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 IN MEMORIAM 
 FLORIAN CAJORl 
 
r^/i^a*i^'<^^B!t^*f>f^ 
 
 ELEMENTS 
 
 or 
 
 GEOMETRY, 
 
 PLANE TRIGONOMETRY. 
 
 WITH AN 
 
 APPENDIX, 
 
 AND COPIOUS NOTES AND ILLUSTRATIONS^ 
 
 BY 
 
 JOHN LESLIE, F.R.S.E. 
 
 PROFESSOE OF MATHEMATICS IN THE UNIVERSITY 
 OP EDINBURGH. 
 
 THIRD EDITION, 
 
 IMPROVED AND ENLARGED. 
 
 EDINBURGH : 
 
 PRINTED FOR ARCHIBALD CONSTABLE & CO. 
 
 AND FOR LONGMAN, HURST, REES, ORME, & BROWN, 
 
 LONDON. 
 
 I8I7. 
 
EDINBURGH : 
 
 Printed by Abernethy ^ Walker. 
 

 PREFACE. 
 
 The volume now laid before the public, is the 
 first of a projected Course of Mathematical Sci- 
 ence. Many compendiums or elementary treatises 
 have appeared — at different times, and of various 
 merit ; but there seemed still wanting, in our lan- 
 guage, a work that should embrace the subject 
 in its full extent, — that should unite theory with 
 practice, and connect the ancient with the mo- 
 dern discoveries. The magnitude and difficulty 
 of such a task might deter an individual from 
 the attempt, if he were not deeply impressed 
 with the importance of the undertaking, and felt 
 his exertions to accomplish it animated by zeal, 
 and supported by active perseverance. 
 
 The study of Mathematics holds forth two capi- 
 tal objects : — While it traces the beautiful rela- 
 tions of figure and quantity, it likewise accustoms 
 the mind to the invaluable exercise of patient at- 
 tention and accurate reasoning. Of these distinct 
 objects, the last is perhaps the most important in 
 a course of liberal education. For this purpose. 
 
 j5^^f)rro^^ 
 
IV PREFACE. 
 
 the Geometry of the Greeks is the most powerful- 
 ly recommended, as bearing the stamp of that 
 acute people, and displaying the finest specimens 
 of logical deduction. Some of its conclusions, 
 indeed, might be reached by a sort of calculation ; 
 but such an artificial mode of procedure gives 
 merely an apparent facility, and leaves no clear or 
 permanent impression on the mind. 
 
 We should form a wrong estimate, however, 
 did we consider the Elements of Euclid, with all 
 its merits, as a finished production. That admi- 
 rable work was composed at the period when Geo- 
 metry was making its most rapid advances, and 
 new prospects were opening on every side. No 
 wonder that its structure should now appear loose 
 and defective. In adapting it to the actual state 
 of the science, I have therefore endeavoured care- 
 fully to retain the spirit of the original, but have 
 sought to enlarge the basis, and to dispose the 
 accumulated materials into a regular and more 
 compact system. By simplifying the order of 
 arrangement, I presume to have materially a- 
 bridged the labour of the student. The nume- 
 rous additions that are incorporated in the text, 
 so far from retarding, will rather facilitate his pro- 
 gress, by rendering more continuous the chain of 
 demonstration. 
 
 The view which I have given of the nature of 
 Proportion, in the Fifth Book, will contribute, I 
 
PREFACE. 
 
 hope, to remove the chief difficulties attending that 
 important subject. The Sixth Book, which ex- 
 hibits the application of the Doctrine of Ratios, 
 contains a copious selection of propositions, not 
 only beq,utiful in themselves, but which pave the 
 way to the higher branches of Geometry, or lead 
 immediately to valuable practical results. The 
 Appendix, without claiming the same degree of 
 utility, will not perhaps be deemed the least in- 
 teresting portion of the volume, since the inge- 
 nious resources which it discloses for the construc- 
 tion of certain problems are calculated to afford 
 a very pleasing and instructive exercise. 
 
 The Elements of Trigonometry are as ample 
 as my plan would allow. I have explained fully 
 the properties of the lines about the circle, and 
 the calculation of the trigonometrical tables ; 
 nor have I omitted any proposition which has a 
 distinct reference to practice. Some of the pro- 
 blems annexed are of essential consequence in 
 marine surveying. 
 
 In the improvement of this edition, I have spa- 
 red no trouble or expence. The text has been 
 simplified and reduced to a shorter compass, by 
 throwing such propositions as were less elemen- 
 tary to the Notes Other Notes of a simpler 
 kind are intended chiefly to engage the attention 
 of the young student. In various parts of the 
 work, the demonstrations are occasionally abbre- 
 
VI PllEFACE. 
 
 viated. The Elements of Trigonometry ai'e much 
 expanded, and now brought to include whatever 
 appears to be most valuable in recent practice. 
 But the greatest additions have been made in the 
 Notes and Illustrations, which will be found to 
 contain a variety of useful and curious infor- 
 mation. The more advanced student may per» 
 use with advantage the historical and critical 
 remarks ; and some of the disquisitions, with the 
 solutions of certain more difficult problems re- 
 lative to trigonometry and geodesiacal operations, 
 in which the modern analysis is but sparingly 
 introduced, are of a nature snfficiently interest- 
 ing to claim the notice of proficients in science. 
 I have simplified, and materially enlarged the for- 
 mulae connected with trigonometrical computa- 
 tion ; explained the art of surveying, in its dif- 
 ferent branches ; and given reduced plans, blend- 
 ed with the narrative of the great operations 
 lately carried on both in England and France. 
 I have likewise shown a very simple method of 
 calculating heights from barometrical observa- 
 tions, accompanied by illustrative sections ; and 
 I have been thence led to state the law of climate, 
 as it is modified by elevation. On this attractive 
 subject, I should have dwelt with pleasure, had 
 the limits of the volume permitted. 
 
 My original design was to exhibit, within per- 
 haps the compass of five volumes, the Elements of 
 
PREFACE. Vll 
 
 Mathematical Science in their full extent, including 
 the principles and application of the Higher Cal- 
 culus. But, after due reflection, I have abandoned 
 that aspiring project. The publication of abstract 
 works in this country procures neither fame nor 
 emolument ; and after having discharged the more 
 pressing obligations which I had contracted, I shall 
 consider my time as more agreeably and perhaps 
 more beneficially employed in pursuing without 
 distraction the labyrinths of physical research. The 
 text of the present volume has, by successive im- 
 provements, arrived at such a state of maturity^ 
 that I shall hardly be tempted in any future edi- 
 tion to alter it. It will be followed, without delay, 
 by another volume, which is to contain the tract 
 on Geometrical Analysis, enlarged and improved ; 
 the Geometry of Lines of the Second Order, ex- 
 panded to three books, and including the more 
 important of the Higher Curves ; and the Geo- 
 metry of Planes and Solids, embracing Spherical 
 Trigonometry, with Perspective and the Projec- 
 tion of the Sphere. I intend likewise to print, 
 with all convenient speed, a short treatise on the 
 Philosophy of Arithmetic. The substance of it is 
 already before the public, in the Supplement to 
 the Encyclopaedia Britannica ; but 1 shall endea- 
 vour to abridge, to modify and improve that article* 
 As a sequel, I wish to give a concise and accurate 
 view of the Elements of Algebra, though I will 
 
Vlll PREFACE. 
 
 not absolutely pledge myself to the performance 
 of a task so much wanted. 
 
 It is the nature ot genuine science to ad- 
 vance in continual progression. Each step car- 
 ries it still higher ; new relations are descried ; 
 and the most distant objects seem gradually to 
 approximate. But, while science thus enlarges 
 its bounds, it likewise tends uniformly to sim- 
 plicity and concentration. The discoveries of 
 one age are, perhaps in the next, melted down 
 into the mass of elementary truths. What are 
 deemed at first merely objects of enlightened 
 curiosity, become, in due time, subservient to the 
 most important interests. Theory soon descends 
 to guide and assist the operations of practice. To 
 the geometrical speculations of the Greeks, we 
 may distinctly trace whatever progress the mo- 
 derns have been enabled to achieve in mechanics, 
 navagation, and the various complicated arts of 
 life. A refined analysis has unfolded the harmo- 
 ny of the celestial motions, and conducted the 
 philosopher, through a maze of intricate pheno- 
 mena, to the great laws appointed for the govern- 
 ment of the Universe. 
 
 College of Edinburgh,' 
 March 1. 1817. 
 
ELEMENTS 
 
 OF 
 
 GEOMETRY 
 
 Ijeometry is that branch of natural science 
 which treats of bounded space. 
 
 Our knowledge concerning external objects is 
 derived entirely from the information received 
 through the medium of the senses. The science 
 of Physics considers Bodies as they actually exist, 
 invested at once with all 'their various qualities, 
 and endued with their peculiar affections : Its re- 
 searches are hence directed by that refined spe- 
 cies of observation which is termed Experiment. 
 But Geometry takes a more limited view ; and, 
 selecting only the generic property of Magnitude y 
 it can safely pursue the most lengthened train of 
 investigation, and arrive with perfect certainty at 
 the reipotest conclusion. It contemplates mere- 
 
2. ELEMENTS OF GEOMETRY. 
 
 ly the forms which bodies present, and the spaces 
 which they occupy. Geometry is thus founded 
 likewise on external observation ; but such obser- 
 vation is so familiar and obvious, that the primary 
 notions which it furnishes might seem intuitive, 
 and have often been regarded as innate. This 
 science, proceeding from a basis of extreme sim- 
 plicity, is therefore supereminently distinguished, 
 by the luminous evidence which constantly attends 
 every step of its progress. 
 
 PRINCIPLES. 
 
 In contemplating an external object, we can, 
 by successive acts of abstraction, reduce the com- 
 plex idea which arises in the mind into others that 
 are successively simpler. Body, divested of all its 
 essential characters, presents the mere idea of sur- 
 face ; a surface, considered apart from its peculiar 
 qualities, exhibits only linear boundaries ; and a 
 line, abstracting its continuity, leaves nothing in 
 the imagination, but the points which form its ex- 
 tremities. A solid is bounded by surfaces ; a sur- 
 face is circumscribed by lines ; and a line is ter- 
 minated by points, A point marks position ; a line 
 measures distance ; and a surface presents ea^te?!- 
 sion. A line has only length ;, a surface has both 
 length and. breadth ; and a solid combines all the 
 tliree dimensipns of lengthy breadth, and thichiess. 
 
*• ELEMENTS OF GEOMETRY, S 
 
 The uniform tracing of a line which through its 
 whole extent stretches in the same directioin, gives 
 the idea of a straight line. No more than one 
 straight line can therefore join two points ; and if 
 a straight line be conceived to turn as an axis about 
 both extremities, none of its intermediate points 
 will change their position. 
 
 From our idea of the straight line is derived 
 that of a jplane surface, which, though more com- 
 plex, has a like uniformity of character. A straight 
 line connecting any two points situate in a plane, 
 lies wholly on the surface ; and consequently 
 planes must admit, in every way, a mutual and 
 perfect application. 
 
 Two points ascertain the position of a straight 
 line ; for the line may continue to turn about one 
 of the points till it falls upon the other. But to 
 determine the position of a plane, it requires three 
 points ; because a. plane touching the straight line 
 which joins two of the points, may be made to re- 
 volve, till it meets the third point. 
 
 The separation or opening of two straight lines 
 at their point of intersection, constitutes an ajigle. 
 If we obtain the idea of distancey or linear extent, 
 from contemplating progressive motion, we derive 
 that of divergence, or angular magnitude, from the 
 consideration of revolving motion. 
 
/ 
 
 ELEMENTS OP GEOMETRY. 
 
 Geometry is divided into Plane and Solid ; the 
 former confining its views to the properties of 
 space figured 6n the same plane ; the latter em- 
 bracing the relations of different planes or sur- 
 faces, and of the solids which these describe or ter- 
 minate. In the following definitions, therefore, 
 the points and lines are all considered as existing^ 
 in the same plane. 
 
BOOK I. 
 DEFINITIONS. 
 
 1. A croohed line is that which con- 
 sists of straight lines not continued in the 
 same direction. 
 
 2. A curved line is that of whicli no 
 portion is a straight line. 
 
 3. The straight lines which contain an angle are term- 
 ed its sides, and their .point of origin or intersection, its 
 vertex. 
 
 To abridge the reference, it is usual to denote an angle by- 
 tracing over its sides ; the letter at the ver- 
 tex, which is common to them both, being 
 placed in the middle. Thus, the angle con- 
 tained by the straight lines AB and BC, or 
 the opening formed by turning BA about the point B into the 
 position BC, is named ABC or CBA. ^ 
 
 4. A right angle is the fourth part of an 
 entire circuit or revolution of a straight line. 
 
 It is manifest that all right angles, being derived from the 
 same measure, must be equal to each other. 
 
 If a straight line CB stand at equal angles CBA and CBD 
 on another straight line AD, and if the surface ACD be con- 
 ceived laid over towards the opposite side, the point B and 
 
6 
 
 ELEMENTS^ 01' GEOMETRY. 
 
 the line AD remaining in the same place ; CB will, in this 
 new position EB, make angles EBA 
 and EBD equal to the former, and 
 therefore all of them equal to each 
 other. But the four angles ABC, 
 CBD, DBE, and EBA constitute, a- 
 bout the point B, a complete revolu- 
 tion ; or the line BA in forming them, 
 
 by its successive openings, would return into its original place, 
 — and consequently each of those angles is a right angle. 
 
 The angle contained by the opposite portions DA and DB 
 of a straight line is hence equal to two 
 right angles ; and, for the same rea- 
 son, all the angles ADC, CDE, EDF 
 and FDB, formed at the point D and 
 on the same side of the straight line 
 AB, are together equal to two right 
 angles. 
 
 5. The sides of a right angle are said to he perpendicu- 
 lar to each other. 
 
 6. An acute angle is less than a right 
 e. 
 
 7. An obfuse angle is greater than a 
 right angle. 
 
 8. One side of an angle forms with 
 the other produced a siippleme7ital or 
 exterior angle. 
 
SOOK I. 
 
 9. A vertical angle is formed by the 
 production of both its sides. 
 
 10. The inverted divergence of the two sides of an an- 
 gle, or the defect of the angle from four right angles, is 
 named the reverse angle. 
 
 The angle DBE is vertical to ABC, ABD is the supplemental 
 or exterior angle, and the angle made 
 up of ABD, DBE, and EBC, or the 
 opening formed by the regression of 
 AB through the points D and E into 
 the position BC, is the reverse angle. 
 
 It is apparent that vertical angles, or those formed by the 
 same lines in opposite directions, must be equal ; for the an- 
 gles CBA and itfiD which stand on the straight line CD, be- 
 ing equal to two right angles, are equal to ABD and DBE, 
 and, omitting the common angle ABD, there remains CBA 
 equal to DBE. 
 
 . 11. Two straight lines are said to 
 be inclined to each other, if they 
 meet when produced ; and the an- 
 gle so formed is called trheir inclination. 
 
 12. Straight lines which have no in- 
 clination, are termed parallel. 
 
 13. K figure is a plane surface included by a hnear boun- 
 dary called its perimeter. 
 
 14j. Of rectilineal figures, the triangle h contained by 
 three straight lines. 
 
ELEMENTS OF GEOMETRY. 
 
 15. An isosceles triangle is that which has 
 two of its sides equal.' 
 
 16. An equilateral triangle is that which 
 has all its sides equal. 
 
 17. A triangle whose sides are une- 
 qual, is named scalene. 
 
 It will be shown (1. 9. cor. ) that every triangle has at least two 
 acute angles. The third angle may therefore, by its charac- 
 ter, serve to discriminate a triangle. 
 
 18. A right-angled trmngle is that which 
 has a right angle. 
 
 19. An obtuse angled triangle is that 
 which has an obtuse angle. 
 
 20. An acute angled triangle is that 
 which has all its angles acute. 
 
 21. Any side of a triangle may be called its base, &M 
 the opposite angular point its vertex, 
 
 22. A quadrilateral figure is contained hyfoiir straight 
 lines. 
 
 2S. Of quadrilateral figures, a trape- 
 zoid ( 1 ) has two parallel sides : 
 
 / 
 
 
BOOK I. 
 
 24. A trapezium (2) has two of its 
 sides parallel, and the other two equal, 
 though not parallel, to each other : 
 
 25. A rhomboid (3) has its oppo- 
 site sides equal : ^ 
 
 26. A rhoTiihus (4-) has all its sides e- 
 qual: 
 
 27. An ohlong, or rectangle, (5) has a 
 right angle, and its opposite sides equal : 
 
 28. A square (6) has a right angle, and all 
 its sides equal. 
 
 29. A quadrilateral figure, of which the opposite sides 
 are parallel, is called a parallelogram. 
 
 30. The straight line which joins 
 obliquely the opposite angular points 
 of a quadrilateral figure, is named a 
 diagonal. 
 
 31. If an angle of a rectilineal figure be less than two 
 right angles, it protrudes, and is called salient ; if it be 
 greater than two right angles, it makes a sinuosity, and is 
 termed re-entrant. 
 
 Thus the angle ABC is re-entrant, and ""^~~ ^ ^ 
 the rest of the angles of the polygon 
 ABCDEF are salient at A, C, D, E and 
 
10 ELEMENTS OP GEOMETRY. 
 
 32. A rectilineal figure having more than four sides, 
 bears the general name of a polygon, 
 
 S3. A circle is a figure described by the revolution of a 
 straight line about one of its extremities : 
 
 •a' 
 
 S^. The fixed point is called the centre 
 of the circle, the describing line its radius^ 
 and the boundary traced by the remote end 
 of that line its circumference. 
 
 35. The diameter of a circle is a straight line drawn 
 through the centre, and terminated both ways by the cir- 
 cumference. 
 
 It is obvious that all radii of the same circle are equal to 
 each other and to a semidiameter. It likewise appears, from 
 the slightest inspection, that a circle can only have one centre, 
 and that circles are equal which have equal diameters. 
 
 36. Figures are said to be equals when, applied to each 
 other, they wholly coincide ; they are equivalent^ if, with- 
 out coinciding, they yet contain the same space. 
 
BOOK I. li 
 
 A Proposition is a distinct portion of abstract science. 
 It is either a problem or a theorem. 
 
 A Problem proposes to effect some combination. 
 
 A Theorem advances some truth, which is to be esta- 
 blished. 
 
 A problem requires solution, a theorem wants demonstra- 
 Hon ; the former implies an operation, and the latter ge- 
 nerally needs a previous construction. 
 
 K direct demonstration proceeds from the premises, by 
 a regular deduction. 
 
 An indirect demonstration attains its object, by showing 
 that any other hypothesis than the one advanced would in- 
 volve a contradiction, or lead to an absurd conclusion. 
 
 A subordinate property, included in a demonstration, is 
 sometimes, for the sake of unity, detached, and then it 
 forms a Lemma. 
 
 A Corollary is an obvious consequence that results 
 from a proposition. 
 
 A Scholium is an excursive remark on the nature and 
 application of a train of reasoning. 
 
 The operatio7is in Geometry suppose the dramming of 
 straight lities and the description of circles, or they require 
 in practice the use of the nde and compasses. 
 
12 ELEMENTS OF GEOMETRY. 
 
 PROPOSITION I. PROBLEM. 
 
 To construct a triangle, of which the three sides 
 are given. 
 
 Let AB represent the base, and G, H two sides of the 
 triangle which it is required to construct. 
 
 From the centre A, with the distance G, describe a cir- 
 cle ; and, from the centre B, with the distance H, describe 
 another circle, meeting the 
 former in the point C : ACB 
 is the triangle required. 
 
 Because all the radii of the 
 same circle are equal, AC is 
 equal to G ; and, for the same 
 reason, BC is equal to H. 
 Consequently the triangle ACB answers the conditions of 
 the problem. The limiting circles, after mutually intersect- 
 ing, must obviously diverge from each other, till, crossing 
 the extension of the base AB, they return again and meet 
 below it j thus marking two positions for the required tri- 
 angle. 
 
 Corollary, If the radii G and H be equal to each other, 
 the triangle will evidently be isosceles ; and if those lines 
 be likewise equal to the base AB, the triangle must be equi- 
 lateral. 
 
BOOK I. IS 
 
 PROP. II. THEOREM. 
 
 Two triangles are equal, which have all the 
 sides of the one equal to those of the other. 
 
 Liet the two triangles ABC and DFE have the side AB 
 equal to DF, AC to DE, and BC to FE : These triangles 
 are equal. 
 
 For conceive the triangle ACB to be applied to DEF : 
 The point A being laid on D, and the side AC on DE, 
 their other extremities C and E must coincide, since AC 
 is equal to DE. And because AB is equal to DF, the 
 point B must occur in the circum- 
 ference of a circle described from 
 D with the distance DF ; and, for 
 the same reason, B must be found 
 in the circumference of a circle de- 
 scribed from E with the distance 
 EF: The vertex of the triangle ACB must, therefore, ap- 
 pear in a point which is common to both those circles, or, 
 by the first proposition, in F the vertex of the triangle 
 DFE. Consequently these two triangles, being rectilineal, 
 must entirely coincide. The angle CAB is equal to EDF, 
 ACB to DEF, and CBA to EFD ; the equal angles be- 
 ing thus always opposite to the equal sides. 
 
 Scholium. This proposition is only the preceding one 
 changed into a theorem. But any rectilineal figure may be 
 divided into triangles, which, being separately constructed 
 with the same corresponding sides, must, by their combi- 
 nation, hence form an equal figure. 
 
M. ELEMENTS OF GEOMETRY. 
 
 PROP. III. THEOR. 
 
 Two triangles are equal, if two sides and the 
 angle contained by these in the one be respective- 
 ly equal to two sides and the contained angle in 
 the other. 
 
 Let ABC and DEF be two triangles, of which the side 
 AB is equal to DE, the side BC to EF, and the angle ABC 
 contained by the former equal to DEF which is contained 
 by the latter : These triangles are equal. 
 
 For let the triangle ABC be applied to DEF : The ver- 
 tex B being placed on E, and the side BA on ED, the ex- 
 tremity A must fall upon D, 
 since AB is equal to DE. 
 And because the angle or 
 divergence ABC is equal to 
 DEF, and the side AB co- 
 incides with DE, the other side BC must lie in the same 
 direction with EF, and being of the same length, must ter- 
 minate with it ; and consequently, the points A and C rest- 
 ing on D and F, the straight lines AC and DF will also 
 coincide. Wherefore, the one triangle being thus per- 
 fectiy adapted to the other, a general equality must obtain 
 between them : The third sides AC and DF are hence 
 equal, and the angles BAC, BCA opposite to BC and BA 
 are equal respectively to EDF and EFD, which the corre- 
 sponding sides EF and ED subtend. 
 
 Schol. By applying this proposition to practice, the mu- 
 tual distance may be found between two remote objects 
 which have their communication obstructed. 
 
BOOK I. 
 
 15 
 
 .PROP. IV. PROB. 
 
 At a point in a straight line, to make an angle 
 equal to a given angle. 
 
 At the point D in the given straight line DE, to form an 
 angle equal to the given angle BAG. 
 
 In the sides AB and AC of the given angle, assume the 
 points G and H, join GH, 
 from DE cut off DI equal 
 to AG, and on DI consti- 
 tute (I. 1.) a triangle DKI, 
 having the sides DK and 
 IK equal to AH and GH: 
 EDK or EDF is the angle required. 
 
 For all the sides of the triangles GAH and IDK being 
 respectively equal, the angles opposite to the equal sides 
 must be likewise equal (I. 2.), and consequently IDK is 
 equal to GAH. 
 
 Cor. If the segments AG, AH be taken equal, the con-, 
 struction will be rendered simpler and more commodious. 
 
 SchoL By the successive application of this problem 
 an angle may be continually multiplied. Two circles CEG 
 and ADF being described from the 
 vertex B of the given angle with radii 
 BC and BA equal to its sides, and the 
 base AC being repeatedly inserted be- 
 tween those circumferences ; a multi- 
 tude of triangles will be thus formed, all 
 of them equal to the original triangle 
 ABC. Consequently the angle ABD is double of ABC, 
 ABE triple, ABF quadruple, ABG quintuple, &c. 
 
16 
 
 ELEMENTS OF GEOMETRY. 
 
 If the sides AB and BC of the given angle be supposed 
 equal, only one circle would be required, 
 a series of equal isosceles triangles being 
 constituted about its centre/ It is evi- 
 dent that this addition is without limit, 
 and that the angle so produced may con- 
 tinue to spread out, and its opening side 
 even make repeated revolutions. 
 
 PROP. V. PROB. 
 
 To bisect a given angle. 
 
 Let ABC be an angle which it is required to bisect. 
 
 In the side AB take any point D, and from BC cut off 
 BE equal to BD ; join DE, on whiph construct (I. 1.) the 
 isosceles triangle DFE, and draw the 
 straight line BF : The angle ABC 
 is bisected by BF. 
 
 For the two triangles DBF and 
 EBF, having the side DB equal to 
 EB, the side DF to EF, and BF 
 common to both, are (I. 2.) equal, 
 and consequently the angle DBF is equal to EBF. 
 
 Cor, Hence the mode of drawing a perpendicular from 
 a given point B in the straight line AC; for the angle 
 ABC, which the opposite seg- 
 ments BA and BC make with 
 each other, being equal to two 
 right angles, the straight line 
 that bisects it must be the per- 
 pendicular required. , Taking 
 BD, therefore, equal to BE, and 
 
 :ij 
 
BOOK I. 17 
 
 constructing the isosceles triangle DFE ; the straight line 
 BF, which joins the vertex of the triangle, is perpendicular 
 to AC. 
 
 ScJioL In the general construction, the isosceles triangle 
 33FE may stand either below or above the base DE; 
 but if it were made equal to DBE, the vertex F would coin- 
 cide with B, and render the construction indeterminate. 
 
 PROP. VI. PROB. 
 
 To let fall a perpendicular upon a straight line, 
 from a given point above it. 
 
 From the point C, to let fall a perpendicular upon the 
 given straight line AB. 
 
 In AB take towards A the point D, and with the 
 distance DC describe a circle; and, in the same line, 
 take towards B another point E, and with the distance 
 EC describe a second circle intersecting the former 
 in F; join CF, crossing the gi- 
 ven line in G : CG is perpendicular 
 toAB. 
 
 For the straight lines DC, DF a-^^ [ U \r ^ 
 and EC, EF being joined, the trian- 
 gles DCE and DFE have the side 
 DC equal to DF, EC to EF, and 
 DE common to them both ; whence (I. 2.) the angle CDE 
 or CDG is equal to FDE or FDG. And because, in the 
 triangles DCG and DFG, the side DC is equal to DF, 
 DG common, and the contained angles CDG and FDG 
 are proved to be equal; these subordinate triangles are 
 (I. 3.) equal, and consequently the angle DGC is equal 
 to DGF, and each of them a right angle, or CG is per- 
 pendicular to AB. 
 
18 ELEMENTS OF GEOMETRY. 
 
 PROP. VII. PROB. 
 
 To bisect a given finite straight line. 
 
 On the given straight line AB, construct two isosceles 
 triangles (I. 1.) ACB and ADB, and join their vertices C 
 and D by a straight line cutting AB in the point E : AB 
 is bisected in E. 
 
 For the sides AC and AD of the triangle 
 CAD being respectively equal to BC and 
 BD of the triangle CBD, and the side 
 CD common to them .both; these triangles 
 (I. 2.) are equal, and the angle ACD or 
 ACE k equal to BCD or B^CE. Again, 
 the inferior triangles ACE and BCE, having the side AC 
 equal to BC, CE common, and the contained angle ACE 
 equal to BCE, are (I. 3.) equal, and consequently the base 
 AE is equal to BE. 
 
 PROP. VIII. THEOR. 
 
 The exterior angle of a triangle is greater than 
 either of its interior opposite angles. 
 
 The exterior angle BCF, formed by producing a side 
 AC of the triangle ABC, is 
 greater than either of the op- 
 posite and interior angles CAB 
 and CBA. 
 
 For bisect the side BC in the 
 point D (I. 7.), draw AD, and 
 produce it until DE be equal to 
 AD, and join EC. 
 
BOOK I. 19 
 
 The triangles ADB and EDC have, by construction, the 
 side DA equal to DE, the side DB to DC, and the vertical 
 angle BDA equal to CDE ; these triangles are, therefore, 
 equal (I. 3.), and the angle DCE is equal to DBA. But 
 the angle BCF is evidently greater than DCE ; it is con- 
 sequently greater than DBA qr CBA. 
 
 In like manner, it may be shown, that if BC be produ- 
 ced, the exterior angle ACG is greater than CAB. But 
 ACG is equal to the vertical angle BCF, and hence BCF 
 must be greater than either the angle CBA or CAB. 
 
 Cor. Hence all the exterior angles of a triangle are 
 greater than the interior, and likewise greater than three 
 right angles. 
 
 PROP. IX. THEOR. 
 
 Any two angles of a triangle are together less 
 than two right angles. 
 
 The two angles BAC and BCA of the triangle ABC are 
 together less than two right angles. 
 
 For produce the common side AC. 
 And, by the last proposition, the ex- 
 terior angle BCD is greater than 
 BAC, add BCA to each, and the "^ C JD 
 
 two angles BCD and BCA are greater than BAC and 
 BCA, or BAC and BCA are together less than BCD and 
 BCA, that is, less than two right angles (Def. 4). 
 
 Cor. Hence a triangle can only have one right or ob- 
 tuse angle, its two remaining angles being always acute. 
 
20 ELEMENTS OF GEOMETRY. 
 
 PROP. X. THEOR. 
 
 The angles at the base of an isosceles triangle 
 are equal. 
 
 The angles BAG and BCA at the base of the isosceles 
 triangle ABC are equal. 
 
 For draw (I. 5.) BD bisecting the vertical angle ABC. 
 
 Because, by hypothesis, AB is equal to 
 BG, the side BD common to the two tri- 
 angles BDA and BDG, and the angles 
 ABD and GBD contained by them are 
 equal ; these triangles are equal (I. 3.), 
 and consequently the angle BAD is equal 
 to BGD. 
 
 Cor. Every equilateral triangle is also equiangular^ 
 
 PROP. XL THEOR. 
 
 If two angles of a triangle be equal, the sides 
 opposite to them are likewise equal. 
 
 Let the triangle ABG have two equal angles BGA and 
 BAG ; the opposite sides AB and BG are also equal. 
 
 For if A B be not equal to GB, let it be equal to GD, 
 and join AD. 
 
 Comparing now the triangles BAG and DGA, the side 
 AB is by supposition equal to CD, AG is 
 common to both, and the contained angle 
 BAG is equal to DGA ; the two triangles 
 (L 3.) are, therefore, equal. But this con- 
 clusion is manifestly absurd. To suppose 
 then the inequality of AB and BG involves 
 
BOOK I. 21 
 
 a contradiction j and consequently those sides must be 
 equal 
 
 Cor, Every equiangular triangle is also equilateral. 
 
 SchoL By the application of this proposition, the dis- 
 tance of an object inaccessible from one side may in some 
 cases be measured. 
 
 PROP. XII. THEOR. 
 
 In any triangle, that angle is the greater which 
 lies opposite to a greater side. 
 
 If a side BC of the triangle ABC be greater than BA ; 
 the opposite angle BAC is greater than BCA. 
 
 For make BD equal to BA, and join AD. The angle 
 CAB is evidently greater than DAB ; 
 but since BA is equal to BD, this angle /s. 
 
 DAB (I. 10.) is equal to ADB, and / \^^ 
 
 consequently CAB is greater than ADB. f^^,^^ — ^ ^ 
 Again, the angle ADB, being an exte- 
 rior angle of the triangle CAD, is (I. 8.) greater than ACD 
 or ACB; wherefore the angle CAB is much greater than 
 ACB. 
 
 PROP. XIII. THEOR. 
 
 That side of a triangle is the greater which sub- 
 tends a greater angle. 
 
22 ELEMENTS OF GEOMETRY. 
 
 If, in the triangle ABC, the angle CAB be greater than 
 ACB ; its opposite side BC is greater than AB. 
 
 For if BC be not greater than AB, it must be either 
 equal or less. But it cannot be equal, be- 
 cause the angle CAB would then be equal 
 to ACB (I. 10.) ; nor can BC be less than 
 AB, for then AB would be greater than 
 BC, and consequently (I. 12.) the angle *^ 
 ACB would be greater than CAB, or CAB less than ACB, 
 which is absurd. The side BC being thus neither equal 
 to AB, nor less than it, must therefore be greater than 
 AB. - 
 
 PROP. XIV. THEOR. 
 
 Two sides of a triangle are together greater thai^ 
 the third side. 
 
 The two sides AB and BC of the triangle ABC are to- 
 gether greater than the third side AC. 
 
 For produce AB until DB be equal to the side BC, 
 and join CD. 
 
 . Because BC is equal to BD, the angle BCD is equal to 
 BDC (I. 10.) ; but the angle ACD 
 is greater than BCD, and therefore 
 greater than BDC, or ADC 5 con- 
 sequently the opposite side AD is 
 greater than AC (I. 13.) ; and since 
 AD is equal to AB and BD, or to 
 AB and BC, the two sides AB and 
 BC are together greater than the diird side AC. 
 
»00K I. 2S 
 
 PROP. XV- THEOR. 
 
 The difference between two sides of a triangle 
 is less than the third side. 
 
 Let the side AC be greater than AB, and from it cut off 
 a part AE equal to AB ; the remainder 
 EC is less than the third side BC. B 
 
 For the two ^ides AB and BC are to- 
 gether greater than AC (I. 14.); take 
 away the equal lines AB and AE, and 
 there remains BC greater than EC, or EC is less than BC. 
 
 PROP. XVI. THEOR. 
 
 Two straight lines drawn to a point within a 
 triangle from the extremities of its base, are toge- 
 ther less than the sides of the triangle, but con- 
 tain a greater angle. 
 
 The straight lines AD and CD, projected to a point D 
 within the triangle ABC from the extremities of the base 
 AC, are together less than the sides AB and CB of the 
 triangle, but contain a greater angle. 
 
 For produce AD to meet CB in E. The two sides AB 
 and BE of the triangle ABE are greater 
 than the third side AE (I. U.) ; add EC 
 to each, and AB, BE, EC, or AB and 
 BC, are greater than AE and EC. But 
 the sides CE and ED of the triangle 
 DEC are (I. 14.) greater than DC, and 
 
24 (ELEMENTS OF GEOMETRY. 
 
 consequently CE, ED, together with DA, or CE and EA, 
 are greater than CD and DA. Wherefore the sides AB 
 and BC, being greater than AE and EC, which are 
 themselves greater than AD and DC, must be still greater 
 than AD and DC, or the lines AD and DC are less than 
 AB and BC, the sides of the triangle. 
 
