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
Let the Line AB Represent the Plain and from the Center of the Earth C let the Line CA be drawn Perpendicular to the Line AB I say that a Boul being placed on the Point B ought not to stand on that Point because no heavy Body will stand still while it may Descend Now the Boul B going towards A is always Descending and cometh nearer and nearer the Center of the Earth C Because in the Triangle CAB the Perpendicular CA is shorter than BC. We also prove in like manner that a liquid Body ought to Descend from B to A and that the surface ought to be â—ound PROPOSITION XX. THEOREM THe Two Sides of a Triangle taken together are greater then the Third I say that the Two Sides TL LV of the Triangle TLV are greater than the Side TV. Some prove this Proposition by the Definition of a streight Line which is the shortest which can be drawn between Two Points Therefore the Line TV is less than the Two Lines TL LV. But we may Demonstrate the same another way continue the Side LV to R and let LR LT be equal then draw the Line RT Demonstration The Sides LT LR of the Triangle LTR are equal Therefore the Angles R and RTL equal by the 5â—h but the Angle RTV is greater than the Angle RTL Therefore the Angle RTV is greater than the Angle R and by the 17th in the Triangle RTV the Side RV that is to say the sum of the Sides LT LV are greater than the Side TV. PROPOSITION XXI THEOREM IF on the same Base you draw a lesser Triangle in a greater the Sides of the lesser shall be shorter than the greater but contain a greater Angle Let the less Triangle ADB be drawn within the greater ACB on the same Base AB I say in the first place that the Sides AC BC are greater than the Sides AD BD. Continue the Side AD unto E. Demonstration In the Triangle ACE the Sides AC CE taken together are greater than the Side AE by the 20th Add thereto the Side EB the Sides AC CEB shall be greater than the Sides AE EB Likewise in the Triangle DBE the Two Sides BE ED taken together are greater than BD and adding thereto AD the Sides AE EB shall be greater than AD BD. Moreover I say that the Angle ADB is greater than the Angle ACB For the Angle ADB is exteriour in respect of the Triangle DEB it is therefore greater than the Interiour DEB by the 16th Likewise the Angle DEB being Exteriour in respect of the Triangle ACE is greater than the Angle ACE Therefore the Angle ADB is greater than the Angle ACB USE Prop. XXI WE Demonstrate in Opticks by this Proposition that if from the Point C one should see the Base AB it would seem less than if one should see the same from the Point D according to this Principle that quantities seen under a greater Angle appear greater for which reason Vitruvius would that the Tops of very high Pillars should be made but little tapering because that their Tops being at a good distance from the Eyes will of themselves appear very much diminished PROPOSITION XXII PROBLEM TO make a Triangle having its Sides equal to Three Right Lines given provided that Two of them be greater than the Third Let it be proposed to make a Triangle having its Three Sides equal to the Three given Lines AB D E take with your Compasses the Line D and putting one Foot thereof in the Point B make an Arch Then take in your Compasses the Line E and putting one Foot in the Point A cross with the other Foot the former Arch in C Draw the Lines AC BC. I say that the Triangle ABC is what you desire Demonstration The Side AC is equal to the Line E since it is the Radius of an Arch drawn on the Center A equal in Length to the Line E likewise the Side BC is equal to the Line D Therefore the Three Sides AC BC AD are equal to the Lines E D AB USE WE make use of this Proposition to make a Figure equal or like unto another for having Divided the Figure into Triangles and making other Triangles having their Sides equal to the sides of the Triangles in the proposed figure we shall have a Figure like unto the same in all Respects But if we desire it should be only like thereunto but lesser for Example if we would have the Form of any Plain or Piece of Ground on Paper having Divided the same into Triangles and measured all their Sides we make Triangles like unto those of the Plain by the help of a Scale of equal parts from which we take the number of Parts which their Sides contain whether Feet Rods or any other measure and applying them as is here Taught PROPOSITION XXIII PROBLEM TO make an Angle equal to an Angle given in any Point of a Line Let it be proposed to make an Angle equal to EDF at the Point A of the Line AB at the Points A and D as Centers draw two Arches BC EF with the same extent of the Compasses then take the Distance EF between your Compasses put one Foot in B and cut off BC and draw AC I say that the Angles BAC EDF are equal Demonstration The Triangles ABC DEF have the Sides AB AC equal to the Sides DE DF since that the Arches BC EF were described with the same extent of the Compass they have also their Bases BC EF equal Therefore the Angles BAC DEF are equal by the 8th USE THis Problem is so necessary in Surveying Fortifications Prospective Dialling and in all other parts of the Mathematicks so that the greatest part of their Practice would be impossible if one Angle could not be made equal to another or of any number of Degrees required PROPOSITION XXIV THEOREM IF Two Triangles which have Two Sides of the one equal to the Two Sides of the other that which hath the greatest Angle hath the greatest Base Let the Sides AB DE AC DF of the Triangles ABC DEF be equal and let the Angle BAC be greater than the Angle EDF I say that the Base BC is greater than the Base EF Make the Angle EDG equal to the Angle BAC by the 23d. and the Line DG equal to AC then draw EG In the first place the Triangles ABC DEG having the Sides AB DE AC DG equal and the Angle EDG equal to BAC their Bases BC EG are equal by the 4th and the Lines DG DF being equal to AC they shall be equal amongst themselves Demonstration In the Triangle DGF the Sides DG DF being equal the Angles DGF DFG are equal by the 5th but the Angle EGF is less than DGF and the Angle EFG is greater than DFG Therefore in the Triangle EFG the Angle EFG shall be greater than the Angle EGF thence by the 18th the Line EG opposite to the greatest Angle EFG shall be
greater than EF thence BC equal to EG is greater than the Base EF. PROPOSITION XXV THEOREM OF Two Triangles which have Two Sides of the one equal to Two Sides of the other that Triangle which hath the greatest Base hath also the greatest Angle Let the Sides AB DE AC DF of the Triangles ABC DEF be equal and let the Base BC be greater than the Base EF. I say that the Angle A shall be greater than the Angle D. Demonstration If the Angle A were not greater than the Angle D it would be equal or less if equal in this case the Bases BC would be equal by the 4th If less then the Base EF would be greater than the Base BC by the 24th both which is contrary to our Hyp. These Propositions are of use to Demonstrate those which follow PROPOSITION XXVI THEOREM A Triangle which hath One Side and Two Angles equal to those of an other shall be equal thereto in every respect Let the Angles ABC DEF ACD DFE of the Triangles ABC DEF be equal and let the Sides BC EF which are between those Angles be also equal to each other I say that their other Sides are equal For Example AC DF but let it be imagined that the Side DF is greater than AC and that GF is equal to AC and draw the Line GE. Demonstration The Triangles ABC GEF have the Sides EF BC AC GF equal the Angle C is also supposed equal to the Angle F thence by the 4th the Triangles ABC GEF are equal in every respect and the Angles GEF ABC are equal but according to our first Hyp. the Angles ABC DEF were equal by this Argument the Angles DEF GEF would be equal that is to say the whole equal to a part which is impossible therefore DF cannot be greater than AC nor AC greater than DF because the same Demonstration might be made in the Triangle ABC Secondly Let us suppose that the Angles A and D C and F are equal and that the Sides BC EF which are opposite to the equal Angle A and D are also equal to each other I say the other Sides are equal for if DF be greater than AC make GF equal to AC and draw the Line GE. Demonstration The Triangles ABC GEF have the Sides EF BC FG CA equal they are then equal in every respect by the 4th and the Angles EGF BAC shall be equal but according to our Hyp. A and D were equal thus the Angles D and EGF should be equal which is impossible since that the Angle EGF being exteriour in respect of the Triangle EGD it ought to be greater than the Interiour Angle D by the 16th Therefore the Side DF is not greater than AC USE Use 26. THales made use of this Proposition to measure inaccessible Distances the Distance AD being proposed from the Point A he draws the Line AC Perpendicular to AD then placing a Semicircle at the Point C he measureth the Angle ACD then he takes an Angle equal thereto on the other side drawing the Line CB untill it meets the Line DA continued to the Point B. He Demonstrates that the Lines AD AB were equal so measuring actually the accessible Line he might know by that means the other For the Two Triangles ADC ABC have the Right Angles CAD CAB equal both the Angles ACD ACB were taken equal to each other and the Side AC is common to both Triangles Therefore by the 26th the Sides AD AB are equal LEMMA Lem. 26. A Line which is Perpendicular to one Parallel is also Perpendicular to the other Let the Lines AB CD be Parallel to each other and let EF be Perpendicular to CD I say that it is Perpendicular to AB Cut off two equal Lines CF FD at the Points C and D erect two Perpendiculars to the Line CD which shall also be equal to FE by the Definition of Parallels and draw the Lines EC ED. Demonstration The Triangles CEF FED have the Side FE common the Sides FD FC are equal the Angles at F are Right and by consequence equal Therefore by the 4th the Bases EC ED the Angles FED FEC FDE FCE are equal and those two last being taken away from the Right Angles ACE BDF leaveth the equal Angles EDB ECA Now the Triangles CAE DBE shall by the 4th have the Angles DEB CEA equal which Angles being added to the equal Angles CEF FED maketh equal Angles FEB FEA Therefore EF is Perpendicular to AB PROPOSITION XXVII THEOREM IF a Right Line falling upon Two Right Lines make the Alternate Angles equal the one to the other then are the Right Lines Parallel Let the Line EH fall on the Right Lines AB CD making therewith the Alternate Angles AFG FGD equal I say in the first place that the Lines AB CD shall not meet although continued as far as one lists For supposing they should meet in I and that FBI CDI are two streight Lines Demonstration If FBI GDI be two streight Lines then FIG is a Triangle then by the 16th the exteriour Angle AFG shall be