Selected quad for the lemma: end_n

Word A	Word B	Word C	Word D	Occurrence	Frequency Band	MI	MI Band	Prominent
end_n	draw_v	line_n	perpendicular_a	3,095	5	14.0786	5	true

Text snippets containing the quad

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

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 Lines AB BC you would have divided them equally and Perpendicularly by so doing This is very necessary to describe Astrolabes and to compleat Circles of which we have but a part That Astronomical Proposition which teacheth to find the Apogeum and the excentricity of the Suns Circle requires this Proposition We often make use there of in the Treatise of cutting of stones PROPOSITION XXVI THEOREM THe equal Angles which are at the Center or at the Circumference of equal Circles have for Base equal Arks. If the equal Angles D and I are in the Center of equal Circles ABC EFG the Arks BC FG shall be equal For if the Ark BC was greater or lesser than the Ark FG seeing that the Arks are the measure of the Angles the Angle D would be greater or lesser than the Angle I. And if the equal Angles A and E be in the Circumference of the equal Circles the Angles D and I which are the double of the Angles A and E being also equal the Arks BC FG shall be also equal PROPOSITION XXVII THEOREM THe Angles which are either in the Center or in the Circumference of equal Circles and which hath equal Arks for Base are also equal If the Angles D and I are in the Centers of equal Circles and if they have for Base equal Arcks BC FG they shall be equal because that their measures BC FG are equal and if the Angles A and E be in the Circumference of equal Circles have for Base equal Arks BC EG the Angles in the Center shall be equal and they being their halfs by the 20th shall be also equal PROPOSITION XXVIII THEOREM EQual Lines in equal Circles correspond to equal Arks. If the Line BC EF are applyed in equal Circles ABC DEF they shall be Chords of equal Arks BC EF. Draw the Lines AB AC DE EF. Demonstration In the Triangles ABC DEF the Sides AB AC DE EF are equal being the Semi-Diameters of equal Circles the Bases BC EF are supposed equal thence by the 8th of the 1st the Angles A and D shall be equal and by the 16th the Arks BC EF shall be also equal PROPOSITION XXIX THEOREM LInes which subtend equal Arcks in equal Circles are equal If the Lines BC EF subtend equal Arks BC EF in equal Circles those Lines are equal Demonstration The Arks BC EF are equal and parts of equal Circles therefore by the 27th the Angles A and D shall be equal So then in the Triangles CAB EDF the Sides AB AC DE DF being equal as also the Angles A and D the Bases BC EF shall be equal by the 4th of the 1st USE THeodosius demonstrateth by the 28th and 29th that the Arks of the Circles of the Italian and Babylonian hours comprehended between Two Parallels are equal We demonstrate also after the same manner that the Arks of Circles of Astronomical hours comprehended between Two Parallels to the Equator are equal these Propositions come almost continually in use in spherical Trigonometry as also in Gnomonicks PROPOSITION XXX PROBLEM TO divide an Ark of a Circle into Two equal parts It is proposed to Divide the Ark AEB into Two equal parts put the Foot of the Compass in the Point A make Two Arks F and G then transporting the Compass without opening or shutting it to the Point B describe two Arks cutting the former in F and G the Line GF will cut the Ark AB equally in the Point E. Draw the Line AB Demonstration You divide the Line AB equally by the construction for imagine the Lines AF BF AG BG which I have not drawn lest I should imbroil the figure the Triangles FGA FGB have all their Sides equal so then by the 8th of the 1st the Angles AFD BFD are equal Moreover the Triangles DFA DFB have the Sides DF common the Sides AF BF equal and the Angles DFA DFB equal whence by the 4th of the 1st the Bases AD DB are equal and the Angles ADF BDF are equal We have then divided the Line AB equally and perpendicularly in the Point D. So then by the 1st the Center of the Circle is in the Line EG Let it be the Point C and let be drawn the Lines CA CB all the Sides of the Triangles ACD BCD are equal Thence the Angles ACD BCD are equal by the 8th of the 1st and by the 27th the Arks AE EB are equal USE AS we have often need to divide an Ark in the middle the practice of this Proposition is very ordinarily in use it is by this means we divide the Mariners Compass into 32 Rumbs for having drawn Two Diameters which cut each other at Right Angles we divide the Circle in Four and sub-dividing each quarter in the middle we have Eight parts and sub-dividing each part twice we come to Thirty Two parts We have also occasion of the same practice to divide a Semi-circle into 180 degrees and because