 Again, the angle ADC, being the exterior angle of the 
 triangle DCE, is greater than DEC (I. 8.) ; and, for the 
 same reason, DEC is greater than ABE, the opposite in- 
 terior angle of the triangle EAB. Consequently ADC is 
 still greater than ABE or ABC. 
 
 PROP. XVII. THEOR, 
 
 If straight lines be drawn from the same point 
 to another straight line, the perpendicular is the 
 shortest of them all ; the lines equidistant from it 
 on both sides are equal ; and those more remote 
 are greater than such as are nearer. 
 
 Of the straight lines CG, CE, CD, andCF drawn from 
 a given point C to the straight line AB, the perpendicular 
 CD is the least, the equidistant lines CE and CF are equal, 
 but the remoter line CG is greater than either of these 
 two. 
 
 For the right angle CDE, equal to CDF, is (I. 8.) great- 
 er than the interior angle CFD of the triangle DCF, and 
 consequently the opposite 
 side CF is (I. 13.) greater 
 than CD, or CD is less than 
 CF. 
 
 But if ED be equal to FD, 
 
 A Gr £ ;d :f Ji 
 
BOOK r. 25 
 
 CD being common to the two triangles ECD and FCD, 
 and the contained angles CDE and CDF equal; these 
 triangles (I. 3.) are equal, and consequently their bases 
 CE and CF are equal. 
 
 Again, because GCD is a right-angled triangle, the an- 
 gle CGD or CGE is (I. 9. cor.) acute, and, for the same 
 reason, the angle CED of the triangle CDE is acute, and 
 consequently its adjacent angle CEG is obtuse. Where- 
 fore CEG, being greater than a right angle, is still greater 
 than CGE, and the opposite side CG must be greater 
 (L 13.) than CE. 
 
 Cot; Hence only a single perpendicular CD can be let 
 fall from the same point C upon a given straight line AB; 
 and hence also a pair only of equal straight lines greater 
 than CD can at once be extended from C to AB, making 
 on the same side, the one an obtuse angle CEA, and the 
 other an acute angle CFA. — As the term distance signi- 
 fies the shortest road, the distance between two points, is 
 the straight line which joins them ; and the distance from a 
 point to a straight line, is the perpendicular let fall upon it. 
 
 PROP. XVIII. THEOR. 
 
 If two sides of one triangle be respectively equal 
 to those of another, but contain a greater angle ; 
 the base also of the former will be greater than 
 that of the latter. 
 
 In the triangles ABC and DEF, let the sides AB and 
 BC be equal tb DE and EF, but the angle ABC greater 
 than DEF ; then is the base AC greater than DF. 
 
26 ELEMENTS OF GEOMETRY. 
 
 For, suppose AB one of the sides to be not greater than 
 BC or EF, and (I. 4.) draw BG equal to EF making an 
 angle ABG equal to DEF, join AG and GC. 
 
 Because AB and BG are equal to DE and EF, and 
 the contained angle ABG is equal to DEF; the triangles 
 ABG and DEF (I. 3.) are equal, and have equal bases AG 
 and DF. 
 
 First, let the triangles ABC an^ DEF be isosceles. 
 Since the side AB is equal 
 to BC, the angle BAC (1. 10.) 
 is equal to BCA ; but (I. 8.) 
 the angle BHC is greater 
 than the interior angle BAH 
 or BCH, and consequently 
 (I. IS.) the side BC or BG is greater than BH, or the 
 point G lies beyond H. 
 
 Next, suppose the side 
 BC or EF to be greater 
 than AB or DE. Where- 
 fore (I. 12.) the angle BAC 
 is greater than BCA ; but 
 (I. 8.) the exterior angle 
 BHC of the triangle ABH 
 is greater than BAH or BAC, and hence still greater than 
 BCA or BCH -, consequently the side BC or EF is (I. 13.) 
 greater than BH. 
 
 In every case, therefore, the point G must lie below the 
 base AC. But the triangle GBC being evidently isosceles, 
 its angles BGC and BCG (I. 10.) are equal. Whence the 
 angle AGC, being greater than BGC or BCG, which a- 
 gain is greater than ACG, must be still greater than ACG; 
 and therefore the opposite side AC is (I. 13.) greater than 
 AG or DF. 
 
ppoK I. ■ 27 
 
 PROP. XIX. THEOR. 
 
 If two sides of one triangle be respectively e- 
 qual to those of another, but stand on a greater 
 base ; the angle contained by the former will be 
 likewise greater than what is contained by the lat- 
 ter. 
 
 Let the triangles ABC and DEF have the sides AB and 
 BC equal to DE and EF, but the base AC greater than 
 PF; the vertical angle ABC is greater than DEF. 
 
 For if ABC be not greater than the angle DEF, it must 
 either be equal or less. But it 
 cannot be- equal to DEF, for 
 tfee sides AB, BC being then e- 
 qual to DE, EF, and contain- 
 ing equal angles, the base AC 
 would (I. S.) be equal to DF, which is contrary to the hy- 
 pothesis. Still more absurd it would be to suppose the an- 
 gle ABC less than DEF, since the triangles BAC and 
 EDF, having their sides AB, BC equal to DE, EF, but 
 the contained angle ABC less than DEF, or DEF great- 
 er than ABC, the base DF would, from the preceding pro- 
 position, be greater than AC, or AC would be less than 
 DF. 
 
 PROP. XX. THEOR. 
 
 Two triangles are equal, which have two angles 
 and a corresponding side in the one respectively 
 equal to those in the other. 
 
28 ELEMENTS OF GEOMETRY. 
 
 Let the triangles ABC and DEF have the angle BAC 
 equal to EDF, the angle BCA to EFD, and a side of the 
 one equal to a side of the other, whether it be interjacent 
 or opposite to those equal angles ; the triangles will be e* 
 qual. 
 
 First, let the equal sides be AC and DF, which are in- 
 terjacent to the equal angles in both triangles. — Apply the 
 triangle ABC to DEF ; the point A being laid on D, and 
 the straight line AC on DF, the other extremities C and 
 F must coincide, since those lines 
 are equal. And because the angle 
 BAC is equal to EDF, and the 
 side AC is applied to DF, the o- 
 ther side AB must lie along DE ; 
 and for the same reason, the an- 
 gles BCA and EFD being equal, the side CB must lie a- 
 long FE. Wherefore the point B, which is common to 
 both the lines AB and CB, will be found likewise in both 
 DE and FE ; that is, it must fall upon the corresponding 
 vertex E. The two triangles ABC and DEF, thus mu- 
 tually adapting, are hence entirely equal. 
 
 Next, let the equal sides be AB and DE, which are op-' 
 posite to the equal angles BCA and EFD. The triangle 
 ABC being laid on DEF, the sides AB and AC of the an- 
 gle BAC will apply to DE and DF, 
 the sides of the equal angle EDF; 
 and since AB is equal to DE, the 
 
 points B and E must coincide ; but 
 
 by hypothesis, the angles BCA and "''' 
 
 EFD being equal, BC must adapt itself to EF, for other- 
 wise one of those angles becoming the exterior angle of a 
 secondary triangle, would (I. 8.) be greater than the other. 
 
BOOK I. 2ft 
 
 Whence the triangles ABC, DEF are entirely coincident, 
 and have those sides equal which subtend equal angles, 
 
 Schol, By the application of the first case, where the 
 sides lying between the equal angles are equal, the distance 
 of an inaccessible object can be measured in all cases. 
 
 PROP. XXI. THEOR. 
 
 Two triangles are equal if two sides and a cor- 
 responding opposite angle be equal in both, and 
 the other opposite angles have the same character. 
 
 In the triangles ABC and DEF, let the side AB be equal 
 to DE, BC to EF, and the angles BAC, EDF, opposite to 
 BC, EF, be also equal j the triangles themselves are equal, 
 if the other angles BCA and EFD opposite to AB and DE 
 be of the same character, or at once right, or acute, ot 
 obtuse. 
 
 For, the triangle ABC being applied to DEF, the an- 
 gle BAC will adapt itself to EDF, since they are equal ; 
 and the point B must coincide with E, because the side AB 
 is equal to DE. But the other equal sides BC and EF, 
 now stretching from the same point 
 E towards DF, must likewise coin- y\ ^ 
 
 cide J for if the angle at C or F be 
 
 right, there can exist no more than . 
 
 one perpendicular EF (I. 17. cor.) 
 
 and, in like manner, if this angle at F be either obtuse or 
 acute, the line EF, which forms it, can, for the same rea- 
 son, have only one corresponding position. — Whence, ia 
 each of these three cases, the triangle ABC admits of a 
 perfect adaptation with DEF. 
 
50 ELEMENTS OF GEOMETRY, 
 
 PROP. XXII. THEOR. 
 
 If a straight line fall upon two parallel straight 
 lines, it will make the alternate angles equal, the 
 exterior angle equal to the interior opposite one, 
 and the two interior angles on the same side to- 
 gether equal to two right angles. 
 
 Let the straight line EFG fall upon the parallels AB 
 and CD ; the alternate angles AGF and DFG are equal, 
 the exterior angle EFC is equal to the interior angle EGA, 
 and the interior angles CFG and AGF, or FGB and GFD, 
 are together equal to two right angles. 
 
 I^or conceive a straight line, produced both ways from 
 F, to turn about that point in the same plane; it 
 will first cut the extended line AB above G and to- 
 wards A, and will in its progress afterwards meet this 
 line on the other side below G and towards B. In 
 the position IFH, the angle EFH is the exterior angle 
 of the triangle FHG, and therefore greater than FGH 
 or EGA (I. 8.) But in the last position LFK, the exte- 
 rior angle EFL is equal to its vertical angle GFK in the 
 triangle FKG, and to which the 
 angle FGA is exterior -, conse- 
 quently (I. 8.) FGA is greater than 
 EFL, or the angle EFL is less 
 than FGA or EGA. When the 
 incident line EFG, therefore, meets 
 AB above the point G, it makes 
 an angle EFH greater than EGA ; 
 and when it meets AB below that 
 point, it makes an angle EFL, 
 which is less than the game angle. But in passing through 
 
ELEMENTS OF GEOMETRY. 3 I 
 
 all the degrees from greater to less, a varying magnitude 
 must evidently encounter the single interfnediate limit of 
 equality. Wherefore, there is a certain position CD, in 
 which the line revolving about the point F makes the ex- 
 terior angle EFC equal to the interior EGA, and at the 
 same instant of time meets AB neither towards the one 
 part nor the other, or is parallel to it. 
 
 And now, since EFC is proved to be equal to £GA, 
 and is also equal to the vertical angle GFD ; the alternate 
 angles FGA and GFD are equal. Again, because GFD 
 and FGA are equal, add the angle FGB to each, and the 
 two angles GFD and FGBi are equal to FGA and FGB ; 
 but the angles FGA and FGB, on the same side of AB, 
 are equal to two right angles, and consequently the inte- 
 rior angles GFD and FGB are likewise equal to two right 
 angles. 
 
 Cor. Since the position CD is individual, or that only 
 one sti'aight line can be drawn through the point F pa- 
 rallel to A B, it follows that the converse of the proposition 
 is hkewise true, and that those three properties of parallel 
 lines are criteria for distinguishing parallels. 
 
 PROP. XXIir. PROB. 
 
 Through a given point, to draw a straight line 
 parallel to a given straight line. 
 
 To draw, through the point C, a straight line parallel 
 to AB. 
 
 In AB take any point D, join CD, 
 and at the point C make (I. 4.) an an- \^ 
 
 gle DCE equal to CDA ; CE is paral- -^^^^^^^^^ ^^ 
 
 leltoAB. ( ^ 
 
S2 ELEMENTS OF GEOMETRY. 
 
 For the angles CDA and DCE, thus formed equal, are 
 the alternate angles which CD makes with the straight 
 lines CE and AB, and, therefore, by the corollary to the 
 last proposition, these lines are parallel. 
 
 PROP. XXIV. THEOR. 
 
 Parallel lines are equidistant, and equidistant 
 straight lines are parallel. 
 
 The perpendiculars EG, FH, let fall from any points 
 E,.Fin the straight line AB upon its parallel CD, are 
 equal ; and if these perpendiculars be equal, the straight 
 lines AB and CD are parallel. 
 
 For join EH : and because each of the interior angles 
 EGH and FHG is a right angle, they are together equal 
 to two right angles, and consequently the perpendiculars 
 EG and FH are (I. 22. cor.) 
 
 parallel to each other ; where- -^ ^^ ^ 
 
 fore (I. 22.) the alternate an- 
 gles HEG and EHF are equal. ^ 
 But, EF being parallel to GH, 
 the alternate angles EHG and HEF are likewise equal ; 
 and thus the two triangles HGE and HFE, having the 
 angles HEG and EHG respectively equal to EHF and 
 HEF, and the side EH common to both, are (I, 20.) e- 
 qual, and hence the side EG is equal to FH. 
 
 Again, if the perpendiculars EG and FH be equal, the 
 two triangles EGH and EFH, having the side EG equal 
 to FH, EH common, and the contained angle HEG equal 
 to EHF, are (I. 3.) equal, and therefore the angle EHG 
 equal to HEF, and (I. 22.) the straight line AB parallel 
 to CD. 
 
 Cr Jrl 
 
ISOOK I. S3 
 
 PROP. XXV. THEOR. 
 
 The opposite sides of a rhomboid are parallel. 
 
 If the opposite sides AB, DC, and AD, BC of the qua- 
 drilateral figure ABCD be equal, they are also parallel. 
 
 For draw the diagonal AC. And because AB is equal 
 to DC, BC to AD, and AC is com- 
 mon ; the two triangles ABC and 
 ADC are (I. 2.) equal. Conse- 
 quently the angle ACD is equal ^ ^ 
 to CAB, and therefore the side AB (I. 22. cor.) parallel 
 to CD ', and, for the same reason, the angle CAD being 
 equal to A CB, the side AD is parallel to BC. 
 
 Cor, Hence the angles of a square or rectangle are ail 
 of them right angles ; for the opposite sides being equal, 
 are parallel ; and if the angle at A be right, the other in- 
 terior one at B is also a right angle (I. 22.), and conse- 
 quently the angles at C and D, opposite to these, are 
 right. — On this proposition depends the construction of 
 the instrument called a Parallel Ruler, 
 
 PROP. XXVI. THEOR. 
 
 The opposite sides and angles of a parallelo- 
 gram are equal. 
 
 Let the quadrilateral figure ABCD have the sides AB 
 and BC parallel to CD and AD ; these are respectively 
 equal, and so are likewise the opposite angles at A and C, 
 and at B and D. 
 
34< ELEMENTS OF GEOMETRY. 
 
 For join AC. Because AB is parallel to CD, the al- 
 ternate angles BAC and ACD are (I. 22.) equal; and 
 since AD is parallel to BC, the alternate angles ACB and 
 CAD are also equal. Where- 
 fore the triangles ABC and ADC, 
 having the angles CAB and ACB 
 equal to ACD and CAD, and the ^ 
 inteijacent side AC common to both* are (I. 20.) equal. 
 Consequently, the side AB is equal to CD, and the side 
 BC to AD ; and these opposite sides being thus equal, the 
 opposite angles (I. 25.) must be likewise equal. 
 
 Cor, Hence the diagonal divides a rhomboid or paral- 
 lelogram into two equal triangles. 
 
 PROP. XXVII. TPIEOR. 
 
 If the parallel sides of a trapezoid be equal, the 
 other sides are likewise equal and parallel. 
 
 Let the sides AB and DC be equal and parallel j the 
 sides AD and BC are themselves equal and parallel. 
 
 For draw the diagonal AC. Because AB is parallel to 
 CD, the alternate angles CAB and ACD are (1. 22.) equal ; 
 and the triangles ABC and ADC, having the side AB 
 equal to CD, AC common to both, 
 and the contained angle CAB equal 
 to ACD, are, therefore, equal (1. 3.). 
 Whence the side BC is equal to AD, 
 and the angle ACB equal to CAD ; but these angles being 
 alternate, BC must also be parallel to AD (I. 22. cor.) 
 
BOOK I. 35 
 
 PROP. XXVIIL THEOR. 
 
 Lines parallel to the same straight line, are pa- 
 rallel to each other. 
 
 If the straight line AB be parallel to CD, and CD pa- 
 rallel to EF; then is AB parallel to EF. 
 For let a straight line GH cut these v 
 
 ^^"^'- a\i B 
 
 And because AB is parallel to CD, 7 \ ^ !^ 
 
 the exterior angle GIA is equal (I. 22.) Z \ -j. ^ 
 
 to the interior GKC ; and since CD is NT 
 
 parallel to EF, this angle GKC is, for \ 
 
 the same reason, equal to GLE. There- 
 fore the angle GIA is equal to GLE, and consequently 
 AB is parallel to EF (I. 22. cor.) 
 
 PROP. XXIX. THEOR. 
 
 Straight lines drawn parallel to the sides of an 
 angle, contain an equal angle. 
 
 If the straight lines AB, AC be pa- 
 rallel to DE, DF i the angle BAC is 
 equal to EDF. 
 
 For draw the straight line GAD 
 through the vertices. And since AC 
 is parallel to DF, the exterior angle 
 GAC is (I. 22.) equal to GDF ; and, 
 for the same reason, GAB is equal to GDE ; there con- 
 sequently remains the angle BAC equal to EDF. 
 
36 ELEMENTS OF GEOMETRY. 
 
 PROP. XXX. THEOR. 
 
 An exterior angle of a triangle is equal to both 
 its opposite interior angles, and all the interior 
 angles of a triangle are together, equal to two 
 right angles. 
 
 The exterior angle BCD, formed by the production of 
 the side AC of the triangle ABC, is equal to the two op- 
 posite interior angles CAB and CB A, and all the interior 
 angles CAB, CB A and BCA of the triangles are together 
 equal to two right angles. 
 
 For, through the point C, draw (I. 23.) the straight 
 line CE parallel to AB. And, AB being parallel to CE, 
 the interior angle BAC is (I. 22.) equal to the exterior one 
 ECD j and, for the same reason, the alternate angle ABC 
 is equal to BCE. Wherefore the two 
 angles CAB and ABC are equal to 
 DCE and ECB, or to the whole exte- 
 rior angle BCD. 
 
 Again, add the adjacent angle BCA 
 to the exterior angle BCD, and to the 
 two interior angles CAB and ABC ; and all the interior 
 angles of the triangle ABC are together equal to the angles 
 BCD and BCA on the same side of the straight line AD, 
 that is, to two right angles. 
 
 Cor, 1. Hence the two acute angles of a right angled 
 triangle are together equal to one right angle ; and hence 
 each angle of an equilateral triangle is two-third parts of 
 a right angle. 
 
 Cor. 2. Hence if a triangle have its exterior angle, and 
 pne of its opposite interior angles, double of those in an- 
 
BOOK I. 37 
 
 other triangle ; its remaining opposite interior angle will 
 also be double of the corresponding angle in the other. 
 
 SchoL On the second corollary depends the construction 
 of that invaluable reflecting angular instrument, called 
 Hadley's quadrant or sextant. 
 
 PROP. XXXI. THEOR. . 
 
 The interior angles of any rectilineal figure are 
 together equal to twice as many right angles (a- 
 bating four from the amount) as the figure has 
 sides. 
 
 For assume a point O within the figure, and draw 
 straight lines OA, OB, OC, OD, and OE, to the several 
 corners. It is obvious, that the figure is thus resolved in- 
 to as many triangles as it has sides, 
 and whose collected angles must, by 
 the last proposition, be equal to twice 
 as many right angles. But the an- 
 gles at the bases of these triangles 
 constitute the internal angles of the fi- 
 gure. Consequently, from the whole amount, there is to 
 be deducted the vertical angles about the point O, and 
 which are (Def. 4.) equal to four right angles. 
 
 Cor, Hence all the angles of a quadrilateral figure are 
 equal to four right angles, those of a pentelateral figure 
 equal to six right angles, and so forth ; increasing the ag- 
 gregate by two right angles, for each additional side. — The 
 same conclusion is derived from the successive apj5lication 
 of triangles, by which the figure, at each accession, has 
 the number of its sides increased by one, and the amount 
 ©f its interior angles augmented by two right angles. 
 
dS 
 
 ELEMENTS Of GEOMETRY. 
 
 PROP. XXXII. THEOR. 
 
 The e^iterior angles of a rectilineal figure are 
 together equal to four right angles. 
 
 The exterior angles DEF, CDG, BCH, ABI, and 
 EAK of the rectilineal figure ABCDE are taken together 
 equal to four right angles. 
 
 For each exterior angle DEF, with its adjacent interior 
 one AED, is equal to two right 
 angles. All the exterior an- 
 gles, therefore, added to the 
 interior angles, are equal ta 
 twice as many right angles as 
 the figure has sides. Conse- 
 quently the exterior angles 
 are equal to the four right an- 
 gles which, by the Proposition 
 immediately preceding, were 
 abated, to form the aggregate of the interior angles. 
 
 Cor, If the figure has a re-entrant angle BCD, the an- 
 gle BCK which occurs in place 
 of an exterior angle, must be 
 subducted in forming the a- 
 mount ; for the corresponding- 
 interior angle BCD, in this 
 case, exceeds two right angles, 
 by the angle BCK. Hence 
 the angles EFG, DEH, CDI, 
 ABL, FAM, diminished by 
 BCK, are equal to four right angles. 
 
BOOK I* 3'J 
 
 Schol. The amount of the exterior angles might be de- 
 duced from the successive deflections which a side would 
 make before it has returned to its first position. Thus, in 
 the first case, AF makes a complete circuit, changing into 
 the positions EG, DH, CI, BK, and finally into AF again. 
 But in the second case, AG, after making similar deflec- 
 tionsy turns backwards at C from the position DK to GL. 
 
 PROP. XXXIII. THEOR. 
 
 If the opposite angles of a quadrilateral figure 
 he equal, its opposite sides will be likewise equal 
 and parallel. 
 
 In the quadrilateral figure ABCD, let the angle at B be 
 equal to the opposite one at D, and the angle at A equal 
 to that at C ; the sides AB and BC are equal and parallel 
 to DC and DA. 
 
 For all the angles of the figure being equal to four right 
 angles (1. 31. cor.), and the opposite angles being mutually 
 equal, each pair of adjacent angles ^ ^ . 
 
 must be equal to two right angles. 
 
 Wherefore ABC and BCD are 
 
 equal to two right angles, and the -A. 1> 
 
 lines AB and DC (I. 22. cor.) parallel; for the same rea- 
 son, ABC and BAD being together equal to two right an- 
 gles, the sides BC and AD, which limit them, are parallel. 
 But (I. 26.) the parallel sides of the figure are also equal. 
 
 Cor, Hence a quadrilateral figure contained by right 
 angles has its opposite sides equal and parallel, 
 t 
 
40 ELEMENTS Of GEOMETRV. 
 
 PROP. XXXIV. PROB. 
 
 To draw a perpendicular from the extremity of 
 a given straight line. 
 
 From the point B, to draw a perpendicular to AB, with- 
 out producing that line. 
 
 In AB take any point C, and on BC (I. 1. cor.) de- 
 scribe an isosceles triangle BDC, produce CD till DF be 
 equal to it; and BF being joined, is the perpendicular re- 
 quired. 
 
 For, since by construction DF is equal to CD or BD, 
 the triangle BDF is isosceles, and (I. 10.) the angle DBF 
 equal to DFB ; whence the angle CDB, , j., 
 
 being equal (I. 30.) to the interior angles y 
 
 DBF and DFB, is double of DBF, or y'\ 
 
 the angle DBF is half of CDB. But / 
 
 the triangle BDC being isosceles, the 
 angle CBD is equal to BCD ; consequently the angles DBF 
 and DBC are the halves of the vertical and base angles of 
 BDC, and therefore (I. 30.) the whole ang:le CBF is the 
 half of two right angles, or it is equal to one right angle. 
 
 SchoL This problem, of which the construction may be 
 slightly modified, is often more convenient in practice than 
 the one given in the corollary to Prop. 5. of this Book. 
 
 PROP. XXXV. PROB. 
 
 On a given finite straight line, to construct a 
 square. 
 
 Let AB be the side of the square which it is required to 
 construct. 
 
BOOK I. 
 
 41 
 
 From the extremity B draw, by the last proposition, 
 BC perpendicular to BA and equal 
 to it, and, from the points A and C 
 with the distance BA or BC describe 
 two circles intersecting each other in 
 the point D, join AD and CD ; the 
 quadrilateral figure ABCD is the 
 square required. 
 
 For, by this construction, the figure has all its sides e- 
 qual, and one of its angles ABC a right angle ; which com- 
 prehends the whole of the definition of a square^ 
 
 PROP. XXXVI. PROB. 
 
 To divide a given straight line into any number 
 of equal parts. 
 
 Let it be required to divide tho straight line AB into a 
 given number of equal parts, suppose five. 
 
 From the point A and at any 
 oblique angle with AB, draw a 
 straight line AC, in which take the 
 portion AD, and repeat it five times 
 from A to C, join CB, and from the 
 several points of section D, E, F, 
 and G draw -the parallels DH, EI, 
 FK, and GL, (I. 23.), cutting AB 
 in H, I, K, and L : AB is divided at these points into 
 five equal parts. 
 
 For (I. 23,) draw DM, EN, FO, and GP parallel to 
 AB. And because DH is parallel to EM, the exterior 
 
 ilLl-II Ki. B 
 
42 ELEMENTS OF GEOMETRY. 
 
 angle ADH is equal to DEM (I. 22.) ; and, for the same 
 reason, since AH is parallel to DM, the angle DAH is 
 equal to EDM. Wherefore the triangles ADH and 
 DEM, having two angles respectively equal and the inter- 
 jacent sides AD, DE — are (I. 20.) equal, and consequent- 
 ly AH is equal to DM. In the same manner, the tri- 
 angle ADH is proved to be equal to EFN, to FGO, and 
 GCP ; and therefore their bases, EN, FO, and GP are 
 all equal to AH. But these Hues are equal to HI, IK, 
 KL, and LB, for the opposite sides of parallelograms are 
 equal (I, 26.). Wherefore the several segments AH, HI, 
 IK, KL, and LB, into which the straight hne AB is di- 
 vided, are all equal to each other. 
 
 Scholium. The construction of this problem may be fa- 
 jcilitated in practice, by drawing from B in the opposite di- 
 rection a straight line parallel to AC, and repeating on 
 both of them portions equal to the assumed segment AD, 
 but only four times, or one fewer than the number of di- 
 visions required ; then joining D, the first section of AC, 
 with the last of its parallel, E with the next, and so on till 
 G, which connecting lines are (I. 27.) all parallel, and con- 
 sequently the former demonstration still holds. 
 
 \ 
 
ELEMENTS 
 
 GEOMETRY 
 
 BOOK IT. 
 DEFINITIONS. 
 
 1 . In a right-angled triangle, the side that subtends the 
 right angle is termed the hypotenuse; either of the sides 
 which contain it, the base ; and the other side, the perpeU" 
 dicular. 
 
 2. The altitude of a triangle is 
 a perpendicular let fall from the 
 vertex upon the base or its ex- 
 tension. 
 
 3. The altitude of a trapezoid is the 
 perpendicular drawn from one of its 
 parallel sides to the other. 
 
 4. The complements of rhomboids about the diagonal of 
 a rhomboid, are the spaces required to 
 complete the rhomboid ; and the defect 
 of each rhomboid from the whole figure, 
 is termed a gnomon. 
 
d* ELEMENTS OF GEOMETRY. 
 
 5. A rhomboid or rectangle is said to be contained hy 
 any two adjacent sides. 
 
 A rhomboid is often indicated merely by the two letters pla- 
 ced at opposite corners. 
 
 PROP. I. THEOR. 
 
 Triangles which have the same altitude, and 
 stand on the same base, are equivalent. 
 
 The triangles ABC and ADC which stand on the same 
 base AC and havewthe same altitude, contain equal spaces. 
 
 For join the vertices B, D by a straight line, which pro- 
 duce both ways ; and from A draw AE (I. 23.) parallel to 
 CB, and from C draw CF parallel to AD. 
 
 Because the triangles ABC, ADC have the same altitude, 
 the straight line EF is parallel to AC (I. 24.), and con- 
 sequently the figures CE and AF 
 are parallelograms. Wherefore 
 EB, being equal to AC (I. 26.) 
 which is equal to DF, is itself e- 
 qual to DF. Add BD to each, 
 and ED is equal to BF; but EA is equal to BC (I. 26.), 
 and the interior angle AED is equal to the exterior angle 
 CBF (I. 22.). Thus the two triangles EDA, BFC have 
 the sides ED, EA equal to BF, BC, and the contained an- 
 gle AED equal to CBF, and are therefore equal (l. 3.). 
 Take these equal triangles CBF and EDA from the whole 
 quadrilateral space AEFC, and there remains the rhom- 
 boid AEBC equivalent to ADFC. Whence the tri- 
 angles ABC and ADC, which are the halves of these rhom- 
 boids (I. 26. cor.), are likewise equivalent. 
 
 Cor. Hence rhomboids on the same base and between 
 the same parallels, are equivalent. 
 
BOOK II. 45 
 
 PROP. II. THEOR. 
 
 Triangles which have the same altitude, and 
 stand on equal bases, are equivalent. 
 
 The triangles ABC, DEF, standing on equal bases AC 
 and DF and having the same altitude, contain equal spaces. 
 For let the bases AC, DF be placed in the same 
 straight line, join BE, and produce it both ways, draw AG 
 and DH parallel to CB and FE (I. 23.), and join AH, 
 CE. 
 
 Because the triangles ABC, DEF are of equal altitude, 
 GE is parallel to AF (I. 24.), and GC, HF are parallelo- 
 grams. But AC, being equal to ^ 
 DF, and DF equal (I. 26.) to 
 HE, must also be equal to HE, 
 and therefore (I. 27.) AE is 
 a rhomboid or parallelogram. 
 Whence the rhomboid GC is equivalent to AE (II. 1. cor.), 
 and this again* is, for the same reason, equivalent to HF j 
 consequently GC is equivalent to HF, and therefore their 
 halves or (I. 26. cor.) the triangles ABC and DEF are 
 equivalent. 
 
 Cor, 1. Hence rhomboids on equal bases and between the 
 same parallels, are equivalent. 
 
 Cor, 2. Hence triangles which have the same vertex, 
 and equal bases in the extension of the same straight line, 
 are equivalent; and hence straight Jines drawn from the 
 vertex of a triangle to equal sections of [the base, will like- 
 wise divide it into equivalent triangles. 
 
46 fiLKMENTS OF OEOMETllY. 
 
 PROP. III. THEOIl. 
 
 Equivalent triangles on the same or equal bases> 
 have the same altitude. 
 
 If the triangles ABC and ADC, standing on the same 
 base AC, contain equal spaces, they have the same alti- 
 , tude, or the straight line which joins their vertices is pa- 
 rallel to AC. 
 
 For if BD be not parallel to AC, draw the parallel BE 
 hieeting AD or that side produced, in E, and join CE. 
 
 Because BE is made parallel to AC, the triangle ABC 
 is (11. 1.) equivalent to AEC ; but 
 ABC is by hypothesis equivalent to 
 ADC, and therefore AEC is equivalent 
 to ADC, which is absurd. The sup- 
 position then that BD is not parallel to "^ ^ 
 AC involves a contradiction. 
 
 The same mode of demonstration, it is obvious, will ap- 
 ply in the case where the equivalent triangles stand on e- 
 qual bases. 
 
 Co7\ Hence equivalent rhomboids on the same or equal 
 bases, have the same altitude. 
 
 PROP. IV. PR OB. 
 
 To find a triangle equivalent to any rectilineal 
 figure. 
 
 Let it be required to reduce the five-sided figure ABCDE 
 to a triangle, or to find a triangle that shall contain ^a e- 
 qual space* ^ 
 
BOOK II. ^T 
 
 Join any two alternate points A, C, and through the 
 intermediate point B, draw BF parallel to AC, meeting 
 either of the adjoining sides AE or CD in F ; which point, 
 when the angle ABC is re-entrant, will lie within the figure: 
 Join CF. Again, join the alter- 
 nate points C, E, and through 
 the intermediate point D draw the 
 parallel DG, to meet in G either 
 of the adjoining sides AE or BC, 
 which, since the angle CDE is 
 salient, must for that eifect be ^ ^' 
 produced ; and join CG. The triangle FCG is equivalent 
 to the five-sided figure ABCDE. 
 
 Because the triangles CFA and CBA have by construc- 
 tion the same altitude and stand on the same base AC| 
 they are (II. 1.) equivalent; take each of them away from 
 the space ACDE, and there remains the quadrilateral fi- 
 gure FCDE equivalent to the five-sided figure ABCDE. 
 Again, because the triangles CDE and CGE are equal, 
 having the same altitude and the same base ; add the tri- 
 angle FCE to each, and the triangle FCG is equivalent 
 to the quadrilateral figure FCDE, and is consequently 
 equivalent to the original figure ABCDE. 
 
 In this manner, any polygon may, by successive steps, 
 be reduced to a triangle ; for an exterior triangle is always 
 exchanged for another equivalent one, which, attaching 
 itself to either of the adjoining sides, coalesces with the 
 rest of the figure. 
 
 Schol. This problem is of singular use in practice, since 
 it enables the surveyor greatly to abridge his computations, 
 by reducing any plan that he has delineated at once to an 
 equivalent triangle. 
 
 t 
 
48 i^EMENTS OF GEOMETRY. 
 
 PROP. V. PROB. 
 
 A triangle is equivalent to a rhomboid which 
 has the same altitude and stands on half the base. 
 
 The triangle ABC is equivalent to the rhomboid DEFC, 
 which stands on half the base DC, but has the same alti- 
 tude. 
 
 For join BD and EC. The triangles ABD and DBC 
 having the same vertex and equal 
 bases, are (11. 2. cor. 2.) equivalent. 
 But the diagonal EC bisects the 
 rhomboid DEFC (I. 26. cor.), and 
 the triangles DBC and DEC, having 
 the same altitude, are equivalent (II. 1.^; consequently 
 their doubles, or the triangle ABC and the rhomboid 
 DEFC, are equivalent. • 
 
 Cor. Hence the area of a triangle is equal to half the 
 rectangle contained under its base and its altitude— from 
 which property is derived the mensuration of any rectili- 
 neal figure. 
 
 PROP. VI. PROB. 
 
 To construct a rhomboid equivalent to a given 
 rectilineal figure, and having its angle equal to a 
 given angle. 
 
