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ID	Title	Author	Corrected Date of Publication (TCP Date of Publication)	STC	Words	Pages
A38722	The elements of Euclid, explained and demonstrated in a new and most easie method with the uses of each proposition in all the parts of the mathematicks / by Claude Francois Milliet D'Chales, a Jesuit ; done out of French, corrected and augmented, and illustrated with nine copper plates, and the effigies of Euclid, by Reeve Williams ...; Huict livres des Eléments d'Euclide rendus plus faciles. English	Dechales, Claude-François Milliet, 1621-1678.; Euclid. Elements.; Williams, Reeve, fl. 1682-1703.	1685 (1685)	Wing E3399; ESTC R10241	136,603	430

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

the Lines being in sub-duplicate Ratio shall be also Proportional USE A B C D. 3. 2. 6. 4. 9. 4. 36. 16 E F G H. THis Proposition may be easily applied to Numbers If the Numbers A B C D are Proportional their Squares EF GH shall be so likewise which we make use of in Arithmetick and yet more use thereof in Algebra PROPOSITION XXIII THEOREM EQuiangular Parallelograms have their Ratio compounded of the Ratio of their sides If the Parallelograms L and M be equiangular the Ratio of L to M shall be compounded of that of AB to DE and that of DB to DF. Joyn the Parallelograms in such a manner that their sides BD DF be on a streight Line as also CD DE which may be if they be equiangular Compleat the Parallelogram BDEH Demonstration The Parallelogram L hath the same Ratio to the Parallelogram BDEH as the Base AB hath to the Base BH or DE by the first the Parallelogram BDEH hath the same Ratio to the Parallelogram DFGE that is to say M as the Base BD hath to the Base DF. Now the Ratio of the Parallelogram L to the Parallelogram M is compounded of that of L to the Parallelogram BDEH and that of BDEH to the Parallelogram M. Thence the Ratio of L to M is compounded of that of AB to DE and of that of BD to EG For example if AB be eight and BH five BD four DF seven say as four is to seven so is five to eight ¾ you shall have these three numbers eight five eight ¾ eight to five shall be the Ratio of the Parallelogram L to BDEH the same as that of AB to DE five to eight ¾ shall be that of the Parallelogram BDEH to M. So then taking away the middle term which is five you shall have the Ratio of eight to eight ● for the Ratio compounded of both or as 4 times 8 or 32 to 5 times 7 or 35. PROPOSITION XXIV THEOREM IN all sorts of Parallelograms those through which the Diameter passeth are like to the greater Let the Diameter of the Parallelogram AC pass through the Parallelograms EF GH I say they are like unto the Parallelogram AC Demonstration The Parallelograms AC EF have the same Angle B and because in the Triangles BCD IF is Parallel to the Base DC the Triangles BFI BCD are equiangular There is therefore by the 4th the same Ratio of BC to CD as of BF to FI and consequently the sides are in the same Ratio In like manner IH is Parallel to BC there shall then be the same Ratio of DH to HI as of DC to BC and the Angles are also equal all the sides being Parallel thence by the 8th Def. the Parallelograms EF GH are like unto the Parallelogram AC USE I Made use of this Proposition in the Tenth Proposition of the last Book of Perspective to shew that an Image was drawn like unto the Original with the help of a Parallelogram Composed of four Lines PROPOSITION XXV PROBLEM TO describe a Polygon like unto a given Polygon and equal to any other Right Lined figure If you would describe a Polygon equal to the Right Lined figure A and like unto the Polygon B make a Parallelogram CE equal to the Polygon B by the 34th of the first and on DE make a Parallelogram EF equal to the Right Lined figure A by the 45th of the first Then seek a mean Proportional GH between CD and DF by the 13th Lastly make on GH a Polygon O like unto B by the 18th It shall be equal to the Right Lined figure A. Demonstration Seeing that CD GH DF are continually Proportional the Right Lined figure B described on the first shall be to the Right Lined figure O described on the second as CD is to DF by the Coroll of the 20th Now as CD is to DF so is the Parallelogram CE to the Parallelogram EF or as B to A seeing they are equal There is thence the same Ratio of B to O as of B to A. So then by the 7th of the 5th A and O are equal USE THis Proposition contains a change of figures keeping always the same Area which is very useful