greater than the interiour FGI Wherefore that the equality of the Angles may subsist the Lines AB CD must never meet each other But because we have Examples of some crooked Lines that never intersect which notwithstanding are not Parallels but approach continually To prove the foregoing I make another Demonstration as followeth First I say that if the Line EH falleth on the Lines AB CD maketh the Alternate Angles AFG FGD equal the Lines AB CD are Parallels that is in every part equidistant from each other for which reason the Perpendiculars shall be equal to each other Draw from G to the Line AB the Perpendicular GA and CD being taken equal to AF draw FD. Demonstration The Triangles AGF FGD have the Side GF common the Side GD was taken equal to AF it is supposed that the Angles AFG FGD are equal Therefore by the 4th AC FD are equal and the Angle GDF is equal to the Right Angle GAF Thence FD is Perpendicular Furthermore that AB is Parallel to CD for the Parallel to CD is to be drawn from the Point F and must pass through the Point A according to our Definition of Parallels which is that the Perpendiculars AG FD are equal PROPOSITION XXVIII THEOREM IF a Right Line falling upon Two Right Lines make the Exteriour Angle equal to the Interiour opposite Angle of the other on the same Side or the Two Interiour on the same Side equal to Two Right then are the Right Lines Parallel Having drawn a Figure like unto the former let the Line EH fall on AB CD make in the first place the Exteriour Angle EFB equal to the Interiour opposite Angle of the other on the same Side FGD I say that the Lines AB CD are Parallel Demonstration The Angle EFB is equal to the Angle AFG by the 15th and it being supposed that EFB is also equal to
the 1st because the Sides BC BD are equal the Angle ABC shall be double of each The Second case is when an Angle encloseth the other and the Lines making the same Angles not meeting each other as you see in the second figure the Angle BID is in the Center and the Angle BAD is at the Circumference Draw the Line AIC through the Center Demonstration The Angle BIC is double to the Angle BAC and CID double to the Angle CAD by the preceding case Therefore the Angle BID shall be double to the Angle BAD USE THere is given ordinarily a practical way to describe a Horizontal Dial by a single opening of the Compass which is grounded in part on this Proposition Secondly when we would determine the Apogaeon of the Sun and the excentricity of his Circle by Three observations we suppose that the Angle at the Center is double to the Angle at the Circumference Ptolomy makes often use of this Proposition to determine as well the excentricity of the Sun as the Moon 's epicycle The first Proposition of the Third Book of Trigonometry is grounded on this PROPOSITION XXI THEOREM THe Angles which are in the same Segment of a Circle or that have the same Arch for Base are equal If the Angles BAC BDC are in the same Segment of a Circle greater than a Semicircle they shall be equal Draw the Lines BI CI. Demonstration The Angles A and D are each of them half of the Angle BIC by the preceding Proposition therefore they are equal They have also the same Arch BC for Base Secondly let the Angles A and D be in a Segment BAC less than a semi-circle they shall notwithstanding be equal Demonstration The Angles of the Triangle ABE are equal to all the Angles of the Triangle DEC The Angles ECD ABE are equal by the preceding case since they are in the same Segment ABCD greater than a Semi-circle the Angles in E are likewise equal by the 15th of the 1st therefore the Angles A and D shall be equal which Angles have also the same Arch BFC for Base USE Prop. XXI IT is proved in Opticks that the Line BC shall appear equal being seen from A and D since it always appeareth under equal Angles We make use of this Proposition to describe a great Circle without having its Center For Example when we would give a Spherical figure to Brass Cauldrons to the end we may work thereon and to pollish Prospective or Telescope Glasses For having made in Iron an Angle BAC equal to that which the Segment ABC contains and having put in the Points B and C two small pins of Iron if the Triangle BAC be made to move after such a manner that the Side AB may always touch the Pin B and the Side AC the Pin C the Point A shall be always in the Circumference of the Circle ABCD. This way of describing a Circle may also serve to make large Astrolabes PROPOSITION XXII THEOREM QVadrilateral figures described in a Circle have their opposite Angles equal to Two Right Let a Quadrilateral or four sided figure be described in a Circle in such manner that all its Angles touch the Circumference of the Circle ABCD I say that its opposite Angles BAD BCD are equal to two Right Draw the Diagonals AC BD. Demonstration All the Angles of the Triangle BAD are equal to Two Right In the place of its Angle ABD put the Angle ACD which is equal thereto by the 21st as being in the same Segment ABCD and in the place of its Angle ADB put the Angle ACB which is in the same Segment of the Circle BCDA So then the Angle BAD and the Angles ACD ACB that is to say the whole Angle BCD is equal to Two Right USE PTolomy maketh use of this Proposition to make the Tables of Chords or Subtendents I have also made use thereof in Trigonometry in the Third Book to prove that the sides of an Obtuse Angled Triangle hath the same reason amongst themselves as the Sines of their opposite Angles PROPOSITION XXIII THEOREM TWo like Segments of a Circle described on the same Line are equal I call like Segments of a Circle those which contain equal Angles and I say that if they be described on the same Line AB they shall fall one on the other and shall not surpass each other in any part For if they did surpass each other as doth the Segment ACB the Segment ADB they would not be like And to demonstrate it draw the Lines ADC DB and BC. Demonstration The Angle ADB is exteriour in respect of the Triangle BDC Thence by the 21th of the 1st it is greater than the Angle ACB and by consequence the Segments ADB ACB containeth unequal Angles which I call unlike PROPOSITION XXIV THEOREM TWo like Segments of Circles described on equal Lines are equal If the Segments of Circles AEB CFD are like and if the Lines AB CD are equal they shall be equal Demonstration Let it be imagined that the Line CD be placed on AB they shall not surpass each other seeing they are supposed equal and then the Segments AEB CED shall be described on the same Line and they shall then be equal by the preceding Proposition USE Use 24. CVrved Lined Figures are often reduced to Right Lined by this Proposition As if one should describe Two like Segments of Circles AEC ADB on the equal sides AB AC of the Triangle ABC It is evident that Transposing the Segment AEC on ADB the Triangle ABC is equal to the figure ADBCEA PROPOSITION XXV PROBLEM TO compleat a Circle whereof we have but a part There is given the Arch ABC and we would compleat the Circle There needeth but to find its Center Draw the Lines AB BC and having Divided them in the middle in D and E draw their Perpendiculars DI EI which shall meet each other in the Point I the Center of the Circle Demonstration The Center is in the Line DI by the Coroll of the 1st It is also in EI it is then in the Point I. USE Use 25. THis Proposition cometh very frequently in use it might be propounded another way as to inscribe a Triangle in a Circle or to make a Circle pass through three given points provided they be not placed in a streight Line Let be proposed the Points A B C put the Point of the Compass in C and at what opening soever describe Two Arks F and E. Transport the foot of the Compass to B and with the same opening describe Two Arcks to cut the former in E and F. Describe on B as Center at what opening soever the Arches H and G and at the same opening of the Compass describe on the Center A Two Arks to cut the same in G and H. Draw the Lines FE and GH which will cut each other in the Point D the Center of the Circle The Demonstration is evident enough for if you had drawn
ABC Draw the touch Line FED by the 17th of the 3d. and make at the Point of touching E the Angle DEH equal to the Angle B and the Angle FEG equal to the Angle C by the 23d of the 1st Draw the Line GH the Triangle EGH shall be equi Angled to ABC Demonstration The Angle DEH is equal to the Angle EGH of the Alternate Segment by the 32d of the 3d. now the Angle DEH was made equal to the Angle B and by consequence the Angles B and G are equal The Angles C and H are also equal for the same reason and by the Coroll 2. of the 32d of the 1st the Angles A and GEH shall be equal Therefore the Triangles EGH ABC are equiangled PROPOSITION III. PROBLEM TO describe about a Circle a Triangle equiangled to another If one would describe about a Circle GKH a Triangle equiangled to ABC one of the Sides BC must be continued to D and E and make the Angle GIH equal to the Angle ABD and HIK equal to the Angle ACE then draw the Tangents LGM LKN NHM through the Points G K H. The Tangents shall meet each other for the Angles IKL IGL being Right if one should draw the Line KG which is not drawn the Angles KGL GKL would be less than two Right therefore by the 11th Axiom the Lines GL KL ought to concur Demonstration All the Angles of the Quadrilateral GIHM are equal to four Right seeing it may be reduced into Two Triangles the Angles IGM IHM which are made by the Tangents are Right thence the Angles M and I are equivalent to Two Right as well as the Angles ABC ABD Now the Angle GIH is equal to the Angle ABD by construction therefore the Angle M shall be equal to the Angle ABC For the same reason the Angles N and ACB are equal and so the Triangles LMN ABC are equiangled PROPOSITION IV. PROBLEM To inscribe a Circle in a Triangle IF you would inscribe a Circle in a Triangle ABC divide into Two equally the Angles ABC ACB by the 9th of the 1st drawing the Lines CD BD which concurr in the Point D. Then draw from the Point D the Perpendiculars DE DF DG which shall be equal so that the Circle described on the point D at the opening DE shall pass through F and G. Demonstration The Triangles DEB DBF have the Angles DEB DFB equal seeing they are Right the Angles DBE DBF are also equal the Angle ABC having been divided into Two equally the Side DB is common therefore by the 26th of the 1st these Triangles shall be equal in every respect and the Sides DE DF shall be equal One might demonstrate after the same manner that the Sides DF DG are equal One may therefore describe a Circle which shall pass through the Points E F G and seeing the Angles E F G are Right the Sides AB AC BC touch the Circle which shall by consequence be inscribed in the Triangle PROPOSITION V. PROBLEM To describe a Circle about a Triangle IF you would describe a Circle about a Triangle ABC