 Let it be required to construct a rhomboid which shall 
 be equivalent to a given rectilineal figure, and contain an 
 angle equal to G. 
 
 Reduce the rectilineal figure to an equivalent triangle 
 
BOOK II. 49 
 
 ABC (II. 4.), bisect the base AC in the point D (I. 7.), and 
 draw DE making an angle CDE equal to the given angle 
 G (I. 4.), through B draw BF parallel to AC (I. 23.), 
 and through C the straight line 
 CF parallel to DE : DEFC is the 
 rhomboid that was required. 
 
 For the figure DF is by con- 
 struction a rhomboid, contains 
 
 an angle CDE equal to G, and is equivalent to the trian- 
 gle ABC (II. 5. J, and consequently to the given rectili- 
 neal figure. 
 
 PROP. VII. THEOR. 
 
 ^he complements of the rhomboids about the 
 diagonal of a rhomboid, are equivalent. 
 
 Let EI and HG be rhomboids about the diagonal of the 
 rhomboid BD 5 their Complements BF and FD contain 
 equal spaces. 
 
 Since the diagonal AF bisects the rhomboid EI (I. 26.. 
 cor.), the triangle AEF is equivalent to AIF; and for the 
 i&ame reason, the triangle FHC is equi- -^ TC c 
 
 valent to FGC. From the whole tri- ^/T 
 angle ABC on the one side of the dia- 
 gonal, take away the two triangles "^ I J3 
 AEF and FHC ; and from the triangle ADC, which is 
 equal to it, take away, on the other side, the two triangles 
 AlF and FGC, and there remains the rhomboid BF equi- 
 valent to FD. 
 
 Cor, The same property will extend to the spaces left on 
 
 both si^es of the diagonal by rhomboids any how combined. 
 
50 » ELEMENTS OF GEOMETRY. 
 
 PROP. VIII. PROB. 
 
 With a given straight line to construct a rhom- 
 boid equivalent to a given rectilineal figure, and 
 having an angle equal to a given angle. 
 
 Let it be required to construct, with the straight line L, a 
 rhomboid, containing a given space, and having an angle 
 equal to K. 
 
 Construct (II. 6.) the rhomboid BF equivalent to the 
 given rectilineal figure, and having an angle BEF equal to 
 K ; produce EF until FG be e- 
 qual to L, through G draw DGC 
 parallel to EB and meeting the 
 extension of BH in C, join GF 
 and produce it to meet the exten- 
 sion of BE in A ; draw AD parallel to EF, meeting CG 
 in D, and produce HF to I : FD is the rhomboid re- 
 quired. 
 
 For FD and FB are evidently complementary rhom- 
 boids, and therefore (II. 7.) equivalent; and, by reason of 
 the parallels AE, IF, the angle FID is equal to EAI(1. 22.), 
 which again is equal to BEF or the given angle K. 
 
 ScJiol. This problem might also be solved by repeated 
 operations ; each triangle, into which the rectilineal figure 
 is divided, being successively converted into a rhomboid, 
 having an angle equal to K, and placed on a line equal to 
 L, or the summit of each preceding rhomboid. These 
 rhomboids will evidently coalesce and fulfil the conditions 
 required. The process is not so direct as when the figure 
 was previously reduced to an equivalent triangle ; but it 
 seems better adapted for the solution of another similar 
 
BOOK 11. 51 
 
 J)rq{)lem— To constitute under the same conditions a rhom- 
 boid equivalent to the difference between given figures. The 
 smaller rhomboid is here placed below the summit of the 
 other, leaving the defect standing on the original base. 
 
 PROP. IX. THEOfe. 
 
 A trapezoid is equivalent to the rectangle con- 
 tained by its altitude and half the sum of its pa- 
 rallel sides. 
 
 The trapezoid ABCD is equivalent to the rectangle con- 
 tained by its altitude and half the sum of the parallel sides 
 BC and AD. 
 
 For draw CE parallel to AB (I. 23.), bisect ED (I. 7.) 
 in Fj and draw FG parallel to AB, meeting the production 
 of BC in G. 
 
 Because BC is equal to AE (I. 26.), BC and AD are 
 together equal to AE and AD, or to twice AE with EDj 
 or to twice AE and twice EF, that is, to twice AF j con-* 
 sequently AF is half the sum of BC 
 
 K C Gr- . 
 
 and AD. Wherefore the rectangle A" "Vv / 
 
 contained by the altitude of the /^ / /X 
 
 trapezoid and half {he sum of its pa- 
 rallel sides, is equivalent to the rhomboid BF (II. 1. cor.),» 
 but the rhomboid EG is equivalent to th6 triangle ECD ^ 
 (II. 5.), add to each the rhomboid BE, and the rhomboid 
 BF is equivalent to the trapezoid ABCD. 
 
 ^hol. Hence the area is found of any rectilineal figure 
 referred to a given base j for it is equal to that of the aggre- 
 gate rectangles under the mean of each pair of perpendi- 
 culars and the interjacent portion of the base This pro- 
 position is of great use in surveying, since it abridges the 
 
52 
 
 ELEMENTS OF GEOMETRY. 
 
 mensuration of the irregular borders of a field, by help of 
 what are called offsets, or perpendiculars branching from, 
 the great line to each remarkable flexure of the extreme 
 boundary. 
 
 PROP. X. THEOR. 
 
 The square described on the hypotenuse of a 
 right-angled triangle, is equivalent to the squares 
 of the two sides. 
 
 Let the triangle ABC be right-angled at B ; the square 
 described on the hypotenuse AC is equivalent to'BF and 
 BI the squares of the sides AB and BC. 
 
 For produce DA to K, and through B draw MBL pa- 
 rallel to DA (I. 23.) and meeting FG produced in L. 
 
 Because the angle CAK, adjacent to CAD, is a right 
 angle, it is equal to BAFr'^from each of these take away 
 the anglfe BAK, and there 
 remains the angle BAC 
 equal to FAK. But the 
 angle ABC is equal to 
 AFK, both of them being 
 right angles. Wherefore 
 the triangles ABC and 
 AFK, thus having two an- 
 gles of the one respectively 
 equal to those of the other, 
 and the interjacent sideAF 
 equal to AB, are equal 
 (1. 20.), and consequently 
 
 the side AC is equal to AK. Hence the rectangle or 
 rhomboid AM is equivalent to ABLK (II. 2. cor.), since 
 they stand on equal bases AD and AK, and between the 
 
BOOK ir, 53 
 
 same parallels DK and ML. But ABLK is (11. 1. cor.) 
 equivalent to the rhomboid or square BF, for it stands on 
 the same base AB and between the same parallels FL and 
 AH. Wherefore the rectangle AM is equivelent to the 
 square of AB. 
 
 And in like manner, by drawing MB to meet the pro- 
 duction of HI, it may be proved, that the rectangle CM 
 is equivalent to the square of BC. Consequently the whole 
 square, ADEC, of the hypotenuse, contains the same space 
 as both together of the squares described on the two sides 
 ABandBG. 
 
 Cor. Hence the square of a side AB is equivalent to the 
 rectangle under the hypotenuse AC and the adjacent seg- 
 ment AN made by ,^ perpendicular. 
 
 Schol. This proposition is deservedly the most celebrated 
 of the whole Elements, and serves as the main link for con- 
 necting Geometry with the modern Algebra. — The de- 
 monstration may be variously modified ; but one of the 
 simplest forms is that in which CAKO is proved to be a 
 square, and the rectangle NK equivalent to the rhomboid 
 AL and to the square BF on the one side, while the remain- 
 ing rectangle NO is equivalent to the rhomboid CL and to 
 the square BI on the other. 
 
 PROP. XI. THEOR. 
 
 If the square of one side of a triangle be equiva- 
 lent to the squares of both the other sides, that 
 side subtends a right angle. 
 
 Let the square described on AC be equivalent to the 
 two squares of AB and BC ; the triangle ABC is right- 
 angled at B. 
 
^4i ELEMENTS OF GEOMETRY. 
 
 For draw BD perpendicular to AB (I. S^.) and equal 
 to BC, and join AD. 
 
 Because BC is equal to BD, the square of BC is equal 
 to the square of BD, and consequently the squares of AB. 
 and BC are equal to the squares of AB and BD. But the 
 squares of AB and BC are, by hypothesis, 
 equivalent to the square of AC ; and sincie 
 ABD is, by construction, a right angle, the 
 squares of AB and BD are, by the preced- 
 ing proposition, equivalent to the square of, 
 AD. Whence the square of AC is equi- 
 valent to that of AD, and the straight line AC equal to 
 AD. The two triangles ACB and ADB, having all the 
 sides in the one respectively equal to those in the other, 
 are therefore equal (I. 2.), and consequently the angle 
 ABC is equal to the corresponding angle.ABD, that is, to 
 a right angle. 
 
 Cor. Hence the numbers 3, 4, and 5 will express the 
 sides and hypotenuse of a right-angled triangle— a proper- 
 ty which readily suggests another method of erecting a per- 
 pendicular at the extremity of a straight line. 
 
 PROP. XII. PROB. 
 
 To find the side of a square equivalent to any 
 number of giyen squares. 
 
 Let A, B, and C be the sides of the squares, to which 
 it is required to find an equivalent square. 
 
 Draw DE equal to A, and from its extremity E erect 
 {I. 34.) the perpendicular EF equal to B, join DF, and 
 again, perpendicular to this, draw FG equal to C, and join 
 PG : DG is the side of the square which was required. 
 
BOOK II. 55 
 
 For, since DEF is a right-angled triangle, 
 the square of DF is equivalent to the squares 
 of DE and EF (II. 10.), or of the lines A and 
 B. Add on both sides the square of FG or 
 of C, and the squares of DF and FG, which 
 are equivalent to the square of DG (II. 10.), 
 are equivalent to the aggregate squares of ^ 
 
 A, B, and C. And, by thus repeating the 
 process, it may be extended to any number of squares. 
 
 PROP. XIII. PROB. 
 
 To find the side of a square equivalent to the 
 difference between two given squares. 
 
 Let A and B be the sides of two squares ; it is required 
 to find a square equivalent to their difference. 
 
 Draw CD equal to the smaller line B, from its extre- 
 mity erect (I. 34-.) the indefinite per- 
 pendicular DE, and about the cen- 
 tre C, with a distance equal to thp 
 greater line A, describe a circle cut- 
 ting DE in F: DF is the side of 
 the square required. 
 
 For join CF. The triangle CDF 
 being right-angled, the square of 
 its hypotenuse CF is equivalent to 
 the squares of CD and DF (II. 10.), and consequently ta- 
 king the square of CD from both, the excess of the square 
 of CF above that of CD is equivalent to the square of DF, 
 or the square of DF is equivalent to the excess of the square 
 of A above that of B. 
 
B& (ELEMENTS OP GEOMETRY. 
 
 PROP. Xiy. THEOR. 
 
 The rectangle contained by two straight lines, 
 is equivalent to the rectangles contained under 
 one of them and the several segments into which 
 the other is divided. 
 
 The rectangle under AC and AB, is equivalent to the 
 rectangles contained by AC and the segments AD, DE, 
 and EB. 
 
 For, through the points D and E, draw DF and EG 
 parallel and equal to AC (I. 23.). 
 
 The figures AF, DG, and EH are evidently rhomboid- 
 al ; they are also rectangular, for the 
 
 angles ADF, AEG, and ABH are A p f b 
 
 each equal to the opposite angle ACF 
 
 (I. 26.). And the opposite sides DF, 
 
 EG, and BH, being equal to AC,— the 
 
 spaces into which the rectangle BC is resolved, are equal 
 
 to the rectangles contained respectively by AC and AD, 
 
 DE and EB. 
 
 PROP. XV. THEOR. 
 
 The square described on the sum of two 
 straight lines, is equivalent to the squares of those 
 lines, together with twice their rectangle. 
 
 If AB and BC be two straight lines placed continuous ; 
 the square described on their sum AC, is equivalent to the 
 two squares of AB, BC, and twice the rectangle contained 
 by them. 
 
 ir 
 
BOOK II. 
 
 51 
 
 
 
 
 G 
 
 For through B draw BI (1. 23.) parallel to AD, make 
 AF equal to AB, and through F draw FH parallel to 
 DE. 
 
 It is manifest that the spaces AG, GE, DG and CG, 
 into which the square of AC is divid- 
 ed, are all rhomboidal and rectangu- 
 lar. And because AB is equal to j^j j__ — |j|; 
 
 AF, and the opposite sides equal, 
 the figure AG is equilateral, and ha- 
 ving a right angle at A, is hence a 
 square. Again, AD being equal to 
 AC, take away the equals AF and AB, and there remains 
 DF equal to BC, and consequently IG equal to GH 
 (I. 26.) : wherefore IH is likewise a square. The rect-" 
 angle DG is contained by the sides FG and DF, which 
 are equal to AB and BC ; and the rectangle CG is con- 
 tained by the sides GB and GH, which are likewise equal 
 to AB and BC. Consequently the whole square of AC is 
 composed of the two squares of AB and BC, together 
 with twice the rectangle contained by these lines. 
 
 PROP. XVI. THEOR. 
 
 The square described on the difference of two 
 straight lines, is equivalent to the squares of those 
 lines, diminished by twice their rectangle. 
 
 Let AC be the difference of two straight lines AB and 
 BC ; the square of AC is equivalent to the excess of the 
 two squares of AB and BC above twice their rectangle. 
 
 For let the squares of AB, BC and AC be completed, 
 and produce CE and DE the sides of the latter to H and I. 
 
58 
 
 ELEMENTS OF GEOMETRY. 
 
 It is evident, that GE is equal to BL or the square of BC ; 
 to each add the intermediate rect- 
 angle EB, and GC isequal to IL ; 
 but the rectangle under AB and 
 BC is equal to the rectangle IL, 
 which is also equal to DG. From 
 the compound surface CAFGBKL, 
 or the squares of AB and BC, take 
 away the space DFGBKLC, or 
 the rectangles IL and DG, that is, 
 twice the rectangle under AB and BC, — and there remains 
 ADEC, or the square of the difference AC of the two lines 
 AB and BC. 
 
 
 i^" 7-r 
 
 n- 
 
 33 
 
 
 
 
 
 K 
 
 c 
 
 -A- C 
 
 
 B 
 
 K 
 
 PROP. XVIL THEOR. 
 
 The rectangle contained by the sum and diffe- 
 rence of two straight lines, is equivalent to the 
 difference of their squares. 
 
 Let AB and BD be two continuous straight lines, of 
 which AD is the sum and AC the difference ; the rectan- 
 gle under AD and AC, is equivalent to the excess of the 
 square of AB above that of BC. 
 
 For, having made AG equal to AC, draw GH parallel 
 to AD (I. 23.)> and CI, DH parallel to AE. 
 
 Because GK is equal to KC or HD, and EG is equal 
 to CB or BD, the rectangle EK is 
 equal to_ LD (IL 2. cor.); and 
 consequently, adding the rectangle 
 BG to each, the space AEIKLB 
 is equivalent to the rectangle AH. 
 But this space AEIKLB is the ex- 
 cess of the square of AB above IL 
 
 :< 
 
 Z J -F 
 
 G 
 
 
 
 
 
 K 
 
 y. 
 
 II 
 
 A 
 
 JB D 
 
BOOK II. 
 
 or the square of BC ; and the rectangle AH is contained by 
 AD and DH or AC. Wherefore the rectangle under 
 AD and AC is equivalent to the difference of the squares 
 of ABandBC. 
 
 Cor. 1. Hence if a straight line AB be bisected in C 
 and cut unequally in D, the rectangle under the unequal 
 segments AD, DB, together with the square of CD, the 
 interval between the points of section, is equivalent to 
 the square of AC, the half line. For 
 AD is the sum of AC, CD, and DB is > <;^ ir b 
 evidently their difference ; whence, by 
 the Proposition, the rectangle AD, DB is equivalent to 
 the excess^of the square of AC above that of CD, and 
 consequently the rectangle AD, DB, with the square of 
 CD, is equal to the square of AC. 
 
 Cor. 2. If a straight line AB be bisected in C and pro- 
 duced to D, the rectangle contained by AD the whole line 
 thus produced, and the produced part DB, together with 
 the square of the half line AC, is equivalent to the square 
 of CD, which is made up of the half line and the produced 
 part. For AD is the sum of AC, CD, 
 and DB is their difference; whence - ^ ^ ? ? 
 the rectangle AD, DB is equivalent to 
 the excess of the square of CD above AC, or the rectangle 
 AD, DB, with the square of AC, is equivalent to the 
 square of CD. 
 
 Scholium. If we consider the distances DA, DB of the 
 point D from the extremities of AB as segments of this 
 line, whether formed by internal or external section ; both 
 corollaries may be comprehended under the same enun- 
 ciation. That, if a straight line be divided equally and 
 unequally, the rectangle contained by the unequal segments 
 is equivalent to the difference of the squares of the hklf 
 line and of the interval between the points of section. 
 
60 ELEMENTS OF GEOMETRY. 
 
 PROP. XVIII. THEOR. 
 
 The sum of the squares of two straight lines, is 
 equivalent to twice the squares of half their sum 
 and of half their difference. 
 
 Let AB, BC be two continuous straight lines, D the 
 middle point bf AC, and consequently AD half the sum of 
 these lines and DB half their difference; the squares of 
 AB and BC are together equivalent to twice the square of 
 AD, with twice the square of DB. 
 
 For erect (I. 5. cor.) the perpendicular DE equal to AD 
 or DC, join AE and EC, through B and F draw (I. 23.) 
 BF and FG parallel to DE and AC, and join AF. 
 
 Because AD is equal to DE, the angle DAE (I. 10.) is 
 equal to DEA, and since (I. 30. cor.) they make up toge- 
 ther one right angle, each of them must be half a right an- 
 gle. In the same manner, the angles DEC and DCE of 
 the triangle EDC are proved to be each half a right an- 
 gle; consequently the angle A EC, composed of AED and 
 CED, is equal to a whole right angle. And in the trian- 
 gle FBC, the angle CBF being equal 
 to CDE (I. 22.) which is a right 
 angle, and the angle BCF being half 
 a right angle — the remaining an- 
 gle BFC is also half a right angle 
 (I. 30.), and therefore equal to the angle BCF ; whence 
 (I. 11.) the side BF is equal to BC. By the same 
 reasoning, it may be shown, that the right angled triangle 
 GEF is likewise isosceles. The square of the hypotenuse 
 
^BOOK II. 61 
 
 EF, which is equivalent to the squares of EG and GF 
 (II. 10.), is therefore equivalent to twice the square of GF 
 or of DB ; and the square of AF|, in the right angled tri- 
 angle ADE, is equivalent to the squares of AD arid DE, 
 or twice the square of AD. But since ABF is a right an- 
 gle, the square of AF is equivalent to the squares of AB 
 and BF, or BC; and because AEF is also a right an- 
 gle, the square of the same line AF is equivalent to the 
 squares of AE and EF, that is, to twice the squares of AD 
 and DB. Wherefore the squares of AB, BC are toge- 
 ther equivalent to twice the squares of AD and DB. 
 
 Cor. Hence if a straight line AB be bisected in C and 
 cut unequally in D, whether by in- 
 terjial or external section, the squares A C p "B 
 of the unequal segments AD and DB ^ 9 ,^ P 
 are together equivalent to twice the 
 square of the half line AC, and twice the square of CD 
 the interval between the points of division. 
 
 PROP. XIX. PROB. 
 
 To cut a given straight line, such that the 
 square of one part shall be equivalent to the rect- 
 angle contained by the whole line and the re- 
 maining part. 
 
 Let AB be the straight line which it is required to di- 
 vide into two segments, such that the square of the one 
 shall be equivalent to the rectangle contained by the whole 
 line and the other. 
 
62 ELEMENTS OF GEOMETKIf. 
 
 Produce AB till BC be equal to 
 it, erect (I. 5. cor.) the perpendicu- 
 lar BD equal to AB or BC, bisect 
 BC in E (I. 7.), join ED and make 
 EF equal to it ; the square of the 
 segment BF is equivalent to the rectangle contained by the 
 whole line BA and its remaining segment AF. 
 
 For on BC construct the square BG (I. 35.), make BH 
 equal to BF, and draw IHK and FI parallel to AC and 
 BD (I. 23.). Since AB is equal to BD, and BF to BH j 
 the remainder AF is equal to HD : and it is farther evi- 
 dent, that FH is a square, and that IC and DK are rect- 
 angles. But BC being bisected in E and produced to F, 
 the rectangle under CF, FB, or the rectangle IC, toge- 
 ther with the square of BE, is equivalent to the square of 
 EF or of DE (II. 17. cor. 2.). But the square of DE is 
 equivalent to the squares of DB and BE (II. 10.) ; whence 
 the rectangle IC, with the square of BE, is equivalent to 
 the squares of DB and_BE; or, omitting the common 
 square of BE, the rectangle IC is equivalent to the square 
 of DB. Take away from both the rectangle BK, and there 
 remains the square BI, or the square of BF, equivalent to 
 the rectangle HG, or the rectangle contained by BA and 
 AF. 
 
 Cor. 1. Since the rectangle under CF and FB is equi- 
 valent to the square of BC, it is evident that the line CF 
 is likewise divided at B in a manner similar to the original 
 line AB. But this line CF is made up, by joining the 
 whoje line AB, now become only the larger portion, to its 
 greater segment BF, which next forms the smaller portion 
 in the new compound. Hence this division of a line being 
 once obtained, a series of other lines all possessing the same 
 property may readily be found, by repeated additions. 
 
BOOK II. 63 
 
 Thus, let AB be so cut, that the square of BC is equiva- 
 lent to the rectangle BA, AC: Make successively, BD 
 equal to BA, DE equal to DC, EF equal to EB, and FG 
 
 AC B T) E. ' ir a 
 
 I — t i 1 i i 1 
 
 equal to FD ; the lines CD, BE, DF, and EG, beginning 
 in succession at the points C, B, D, and E, are divided 
 at the points B, D, E, and F, such that, in each of them, 
 the square of the larger part is equivalent to the rectangle 
 contained by the whole and the smaller part. — It is obvious, 
 that this procedure might likewise be reversed. If FD, 
 EB, and DC be made successively equal to FG, EF and 
 DE, the lines DF, BE, and CD will be divided in the same 
 manner at the points E, D and B. 
 
 Cor. 2. Hence also the construction of anotlier problem 
 of the same nature ; in which it is required to produce a 
 straight line AB, such that the rectangle contained by the 
 whole line thus produced and the part produced, shall be 
 equivalent to the square of the line AB itself. Divide AB, 
 by this proposition, in C, so that the 
 rectangle BA, AC is equivalent to the - ^ ^ ^ ^ 
 square of BC, and produce AB until 
 BD be equal to BC : Then, from what has been demon- 
 strated, it follows that the rectangle under AD and DB is 
 equivalent to^^Vje square of the whole line AB. 
 
 It will be convenient y for the sake of conciseness, to desig- 
 nate in future this remarJcahle. division of a line, where the 
 rectangle under the whole and one part is equivalent to the. 
 square of the other, hi/ the term Medial Section. 
 
64- ELEMENTS OF GEOMETRY. 
 
 PROP. XX. THEOR. 
 
 The square of the side of an isosceles triangle is 
 greater or less than the square of a straight line 
 drawn from the Vertex to the base or its exten- 
 sion, by the rectangle contained under its inter- 
 nal or external segments. 
 
 1. If BD be drawn from the vertex of the isosceles tri- 
 angle ABC to a point D in the base ; the square of AB 
 exceeds the square of BD, by the rectangle under the 
 segments AD, DC. 
 
 t'or (I. 7.) bisect the base AC in E, and join BE. Be- 
 cause the triangles ABE and CBE have ^ 
 the sides AB, AE equal to BC, CE, and 
 the side BE common, they are equal 
 (1. 2.), and consequently the correspond- 
 ing angles BEA, BEC are equal, and 
 each of them (Def. 4.) a right angle. 
 Wherefore the square of AB is equi- ^ 
 valent to the squares of AE and BE (II. 10.); and since 
 AC is cut equally in E and unequally in D, the square 
 of AE is equivalent to the square of DE, together with the 
 rectangles AD, DC (II. 17. cor. 1.) ; and consequently the 
 square of AB is equivalent to the squares of BE and DE, 
 together with the rectangle AD, DC. Bi;^the square of 
 BD is equivalent to the squares of BE and DE (II. 10.); 
 whence the square of AB is equivalent to the square of 
 BD, together with the rectangle AD, DC. 
 
 2. But the square of the straight line BD drawn from 
 the vertex to any point in the base produced, is greatefr 
 
BOOK It. 
 
 65 
 
 than the square of AB by the rectangle contained under 
 AD and DC, the external seg- 
 ments of the base. 
 
 For draw BE, as before, to bi- 
 sect the base AC. The square of 
 DE is equivalent to the square of 
 AE, together with the rectangle 
 AD, DC, (IL 17. cor. 2.); to 
 each of these, add the square of 
 BE, and the squares of DE and BE, — that is, the square 
 of BD (II. 10.) — are equal to the squares of AE and BE, 
 or the square of BA, together with the rectangle AD, DC. 
 
 PROP. XXI. THEOR. 
 
 The difference between the squares of the 
 sides of a triangle, is equivalent to twice the 
 rectangle contained by the base and the distance 
 of its middle point from the perpendicular. 
 
 Let the side AB of the triangle ABC be greater than 
 BC ; and, having let fall the perpendicular BE, and bisect- 
 ed AC in D, the excess of the square of AB above that 
 of BC is equivalent to twice the rectangle contained by AC 
 and DE. 
 
 For the square of AB is equivalent to the squares of AE 
 and BE (II. 10.), and the square of BC is equivalent to 
 the squares CE and BE ; wherefore the excess of the square 
 of AB above that of BC is equivalent 
 to the excess of the square of AE above . 
 that of CE. But the excess of the 
 square of AE above that of CE, is (IL 
 17.) equivalent to the rectangle con- 
 tained by theiy sum AC and their difference, which is evi- 
 
66 ELEMENTS OF GEOMETRY. 
 
 dently the double of DE ; and consequently the difference 
 between the squares of AE and CE, being equivalent to 
 the rectangle contained by AC and the double of DE, is 
 equivalent to twice the rectangle under AC and DE. 
 ' Cor, The difference between the squares of the sides of 
 a triangle, is equivalent to the difference between the 
 squares of the segments of the base made by a perpendicu- 
 lar ; — a property likewise easily derived from the preceding 
 proposition. 
 
 PROP. XXII. THEOR. 
 
 In any triangle, the sum of the squares of the 
 sides, is equivalent to twice the square of half the 
 base and twice the square of the straight line 
 which joins the point of its bisection with the ver- 
 tex. 
 
 Let BD be drawn from the vertex B of the triangle ABC 
 to bisect the base; the squares of the sides AB and BC are 
 together equivalent to twice the squares of AD and DB. 
 
 For let fall the perpendicular BE (I. 6.) ; and if the 
 point D coincide with E, the triangle ABC 
 being evidently isosceles, the squares of AB 
 and BC are the same with twice the square 
 of AB, or twice the squares of AE and EB, 
 or of AD and DB (II. 10.) -^ dk c 
 
 But if the perpendicular fall upon C, the triangle is right 
 angled, and the squares of AB and BC 
 are then equivalent to the square of AC, 
 and twice the square of BC, or to twice 
 the squares of AD, DC and BC ; but (II. 
 10.) twice the squares of DC and BC are 
 
BOOK II. 67 
 
 equivalent to twice the square of DB, and consequently the 
 squares of AB and BC are equivalent to twice the squares 
 of AD and DB. 
 
 In every other case, whether the perpendicular BE fall 
 within or without the base AC, the 
 squares of AE, EC, the unequal seg- 
 ments of AC, are (II. 19. cor.) equiva- 
 lent to twice the square of AD and twice 
 the square of DE ; add twice the square -^ X) E c 
 of EB to both, and the squares of AE, 
 EB and of CE, EB — or the squares of 
 the hypotenuses AB, BC — are equiva- 
 
 H C T2 
 
 lent to twice the square of AD, and 
 
 twice the squares of DE, EB, that is, (II. 10.) to twice the 
 
 square of DB. 
 
 PROP. XXIII. THEOR. 
 
 The square of the side of a triangle is greater 
 or less than the squares of the base and the other 
 side, according as the opposite angle is obtuse or 
 acute,- — by twice the rectangle contained by the 
 base and the distance intercepted between the ver- 
 tex of that angle and the perpendicular. 
 
 In the oblique-angled triangle ABC, where the perpen- 
 dicular BD falls without the base j the square of the side 
 AB which subtends the oblique angle exceeds the squares 
 of the sides AC and BC which contain it, by twice the 
 rectangle under AC and CD. 
 
 For die square of AD, or of the sum of AC and CD, is 
 (II. 15.) equivalent to the squares of these lines AC, CD, 
 
68 ELEMENTS OF GEOMETRY. 
 
 together with twice their rectangle. Add the square of 
 DB to each side, and the squares of AD, -^ 
 
 DB, or (II. 10.) the square of AB is equi- 
 valent to the square of AC^ and the 
 
 squares of CD, DB, together with twice - — ^ ^ 
 
 the rectangle AC, CD ; but the squares 
 of CD, DB are (II. 10.) equivalent to the square of CB ; 
 whence. the square of AB exceeds the squares of AC, BC, 
 by twice the rectangle under AC and CD. 
 
 Again, in the acute-angled triangle ABC, where the 
 perpendicular BD falls within the triangle; 
 the square of the side AB that subtends the U 
 
 acute angle, is less than the squares of the 
 containing sides AC, BC, by twice the rect- 
 
 angle under the base AC and its intercept- -A. u c; 
 ed portion CD. 
 
 For the square of AD, or of the difference between AC 
 and CD, is (II. 16.) equivalent to the squares of AC and 
 CD, diminished by twice their rectangle. Add to each the 
 square of DB, and the squares of AD and DB — or the 
 square of AB— -are equivalent to the square of AC, with 
 the squares of CD and DB, or the square of BC, diminish- 
 ed by twice the rectangle under AC and CD. Conse- 
 quently the square of AB is less than the squares of AC 
 and BC, by twice the rectangle under AC and CD» 
 
 Cor. If the triangle ABC be isosceles, having equal sides 
 AC and BC, the square of the base AB is equivalent to 
 twice the rectangle under the side AC, and the adjacent 
 segment AD made by the perpendicular BD, whether the 
 vertical angle be obtuse or acute. For the square of AB 
 is equivalent to the sqnai'es of AC and BC, or twice the 
 square of AC increased or diminished by twice the rect- 
 
BOOK II. 69 
 
 angle under AC and CD ; that is, equivalent to twice the 
 rectangle under AC and AD, the sum or difference of AC 
 and CD — This might also be demonstrated from the co- 
 rollary to Prop. 10. 
 
 ScJioL When the three sides of a triangle are given, the 
 segments of the base made by a perpendicular may be found 
 either by Prop. 21. or Prop. 23., and thence the perpen- 
 dicular can easily be determined from the application of 
 Prop. 10. But half the rectangle under this perpendicu- 
 lar and the base will, by corollary to Prop. 5., express the 
 area of the triangle. 
 
 PROP. XXIV. THEOR. 
 
 The squares of the sides of a rhomboid, are to- 
 gether equivalent to the squares of its diagonals. 
 
 Let ABCD be a rhomboid : The squares of all the sides 
 AB, BC, CD, and AD, are together equivalent to the 
 squares of the diagonals AC, BD. 
 
 For the angles BCE and CBE are equal to the alternate 
 angles DAE and x\DE, and the 
 interjacent sides BC and AD are 
 equal ; wherefore (I. 20.) the tri- 
 angles BEC and DEA are equal. 
 Consequently CE being equal to 
 EA,the squares of AB,?BC, are (II. 22.) equivalent to twice 
 the square of AE and twice the square of BE ; whence 
 twice the squares of AB, BC, or the squares of all the sides 
 of the rhomboid, are equivalent to four times the square of 
 AE and four times the square of BE, that is, to the squares 
 of AC and BD. 
 
ELEMENTS 
 
 OP 
 
 GEOMETRY. 
 
 BOOK III. 
 DEFINITIONS. 
 
 1. Any portion of the cir- 
 cumference of a circle is called 
 an «rc, and the straight line 
 which joins the two extremi- 
 ties, a chord. 
 
 2. The space included between an arc and its chord, is 
 named a segment. 
 
 S. A sector is the portion of a circle con- 
 tained by two radii and the arc lying be- 
 tween them. 
 
 4. The tangent to a circle is a straight 
 line which touches the circumference, or 
 meets it only in a single point. 
 
75 
 
 ELEMENTS OF GEOMETRY. 
 
 5. Circles are said to touch 
 mutually, if they meet, but do 
 not cut each other. 
 
 6. The point where a straight line touches^ a circle, or 
 one circle touches another, is called the point of contact. 
 
 v. A straight line is said to be inflected 
 from a point, when it terminates in another 
 straight line, or a( the circumference of a 
 circle. 
 
BOOK IH. 73 
 
 PROP. I. THEOR. 
 A circle is bisected by its diameter. 
 
 The circle ADBE is divided into two equal portions, 
 by the diameter AB. 
 
 For let the portion ADB be reversed and applied to 
 AEB, the straight line AB and its 
 middle po^nt, or the centre C, re- 
 maining the same. And since the 
 radii of the circle are all equal, or 
 the distance of C from any point in 
 the boundary ADB is equal to its 
 distance from any point of the op- 
 posite boundary AEB, every point 
 
 D of the former must meet with a corresponding point of 
 the latter, and consequently the two portions ADB and 
 AEB will entirely coincide. 
 
 Cor, The portion ADB limited by a diameter, is thus a 
 semicircle, and the arc ADB la a semicircwmference, 
 
 PROP. II. THEOR. 
 
 A straight line cuts the circumference of a cir- 
 cle only in two points. 
 
 If the straight line AB cut the 
 circumference of a circle in D, it 
 can only meet it again in another 
 point E. 
 
 JFor join D and the centre C; 
 and because from the point C on- 
 
74 ELEMENTS OF GEOMETRY. 
 
 ly two equal straight lines, s».ich as CD and CE, can be 
 drawn to AB (I. 17. qor.) the circle described from C 
 through the point D will cross AB again only at E. 
 
 PROP. III. THEOR. 
 
 The chord of an arc lies wholly within the cir- 
 cle. 
 
 The straight line AB which joins two points A, B in 
 
 t circumference of a circle, lies wholly within the figure. 
 
 For, from the centre C, draw CD 
 to any point in AB, and join CA and 
 CB. 
 
