together with the Square of CD shall be equal to the Square CF that is of half of AB viz. CB. Make an end of the Figure as you see it the Rectangles LG DI shall be Squares by the Coroll of the 4th I prove that the Rectangle AH comprehended under AD and DH equal to DB with the Square LG is equal to the Square CF. Demonstration The Rectangle AL is equal to the Rectangle DF the one and the other being comprehended under half the Line AB and under BD or DH equal thereto Add to both the Rectangle CH the Rectangle AH shall be equal to the Gnomon LBG Again to both add the Square LG the Rectangle AH with the Square LG shall be equal to the Square CF. ARITHMETICALLY LEt AB be 10 AC is 5 as also CB. Let CD be 2 and DB 3 the Rectangle comprehended under AD 7 and DB 3 that is to say 21 with the Square of CD 2 which is 4 shall be equal to the Square of CB 5 which is 25. USE THis Proposition is very useful in the Third Book we make use thereof in Algebra to Demonstrate the way of finding the Root of an affected Square or Equation PROPOSITION VI. THEOREM IF one add a Line to another which is Divided into Two equal parts the Rectangle comprehended under the Line compounded of both and under the Line added together with the Square of half the Divided Line is equal to the Square of a Line compounded of half the Divided Line and the Line added If one add the Line BD to the Line AB which is equally Divided in C the Rectangle AN comprehended under AD and under DN or DB with the Square of CB is equal to the Square of CD Make the Square of CD and having drawn the Diagonal FD draw BG Parallel to FC which cuts FD in the Point H through which passeth HN Parallel to AB KG shall be the Square of BC and BN that of BD. Demonstration The Rectangles AK CH on equal Bases AC BC are equal by the 38th of the 1st The Complements CH HE are equal by the 43d of the 1st Therefore the Rectangles AK HE are equal Add to both the Rectangle CN and the Square KG the Rectangles AK CN that is to say the Rectangle AN with the Square KG shall be equal to the Rectangles CN HE and to the Square KG that is to say to the Square CE. Arithmetically or by Numbers LEt AB be 8 AC 4 CB 4 BD 3 then AD shall be 11. It is evident that the Rectangle AN three times 11 that is to say 33 with the Square of KG 16 which together are 49 is equal to the Square of CD 7 which is 49 for 7 times 7 is 49. USE Fig. 6. MAurolycus measured the whole Earth by one single
makes the Angles ABC ABD equal that is if having on B as a Center described a Semicircle CAD the Arches AC DA are found equal the Angle ABC ABD are called Right Angles and the Line AB a Perpendicular Now because the Arch CAD is a Semicircle the Arch CA AD are each of them a Quadrant that is to say the Fourth part of the Three Hundred and Sixty Degrees which is Ninety 11. An Obtuse Angle is greater than a Right Angle Fig. 5. As the Angle EBD is Obtuse or Blunt because its Arch EAD contains more than a Quadrant 12. An Acute Angle is less than a Quadrant Fig. 5. As the Angle EBC is Acute because the Arch EC which is the measure thereof is less than Ninety Degrees 13. A Term is the extremity or end of a Quantity 14. A Figure is a Quantity termined by one or many Terms It ought to be bounded and inclosed on every side to be called a Figure 15. A Circle is a Plain Figure whose bounds are made by the winding or turning of a Line which is called Circumference and which is equally distant from the middle Point Fig. 6. The figure RVSX is a Circle because all the Lines TR T S TV TX drawn from the Point T to the Line RVSX are equal 16. This middle Point is called the Center 17. Fig. 7. The Diameter of a Circle is any Line whatsoever which passeth through the Center and which ends at the Circumference cutting the same into two equal parts As the Line VTX or RTS some one may perhaps doubt whether the Line VTX doth effectually divide the Circle into Two equally that is whether the part VSX be equal to the part VRX that they are I thus prove Fig. 7. Let it be imagined that the part VRX be placed on VSX I say that the one shall not exceed the other for if the one as VSX should exceed the other VRX the Line TR would then be shorter than T S the same of TY in respect of TZ which is contrary to the Definitions of a Circle that is that all the Lines drawn from the Center of a Circle to the Circumference be all equal 18. Fig. 7. A Semicircle is a Figure Termined by the Diameter and half the Circumference as VSX 19. Right Lined Figures are Termined by streight Lines there are some of Three of Four of Five or as many sides as you would have them Euclide distinguisheth Triangles by their Angles or by their sides 20. Fig. 8. An Equilater Triangle is that whose Three sides are equal as ABC 21. Fig. 8. An Isosceles Triangle is that which hath two equal Sides As if the Sides AB AC are equal the Triangle ABC is an Isosceles Triangle 22. Fig. 9. Scalenum is a Triangle whose three Sides are unequal as DEF 23. Fg i 9. A Right Angled Triangle is that which hath one Angle right as DEF