divide the Sides AB BC into Two equally in D and E drawing the Perpendiculars DF EF which concurr in the Point F. If you describe a Circle on the Center F at the opening FB it shall pass through A and C that is to say that the Lines FA FB FC are equal Demonstration The Triangles ADF BDF have the Side DF common and the Sides AD DB equal seeing the Side AB hath been divided equally and the Angles in D are equal being Right Thence by the 4th of the 1st the Bases AF BF are equal and for the same reason the Bases BF CF. USE WE have often need to inscribe a Triangle in a Circle as in the first Proposition of the Third Book of Trigonometry This practice is necessary for to measure the Area of a Triangle and upon several other occasions PROPOSITION VI. PROBLEM To inscribe a Square in a Circle TO inscribe a Square in the Circle ABCD draw to the Diameter AB the Perpendicular DC which may pass through the Center E. Draw also the Lines AC CB BD AD and you will have inscribed in the Circle the Square ACBD Demonstration The Triangles AEC CEB have their Sides equal and the Angles AEC CEB equal seeing they are Right therefore the Bases AC CB are equal by the 4th of the 1st Moreover seeing the Sides AE CE are equal and the Angle E being Right they shall each of them be semi-right by the 32d of the 1st So then the Angle ECB is semi-right And by consequence the Angle ACB shall be Right It is the same of all the other Angles therefore the Figure ACDB is a Square PROPOSITION VII PROBLEM To describe a Square about a Circle HAving drawn the Two Diameters AB CD which cut each other perpendicularly in the Center E draw the touch Lines FG GH HI FI through the Points A D B C and you will have described a Square FGHI about the Circle ACBD Demonstration The Angle E and A are Right thence by the 29th of the 1st the Lines FG CD are parallels I prove after the same manner that CD HI FI AB AB GH are parallels thence the Figure FCDG is a parallelogram and by the 34th of the 1st the Lines FG CD are equal as also CD IH FI AB AB GH and by consequence the Sides of the Figure FG GH HI IF are equal Moreover seeing the Lines FG CD are parallels and that the Angle FCE is Right the Angle G shall be also Right by the 29th of the 1st I demonstrate after the same manner that the Angles F H and I are Right Therefore the Figure FGHI is a Square and its Sides touch the Circle PROPOSITION VIII PROBLEM To inscribe a Circle in a Square IF you will inscribe a Circle in the Square FGHI divide the Sides FG GH HI FI in the middle in A D B C and draw the Lines AB CD which cutteth each other in the Point E. I demonstrate that the Lines EA ED EC EB are equal and that the Angles in A B C D are Right and that so you may describe a Circle on the Center E which shall pass through A D B C and which toucheth the Sides of the Square Demonstration Seeing the Lines AB GH conjoyns the Lines AG BH which are parallel and equal they shall be also parallel and equal therefore the Figure AGHB is a parallelogram and the Lines AE GD AG ED being parallel and AG GD being equal AE ED shall be also equal It is the same with the others AE EC EB Moreover AG ED being parallels and the Angle G being Right the Angle D shall be also a Right Angle One may then on the Center E describe the Circle ADBC which shall pass through the Points A D B C and which shall touch the Sides of the Square PROPOSITION IX PROBLEM To describe a Circle about a Square
Triangle ABC the Line DE is Parallel to the Base BC the Sides AB AC shall be divided proportionally that is to say that there shall be the same Reason of AD to DB as of AE to EC Draw the Lines DE BE. The Triangles DBE DEC which have the same Base DE and are between the same Parallels DE BC are equal by the 37th of the 1st Demonstration The Triangles ADE DBE have the same point E for their vertical if we take AD DB for their Bases and if one should draw through the point E a Parallel to AB they would be both between the same Parallels they shall have thence the same Reason as their Bases by the 1st that is to say that there is the same Reason of AD to DB as of the Triangle ADE to the Triangle DBE or to its equal CED Now there is the same Reason of the Triangle ADE to the Triangle CED as of the Base AE to EC There is therefore the same Reason of AD to DB as of AE to EC And if there be the same Reason of AE to EC as of AD to DB I say that the Lines DE BC would then be Parallels Demonstration There is the same Reason of AD to DB as of the Triangle ADE to the Triangle DBE by the 1st there is also the same Reason of AE to EC as of the Triangle ADE to the Triangle DEC consequently there is the same Reason of the Triangle ADE to the Triangle BDE as of the same Triangle ADE to the Triangle CED So then by the 7th of the 5th the Triangles BDE CED are equal And by the 39th of the 1st they are between the same Parallels USE THis Proposition is absolutely necessary in the following Propositions one may make use thereof in Measuring as in the following figure If it were required to measure the height BE having the length of the staff DA there is the same Reason of CD to DA as of BC to BE. PROPOSITION III. THEOREM THat Line which divideth the Angle of a Triangle into two equal parts divideth its Base in two parts which are in the same Reason to each other as are their Sides And if that Line divideth the Base into parts proportional to the Sides it shall divide the Angle into Two equally If the Line AD divideth the Angle BAC into Two equal parts there shall be the same Reason of AB to AC as of BD to DC Continue the Side CA and make AE equal to AB then draw the Line EB Demonstration The exterior Angle CAB is equal to the Two interior Angles AEB ABE which being equal by the 5th of the 1st seeing the Sides AE AB are equal the Angle BAD the half of BAC shall be equal to one of them that is to say to the Angle ABE Thence by the 27th of the 1st the Lines AD EB are parallel and by the 2d there is the same Reason of EA or AB to AC as of BD to DC Secondly If there be the same Reason of AB to AC as of BD to DC the Angle BAC shall be divided into Two equally Demonstra There is the same reason of AB or AE to AC as of BD to DC thence the Lines EB AD are parallel and by the 29th of the 1st the Alternate Angles EBA BAD the internal BEA and the external DAC shall be equal and the Angles EBA AEB being equal the Angles BAD DAC shall be so likewise Wherefore the Angle BAC hath been divided equally USE WE make use of this Proposition to attain to the Proportion of the sides PROPOSITION IV. THEOREM EQuiangular Triangles have their Sides Proportional If the Triangles ABC DCE are equiangular that is to say that the Angles ABC DCE BAC CDE be equal There will be the same Reason of BA to BC as of CD to CE. In like manner the reason of BA to AC shall be the same with that of CD to DE. Joyn the Triangles after such a manner that their Bases BC CE be on the same Line and continue the sides ED BA seeing the Angles ACB DEC are equal the Lines AC EF are parallel and so CD BF by the 29th of the 1st and AF DC shall be a parallelogram Demonstration In the Triangle BFE AC is parallel to the Base FE thence by the 2d there shall be the same reason of BA to AF or CD as of BC to CE and by exchange there shall be the same reason of AB to BC as of DC to CE. In like manner in the same Triangle CD being parallel to the Base BF there shall be the same Reason of FD or AC to DE as of BC to GE by the 2d and by exchange there shall be the same reason of AC to BC as of DE to CE. USE THis Proposition is of a great extent and may pass for a universal Principle in all sorts of Measuring For in the first place the ordinary practice in measuring inaccessible Lines by making a little Triangle like unto that which is made or imagined to be made on the ground is founded on this Proposition as also the greatest part of those Instruments on which are made Triangles like unto those that we would measure as the Geometrical Square Sinical Quadrant Jacobs Staff and others Moreover we could not take the plane of a place but by this Proposition wherefore to explain its uses we should be forced to bring in the first Book of practical Geometry PROPOSITION V. THEOREM TRiangles whose sides are proportional are equianguler If the Triangles ABC DEF have their sides proportional that is to say if there be the same reason of AB to BC as of DE to EF as also if there be the same reason of AB to AC as of DE to DF the Angles ABC DEF A and D C and F shall be equal Make the Angle FEG equal to the Angle B and EFG equal to the Angle C. Demonstration The Triangles ABC EFG have two Angles equal they are thence equiangled by the Cor. of the 32d of the 1st and by the 4th there is the same reason of DE to EF as of EG to EF. Now it is supposed that there is the same reason of DE to EF as of EG to EF. Thence by the 7th of the 5th DE EG are equal In like manner DF FG are also equal and by the 8th of the 1st the Triangles DEF GEF are equiangular Now the Angle GEF was made equal to the Angle B thence DEF is equal to the Angle B and the Angle DFE to the Angle C. So that the Triangles ABC DEF are equiangular PROPOSITION VI. THEOREM TRiangles which have their sides proportional which include an equal Angle are equiangular If the Angles B and E of the Triangles ABC DEF being equal there be the same reason of AB to BC as of DE to EF the Triangles ABC DEF shall be equiangular Make the Angle FEG equal to the Angle B and