for the performing the same Division throughout we are obliged to divide an Ark into Three all the Ancient Geometricians have endeavoured to find a method to divide an Angle or an Ark into Three equal parts but it is not yet found PROPOSITION XXXI THEOREM THe Angle which is in a Semi-circle is Right that which is comprehended in a greater Segment is Acute and that in a lesser Segment is Obtuse If the Angle BAC be in a Semi-circle I demonstrate that it is Right Draw the Line DA. Demonstration The Angle ADB exteriour in respect of the Triangle DAC is equal by the 32d of the 1st to the Two Interiours DAC DCA and those being equal by the 5th of the 1st seeing the Sides DA DC are equal it shall be double to the Angle DAC In like manner the Angle ADC is double to the Angle DAB therefore the Two Angles ADB ADC which are equal to Two Right are double to the Angle BAC and by consequence the Angle BAC is a Right Angle Secondly the Angle AEC which is in the Segment AEC is obtuse for in the Quadrilateral ABCE the Opposite Angles E and B are equal to Two Right by the 22d the Angle B is Acute therefore the Angle E shall be Obtuse Thirdly the Angle B which is in the Segment ABC greater than a Semi-circle is Acute seeing that in the Triangle ABC the Angle BAC is a Right Angle USE Use 31. THe Workmen have drawn from this Proposition the way of trying if their Squares be exact for having drawn a Semi-circle BAD they apply the Point A of their Square BAD on the Circumference of the Circle and one of its Sides AB on the Point B of the Diameter the other Side AD must touch the other Point D which is the other end of the Diameter Ptolomy makes use of this Proposition to make the Table of Subtendants or Chords of which he hath occasion in Trigonometry Use 31. We have also a practical way to erect a perpendicular on the end of

B. By consequence the whole Angle ACD is equal to the Two Angles ACE ECD of whom it is composed it shall then be equal to A and B taken together In the Second place the Angles ACD ACB are equal to Two Right by the 13th And I have Demonstrated that the Angle ACD was equal to the Angles A and B taken together Therefore the Angle ACD is equal to A and B that is to say all the Angles of the Triangle ABC are equal to Two Right or 180 degrees Corollariy 1. The Three Angles of one Triangle are equal to the Three Angles of another Triangle Coroll 2. If Two Angles of a Triangle be equal to Two Angles of another Triangle the remaining Angle in the one shall be equal to the remaining Angle in the other Coroll 3. If a Triangle have one of its Angles Right the other Two shall be Acute and being taken together shall be equal to One Right Angle Coroll 4. From one and the same Point of a Line there can be drawn but one Perpendicular because a Triangle cannot have Two Right Angles Coroll 5. The Perpendicular is the shortest Line which can be drawn from a Point to a Line Coroll 6. In a Right Angled Triangle the greatest Angle is a Right Angle and the longest Side is opposite thereto Coroll 7. Each Angle of an Equilateral Triangle containeth 60 degrees that is to say the third of 180. USE Use 32. THis Proposition is of use to us in Astronomy to determine the Parallax Let the Point A represent the Center of the Earth and from the Point B on the Superficies of the Earth let there be taken by observation the Angle DBC that is to say the apparent Distance of a Plannet or Commet from the Zenith D. I say if the Earth were transparent this Planet or Commet viewed from the Center of the Earth A would appear distant from the Zenith D equal to the quantity of the Angle CAD which is less than the Angle CBD For the Angle CBD being Exteriour in respect of the Triangle ABC is equal by the 32d. to the opposite Angles A and C. Whence the Angle C shall be equal to the excess of the Angle CBD above the Angle A. From whence I conclude that if I know by an Astronomical Table how far that Planet or Commet ought to appear Distant from the Zenith to one which should be in the Center of the Earth and if I observe at the same time the difference of those Two Angles shall be the Parallax viz. The Angle BCA PROPOSITION XXXIII THEOREM IF Two equal and Parallel Lines be joyned together with Two other Right Lines then are those Lines also equal and Parallel Let the Lines AB GD be Parallel and equal and let them be joyned with AG BD. I say that the Lines AG BD are equal and Parallel draw the Diagonal BG Demonstration Seeing the Lines AB GD are parallel the Alternate Angles ABG BGD shall be equal by the 29th and the Side GB being common to both the Triangles ABG BGD and the Sides AB GD equal with the Angles ABG BGD equal as before the Bases of those Triangles AG BD shall be equal by the 4. as also the Angles DBG BGA which because Alternate the Lines AG BD shall be