supposing the Angle E to be right 24. Fig. 10. An Amblygonium Triangle hath one Angle Obtuse as GHI 25. Fig. 10. An Oxygonium Triangle hath all its Angles Acute as ABC 26. Fig. 8. A Rect-Angle is a figure having four Sides which hath all its Angles right 27. Fig. 11. A Square hath all its Sides equal and all its Angles right as AB 28. Fig. 12. A Rect-Angular figure whose Lines are unequal and Angles right is called an Oblong or long Square as CD 29. Fig. 13. A Rhombus or Diamond figure is that which hath four equal Sides but is not right Angled as EF. 30. Fig. 14. A Rhomboides or Diamond like figure is that whose opposite Sides and opposite Angles are equal but hath neither equal Sides nor Right Angles as GH 31. Other Irregular figures having four sides are called Trapeziums 32. Fig. 15. Parallel or Equidistant Right Lines are such which being in the same Superficies if infinitely produced would never meet as the two Lines AB CD 33. Fig. 15. A Parallelogram is a figure whose opposite Sides are parallel as the figure ABCD of which the Sides AB CD AC BD are parallel 34. Fig. 16. The Diameter of a Parallelogram is a streight Line drawn from Angle to Angle as BC. 35. Fig. 16. The Complements are the two lesser Parallelograms through which the Diameter passeth not as AFEH GDIE Postulata or SVPPOSITIONS 1. IT is supposed that a streight Line may be drawn from any Point whatsoever to any other Point 2. That a streight Line may be infinitely continued 3. That on a Center given a Circle may be described with any distance taken between the Compasses AXIOMS 1. QUantities or things equal to the same third are also equal among themselves 2. If to two equal quantities be added two other Quantities which are also equal those Quantities which are produced shall be equal 3. If from two equal Quantities be taken away two other equal Quantities those Quantities remaining shall be equal 4. If equal Quantities be added to unequal Quantities their sums shall be unequal 5. If from equal Quantities be taken away unequal Quantities the Quantities remaining shall be unequal 6. The Quantities which are double treble quadruple to the same are equal among themselves 7. Quantities are equal when applyed together the parts of the one are not extended beyond those of the other 8. Lines and Angles are equal when the one being applied to the other they agree 9. Every whole is greater than its part and equal to all its parts taken together 10. All Right Angles are equal amongst themselves Fig. 17. Let there be proposed two Right Angles ABC EFH I say they are equal to each other for if two equal Circles be described CAD HEG on the Centers B and F the quarters of Circles CA HE which are the measures of the Angles ABC EFH shall be equal therefore the Angles ABC EFH whose measures being equal are consequently equal Fig. 18. The Eleventh maxim of Euclid beareth this signification that if the Lines AB CD maketh with the Line EF which cutteth both the said Lines the Internate Angle BEF DFE less than two Right Angles the Lines AB CD being continued shall meet each other on the same side with B and D. Although this maxim be true yet it is not evident enough to be received for a maxim wherefore I substitute another in the place thereof 11. If two Lines are parallel all the perpendiculars inclosed between them shall be equal Fig. 19. As if the Line AB CD are parallels the perpendicular Lines FE HG are equal for if EF was greater than GH the Lines AB and CD should be farther distant from each other towards the Points E and F than towards G and H which would be contrary to the Definition of parallels which sayeth that they are equidistant from each other 12. Two Right Lines do not comprehend a Superficies that is to say do not incompass a space on every
the Squares of the other two Sides AB AC Draw the Line AH Parallel to BD CE and draw also the Lines AD AE FC BG I prove that the Square AF is equal to the Right Angled Figure or long square BH and the Square AG to the Right Angled Figure CH and that so the Square BE is equal to the Two Squares AF AG. Demonstration The Triangles FBC ABD have their Sides AB BF BD BC equal and the Angles FBC ABD are equal Seeing that each of 'em besides the Right Angle includes the Angle ABC Thence by the 4th the Triangles ABD FBC are equal Now the Square AF is double to the Triangle FBC by the 41st because they have the same Base BF and are between the same Parallels BF AC Likewise the Right Lined Figure BH is double to the Triangle ABD seeing they have the same Base BD and are between the same Parallels BD AH Therefore the Square AF is equal to the Right Lined Figure BH After the same manner the Triangles ACE GCB are equal by the 4th the Square AG is double the Triangle BCG and the Right Lined Figure CH is double the Triangle ACE by the 41st Thence the Square AG is equal to the Right Lined Figure CH and by consequence the Sum of the Squares AF AG are equal to the Square BDEC USE Use 47. IT is said that Pythagoras having found this Proposition Sacrificed