the Angle EFG equal to the Angle C. Demonstra The Triangles ABC EGF are equiangular by the Cor. of the 32d of the 1st there is thence the same reason of AB to BC as of EG to EF by the 4th Now as AB to BC so is DE to EF there is then the same reason of DE to EF as of GE to EF. So then by the 7th of the 5th DE EG are equal and the Triangles DEF GEF which have their Angles DEF GEF each of them equal to the Angle B and the Sides DE EG equal with the Side EF common they shall be equal in every respect by the 4th of the 1st they are thence equiangular and the Triangle EGF being equal to the Triangle ABC the Triangles ABC DEF are equiangular The Seventh Proposition is unnecessary PROPOSITION VIII THEOREM A Perpendicular being drawn from the Right Angle of a Right Angled Triangle to the opposite side divideth the same into Two Triangles which are a like thereto If from the Right Angle ABC be drawn a perpendicular BD to the opposite side AC it will divide the Right Angled Triangle ABC into Two Triangles ADB BDC which shall be like or equiangular to the Triangle ABC Demonstration The Triangles ABC ADB have the same Angle A the Angle ADB ABC are right they are thence equiangular by the Cor. 2. of the 32d of the 1st In like manner the Triangles BDC ABC have the Angle C common and the Angles ABC BDC being right they are also equal Thence the Triangles ABC DBC are like USE WE measure inaccessable distances by a Square according to this Proposition For example if we would measure the distance DC having drawn the perpendicular DB and having put a Square at the Point B in such manner that by looking over one of its Sides BC I see the Point C and over its other Side I see the Point A it is evident that there will be the same reason of AD to DB as of DB to DC So that multiplying DB by its self and dividing that product by AD the Quotient shall be DC PROPOSITION IX PROBLEM To cut off from a Line any part required LEt there be proposed the Line AB from which it is required to cut off three Fifths Make the Angle ECD at discretion take in one of those Lines CD five equal parts and let CF be three of the same and CE be equal to AB Then draw the Line DE after which draw FG parallel to DE the Line CG will contain three Fifths of CE or AB Demonstration In the Triangle ECD FG being parallel to the Base DE there will be the same Reason of CF to FD as of CG to GE by the second and by composition by the 18th of the 5th there shall be the same Reason of CG to CE as of CF to CD Now CF contains three fifths of CD wherefore CG shall contain three fifths of CE or AB PROPOSITION X. PROBLEM TO divide a Line after the same manner as another Line is divided If one would divide the Line AB after the same manner as the Line AC is divided Joyn those Lines making an Angle at pleasure as CAB Draw the Line BC and the parallels EO FV and the Line AB shall be divided after the same manner as AC Demonstration Seeing that in the Triangle BAC the Line HX hath been drawn parallel to the Base BC it will divide the Sides AB AC proportionally by the second it is the same with all the other parallels To do the same with more facility one may draw BD parallel to AC and put off the same Divisions of AC on BD then draw the Lines from the one to the other PROPOSITION XI THEOREM TO find a third Proportional to Two given Lines It is required to find a third proportional to the Lines AB BC that is to say that there may be the same reason of AB to BC as of BC to the Line required Make at discretion the Angle EAC put off one after the other the Lines AB BC and let AD be equal to BC Draw the Lines BD and its parallel CE. The Line DE shall be that which you require Demonstration In the Triangle EAC the Line DB is parallel to the Base CE There is thence by the 2d the same reason of AB to BC as of AD or BC to DE. PROPOSITION XII PROBLEM TO find a fourth Proportional to three Lines given Let there be proposed three Lines AB BC DE to which must be found a fourth proportional make an Angle as FAC at discretion take on AC the Lines AB BC and on AF the Line AD equal to DE then draw DB and its parallel FC I say that DF is the Line you seek for that is to say that there is the same Reason of AB to BC as of DE or AD to DF. Demonstration In the Triangle FAC the Line DB is parallel to the Base FC there is thence the same reason of AB to BC as of AD to DF by the 2d USE THe use of the Compass of Proportion or Sector is established on these Propositions for we divide a Line as we please by the Compass of Proportion we do the Rule of Three without making use of Arithmetick we extract the Square Root and Cube Root we double the Cube we measure all sorts of Triangles we find the Content of Superficies and the solidity of Bodies we augment or diminish any figure whatever according to what Proportion we please and all those uses are Demonstrated by the foregoing Propositions PROPOSITION XIII PROBLEM TO find a mean Proportional between Two Lines If you would have a mean Proportional between the Lines LV VR having joyned them together on a strait Line divide the Line LR into two equal parts in the point M and having described a Semi-circle LTR on the Center M draw the perpendicular VT it shall be a mean Proportional between LV VR Draw the Lines LT TR. Demonstration The Angle LTR described in a Semi-circle is right by the 31st of the 3d. and by the 8th the Triangles LVT TVR are like there is thence the same Reason in the Triangle LVT of LV to VT as of VT to VR in the Triangle TVR by the 4th So then VT is a mean Proportional between LV and VR USE WE Reduce to a Square any Rectangular Parallelogram whatever by this Proposition For example in the Rectangle comprehended under LV VR I will demonstrate hereafter that the Square of VT is equal to a Rectangle comprehended under LV and VR PROPOSITION XIV THEOREM EQuiangular and equal Parallelograms have their Sides reciprocal and equiangular Parallelograms whose sides are reciprocal are equal If the Parallelograms L and M be equiangular and equal they shall have their sides reciprocal that is to say that there shall be the same Reason of CD to DE as of FD to DB. For seeing they have their Angles
greater into a lesser wherefore this Proposition extends almost to all Arts in which it is necessary to take a design or Model PROPOSITION XIX THEOREM LIke Triangles are in duplicate Ratio to their homologous sides If the Triangles ABC DEF be like or equiangular they shall be in duplicate Ratio of their homologous Sides BC EF that is to say that the Ratio of the Triangle ABC to the Triangle DEF shall be in duplicate Ratio of BC to EF wherefore by seeking the third Proportional HI to the Lines BC EF or so making it that there may be the same Ratio of BC to EF as of EF to HI the Triangle ABC shall have the same Ratio to the Triangle DEF as the Line BC hath to the Line HI Which is called a duplicate Ratio or doubled Reason by the 11th Def. of the fifth Let BG and HI be equal and let the Line AG be drawn Demonstration The Angle B and E of the Triangles ABG DEF are equal also seeing the Triangle ABC DEF are like there shall be the same Ratio by the 4th of AB to DE as of BC to EF Now as BC is to EF so is EF to HI or BG thence as AB is to DE so is EF to BG and consequently the sides of the Triangles ABG DEF being reciprocal the Triangles shall be equal by the 15th Now by the first the Triangle ABC hath the same Ratio to the Triangle ABG as BC to BG or HI thence the Triangle ABC hath the same Ratio to the Triangle DEF as BC to HI USE THOSE Propositions do correct the opinions of many who easily imagine that like figures have the same Ratio as their Sides For example let there be proposed two Squares Two Pentagons two Hexagons two Circles and let the sides of the first be double to the side of the second the first figure shall be quadruple to the second If the side of the first be triple to that of the second the first figure shall be nine times greater than the second So that to find a Square triple to another there must be sought a mean Proportional between one and three which would be almost 1¾ for the side of the tripple figure PROPOSITION XX. THEOREM LIke Poligons may be divided into so many like Triangles and they are in duplicate Ratio to their homologous sides If the Poligons ABCDE GHIML be like they may be divided into so many like Triangles and which shall be like parts of their whole Draw the Lines AC AD GI GL Demonstration Seeing the Poligons are alike their Angles B and H shall be equal and there shall be the same Ratio of AB to BC as of GH to HI by the 15th thence the Triangles ABC GHI are alike and by the 4th there shall be the same Ratio of BC to CA as of HI to GI Moreover seeng there is the same Ratio of CD to BC as of IL to IH and the same Ratio of BC to CA as of HI to GI There shall be by equality the same Ratio of CD to CA as of IL to GI Now the Angles BCD and HIL being equal if you take away the equal Angles ACB GIH the Angles ACD GIL shall be equal Thence the Triangles ACD GIL shall be like by the 15th So that it is easie after the same manner to go round about the Angles of the Polygons and to prove they are circular or alike I further add that the Triangles are in the same Ratio as are the Polygons Demonstration Seeing all the Triangles are like their sides shall be in the same Ratio by the 4th Now each Triangle is to its like in duplicate Ratio of its homologous sides by the 19th thence each Triangle of one Polygon to each Triangle of the other Polygon is in duplicate Ratio to the sides which being the same there shall be the same Ratio of each Triangle to its like as of all the Triangles of one Polygon to all the Triangles of the other Polygon by the 12th of the 5th that is to say as of the one Poligon to the other Coroll 1. Like Polygons are in duplicate Ratio to their sides Coroll 2. If three Lines are continually Proportional the Polygon described on the first shall have the same Ratio to the Polygon described on the second as the first hath to the third that is to say in duplicate Ratio of that of the first Line to the second USE from the Learned Dr. Barrow BY this is found a method of enlarging or diminishing a Right Lined figure in a Ratio given as if you would have a Pentagon Quintuple of that Pentagon whereof CD is the side then betwixt AB and 5 AB find out a mean Proportional upon this raise a Pentagon like unto that given and it shall be quintuple also of the Pentagon given Hence also if the Homologous sides of like figures be known then will the Proportion of the figure be evident viz. by finding out a third Proportional PROPOSITION XXI THEOREM POlygons which are like to a Third Polygon are also like one to the other If Two Polygons are like to a third they shall be like one to the other for each of them may be divided into as many like Triangles as there is in the third Now the Triangles which are like unto the same Third are also like one to the other because the Angles which are equal to a Third are equal one to the other and the Angles of the Triangles being equal those of the Polygons of whom it is compounded are so likewise I farther add that the sides of Triangles being in the same Ratio those of the Polygons shall be so likewise seeing they are the same PROPOSITION XXII THEOREM LIke Polygons described on four Lines which are Proportional are also Proportional And if the Polygons be in the same Ratio the Lines on which they are described shall be so also If there be the same Ratio of BC to EF as of HT to MN there shall also be the same Ratio of the Polygon ABC to the like Polygon DEF as of the Polygon HL to the like Polygon MO. Seek to the Lines BC EF a third Proportional G and to the Lines HT MN a third Proportional P by the 11th seeing there is the same Ratio of BC to EF as of HT to MN and of EF to G as of MN to P there shall be by equality the same Ratio of BC to G as of HT to P and this Ratio shall be duplicate or doubled of the Ratio of BC to EF or of HT to MN Demonstration The Polygon ABC to the Polygon DEF is in duplicate or doubled Ratio of the Ratio of BC to EF by the 21st that is to say as BC is to G and the Polygon HL to MO hath the same Ratio as HT to P. There is therefore the same Ratio of ABC to DEF as of HL to MO. And if like Polygons are proportional