principally in Practical Geometry to reduce an irregular figure into a Square PROPOSITION XXVI THEOREM IF in one of the Angles of a Parallelogram there be described a lesser Parallelogram like unto the greater the Diameter of the greater shall meet the Angle of the lesser If in the Angle D of the Parallelogram AC be described another lesser DG like thereto The Diameter DB shall pass through the point G. For if it did not but it passed through I as doth the Line BID Draw the Line IE Parallel to HD Demonstration The Parallelogram DI is like unto the Parallelogram AC by the 24th Now it is supposed that the Parallelogram DG is also like thereto thence the Parallelograms DI DG would be like which is impossible otherwise there would be the same Ratio of HI to IE or GF as of HG to GF and by the 7th of the 5th the Lines HI HG would be equal The Twenty Seventh Twenty Eighth and Twenty Ninth Propositions are unnecessary PROPOSITION XXX PROBLEM TO cut a Line given into extream and mean Proportion It is proposed to cut the Line AB in extream and mean Proportion that is to say in such a manner that there may be the same Ratio of AB to AC as of AC to CB. Divide the Line AB by the 11th of the second in such a manner that the Rectangle comprehended under AB CB be equal to the Square of AC Demonstration Seeing the Rectangle of AB BC is equal to the Square of AC there will be the same Ratio of AB to AC as of AC to BC by the 17th USE THis Proposition is necessary in the Thirteenth Book of Euclid to find the length of the Sides of some of the five Regular Bodies Father Lucas of St. Sepulchres hath Composed a Book of the Proprieties of a Line which is cut into extream and mean Proportion PROPOSITION XXXI THEOREM A Polygon which is described on the Base of a Right Angled Triangle is equal to the other Two like Polygon described on the other Sides of the same Triangle If the Triangle ABC hath a Right Angle BAC the Polygon D described on the Base BC is equal to the like Polygons F and E described on the Sides AB AC Demonstration The Polygons D E F are amongst themselves in duplicate Ratio of their homologous Sides BC AC AB by the 20th If there were described a Square on those sides they would be also in duplicate Ratio to their sides Now by the 47th of the first the Square of BC would be equal to the Squares of AC AB thence the Polyligon D described on BC is equal to the like Polygons E and F described on AB AC USE THis Proposition is made use

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 Prop●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

of to make all sorts of figures greater or lesser for it is more universal then the 47th of the 1st which notwithstanding is so useful that it seemeth that almost all Geometry is established on this Principle The 32d Proposition is unnecessary PROPOSITION XXXIII THEOREM IN equal Circles the Angles as well those at the Center as those at the Circumference as also the Sectors are in the same Ratio as the Arks which serve to them as Base If the Circles ANC DOF be equal there shall be the same Ratio of the Angle ABC to the Angle DEF as of the Ark AC to the Ark DF. Let the Ark AG GH HC he equal Arks and consequently Aliquot parts of the Ark AC and let be divided the Ark DF into so many equal to AG as may be found therein and let the Lines EI EK and the rest be drawn Demonstration All the Angles ABG GBH HBC DEI IEK and the rest are equal by the 37th of the third so then AG an Aliquot part of AC is found in the Ark DF as many times as the Angle ABG an Aliquot part of the Angle ABC is found in DEF there is therefore the same Ratio of the Ark AC to the Ark DF as of the Angle ABC to the Angle DEF And because N and O are the halfs of the Angles ABC DEF they shall be in the same Ratio as are those Angles there is therefore the same Ratio of the Angle N to the Angle O as of the Ark AC to the Ark DF. It is the same with Sectors for if the Lines AG GH HC DI IK and the rest were drawn they would be equal by the 28th of the Third and each Sector would be divided into a Triangle and a Segment The Triangles would be equal by the 8th of the first and the little Segments would be also equal by the 24th of the third thence all those little Sectors would be equal and so as many as the Ark BF containeth of Aliquot parts of the Ark of AC so many the Sector DKF would contain Aliquot parts of the Sector AGC There is therefore the same Reason of Ark to Ark as of Sector to Sector The end of the Sixth