One Hundred Oxen in thanks to the Muses it was not without reason seeing this Proposition serves for a Foundation to a great part of the Mathematicks For in the First place Trigonometry cannot be without it because it is necessary to make the Table of all the Lines that can be drawn within a Circle that is to say of Chords of Sines Also Tangents and Secants which I shall here shew by one Example Let it be supposed that the Semi-Diameter AB be Divided into 10000 parts and that the Arch BC is 30 degrees Seeing the Chord or subtendent of 60 Degrees is equal to the Semi-diameter AC BD the Sine of 30 degrees shall be equal to the half of AC it shall therefore be 5000 in the Right Angled Triangle ADB The Square of AB is equal to the Squares of BD and AD make then the Square of AB by Multiplying 10000 by 10000 and from that Product Subtract the Square of BD 5000 there remains the Square of AD or BF the Sine of the Complement and extracting the Square Root there is found the Line FB Then if by the Rule of Three you say as AD is to BD so is AC to CE you shall have the Tangent CE and adding together the Squares of AC CE you shall have by the 47th the Square of AE and by extracting the Root thereof you shall have the Length of the Line AE the Secant Use 47. We augment Figures as much as we please by this Proposition Example to double the Square ABCD continue the Side CD and make DE equal to AD the Square of AE shall be the double of the Square of ABCD seeing that by the 47th it is equal to the Squares of AD and DE. And making a Right Angle AEF and taking EF equal to AB the Square of AF shall be Triple to ABCD. And making again the Right Angle AFG and FG equal to AB the Square of AG shall be Quadruple to to ABCD. What I here say of a Square is to be understood of all Figures which are alike that is to say of the same species PROPOSITION XLVIII THEOREM IF the Two Squares made upon the Side of a Triangle be equal to the Square made on the other Side then the Angle comprehended under the Two other Sides of the Triangle is a Right Angle If the Square of the Side NP is equal to the Squares of the Sides NL LP taken together the Angle NLP shall be a Right Angle draw LR Perpendicular to NL and equal to LP then draw the Line NR Demonstration In the Right Angled Triangle NLR the Square of NR is equal to the Squares of NL and of LR or LP by the 47th now the Square of NP is equal to the same Squares of NL LP therefore the square of NR is equal to that of NP and by consequence the Lines NR NP are equal And because the Triangles NLR NLP have each of them the Side NL common and that their Bases RN NP are also equal the Angles NLP NLR shall be equal by the 8th and the Angle NLR being a Right Angle the Angle NLP shall be also a Right Angle The End of the First Book THE SECOND BOOK OF Euclid's Elements EUclid Treateth in this Book of the Power of Streight Lines that is to say of their Squares comparing the divers Rectangles which are made on a Line Divided as well with the Square as with the Rectangle of the whole Line This part is very useful seeing it serveth for a Foundation to the Practical Principles of Algebra The Three first Propositions Demonstrateth the Third Rule of Arithmetick The Fourth teacheth us to find the Square Root of any number whatsoever those which follow unto the Eighth serveth in several accidents happening in Algebra The remaining Propositions to the end of this Book are conversant in Trigonometry This Book appeareth at the first sight very difficult because one doth imagine that it contains mysterious or intricate matters notwithstanding the greater part of the Demonstrations are founded on a very evident Principle viz. That the whole is equal to all its parts taken together therefore one ought not to be discouraged although one doth not Apprehend the Demonstrations of this Book at the First Reading DEFINITIONS Def. 1. of the Second Boook A Rectangular Parallelogram is Comprehended under Two Right Lines which at their Intersection containeth a Right Angle It is to be noted henceforward that we call that Figure a Rectangular Parallelogram which hath all its Angles Right and that the same shall be distinguished as much at is requisite if we give thereto Length and Breadth naming only Two of its Lines which comprehendeth any one Angle as the Lines AB BC For the Rectangular Parallelogram ABCD is comprehended under the Lines AB BC having BC for its Length and AB for its Breadth whence it is not necessary to mention the other Lines because they are equal to those already spoken of I have already taken notice that the Line AB being in a Perpendicular Position in respect of BC produceth the Rectangle ABCD if moved along the Line BC and that this Motion Representeth Arithmetical Multiplication in this manner as the Line AB moves along the Line BC that is to say taken as many times as there are Points in BC Composeth the Rectangle ABCD wherefore Multiplying AB by BC I shall have the Rectangle ABCD. As suppose I know the Number of Mathematical Points there be in the Line AB for Example let there be 40 and that in BC