 Because CDA is the exterior an- 
 gle of the triangle CDB, it is greater 
 (I. 8.) than the interior CBD or 
 CBA; but CBA, being (I. 10.) e- 
 <]ual to CAB or CAD, the angle CDA is consequently 
 greater than CAD, a!;id its opposite side CA (J. 13.) great- 
 er than CD, or CD is less than CA, and therefore the 
 point D must lie within the circle. 
 
 Cor. Hence a circle is concave towards its centre. 
 
 PROP. IV. THEOR. 
 
 A straight line drawn from the centre of a cir- 
 cle at right angles to a chord, likewise bisects it ; 
 and, conversely, the straight line which joins the 
 centre with the middle of a chord, is perpendicu- 
 lar to it. 
 
BOOK III. 
 
 75 
 
 The perpendicular let fall from the centre C upon the 
 chord AB, cuts it into two equal parts AD, DB. 
 
 For join CA, CB : And, in the triangles ACD, BCD, 
 the side AC is equal to CB, CD is 
 common to both, and the right angle 
 ADC is equal to BDC ; these trian- 
 gles having thus their corresponding 
 angles at A and B both acute, are equal 
 (I. 21.) and consequently the side AD 
 is equal to BD. 
 
 Again, let AD be equal to BD ; the bisecting line CD 
 is at right angles to AB. 
 
 For join CA, CB. The triangles ACD and BCD, ha- 
 ving the sides AC, AD equal to CB, BD, and the remain- 
 ing side CD common to both, are equal (I. 2.), and con- 
 sequently the angle CDA is equal to CDB, and each of 
 them a right angle. 
 
 Cor. Hence a straight line cutting two concentric cir- 
 cles has equal portions intercepted by their circumferences. 
 
 PROP. V. THEOR. 
 
 A straight line which bisects a chord at right 
 angles, passes through the centre of the circle. 
 
 If the perpendicular FE bisect a 
 chord AB, it will pass through G 
 the centre of the circle. 
 
 For in FE take any point D, and 
 join DA and DB. The triangles 
 ADC and BDC, having the side 
 AC equal to BC, CD common, and 
 
76 ELEMENTS OF GEOMETRY. 
 
 the right angle ACD equal to BCD, are '^qual (I. 3.), and 
 CO! sequently the base AD is equal to BD. The point D 
 is, therefore, the centre of a circle described through A 
 and B j and thus the centres of the circles that can pass 
 through A and B are all found in the straight line EF. 
 The centre G of the circlt AEBF must hence occur in that 
 perpendicular. 
 
 Cor, The centre of a circle may hence be found by bi- 
 secting the chord AB by the diameter EF (I. 7,), and bi- 
 secting this again in G, 
 
 PROP. VI. THEOR. 
 
 The diameter is ihe greatest line that can be 
 inflected within a circle, 
 
 The diameter AB is greater than 
 any chord DE. 
 
 For join CD and CE. The two 
 sides DC and EC of the triangle 
 DCE are together greater than the 
 third side DE (I. 14.); but DC 
 and CE are equal to AC and CB, 
 or to the whole diameter AB. Wherefore AB is great- 
 er than DE. 
 
 PROP. VII. THEOR. 
 
 If from any eccentric point, two straight lines 
 be drawn to the circumference of a circle ; the 
 one which passes nearer the centre, is greater 
 than that which lies more remote. 
 
]gOOK III. 
 
 77 
 
 Let C be the centre of a circle, 
 and A a different point, from 
 which two straight lines AD and 
 AE are drawn to the circumfe- 
 rence ; of these lines, AD, which 
 lies nearer to B the opposite ex- 
 tremity of the diameter, is greater 
 than AE. 
 
 For the triangles ADC and 
 AEC have the side CD equal to 
 CE, the side CA common to both, 
 but the contained angle DCA 
 greater than EC A 5 wherefore (I. 
 18.) the base AD is likewise great- 
 er than the base AE. 
 
 Cor. 1. Hence the straight line 
 ACB, which passes through the 
 centre, is the greatest of all those 
 that can be drawn to the circum- 
 ference of the circle from the ec- 
 centric point A. For it is evident fropi the Proposition, 
 that the nearer the point D approaches to B, the greater 
 is AD ; consequently the point B forms the extreme limit 
 of majority, or AB is the greatest line that can be drawn 
 from A to the circumference. 
 
 Cor. 2. Hence also, whether the eccentric point be with- 
 in or without the circle, the straight line AH is the short- 
 est that can be drawn from A to the circumference. For 
 AE is less than AD, and AG less than AF ; and the near- 
 er the terminating point approaches to H, which is obvi- 
 ously the most remote from B, the sshorter must be its dis- 
 tance from A. Wherefore the point H marks the limit of 
 
IB 
 
 ELEMENTS OF GEOMETRY. 
 
 minority, and AH is the shortest line that can be drawn 
 from A to the circumference of the circle. 
 
 PROP. VIIL THEOR. 
 
 From any eccentric point, not more than two 
 equal straight lines can be drawn to the circum- 
 ference, one on each side of the diameter. 
 
 Let A be a point which is not 
 the centre of the circle, and AD a 
 straight line drawn from it to the 
 circumference. 
 
 Find the centre C (III. 5. cor.) 
 join CA and CD, draw (I. 4.) CE 
 making an angle ACE equal to 
 ACD and cutting the circumfe- 
 rence in E, and join AE : The 
 straight lines AE, AD are equal. 
 
 For the triangles ADC, AEC, 
 having the side CD equal to CE, 
 the side AC common, and the con- 
 tained angle ACD equal to ACE, are equal (I. 3.), and 
 consequently the base AD is equal to AE. 
 
 But, except AE, no straight line can be drawn from A 
 on the same side of the diameter HB, that shall be equal 
 to AD : For if the line terminate in a point F between E 
 and B, it will be greater than AE (III. 7.) ; and if the line 
 terminate in G between E and H, it will, for the same rea- 
 son, be less than AE. 
 
 Cor, 1. That point from which more than two equal 
 
BOOK III. t9 
 
 straight lines can be drawn to the circumference, is the cen- 
 tre of the circle. 
 
 Cor. 2. Hence a circle will not cut another in more 
 than two points. 
 
 PROP. IX. THEOR. 
 
 A circle may be described through three points 
 which are hot in the same straight line. 
 
 Let A, B, C, be three points not lying in the same di- 
 rection ; the circumference of a circle may be made to pass 
 through them. 
 
 For (I. 7.) bisect AB by the per- 
 pendicular DF, and BC by the per- 
 pendicular EF. These straight lines 
 DF, EF will meet ; because, DE be- 
 ing joined, the angles EDF, DEF 
 are less than BDF, BEF, and con- 
 sequently are together less than two 
 right angles, and DF, EF are not parallel (I. 22.), but 
 concur to form a triangle whose vertex is F. 
 
 Again, every circle that passes through the two points 
 A and B, has its centre in the perpendicular DF (III. 5.) ; 
 and, for the same reason, every circle that passes through 
 Band C has its centre in EF; consequently the circle 
 which would pass through all the three points, must have 
 its centre in F, the point common to both perpendicular* 
 DF and EF. 
 
 It is farther manifest, that there is only one circle which 
 can be made to pass through the three points A, B, C j 
 
80 ELEMENTS OF GEOMETRY. 
 
 for the intersection of the straight lines DF and EF, which 
 marks the centre, is a single point. 
 
 Cor, Hence the mode of describing a circle about a gi- 
 ven triangle ABC. 
 
 PROP. X. THEOR. 
 
 Equal chords are equidistant from the centre 
 of a circle \ and chords which are equidistant from 
 the centre, are likewise equal. 
 
 Let AB, DE be equal chords inflected within the same 
 circle ; their distances from the centre, or the perpendicu- 
 lars CF, CG, let fall upon them, are equal. 
 
 For the perpendiculars CF and CG bisect the chords 
 AB and DE (III. 4.), and consequently BF, DG, the 
 halves of these, are likewise equal. The right-angled tri- 
 angles CBF and CDG, which are 
 thus of the same character, having 
 the two sides BC, BF equal respec- 
 tively to DC, DG, and the corre- 
 sponding angle BFC equal to DGC, 
 are equal (I. 21.), and consequently 
 the side FC is equal to GC. 
 
 Again, if the chords AB, DE be equally distant from 
 the centre, they are themselves equal. 
 
 For the same construction remaining : The triangles 
 CBFandCDG are still right-angled, or of the same charac- 
 ter, and have now the two sides CB, CF equal to CD, CG, 
 and the angle BFC equal to DGC ; consequently they are 
 equal, and the side BF equal to DG ; the doubles of these, 
 therefore, or the whole chords AB, DE, are equal. 
 
BOOK III. 
 
 81 
 
 PROP. XI. THEOR. 
 
 The greater chord is nearer the centre of the 
 circle ; and that chord which is nearer the centre 
 is also the greater. 
 
 Let the chord DE be greater than AB ; its distance from 
 the centre, or the perpendicular CG let fall upon it, is less 
 than the distance CF. 
 
 For in the right-angled triangle BCF, the square 
 of the hypotenuse BC is equivalent 
 to the squares of BF and FC (II. 
 10.) ; and, for the same reason, the 
 square of the hypotenuse DC of the 
 right angled triangle DCG is equiva- 
 lent to the squares of DG and GC. But 
 the radii BC and DC are equal, and , 
 consequently their squares ; wherefore the squares of DG 
 and GC are equivalent to the squares of BF and FC. And 
 since BE is greater than AB, its half DG, made by the 
 perpendicular from the centre, is greater than BF, and con- 
 sequently the square of DG is greater than the square 
 of BF ; the square of GC is, therefore, less than the square 
 of FC, because, when conjoined xyith the squares of DG 
 and BF, they produce the same amount, or the square of 
 the radius of the circle. Hence the perpendicular GC it- 
 self is less than FC. 
 
 Again, if the chord DE be nearer the centre than AB, 
 it is also greater. 
 
 For the same construction remaining : It has been proved 
 that the squares of BF and FC are together equivalent to the 
 squares of DG and GC ; but GC being less than FC, the 
 
82 ELEMENTS OP GEOMETRY. 
 
 square of GC is less than the square of FC, and consequently 
 the square of DG is greater than the square of BF ; whence 
 the side DG is greater than BF, and its double, or the 
 chord DE, greater than AB. 
 
 PROP. XII. THEOR. 
 
 In the same or equal circles, equal angles at the 
 centre are subtended by equal chords, and termi- 
 nated by equal arcs. 
 
 If the angle ACB at the centre C be equal to DCE, the 
 chord AB is equal to DE, and the arc AFB equal to 
 DGE. 
 
 For let the sector ACB be applied to DCE. The cen- 
 tre remaining in its place, the radius CA will lie on CD ; 
 and the angle ACB being equal to DCE, the radius CB 
 will adapt itself to CE. And be- 
 cause all the radii are equal, their 
 extreme points A and B must co- 
 incide with D and E ; wherefore 
 the straight lines which join those 
 points, or the chords AB and DE, 
 must coincide. But the arcs AFB 
 and DGE that connect the same points, will also coincide ; 
 for any intermediate point F in the one, being at the same 
 distance from the centre as every point of the other, must, 
 on its application, find always a corresponding point G. 
 
 The same mode of reasoning is applicable to the case of 
 equal circles. 
 
 Cor, 1 . Hence, in the same or equal circles, equal arcs 
 
BOOK III. 83 
 
 are subtended by equal chords, and terminate equal angles 
 at the centre. 
 
 Cor, 2. Hence also, in the same or equal circles, equal 
 chords must subtend equal arcs of a like kind, that is, arcs 
 which are both greater or both less than a semicircumfe- 
 rence. 
 
 Schol. The length of a chord in a circle is thus insufficient 
 alone to determine the magnitude of the angle which it 
 subtends at the centre. To remove the ambiguity, it is 
 requisite to know, whether this angle be greater or less than 
 two right angles. 
 
 PROP. XIII. PROB. 
 
 To bisect a given arc of a circle. 
 
 Let it be required to divide the arc AEB into two equal 
 portions. 
 
 Draw the chord AB, and bisect it (I. 7.) by the perpen- 
 dicular EF cutting the circumference AB in E: The arc 
 AE is equal to EB. 
 
 For the triangles ADE, BDE, have the side AD equal 
 to BD, the side DE common, and 
 the containing right angle ADE e- 
 qual to BDE; they are (I. 3.) con- 
 sequently equal, and the base AE 
 equal to BE. But these equal chords 
 AE, BE must subtend equal arcs of 
 a hke kind (III. 12. cor. 2.), and the 
 arcs AE, BE are evidently each of 
 them less than a semicircumference. 
 
 Cor. The correlative arc AFB is also bisected, by the 
 perpendicular EDF at the opposite point F. 
 
8't 
 
 ELEMENTS OF GEOMETRy. 
 
 PROP. XIV. PROB. 
 An arc being given, to complete its circle. 
 
 Let ADB be an arc *, it is required to trace out the circle 
 to which it belongs. 
 
 Draw the chord AB, and bisect it 
 by the perpendicular CD (I. 7.), cut- 
 ting the arc in D, join AD, and from 
 A draw AC making an angle DAC 
 equal to ADC (I. 4.): The intersec- 
 tion C of this straight line with the 
 perpendicular, is the centre of the 
 circle required. 
 
 For join CB. The triangles ACE 
 and BCE, having the side EA equal 
 to EB, the side EC common, and the 
 contained angle AEC equal to BEC, 
 are equal (I. 3.)) and consequently AC 
 is equal to BC. But (I. 1 1.) AC is also 
 equal to CD, because the angle DAC 
 was made equal to ADC. Wherefore (III. 8. cor. 1.) 
 the three straight lines CA, CD, and CB being all equal, 
 the point C is the centre of the circle. 
 
 PROP. XV. THEOR. 
 
 The angle at the centre of a circle is double of 
 the angle which, standing on the same arc, has its 
 vertex in the circumference. 
 
 Let AB be an arc of a circle ; the angle which it termi- 
 
BOOK III. 
 
 85 
 
 nates at the centre, is double of ADB the corresponding 
 angle at the circumference. 
 
 For join DC and produce it to the opposite circumfeo 
 rence. This diameter DCE, if it lie not on one of the sides 
 of the angle ADB, must either fall within that angle or 
 without it. 
 
 First, let DC coincide with DB. And because AC is 
 equal to DC, the angle ADC is equal 
 to D AC (I. 10.) ; but the exterior an- 
 gle ACB is equal to both of these (I. 
 30.)j and therefore equal to double of 
 either, or the angle ACB at the centre 
 is double of the angle ADB at the . 
 circumference. 
 
 Next, let the straight line DCE lie within the angle 
 ADB. Froni what has been demon- 
 strated, it is apparent, that the angle 
 ACE is double of ADE, and the an- 
 gle BCE double of BDE ; wherefore 
 the angles ACE, BCE taken together, 
 or the whole angle ACB, are double 
 of the collected angles ADE, BDE, 
 or the angle ADB at the circumference. 
 
 Lastly, let DCE fall without the angle ADB. 
 the angle BCE is double of BDE, and 
 the angle ACE is double of ADE ; the 
 excess of BCE above ACE, or the an- 
 gle ACB at the centre, is double of the 
 excess of BDE above ADE, that is, of 
 the angle ADB at the circumference. 
 
 Cor. Hence if an equal circle be described from any 
 point D in the circumference, its arc intercepted by the 
 
 Because 
 
86 
 
 ELEMENTS OF GEOMETRY. 
 
 lines DA and DB will hejhe half of AB, and the whole 
 of the interior arc half of the exterior. 
 
 PROP. XVI. THEOR. 
 
 The angles in the same segment of a circle are 
 equal. 
 
 Let ADB be the segment of a cir- 
 cle ; the angles AFB, AGB contain- 
 ed in it, or which stand on the same 
 opposite portion AEB of the cir- 
 cumference, are equal to each other. 
 
 For join CA, CB. The angle 
 ACB, or its reverse at the centre, 
 and terminated by the arc AEB, is 
 double of the angle AFB or AGB 
 at the circumference (III. 15.); these 
 angles AFB, AGB, which stand on 
 the same arc AEB, are, therefore, 
 in every case, the halves of the same 
 central angle ACB, and are consequently equal to each 
 other. 
 
 Co7\ Hence equal angles at the circumference must stand 
 on equal arcs ; for their doubles or the central angles, be- 
 ing equal, are terminated by equal arcs (III. 12.). Hence 
 also equal angles that stand on the same base, have their 
 vertices in the same segment of a circle. 
 
 Schol. Hence the ordinary construction of theatres, the 
 seats being disposed in large arcs of a circle, so that the 
 stage may to each spectator subtend an equal angle, or 
 present always the same visual magnitude. 
 
BOOK III. 87 
 
 PROP. XVII. THEOR. 
 
 •The opposite angles of a quadrilateral figure 
 contained within a circle, are together equal to 
 two right angles. 
 
 Let ABCD be a quadrilateral figure described in a cir- 
 cle ; the angles A and C are together equal to two right 
 angles, and so are those at B and D. 
 
 For join EB and ED. The angle BED at the centre 
 is double of the angle BCD at the 
 circumference (III. 15.) •, and for the 
 same reason, the reverse angle BED 
 is double of BAD. Consequently 
 the angles BCD and BAD are the 
 halves of angles about the point E, 
 which make up four right angles; 
 wherefore the angles BCD and BAD are together equal 
 to two right angles. 
 
 In the same manner, by joining EA and EC, it may be 
 proved, that the angles ABC and ADC are together equal 
 to two right angles. 
 
 i Cor, 1. Hence it is evident from Prop. I. 16., that a cir- 
 cle may be described about a quadrilateral figure which has 
 its opposite angles equal to two right angles. 
 
 Cor» 2. Hence if one side of a quadrilateral figure in- 
 scribed in a circle be produced, it will form an exterior e- 
 qual to the opposite angle. 
 
 Cor. 3. Hence the angles at the base of a triangle in- 
 scribed in a circle, are together equal to an angle contain- 
 ed in the segment opposite to its vertex. 
 
88 ELEMENTS OF GEOMETRY. 
 
 PROP. XVIII. THEOR. 
 
 Parallel chords intercept equal arcs of a circle. 
 
 Let the chord AB be parallel to CD ; the intercepted 
 arc AC is equal to BD. 
 
 For join AD. And because the 
 straight lines AB and CD are pa- 
 rallel, the alternate angles BAD and 
 ADC are equal (I. 22.) ; wherefore 
 these angles, having their vertices in 
 the circumference of the circle, must 
 stand on equal arcs (III. 16. cor.), 
 
 and consequently the arcs AC and BD are equal to each 
 other. ^ 
 
 Cor, Hence, conversely, the straight lines which inter- 
 cept equal arcs of a circle are parallel; and hence ano- 
 ther mode of drawing a parallel through a given point to 
 a given straight line. 
 
 PROP. XIX. THEOR. 
 
 The angle in a semicircle is a right angle, the 
 angle in a greater segment is acute, and the angle 
 in a smaller segment is obtuse. 
 
 Let ABD be an angle in a semicircle, or that stands on 
 the semicircumference AED ; it is a right angle. 
 
 For ABD, being an angle at the 
 circumference, is half of the angle 
 at the centre on the same base 
 AED (III. 15.); it is, therefore, 
 half of the angle ACD formed by 
 the diverging of the opposite por- 
 tions CA, CD of the diameter, or 
 
BOOK III. 89 
 
 or half of two right angles, and is consequently equal to 
 one right angle. 
 
 Again, let ABD be an angle in a segment greater than 
 a semicircle, or which stands on a less arc AED than the 
 semicircumference ; it is an acute angle. 
 
 For join CA, CD. The angle ABD is half of the cen- 
 tral angle ACD, which is evidently less than two right an- 
 gles ; wherefore ABD is less than 
 one right angle, or it is acute. 
 
 But the angle AED, in the small- 
 er segment, is obtuse. For AED 
 stands on the arc ABD, which is 
 greater than a semicircumference, 
 and is the base of an angle at the 
 centre, the reverse of ACD, and greater, therefore, than 
 two right angles ; AED is hence an obtuse angle. 
 
 Cor, Hence conversely the arc which contains a right 
 angle must be a semicircle. 
 
 SchoL From the remarkable property, that the angle in 
 a semicircle is a right angle, may be derived an elegant me- 
 thod of drawing perpendiculars. 
 
 PROP. XX. THEOR. 
 
 The perpendicular at the extremity of a diame- 
 ter is a tangent to the circle, and the only tangent 
 which can be applied at that point. 
 
 Let ACB be the diameter of a circle, to which the 
 straight hne EBD is drawn at right angles from the extre- 
 
90 
 
 ELEMENTS OF GEOMETRY. 
 
 mity B ; it will touch the cir- 
 cumference at that point. 
 
 For CB, being perpendicu- 
 lar, is the shortest distance of 
 the centre C from the straight 
 lineEBD (I. 17.) j wherefore 
 every other point in this line iis 
 farther from the centre than B, 
 and consequently, falls without the circle. 
 
 But EBD, drawn at right angles to the diameter, is the 
 only straight line which can pass through the point B 
 and not cut the circle. For were HBF such a line, the 
 perpendicular CG let fall upon it from the centre, would 
 be less than CB (I. 17.), and must therefore lie within the 
 circle ; consequently HBG, being extended, would again 
 meet the circumference. 
 
 Cor\ Hence a straight line drawn from the point of con- 
 tact at right angles to a tangent, must be a diameter, or 
 pass through the centre of the circle. 
 
 Scholium. The nature of a tangent to the circle is easily 
 discovered from the consideration of limits. For suppose 
 the straight line DE, extending 
 both ways, to turn about the extre- 
 mity B of the diameter AB ; it will 
 cut the circle first on the one side 
 of AB, and afterwards on the o- 
 ther. But the arc AH being less 
 than a semicircumference, the an- 
 gle HBA which the line D'E^ 
 makes with the diameter is acute 
 (III. 19.); and, for the same reason, the angle KBA isa- 
 cute, and consequently its adjacent angle D'BA is obtuse. 
 Thus the revolving line DE, when it meets the semicir- 
 
BOOK III. 91 
 
 cumference AHB, makes an acute angle with the diame- 
 ter J but when it comes to meet the opposite semicircum- 
 ference, it makes an obtuse angle. In passing, therefore, 
 through all the intermediate gradations from minority to 
 majority, the line DE must find a certain individual posi- 
 tion in which it is at right angles to the diameter, and cuts 
 the circle neither on the one side nor the other. 
 
 PROP. XXI. THEOR. 
 
 If, from the point of contact, a straight line be 
 drawn to cut the circumference, the angles which 
 it makes with the tangent are equal to those in the 
 alternate segments of the circle. 
 
 Let 'CD be a tangent, and BE a straight line drawn from 
 the point of contact, cutting the circle into two segments 
 BAE and BFE ; the angle EBD is equal to EAB, and the 
 angle EBC to EFB. 
 
 ,For draw BA perpendicular to CD (I. 5. cor.), join AE, 
 and from any point F in the opposite arc, draw FB and 
 FE. 
 
 Because BA is perpendicular to the tangent at B, it is 
 a diameter (III. 20. cor.), and con- 
 sequently AEFB is a semicircle ; 
 wherefore AEB is a right angle (III. 
 19.), and the remaining acute an- 
 gles BAE, ABE of the triangle, be- 
 ing together equal to another right 
 angle, are equal to ABE and EBD, 
 which compose the right angle ABD. 
 Take the angle ABE away from both, and the angle BAE 
 remains equal to EBD. 
 
92 ELEMENTS OF GEOMETRY. 
 
 Again, the opposite angles BAE and BFE of the qua- 
 drilateral figure BAEF, being equal to two right angles 
 (III. 17.), are equal to the angle EBD with its adjacent 
 angle EEC ; and taking away the equals BAE and EBD, 
 there remains the angle BFE equal to EBC. 
 
 Co7\ If a straight line meet the circumference of a cir- 
 cle, and make an angle with an inflected line equal to that 
 in the alternate segment, it touches the circle. 
 
 Schol. A tangent may be considered as only a secant 
 arrived at its ultimate position, when the two points through 
 which it is drawn come to coincide. Suppose the straight 
 line joining B and F were extended, it would make with 
 the chord BE an angle EBF, equal to what the arc EF 
 subtends from any point in the opposite circumference. 
 But, when the point F is brought into the situation B, and 
 BF merges into a tangent, the angle EBF passes into 
 EBD, and the angle of the opposite or alternate segment 
 becomes BAE. 
 
 PROP. XXII. PROB. 
 
 To draw a tangent to a circle, from a given 
 point without it. 
 
 Let A be a given point, from which it is required to draw 
 a straight line that shall touch the circle DGH. 
 
 Find the centre C (III. 5- cor.), 
 join AC, and on this as a diame-, 
 ter describe the circle AGCK, cut- 
 ting the given circle in the points 
 G, K : Join AG, AK ; either of 
 these lines is the tangent required. 
 
 For join CG, CK. And the angles CGA, CKA, be- 
 
BOOK III. 
 
 93 
 
 ing each in a semicircle, are right angles (III. 19.), and 
 consequently AG, AK, touch the circle DGHK at the 
 points G, K (III. 20.). 
 
 Cm; Hence tangents drawn from the same point to a 
 circle are equal ; for the right angled triangles AGG and 
 ACK having the side CG equal to CK, CA common, are 
 equal (I. 21.), and consequently AG is equal to AK. 
 
 PROP. XXIII. PROB. 
 
 On a given straight line, to describe a segment 
 of a circle, that shall contain an angle equal to a 
 given angle. 
 
 Let AB be a straight line, on which it is required to de- 
 scribe a segment of a circle containing an angle equal to C. 
 
 If C be a right angle, it is evident that the problem will 
 be performed, by describing a semicircle on AB. But if 
 the angle C be either acute or 
 obtuse j draw AD (I. 4.) making 
 an angle BAD equal to C, 
 erect AE (I. 34.), perpendicular 
 to AD, draw EF (I. 5. cor.) to 
 bisect AB at right angles and 
 meeting AE in E, and, from 
 this point as a centre and with 
 the distance EA, describe the re- 
 quired segment AGB. 
 
 Because EF bisects AB at 
 right angles, the circle described through A must also pas$ 
 through (III. 5.) the point B ; and since EAD is a right 
 
91 
 
 ELEMENTS OF GEOMETRY. 
 
 angle, AD touches the circle at A (III. 20.)> and the an- 
 gle BAD, which was made equal to C, is equal (III. 21.) 
 to the angle in the alternate segment AGB. 
 
 PROP. XXIV. THEOR. 
 
 Two straight lines drawn through the point of 
 contact of two circles, intercept arcs of which the 
 chords are parallel. 
 
 Let the circles ACE and ABD touch mutually in A, 
 and from this point the straight lines AC, AE be drawn to 
 cut the circumferences ; the chords CE and BD are paral- 
 lel. 
 
 For draw the tangent FAG, (III. 20.), which must 
 touch both circles. 
 
 In the case of internal contact, the 
 angle GAE is equal to ACE in the al- 
 ternate segment, (III. 21.); and, for 
 the same reason, GAE or GAD is 
 equal to ABD ; consequently the an- 
 gles ACE and ABD are equal, and 
 therefore (I. 22.) the straight lines CE 
 and BD are parallel. 
 
 When the contact is exter- 
 nal, the angle GAE is still e- 
 qual to ACE, and its vertical 
 angle FAD is, for the same 
 reason, equal to ABD; whence 
 ACE is equal to ABD ; and 
 these being alternate angles, the straight line CE (I. 22.) 
 is parallel to BD. 
 
BOOK III. 
 
 95 
 
 PROP. XXV. THEOR. 
 
 If through a point, within or without a circle, 
 two perpendicular lines be drawn to meet the cir- 
 cumference, the squares of all the intercepted dis- 
 tances are together equivalent to the square of the 
 diameter. 
 
 Let E be a point within or without the circle, and AB, 
 CD two straight lines drawn through it at right angles to 
 the circumference ; the squares of the four segments EA, 
 EB, ED, and EC, are together equivalent to the square 
 of the diameter of the circle. 
 
 For draw BF parallel to CD, and join AF, AD, CB, 
 and DF. 
 
 Because BF is parallel to CD, the 
 arc BC is equal to the arc FD (III. 
 18.), and consequently the chord BC 
 is also equal to the chord FD (III. 12. 
 cor. 1.) ; but BC being the hypotenuse 
 of the right-angled triangle BEC, its 
 square, or that of FD is equivalent to 
 the squares of EB and EC (II. 10.), 
 and AED being likewise right-angled, 
 the square of AD is equivalent to the 
 squares of E A and ED. Whence the 
 squares of AD and FD are equivalent 
 to the four squares of EA, EB, ED, 
 and EC. But since ED is parallel to 
 BF, the interior angle ABF is equal to AED (I. 22.), and 
 
96 ELEMENTS OF GEOMETRY. 
 
 therefore a right angle ; consequently ACBF is a semicir- 
 cle (III. 19. cor.) and AF the diameter. The angle ADF 
 in the opposite semicircle is hence a right angle (III. 19.), 
 and the square of the diameter AF is equal to the squares 
 of AD and FD, or to the sum of the squares of the four 
 segments EA, EB, ED, and EC intercepted between the 
 circumference and the point E. 
 
 PROP. XXVI. THEOR. 
 
 If through a point, within or without a circle, 
 two straight lines be drawn to cut the circumfe- 
 rence ; the rectangle under the segments of the 
 one, is equivalent to that contained by the seg- 
 ments of the other. 
 
 Let the two straight lines AD and AF be extended 
 through the point A, to cut the circumference BFD of a 
 circle ; the rectangle contained by the segments AE and 
 AF of the one, is equivalent to the rectangle under AB 
 and AD, the distances intercepted from A in the other. 
 
 For draw AC to the centre, and produce it both ways 
 to terminate in the circumference at G and H ; let fall the 
 perpendicular CI upon BD (I. 6.), and join CD. 
 
 Because CI is perpendicular to AD, the difference be- 
 tween the squares of CA and CD, the sides of the trian* 
 gle ACD is equivalent to the diffe- 
 rence between the squares of the seg- 
 ments AI and ID the segments of 
 the base (II. 21. cor.) ; and the dif- 
 ference between the squares of two 
 straight lines being equivalent to the 
 rectangle under their sum and their 
 
BOOK III. 
 
 97 
 
 difference (II. 17.), the rectangle 
 contained by the sum and differ- 
 ence of AC, CD is equivalent to 
 the rectangle contained by the 
 sum and difference of AI, ID. 
 But since the radius CG is equal 
 to CH, the sum of AC and CD 
 is AH, and their difference is 
 
 AG 5 and because the perpendicular CI bisects the chord 
 BD (III. 4.), the sum of AI and ID is AD, and their dif- 
 ference AB. Wherefore the rectangle AH, AG is equi- 
 valent to the rectangle AB, AD. In the same way it is 
 proved, that the rectangle AH, AG is equivalent to the 
 rectangle AE, AF; and consequently the rectangle AE, 
 AF is equivalent to the rectangle AB, AD. 
 
 Cor. 1. If the vertex A of the straight lines lie within 
 the circle and the point I coincide with 
 it, BD, being then at right angles to CA, 
 is bisected at A (III. 4.), and the rect- 
 angle AB, AD is the same as the square 
 of AB. Consequently the square of a 
 perpendicular AB limited by the circum- 
 ference is equivalent to the rectangle under the segments 
 AG, AH of the diameter. 
 
 Cor, 2. If the vertex A lie without the circle and the 
 point I coincide with B or D, the an- 
 gle ABC being then a right angle, the 
 incident line AB must be a tangent 
 (III. 20,), and consequently the two 
 points of section B and D must coa- 
 lesce in a single point of contact. 
 Wherefore the rectangle under the distances AB, AD be- 
 comes the same as the square of AB j and consequently 
 
 H 
 
 H 
 
98 ELEMENTS OF GEOMETRY. 
 
 the rectangle contained by the segments AG, AH of the 
 diameter, is equivalent to the square of the tangent AB. 
 
 PROR XXVII. PROB. 
 
 To construct a square equivalent to a given 
 rectilineal figure. 
 
 Let the rectilineal figure be reduced by Proposition 6. 
 Book II. to an equivalent rectangle, of 
 which A and B are the two contain- 
 ing sides ; draw an indefinite straight 
 line CE, in which take the part CD 
 equal to A and DE to B, on C de- 
 scribe a semicircle, and erect the per- 
 pendicular DF from the diameter to meet the circumfe- 
 rence : DF is the side of the square equivalent to the given 
 rectilineal figure. 
 
 For, by Cor. 1 . to the last Proposition, the square of the 
 perpendicular DF is equivalent to the rectangle under the 
 segments CD, DE of the diameter, and is consequently e- 
 quivalent to the rectangle contained by the sides A and B 
 of a rectangle that was made equivalent to the rectilineal 
 figure. 
 
 PROP. XXyill. THEOR. 
 
 A quadrilateral figure may have a circle describ- 
 ed about it, if the rectangles under the segments 
 made by the intersection of its diagonals be equi- 
 
BOOK HI. 99 
 
 valent, or if those rectangles are equivalent which 
 are contained by the external segments formed by 
 producing its opposite sides. 
 
 Let ABCD be a quadrilateral figure, of which AC and 
 BD are the diagonals, and such that the rectangle AE, 
 EC is equivalent to the rectangle BE, ED ; a circle may 
 be made to pass through the four points A, B, C, and D. 
 
 For describe a circle through 
 the three points A, B, C (III. 9. 
 cor.), and let it cut BD in G. Be- 
 cause AC and BG intersect each 
 other within a circle, the rectan- 
 gle AE, EC is equivalent to the 
 rectangle BE, EG (III. 26.) ; but 
 the rectangle AE, EC is by hypothesis equivalent to the 
 rectangle BE, ED. Wherefore BE, EG is equivalent to 
 BE, ED ; and these rectangles have a common base BE, 
 consequently (II. 3. cor.) their altitudes EG and ED are 
 equal, and hence the point G is the same as D, or the cir- 
 cle passes through all the four points A, B, C, and D. 
 
 Again, if the opposite sides CB and DA be produced' 
 to meet at F, and the rectangle CF, FB be equal to DF, 
 FA, a circle may be described about the figure. 
 
 For, as before, let a circle pass through the three 
 points A, B, C, but cut AD in H. And from the proper- 
 ty of the circle, the rectangle CF, FB is equivalent to HF, 
 FA ; but the rectangle CF, FB is also equivalent to DF, 
 FA ; whence the rectangle HF, FA is equivalent to DF, 
 FA, and the base HF equal to DF, or the point H is the 
 same as D. 
 

 ELEMENTS 
 
 OJP 
 
 GEOMETRY. 
 
 BOOK IV. 
 