AED by drawing the Lines AD BD and the Perpendiculars CG BF EI. For Multiplying the half of BD by CG and the half of AD by BF and by EI we have the Area of all those Triangles which added together is equal to the content of the Right Lined Figurâ— ABCDE Use 39. We find the Area of Regular Polygons by Multiplying one half of their peripheries by the Perpendicular drawn from their Centers to one of their Sides for Multiplying IG by AG we have the Rectangle HKLM equal to the Triangle AIB And doing the same for every Triangle taking always the Semi Base we have the Rectangle HKON whose Side KO Composed of the Semi-Bases and consequently equal to the Semi-Periphery and the Side HK equal to the Perpendicular IG According to this Principle Archimedes hath Demonstrated that a Circle is equal to a Rectangle comprehended under the Semi-Diameter and a Line equal to the Semi-Circumference PROPOSITION XLII PROBLEM TO make a Parallelogram equal to a Triangle given in an Angle equal to a Right Lined Angle given It is required to make a Parallelogram equal to the Triangle ABC having an Angle equal to the Angle E. Divide the Base BC into two equal parts in D. Draw AG Parallel to BC by the 31st make also the Angle CDF equal to the Angle E by the 23d. Then draw the Parallel CG the figure FDCG is a Parallelogram seeing the Lines FG DC DF CG are Parallel The Angle CDF is equal to the Angle E. I say that the Parallelogram is equal to the Triangle ABC Demonstration The Triangle ADC is half of the Parallelogram FDCG by the 41st it is also half the Triangle ABC because the Triangles ADC ADB are equal by the 37th Therefore the Triangle ABC is equal to the Parallelogram FDCG PROPOSITION XLIII PROBLEM THe complements of a Parallelogram are equal In the Parallelogram ABDC the complements AFEH EGDI are equal Demonstration The Triangles ABC BCD are equal by the 33d Whence if we Substract the Triangles HBE BIE FEC CGE which are also equal by the same the complements AFEH GDIE which remains shall be equal PROPOSITION XLIV PROBLEM UPon a Right Line given to make a Parallelogram at a Right Lined Angle given equal to a Triangle given It is proposed to make a Parallelogram having one of its Sides equal to the Line D and one of its Angles equal to the Right Lined Angle E which must be equal to the Triangle ABC Make by the preceding Problem the Parallelogram BFGH having the Angle HBF equal to the Angle E and which may be equal to the Triangle ABC Continue the Sides GH CF and make HI equal to the Line D then draw the Line IBN and by the Points I and N draw IL Parallel to GN and NL Parallel to GI and continue HB to M and FB to K. The Parallelogram KLBM is the Parallelogram required Demonstration The Angle HBF equal to the given Angle E is also equal to the Angle KBM by the 15th and the Side KB is equal to the Line HI or D. Lastly the Parallelogram MK is equal by the foregoing to the Parallelogram GFBH and this was made equal to the Triangle ABC Therefore the Parallelogram MK is equal to the Triangle ABC USE Use 44. IN this Proposition is contained a sort of Geometrical Division For in Arithmetical Division there is proposed a number which may be imagined to be like a Rectangle For example the Rectangle AB containing Twelve square Feet which is to be Divided by another Number as by two that is to say that there must be made another Rectangle equal to the Rectangle AB which may have BD too for one of its Sides and to find how many Feet the other Side ought to be that is to say the Quotient One may attain it Geometrically by a Rule and Compass Take BD of the Length of Two Feet and draw the Diagonal DEF the Line AF is that which you seek for For having made the Rectangle DCFG the Complements EG EC are equal by the 43d. and EG hath for one of its Sides EH equal to BD two Foot and EI equal to AF. This way of Division is called Application because the Rectangle AB is applyed to the Line BD or EH and this is the reason why Division is called Application for the Ancient Geometricians did choose rather to make use of a Ruler and Compass than Arithmetick PROPOSITION XLV PROBLEM UPon a right-Right-Line to make a Parallelogram equal to a Right-Lined Figure given at a Right-Lined Angle given There is proposed or given the Right Lined Figure ABCD and it is required to make a Parallelogram equal thereto having an Angle equal to the Angle E. Reduce the Right Lined Figure into Triangles by drawing the Right Line BD and make by the 42d a Parallelogram FGHI having the Angle FGH equal to the Angle E and being in content equal to the Triangle ABD Make also by the 44th a Parallelogram IHKL equal to the Triangle BCD and having a Line equal to IH and the Angle IHK equal to the Angle E. The Parallelogram FGKL shall be equal to the Right Lined Figure ABCD. Demonstration It remains to prove that the Parallelograms FGHI HKLI are but one that is to say that GH HK make but one streight Line the Angles GHI and K are equal to the Angle E the Angle K and KHI are equal to Two Right seeing we have made a Parallelogram KHIL Whereof the Angles GHI KHI are equal to Two Right and by the 14th GH HK is one streight Line USE THis Proposition is in the Practice much like to the former and serveth to Measure the Content of any Figure whatsoever by Reducing it first into Triangles then making a Right Angled Parallelogram equal thereto We may also make a Right Angled Parallelogram on a determined or given Line and which shall be equal to several irregular Figures Likewise having several Figures we may make another equal to their difference PROPOSITION XLVI PROBLEM UPon a Right Line to describe a Square To describe a Square on the Right Line AB erect two Perpendiculars AC BD equal to AB and draw the Line BD. Demonstration The Angles at A and B being Right the Lines AC BD are Parallel by the 28th they are also equal Therefore the Lines AB CD are Parallel and equal by the 33d. and the Angles at A and C B and D equal to Two Right by the 29th and seeing A and B are Right Angles the Angles C and D shall be also Right Thence the Figure AD hath all its Sides equal and all its Angles Right and consequently is a Square PROPOSITION XLVII THEOREM THe Square of the Base of a Right Angled Triangle is equal to the Sum of the Squares of the other two Sides It is supposed that the Angle BAC is a Right Angle and that there be described Squares on the several Sides BC AB AC the Square of the Base BC shall be equal to
to EFG Demonstration We have already demonstrated that there is a greater Reason of A B C to E F G than of the part BC to the part FG There shall thence be a greater Reason of A to E than of A B C to E F G by the 32d The end of the Fifth Book THE SIXTH BOOK OF Euclid's Elements THis Book explaineth and beginneth to apply perticular matters of the Doctrine of Proportions which the preceding Book Explaineth but in General It beginneth with the most easiest figures that is to say Triangles giving Rules to determine not only the Proportion of their Sides but also that of their capacity area or superficies Then it teacheth to find Proportional Lines and to augment or diminish any figure whatever according to a given Ratio It Demonstrateth the Rule of Three it applieth the forty seventh of the first to all sorts of Figures In fine it giveth us the most easie and most certain principles to conduct us in all sorts of Measuring The DEFINITIONS 1. Def. 1. Fig. I. Plate 6. RIght Lined Figures are like when that they have all their Angles equal and the sides which formeth those Angles Proportional Fig. I. As the Triangles ABC DEF shall be like if the Angles A and D B and E C and F are equal and if there be the same Ratio of AB to AC as of DE to DF and of AB to CB as of DE to EF. 2. Fig. II. Figures are reciprocal when they may be compared after such sort that the Antecedent of one Reason and the Consequent of the other be found in the same Figure That is when the Analogie beginneth in one figure and endeth in the same As if there be the same Reason of AB to CD as of DE to BF 3. Fig. III. A Line is divided into extream and mean Proportion or Reason when there is the same Reason of the whole Line to its greatest part as of its greatest part to its lesser part As if there be the same Reason of AB to AC as of AC to CB the Line AB shall be divided in the point C in extream and mean proportion 4. The Altitude or height of a figure is the Length of the Perpendicular drawn from the top thereof to its Base Fig. IV. As in the Triangles ABC EFG the Perpendicular EH AD whether it fall without or within the Triangles is their height or Altitude Triangles and Parallelograms whose heights are equal may be placed between the same Parallels for having placed their Bases on the same Line HC if the Perpendiculars DA HE are equal the Lines EA HC shall be parallel 5. A Reason is composed of several Reasons when the Quantities of those Reasons being Multiplied make a Third Reason It is to be taken notice of that a Reason at least the rational hath its name taken from some number which specifieth the Reason or habitude of the Antecedent of that Reason to its Consequent As when there is proposed Two Magnitudes or Quantities the one of Twelve Foot and the other of Six we say that the Reason of Twelve to Six is double In like manner when there is proposed Two Magnitudes Four and Twelve we would say it is a subtriple Reason and â…“ is the Denominator thereof which specifieth that there is the same Reason of Four to Twelve as of â…“ to One or as of One to Three One may call this Denominaotr the Quantity of the Reason Let there be proposed Three Terms Twelve Six Two The First Reason of Twelve to Six is double its Denominator is Two the Reason of Six to Two is Triple its Denominator is Three the Reason of Twelve to Two is compounded of the Reason of Twelve to Six and of that of Six to Two to have thâ—… Denominator or the Reason of Twelve to Two which is compounded of Double and of Triple multiply Three by Two and you shall have Six thence the Reason of Twelve to Two is Sextuple This is that which Mathematicians understand by composition of Reason although it ought to be called Multiplication of Reason PROPOSITION I. THEOREM PArallelograms and Triangles of equal height have the same Reason as their Bases Plate VI. Let there be proposed the Triangles ACG DEM of equal height in such sort that they may be placed between the same Parallels AD GM I say that there shall be the same Reason of the Base GC to the Base EM as of the Triangle AGC to the Triangle DEM Let the Base EM be divided into as many equal parts as you please and let there be drawn to each Division the Lines DF DH c. Let also the Line GC be divided into parts equal to those of EM and let be drawn Lines to each division from the top A All those little Triangles which are made in the Two great Triangles are between the same Parallels and they have equal Bases they are thence equal by the 28th