the Third proportional to Two Lines PROPOSITION XXXVI THEOREM IF from a Point taken without a Circle one draw a touch Line and another Line to cut the Circle the Square of the Tangent Line shall be equal to the Rectangle comprehended under the whole Secant and under the exteriour Line Let from the Point A taken without the Circle be drawn a Line AB to touch the same in B another AC or AH to cut the Circle The Square of AB shall be equal to the Rectangle comprehended under AC AO as also to the Rectangle comprehended under AH AF. If the Secant pass through the Center as AC draw the Line EB Demonstration Seeing the Line OC is divided in the middle in E and that thereto is added the Line AO the Rectangle comprehended under AO AC with the Square of OE or EB shall be equal to the Square of AE by the 6th of the 2d Now the Line AB toucheth the Circle in the Point B so by the 17th the Angle B is Right and by the 47th of the 1st the Square of AE is equal to the Squares of AB EB Therefore the Rectangle under AC AO with the Square of EB is equal to the Squares of AB EB and taking away the Squares of EB the Rectangle under AC AO shall be equal to the Square of AB Secondly let the Secant AH not pass through the Center Draw the Line AH the perpendicular EG which shall divide in the middle in G the Line FH draw also the Line EF. Demonstration The Line FH being divided equally in the point G and the Line AF being added thereto the Rectangle comprehended under AH AF with the Square of FG shall be equal to the Square of AG. Add to both the Square of EG the Rectangle AH AF with the Square of FG GE that is to say by the 47th of the 1st with the Square of FE or EB shall be equal to the Square of AG GE that is to say by the 47th of the 1st to the Square of AE Moreover the Square of AE by the same is equal to the Square of EB AB therefore the Rectangle compprehended under AH AF with the Square of BE is equal to the Squares of BE AB and taking from both the Square of BE the Rectangle comprehended under AH AF shall be equal to the Square of AB Coroll 1. If you draw several Secants AC AH the Rectangles AC AO and AH AF shall be equal amongst themselves seeing that the one and the other are equal to the Square of AB Coroll 2. If there be drawn Two Tangents AB AI they shall be equal because their Squares are equal to the same Rectangle AC AO and by consequence amongst themselves as also the Lines PROPOSITION XXXVII THEOREM IF the Rectangle comprehended under the Secant and under the exteriour Line is equal to a Line which falls on the Circle that Line toucheth the same Let be drawn the Secant AC or AH and let the Rectangle AC AO or the Rectangle AH AF be equal to the Square of the Line AB that Line shall touch the Circle draw the touch Line AI by the 17th and the Line IE Demonstration Seeing the Line AI toucheth the Circle the Rectangle AC AO or AH AF shall be equal to the Square of AI. Now the Square of AB is supposed equal to each Rectangle therefore the Squares of AI and of AB are equal and by consequence the Lines AI AB So then the Triangles ABE AEI which have all their Sides equal to each other shall be equiangled by the 8th of the 1st and seeing the Angle AIE is Right by the 17th the Line AI being a touch Line the Angle ABE shall be Right and the Line AB a touch Line also by the 16th USE Prop. XXXVII MAurolicus maketh use of this Proposition to find the Diameter of the Earth For looking from the top of a Mountain OA to the extremity of the Earth along the Line BA he observeth the Angle OAB which the Line AB maketh with a plumb Line AC and he concludes the Length of the Line AB by a Trigonometrical Calculation He Multiplieth AB by AB to have its Square which he divides by AO the height of the Mountain from which having taken away AO there remains OC the Diameter of the Earth this Proposition serveth also to prove the Fifth Proposition of the Third Book of Trigonometry The end of the Third Book THE FOURTH BOOK OF Euclid's Elements THis Book is very useful in Trigonometry seeing that by inscribing of Polygons in a Circle we have the practice of making the Table of Subtendants of Sines of Tangents and of Secants which are very necessary in all sorts of measuring Secondly in inscribing of Polygons in a Circle we have the diversity of the Aspects of the Planets which take their Names from the same Polygons Thirdly in the practice hereof we have a method which giveth the Quadrature of the Circle as near the truth as shall be needful We demonstrate also that a Circle is in duplicate reason to its Diameter Fourthly Military Architecture hath need of Inscribing Polygons into Circles in the designing or drawing of regular Fortifications The DEFINITIONS 1. Fig. I. Plate V. A Right Lined figure is inscribed in a Circle or the Circle is describ'd about a figure when all its Angles are in the Circumference of the same Circle As the Triangle ABC is inscribed in a Circle and the Circle is described about the Triangle