 DEFINITIONS. 
 
 1. A rectilineal figure is said to be in* 
 scribed in a circle, when all its angular 
 points lie on the circumference* 
 
 2. A rectilineal figure circumscribes a 
 circle, when each of its sides is a tan- 
 gent. 
 
 5. A circle is inscribed, in a rectilineal 
 figure, when it touches all the sides. 
 
fb2 ELEMENTS OF GEOMETRY. 
 
 4, A circle is described about a rectili- 
 neal figure or circumscribes it, when the cir- 
 cumference passes through all the angular 
 points of the figure. 
 
 5. Polygons are equilateral, when their sides, in the 
 same order, are respectively equal : They are equiangular, 
 if an equality obtains between their corresponding angles. 
 
 6. Polygons are said to be regular, when all their sides 
 and their angles are equal. 
 
BOOK IV. 
 
 103 
 
 PROP. I. PROB. 
 
 Given an isosceles triangle, to construct ano- 
 ther on the same base, but with only half the ver- 
 tical angle. 
 
 Let ABC be an isosceles triangle standing on AC; it 
 is required, on the same base, to construct another isosce- 
 les triangle, that shall have its vertical angle equal to half 
 of the angle ABC. 
 
 Bisect AC in D (I. Y.), join DB, 
 which produce till BE be equal to BA 
 or BC, and join AE, CE : AEC is the 
 isosceles triangle required. 
 
 For, the straight line BE being e- 
 qual to BA and BC, the point B is the 
 centre of a circle which passes through 
 the points A, E, and C ; and consequently the angle ABC 
 is the double of AEC at the circumference (III. 15.), or 
 the vertical angle AEC is half of ABC. But the triangles 
 AED and CED, having the side DA equal to DC, the 
 side DE common to both, and the right angle ADE 
 (III. 4'.) equal to CDE are (I. 3.) equal, and consequent- 
 ly AE is equal to CE. Wherefore the triangle AEC is 
 likewise isosceles. 
 
 PROP. ti. PROBi. 
 
 Given an acute-angled isosceles triangle, to con- 
 struct another on the same base, which shall have 
 double the vertical angle. 
 
104? ELEMENTS OF GEOMETRY. 
 
 Let ABC be an acute-angled isos- 
 celes triangle; it is required, on the 
 base AC, to construct another isosce- 
 les triangle, having its vertical angle 
 double of the angle ABC. 
 
 Describe a circle through the three 
 points A, B, and C (III. 9. cor.), and draw AD, CD to 
 the centre D ; the triangle ADC is the isosceles triangle 
 required. For the angle ADC, being at the centre of the 
 circle, is (III. 15.) double of ABC, the angle at the cir- 
 cumference. 
 
 PROP. III. THEOR. 
 
 If an isosceles triangle have each angle at the 
 base' double of the vertical angle, its base will be 
 equal to the greater segment of one of its sides 
 divided by a medial section. 
 
 Let ABC be an isosceles triangle which has each of the 
 angles BAC, BC A double of the vertical angle J\.BC \ the 
 base AC is equal to the greater segment of the side Bx4i 
 formed by a medial section. 
 
 For draw CD to bisect the angle BCA (I. 5.), and 
 about the triangle BDC describe a circle (III. 9. cor.). 
 
 Because the angle BCA is double 
 of ABC and has been bisected by 
 CD, the angles ACD, BCD are 
 each of them equal to CBD, and 
 consequently the side BD is equal 
 to CD (I. 11.). But the triangles 
 BAC and DAC, having the angle 
 
BOOK IV. 105 
 
 ACD equal to ABC, and the angle at A common to both, 
 must have also (I. 30.) the remaining angle CD A equal 
 to BCA or CAD; whence (I. 11.) the triangle DAC is 
 likewise isosceles, and the side AC equal to CD ; but CD 
 being equal to BD, therefore AC is also equal to it. And 
 since the angle ACD -is equal to CBD in the alternate 
 segment of the circle, the straight line AC touches the 
 circumference at C (III. 21. cor.); wherefore the rectan- 
 gle contained by AB and AD (III. 26. cor. 2.) is equiva-^ 
 lent to the square of AC, or the square of BD. Conse- 
 quently the base AC of this isosceles triangle is equal to 
 the greater segment BD of the side AB cut by a medial 
 section. 
 
 Cor, Hence the interior triangle ACD is likewise isos- 
 celes and of the same nature with ABC, having the great- 
 er segment of AB for its side, and the smaller segment for 
 its baso. 
 
 PROP. IV. PROB. 
 
 Given either one of the sides, or the base, to 
 construct an isosceles triangle, so that each of the 
 angles at the base may be double of its vertical 
 angle. 
 
 First, let one of the sides AB be given, to construct such 
 an isosceles triangle. 
 
 Divide AB by a medial section at C (II. 19.), and on 
 CB, as a base with the distance AB for each of the sides, 
 describe an isosceles triangle (I. 1.) 
 
106 ELEMENTS OF GEOMETRY. 
 
 Next, let the base AB be given, AC B 
 
 to construct an isosceles triangle of J[ ' ^g ^ 
 
 this nature. 
 
 Produce AB to C, such that the rectangle AC, CB be 
 equal to the square of AB (II. 19. cor. 2.), and on the 
 base AB, with the distance AC for each of the sides, de- 
 scribe an isosceles triangle. 
 
 These isosceles triangles will fulfil the conditions re- 
 quired. For it is evident, from the last Proposition, that 
 isosceles triangles constituted on CB or AB, with each of 
 the angles at the base double the vertical angle, would 
 have AB or AC for their sides, and consequently (I. 2.) 
 must coincide with the triangles now described. 
 
 Cor. Hence of such an isosceles triangle the vertical 
 angle is equal to the fifth , part of two right angles ; for 
 each of the angles at the base being double of the vertical 
 angle, they are both equal to four times it, and consequent- 
 ly this vertica} angle is the fifth part of all the angles of 
 the triangle, or of two right angl^. 
 
 PROP. V. PROB. 
 
 On a given finite straight line, to describe a re- 
 gular pentagon. i 
 
 Let AB be the straight line, on which it is required to 
 describe a regular pentagon. 
 
 On AB erect (IV. 4.) the isosceles triangle ACB, ha- 
 ving each of the angles at its base double of its vertical an- 
 (jle, from the centre A with the distance AB describe an 
 
BOOK IV. 
 
 lo: 
 
 arc of a circle, and from the cen- 
 tre B with the same distance de- 
 scribe another arc, and from C 
 inflect the straight lines CE, CD 
 equal to AB: The points C, D, 
 E mark out the pentagon. 
 
 For it is evident from this con- 
 struction that BF and AG bisect 
 the angles at the base of the triangle ACB, and conse- 
 quently (IV. 3.) AB is equal to BF and FC, or AG and 
 GC. Again, the triangles BAD and BFC, having the 
 sides AB, BD equal to BF, BC, and the contained angles 
 equal, are themselves equal (I. 3.), and consequently AB 
 is equal to AD, and the angle BAD equal to BFC, or 
 three times ACB. In the same way it is shewn that' AB 
 is equal to BE, and that the angles round the figure are 
 each equal to thrice the vertical angle of the original isos- 
 celes triangle. 
 
 PROP. VI. PROB. 
 
 On a given finite straight line, to describe a re- 
 gular hexagon. 
 
 Let AB be the given straight line, on which it is re- 
 quired to describe a regular hexagon. 
 
 On AB construct (I. 1.) the equilateral triangle AOB, 
 and repeat equal triangles about the vertex O ; these tri- 
 angles will together compose the hexagon required. 
 
 Because AOB is an equilateral triangle, each of its an- 
 
108 
 
 ELEMENTS OF GEOMET^IY. 
 
 gles is equal to the third part of 
 two right angles (1. 30. cor. 1.) ; 
 wherefore the vertical angle 
 AOB is the sixth part of four 
 right angles, or six of such an- 
 gles may be placed about the 
 point O. But the bases of the 
 triangles AOB, AOC, COD, 
 DOE, EOF, and BOF are all equal ; and so are the an- 
 gles at the bases, and which, taken by pairs, form the inter- 
 nal angles of the figure B ACDEF. This figure is, there- 
 fore, a regular hexagon. 
 
 PROP. VII. PROB. 
 
 On a given finite straight line, to describe a re- 
 gular octagon. 
 
 Let AB be the given straight line, on which it is re- 
 quired to describe a regular octagon. 
 
 Bisect AB (I. 7.) by the perpendicular CD, which make 
 equal to CA or CB, join DA and DB, produce CD until 
 DO be equal to DA or DB, 
 draw AO and BO, thus forming 
 (IV. 1.) an angle equal to the 
 half of ADB, and, about the ver- 
 tex O, repeat the equal trian- 
 gles AOB, AOE, EOF, FOG, 
 GOH, HOI, lOK, and KOB 
 to compose the octagon. 
 
 For the distances AD, BD are 
 evidently equal ; and because CA, CD, and CB are all 
 
BOOK IV. 109 
 
 equal, the angle ADB is contained in a semicircle, and 
 is therefore a right angle (III. 19.)» Consequently AOB 
 is equal to the half of a right angle, and eight such angles 
 will adapt themselves about the point O. Whence the fi- 
 gure BAEFGHIK, having eight equal sides and equal 
 angles, is a regular octagon. 
 
 PROP. VIII. PROB. 
 
 On a given finite straight line, to describe a re- 
 gular decagon. 
 
 Let AB be the straight line, on which it is required to 
 describe a regular decagon. 
 
 On AB construct (IV. 4.) an isosceles triangle having 
 each of the angles at its base double of the vertical angle, 
 and, about the point O, place 
 a series of triangles all equal 
 to AOB : A regular decagon 
 will result from this composi- 
 tion, ni 
 
 For the vertical angle AOB 
 of the isosceles triangle is e- 
 qual to the fifth part of two 
 right angles (IV. 4. cor.), or 
 
 to the tenth part of four right angles ; whence ten such 
 angles may be formed about the point O. The figure 
 BACDEFGHIK, having therefore ten equal sides and 
 eq^ual angles, is a regular decagon. ' 
 
 '^/v 
 
 F 
 
 
 SIC 
 
 k 
 
 \/ 
 
 t 
 
 yK 
 
ii^ - ELEMENTS OF GEOMETRY. 
 
 PROP. IX. PROB. 
 
 On a given finite straight line, to describe a re- 
 gular dodecagon. 
 
 Let AB be the straight line, on which it is required to 
 describe a regular twelve-sided figure. 
 
 On AB construct (I. 1.) the equilateral triangle ACB, 
 and again (IV. 1.) the isosceles triangle AOB, having its 
 vertical angle equal to the half of ACB, and repeat this 
 triangle AOB about the point 
 O; a regular dodecagon will 
 be thus formed. 
 
 For ACB being an equilate- 
 ral triangle, each of its angles is 
 the third part of two right an- 
 gles (I. 30. cor. 1.) ; conse- 
 quently the angle AOB is the 
 sixth part of two right angles 
 
 or the twelfth part of four right angles, and twelve such an- 
 gles can, therefore, be placed about the vertex O. 
 
 Scholium. Hence a regular twenty- sided figure may be 
 described on a given straight line, by first constructing on 
 it an isosceles triangle having each of the angles at the 
 base double of the vertical angle, and then erecting ano- 
 ther isosceles ti'iangle with its vertical angle equal to the 
 half of this. And, by thus changing the elementary tri- 
 angle, a regular polygon maybe always described, with 
 twice the number of sides. 
 
BOOK IV. Ill 
 
 PROP. X. PROB. 
 In a given triangle, to inscribe a circle. 
 
 Let ABC be a triangle, it which it is required to in- 
 scribe a circle. 
 
 Draw AD and CD (I. 5.) to bisect the angles CAB and 
 ACB, and from their point of concourse D, with its dis- 
 tance DE from the base, describe the circle EFG : This 
 circle will touch the triangle internally. 
 
 For let fall the perpendiculars DG and DF upon the sides 
 AB and BC (I. 6.). The trian- 
 gles ADE, ADG, having the an- 
 gle DAE equal to DAG, the right 
 angle DEA equal to DGA, and 
 the interjacent side AD common, 
 are equal (I. 20.), and therefore 
 the side DE is equal to DG. In 
 the same manner, it is proved, from the equality of the 
 triangles CDE, CDF, that DE is equal to DF; conse- 
 quently DG is equal to DF, and the circle passes through 
 the three points E, G, and F. But it also touches (III. 20.) 
 the sides of the triangle in those points, for the angles 
 DEA, DGA, and DEC are all of them right ano-les. 
 
 PROP. XL PROB. 
 
 In a given circle, to inscribe a triangle equian- 
 gular to a given triangle. 
 
 Let GDH be a circle, in which it is required to inscribe 
 
112 ELEMENTS OF GEOMETRY. 
 
 a triangle that shall have its angles equal to those of the 
 triangle ABC. 
 
 Assuming any point D in the circumference of the cir- 
 cle, draw (III. 22.) the tan^ 
 gent EDF, and make the an- 
 gles EDG, FDH equal to 
 BCA, BAC, and join GH : 
 The triangle GDH is equi- 
 angular to ABC. 
 
 For EF being a tangent, 
 and DG drawn from the 
 
 point of contact, the angle EDG, which was made equal 
 to BCA, is equal to the angle DHG in the- alternate seg- 
 ment (III. 21.); consequently DHG is equal to BCA. 
 And for the same reason, the angle DGH is equal to BAC j 
 wherefore (I. 30.) the remaining angle GDH of the trian- 
 gle GHD is equal to the remaining angle ABC of the 
 triangle ACB, and these triangles are mutually equian- 
 gular. 
 
 PROP. XXL PROB. 
 
 About a given circle, to describe a triangle ^f 
 quiangular to a given triangle. 
 
 Let GIH be a circle, about which it is required to de- 
 scribe a triangle, having its angles equal to those of the 
 triangle ABC. 
 
 Draw any radius FG, and with it make (I. 4.) the an- 
 gles GFI, GFH equal to the exterior angles BAE, BCD 
 of the triangle ABC, and, from the points G, I, and H 
 
BOOK IV. n^ 
 
 draw the tangents KM, KL, and LM to form the trian- 
 gle KLM : This triangle is equiangular to ABC. 
 
 For all the angles of the quadrilateral figure KIFG be- 
 ing equal to four right 
 angles, and the angles 
 KIF and KGF being 
 
 €ach a right angle (III. y/^\/^ \\^a; cHb 
 
 20.), the remaining an- 
 gles GKI and GFI are 
 together equal to two 
 right angles, and consequently equal to the angles BAG 
 and B AE on the same side of the straight line ED. But 
 the angle GFI was made equal to BAE ; whence GKI is 
 equal to CAB. In Hke manner, it may be proved that the 
 angle GMH is equal to ACB ; and the angles at K and M 
 being thus equal to BAC and BCA, the remaining angle 
 at L is (I. 30.) equal to that at B, and the two triangles 
 are therefore equiangular. 
 
 PROP. XIII. THEOR. 
 
 A straight line drawn from the vertex of an 
 equilateral triangle inscribed in a circle to any 
 point in the opposite circumference, is equal to 
 the two chords inflected from the same point to 
 the extremities of the base. 
 
 Let ABC be an equilateral triangle inscribed in a cir- 
 cle, and BDj AD, and CD chords drawn from it to a 
 point D in the circumference ; BD is equal to AD and 
 CD taken together. 
 
 If 
 
114? ELEMENTS OF GEOIHETRY. 
 
 For, make DE equal to DA, and join AE. The angle 
 ADB is (III. 16) equal to ACB in 
 the same segment, which, being the 
 angle of an equilateral triangle, is e- 
 qual (I. SO. cor. 1.) to the third part 
 of two right angles. Bnt the triangle 
 ADE being isosceles by construction, ^ 
 
 the angles DAE, DEA at its base are equal (I. 10.), and 
 each of them is, therefore, equal to half of the remaining 
 two-thirds of two right-angles, or to one-third part. Con- 
 sequently ADE is likewise an equilateral triangle (I. 11. 
 cor.), and the angle DAE equal to CAB ; take CAE from 
 both, and there remains the angle DAC equal to EAB ; 
 but the angle ABD is equal to ACD in the same segment. 
 And thus the triangles ADC and AEB have the angles 
 DAC, DC A equal to EAB, EBA, and the interjacent side 
 AC equal to AB ; they are consequently equal (I. 20.), 
 and the side DC is equal to EB. But DE was made equal 
 to DA ; wherefore DA and DC are togethei: equal to DE 
 and EB, or to DB. 
 
 PROP. XIV. THEOR. 
 
 About and in a given square, to circumscribe 
 and inscribe a circle. 
 
 J^ct ABCD be a square, about which it is required to 
 circumscribe a circle. 
 
 Draw the diagonals AC, DB intersecting each other in 
 Q, and, from that point with the distance AO, describe 
 the circle ABCD : This circle will circumscribe the 
 square. 
 
BOOK IV. 
 
 115 
 
 Because the diagonals of the square ABCD are equal and 
 bisect each other, the straight 
 lines OA, OB, OC, and OD are 
 all equal, and consequently the 
 circle described through A 
 passes through the other points 
 B, C, and D. 
 
 Again, let it be required to 
 inscribe a circle in the square 
 ABCD. 
 
 From O the intersection of the diagonals and with its 
 distance from the side AD, describe the circle EGHF 
 This circle will touch the square internally. 
 
 For let fall the perpendiculars OG, OH, and OF (I. 6.). 
 And because the straight lines AB, BC, CD, and DA are 
 equal, they are equally distant from the centre O of the 
 exterior circle (III. 10.) ; wherefore the perpendiculars 
 OE, OG, OH, and OF are all equal, and the interior 
 circle passes through the points G, H, and F; but (III. 20.) 
 it likewise touches the sides of the square, since they are 
 perpendicular to the radii drawn from O. 
 
 Cor, Hence an octagon may be inscribed within a given 
 square. For let tangents be applied at the points I, K, 
 L, and M, where the diagonals cut the interior circle. It 
 is evident, that the triangle AOE is equal to DOE, lOP 
 to EOF, and EOZ to MOZ ; whence the angles POE 
 and ZOE are equal, being the halves of EOA and EOD, 
 and consequently the triangles PEO and ZEO are equal. 
 Wherefore PZ, the double of PE, is equal to PQ, the 
 double of PI ; and the angle EZM is, for a like reason, 
 equal to EPI. And, in this manner, all the sides and all 
 the angles about the eight-sided figure PQRSTWYZ are 
 proved to be equal. 
 
1 1 EJ.EMENTS OF GEOMETRY. 
 
 PROP. XV. PROB. 
 
 In and about a given circle, to inscribe and cir- 
 cumscribe a square. 
 
 Let EADB be a circle in which it is required to in- 
 scribe a square. 
 
 Draw the diameter AB, the perpendicular ED through 
 the centre, and join AD, DB, BE, and EA : The inscribed 
 figure ADBE is a square. 
 
 The angles about the centre C, being right angles, are 
 equal to each other, and are, therefore, subtended by equal 
 chords AD, DB, BE, and AE, but 
 one of the angles ADB, being in a se- 
 micircle, is (III. 19.) a right angle, 
 and consequently ADBE is a square. 
 
 Next, let it be required to circum- 
 scribe a square about the circle. 
 
 Apply tangents FG, GH, HI, and 
 FI at the extremities of the perpendicular diameters: 
 These will form a square. 
 
 For all the angles of the quadrilateral figure CG, being 
 together equal to four right angles, and those at C, A, and 
 D being each a right angle, the remaining angle at G is 
 also a right angle, CG is a rectangle ; and AC being equal 
 to CD, it is likewise a square. In the same manner, CH, 
 CI, and CF are proved to be squares ; the sides FG, GH, 
 HI, and IF of the exterior figure, being therefore the 
 doubles of equal lines, are mutually equal, and the angle 
 at G being a right angle, FH is consequently a square. 
 
 Co?\ Hence the circumscribing square is double of the 
 inscribed square, and this again is double of the square de- 
 scribed on the ra^us of the circle. 
 
BOOK IV. 117 
 
 PROP. XVI. PROB. 
 
 In and about a given circle, to inscribe and cir- 
 cumscribe a regular pentagon. 
 
 Let ABCDE be a circle in which it is required to in- 
 scribe a regular pentagon. 
 
 Construct an isosceles triangle having each of its an- 
 gles at the base double of its vertical angle (IV. 4.), and 
 equiangular to tliis, inscribe the triangle ACE within the 
 circle (IV. 11.), draw AD, EB bisecting the angles CAE, 
 CEA (I. 5.), and join AB, BC, CD, and DE; The figure 
 ABCDE is a regular pentagon. 
 
 For the angles AEB, BEC are each the half of 
 CEA, and therefore equal 
 to ACE; but the angles 
 EAD, DAC are likewise 
 equal to ACE. tlence these 
 angles, being all equal, must 
 stand on equal arcs (III. 16. 
 cor.); and the chords of 
 these arcs, or the sides AB, 
 BC, CD, DE, and AE are 
 equal (III. 12. cor.). And because the segments EAB, 
 ABC, BCD, CDE, and DEA are evidently equal, (III. 
 16.), the interior angles of the figure are all equal, and it 
 is, therefore, a regular pentagon. 
 
 Next, let it be required to circumscribe a regular pen- 
 tagon about the circle. 
 
 At the points A, B, C, D, and E apply tangents; these 
 will form a regular pentagon. 
 
 For FAK being a tangent, the angle KAE is equal to 
 ACE (III. 21.); and in like manner it is shown that the 
 angles AEK, DEI, EDI, CDH, DCH, BCG, CBG, 
 
1 ^ 8 ELEfylENTS OF GEOMETRY. 
 
 ABF, BAF are all equal to ACE. The isosceles trian- 
 gles AKE, BFA, having, therefore, the angles at the base 
 equal and the bases themselves AE, AB,— -are equal 
 (L 20.); for the same reason, the triangles BGC, CHD, 
 DIE, EKA, are equal. Whence the internal angles of 
 the figure are equal, and its sides, being double of those of 
 the annexed triangles, are likewise equal : The figure is, 
 therefore, a regtilar pentagon. 
 
 PROP. XVII. PROB. 
 To inscribe a regular hexagon in a given circle. 
 
 Let it be required, in the circle FBD, to inscribe a hexa- 
 gon. 
 
 Draw the radius OA, on which construct the equilate- 
 ral triangle ABO (I. 1. cor.), and repeat the equal trian- 
 gles about the vertex O : These triangles will compose a 
 hexagon. 
 
 For the triangle ABO, being equilateral, each of its an- 
 gles AOB is the third part of 
 two right angles; and conse- 
 quently six of such angles may 
 be placed about the centre O. 
 But the bases of the triangles 
 AOB, BOC, COD, DOE, 
 EOF, and FOA form the sides 
 of the figure, and the angles at 
 
 those bases its internal angles ; wherefore it is a regular 
 hexagon. 
 
 Cor, 1. Tangents applied at the points A, B, C, D, E, 
 F, would evidently form a regular circumscribing hexa- 
 gon. — An equilateral triangle might be inscribed by join- 
 ing the alternate points 5 and, by applying tangents at 
 
BOOK IV. 119 
 
 those points> an equilateral triangle would be made to cir- 
 cumscribe the circle. 
 
 Cor, 2. The side AB of the inscribed hexagon is equal 
 to the radius ; and since ABD is a right-angled triangle, 
 and the squares of AB and BD are equal to the square of 
 AD or to four times the square of AO, the square of BD 
 the side of an inscribed equilateral triangle is triple the 
 square of the radius. 
 
 Cor. 3. The perimeter of the inscribed hexagon is equal 
 to six times the radius, or three times the diameter, of the 
 circle. Hence the circumference of a circle being, from 
 its perpetual curvature, greater than any intermediate sys- 
 tem of straight lines, is more than triple its diameter. 
 
 PROP. XVIII. PROB. 
 To inscribe a regular decagon in a given circle. 
 
 Let ADH be a circle, in which it is required to inscribe 
 a regular decagon. 
 
 Draw the radius OA, and with OA as its side describe 
 the isosceles triangle AOB, having each of its angles at 
 the base double of its vertical angle (IV. 4.)y repeat the 
 equal triangles about the centre O : These triangles will 
 compose a decagon. 
 
 For the vertical angle AOB 
 of the component isosceles tri- 
 angle, is the fifth part of two " 
 right angles (IV. 4. cor.), and 
 consequently ten such angles 
 can be placed about the point 
 O. But the sides and angles 
 of the resulting figure are all "f^=Hffi^ 
 
 evidently equal; it is, therefore, a regular decagon. 
 
120 ELEMENTS OF GEOMETRY. 
 
 Co7\ Hence a regular pentagon will be formed, by join- 
 ing the alternate points A, C, E, G, I, and A. It is also 
 manifest, that a decagon and a pentagon may be circum- 
 scribed about the circle, by applying tangents at their se- 
 veral angular points. 
 
 PROP. XIX. THEOR. 
 
 The square of the side of a pentagon inscribed 
 in a circle, is equivalent to the squares of the sides 
 of the inscribed hexagon and decagon. 
 
 Let ABCDEF be half of a decagon inscribed in a cir- 
 cle whose diameter is AF ; the square of AC, the side of 
 an inscribed pentagon, is equivalent to the square of AB 
 the side of the inscribed decagon, and of the square of the 
 radius AO, or the side of an inscribed hexagon. 
 
 For join AD, and draw OB, OC, and OD. Since the 
 arc DEF is double of AB, the angle AOB at the centre is 
 (III. 15.) evidently equal to OAD or OAG at the circum- 
 ference •, and because the arc BCDEF again is double of 
 DEF, the angle OAB at the circumference is likewise e- 
 qual to AOG at the centre. Whence the triangles AOB 
 and AGO, having the an^ 
 gles OAB and AOB equal 
 to AOG and OAG, and 
 the interjacent side AO 
 common, are equal (I. 20.), 
 and therefore the base AB 
 is equal to OG. Conse- 
 quently, (IV. 18.) GAO is au isosceles triangle having 
 each of the angles at its base double the vertical angle; 
 
BOOK IV. 
 
 121 
 
 wherefore (IV. 3.) OG is equal to the greater segment of 
 side AO divided by a medial section. But (II. 20.) the 
 square of AC, drawn from the vertex to a point in the 
 extension of the base of the triangle OAG, is equivalent 
 to the square of AG, together with the rectangle under OC 
 and CG, or the square of OG ; that is, the square of the 
 side of the inscribed pentagon is equivalent to the squares 
 of AO and of AB, the sides of the hexagon and decagon. 
 
 Cor. The triple chord AD of the decagon is equal 
 to the combined sides AO and AB of the inscribed hexa- 
 gon and decagon. For the triangle OAG, being equal 
 to AOB or COD, the angle DCO or DCG is equal to 
 AGO or DGC, and consequently (I. 11.) CD is equal 
 to GD. Wherefore AD being equal to AG and GD, is 
 equal to AO with OG or AB. 
 
 Scholium. Hence the sides of the inscribed decagon 
 and pentagon may be found by a single construction. For 
 draw the perpendicular dia- 
 meters AC and EF, bisect OC 
 in D, join DE, make DG e- 
 qual to it, and join GE. It 
 is evident, that AO is cut me- 
 dially in G(II. 19.), and con- 
 sequently that OG is equal to 
 a side of the inscribed deca- 
 gon. But GOE being a right- 
 angled triangle, the square of GE is equivalent to the 
 squares of GO and OE (II. 10.), or the squares of the 
 sides of the decagon and hexagon ; whence GE is equal to 
 the side of the inscribed pentagon. It also follows, that 
 CG is equal to CI or CP, the triple chords of the inscri- 
 bed decagon. 
 
 H^ y 
 
 p^ 
 
 X 
 
 M ' ^ 
 
 
 ^ 
 
 *^\ / 
 
 !f\ D 
 
 '1 
 
 
 ■M. 
 
 i- 
 
 2ir 
 
J 22 ELEMENTS OF GEOMETRY. 
 
 PROP. XX. PROB. 
 
 In a given circle, to inscribe regular polygons 
 of fifteen and of thirty sides. 
 
 Let AB and BC be die sides of an inscribed decagon, 
 and AD the side of a hexagon inscribed ; the arc BD will 
 be the fifteenth part of the circumference of the circle, and 
 DC the thirtieth part. 
 
 For, if the circumference were divided into thirty equal 
 portions, the arc AB would be 
 equal to three of these, and the 
 arc AD to five ; consequently the 
 excess BD is equal to two of these 
 portions, or it is the fifteenth part 
 of the whole circumference. A- 
 
 gain, the double arc ABC being equal to six portions, and 
 ABD to five, the defect DC is equal to one portion, or to 
 the thirtieth part of the circumference. 
 
 Scholium. From the inscription of the square, the pen- 
 tagon, and the hexagon, — may be derived that of a variety 
 of other regular polygons : For, by continually bisecting 
 the intercepted arcs and inserting new chords, the inscribed 
 figure will, at each successive operation, have the number 
 of its sides doubled. Hence polygons will arise of 6, 8, 
 and 10 sides; then of 12, 16, and 20; next of 24', 32, 
 and 40 ; again, of 48, 64, and 80 ; and so forth repeated- 
 ly. The excess of the arc of the hexagon above that of 
 the decagon, gives the arc of a fifteen-sided figure ; and 
 the continued bisection of this arc will mark out polygons 
 with 30, 60, or 120 equal sides, in perpetual succession. 
 
BOOK IV. 123 
 
 The same results might also be obtained from the diffe- 
 rences of the preceding arcs. 
 
 Of the regular polygons, three only are susceptible of 
 perfect adaptation, and capable therefore of covering, by 
 their repeated addition, a plane surface. These are the 
 equilateral triangle, the square, and the hexagon. The 
 angles of an equilateral triangle are each two-thirds of a 
 right angle, those of a square are right angles, and the 
 angles of a hexagon are each equal to four-third parts of 
 a right angle. Hence there may be constituted about a 
 point, six equilateral triangles, four squares, and three 
 hexagons. But no other regular polygon can admit of a 
 like disposition. The pentagon, for instance, having each 
 of its angles equal to six-fifths of a right angle, would not 
 fill up the whole space about a point, on being repeated 
 three times ; yet it would do more than cover that space, 
 if added four times. On the other hand, since each angle 
 of a polygon which has more than six sides must exceed 
 four-third parts of a right angle, three such polygons can- 
 not stand round a point. Nor can the space about a point 
 ever be bisected by the application of any regular polygons, 
 of whatever number of sides ; for their angles are always 
 necessarily each less than two right angles. 
 
ELEMENTS 
 
 OF 
 
 GEOMETRY. 
 
 BOOK V. 
 
 OF PROPORTION. 
 
 The prieceding Books treat of magnitude as 
 concrete, or having mere extension 5 and the sim- 
 pler properties of lines, of angles, and of surfaces, 
 were deduced, by a continuous process of reason- 
 ings grounded on the principle of superposition* 
 But this mode of investigation, how satisfactory so- 
 ever to the mind, is by its nature very limited and 
 laborious. By introducing the idea of Number 
 into geometry, a new scene is opened, and a far 
 wider prospect rises into view. Magnitude, being 
 considered as discrete, or composed of integrant 
 parts, becomes assimilated to multitude ; and un- 
 der this aspect, it presents a vast system of rela- 
 
126 EtEMENTS OF GEOxMETRY. 
 
 tions, which may be traced out with the utmost 
 facility. - 
 
 Numbers were at first employed, to denote the 
 aggregation of separate, though kindred, ob- 
 jects ; but the subdivision of extent, whether ac- 
 tually effected or only conceived to exist, bestow- 
 ing on each portion a sort of individuality, 
 they came afterwards to acquire a more com- 
 prehensive application. In comparing together 
 two quantities of the same kind, the one may 
 contain the other, or be contai?ied by it ; that is, 
 the one may result from the repeated addition 
 of the other, or it may in its turn produce this 
 other by a successive composition. The one quan- 
 tity is, therefore, equal, either to so many times 
 the other, or to a certain aliquot part of it. 
 
 Such seems to be the simplest of the numerical 
 relations. It is very confined, however, in its ap- 
 plication, and is evidently, in this shape, insufficient 
 altogether for the purpose of general comparison. 
 But that object is attained, by adopting some in- 
 termediate term of reference. Though a quantity 
 neither contain another exactly, nor be contained 
 by it ; there may yet exist a third and smaller 
 quantity, which is at once capable of measuring 
 them both. This measure corresponds to the 
 arithmetical unit ; and as number denotes the col- 
 lection of units, so quantity/ may be viewed as the 
 aggregate of its component measures. 
 
 But mathematical quantities are not ajl suscep- 
 
BOOK V. 127 
 
 tible of such perfect mensuration. Two quantities 
 maybe conceived to be so constituted, as not to ad- 
 mit of any other quantity that will measure them 
 completely, or be contained in both without leaving 
 a remainder. Yet this apparent imperfection, which 
 proceeds entirely from the infinite variety ascri- 
 bed to possible magnitude, creates no real obstacle 
 to the progress of accurate science. The mea- 
 sure or primary element, being assumed succes- 
 sively still smaller and smaller, its corresponding 
 remainder must be perpetually diminished. This 
 continued exhaustion will hence approach nearer 
 than any assignable difference to its absolute term. 
 
 Quantities in general can, therefore, either ex- 
 actly or to any required degree of precision, be 
 represented abstractly by numbers ; and thus the 
 science of Geometry is at last brought under the 
 dominion of Arithmetic. 
 
 It is obvious, that quantities of any kind must 
 have the same composition, when each contains 
 its measure the same number of times. But quan- 
 tities, viewed in pairs, may be considered as ha- 
 ving a similar composition, if the corresponding 
 terms of each pair contain its measure equally. 
 Two pairs of quantities of a similar composition, 
 being thus formed by the same distinct aggrega- 
 tions of their elementary parts, constitute a Pro- 
 portion, 
 
12^ ELEMENTS OF GEOMETRY. 
 
 DEFINITIONS. 
 
 1 . Quantities are homogeneous^ which can be added to- 
 gether. 
 
 2. One quantity is said to contain another, when the 
 subtraction of the smaller— continued if necessary — leaves 
 no remainder. 
 
 3. A quantity which is contained in another, is said to 
 measure it. 
 
 4. The quantity which is measured by another, is called 
 its rhultiple ,• and that which measures the other, its sub- 
 multiple, 
 
 5. Lilce multiples and submultiples are those which 
 contain their measures equally, or which equally measure 
 their corresponding compounds. 
 
 6. Quantities are commensurable, which have a finite 
 common measure ; they are incommensurable^ if they will 
 admit of no such measure. 
 