Demonstration The Base GC contains as many Aliquot parts of the Line EM as could be found parts equal to EF Now as many times as there are in the Base GC parts equal to EF so many times the Triangle AGC containeth the little Triangles equal to those which are in the Triangle DEM which being equal among themselves are its Aliquot parts thence as many times as the Base GC containeth Aliquot parts of EM so many times the Triangle AGC containeth Aliquot parts of the Triangle DEM which will happen in all manner of Divisions There is thence the same Reason of the Base GC to the Base EM as of the Triangle AGC to the Triangle DEM Coroll Parallelograms drawn on the same Bases and that are between the same Parallel Lines are double to the Triangle by the 41st they are thence in the same Reason as Triangles that is to say in the same Reason as their Bases USE Fig. I. THis Proposition is not only necessary to Demonstrate those which follow but we may make use thereof in Dividing of Land Let there be proposed a Trapezium ABCD which hath its sides AD BC parallel and admit one would cut of a third part Let CE be made equal to AD and BG the third part of BE. Draw AG. I say that the Triangle ABG is the Third of the Trapezium ABCD. Demonstration The Triangles ADF FCE are equiangled because of the Parallel AD CE and they have their Sides AD CE equal They are thence equal by the 26th of the 1st and by consequence the Triangle ABE is equal to the Trapezium Now the Triangle ABG is the third part of the Triangle ABE by the preceding Proposition thence the Triangle ABG is the third part of the Trapezium ABCD. PROPOSITION II. THEOREM A Line being drawn in a Triangle Parallel to its Base divideth its sides proportionally and if a Line divideth proportionally the sides of a Triangle it shall be parallel to its Base If in the
equal they may be joyned after such manner that their joyned sides CD DE be on one streight Line by the 15th of the 1st continue the Sides AB GE you will have compleated the Parallelogram BDEH Demonstration Seeing the Parallelogram L and M are equal they shall have the same reason to the Parallelogram BDEH Now the reason of the parallelogram L to the Parallelogram BDEH is the same with that of the Base CD to the Base DE by the 1st and that of the Parallelogram M or DFGE is the same with that of the Base FD to the Base BD. Thence there is the same reason of CD to DE as of FD to BD. Secondly if the equiangular Parallelograms L and M have their Sides reciprocal they shall be equal Demonstration The Sides of the Parallelograms are reciprocal that is to say that there is the same Reason of CD to DE as of FD to BD Now as the Base CD is to DE so is the parallelogram L to the parallelogram BDEH by the first and as FD is to DB so is the parallelogram M to BEDH there is thence the same Reason of L to BDEH as of M to the same BDEH so then by the 7th of the 5th the parallelograms L and M are equal PROPOSITION XV. THEOREM EQual Triangles which have one Angle equal have the Sides which form that Angle reciprocal and if their sides be reciprocal they shall be equal If the Triangles F and G being equal have their Angles ACB ECD equal their sides about that Angle shall be reciprocal that is to say that there shall be the same Ratio of BC to CE as of CD to CA. Dispose the Triangles after such a manner that the Sides CD CA be one straight Line seeing the Angles ACB ECD are supposed equal the Lines BC CE shall be also a straight Line by the 14th of the 1st Draw the Line AE Demonstration There is the same Ratio of the Triangle ABC to the Triangle ACE as of the Triangle ECD equal to the first to the same Triangle ACE by the 7th of the 5th Now as ABC is to ACE so is the Base BC to the Base CE by the 1st seeing they have the same vertical A and as FCD is to ACE so is the Base CD to CA. Now if it be supposed that the sides are reciprocal that is to say that there be the same Ratio of BC to CE as of CD to CA the Triangles ABC CDE shall be equal because they would then have the same Ratio to the Triangle ACE PROPOSITION XVI THEOREM IF four Lines be proportional the Rectangle comprehended under the first and fourth is equal to the Rectangle comprehended under the second and third and if the Rectangle comprehended under the extrems be equal to the Rectangle comprehended under the means then are the four Lines proportional if the Lines AB CD be proportional that is to say that there is the same Ratio of A to B as of C to D the Rectangle comprehended under the first A and the fourth D shall be equal to the Rectangle comprended under B and C. Demonstration The Rectangles have their Angles equal seeing it is Right they have also their Sides reciprocal they are thence equal by the 16th In like manner if they be equal their sides are reciprocal that is to say that there is the same Ratio of A to B as of C to D. PROPOSITION XVII THEOREM IF three Lines be proportional the Rectangle comprehended under the first and third is equal to the Square of the mean and if the Square of the mean be equal to the Rectangle of the extrems the three Lines are then proportional If the three Lines A B D be proportional the Rectangle comprehended under A and under D shall be equal to the Square of B. Take C equal to B there shall be the same Ratio of A to B as of C to D thence the four Lines A B C D are proportional Demonstration The Rectangle under A and D shall be equal to the Rectangle under B and C by the foregoing Now this last Rectangle is a Square seeing the Lines B and C are equal thence the Rectangle under A and D is equal to the Square of B. In like manner if the Rectangle under A and D be equal to the Square of B there shall be the same Ratio of A to B as of C to D and seeing that B and C are equal there shall be the same Ratio of A to B as of B to D. USE THose Four Propositions demonstrateth that Rule of Arithmetick which we commonly call the Rule of Three and consequently the Rule of Fellowship false position and all other which are performed by Proportion For example let there be proposed these three Numbers A Eight B six C four and it is required to find the fourth proportional Suppose it to be found and let it be D. The Rectangle comprehended under A and D is equal to to the Rectangle comprehended under B and C. Now I can have this Rectangle by Multiplying B by C that is to say six by four and I shall have twenty four thence the Rectangle comprehended under A and D is twenty four wherefore dividing the same by A eight the Quotient three is the number I look for PROPOSITION XVIII THEOREM TO describe a Poligon like to another on a Line given There is proposed the Line AB on which one would describe a Poligon like unto the Poligon CFDE Having divided the Poligon CFDE into Triangles make on the Line AB a Triangle ABH like unto the Triangle CFE that is to say make the Angle ABH equal to the Angle CFE and BAH equal to FCE So then the Triangles ABH CFE shall be equiangled by the 32d of the first make also on BH a Triangle equiangled to FDE Demonstration Seeing the Triangles which are parts of the Poligons are equiangular the two Poligons are equiangular Moreover seeing the Triangles ABH CFE are equiangular there is the same Ratio of AB to BH as of CF to FE by the 4th In like manner the Triangles HBG EFD being equiangular there shall be the same Ratio of BH to BG as of FE to FD and by equality there shall be the same Ratio of AB to BG as of CF to FD. And so of the rest of the sides Thence by the first Definition the Poligons are like to each other USE IT is on this Proposition we establish the greatest part of the practical ways to take the plane of a place of an Edifice of a Field of a Forrest or of a whole Country for making use of the equal parts of a Line for Feet or for Chains we describe a figure like unto the Prototype but lesser in which we may see the Proportion of all its Lines And because it is easier on paper than on the ground we may comprehend in this Proposition all Geodesâ—… all Chorography all Geographical Charts and ways of reducing of the
BED DEC by the 4th I might prove by the same manner of Argument that BF CF are equal Demonstration The Angles BED DEC are equal it is also supposed that the Angles BED AEF are equal Thence the opposite Angles DEC AEF shall be equal and by the Cor. of the 15th Prop. AEC is a streight Line By consequence AFC is a Triangle in which the Sides AF FC are greater than AEC that is to say AE EB For the Lines AE FC are equal to AF EB Therefore the Lines AF FB are greater then AE BE and since natural Causes do act by the most shortest Lines therefore all reflections are made after this sort that the Angles of Reflection and of Incidence are equal Prop. XV. Moreover because we can easily prove that all the Angles which are made upon a Plain by the meeting of never so many Lines in the same Point are equal to Four Right Angles since that in the first Figure of this Proposition the Angles AEC AED are equal to two Right as also BED BEC we make this General Rule to determine the number of Polygons which may be joyned together to Pave a a Floor so we say that four squares six Triangles three Hexigones may Pave the same and that it is for this reason the Bees make their cellules Hexagonal PROPOSITION XVI THEOREM THe exteriour Angle of a Triangle is greater then either of the inward and Opposite Angles Continue the Side BC of the Triangle ABC I say that the exteriour Angle ACD is greater then the interiour opposite Angle ABC or BAC Imagine that the Triangle ABC moveth along BD and that it is Transported to CED Demonstration It is impossible that the Triangle ABC should be thus moved and that the Point A should not alter its place moving towards E Now if it moveth towards E the Angle ECD that is to say ABC is less than the Angle ACD therefore the Interiour Angle ABC is less than the Exteriour ACD It is easy to prove that the Angle A is also less than the Exteriour ACD For having continued the Side AC to F the opposite Angles BCF ACD are equal by the 15th and making the Triangle ABC to slide along the Line ACF it is demonstrable that the Angle BCF is greater than the Angle A. USE Use 16. WE draw from this Proposition several very useful conclusions The First that from a given Point there cannot be drawn no more than one Perpendicular to the same Line Example Let the Line AB be Perpendicular to the Line BC I say AC shall not be Perpendicular thereto because the Right Angle ABC shall be greater than the Interiour ACB Therefore ACB shall not be a Right Angle neither AC a Perpendicular Secondly that from the same Point A there can only be drawn two equal Lines for Example AC AD and that if you draw a Third AE it shall not be equal For since AC AD are equal the Angles ACD ADC are equal by the 5th now in the Triangle AEC the Exteriour Angle ACB is greater than the Interiour AEC and thus is the Angle ADE