because the Angles A B C do touch the Circumference the Triangle DEF is not inscribed in the Circle because the Angle D doth not touch the Circumference of the Circle 2. Fig. I. A Right Lined Figure is described about a Circle and the Circle is inscribed within the same figure when all the sides of the Figure touch the Circumference of the Circle As the Triangle GHI is described about the Circle KLM because its Sides touch the Circle in K L M. 3. A Line is fitted or inscribed in a Circle when it is terminated at both â—…s by the Circumference of the Circle As the Line NO The Line RP is not inscribed in the Circle PROPOSITION I. PROBLEM TO Inscribe in a Circle a Line which surpasseth not its Diameter It is proposed to inscribe in a Circle AEBD a Line which surpasseth not its Diameter take the length thereof on the Diameter and let it be for Example BC. Put the Foot of the Compass in B and describe a Circle at the extent of BC which cutteth the Circle AEBD in D and E. Draw the Line BD or BE. It is evident they are equal to BC by the Definition of a Circle USE THis Proposition is necessary for the practice of those which follow PROPOSITION II. PROBLEM TO Inscribe in a Circle a Triangle equiangled to another Triangle The Circle EGH is proposed in which one would inscribe a Triangle equiangled to the Triangle
Superficies 18. A Cone is a figure made when one Side of a Right Angled Triangle viz. one of those that contain the Right Angle remaining fixed the Triangle is turned round about till it return to the place from whence it first moved And if the fixed Right Line be equal to the other which containeth the Right Angle then the Cone is a Rectangled Cone but if it be less it is an Obtuse Angled Cone if greater an Acute Angled Cone 19. The Axis of a Cone is that fixed Line about which the Triangle is moved 20. A Cylinder is a figure made by the moving round of a Right Angled Parallelogram one of the sides thereof namely which contains the Right Angle abiding fixed till the parallelogram be turned about to the same place whence it began to move 21. Like Cones and Cylinders are those whose Axes and Diameters of their Bases are Proportional Cones are right when the Axis is perpendicular to the Plain of the Base and they are said to be Scalene when the Axis is inclined to the Base and the Diameter of their Bases are in the same Ratio We add that inclined Cones to be like their Axes must have the same inclination to the Planes of their Bases PROPOSITION I. THEOREM Plate VII Prop. I. A Strait Line cannot have one of its parts in a Plane and the other without it If the Line AB be in the Plane AD it being continued shall not go without but all its parts shall be in the same Plane For if it could be that BC were a part of AB continued Draw in the Plane CD the Line BD perpendicular to AB draw also in the same Plane BE perpendicular to BD. Demonstration The Angles ABD BDE are both Right Angles thence by the 14th of the first AB BE do make but one Line and consequently BC is not a part of the Line AB continued otherwise two strait Lines CB EB would have the same part AB that is AB would be part of both which we have rejected as false in the Thirteenth Maxim of the first Book USE WE establish on this Proposition a principle in Gnomonicks to prove that the shadow of the stile falleth not without the Plane of a great Circle in which the Sun is Seeing that the end or top of the stile is taken for the Center of the Heavens and consequently for the Center of all the great Circles the shadow being always in a streight Line with the Ray drawn from the Sun to the Opaque Body this Ray being in any great Circle the shadow must also be therein PROPOSITION II. THEOREM LInes which cut one another are in the same Plane as well as all the parts of a Triangle If the Two Lines BE CD cut one another in the Point A and if there be made a Triangle by drawing the Base BC I say that all the parts of the Triangle ABC are in the same plane and that the Lines BE CD are likewise therein Demonstration It cannot be said that any one part of the Triangle ABC is in a Plane and that the other part is without without saying that one part of a Line is in one Plane and that the other part of the same Line is not therein which is contrary to the first Proposition and seeing that the sides of the Triangle are in th same Plane wherein the Triangle is the Lines BE CD shall be in the same Plane USE THis Proposition doth sufficiently determine a Plane by two streight Lines mutually intersecting each other or by a Triangle I have made use thereof in Opticks to prove that the objective parallel Lines which fall on the Tablet ought to be Represented by Lines which concur in a Point PROPOSITION III. THEOREM THe common section of two Places is a streight Line If Two Planes AB CD cut one another their common section EF shall be a streight Line For if it were not take Two Points common to both Planes which let be E and F and draw a strait Line from the point E to the point F in the Plane AB which let