 7. That relation which one quantity is conceived to bear 
 to another in regard to their composition, is named a ra- 
 tio. 
 
 8. When both terms of comparison are equal, it is call- 
 ed a ratio of equality ; if the first of these be greater than 
 
BOOK V. 129 
 
 the second, it is a ratio of majority ; apd if the first be less 
 than the second, it is a ratio of minority, 
 
 9. A proportion or analogy consists in the identity of ra- 
 tios. 
 
 10. Four quantities are said to he proportional^ when a 
 submultiple of the first is contained in the second as often 
 as a hke submultiple of the third is contained in the fourth. 
 
 1 1 . Of proportional quantities, the first of each pair is 
 named the antecedent^ and the second the consequent, 
 
 12. The antecedents are homologous terms; and so are 
 the consequents. 
 
 13. One antecedent is said to be to its consequent as 
 another antecedent to its consequent. 
 
 14. The first and last terms of a proportion are called 
 the extremes^ and the intermediate ones, the means. 
 
 15. A ratio is direct^ if it follows the order of the terms 
 compared ; it is inverse or reciprocal^ when it holds a re- 
 versed order. 
 
 Thus, if the ratio of A to B be direct^ that of B to A is the 
 inverse or reciprocal ratio. 
 
 16. Quantities form a continued proportion, when the 
 intervening terms stand in the double relation of conse- 
 quents and antecedents. 
 
130 ELEMENTS OF GEOMETRY. 
 
 17. When a proportion consists of three terms, the 
 middle one is said to be a mean proportional between the 
 two extremes. 
 
 18. The ratio which one quantity has to another may 
 be considered as compounded of all the connecting ratios 
 among any interposed quantities. 
 
 Thus, the ratio of A to D is viewed as compounded of that of 
 A to B, that of B to C, and that of C to E). 
 
 19. Of quantities in a continued proportion, the first is 
 said to have to the third, the duplicate ratio of what it 
 has to the second; to have to the fourth, a triplicate ratio ; 
 to the fifth, a quadruplicate ratio ; and so forth, according 
 to the number of ratios introduced between the extreme 
 terms. 
 
 20. If quantities be continually proportional, the ratio 
 of the first to the second is called the suhduplicate of the 
 ratio of the first to the third, the suhtriplicate of the ratio 
 of the first to the fourth, &c, 
 
 To facilitate the language of demonstration relative to 
 numbers or abstract quantities, it is expedient to adopt a 
 clear and concise mode of notation. 
 
 1, The sign = expresses equality^ t^ majority^ and .^^n 
 minority: Thus A = B denotes that A is equal to B, 
 
BOOK V. 131 
 
 A'liP'B signifies that A is greater than B, and A.,^^ im- 
 ports that A is less than B. 
 
 2. The signs + and — mark the addition and subtrac- 
 tion of the quantities to which they are prefixed : Thus, 
 A-f-B denotes that B is to be joined to A, and A — B sig- 
 nifies that B is to be taken away from A. Sometimes 
 these two symbols are combined together : Thus, Ar±=B 
 represents either the sum of A and B, or the excess of A 
 above B. 
 
 3. To express multiplication, the quantities are placed 
 close together ; or they may be connected by the point (.), 
 or the cross X : Thus, AB, or A.B, or A X B, denotes the 
 product of A by B ; and ABC indicates the result of the 
 continued multiplication of A by B, and of this product 
 again by C. 
 
 4. When the same number is repeatedly multiplied, the 
 product is termed its iponsoer ,- and the number itself, in re- 
 ference to that power, is called the root. The notation is 
 here still farther abridged, by retaining only a single letter 
 wfth a small figure over it, to mark how often it is under- 
 stood to be repeated : This figure serves also to distinguish 
 the order of the power. Thus AA, or A% signifies that 
 A is multiplied by A, and that the product is the second 
 poxioer of A j and AAA, or A^, in like manner, imports 
 that AA is again multiplied by A, and that the result is 
 the third power o^ K. 
 
 5. The roots are denoted, by prefixing a contracted t 
 
 or the symbol V , Thus VA or VA marks the second 
 root of A, or that number of which A is the second power ; 
 
 V A signifies the third y^oot of A, or the number which 
 has A for its third power. 
 
132 ELEMENTS OF GEOMETRY. 
 
 6. To represent the multiplication of complex quantities, 
 they are included by a parenthesis. Thus, A(BH-C — D) 
 denotes that the amount of B + C — D, considered as a 
 single quantity, is multiplied into A. 
 
 7. Ratios and analogies are expressed, by inserting points 
 in pairs between the terms. Thus A : B denotes the ra- 
 tio of A to B i and the compound symbols A : B : ; C ; D,. 
 signify that the ratio of A to B is the same as that of C to 
 D, or that A is to B as C to D. 
 
PROP. I. THEOR. 
 
 The pi^oduct of a number into the sum or dif- 
 ference of two numbers, is equal to the sum or 
 difference of its products by those numbers. 
 
 Let A, B, and C be three numbers ; the product of the 
 sum or difference of B and C by the number A, is equal 
 to the sum or difference of the separate products AB and 
 AC. 
 
 For the product AB is the same as each unit contained 
 in B repeated A times, and the product AC is the same as 
 the units in C likewise repeated A times ; whence the sum 
 of the products AB and AC is equal to the units contain- 
 ed in both B and C, all repeated A times, or it is equal to 
 the sum of the numbers B and C multiplied by A. 
 
 Again, for the same reason, the difference between the 
 products AB and AC must be equal to the difference be- 
 tween the units contained in B and in C, repeated A times ; 
 that is, it must be equal to the difference between the num- 
 bers B and C multiplied by A. 
 
 Cor, 1. Hence a number which measures any two num- 
 bers, will measure also their sum and their difference. 
 
 Cor, 2. It is hence manifest, that the first part of the 
 proposition may be extended to more numbers than two ; 
 or that AB+AC-I-AD+, &c. = A(B+C+D+, &c.) 
 
134/ ELEMENTS OF GEOMETRY. 
 
 PROP. Ii: THEOR. 
 
 The product which arises from the continued 
 multiplication of any numbers, is the same in what- 
 ever order this operation be performed. 
 
 Let A and B be two numbers ; the product AB is equal 
 to BA. 
 
 For the product AB is the same as each unit in B add- 
 ed together A times, that is, the same as A itself repeated 
 B times, or BA. 
 
 Next, let there be three numbers A, B, and C; the 
 products ABC, ACB, BAG, BCA, CAB, and CBA are 
 all equal. 
 
 For put D = AB or BA ; then DCzrCD, that is, ABC 
 = CAB, and BAC = CBA. 
 
 Again, put E = AC or'CA; then EB = BE, that is, 
 ACB = BAC, and CAB = BCA. 
 
 Lastly, put F = BC or CB ; then FA = AF, that is, 
 BCA = ABC, and CBA = ACB. 
 
 And thus the several products are all mutually equal. 
 
 It is also manifest, that the same mode of reasoning 
 might be extended to the products of any multitude of 
 numbers. 
 
 PROP. IIL THEOR. 
 
 Homogeneous quantities are proportional to 
 their like multiples or submultiples. 
 
BOOK V. 
 
 "^155 
 
 het A, B be two quantities of the same kind, andj^A, 
 j^B their Hke multiples ; then A : B : : pA : ^B. 
 
 For, since A and B are capable of being measured to 
 any required degree of precision, suppose a to be the mea- 
 sure which A and B contain m and 7i times, or that A = 
 9n.a and Bzzn.a ; consequently pA=p.via, and p3=p,na. 
 But (V. 2.) p.ma = m.pa, and p,na=i7i.pa. Wherefore a 
 andi pa are like submultiples of A and of pA, which con- 
 tain them respectively m times ; and these like submultiples 
 are both contained equally, or n times, in B and in /?B. 
 Consequently (V. def. 10.) the quantities A, B, and pA, 
 ^B are proportional ; and A, /;A are the antecedents, and 
 B, ^B, the consequents, of the analogy. 
 
 Again, because the ratio of pA. to ^B is thus the same 
 as that of A to B, which, in reference to pA and^B, are 
 only like submultiples, it follows that homogeneous quanti- 
 ties are also proportional to their like submultiples. 
 
 PROP. IV. THEOR. 
 
 In proportional quantities, according as tlie 
 first term is greater, equal, or less than the se- 
 condi the third term is greater, equal, or less than 
 the fourth. 
 
 Let A: B : : C: D; if A^^B, then C:::^D ; if A = B, 
 then C = D ; or if A.^B, then C^-nD. 
 
 For, if A be greater than B, then the measure or submul- 
 tiple of A must be contained oftener in B, and hence the like 
 submultiple of C will be contained oftener in D ; where- 
 fore C is greater than D. 
 
IS»6 ELEMENTS OF GEOMETRY. 
 
 If A be equal to B, the measure of A is contained e- 
 qually in B, and hence that of C in D, or C is equal to D. 
 
 But, if A be less than B, the measure of A is not con- 
 tained so often in B, and hence the measure of C is not 
 contained so often in D, or C is less than D. 
 
 Scholium, On this proposition is grounded the mode of 
 stating a proportion in the Rule of Three, while the arith- 
 metical operation will be found to depend on Prop. VI. 
 
 PROP. V. THEOR. 
 
 Of four proportionals, if the first be a multiple 
 or submultiple of the second, the third is a like 
 multiple or submultiple of the fourth. 
 
 Let A : B : : C : D ; if A=:;?B, then C=^D. 
 
 For, suppose the approximate measures of A and C to 
 be a and c, and let Kzzzmp.a^ and C = »zp.r, It is evident, 
 from the hypothesis, that A=:j)'B=:mp,a, or B = W2.ff ,• but 
 the consequents B and D must contain their measures 
 equally (V. def. 10. )> and therefore D=»2.c. Whence C 
 =zmp.czz(y, 2,) p,mc=zp'D. 
 
 Again, if 5'A = B; then will qC = T>, 
 
 For, let A = na, and C^=wc; therefore B = g'A=rgrw«= 
 (V. 2.) nq.af and, from the definition of proportion, D = 
 nq.c=z (V. 2.) q.7lc:=:qC, 
 
BOOK V. 1S7 
 
 PROP. VI. THEOR. 
 
 If four numbers be proportional, the product 
 of the extremes is equal to that of the means j 
 and of two equal products, the factors are con- 
 vertible into an analogy, of which these form se- 
 verally the extreme and the mean terms. 
 
 Let A : B : : C ; D-, then AD = BC. 
 
 For (V. 3.) A.D : B.D : : B-C : B.D ; and the second 
 term of this analogy being equal to the fourth, therefore 
 {V. 4.) AD = BC. 
 
 Again, let AD = BC ; then A : B : : C r D. 
 
 For, by identity of ratios, AD : BD : : BC : BD, and 
 hence (V. 3.) A : B : : C : D. 
 
 Cor. 1. Hence the greatest and least terms of a propor- 
 tion, are either extremes or means. 
 
 Cor. 2. Hence also a proportion is not affected, by trans- 
 posing or interchanging its extreme and mean terms. On 
 
 this principle are founded the two following theorems. 
 
 PROP. VII. THEOR. 
 
 The terms of an analogy are proportional by 
 inversion^ or the second is to the first, as the fourth 
 to the third. 
 
 Let A : B ; : C : D ; then imersely B : A ; : D : C. 
 
 For the extreme and mean terms are thus only mutual* 
 ly interchanged, and consequently the same equality of 
 products AD and BC still obtains. 
 
138 ELEMENTS OF GEOMETRY. 
 
 PROP. VIII. THEOR. 
 
 Numbers are proportional by alternation^ or the 
 iirst is to the third, as the second to the fourth. 
 
 Let A : B : : C : D J then alternately A : C : : B : D. 
 
 For the extreme terms being still retained, the mean 
 terms are merely transposed with respect to each other ; 
 the same equality of products, therefore, also here subsists. 
 
 I»ROP. IX. THEOR. 
 
 The terms of an analogy are proportional by 
 composition ; or the sum of the first and second is 
 to the second, as the sum of the third and fourth 
 to the fourth. 
 
 Let A : B : : C : D ; then by composition A + B : B : : 
 C+D : D. 
 
 Because A : B : : C : D, the product AD = BC (V. 6.) ; 
 add to each of these the product BD, and AD + BD = 
 BC + BD. But (V. 1.) AD + BD = D(A+B), and 
 BC + BD = B(C+D) ; wherefore (V. 6.) assuming the fac- 
 tors of these equal products for the extreme and mean 
 terms, A+B : B : : C+D : D. 
 
BOOK V. isy 
 
 PROP. X. THEOR. 
 
 The terms of an analogy are proportional by 
 division ; or the difference of the first and second 
 is to the second, as the difference of the third and 
 fourth to the fourth. 
 
 Let A : B : : C : D ; suppose A to be greater than B, 
 then will C be greater than D (V. 4.) : It is to be proved 
 that A— B : B : : C— D : D. 
 
 For, since A : B : : C : D, the product AD=:BC 
 (V. 6.), and, taking BD from both, the compound product 
 AD — BD is equal to BC — BD; wherefore* by resolution,. 
 (A— B)D = B(C-D), and consequently A—B : B : : 
 C— D : D. 
 
 If B be greater than A, then BD— AD = BD— BC, and, 
 by resolution, (B— A) D = B (D— C) ; whence B— A : B 
 : : D-C : D. 
 
 PROP. XI. THEOR. 
 
 The terms of an analogy are proportional by 
 tonversioii ; that is, the first is to the' sum or dif- 
 ference of the first and second, as the third to the 
 sum or difference of the third and fourth. 
 
 Let A : B : : C : D, and suppose A^^B 5 then A : A=i=B 
 : : C : C=i=:D. 
 
 For, since (V. 6.) the product AD = BC, add or sub- 
 
140 :fiLEMENTS OF GEOMETRY. 
 
 stract these to or from the product AC ; and AC=t:AD 
 = AC=i=BC. Wherefore, by resolution, A(C=t:D) = 
 C(A=±=B), and consequently A : A=1=B : : C : C=±:D. 
 
 If A--^B, then AD—AC = BC— AC, and, by resolu- 
 tion, A(D— C)=rC(B— A), whence A : B— A: : C: D-C 
 
 Cor, Hence, by inversionj A=t:B : A : : C=izD : C, or 
 B--A : A : : D— C : C. 
 
 PROP. XII. THEOR. 
 
 The terms of an analogy are proportional by 
 mixing ; or the sum of the first and second is to 
 the difference, as the sum of the third and fourth 
 to their difference. 
 
 Let A : B : : C : D, and suppose kr^^ ; then A + B : 
 A—B::C + D:C— D. 
 
 For, by conversion, A : A + B : : C : C + D, and alter- 
 nately A : C : : A + B : C + D. 
 
 Again, by conversion, A : A — B : : C : C — D, and al- 
 ternately A : C : : A — B : C — D. Whence, by identity 
 of ratios, A + B : C + D : : A— B : C~ D, and alternately 
 A + B : A— B : : C + D : C--D. 
 
 The same reasoning will hold if A be less than B, the 
 order of these terms being only changed. 
 
BOOK V. HI 
 
 PKOP. XIII. THEOR. 
 
 A proportion will subsist, if the homologous 
 terms be multiplied by the same numbers. 
 
 Let A : B : : C : D 5 then pK : qBi-.pC : qD. 
 
 For, since A : B : : C : D, alternately A : C : : B : D; 
 but the ratio of A to C is the same as pA : pC (V. 3.), 
 and the ratio of B to D is the same as qB : ^D Where- 
 fore^ A : pC : : ^^B : g'D, and, by alternation, jpA : g'B : : 
 ^C : ^D. 
 
 Car. The Proposition may be extended likewise to the 
 division of homologous terms, by employing submultiples. 
 
 PROP. XIV. THEOR. 
 
 The greatest and least terms of a proportion, are 
 together greater than the intermediate ones. 
 
 Let A : B : : C : D ; and A being supposed to be the 
 greatest term, the other extreme D is the least (V. 6. cor. 
 1.) : The sum of A and D is greater than the sum of B 
 and C. 
 
 Because A : B : : C : D, by conversion A : A — B : ; 
 € : C— D, and alternately A : C : : A— B : C— D; but 
 A, being the greatest term, is therefore greater than C, 
 and consequently (V. 4.) A — B is greater than C — D ; to 
 each add B+D, and A+D is greater than B + C. 
 
342 ELEMENTS OF GEOMETRY. 
 
 The same reasoning is. applicable, if any other term of 
 the analogy be supposed the greatest. 
 
 Cor, Hence the mean term of three proportionals, is less 
 than half the sum of both extremes. 
 
 PROP. XV. THEOR. 
 
 If two analogies have the same antecedents, 
 another analogy may be formed, having the con- 
 sequents of the one for its antecedents, and the 
 consequents of the other for its consequents. 
 
 Let A : B : : C : Dand A : E : : C : F ; then B : E : : 
 D : F. 
 
 For, alternating the first analogy, A : C : : B : D, and 
 alternating the second, A : C : : E : F ; whence, by iden- 
 tity of ratios, B : D : : E : F. This inference is named a 
 direct equality. 
 
 PROP. XVI. THEOR. 
 
 If the consequents of one analogy be antece- 
 dents in another, a third analogy will arise, ha- 
 ving the same antecedents as the former, and the 
 same consequents as the latter. 
 
 Let A : B : : C : D, and B : E : : D : Fj then A : E : : 
 C: F. 
 
 For, alternating both analogies, A : C : : B : D, and 
 
BOOK V. 143 
 
 B : D : : E : F; whence, by identity of ratios, A : C : ; 
 E : F. This conclusion is also named a direct equality. 
 
 PROP. XVII. THEOR. 
 
 If two analogies have the same means, the ex- 
 tremes of the one, with those of the other as the 
 mean terms, will form a third analogy. 
 
 Let A:B: : C:D,andE: B;: C: F;thenA;E: : 
 F: D. 
 
 For, since A : B : : C : D, AD = BC (V. 6.) ; and be- 
 cause E : B : : C : F, EF = BC. Whence AD = EF, and 
 A : E : : F : D. 
 
 Co7\ Hence the extreme and mean terms being inter- 
 changeable, it likewise follows, that, if A : B : : C : D, and 
 ,A:E::F:D, thenBiE:: F:C. 
 
 PROP. XVIII. THEOR. 
 
 If the extremes of one analogy are the mean 
 terms in another, a third analogy will subsist, ha- 
 ving the means of the former as its extremes, and 
 the extremes of the latter as its means. 
 
 Let A : B : : C : D, and E : A : : D : F; then B : E : : 
 F:C. 
 
 For, from the first analogy AD=:BC, and, from the ser- 
 
Hi- ELEMENTS OF GEOMETRY. 
 
 cond, EF=AD ; whence BC = EF, and consequently 
 B : E : : F : C. 
 
 Cor, Hence also, if A : B : : C : D and B : E : : F : C j 
 then E : A ; : D : F. The principle of this and the pre- 
 ceding Proposition is named inverse, or perturbatey eqiia- 
 lity, ■ 
 
 PROP. XIX. THEOR. 
 
 If there be any number of proportionals, as one 
 antecedent is to its consequent, so is the sum of 
 all the antecedents to the sum of all the conse^ 
 quents, 
 
 Let A : B : : C : D : : E : F : : G : H; then A : B : : 
 A+C+E+G : B+D+F+H. 
 
 Because A : B : : C : D, (V. 6.) AD = BC; and, since 
 A : B : : E : F, AF=BE, and, for the same reason, AH 
 = BG. Consequently, the aggregate products, AB + AD 
 + AF+AH=BA+BC + BE4-BG; and, by resolution, 
 A(B + D + F+H) = B(A + C+E+G;; whence A : B : : 
 A + C+E+G: B+D+F + H. 
 
 C(yi\ I. It is obvious, that the Proposition will extend 
 likewise to the difference of the homologous terms, and 
 may, therefore, be more generally expressed thus : A : B : : 
 A=±=CdtE=±=G : B=i=D=±=F=t=H. 
 
 Cor, 2. Hence, in continued proportionals, as one ante- 
 cedent is to its consequent, so is the sum or difference of 
 the several antecedents^ to the corresponding sum or diffe- 
 rence of the consequents. For, if A : B ; ; B : C ; : C : 
 
BOOK V. 145 
 
 then, by corollary 1, A : B : : AdtB=i=E : BdfcDdfcF, or 
 (V. 8.) A : Adt:C=i=E : : B : B=t=D=i=F; wherefore 
 (V. 11.) A : C=S=E : : B : D=4=F, and (V. 8.) A : B : : 
 C=t:E : D=i=F. 
 
 PROP. XX. THEOR. 
 
 If two analogies have the same antecedents, 
 another analogy may be formed of these antece- 
 dents, and the sum or diiFerence of the conse- 
 quents. 
 
 LetA:B::C:D, and A:E: : C : F; then A: BdfcE 
 : : C : D=±=F. For, by alternation, these analogies become 
 A : C : : B : D, and A : C : : E : F •, whence (V. 19. 
 cor. 2.) A : C : : B=i=E : D=fcF, and alternately, A : B=±:E 
 : : C : D=±zF. 
 
 Cor. If A : B : : C : D, and E : B : : F : D -, then 
 A=i=:E : B : : C=t:F : D. For, by alternating the analo- 
 gies, A : C : : B : D, and E : F : : B : D ; whence 
 (V. 19. cor. 2.) B : D : : A=t=E : C=±=F, and, by alterna- 
 tion and inversion, A=±=E : B : : C=i=F : D. 
 
 PROP. XXI. THEOR. 
 
 In continued proportionals, the difference be- 
 tween the first and second is to the first, as the 
 difference between the first and last terms to the 
 sum of all the terms excepting the last. 
 
14f6 Elements of geometry. 
 
 Let A : B : : B : C : : C : D : : D : E; then if A-::^B, 
 A— B : A : : A—E : A + B+C + D. 
 
 For (V. 19.), A : B : : A+B+C+D : B+C+D+E, 
 and consequently (V. 11. cor.), A— B : A :: (A + B + C+D) 
 ^(B + C+D+E) : A+B + C+D; that is, omitting 
 B+C+Dinthe third term, A-B:A::A-E: A + B+C+D. 
 
 If A^B, then B— A : A : : (B+C+D+E)— (A + B 
 + C+D) : A+B+C+D, that is, B— A : A : : E— A : A 
 + B+C+D. 
 
 The same reasoning, it is evident, will hold for any num- 
 ber of terms. 
 
 Scholium. Hence the summation of continued progres- 
 sions, whether ascending or descending, is easily derived. 
 
 PROP. XXII. THEOR. 
 
 The products of the like terms of any numeri- 
 cal proportions, are themselves proportional. 
 
 
 Let A 
 
 :B: 
 
 :C: 
 
 :D 
 
 
 
 E: 
 
 :F:: 
 
 ;G; 
 
 :H 
 
 
 
 I: 
 
 K:: 
 
 :L: 
 
 :M; 
 
 then AEI ; 
 
 :BFK 
 
 : : CGL 
 
 :DHM. 
 
 For (V. 6.), from the first analogy AD = BC, from the 
 second analogy EH = FG, and from the third analogy IM 
 = KL; whence the compound product AD.EH.IMi= 
 BC.FG KL. But AD.EH.IM= AEI.DHM (V. 2.), and 
 BC.FG.KL = BFK.CGL ; wherefore AEI.DHM = 
 BFK.CGL, and consequently (V. 6.), AEI : BFK : : 
 CGLiDHM. 
 
BOOK v: 147 
 
 The sAme reason, it is obvious, will apply to any number 
 of proportionals. 
 
 Cor. 1. Hence the powers of the successive terms of nu- 
 merical proportions, are likewise proportional. For, if A : 
 B : : C : D, and, repeating the analogy, A ; B : : C : D ; 
 then, by multiplication, AA : BB : : CC : DD, or A* : 
 B» : : C* : D^ 
 
 Again, let A : B : : C : D, and, repeating the analogy, 
 A:Br:C:D, 
 and A : B : : C : D i whence, by multiplying the 
 corresponding terms, 
 
 AJ : B» :: CJ : D. 
 And so the induction may be pursued generally. 
 
 Cor. 2. Hence also the roots of the terms of a numeri- 
 cal proportion, are proportional. If A : B : : C : D, then 
 VA: VB: : VC: VD. For let VA:>/B:: VC: VE, 
 and, by the last corollary, A : B : : C : E ; but A : B 
 : : C : D, whence C : E : : C ; D, and consequently 
 E = D, or VA : VB : : VC : VD. — In the same man- 
 ner, it may be shewn in general that, if A : B : : C : D, 
 VA: VB:; VC: VD. 
 
 PROP. XXIII. THEOR. 
 
 The ratio which is conceived to be compounded 
 of other ratios, is the same as that of the products. 
 of their corresponding numerical expressions. 
 
 Suppose the ratio of A : D is compounded of A : B, of 
 B : C, and of C : D, and let A : B : : K : L, B : C : : 
 
148 ELEMENTS OF GEOMETRY. 
 
 M : N, and C : D : : O : P ; then will A : D : : 
 KMO : LNP. 
 
 For, since A : B : : K : L, 
 B : C : : M : N, 
 and C : D : : O : P, 
 the products of the similar terms are proportional (V. 22.), 
 or ABC : BCD:: KMO : LNP. But A :D: : ABC: BCB 
 (V. 3.), and consequently A : D : : KMO : LNP. 
 
 The same mode of reasoning is applicable to any num- 
 ber of component ratios. 
 
 PROP. XXIV. THEOR. 
 
 A duplicate ratio is the same as the ratio of the 
 second powers oT the terms of its numerical ex- 
 pression, and a triplicate ratio is the same as that 
 of the third powers of those terms. 
 
 Let A : B : : B : C : : C : D ; then A* : B* : : A : C, 
 
 and A^ : B^ : : A : D. 
 For, since A : B : : B : C, 
 
 and A : B : : A : B, the products of the corre- 
 sponding terms are proportional (V. 22.), or A* : B* : : 
 BA : CB. Whence (V. 3.) A* : B^ : : A : C. 
 Again, since A : B : : B : C, 
 and A : B : : C : D, 
 and A : B : : A : B, as before, (V. 
 22.), A' : B^ : BCA : CDB. And consequently (V. 3.) 
 A':B^::A:D. 
 
BOOK V. 
 
 149 
 
 A 
 
 T> 
 
 
 
 
 
 
 
 
 
 
 
 
 
 J3 
 
 
 
 C 
 
 oD.- 
 
 PROP. XXV. THEOR. 
 
 The product of the numbers expressing the sides 
 of a rectangle, will represent its quantity of sur- 
 face, as measured by a square described on the li- 
 near unit. 
 
 Let ABCD be a rectangle and OP the linear measure ; 
 and suppose the side AB to contain OP, m times, and the 
 side BC to contain it, n times* 
 Divide these sides accordingly 
 (1. 36.), and, through the points 
 of section, draw straight lines 
 (I. 23.) parallel to AD and 
 DC : the whole rectangle will 
 thus be divided into cells, each 
 
 of them equal to the square of OP. It is evident, that there 
 stand on BC, w columns, and that each of these columns 
 contains, m cells ; consequently the entire space includes, 
 m.n cells, or is equal to the square of OP repeated »m times. 
 
 Cor. 1. If »2 = w, then AB = BC, and the rectangle be- 
 comes a square ; but 7nn is in that case equal to 7iny or »*, 
 Whence the surface of a square is expressed by the second 
 power of the number denoting its side. 
 
 Cor. 2. Rectangles which have the same altitude m are 
 as their bases n and p ; for (V. 3.) mn : mp : : n : p. And 
 triangles having the same altitude, being (II. 5. cor.) the 
 halves of these rectangles, must likewise be as their bases. 
 
 Cor, 3. If two rectangles be equal, their respective sides 
 are reciprocally proportional, or form the extremes and 
 means of an analogy. For if mn ^pq^ then (V. 6.) w : jp : : 
 q '. n. 
 
X 50 ELEMENTS OP GEOMETRY. 
 
 PROP. XXVL PROB. 
 
 Given two homogeneous quantities^ to find, if 
 possible, their greatest common measure. 
 
 Let it be required to find the greatest common measure, 
 which two quantities A and B, of the same kind, will admit. 
 
 Supposing A to be greater than B, take B out of A, till 
 the remainder C be less than it ; again, take C out of B> 
 till there remain only D ; and continue this alternate ope- 
 ration, till the last divisor, suppose E, leave no remainder 
 whatever ; E is the greatest common measure of the quan- 
 tities proposed. 
 
 For, the quantity sought, as it measures B, will measure 
 its multiple ; and since it also measures A, it must mea- 
 sure the difference between the multiple of B and A 
 (V. l.cor. 1.), that is, C; the required measure, there- 
 fore, measures the multiple of C, and consequently the 
 difference of this multiple and B, which it measured,— that 
 is D : And lastly, this measure, as it measures the multiple 
 of D, must consequently measure the difference of this from 
 
 C, or it must measure E. Supposing the decomposition 
 to terminate here, the common measure of A and B, since 
 it measures E, must be E itself; and it is also the greatest 
 possible measure, for nothing greater than E can be con- 
 tained in this quantity. 
 
 By retracing the steps likewise, it might be shown, that 
 E actually measures, in succession, all the preceding terms 
 
 D, C, B, and A. 
 
 If the process of decomposition should never terminate, 
 the quantities A and B do not admit of a common mea- 
 
BOOK V. 151 
 
 sure,— or they are incommensurable. But, as the residue of 
 the subdivision is necessarily diminished at each step of this 
 operation, it is evident that some element may always be 
 discovered, which will measure A and B nearer than any 
 ^issignable limit. 
 
 PROP. XXVII. PROB. 
 
 To express by numbers, either exactly or ap- 
 proximately, the ratio of two given homogeneous 
 quantities. 
 
 Let A and B be two quantities of the same kind, whose 
 numerical ratio it is required to discover. 
 
 Find, by the last proposition, the greatest common mea- 
 sure E of the two quantities ; and let A contain this mea- 
 sure K times, and B contain it I^ times : Then will the ra- 
 tio K : L express the ratio of A : B. 
 
 For the numbers K and L severally consist of as many 
 units, as the quantities A and B contain their measure E. 
 It is also manifest, since E is the greatest possible divisor, 
 that K and L are the smallest numbers capable of express- 
 ing the ratio of A to B. 
 
 If A and B be incommensurable quantities, their decom- 
 position is capable at least of being pushed to an unlimited 
 extent ; and, consequently, a divisor can always be found 
 so extremely minute, as to measure them both to apy de- 
 gree of precision. 
 
152 ELEMENTS OF GEOMETRY. 
 
 PROP. XXVIII. THEOR. 
 
 A straight line is incommensurable with its seg- 
 ments formed by medial section. 
 
 If the straight line AB be cut in C, such that the rect- 
 angle AB, BC is equivalent to the square of AC ; no part 
 of AB, however small, will measure the segments AC, BC. 
 
 For (V. 26.) take AC out 
 
 of AB, and again the re- a T'f V C ^ 
 
 mainder BC out of AC. But 
 
 AD, being made equal to BC, the straight line AC is like- 
 wise divided in D, by medial section (11. 19. cor. 1.); and, 
 for the same reason, taking away the successive remainders 
 CD, or AE, from AD, and DE or AF from AE, the sub- 
 ordinate lines AD and AE are also divided medially in 
 the points E and F. This operation produces, therefore, 
 a series of decreasing lines, all of them divided by medial 
 section : Nor can such a process of decomposition ever ter- 
 minate ; for though the remainders BC, CD, DE, and 
 EF continually diminish, they must still constitute the seg_ 
 ments of a similar division. Consequently there exists no 
 final quantity capable of measuring both AB and ACc 
 
 Cor, Since (V. 6. and V. 24.) the whole line is to its 
 smaller segment in the duplicate ratio of the same line to 
 its greater segment, it evidently follows that the squares of 
 the parts of a line divided by medial section are likewise 
 mutually incommensurable. 
 
BOOK V. 
 
 153 
 
 PROP. XXIX. THEOR. 
 
 The side of a square is incommensurable with 
 its diagonal. 
 
 Let ABCD be a square and AC its diagonal; AC and 
 AB are incommensurable. 
 
 For make CE equal to AB or BC, draw (I. 5. cor.) the 
 perpendicular EF, and join BE. 
 
 Because CE is equal to BC, the angle CEB (I. 10.) is 
 equal to CBE ; and since CEF and CBF are right angles, 
 the remaining angle BEF is e- 
 qual to EBF, and the side EF 
 (I. 11.) equal to BF; butEFis 
 also equal to AE, for the angles 
 EAF and EFA of the triangle 
 AEF are evidently each of them 
 half a right angle. Whence, 
 making FH equal to FB, FE 
 or AE, — the excess AE of the 
 
 diagonal AC above the side AB, is contained twice in AB, 
 with a remainder AH ; and AEl again, being the excess of 
 the diagonal AF of the derived or secondary square GE 
 above the side AE, must, for the same reason, be contain- 
 ed twice in AG, with a new remainder AL ; and this re- 
 mainder will likewise be contained twice with a correspond- 
 ing remainder in AH, the side of the ternary square KH. 
 This process of subdivision is, therefore, interminable, and 
 the same relations are continually reproduced. 
 
ELEMENTS 
 
 or 
 
 GEOMETRY. 
 
 BOOK VL 
 
 The doctrine of Proportion, grounded on the 
 simplest theory of numbers, furnishes a most 
 powerful instrument, for abridging and extending 
 mathematical investigations. It easily unfolds 
 the primary relations subsisting among figures, and 
 those of the sections of lines and circles ; but it 
 also discloses with admirable felicity that vast con- 
 catenation of general properties, not less impor- 
 tant than remote, which, without such aid, might 
 for ever have escaped the penetration^ of the geo- 
 meter. The application of Arithmetic to Geome- 
 try forms, therefore, one of those grand epochs 
 which occur, in the lapse of ages, to mark and 
 accelerate the progress of scientific discovery. 
 
156 JILEMENTS OF GEOMETRY, 
 
 DEFINITIONS. 
 
 1. Straight lines drawn from the Same point, are termed 
 diverging lines. 
 
 2. Straight lines are divided similarli/f when their cor- 
 responding segments have the same ratio. 
 
 3. A straight line is cut in extreme and mean ration when 
 the one segment is a mean proportional between the other 
 segment and the whole line. 
 
 4. A straight line is said to be cut harmonically, if it 
 consist of three segments, such that the whole line is to 
 one extreme, as the other extreme to the middle part. 
 
 5. The area of a figure is the quantity of space which 
 its surface occupies. 
 
 6. Similar figures are such as have their angles respec- 
 tively equal, and the containing sides proportional. 
 
 7. If two sides of a rectilineal figure be the extremes of 
 an analogy, of which the means are two corresponding 
 sides in another rectilineal figure ; those figures are said 
 to have their sides reciprocally proportional. 
 