Parallel by the 27th USE Use 33. THis Proposition is put in practice to measure as well the Perpendicular height AG of Mountains as their Horizontal Lines C G which are hidden in their thickness To effect which we make use of a very large square ABD putting one end thereof in A in such manner that the other Side thereof BD may be Perpendicular to the Horizon then we measure the Sides AD BD then we do the like again at the Point B and measure BE EC the Sides Parallel to the Horizon that is to say AD BE being added together gives the Horizontal Line CG and the Perpendicular Sides BD EC being added gives the Perpendicular Height AG. PROPOSITION XXXIV THEOREM THe Sides and the opposite Angle in a Parallelogram are equal and the Diamter doth divide the same into Two equal parts Let the Figure AB CD be a Parallelogram that is to say let the Sides AB CD AC BD be Parallel I say that the opposite Sides AB CD AC BD are equal as also the Angles A and D ABD ACD and that the Diameter BC doth divide the Figure into Two equal parts Demonstration The Lines AB CD are supposed Parallel Therefore the Alternate Angles ABC BCD shall be equal by the 29th Likewise the Sides AC BD being supposed Parallel the Alternate Angles ACB CBD are equal Both which Triangles have the same Side BC and the Angles ABC BCD ACB CBD equal they shall be equal in every respect by the 26th Therefore the Sides AB CD AC BD and the Angles A and D are equal and the Diameter divides the Figure into Two equal parts and seeing the Angles ABC BCD ACD CBD are equal adding together ABC CBD BCD ACD we conclude that the opposite Angles ABD ACD shall be equal USE Use 34. SVrveyers have sometimes occasion to make use of this Proposition to part or divide Lands If the Field be a Parallelogram it may be divided into Two equal parts by the Diameter AD. But if one be obliged to divide the same into Two equal parts from the Point E divide the Diameter into two equal parts in the Point F and draw the Line EFG that Line shall divide the Field into Two equal parts for the Triangle AEF GFD which have their Alternate Angles EAF FDG AEF FGD and the Sides AF FD equal are equal by the 26th And since the Trapezium BEFD together with the Triangle AFE that is to say the Triangle ADB is one half of the Parallelogram by the 34●h. the same Trapezium EFBD together with the Triangle DFG shall be one half of the Figure or Field and the Line EG divideth the same into Two equal parts PROPOSITION XXXV THe Parallelograms are equal when having the same Base they are between the same Parallel Lines Let the Parallelograms ABEC ABDF have the same Base AB and be between the same Parallels AB CD I say they are equal Demonstration The Sides AB CE are equal by the 34th as also AB FD wherefore CE FD are equal and adding EF to each the Lines CF ED shall be equal the Triangles CFA EBD have the Sides CA EB CF ED equal with the Angles DEB FCA by the 29th the one being Exteriour and the other Interiour on the same Side whence by the 4th the Triangles ACE BED are equal and taking away from both that which is common to both that is to say the little Triangle EGF the Trapezium FGBD shall be equal to the Trapezium CAGE and adding to both the little Triangle AGB the Parallelograms ABEC ABDF shall be equal Demonstration by the method of Indivisibles THis method is new invented by Cavalerius which is approved

a Line which is founded on this Proposition For Example to erect a Perpendicular from the Point A of the Line AB I put the Foot of the Compass on the Point C taken at discretion and extending the other to A I describe a Circle which may cut the Line AB in the Point B. I draw the Line BCD It is evident that the Line DA shall be perpendicular to the Line AB seeing the Angle BAD is in the Semi-circle PROPOSITION XXXII THEOREM THe Line which cutteth the Circle at the Point of touching maketh with the touch Line the Angles equal to those of the Alternate Segments Let the Line BD cut the Circle in the Point B which is the Point where the Line AB doth touch the same I say that the Angle CBD which the Line BD Comprehendeth with the touch Line ABC is equal to the Angle E which is that of the Alternate Segment BED and that the Angle ABD is equal to the Angle F of the Segment BFD In the first place if the Line passeth through the Center as doth the Line BE It would make with the touch Line AB two Right Angles by the 17th and the Angle of the Semi-circles would be also Right by the preceding So the Proposition would be true If the Line passeth not through the Center as doth the Line BD Draw the Line BE through the Center and joyn the Line DE Demonstration The Line BE maketh two Right Angles with the touch Line and all the Angles of the Triangle BDE are equal to Two Right by the 32d of the 1st So taking away the Right Angles ABE