be EHF Draw also in the Plane CD a streight Line from E to F if it be not the same with the former let it be EGF Demonstration Those Lines drawn in the Two Planes are two different Lines and they comprehend a space whch is contrary to the Twelfth Maxim Thence they are but one Line which being in both Planes shall be their common section USE THis Proposition is fundamental We do suppose it in Gnomonicks when we represent in a Dial the Circles of the hours marking only the common section of their Planes and that of the Wall PROPOSITION IV. THEOREM IF a Line be perpendicular to two other Lines which cut one another it shall be also perpendicular to the Plane of those Lines If the Line AB be perpendicular to the Lines CD EF which cut one another in the point B in such manner that the Angles ABC ABD ABE ABF be right which a flat figure cannot represent it shall be perpendicular to the Plane CD EF that is to say that it shall be Perpendicular to all the Lines that are drawn in that Plane through the point B as to the Line GBH Let equal Lines be cut BC BD BE BF and let be drawn the Lines EC DF AC AD AE AF AG and AH Demonstration The four Triangles ABC ABD ABE ABF have their Angles Right in the Point B and the Sides BC BD BE BF equal with the side AB common to them all Therefore their Bases AC AD AE AF are equal by the 4th of the 1st 2. The Triangles EBC DBF shall be equal in every respect having the Sides BC BD BE BF equal and the Angles CBE DBF opposite at the vertex being equal so then the Angles BCE BDF BEC BFD shall be equal by the 4th of the first and their Bases EC DF equal 3. The Triangles GBC DBH having their opposite Angles CBG DBH equal as also the Angles BDH BCG and the sides BC BD they shall then have by the 26th of the 1st their Sides BG BH CG DH equal 4. The Triangles ACE AFD having their sides AC AD AE AF equal and the Bases EC DF equal they shall have by the 8th of the 1st the Angles ADF ACE equal 5. The Triangles ACG ADH have the Sides AC AD CG DH equal with the Angles ADH AGC Thence they shall have their Bases AG AH equal Lastly the Triangles ABH ABG have all their sides equal thence by the 27th of the 1st the Angles ABG ABH shall be equal and the Line AB perpendicular to GH So then the Line AB shall be perpendicular to any Line which may be drawn through the point B in the Plane of the Lines CD EF which I call perpendicular to the Plane USE THis Proposition cometh often in use in the first Book of Theodosius for example to Demonstrate that the Axis of the World is
the Lines AB BC you would have divided them equally and Perpendicularly by so doing This is very necessary to describe Astrolabes and to compleat Circles of which we have but a part That Astronomical Proposition which teacheth to find the Apogeum and the excentricity of the Suns Circle requires this Proposition We often make use there of in the Treatise of cutting of stones PROPOSITION XXVI THEOREM THe equal Angles which are at the Center or at the Circumference of equal Circles have for Base equal Arks. If the equal Angles D and I are in the Center of equal Circles ABC EFG the Arks BC FG shall be equal For if the Ark BC was greater or lesser than the Ark FG seeing that the Arks are the measure of the Angles the Angle D would be greater or lesser than the Angle I. And if the equal Angles A and E be in the Circumference of the equal Circles the Angles D and I which are the double of the Angles A and E being also equal the Arks BC FG shall be also equal PROPOSITION XXVII THEOREM THe Angles which are either in the Center or in the Circumference of equal Circles and which hath equal Arks for Base are also equal If the Angles D and I are in the Centers of equal Circles and if they have for Base equal Arcks BC FG they shall be equal because that their measures BC FG are equal and if the Angles A and E be in the Circumference of equal Circles have for Base equal Arks BC EG the Angles in the Center shall be equal and they being their halfs by the 20th shall be also equal PROPOSITION XXVIII THEOREM EQual Lines in equal Circles correspond to equal Arks. If the Line BC EF are applyed in equal Circles ABC DEF they shall be Chords of equal Arks BC EF. Draw the Lines AB AC DE EF. Demonstration In the Triangles ABC DEF the Sides AB AC DE EF are equal being the Semi-Diameters of equal Circles the Bases BC EF are supposed equal thence by the 8th of the 1st the Angles A and D shall be equal and by the 16th the Arks BC EF shall be also equal PROPOSITION XXIX THEOREM LInes which subtend equal Arcks in equal Circles are equal If the Lines BC EF subtend equal Arks BC EF in equal Circles those Lines are equal Demonstration The Arks BC EF are equal and parts of equal Circles therefore by the 27th the Angles A and D shall be equal So then in the Triangles CAB EDF the Sides AB AC DE DF being equal as also the Angles A and D the Bases BC EF shall be equal by the 4th of the 1st USE THeodosius demonstrateth by the 28th and 29th that the Arks of the Circles of the Italian and Babylonian hours