BOOK VI. 157 
 
 PROP. I. THEOR. 
 
 Parallels cut diverging lines proportionally. 
 
 The parallels DE and BC cut the diverging lines AB 
 and AC into proportional segments. 
 
 Those parallels may lie on the same side of the vertex, 
 or on opposite sides ; and they may consist of two, or of 
 more straight lines. 
 
 1. Let the two parallels DE and BC intersect the di- 
 verging lines AB and AC, on the same side of the vertex 
 A ; then are AB and AC cut proportionally, in the points 
 D and E,~or AD : AB : : AE : AC. 
 
 For if AD be commensurable with AB, find (V. 26.) 
 their common measure M, which repeat from the vertex A 
 to B, and, from the corresponding points of section in AD 
 and AB, draw (I. 23.) the parallels FI, GK, and HL. It 
 is evident, from Book I. Prop. 36. 
 that these parallels will also divide 
 the straight lines AE and AC equal- 
 ly. Wherefore the measure M, or 
 AF the submultiple of AD, is con- X v o -P H B 
 
 tained in AB, as often as AI, the m 
 
 like submultiple of AE, is contained in AC ; consequently 
 (V. def. 10.) the ratio of AD to AB is the same with that 
 ofAEtoAC. 
 
 But if the segments AD and AB be incommensurable, 
 they may still bie expressed numerically, to any required de- 
 gree of precision. For AD being divided (I. 36.) into e-r 
 qual sections, these parts, continued towards B, will, to- 
 gether with some residuary portion, compose the whole of 
 
158 ELEMENTS OF GEOMETRY. 
 
 AB. Let this division of AD extend through DB as far as h^ 
 and draw the parallel he. Let the parts of AD and AB he 
 again subdivided, and the corresponding residue will evi- 
 dently be diminished ; consequently, at each successive sub- 
 division, the terminating parallel he 
 will approximate perpetually to BC. 
 Wherefore, by continuing this pro- K 
 
 cess of exhaustion, the divided lines 
 Kb and Arwill approach their limits 
 AB and AC, nearer than^any finite 
 
 or assignable interval. Consequently, from the preceding 
 demonstration, AD : AB : : AE : AC. 
 
 And since AD : AB : : AE : AC, it follows, by conver- 
 sion (V. 11.), that AD ; DB ; : AE : EC, and again, by 
 composition (V. 9.), that AB : DB : : AC : EC. 
 
 2. Let the two parallels DE and BC cut the diverging 
 lines DB and EC, on opposite sides of A ; the segments 
 AB, AD have the same ratio with AC, AE, — or AB : AD ; : 
 AC : AE. 
 
 For, make AO equal to AD, AP to AE, and join OP. 
 The triangles APO and 
 AED, having the sides AO, 
 AP equal to AD, AE, and 
 the contained vertical an- 
 gle OAP equal to DAE, 
 are equal (L 3.), and con- 
 sequently the angle AOP 
 is equal to ADE ; but these 
 being alternate angles, the straight line OP (I. 22.) is pa- 
 rallel to DE or BC, and hence, from what was already de- 
 monstrated, AB : AO or AD : : AC : AP or AE. 
 
 And since AB : AD : : AC ; AE, by composition BD : 
 
BOOK VI. 159 
 
 AD : : CE : AE, and, by conversion, BD : AB : : CE : 
 AC. 
 
 3, Lastly, let more than two parallels, BC, DE, FH, 
 and GI, intersect the diverging lines AB and AC ; the seg- 
 ments DA, AF, FG, and GB, in DB, are proportional re- 
 spectively to EA, AH, HI, and IC, the corresponding 
 segments in EC. 
 
 For, from the second case, 
 AD : AF : : AE : AH ; and, 
 from the first case, AF : FG : : 
 AH : HI. But from the same 
 case, AG : FG : : AI : HI, 
 and AG : GB : : AI : IC; 
 whence (V. 15.) FG : GB : : 
 HI:IC. 
 
 Cor. 1 . Hence the converse of the proposition is also 
 true, or straight lines which cut diverging lines pro- 
 portionally are parallel ; for it would otherwise follow, that 
 a new division of the same line would not alter the rela- 
 tion among the segments, which is evidently absurd. 
 
 Cor, 2. Hence, if the segments of one diverging line be 
 equal to those of another, the straight lines which join them 
 are parallel. 
 
 PROP. II. THEOR. 
 
 t)iverging lines are proportional to the corre- 
 sponding segments into which they divide paral- 
 lels. 
 
160 
 
 ELEMENTS OF GEOMETEY. 
 
 Let two diverging lines 
 AB and AC cut the parallels 
 BC and DE; then AB : 
 AD : : BC : DE. 
 
 For draw DF parallel to 
 AC. And, by the last Pro* 
 position, the parallels AC 
 and DF must cut the straight 
 lines AB and BC proportion- 
 ally, or AB : AD :: BC : CF. 
 But CF is equal (I. 26.) 
 to the opposite side DE of 
 the parallelogram DECF; 
 and consequently AB : AD:; 
 BC : DE. 
 
 Next, let more than two diverging lines, AB, AF and 
 AC intersect the parallels BC and DE j the segments BF 
 and FC have respectively to DG 
 and GE the same ratio as AB has 
 to AD. 
 
 ' From what has been already 
 demonstrated, it appears, that 
 AB : AD : : BF : DG, and al- 
 so that AF : AG : : FC : GE. 
 But by the last Proposition, AB : 
 AD : : AF : AG ; wherefore AB : AD : : FC : GE. 
 The same mode of reasoning, it is obvious, might be ex- 
 tended to any number of sections. "Whence AB : AD : : 
 BF : DG : : FC : GE. 
 
 Cor. 1. Hence the straight lines which cut diverging lines 
 equally, being parallel (VI. 1. cor. 2.), are themselves pro- 
 portional to the segments intercepted from the vertex. 
 
BOOK VI. 161 
 
 Cor, 2. Hence parallels are cut proportionally by diver- 
 ging lines. 
 
 PROP. III. PROB. 
 
 To find a fourth proportional to three given 
 straight lines. 
 
 Let A, B, and C be three straight lines, to which it is 
 required to find a fourth proportional. 
 
 Draw the diverging lines 
 DG and DH^ make DE equal 
 to A, DF to B, and DG to 
 C, join EF, and through G 
 draw (I. 23.) GH parallel to 
 EF and meeting DH in H; 
 DH is a fourth proportional 
 to the straight lines A, B, and 
 C. 
 
 For the diverging lines DG and DH are cut propor- 
 tionally by the parallels EF and GH (VI. 1.), or DE : DF 
 : : DG : DH, that is, A : B^- : C : DH. 
 
 Cor, If the mean terms B and C be equal, it is obvious 
 that DG will become equal to DF, and that DH will be 
 jfbund a third proportional to the two given terms A and B. 
 
 PROP. IV. PROB. 
 
 To cut a given straight line into segments, 
 which shall be proportional to those of a divided, 
 straight line. 
 
162 ELEMENTS OF GEOMETRY. 
 
 Let AB be a straight line, which it is required to cut 
 into segments proportional to those of a given divided 
 straight line. 
 
 Draw the diverging line AC, 
 and make AD, DE, and EC, 
 equal respectively to the seg- 
 ments of the divided line, join 
 CB, and draw EG and DF 
 parallel to it (I. 23.) and meet- 
 ing AB in G and Fj AB is 
 cut in those points proportionally to the segments oi AC. 
 
 For the parallels DF, EG, and CB must cut the di- 
 verging lines AB and AC proportionally (VI. l.)> or 
 AF : FG : : AD : DE, and FG : GB : : DE ; EC. 
 
 PROP. V. PROB. 
 
 To cut off the successive parts of a given straight 
 line. 
 
 Let AB be a straight line, from which it is required to 
 cut off successively the half, th^e third, the fourth, the fifth, 
 Stc. 
 
 Oil AB describe (I. 23.) the rhomboid ABCD, and 
 through E, the intersection of its diagonals AC and BD, 
 draw EF parallel to AD, join DF, and through G, where 
 it cuts AC, draw GH likewise parallel to AD, again join 
 DH and draw the parallel IK, and so repeat the opera- 
 tion : Then will AF be the half of AB, AH the third, 
 AK the fourth, and AM the fifth part of it. 
 
 The triangles AED and CEB are equal (I. 20.), since 
 
BOOK vr. 
 
 163 
 
 KKH Jf 
 
 they have (1. 23.) the angles DAE and ADE equal to BCE 
 and CBE, and the interjacent sides AD and CB (I. 26.) 
 likewise equal ; and therefore DE = EB. But AD and EF 
 being parallel, DE : EB: : AF : FB (VI. 1.)'; whence (V. 4.) 
 AF = FB, or AF is the half of AB. And AD and EF being 
 intercepted parallels, AD : EF : : AB : BF (VI. 2.) ; con- 
 sequently, since AB is double of BF, AD is likewise double 
 
 ofEF(V.5.) Again, the ^y C 
 
 diverging lines AGE and 
 DGF are proportional to 
 the intercepted parallels 
 AD and EF (VI. 2.), or 
 AD : EF : : AG : GE ; 
 and GH being parallel to 
 EF, AG : GE : : AH : HF (VI. 1.), whence AD : EF : : 
 AH : HF; but AD was shown to be double of EF, where- 
 fore AH is double of HF (V. 5), or AH is two-thirds of 
 AF, or of the half of AB, and is consequently the third 
 part of the whole AB. Now, since AF : HF : : AD : GH, 
 (VI. 2.) and AFis triple of HF, it is evident that AD is triple 
 of GH; but AD : GH : : AI : IG : : AK : KH, and, AD 
 being triple of GH, AK must also be triple of KH ; or 
 AK is three-fourths of AH, which was proved to be the 
 third of AB, whence the segment AK is the fourth part of 
 the whole line AB. By a like process, it is shown that 
 AM is the fifth part of AB. 
 
 Cor. This construction likewise exhibits other portions 
 of the line AB. For, since AF is the half, and AH the 
 third, their difference FH must be the sixth part. Again, 
 AH and AK being the third and fourth parts, the inter- 
 val HK is the twelfth. In like manner, it is shown that 
 KM i§ the twentieth part of AB. 
 
164 
 
 ELEMENTS OF GEOMETRY. 
 
 / 
 
 
 1 
 
 
 J 
 
 "^ 
 
 X^:- 
 
 
 1 
 
 / 
 
 r 
 
 \ 
 
 ^h" "■■" 
 
 /K 
 
 Cr 
 
 i 
 
 PROP. VI. PROB. 
 
 To divide a straight line harmonically, and in a 
 given ratio. 
 
 Let AB be a straight line, which it is required to cut 
 harmonically, in the ratio of K to L. 
 
 Through A draw the diverging line AC, and produce 
 it both ways till AC and AD be each equal to K, make 
 AE equal to L, join 
 CB, draw EF parallel 
 to AB, and FG paral- 
 lel to CA, and join 
 DF cutting AB in H; 
 the straight line AB is 
 divided harmonically 
 in the points H and G, 
 such that K : L : : AB : 
 BG : : AH : HG. 
 
 For the parallels AC 
 and GF, being intercepted by the diverging lines AB and 
 CB, AC : GF : : AB : BG (VL2.). Again, the diverging 
 lines AG and DF are cut by the parallels AD and FG, 
 whence (VI. 2.) AD or AC : GF : : AH : HG. Where- 
 fore, AB : BG : : AH : HG; and each of these ratios is 
 the same as that of AC or AD to GF, or that of K to L. 
 
 Cor. Hence AG is divided, internally in H and exter- 
 nally in B, in the same ratio. In like manner, BH is di- 
 vided proportionally, by an external and internal section 
 in A and G ; for AB : BG : : AH : HG, and alternately 
 AB : AH : : BG : HG. 
 
BOOK V|. 165 
 
 PROP. VII. THEOR. 
 
 If a straight line be divided internally and ex- 
 ternally in the same ratio, half the line is a mean 
 proportional between the distances of the middle 
 from the two points of unequal section. 
 
 Let the straight line AB be divided in the same ratio, 
 internally and externally in C and D, and also be bisected 
 in E ; the half EB is a mean 
 
 proportional between EC and ■ )■ "-^ ^ ^f ^ 
 
 ED, or EC : EB : : EB : ED. 
 
 For since AC : CB : : AD : DB, by mixing and inver- 
 sion AC— CB : AC+CB : : AD~DB : AD+DB, that 
 is, 2EC : AB : : AB : 2ED, and, halving all the terms of 
 the analogy, (V. 3.) EC : EB : : EB : ED. 
 
 Cor, Hence if a straight line be cut internally and exter- 
 nally in the same ratio, the square of the interval between 
 the points of section is equivalent to the difference between 
 the rectangles under the internal and external segments. 
 For (II. 17.) AD.DB=ED^--EB% and AC.CB = EB*^ 
 EC^; consequently AD.DB— AC.CB = ED*— 2EB*+ 
 ECS or (V. 6. and VI. 7.) ED*— 2ED.EC + EC% which 
 (IL 16.) is the square of ED— EC or of CD. 
 
 PROP. VIII. THEOR. 
 
 If diverging lines divide a straight line harmo- 
 nically, they will cut every intercepted straight 
 line also in harmonic proportion. 
 
166 
 
 ELEMENTS OF GEOMETRY. 
 
 Let the diverging lines EA, EC, EB, and ED termi- 
 nate in the harmonic section of the straight line AD ; any 
 intercepted straight line FG will be likewise cut by them 
 harmonically, or FG : GI : : FH : HI. 
 
 For, through the points B and I, draw (I. 23.) KL and 
 MN parallel to AE. 
 
 Because the parallels AE and BL are intercepted by 
 the diverging lines DA and DE, AD : DB : : AE : BL 
 (YI. 2.) ; and for the same reason, the parallels AE and 
 BK being intercepted by the diverging lines AB and EK> 
 AC : CB : : AE : BK. 
 And since AD is divided 
 harmonically, AD : DB : : 
 AC : CB ; wherefore AE : 
 BL : : AE : BK, and con- 
 sequently (V. 8. and 4;.) BL 
 = BK. But, KL being 
 parallel to MN, BL ; BK 
 ::IN:IM(VI. 2. cor. 2.); 
 consequently, BL being 
 equal to BK, IN must also 
 be equal to IM; whence 
 FE : IN : : FE : IM. Again, FE : IN : : FG : GI, for 
 the parallels FE and IN are cut by the diverging lines GF 
 and GE ; and FE : IJVI : : FH : HI, since the parallels 
 FE and IM are cut by the diverging lines FI and EM. 
 Wherefore, by identity of ratios, FG : GI : : FH : HI; 
 or the intercepted straight line FG is cut harmonically in 
 the points H and I. 
 
BOOK VI. 167 
 
 PROP. IX. THEOR. 
 
 A straight line drawn from the concourse of 
 two tangents to the concave circumference of a 
 circle, is divided harmonically, by the convex cir^ 
 cumference and the chord which joins the points 
 of contact. 
 
 Let ED and FD be two tangents applied to the circle 
 AEBF; the secant DA, drawn from their point of con- 
 course, will be cut in harmonic proportion, by the convex 
 circumference EBF and the chord EF which joins the 
 points of contact, — or AD : DB : : AC : CB. 
 
 For the tangents ED and FD are equal (III. 22. cor.), 
 and EDF being thus an 
 isosceles triangle, DE* 
 = DC* + EC.CF (II. 
 20.); (but III. 26. cor. 2.) 
 DE* is also equal to 
 AD.DB, and the chords 
 ABandEF, by their mu- 
 tual intersection, make 
 
 the rectangle EC, CF equal to AC, CB. Whence 
 DC* = AD.DB— AC.CB, and therefore (VI. 7. cor.) 
 AC : CB : : AD : DB. 
 
 Cor. Hence by applying Prop. 7, it follows, that half 
 the chord AB is a mean proportional between the distan- 
 ces of its middle point from C and D ; and that, when AD 
 passes through the centre of the circle, the square of the 
 radius is equivalent to the rectangle under the distances of 
 the chord and of the intersection of the tangents from the 
 centre. 
 
168 ELEMENTS OP GEOMETRY. 
 
 PROP. X. THEOR. 
 
 A straight line which bisects, either internally 
 or externally, the vertical angle of a triangle, will 
 divide its base into segments, internal or external, 
 that are proportional to the adjacent sides of the 
 triangle. 
 
 Let the straight line BD bisect the vertical angle of the 
 triangle ABC ; it will cut the base AC into segments 
 which have the same ratio as the adjacent sides, or 
 AD : DC : : AB : BC. 
 
 For through C draw CE parallel to DB (I. 23.), and 
 meeting the production of AB in E. 
 
 Because DB and CE are parallel, the exterior angle 
 ABD is equal to BEC, and the 
 
 alternate angle DBC equal to J^ 
 
 BCE (I. 22.) > wherefore the 
 angle ABD being equal by hy- 
 pothesis to DBC, the angle 33/ 
 BEC is equal to BCE, and con- 
 sequently (I. 11.) the triangle 
 CBE is isosceles, or BE is e- 
 qual to BC. But the parallels 
 DB and CE cut the diverging 
 lines AC and AE proportionally (VI. 1.), or AD: DC : : 
 AB : BE ; that is, since BE = BC, AD : DC : : AB : BC. 
 
 Again, let the vertical line BD bisect the exterior angle 
 CBG of the triangle ; it will divide the base into external 
 segments AD and DC, which are also proportional to the 
 adjacent sides AB and BC. 
 
BOOK VI. 
 
 169 
 
 For through C draw CE parallel to DB, and meeting 
 AB in E. 
 
 The equal angles 
 GBD and DBC 
 are, from the pro- 
 perties of parallel 
 straight lines, re- 
 spectively equal to 
 
 BEC and BCE, ^ 
 
 and consequently ^ C 
 
 the triangle C'BE is isosceles, or the side BC is equal to 
 BE. And since the diverging lines AD and AB are cut 
 by the parallels DB and CE proportionally, AD : DC : : 
 AB : BE or BC. 
 
 Cor, Hence the converse of the Proposition is likewise 
 true, or if a straight line be drawn from the vertex of a 
 triangle to cut the base in the ratio of the adjacent sides, 
 it will bisect the vertical angle •, for it is evident, from 
 VI. 6. cor., that a straight line is only capable of a single 
 section, whether internal or external, in a given propor- 
 tion. 
 
 Scholium, The vertical line BD must bisect the base 
 AC of the triangle, when the sides AB and BC are equal. 
 In the case where BD bisects the exterior angle CBG, if 
 AB be supposed to approach to an equality with BC, the 
 straight Hne EC will come nearer to AC, and consequent- 
 ly the incidence D of the parallel BD with AC will be 
 thrown continually more remote. But when the side AB 
 is equal to BC, the straight line BD, being now parallel 
 to AC, will never meet it, or there can be no equality of 
 external section ; for though the ratio of AD to CD tends 
 towards the ratio of equality as the point D retires, yet the 
 
170 
 
 ELEMENTS OF GEOMETRY. 
 
 constant difference AC betwreen those distances must al- 
 ways bear a sensible relation to them. After BD, in turn- 
 ing about the point B, has passed the limits of distance 
 beyond C, it re-appears in an opposite direction beyond 
 A, when AB, receding from equality, has become less than 
 BC. 
 
 PROP. XL THEOR. 
 
 Triangles are similar, which have their corre- 
 sponding angles equal. 
 
 Let the triangles ABC and DEF have the angle CAB 
 equal to FDE, CBA to FED, and consequently (L 30.) 
 the remaining angle BCA equal to EFD ; these triangles 
 are similar, or the sides in both which contain equal an- 
 gles are proportional. 
 
 For make BG equal to ED, and draw GH parallel to 
 AC. 
 
 Because GH is parallel 
 to AC, the exterior angle 
 BGH is equal (I. 22.) to 
 BAC, that is to EDF; 
 and the angle at B is, by 
 hypothesis, equal to that 
 at E, and the interjacent 
 side BG was made equal to 
 
 ED; wherefore (I. 20.) the triangle GBH is equal to 
 DEF. But, the diverging lines BA and BC being cut 
 proportionally by the parallels AC and GH (VI. J.), AB 
 
BOOK VI. 171 
 
 is to BC as BG to BH, or as ED to EF. Again, thosi^ 
 diverging lines being proportional to the intercepted seg- 
 ments AC and GH of the parallels (VI. 2.), AB is to 
 BG as AC is to GH, and alternately AB is to AC as BG 
 is to GH, or as ED to DF. In the same manner, as BC 
 is to BH so is AC to GH, and alternately, as BC is to 
 AC so is BH or EF to GH or DF. And thus, the sides 
 opposite to equal angles in the triangles ABC and DEF 
 are the homologous terms of a proportion. 
 
 Cor, Isosceles triangles are similar which have their ver- 
 tical angles equal. For the supplementary angles at the 
 base, forming (I. 30.) the same amount, must consequently 
 be equal to each other. 
 
 Scholium, It is obvious that the twentieth Proposition of 
 Book I. is but a particular case of this theorem. 
 
 PROP. XII. THEOR. 
 
 Triangles which have the sides about two of 
 their angles proportional, are similar. 
 
 In the triangles ABC and DEF, let AB: AC : : DE : DF 
 and BC : AC ; : EF : DF; then is the angle BAC equal 
 to EDF, and the angle BCA equal to EFD. 
 
 For (1. 4.) draw DG and FG, making angles FDG and 
 DFG equal to CAB and ACB. 
 
 B}" the last Proposition, the triangle ABC is similar to 
 DGF, and consequently AB : AC : : DG : DF; but by 
 hypothesis, AB : AC : : DE : DF, and hence, from iden- 
 
172 ELEMENTS OF GEOMETRY. 
 
 tity of ratios, DG : DF : : 
 
 DE :DF, or DG.is e- 
 
 qual to DE. In the same 
 
 manner, BC : AC : : 
 
 EF : DF, and BC AC : : 
 
 GF:DF;whenceEF:DF 
 
 : ; GF : DF, and EF is e- 
 
 qual to FG. Wherefore 
 
 the triangles DEF and DGF, having thus the sides DE 
 
 and EF equal to DG and FG, and the side DF common 
 
 to both, are (I. 2.) equal ; consequently the angle EDF is 
 
 equal to FDG or BAC, and the angle EFD is equal to 
 
 DFG or BCA. 
 
 Cor. Hence isosceles triangles which have either side 
 proportional to the base, are similar. 
 
 Scholium. The second Proposition of Book I. may be 
 considered as only a particular case of this theorem. 
 
 PROP. XIII. THEOR. 
 
 i 
 Triangles are similar, if each have an equal an- 
 gle, and its containing sides proportional. 
 
 In the triangles BAC and EDF, let the angle ABC 
 be equal to DEF, and the sides which contain the one 
 be proportional to those which contain the other, or 
 AB : BC : : DE : EF; the triangles BAC and EDF are 
 similar. 
 
 For, from the points E and F, draw EF and FG, ma- 
 king the angles FEG and EFG equal to CBA and BCA. 
 
IJOOK \u 1 78 
 
 The triangles BAG and EGF, having thus their corre- 
 sponding angles equal, are similar (VI. 11. )j and therefore 
 AB : BC : : EG : EF. But by hypothesis, AB : BC : : 
 ED : EF; where- 
 fore EG : EF : : ^ ^^ 
 ED : EF, and con- ^ ^ ^ 
 
 sequently EG is e- 
 qual to ED. Hence 
 the triangles GFE 
 and DFE, having the side EG equal to ED, EF common 
 to both, and the contained angle GEF equal to ABC or 
 DEF, are equal (I. 3.), and therefore the angle EFG or 
 BCA is equal to EFD; consequently the remaining angles 
 BAG and EDF of the triangles ABG and DEF are equal 
 (I. 30.), and these triangles are (VI. 11.) similar. 
 
 Scholiwn. The third Proposition of Book I. is merely a 
 particular case of this general theorem. 
 
 PROP. XIV. THEOR. 
 
 Triangles are similar, which have each an equal 
 angle, and the sides containing another angle of 
 the same character proportional. 
 
 Let the triangles GAB and FDE have the angle ABC 
 equal to DEF, and the sides that contain the angles at C 
 and F proportional, or BC : AG : : EF : FD ; while those 
 angles are both of them either acute or obtuse, the trian- 
 gles ABG and DEF are similar. 
 
 For, from the points E and F draw EG and FG, 
 
174! ' ELEMENTS OF GEOMETRY. 
 
 making the an- 
 gles FEG and 
 EFG equal to 
 ABC and BCAv 
 The triangle' 
 ABC is evidently 
 similar to GEF, 
 
 and BC : CA : : EF : FG; but, by hypothesis, BC : CA : : 
 EF : FD, and therefore £F : FG : : EF : FD, and FG is 
 equal to FD. Whenee the triangles EGF and EDF, ha- 
 ving the angle FEG equal to FED, the side FG equal to 
 FD, and the side EF common, and being both of the sanie 
 character with CAB, are equal (I. 21.) ; consequently the 
 angle GFE or ACB is equal to DFE, and therefore 
 (VI. 11.) the triangles ABC and DEF are similar. 
 
 Scholium, This Theorem exhibits the general property 
 of which Prop. 2. Book II. is only a particular case. 
 
 PROP. XV. THEOR. 
 
 A perpendicular let fall upon the hypotenuse 
 of a right-angled triangle from the opposite ver- 
 tex, will divide it into two triangles that are simi- 
 lar to the whole and to each other. 
 
 Let the triangle ABC be right-angled at B, from which 
 the perpendicular BD falls upon the hypotenuse AC; the 
 triangles ABD and DBC, thus formed, are similar to each 
 other, and to the whole triangle ACB. 
 
 For the triangles ABD and ACB, having the angle 
 BAC common, and the right angle ADB equal to ABC, 
 
li ooK vi. ' 175 
 
 are similar (VI. 1 1 .).] Again, the 
 triangles DBC and ACB are si- 
 milar, since they have the angle 
 BCD common, and the right an- 
 gle BDC equal to ABC. The tri- 
 angles ABD and DBC being, 
 
 therefore, both similar to the same triangle ABC, are evi- 
 dently similar to each other (VI. 11.). 
 
 Cor, Hence the side of a right-angled triangle is a mean 
 proportional between the hypotenuse and the adjacent seg- 
 ment, formed by a perpendicular let fall upon it from the 
 opposite vertex; and the perpendicular itself is a mean 
 proportional between those segments of the hypotenuse. 
 For the triangles ABC and ADB being similar, AC : AB : ; 
 AB : AD ; and the triangles ABC and BDC being si- 
 milar, AC : BC : : BC : CD; again, the triangles ADB 
 and BDC are similar, and therefore AD : DB : : DB : DC. 
 
 Scholium, This corollary affords an easy demonstration 
 6f the celebrated theorem contained in Prop. 10. Book 1.;. 
 
 PROP. XVI. PROB. 
 
 To find the mean proportional between two gl* 
 ven straight lines. 
 
 Let it be required to find the mean proportional between 
 the straight lines A and B. 
 
 Find C (III. 27.) the side of a 
 square which is equivalent to the rect- -g, , 
 
 angle contained by A and B ; C is C\ — ' « 
 
 the mean proportional required. 
 
 For since C^ = AB, it follows (V. 6.) that A : C : : C : B. 
 
176 
 
 tXEMENTS or GF.0METIl5f. 
 
 PROP. XVIL PROB. 
 
 To divide a straight line, whether internally or 
 externally, so that the rectangle under its seg- 
 ments shall be equivalent to a given rectangle. 
 
 Let AB be the straight line which it is required to cut, 
 so that the rectangle under its segments shall be equivalent 
 to a given rectangle. 
 
 From the extremities of AB, erect the perpendiculars AD 
 and BE, equal to the sides of the given rectangle, and 
 in the same or in opposite directions, according as the 
 line is to be cut inter- 
 nally or externally"; join 
 DE, on which, as a dia- 
 meter, describe a circle, 
 meeting AB or its exten- 
 sion in the point C : AC 
 and CB are the segments 
 required. 
 
 For join DC and CE. 
 The angle DCE, being 
 contained in a semicircle, 
 is a right angle (III. 19.), 
 and therefore, in both 
 cases, the angles ACD 
 and BCE are together 
 equal to a right angle. 
 But the angles ACD and 
 CDA are likewise toge- 
 ther equal to a right angle (1. 30. cor. 1.) ; and consequent- 
 
BOOK VT. 177 
 
 \y the angles BCE and CDA are equal. Wherefore the 
 right-angled triangles GBE aYid CAD, having the acute 
 angle ADC equal to BCE, are similar (VI. 11.); whence 
 AC : AD : : BE : CB, and (V. 6.) AC.CB = AD.BE. 
 
 Scholium. It is obvious that, in the second case, the cir- 
 cle, lying on both sides of the given line AB, must always 
 intersect its extension in two points C and C. But, in the 
 first case, the circle may either cut AB in two points C and 
 C, or touch it in a single point, which will hence mark a 
 limitation of the problem. A straight line drawn from the 
 centre of the circle parallel to AD or BE, must (Vl. 1.). 
 divide AB proportionally, and hence bisect it -, but that 
 parallel would also be perpendicular (I. 22.) to AB, and 
 therefore (III. 4.) bisect the chord CC^ Consequently the 
 points C and C are equally distant from the middle of AB, 
 and the portion AC is equal to BC. When these points 
 come to coincide, they rhust therefore pass into the middle 
 point of AB, or that of its contact with the circle. When 
 the circle does not reach AB, the problem fails, because 
 (11. 17. cor. 1.) no straight litie can be divided internally, 
 such that the rectangle under the segments shall exceed the 
 square of its half. This impossibility is indicated by the 
 circle not reaching the straight line AB. 
 
 This proposition furnishes one of the simplest and most 
 elegant methods for constructing quadratic equations ; the 
 segments of the line denoting the roots, and indicating by 
 position their character. The first case has two additive 
 roots, which may become equal or merge in a single root, 
 then limiting the possibility of the equation ; the second 
 case has always two unequal roots, the one additive and the 
 other subtractive. In both cases, those roots, conjoined 
 in their actual position, complete the line AB. 
 
 N 
 
178 
 
 ELEMENTS OF GEOMETRY. 
 
 PROP. XVIII. THEOR. 
 
 The rectangle under any two sides of a trian- 
 gle is equivalent to the rectangle under the per- 
 pendicular let fall on the base and the diameter of 
 the circumscribing circle. 
 
 Let ABC be a triangle, about which is described a cir- 
 cle having the diameter BE ; the rectangle under the sides 
 AB and BC is equivalent to the rectangle under BE and 
 the perpendicular BD let fall from the vertex of the trian- 
 gle upon the base AC. 
 
 For join CE. The angle BAD is 
 equal to BEC (III. 16.), since they 
 both stand upon the same arc BCj 
 and the angle ADB, being a right an- 
 gle, is (III. 19.) equal to ECB, 
 which is contained in a semicircle. 
 Wherefore the triangles ABD and 
 EBC, being thus similar (VI. 11.), AB : BD : : EB : BC, 
 and consequently (V. 6.) AB.BC=EB.BD. 
 
 PROP. XIX. THEOR. 
 
 The square of a straight line that bisects, whe- 
 ther internally or externally, the vertical angle of 
 a triangle, is equivalent to the difference between 
 the rectangle under the sides, and the rectangle 
 under the segments into which it divides the base. 
 
fiOOK VI. 
 
 179 
 
 In the triangle ABC, let BE bisect the vertical angle 
 CBA or its adjacent angle CBF ; then BE^ = AB.BC— 
 AE.EC, or AE.EC—AB.BC. 
 
 For (III. 9. cor.) about the triangle describe a circle, 
 produce BE to the circumference, and join CD. 
 
 TheanglesBAE and BDC, 
 standing upon the same arc 
 BC, are (III. 16.) equal, audi 
 the angle ABE is, by hypo- 
 thesis, equal to DBC; where- 
 fore (VI. 11.) the triangles 
 AEB ^nd DCB are similar, 
 and AB : BE : : DB : BC. 
 Consequently (V. 6.) 
 AB.BC = BE.BD; but 
 BE.BD = BE.liD + BE», 
 or BE.ED^BE^ and (III. 
 26.) BE.ED = AE.EC; 
 
 wherefore AB.BCn AE.EC + BE% or AE.EC— BE* ; 
 and consequently BE* = AB.BC— AE.EC, or AE.EC— 
 AB.BC. 
 
 PROP. XX. THEOR. 
 
 The rectangles under the opposite sides of a 
 quadrilateral figure inscribed in a circle, are toge- 
 ther equivalent to the. rectangle under its diago- 
 nals. 
 
 In the circle ABCD, let a quadrilateral figure be in- 
 scribed, and join the diagonals AC, BD ; the rectangles 
 AB, CD and BC, AD, are together equivalent to the rect- 
 angle AC, BD. 
 
180 
 
 ELEMENTS OF GEOMETRY. 
 
 For (I. 4.) draw BE, making an angle ABE equal to 
 CBD. 
 
 The triangles AEB and DCB, having thus the angle 
 ABE equal to DBC, and the an- 
 gle BAE or BAG equal (III. 16.) 
 to BDC, are similar (VI. 11.), 
 and hence AB : AE : : BD : CD ; 
 whence (V. 6.) AB.CD = AE.BD. 
 Again, because the angle ABE 
 is equal to DBC, add EBD to 
 each, and the whole angle ABD 
 
 is equal to EBC ; and the angle ADB is equal to ECB 
 (III. 16.) i wherefore the triangles DAB and CEB are 
 similar (VI. 11.), and AD : BD : : EC : BC, and con- 
 sequently BC.ADrzEC.BD. V^hence the rectangles 
 
 AB, CD and BC, Al) are together equal to the rectangles 
 AE, BD and EC, BD, that is, to the whole rectangle 
 
 AC, BD. 
 
 PROP. XXI. THEOR. 
 
 Triangles which have a common angle, are to 
 each other in the compound ratio of the contain- 
 ing sides. 
 
 Let ABC and DBE be two triangles, having the same 
 or an equal angle at B ; ABC 
 is to DBE in the ratio com- 
 pounded of that of BA to BD, 
 and of BC to BE. 
 
 For join AE and CD. The 
 ratio of the triangle ABC to 
 DBE may be conceived as com- 
 pounded of that of ABC to 
 
BOOK VJ. 
 
 181 
 
 DBC, and of DBC to DBE. But (V. 25. cor. 2.) the 
 triangle ABC is to DBC, as the base BA to BD ; and, 
 for the same reason, the triangle DBC is to DBE, as the 
 base BC to BE ; consequently the triangle ABC is to DBE 
 in the ratio compounded of that of BA to BD, and of BC 
 to BE, or (V. 23.) in the ratio of the rectangle under BA 
 and BC to the rectangle under BD and BE. 
 