Book THE ELEVENTH BOOK OF Euclid's Elements THis Book comprehendeth the first Principles of Solid Bodies whence it is impossible to establish any thing touching the third species of quantity without knowing what this Book teacheth us which maketh it very necessary in the greatest part of the Mathematicks in the first place Theodosius's Sphericks doth wholly suppose it Spherical Trigonometry the third part of Practical Geometry several Propositions of Staticks and of Geography are established on the principles of Solid Bodies Gnomonicks Conical Sections and the Treatise of Stone-cutting are not difficult but because one is often obliged to represent on paper figures that are raised and which are comprised under several Surfaces I leave out the Seventh the Eighth the Ninth and the Tenth Books of Euclid's Elements because they are unnecessary in almost all the parts of the Mathematicks I have often wondred that they have been put amongst the Elements seeing it is evident that Euclid did Compose them only to establish the Doctrine of incommensurables which being only a curiosity ought not to have been placed with the Books of Elements but ought to make a particular Treatise The same may be said of the Thirteenth and of the rest for I believe that one may learn almost all parts of the Mathematicks provided one under●… well these Eight Books of Euclid's Elements DEFINITIONS 1. Def. 1. Plate VII Fig. I. A Solid is a Magnitude which hath length breadth and thickness As the figure LT its Length is NX its Breadth NO its Thickness LN 2. The extremities or terms of a Solid are Superficies 3. A Line is at Right Angles or Perpendicular to a Plane when it is Perpendicular to all the Lines that it meeteth with in the Plane Fig. II. As the Line AB shall be at Right Angles to the Plane CD if it be Perpendicular to the Lines CD FE which Lines being drawn in the Plane CD pass through the Point B in such sort that the Angles ABC ABD ABE ABF be Right 4. A Plane is perpendicular to another when the Line perpendicular to the common section of the Planes and drawn in the one is also perpendicular to the other Plane Fig. III. We call the common Section of the Planes a Line which is in both Planes as the Line AB which is also in the Plane AC as well as in the Plane AD. If then the Line DE drawn in the Plane AD and perpendicular to AB is also perpendicular to the Plane AC the Plane AD shall be Right or perpendicular to the Plane AC 5. Fig. IV. If the Line AB be not perpendicular to the Plane CD and if there be drawn from the Point A the Line AE perpendicular to the Plane CD and then the Line BE the Angle ABE is that of the inclination of the Line AB to the Plane CD 6. Fig. V. The inclination of one Plane to another is the Acute Angle comprehended by the two perpendiculars to the common section drawn in each Plane As the inclination of the Plane AB to the Plane AD is no other than the Angle BCD comprehended by the Lines BCCD drawnin both Planes perpendicular to the common Section AE 7. Plains shall be inclined after the same manner if the Angles of inclination are equal 8. Planes which are parallel being continued as far as you please are notwithstanding equidistant 9. Like solid figures are comprehended or terminated under like Planes equal in number 10. Solid figures equal and like are comprehended or termined under like Plains equal both in Multitude and Magnitude 11. Fig. VI. A solid Angle is the concourse or inclination of many Lines which are in divers Planes As the concourse of the Lines AB AC AD which are in divers Planes 12. Fig. VI. A Pyramid is a solid figure comprehended under divers Planes set upon one Plane and gathered together to one point As the figure ABCD. 13. Fig. VII A Prism is a solid figure contained under Planes whereof the two opposite are equal like and parallel but the others are parallelograms As the figure AB its opposite Plains may be Polygons 14. A Sphere is a solid figure terminated by a single Superficies from which drawing several Lines to a point taken in the middle of the figure they shall be all equal Some define a Sphere by the motion of a semi-circle turned about its Diameter the Diameter remaining immoveable 15. The Axis of a Sphere is that unmoveable Line about which the semia-circle turneth 16. The Center of the Sphere is the same with that of the semi-circle which turneth 17. The Diameter of a Sphere is any Line whatever which passeth through the Center of the Sphere and endeth in its