comprehended between Two Parallels are equal We demonstrate also after the same manner that the Arks of Circles of Astronomical hours comprehended between Two Parallels to the Equator are equal these Propositions come almost continually in use in spherical Trigonometry as also in Gnomonicks PROPOSITION XXX PROBLEM TO divide an Ark of a Circle into Two equal parts It is proposed to Divide the Ark AEB into Two equal parts put the Foot of the Compass in the Point A make Two Arks F and G then transporting the Compass without opening or shutting it to the Point B describe two Arks cutting the former in F and G the Line GF will cut the Ark AB equally in the Point E. Draw the Line AB Demonstration You divide the Line AB equally by the construction for imagine the Lines AF BF AG BG which I have not drawn lest I should imbroil the figure the Triangles FGA FGB have all their Sides equal so then by the 8th of the 1st the Angles AFD BFD are equal Moreover the Triangles DFA DFB have the Sides DF common the Sides AF BF equal and the Angles DFA DFB equal whence by the 4th of the 1st the Bases AD DB are equal and the Angles ADF BDF are equal We have then divided the Line AB equally and perpendicularly in the Point D. So then by the 1st the Center of the Circle is in the Line EG Let it be the Point C and let be drawn the Lines CA CB all the Sides of the Triangles ACD BCD are equal Thence the Angles ACD BCD are equal by the 8th of the 1st and by the 27th the Arks AE EB are equal USE AS we have often need to divide an Ark in the middle the practice of this Proposition is very ordinarily in use it is by this means we divide the Mariners Compass into 32 Rumbs for having drawn Two Diameters which cut each other at Right Angles we divide the Circle in Four and sub-dividing each quarter in the middle we have Eight parts and sub-dividing each part twice we come to Thirty Two parts We have also occasion of the same practice to divide a Semi-circle into 180 degrees and because for the performing the same Division throughout we are obliged to divide an Ark into Three all the Ancient Geometricians have endeavoured to find a method to divide an Angle or an Ark into Three equal parts but it is not yet found PROPOSITION XXXI THEOREM THe Angle which is in a Semi-circle is Right that which is comprehended in a greater Segment is Acute and that in a lesser Segment is Obtuse If the Angle BAC be in a Semi-circle I demonstrate that it is Right Draw the Line DA. Demonstration The Angle ADB exteriour in respect of the Triangle DAC is equal by the 32d of the 1st to the Two Interiours DAC DCA and those being equal by the 5th of the 1st seeing the Sides DA DC are equal it shall be double to the Angle DAC In like manner the Angle ADC is double to the Angle DAB therefore the Two Angles ADB ADC which are equal to Two Right are double to the Angle BAC and by consequence the Angle BAC is a Right Angle Secondly the Angle AEC which is in the Segment AEC is obtuse for in the Quadrilateral ABCE the Opposite Angles E and B are equal to Two Right by the 22d the Angle B is Acute therefore the Angle E shall be Obtuse Thirdly the Angle B which is in the Segment ABC greater than a Semi-circle is Acute seeing that in the Triangle ABC the Angle BAC is a Right Angle USE Use 31. THe Workmen have drawn from this Proposition the way of trying if their Squares be exact for having drawn a Semi-circle BAD they apply the Point A of their Square BAD on the Circumference of the Circle and one of its Sides AB on the Point B of the Diameter the other Side AD must touch the other Point D which is the other end of the Diameter Ptolomy makes use of this Proposition to make the Table of Subtendants or Chords of which he hath occasion in Trigonometry Use 31. We have also a practical way to erect a perpendicular on the end of
a Line which is founded on this Proposition For Example to erect a Perpendicular from the Point A of the Line AB I put the Foot of the Compass on the Point C taken at discretion and extending the other to A I describe a Circle which may cut the Line AB in the Point B. I draw the Line BCD It is evident that the Line DA shall be perpendicular to the Line AB seeing the Angle BAD is in the Semi-circle PROPOSITION XXXII THEOREM THe Line which cutteth the Circle at the Point of touching maketh with the touch Line the Angles equal to those of the Alternate Segments Let the Line BD cut the Circle in the Point B which is the Point where the Line AB doth touch the same I say that the Angle CBD which the Line BD Comprehendeth with the touch Line ABC is equal to the Angle E which is that of the Alternate Segment BED and that the Angle ABD is equal to the Angle F of the Segment BFD In the first place if the Line passeth through the Center as doth the Line BE It would make with the touch Line AB two Right Angles by the 17th and the Angle of the Semi-circles would be also Right by the preceding So the Proposition would be true If the Line passeth not through the Center as doth the Line BD Draw the Line BE