and D which is in a Semi-circle and taking again away the Angle EBD which is common to both the Angle CBD shall be equal to the Angle BED Thirdly the Angle ABD is equal to the Angle F because in the Quadrilateral BFDE which is inscribed in a Circle the opposite Angles E and F are equal to two Right by the 22d the Angles ABD DBC are also equal to Two Right by the 13th of the 1st and the Angle DBC and E are equal as just now I did demonstrate therefore the Angles ABD and BFD are equal USE THis Proposition is necessary for that which followeth PROPOSITION XXXIII PROBLEM TO describe upon a Line a Segment of a Circle which shall contain a given Angle It is proposed to describe on the Line AB a Segment of a Circle to contain the Angle C. Make the Angle BAD equal to the Angle C and draw to AD the Perpendicular AE Make also the Angle ABF equal to the Angle BAF and lastly describe a Circle on the Point F as Center at the opening BF or FA the Segment BEA containeth an Angle equal to the Angle C. Demonstration The Angles BAF ABF being equal the Lines FA FB are equal by the 6th and the Circle which is described on the Center F passeth through A and B Now the Angle DAE being Right the Line DA toucheth the Circle in the Point A by the 16th therefore the Angle which the Segment BEA comprehendeth as the Angle E is equal to the Angle DAB that is to say to the Angle C. But if the Angle was obtuse we must take the Acute Angle which is its Complement to 180 degrees PROPOSITION XXXIV PROBLEM A Circle being given to cut there-from a Segment to contain an assigned Angle To cut from the Circle BCE a Segment to contain the Angle A. Draw by the 17th the touch Line BD and make the Angle DBC equal to the Angle A. It is evident by the 32d that the Segment BEC contains an Angle equal to DBC and by consequence to the Angle A. USE I Have made use of this Proposition to find Geometrically the excentricity of the Annual Circle of the Sun and his Apogeon Three observations being given It is also made use of in Opticks Two unequal Lines being proposed to find a Point where they shall appear equal or under equal Angles making on each Segments which may contain equal Angles PROPOSITION XXXV THEOREM IF Two Lines cut each other in a Circle the Rectangle comprehended under the parts of the one is equal to the Rectangle comprehended under the parts of the other In the first place if Two Lines cut each other in the Center of the Circle they shall be equal and divided equally so then it is evident that the Rectangle comprehended under the parts of the one is equal to the Rectangle comprehended under the parts of the other Secondly let one of the Lines pass through the Center F as AC and divide the Line BD in two equally in the Point E I say that the Rectangle comprehended under AE EC is equal to the Rectangle comprehended under BE ED that is to say to the Square of BE. The Line AC is perpendicular to BD by the Third Demonstration Seeing that the Line AC is divided equally in F and unequally in F the Rectangle comprehended under AE EC with the Square of FE is equal to the Square of FC or FB by the 5th of the 2d Now the Angle E being Right the Square of FB is equal to the Squares of BE EF therefore the Rectangle comprehended under AE EC with the Square of EF is equal to the Squares of BE EF and taking away the Square of EF there remains that the Square of BE is equal to the Rectangle under BE ED. Thirdly let the Line AB pass through the Center F and let it divide the Line CD unequally in the Point E draw FG perpendicularly to CD and by the 3d. the Lines CG GD shall be equal Demonstration Seeing the Line AB is divided equally in F and unequally in E the Rectangle comprehended under AE EB with the Square of EF is equal to the Square of BF or FC by the 5th of the 2d In the place of EF put the Squares of FG GE which is equal thereto by the 47th of the 1st In like manner the Line CD being equally divided in G and unequally in E the Rectangle CED with the Square of GE shall be equal to the Square of GC Add the Square of GF the Rectangle of CE ED with the Squares of GE FG shall be equal to the Squares of GC GF that is to say by the 47th of the 1st to the Square of CF. Therefore the Rectangle AEB with the Squares of GE GF and the Rectangle of CE ED with the same Squares are equal and by consequence taking away the same Squares the Rectangle AEB is equal to the Rectangle CFD Fourthly let the Lines CD HI cut each other in the Point E so that neither of them pass through the Center I say that the Rectangle CED is equal to the Rectangle HEI For drawing the Line AFB the Rectangles CED HEI are equal to the Rectangle AEB by the preceding case therefore they are equal USE ONe might by this Proposition have a practical way to find the Fourth proportional to Three given Lines or