greater than AED Therefore the Line AE greater than AD and by consequence AC AE are not equal Thirdly if the Line AC maketh the Angle ACB acute and ACF obtuse the Perpendicular drawn from A shall fall on the same Side which the Acute Angle is of for if one should say that AE is a Perpendicular and the Angle AEF is Right then the Right Angle AEF would be greater than the Obtuse Angle ACE Those conclusions we make use of to measure all Parallelograms Triangles and Trapeziams and to Reduce them into Rectangular Figures PROPOSITION XVII THEOREM TWo Angles of any Triangle are less than Two Right Angles Let ABC be the proposed Triangle I say that Two of the Angles taken together BAC BCA are less than Two Right Angles Continue the Side CA to D. Demonstration The Interiour Angle C is less than the Exteriour BAD by the 16th Add to both the Angle BAC the Angles BAC BCA shall be less than the Angles BAC BAD which latter are equal to Two Right by the 13th Therefore the Angles BAC BCA are less than Two Right I might Demonstrate after the same manner that the Angles ABC ACB are less than two Right by continuing the Side BC. Coroll If one Angle of a Triangle be either Right or Obtuse the other Two shall be Acute This Proposition is necessary to Demonstrate those which follow PROPOSITION XVIII THEOREM THe greatest Side of every Triangle subtends the greatest Angle Let the Side BC of the Triangle ABC be greater then the Side AC I say that the Angle BAC which is opposite to the Side BC is greater than the Angle B which is opposite to the Side AC Cut off from BC the Line CD equal to AC and draw AD. Demonstration Seeing that the Sides AC CD are equal the Triangle ACB is an Isosceles Triangle by the 5th the Angles CDA CAD are therefore equal Now the whole Angle BAC is greater than the Angle CAD Thence the Angle BAC is greater than the Angle CDA which being exteriour in regard of the Triangle ABD is greater than the interiour B by the 16th Therefore the Angle BAC is greater than the Angle B. PROPOSITION XIX THEOREM IN every Triangle the greatest Angle is opposite to the greatest Side Let the Angle A of the Triangle BAC be greater than the Angle ABC I say that the Side BC opposite to the Angle A is greater then the Side AC opposite to the Angle B. Demonstration If the Side BC was equal to the Side AC in this case the Angles A and B would be equal by the 5th which is contrary to the Hypothesis If the Side BC was less than AC then the Angle B would be greater than A which is also contrary to the Hypothesis Wherefore I conclude that the Side BC is greater than AC USE Use 19. WE prove by these Propositions not only that from the same Point to a Line given there can be but one Perpendicular drawn but also that that Perpendicular is the shortest Line of all those Lines which might be drawn to the said Line As for instance if the Line RV be Perpendicular to ST it shall be shorter then RS because the Angle RVS being Right the Angle RSV shall be Acute by the Cor. of the 17th and the Line RV shall be shorter than RS by the Preceding For this Reason Geometricians always make use of the Perpendiculars in their measuring and Reducing irregular Figures into those whose Angles are Right I further add that seeing there can only be drawn Three Perpendiculars to one and the same Point it cannot be imagined that there are more than Three Species of quantity viz. A Line a Surface and a Solid We also prove by these Propositions that a Boul which is exactly round being put on a Plain cannot stand but on one determinate Point As for Example
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
for Pentagons are the most ordinary You must also take notice that these ways of describing a Pentagon about a Circle may be applyed to the other Polygons I have given another way to inscribe a Regular Pentagon in a Circle in Military Architecture PROPOSITION XV. PROBLEM TO inscribe a Regular Hexagon in a Circle To inscribe a Regular Hexagon in the Circle ABCDEF draw the Diameter AD and putting the Foot of the Compass in the Point D describe a Circle at the opening DG which shall intersect the Circle in the Points EC then draw the Diameters EGB CGF and the Lines AB AF and the others Demonstration It is evident that the Triangles CDG DGE are equilateral wherefore the Angles CGD DGE and their opposites BGA AGF are each of them the third part of two Right and that is 60 degrees Now all the Angles which can be made about one Point is equal to four Right that is to say 360. So taking away four times 60 that is 240 from 360 there remains 120 degrees for BGC and FGE whence they shall each be 60 degrees So all the Angles at the Center being equal all the Arks and all the Sides shall be equal and each Angle A B C c. shall be composed of two Angles of Sixty that is to say One Hundred and Twenty degrees They shall therefore be equal Coroll The Side of a Hexagon is equal to the Semi diameter USE BEcause that the Side of an Hexagon is the Base of an Ark of Sixty degrees and that is equal to the Semi-Diameter its half is the Sine of Thirty and it is with this Sine we begin the Tables of Sines Euclid treateth of Hexagons in the last Book of his Elements PROPOSITION XVI PROBLEM TO inscribe a Regular Pentadecagon in a Circle Inscribe in a Circle an equilateral Triangle ABC by the 2d and a Regular Pentagon by the 11th in such sort that the Angles meet in the Point A. The Lines BF BI IE shall be the Sides of the Pentadecagon and by inscribing in the other Arks Lines equal to BF BI you may compleat this Polygon Demonstration Seeing the Line AB is the Side of the Equilateral Triangle the Ark AEB shall be the third of the whole Circle or 5 fifteenths And the Ark AE being the fifth part it shall contain 3 15 thence EB contains two and if you divide it in the middle in I each part shall be a fifteenth USE THis Proposition serveth only to open the way for other Polygons We have in the compass of Proportion very easie methods to inscribe all the ordinary Polygons but they are grounded on this For one could not put Polygons on that Instrument if one did not find their sides by this Proposition or such like The end of the Fourth Book THE FIFTH BOOK OF Euclid's Elements THis Fifth Book is absolutely necessary to demonstrate the Propopositions of the Sixth Book It containeth a most universal Doctrine and a way of arguing by Propâ—rtion which is most subtile solid and Brief So that all Treatises which are founded on Proportions cannot be without this Mathematical Logick Geometry Arithmetick Musick Astronomy Staticks and to say in one word all the Treatises of the Sciences are demonstrated by the Propositions of this Book The greatest part of Measuring is done by Proportions and in practisal Geometry And one may demonstrate all the Rules of Arithmetick by the Theorems hereof wherefore it is not necessary to have recourse to the Seventh Eighth or Ninth Books The Musick of the Ancients is scarce any thing else but the Doctrine of Proportion applyed to the Senses It is the same in Staticks which considers the Proportion of Weights In fine one may affirm that if one should take away from Mathematicians the knowledge of the Propositions that this Book giveth us the remainder would be of little use DEFINITIONS The whole corresponds to its part and this shall be the greater quantity compared with the lesser whether it contains the same in effect or that it doth not contain the same Parts or quantities taken in general are divided ordinarily into Aliquot parts and Aliquant parts 1. An Aliquot part which Euclid defines in this Book is a Magnitude of a Magnitude the lesser of the greater when it measureth it exactly That is to say that it is a lesser quantity compared with a greater which it measureth precisely As the Line of Two Foot taken Three times is equal to a Line of Six Foot 2. Multiplex is a Magnitude of a Magnitude the greater of the lesser when the lesser measureth the greater That is to say that Multiplex is a great quantity compared with a lesser which it contains precisely some number of times For Example the Line of Six Foot is treble to a Line of Two Foot Aliquant parts is a lesser quantity compared with a greater which it measureth not exactly So a Line of 4 Foot is an Aliquant part of a Line of 10 Foot Equimultiplexes are Magnitudes which contain equally their Aliquot parts that is to say the same number of times 12. 4. 6. 2. A B C D For example if A contains as many times B as C contains D A and C shall be equal Multiplexes of B and D. 3. Reason or Ratio is a mutual habitude or respect of one Magnitude to another of the same Species I have added of the same Species 4. For Euclid saith that Magnitudes have the same reason when being multiplied they may surpass each other To do which they must be of the same Species In effect a Line hath no manner of Reason with a Surface because a Line taken Mathematically is considered without any Breadth so that if it be multiplyed as many times as you please it giveth no Breadth and notwithstanding a Surface hath Breadth Seeing that Reason is a mutal habitude or respect of a Magnitude to another it ought to have two terms That which the Philosophers would call foundation is named by the Mathematicians Antecedent and the term is called Consequent As if we compare the Magnitude A to the Magnitude I this habitude or Reason shall have for Antecedent the quantity A and for consequent the quantity B. As on the contrary if we compare the Magnitude B with A this Reason of B to A shall have for Antecedent the Magnitude B and for consequent the Magnitude A. The Reason or habitude of one Magnitude to another is divided into rational Reason and irrational Reason Rational Reason is a habitude of one Magnitude to another which is commensurable thereto that is to say that those Magnitudes have a common measure which measureth both exactly As the reason of a Line of 4 Foot to a Line of 6 is rational because a Line of two Foot measureth both exactly and when this happeneth those Magnitudes have the same Reason as one Number hath to another For Example because that the Line of two Foot which is the common measure is found twice in the Line