through the Center and joyn the Line DE Demonstration The Line BE maketh two Right Angles with the touch Line and all the Angles of the Triangle BDE are equal to Two Right by the 32d of the 1st So taking away the Right Angles ABE and D which is in a Semi-circle and taking again away the Angle EBD which is common to both the Angle CBD shall be equal to the Angle BED Thirdly the Angle ABD is equal to the Angle F because in the Quadrilateral BFDE which is inscribed in a Circle the opposite Angles E and F are equal to two Right by the 22d the Angles ABD DBC are also equal to Two Right by the 13th of the 1st and the Angle DBC and E are equal as just now I did demonstrate therefore the Angles ABD and BFD are equal USE THis Proposition is necessary for that which followeth PROPOSITION XXXIII PROBLEM TO describe upon a Line a Segment of a Circle which shall contain a given Angle It is proposed to describe on the Line AB a Segment of a Circle to contain the Angle C. Make the Angle BAD equal to the Angle C and draw to AD the Perpendicular AE Make also the Angle ABF equal to the Angle BAF and lastly describe a Circle on the Point F as Center at the opening BF or FA the Segment BEA containeth an Angle equal to the Angle C. Demonstration The Angles BAF ABF being equal the Lines FA FB are equal by the 6th and the Circle which is described on the Center F passeth through A and B Now the Angle DAE being Right the Line DA toucheth the Circle in the Point A by the 16th therefore the Angle which the Segment BEA comprehendeth as the Angle E is equal to the Angle DAB that is to say to the Angle C. But if the Angle was obtuse we must take the Acute Angle which is its Complement to 180 degrees PROPOSITION XXXIV PROBLEM A Circle being given to cut there-from a Segment to contain an assigned Angle To cut from the Circle BCE a Segment to contain the Angle A. Draw by the 17th the touch Line BD and make the Angle DBC equal to the Angle A. It is evident by the 32d that the Segment BEC contains an Angle equal to DBC and by consequence to the Angle A. USE I Have made use of this Proposition to find Geometrically the excentricity of the Annual Circle of the Sun and his Apogeon Three observations being given It is also made use of in Opticks Two unequal Lines being proposed to find a Point where they shall appear equal or under equal Angles making on each Segments which may contain equal Angles PROPOSITION XXXV THEOREM IF Two Lines cut each other in a Circle the Rectangle comprehended under the parts of the one is equal to the Rectangle comprehended under the parts of the other In the first place if Two Lines cut each other in the Center of the Circle they shall be equal and divided equally so then it is evident that the Rectangle comprehended under the parts of the one is equal to the Rectangle comprehended under the parts of the other Secondly let one of the Lines pass through the Center F as AC and divide the Line BD in two equally in the Point E I say that the Rectangle comprehended under AE EC is equal to the Rectangle comprehended under BE ED that is to say to the Square of BE. The Line AC is perpendicular to BD by the Third Demonstration Seeing that the Line AC is divided equally in F and unequally in F the Rectangle comprehended under AE EC with the Square of FE is equal to the Square of FC or FB by the 5th of the 2d Now the Angle E being Right the Square of FB is equal to the Squares of BE EF therefore the Rectangle comprehended under AE EC with the Square of EF is equal to the Squares of BE EF and taking away the Square of EF there remains that the Square of BE is equal to the Rectangle under BE ED. Thirdly let the Line AB pass through the Center F and let it divide the Line CD unequally in the Point E draw FG perpendicularly to CD and by the 3d. the Lines CG GD shall be equal Demonstration Seeing the Line AB is divided equally in F and unequally in E the Rectangle comprehended under AE EB with the Square of EF is equal to the Square of BF or FC by the 5th of the 2d In the place of EF put the Squares of FG GE which is equal thereto by the 47th of the 1st In like manner the Line CD being equally divided in G and unequally in E the Rectangle CED with the Square of GE shall be equal to the Square of GC Add the Square of GF the Rectangle of CE ED with the Squares of GE FG shall be equal to the Squares of GC GF that is to say by the 47th of the 1st to the Square of CF. Therefore the Rectangle AEB with the Squares of GE GF and the Rectangle of CE ED with the same Squares are equal and by consequence taking away the same Squares the Rectangle AEB is equal to the Rectangle CFD Fourthly let the Lines CD HI cut each other in the Point E so that neither of them pass through the Center I say that the Rectangle CED is equal to the Rectangle HEI For drawing the Line AFB the Rectangles CED HEI are equal to the Rectangle AEB by the preceding case therefore they are equal USE ONe might by this Proposition have a practical way to find the Fourth proportional to Three given Lines or