when the Terms between them are taken twice that is to say as antecedent and as consequent As if there be the same Reason of A to B as of B to C and of C to D. 11. Then A to C shall be in duplicate Ratio of A to B and the Ratio of A to D shall be in triplicate Ratio to that of A to B. It is to be taken notice that there is a great deal of difference between double Ratio and duplicate Ratio We say that the Ratio of four to two is double that is to say four is the double of two whence it followeth that the number two is that which giveth the Name to this Ratio or rather to the Antecedent of this Ratio So we we say double triple quadruple quintuple which are Denominations taken from those numbers duo tres quatuor quinque compared with unity for we better conceive a Reason when its terms are small But as I have already taken notice those Denominations fall rather on the Antecedent than on the Reason it self we call that double triple Reason or Ratio when the Antecedent is double or triple to the consequent but when we say the Reason is duplicate we mean a Reason compounded of two like Reasons as if there be the same Reason of two to four as of four to eight the Reason of two and eight being compounded of the Reason of two and four and of that of four and eight which are alike and as equal the Reason or Ratio of two to eight shall be duplicated by each Three to twenty seven is a duplicated Reason of that of three to nine The Reason of two to four is called subduple that is to say two is the half of four but the reason of two to eight is duple of the sub-duple that is to say that two is the half of the half of eight as three is the third of the third of twenty seven where you see there is taken twice the Denominator ½ and ⅓ In like manner eight to two is a duplicate reason of eight to four because eight is double to four but eight is the double of the double of two If there be four terms in the same continued Reason that of the first and last is triple to that of the first and second as if one put these four Numbers two four eight sixteen the reason of two to sixteen is triple of two to four for two is the half of the half of the half of sixteen As the reason of sixteen to two is triple of sixteen to eight for sixteen is the double of eight and it is the double of the double of the double of two 12. Magnitudes are homologous the Antecedents to the Antecedents and the Consequents to the Consequents As if there be the same Reason of A to B as of C to D A and C are homologous or Magnitudes of a like Ratio The following Definitions are ways of arguing by Proportion and it is principally to demonstrate the same that this Book is composed 13. Alternate Reason or by Permutation or Exchange is when we compare the Antecedents one with the other as also the consequents For example if because there is the same reason of A to B as of C to D I conclude there is the same reason of A to C as of B to D this way of reasoning cannot take place but when the four terms are of the same Specie that is to say either all four Lines or Superficies or Solids Proposition 16. 14. Converse or Inverse Reason is a comparison of the Consequents to the Antecedents As if because there is the same reason of A to B as of C to D I conclude there is the same reason of B to A as there is of D to C. Proposition 16. 15. Composition of Reason is a comparison of the Antecedent and Consequent taken together to the Consequent alone As if there be the same Reason of A to B as of C to D I conclude also that there is the same reason of AB to B as of CD to D. Prop. 18. 16. Division of Reason is a comparison of the excess of the Antecedent above the Consequent to the same Consequent As if there be the same reason of AB to B as of CD to D I conclude that there is the same reason of A to B as of CD Prop. 17. 17. Conversion of Reason is the comparison of the Antecedent to the difference of the Terms As if there be the same reason of AB to B as of CD to D I conclude that there is the same reason of AB to A as of CD to C. Proposition 18. 18. Proportion of Equality is a comparison of the extream Quantities in leaving out those in the middle A B C D E F G H. As if there were the same reason of A to B as of E to F and of B to C as of F to G and of C to D as of G to H. I draw this Consequence that there is therefore the same reason of A to D as of E to H. 19. Proportion of Equality well ranked is that in which one compareth the Terms in the same manner of Order as in the preceding Example Prop. 22. 20. Proportion of Equality ill ranked is that in which one compareth the Terms with a different Order As if there were the same reason of A to B as of G to H and of B to C as of F to G and of C to D as of E to F. I draw this Conclusion that there is the same reason of A to D as of E to H. Prop. 28. Here is all the ways of arguing by Proportion There is the same reason of A to B as of C to D therefore by alternate reason there is the same reason of A to C as of B to D and by inversed reason there is the same Reason of B to A as of D to C and by composition there is the same reason of AB to B as of CD to D. By Division of Reason if there be the same reason of AB to B as of CD to D there is the same reason of A to B as of C to D and by Conversion there is the same reason of AB to A as of CD to C. By reason of Equality well ranked if there be the same reason of A to B as of C to D and also the same reason of B to E as of D to F there will be the same reason of A to E as of C to F. By reason of Equality ill ranked if there be the same reason of A to B as of D to F and also the same reason of B to E as of C to D there will be the same reason of A to E as of C to F. This Book contains twenty five Propositions of Euclid to which there has been added ten which are received The first six of this Book are useful only to prove the following Propositions by the method
fourth of BD as A contains the fourth of B and what I say of the fourth verifieth it self of all other aliquot parts There is therefore the same reason of A to B as of AC to BD. PROPOSITION XIII THEOREM IF of two equal Reasons the one is greater than a third the other shall be so likewise AB CD EF. If there be the same reason of A to B as of C to D and that there is a greater reason of A to B than of E to F I say that there shall be a greater reason of C to D than of E to F. Demonstration Seeing there is a greater reason of A to B than of E to F A shall contain more times any aliquot part of B than E contains a like aliquot part of F by the 6th Definition Now C contains a like aliquot part of D as many times as A contains that of B seeing there is the same reason of A to B as of C to D so then C contains an aliquot part of D more times than E containeth a like aliquot part of F thence there is a greater reason of C to D than of E to F. PROPOSITION XIV THEOREM IF there be the same Reason of the first Magnitude to the second as of the Third to the fourth if the first be greater equal or less than the third the second shall be greater equal or less than the fourth A B C D If there be the same reason of A to B as of C to D I say in the first place that if A be greater than C B shall also be greater than D. Demonstration Seeing that A is greater than C there will be by the 9th a greater Reason of A to B than of C to B. Now as A is to B so is C to D therefore there shall be a greater reason of C to D than of C to B and by consequence according to the tenth B shall be greater than D. I say in the second place that if A be equal to C B shall be also equal to D. Demonstration Seeing that A and C are equal there will be the same reason of A to B as of C to B by the 7th Now as A is to B so is C to D thence there is the same reason of C to B as of C to D and by consequence B and D are equal by the ninth and in the third place if A be less than C B shall be also less than D. Demonstration Seeing that A is less than C there will be a lesser reason of A to B than of C to B by the 8th Now as A is to B so is C to D thence there will be a lesser reason of C to D than of C to B and by consequence according to the 10th B shall be less than D. PROPOSITION XV. THEOREM THe Equimultiplices and like aliquot parts are in the same reason A B C D. 2. 3. 6. 9. E 2. H 3. F 2. I 3. G 2. K 3. If the Magnitudes C and D are Equimultiplices of A and B their aliquot parts there shall be the same reason of A to B as of C to D let the Magnitude C be divided into parts equal to A which shall be E F G let the Magnitude D be divided into parts equal to B seeing that C and D are Equimultiplices of A and B there will be as many parts in the one as in the other Demonstration There is the same reason of E to H of F to I of G to K as of A to B seeing they are equal thence by the 12th there shall be the same reason of E F G to HIK that is to say of C to D as of A to B. Coroll The same number of aliquot parts of two Magnitudes are in the same reason with those Magnitudes for seeing there is the same reason of E to H as of C to D and F to I there will be the same reason of EF to HI as of C to D. PROPOSITION XVI THEOREM Alternate Reason IF four Magnitudes of the same Species are proportional they shall also be proportional alternately A B C D. 12. 8. 9. 6. If there be the same reason of A to B as of C to D and if the four Magnitudes be of the same Species that is to say if all four be Lines or all four Superficies or all four Solids there will be the same reason of A to C as of B to D. Now suppose that there is a greater reason of A to C than of B to D. Demonstration Seeing one would have it that there is a greater reason of A to C than of B to D the Magnitude A will contain an aliquot part of C for Example the third more times than B contains the third of D. Let A contain the third of C four times and B the third of D only three times having divided A into four parts the third of C shall be once in each having also divided B into four parts the third of D shall not be in each Thence the three Fourths of A shall contain the three Thirds of C that is to say the Magnitude C and the three Fourths of B will not contain the three Thirds of D that is to say the Magnitude D. Again seeing there is the same reason of A to B as of C to D there will be also the same reason of the three Fourths of A to the three Fourths of B as of C to D by the Coroll of the 15th and by the 14th if the three Fourths of A be greater than C the three fourths of B shall be greater than D although we have demonstrated the contrary LEMMA IF there be the same reason of the first Magnitude to the second as of the third to the fourth an aliquot part of the first shall have the same reason to the second as a like aliquot part of the third hath to the fourth 16 3. 32. 6. A B C D. E Â F Â 4. Â 8. Â If there be the same reason of A to B as of C to D and that E be an aliquot part of A and F a like aliquot part of C I say there is the same reason of E to B as of F to D. Demonstration If there were a greater reason of E to B than of F to D E would contain an aliquot part of B more times than F contains a like aliquot part of D. Then E taken twice thrice or four times would contain an aliquot part of B more times than F taken twice thrice or four times would contain an aliquot part of D. Now E taken four times is equal to A as F is to C. So A would contain an aliquot part of B more times than C containeth a like aliquot part of D thence there would be a greater reason of A to B than of C to D which is contrary to the Supposition COROLLARY Which in Euclid is after the fourth Proposition