 Cor. 1. Hence similar triangles are in the duplicate ra- 
 tio of their homologous sides. For, if the angle at B be 
 equal to that at E, the triangle ABC is to DEF in the 
 
 A. 
 
 ratio compounded of that of AB to DE, ^nd of CB to 
 FE; but, these triangles being similar, the ratio of AB to 
 DE is the same as that of CB to FE (VI. 11.), and con- 
 sequently the triangle ABC is to DEF in the duplicate ra- 
 tio of AB to DE, or (V. 24:.) as the square of AB to the 
 square of DE. 
 
 Cor, 2. Hence triangles which have the sides that con- 
 tain an equal angle re- 
 ciprocally proportional, 
 are equivalent. For, the 
 angle at B being equal 
 to that at E, the triangle 
 ABC is to DEF as 
 AB.CBtoDE.FE;but 
 AB : DE::FE:CB, 
 andAB.CB = DE.FE; 
 
182 ELEMENTS OP GEOMETRV. 
 
 consequently (V. 4.), the third and fourth terms of the 
 analogy being equal, the first and second must also be 
 equal. 
 
 PROP. XXII. THEOR. 
 
 Similar rectilineal figures may be divided into 
 corresponding similar triangles. 
 
 Let ABCDE and FGHIK be similar rectilineal figures, 
 of which A and F are corresponding points ; these figures 
 ma}" be resolved into a like number of triangles respective- 
 ly similar. 
 
 For, from the point A in the one figure, draw the 
 straight lines AC, AD, and, from F in the other, draw 
 FH, FI ; the triangles BAG, CAD, and DAE are re- 
 spectively similar to GFH, HFI, and IFK. 
 
 Because the polygon ABCDE is similar to FGHIK, 
 the angle ABC is e- 
 qual to FGH, and 
 AB:BC::FG:GH; 
 wherefore (VI. 13.) 
 the triangle BAC is 
 similar to GFH. 
 Hence the angle 
 BCA is equal to GHF j and the whole angle BCD being 
 equal to GHI, the remaining angle ACD must be equal 
 to FHI. But BC : AC : : GH : FH, and BC : CD : : 
 GH : HI, consequently (V. 15.) AC : CD : : FH : HI, 
 and the triangles CAD and HFI (VI. 13.) are similar. 
 Whence, the angle CDA being equal to HIF and the 
 
BOOK VI. 18S 
 
 angle CDE to HIK, the angle ADE is equd to FIK; 
 and since CD : DA : : HI : IF, and CD : DE : : HI : IK, 
 therefore (V. 15.) DA : DE : : IF : IK, and the tri- 
 angles DAE and IFK are similar. 
 
 The same train of reasoning, it is obvious, would apply 
 to polygons of any number of sides. 
 
 PROP. XXIII. PROB. 
 
 On a given straight line, to construct a rectili- 
 neal figure similar to a given rectilineal figure. 
 
 Let FIC be a straight line, on which it is required to 
 construct a rectilineal figure similar to the figure ABCDE. 
 
 Join AC and AD, dividing the given rectilineal figure 
 into its component triangles. From the points F and K 
 draw FI and KI, making the angles KFI and FKI equal 
 to EAD and AED ; from F and I draw FH and IH 
 making the angles IFH and FIH equal to DAC and 
 ADC ; and lastly from F and H draw FG and HG ma- 
 king the angles HFG and FHG equal to CAB and ACB. 
 The figure FGHIK is similar to ABCDE. 
 
 For the several triangles KFI, IFH, and HFG, which 
 compose the figure FGHIK, are, by the construction, evi- 
 dently similar to the triangles EAD, DAC, and CAB, in- 
 
 I> 
 
 B 
 
 IT 
 
 K F K 
 
 to which the figure ABCDE was resolved. Whence 
 FK : KI : : AE : ED ; also KI : IF : : ED : DA, and 
 
iS4i ELExAIENTS OF GEOMETRY. 
 
 IF : IH : : DA : DC, and consequently (V. 16.) KI : IH : : 
 ED : DC. Again, IH : HF : : DC : CA, and HF : HG : . 
 CA : CB ;— and hence (V. 16.) IH : HG : : DC : CB. But 
 HG : GF: : CB : BA ; and the ratio of GF to FK, being 
 compounded of that of GF to FH, of FH to FI, and of FI 
 to FK, is the same with the ratio of BA to AE, which is 
 compounded of the like ratios of BA to AC, of AC to 
 AD, and AD to AE. Wherefore all the sides about the 
 figure FGHIK are proportional to those about ABCDE; 
 but the several angles of the former, having a like com- 
 position, are respectively equal to those of the latter. 
 "Whence the figure FGHIK is similar to the given figure. 
 
 The same reasoning, it is manifest, would extend to po- 
 lygons of any number of sides. 
 
 Scholium. The general solution of this problem is derived 
 from the principle, that similar triangles, by their compo- 
 sition, form similar polygons. The mode of construction, 
 however, admits of some variation. For instance, if the 
 straight line FK be parallel to AE, or in the same exten- 
 sion with that homologous side, — the several triangles 
 FIK, FHI, and FGH may be more easily constituted in 
 succession, by drawing the straight lines FI and KI, FH 
 and IH, and FG and GH parallel to the corresponding 
 sides in the original figure ABCDE ; because (I. 29.) a 
 correspond ing equality of angles will be thus produced. 
 
 But, if FK have no determinate position, the construc- 
 tion may be still farther sim- 
 plified ; For, having made AK 
 / equal to that base and joined 
 AD and AC, draw KI, IH, 
 and HG parallel to ED, DC, 
 andCB. The figure AKIHG 
 js evidently similar to AEDCB, 
 
BOOK VI. 
 
 18' 
 
 since its component triangles have the same vertical angles 
 as those of the original figure, and the angles at the bases 
 equal (I. 22.). 
 
 If the given base FK be parallel to the corresponding 
 side AE of the original figure, a more general construction 
 will result. Jom AF, EK, and produce them to meet in 
 O; join OB, OC, and GD, and draw FG, GH, HI, 
 and therefore IK, parallel to AB, BC, CD, and DE : 
 The figure FGHIK thus formed is similar to ABCDE. 
 For the triangles KOF, FOG, GOH, HOI, and lOKare 
 evidently similar to the triangles EGA, AOB, BOC, 
 COD, and DOE. But these triangles compose severally 
 
 the two polygons, when the point O lies within the ori- 
 ginal figure; and when that point of concurrence lies 
 without the figure ABCDE, the similar triangles lOK 
 and DOE being taken away from the similar compound 
 polygons FGHIOK and ABCDOE, there remains the 
 figure FGHIK similar to the original one. 
 
 It farther appears, from these investigations, that a rec- 
 tilineal figure may have its sides reduced or enlarged in a 
 
186 ELEMENTS OF GEOMETRY. 
 
 given ratio, by assuming any point O and cutting thd di- 
 verging lines OE, OA, OB, OC, and OD in that ratio ; 
 the corresponding points of section being joined, will ex-j' 
 hibit the figure recjuired. 
 
 PROP. XXIV, THEOR. 
 
 Of similar figures, the perimeters are propor- 
 tional to the corresponding sides, and the areas 
 are in the duplicate ratio of those homologous 
 terms. 
 
 Let ABCDE and FGHIK be similar polygons, which 
 bave the corresponding sides AB and FG ; the perimeter, 
 or linear boundary, ABCDE is to the perimeter FGHIK, 
 as AB to FG, BC to GH, CD to HI, DE to IK, or EA 
 to KF; but the area of ABCDE, or the contained surr 
 face, is to the area of FGHIK, in the duplicate ratio of 
 AB to FG, of BC to GH, of CD to HI, of DE to IK, 
 or of EA to KF. 
 
 For, by drawing the diagonals AC, AD in the one, and 
 FH, FI in the other, 
 these polygons will be 
 resolved into similar 
 triangles. Whence 
 the several analogies 
 AB:BC::FG:GH, 
 BC:AC::GH:FH, 
 AC : CD : : FH : HI, CD : AD : : HI : FI, and 
 AD : DE : : FI : IK ; wherefore, by equality and alterna- 
 ]tion, AB : FG : : BC : GH : : CD : HI : : DE : IK : : 
 
 K :\r K. 
 
BdOKVl. 187 
 
 AE : FK, and consequently (V. 19.) as one of the antece- 
 dents AB, BC, CD, DE or AE, is to its consequent FG, 
 GH, HI, IK or FK, so is the amount of all those ante- 
 cedents, or the perimeter ABCDE, to the amount of all 
 the consequents, or the perimeter FGHIK. 
 
 Again, the triangle CAB is to the triangle HFG (VI. 
 21. cor. 1.) in the duplicate ratio of AB to FG, — the tri- 
 angle DAC is to the triangle IFH in the duplicate ratio 
 of AC to FH, orof AB to FG,— and the triangle EAD 
 is to KFI in the duplicate ratio of AD to FI or of AB to 
 FG; wherefore (V. 19.) the aggregate of the triangles 
 CAB, DAC, and EAD, or the area of the polygon 
 ABCDE, is to the aggregate of the triangles HFG, IFH, 
 and KFI, or the area of the polygon FGHIK, in the du- 
 plicate ratio of AB to FG, of BC to GH, of CD to HI, 
 or of DE to IK. 
 
 Cor, Hence also the perimeter ABCDE is to the peri- 
 meter FGHIK, as any diagonal AD to the correspondr 
 ing diagonal FI, and the area ABCDE is to the area 
 FGHIK in the duplicate ratio of AD to FI. 
 
 PROP. XXV. PROB. 
 
 To constriijct a rectilineal figure that shall be 
 similar to one, and equivalent to another, given 
 rectilineal figure. 
 
 Let it be required to describe a rectilineal figure similar 
 to A, and equivalent to B. 
 
 On CD, a side of A, describe (II. 8.) the rectanglq 
 CDFE, equivalent to that figure, and on DF describe the 
 
188 
 
 ELEMENTS OF GEOMETRY. 
 
 B 
 
 rectangle DGHF equivalent to the figure B ; find 
 (VI.16.)IKamean 
 proportional be- 
 tween CD and 
 DG, and on IK 
 construct, in the 
 same position, a fi- 
 gure X similar to 
 the rectilineal fi- X .IT 
 
 A. 
 
 
 « 
 
 G 
 
 K 
 
 TI 
 
 gure A ; this will be likewise equivalent to B. 
 
 For the figures A and X, being similar, must (VI. 24.) 
 be in the duplicate ratio of their homologous sides CD and 
 IK ; and since IK is a mean proportional between CD and 
 DG, the duplicate ratio of CD to IK is the same as the 
 ratio of CD to DG (V. 24.) ; consequently the figure A is 
 to the figure X as CD to DG, or (Y. 25. cor. 2.) as the 
 rectangle CF to the rectangle DH ; but the figure A is 
 equivalent to the rectangle CF, and therefore (V. 4.) the 
 figure X is equivalent to the rectangle DH, that is, to the 
 figure B. 
 
 PROP. XXVI. THEOR. 
 
 A rectilineal figure described on the hypote- 
 nuse of a right-angled triangle, is equivalent to 
 similar figures described on the two sides. 
 
 Let ABC be a right-angled triangle ; the figure ACFE 
 described on the hypotenuse is equivalent to the similar 
 figures AGHB and BIKC, described on the sides AB 
 and BC. 
 
BOOK VI. 
 
 189 
 
 Foi' draw BD perpendicular to the hypotenuse. And since 
 (VI. 15. cor. 1.) AC:AB:: 
 AB : AD, therefore AC is 
 to AD in the duplicate ra- 
 tio of AC to AB, that is, 
 (VI. 24.), as the figure on 
 AC to the figure on AB. 
 For the same reason, AC is 
 to CD in the duplicate ratio 
 of AG to BC, or as the figure 
 on AC to the figure on BC. 
 
 Whence (V. 19. cor. 2.) AC is to the two segments AD 
 and CD taken together, as the figure on AC to both the 
 figures on AB and BC ; and the first term of the analogy 
 being thus equal to the second, the third must be equal to 
 the fourth (V. 4.), or the figure described on the hypote- 
 nuse is equivalent to the similar figures described on the 
 two sides. 
 
 PROP. XXVII. THEOR. 
 
 The arcs of a circle are proportional to the an- 
 gles which they subtend at the centre. 
 
 Let the radii CA, CB, and CD intercept arcs AB and 
 BD ; the arc AB is to BD, as the angle ACB to BCD. 
 
 For (I. 5.) bisect the angle ACB, bisect again each of 
 its halves, and repeat the operation indefinitely. An angle 
 ACa will be thus obtained less than any assignable angle. 
 Let this angle ACa or BCb (I. 4.) be repeatedly applied 
 
190 
 
 Elements or geometry. 
 
 about the point C, from BC towards DC ; it must hence, 
 by its multiplication, fill up the 
 angle BCD, nearer than any pos- 
 sible difference. But the elemen- 
 tary angle ACa being equal to 
 BC6, the corresponding arc A« 
 is (III. 12.) equal to BL Conse- 
 quently this arc Aa and its angle 
 ACa, are like measures of the 
 arc AB and the angle ACB, and they are both contained 
 equally in the arc BD and its corresponding angle BCD. 
 Wherefore AB : BD : : ACB : BCD. 
 
 Cor. Hence the arc AB is also to BD, as the sector 
 ACB to the sector BCD ; for these sectors may be viewed 
 as alike composed of the elementary sector ACa, 
 
 PROP. XXVIII. THEOR. 
 
 The circumference of a circle is proportional 
 to the diameter, and its area to the square of that 
 diameter. 
 
 Let AB and CD be the diameters of two circles ;-*the 
 circumference AFG is to the circumference CKL, as AB 
 to CD ; and the area contained by AFG is to the area 
 contained by CKL, as the square of AB to the square of 
 CD. 
 
 For inscribe the regular hexagons AEFBGH and 
 CIKDLM. Because these polygons are equilateral and 
 equiangular, they are similar ; and consequently (VL 24. 
 
BOOK VI. 191 
 
 cor.) the diagonal AB is to the corresponding diago- 
 nal CD, as the perimeter AEFBGH to the perimeter 
 CIKDLM. But this proportion must subsist, whatever 
 be the number of chords inscribed in either circumference. 
 Insert a dodecagon in each circle between the hexagon 
 and the circumference, and its perimeter will evidently ap- 
 
 proach nearer to the length of that circumference. Pro- 
 ceeding thus, by repeated duplications,— the perimeters of 
 the series of polygons that arise in succession, will conti- 
 nually approximate to the curvilineal boundary, which 
 forms their ultimate limit. Wherefore this extreme term, 
 or the circumference AEFBGH, is to the circumference 
 CIKDLM, as the diameter AB to the diameter CD. 
 
 Again, the hexagon AEFBGH (VI. 24. cor.) is to the 
 hexagon CIKDLM in the duplicate ratio of the diagonal 
 AB to the corresponding diagonal CD, or (V. 24.) as the 
 square of AB to the square of CD. Wherefore the suc- 
 cessive polygons which arise from a repeated bisection of 
 the intermediate arcs, and which approach continually to 
 the areas of their containing circles, must have still that 
 same ratio. Consequently the limiting space, or the circle 
 AEFBGH, is to the circle CIKDLM, as the square of 
 AB to the square of CD. 
 
 Cor, L It hence follows, that if semicircles be described. 
 
192 
 
 ELEMENTS OF GEOMETRY. 
 
 on the sides AB, BC of a right-angled triangle, and on the 
 hypotenuse AC another semicircle be described, passing 
 (III. 19.) through the vertex B, the crescents AFBD and 
 BGCE are together equivalent to the triangle ABC. For, 
 by the Proposition, the square of AC is to the square of 
 AB, as the circle on AC to the circle on AB, or (V. 3.) 
 as the semicircle ADBEC to the semicircle AFB ; and, 
 for the same reason, the square of AC is to the square of 
 BC, as the semicircle ADBEC 
 to the semicircle BGC. Whence 
 (V.8. and 19.) the square of AC 
 is to the squares of AB and BC, 
 as the semicircle ADBEC to the 
 semicircles AFB and BGC. But 
 (II. 10.) the square of AC is equivalent to the squares of 
 AB and BC, and therefore (V. 4.) the semicircle ADBEC 
 is equivalent to the tvvo semicircles AFB and BGC ; take 
 away the common segments ADB and BEC, and there re- 
 mains the triangle ABC equivalent to the tvvo crescents or 
 lunes AFBD and BGCE. 
 
 Cor, 2. Hence the method of dividing a circle into 
 equal portions, by means of concentric circles. Let it be 
 required, for instance, to tri- 
 sect the circle of which AB 
 is a diameter. Divide the 
 radius AC into three equal 
 parts, from the points of sec- 
 tion draw perpendiculars 
 DF, EG meeting the cir- 
 cumference of a semicircle 
 described on AC, join CF, 
 
 CG, and from C as a centre, with the distances CF, CG, de- 
 
 s 
 
BOOK VI. 193 
 
 scribe the circles FHI, GKL: The circle on AB will be 
 divided into three equal portions, by those interior circles. 
 For join AF and AG : Because AFC, being in a semicircle, 
 is a right angle (III. 19.), AC is to CD (VI. 15, cor. 1. 
 and V. 21-.), as the square of AC to the square of CF, that 
 is, as the circle on AB to the circle FHI ; but CD is the 
 third-part of AC ; wherefore (V. 5.) the circle FHI is the 
 third part of the circle on AB. In like manner, it is pro- 
 ved, that the circle GKL is two third-parts of the circle on 
 AB. Consequently, the intervening annular spaces, and 
 the circle FHI, are all equal. 
 
 PROP. XXIX. THEOR. 
 
 The area of any triangle is a mean proportional 
 between the rectangle under the semiperimeter 
 and its excess above the base, and the rectangle 
 under the separate excesses of that semiperimeter 
 above the two remaining sides. 
 
 The area of the triangle ABC is a mean proportional 
 between the rectangle under half the sum of all the sides 
 and its excess above AC, and the rectangle under the ex- 
 cess of that semiperimeter above AB and its excess above 
 BC. 
 
 For produce the sides BA and BC, draw the straight 
 lines BE, AD, and AE bisecting the angles CBA, BAC,^ 
 and CAI, join CD and CE, and let fall the perpendicu- 
 lars DF, DG, and DH within the triangle, and the per- 
 pendiculars EI, EK, and EL without it. 
 
194« 
 
 ELEMENTS OF GEOMETRY. 
 
 The triangles ADF and ADG, having the angle DAF 
 equal to DAG, the angles F and G right angles, and the 
 common side AD,— are (I. 20.) equal; for the same rea- 
 son, the triangles BDG and BDH are equal. In like 
 manner, it is proved, that the triangles AEI and AEK 
 are equal, and the triangles BEI and BEL. Whence the 
 triangles CDH and CDF, having the side DH equal to 
 DF, the side DC common, and the right angle CHD 
 equal to CFD, — are (I. 21.) equal; and, for the same rea- 
 son, the triangles CEK and CEL are equal. The peri- 
 meter of the triangle ABC 
 is therefore equal to twice 
 the segments AF, FC, and 
 BG ; consequently BG is 
 the excess of the semiperi- 
 meter above the base AC, 
 and AG is the excess of 
 that semiperimeter — or of 
 the segments BH, HC, 
 and AG,— above the side 
 BC. But the sides AB 
 and BC, with the segments AK and CK, or AI and CL, 
 also form the perimeter ; whence, BI being equal to BL, 
 the part AI is the excess of the semiperimeter above the 
 side AB* 
 
 Now, because DG and EI, being perpendicular to BI, 
 are parallel, BG : DG : : BI : EI (VI. 2.), and consequently 
 (V. 25. cor. 2.) BI X BG : BI xDG : :DG X BI : DG X EI. 
 But since AD and AE bisect the angle BAC and its ad- 
 jacent angle CAI, the angles GAD and EAI are together 
 equal to a right angle, and equal, therefore, to lEA and 
 EAI ; whence the angle GAD is equal to lEA, and the 
 
BOOK VI. 195 
 
 right-angled triangles DGA and AIE are similar. Where- 
 fore (VI. ll.)DG:AG:: AI:EI, and (V. 6.)DGxEl 
 = AG X AI ; consequently BI X BG : DG X BI : : 
 DG X BI : AG X AI. But the triangle ABC is composed 
 of three triangles ADB, BDC, and CD A, which have the 
 same altitude j and therefore its area is equal to the rect- 
 angle under DG and half their bases AB, BC, and AC, 
 or the semiperimeter BI. Whence the area of the tri- 
 angle ABC is a mean proportional between the rectangle 
 under BI and its excess above AC, and the rectangle un- 
 der its excess above BC and that above AB. 
 
 Co7\ Hence the area of a triangle will be expressed nu- 
 merically, by the square root of the continued product of 
 the semiperimeter into its excesses above the three sides* 
 
 PROP. XXX. PROB. 
 
 To convert a given regular polygon into an- 
 other, which shall have the same perimeter, but 
 double the number of sides. 
 
 It is evident that, by lines radiating from the centre of 
 the inscribed or circumscribing circle, a regular polygon 
 may be divided into as many equal and isosceles triangles 
 as it has sides. Let AOB be such a sector of the given 
 polygon ; from the centre O let fall the perpendicular OC, 
 and produce it to D> till OD be equal to OA or OB, and 
 join AD and BD. The isosceles triangle ADB is there- 
 fore (IV. 1 .) constructed on the same base with AOB, and 
 has only half the vertical angle. Consequently twice as 
 many of such angles could be constituted about D, as were 
 
196 
 
 ELEMENTS OF GEOxMETKV. 
 
 placed about O. Bisect 
 
 AD and BD in E and F, 
 
 and the straight line joining 
 
 these points must (VI. 2.) 
 
 be equal to half the base AB. 
 
 Wherefore the triangle 
 
 EDF, repeated about the 
 
 vertex D, would form a re- 
 gular polygon with twite as 
 
 many sides as before, but 
 
 under the same extent of 
 
 perimeter, since each of 
 
 those sides has only half 
 the former length. 
 
 Cor. 1. Hence DG, the radius of the circle inscribing 
 the derived polygon, is half of CD, that is, half of the sum 
 of OC and OA, the radii of the circles inscribing and cir- 
 cumscribing the given polygon. Again, since AOD is evi- 
 dently isosceles, AD^ = 20A.CD (II. 23. cor.), and conse- 
 quently DE the radius of the circumscribing derived poly- 
 gon, being the half of AD, is a mean proportional between 
 OA and DG, the radius of the circle circumscribing the 
 given polygon, and the radius of the circle inscribing the 
 derived polygon. 
 
 Cor. 2. Hence the area of a circle is equivalent to the 
 rectangle under its radius, and a straight line equal to half 
 its circumference. For the surface of any regular circum- 
 scribing polygon, being composed of triangles such as EDF, 
 which have all the same altitude DG, is equivalent (11. 5.) 
 to the rectangle under DG, and half the sum of their 
 bases, or the semiperimeter of the polygon. Therefore the 
 circle itself, since it fgrras the ultimate limit of the polygon, 
 
BOOK VI. 19*1 
 
 must have its area equivalent to the rectangle under the ra- 
 dius or the limit of all the successive altitudes and the se- 
 micircumference, which limits also the corresponding semi- 
 perimeters. 
 
 Scholium, From this proposition is derived a very sim- 
 ple and elegant method of approximating to the numeri- 
 cal expression for the area of a circle. Let the original po- 
 lygon be a square, each side of which is denoted by unit ; 
 the component sector AOB is therefore a right-angled 
 isosceles triangle, having the perpendicular OC, or the 
 radius of the inscribed circle equal to .5, and the side OA 
 of the circumscribing circle equal to V.5 or .7071067812. 
 But DG, the radius of a circle inscribed in an octagon of 
 
 OA + OC .5H-.707106812 
 the same perimeter, is = - — ^ = = 
 
 .6035533906 ; and DE the radius of the circle circumscri- 
 bing that octagon, is = V(OA.DG)= V(.603533906X 
 .707 1068 12) = .65328 14824. Again, the radius of the cir- 
 cle inscribed in a polygon of 16 sides with the same peri- 
 
 .603533906 + .65328 1 4824 
 meter, is = — = .62844174365 ; 
 
 and the radius of the circle circumscribing that polygon, 
 is =V(.6284174365 X .6532814824) = .64072SS619. In 
 like manner, the radii of circles inscribing and circum- 
 scribing the polygons of 32, 64, 128, &c. sides, under the 
 same perimeter, are successively found, by an alternate se- 
 ries of arithmetical and geometrical means. It may be ob- 
 served, that these radii mutually approximate about four 
 times nearer at each step : For (II. 10.) CA*= OA"*-— OC* 
 = (II. 17) (OA~OC)(OA + OC); and, for the same rea- 
 son, GE^= DE*— DG^ = (DE— DG)(DE+DG) But, 
 C A being double of GE, and C A* = 4GE*, it is evident 
 that (OA— OC) (OA+OC)=4(DE— PG) (DE+DG) ; 
 
19S ELEMENTS OF GEOMETRY. 
 
 and since the successive radii must approach on both sides 
 to form the same amount, or OA + OC = D£4-DG near- 
 ly, it follows tJiatOA—OCrz 4(DE— DG) nearly. In the 
 subjoine4 table, where the computation is carried to ten 
 decimal places, this rate of mutual approximation will be 
 found true to the last figure, in the expressions for the 
 radii of the circles attached to all the polygons beyond 
 that of 256 sides. Thus, for the polygon of 512 sides, 
 
 .6366237671-^.6366117828 ^^^^^^^^^^ ti • i 
 
 = .0000029960, which is the 
 
 4 
 
 difference between .6366207710 and .6366177750, the 
 
 radii of the circles described about and within the polygon 
 
 of 1024 sides. 
 
 After five or six terms have been computed, the rest 
 may be found by a simple process, because the mean 
 proportional between two proximate lines is very nearly 
 equal to half their sum, or the arithmetical mean. While 
 each number in the first column, therefore, is always equal 
 to half the sum of the preceding terms in both columns, 
 the corresponding number in the second column may be 
 considered as equal to half the sum of that number and 
 of the term immediately above itself. Thus, .6366207710, 
 the radius of the circle circumscribing the polygon of 1024? 
 sides, is equal to half the sum of .6366177750, the radius 
 of its inscribed circle, and of .6366237671, the radius of 
 the circle circumscribing the polygon of 5 1 2 sides. 
 
 But the final term may be discovered still more expedi- 
 tiously ; for, since the numbers in both columns are formed 
 by taking successive means, tho^e of the second column must 
 each time be diminished by the fourth-part of the common 
 difference, and consequently (V. 21.) the continued dimi- 
 nution will accumulate to one-third of that difference. 
 Wherefore the ultimate radius of the inscribed and cir- 
 
BOOK vr. 
 
 199 
 
 cumscribing circles, is the third-part of the sum of a radius 
 of inscription and of double the corresponding radius of cir- 
 cumscription. Thus, stopping at the polygon of 256 sides, 
 
 .6366587814-1 +2(.6366357516) ^o^^1V^►,^^. xi. n i 
 -^ ^ =.6366197724, the final 
 
 result. 
 
 NO. of sides of 
 
 Radius of Inscribed 
 
 Radius of Circum- 
 
 the Polygon. 
 
 Circle. 
 
 scribing Circle. 
 
 4 
 
 ■5000000000 
 
 •707 10678 12 
 
 8 
 
 •6035533906 
 
 •6532814824 
 
 16 
 
 •6284174365 
 
 •6407288619 
 
 32 
 
 •6345731492 
 
 •6376435773 
 
 64 
 
 •6361 083633 
 
 '636S755077 
 
 128 
 
 •6364919355 
 
 '63668369^ 
 
 Q56 
 
 •6365878141 
 
 '6366357516 
 
 512 
 
 •6366117828 
 
 •6366237671 
 
 1024 
 
 •6366177750 
 
 •6366207710 
 
 2048 
 
 •6366192730 
 
 •6366200220 
 
 4096 
 
 '6366l964>75 
 
 •6366198348 
 
 8192 
 
 •6366197411 
 
 •6366 197880 
 
 16384 
 
 •6366i976i<5 
 
 '6366197763 
 
 32768 
 
 •6366197704 
 
 •6366107733 
 
 65536 
 
 '63661977^9 
 
 •6366197726 
 
 131072 
 
 •6366197722 
 
 •6366197724 
 
 262144 
 
 •6366197723 
 
 •6366197724 
 
 Hence the radius of a circle, whose circumference is 
 4, or the diameter of a circle whose circumference is 2, 
 will be denoted by .6366197724 ; wherefore, reciprocally, 
 the circumference of a circle whose diameter is 1, will be 
 expressed by 3.1415926536, and its area, or that of the 
 ultimate polygon, by .7853981434. 
 
 In most cases, however, it will be sufficiently accurate 
 to retain only the first four figures. Wherefore 3.1416, 
 multiplied into the diameter of a circle, will denote its cir- 
 cumference, and .7854, multiplied into the square of the 
 diameter, will give the numerical expression for its area, 
 
APPENDIX. 
 
 The constructions used In Elementary Geome- 
 try, were effected, by the combination of straight 
 lines and circles. Many problems, however, can 
 be resolved, by the single application of the straight 
 line or the circle 5 and such solutions are not on- 
 ly interesting, from the ingenuity and resources 
 which they display, but may, in a variety of in- 
 stances, be employed with manifest advantage. 
 This Appendix is intended to exhibit a selection 
 of Geometrical Problems, resolved by either of 
 those methods singly. It is accordingly divided 
 into Two Parts, corresponding to the rectilineal 
 and the circular constructions. 
 
200 
 
 APPENDIX. 
 
 PART L 
 
 Problems resolved hy help of the Ruler^ or by 
 Straight Lines only, 
 
 PROP. I. PROB. 
 
 To bisect a given angle. 
 
 Let BAG be an angle, which it is required to bisect, hy 
 drawing only straight lines. 
 
 In AB take any two points D and E, from AC cut off 
 AF equal to AD and AG to AE, draw EF and DG, cross- 
 ing in the point H : AH will bisect the angle BAG. 
 
 For the triangles EAF and DAG, haying the sides EA 
 and AF equal by construc- 
 tion to GA and AD, and 
 the contained angle DAG 
 common to both, are equal 
 (I. 3.), and consequently the 
 angle AEF is equal to AGD. 
 And since AE is equal to 
 AG, and the part AD to AF, 
 the remainder DE must be . 
 equal to FG ; wherefore the 
 
 triangles DEH and HGF, having the angle at E equal to 
 that at G, the vertical angles at H equal, and also their op- 
 posite sides DE and FG, are equal (1. 20.) ; and hence the 
 side DH is equal to FH. Again, the sides AD and DH 
 
PART 
 
 201 
 
 are equal to AF and FH, and AH is common to the two 
 triangles AHD and AHF, which are therefore equal (1. 2.), 
 and consequently the angle DAH is equal to FAH. 
 
 PROP, II. PROB. 
 
 To bisect a given finite straight line. 
 
 Let it be required to bisect AB, by a rectilineal construc- 
 tion. 
 
 Draw AK diverging from AB, and make AC = CD = DE, 
 join EBj and continue it beyond B till BF be equal to BE, 
 and lastly join FC ; which will bisect AB in the point G. 
 
 For draw BH parallel to AE. 
 And because BD evidently bi- 
 sects the sides EC and EF of 
 the triangle CEF, it is parallel 
 to the base CF (VI. 1. cor. 2.) ; 
 wherefore BDCH is a parallelo- 
 gram, which has (I. 26.) its op- 
 posite sides BH and CD equal. 
 But AC being parallel to BH, 
 the angles GAC and GCA are 
 equal to GBH and GHB, and 
 the side AC, being made equal 
 to CD, is hence equal to its cor- 
 responding interjacent side BH; whence the triangles 
 AGC and BGH -are equal (I. 20.), and therefore AG i^ 
 equal to BG. 
 
202 APPENDIX. 
 
 PROP. III. PROB. 
 
 Through a given point, to draw a line parallel 
 to a given straight line. 
 
 Let it be required, by a rectilineal construction, to draw- 
 through C a straight line parallel to AB, 
 
 In AB take any two points D and F, join CD, which 
 
 produce till DE be equal to it ; ^ ^ 
 
 again Join E with the point F, 
 
 and continue this till FG be e- 
 
 qual to EF: Then CG, being ^ B'^ /f B 
 
 joined, will be parallel to AB. 
 
 For, since AB or DF evident- 
 ly bisects the sides EC and EG 
 
 of the triangle CEG, it must be parallel to the base CG 
 (VI. l.cor. 2.). 
 
 PROP. IV. PROB. 
 
 From a point in a given straight line, to erect 
 a perpendicular. 
 
 Let C be a given point, from which it is required, by 
 help of straight lines merely, to erect a perpendicular to 
 AB. 
 
 In AB, having taken any point D, draw DE equal to 
 DC and inclined to AB, join EC and produce it until CG 
 be equal to CD or DE, make GF equal to CE, join FG 
 
PART 
 
 203 
 
 and produce this till GH be equal to GC : Then CH will 
 be perpendicular to AB. 
 
 For the triangles DCE 
 and GCF, having the sides 
 DC, CE equal to GC, CF, 
 and the contained angles 
 vertical at C, are equal 
 (I. 3.); whence FG = CD 
 = CG=GH. The point 
 G is therefore the centre of 
 a semicircle which would 
 pass through F, C, H, and 
 consequently the angle FCH 
 is a right angle (III. 19.), or CH is perpendicular to AB. 
 
 A B 
 
 PROP. V. PROB. 
 
 To let fall a perpendicular upon a given straight' 
 line, from a point -without it. 
 
 Let C be a given point, from which it is required, by a 
 rectilineal construction, to let fall a perpendicular to AB. 
 
 In AB take any 
 point D, draw DF ob- 
 liquely, and make DE 
 = DF=DG, join FE 
 and produce it until 
 EH be equal to EG, 
 make EI = EF, join 
 HI, and (Appendix, 
 Part I. Prop. 3.) draw 
 CK parallel to it : CK is the perpendicular required. 
 
 A.G 
 
 IB 
 
^04? APPENDIX. 
 
 For the point D being obviously the centre of a semi- 
 circle passing through G, F, and E, the angle GFE is a 
 right angle ; and the triangles EOF, EHI, having the 
 sides GE, EF equal to HE, EI, and their contained an- 
 gles vertical, — are equal (I. 3.), and consequently the an- 
 gle HIE is equal to GFE, or is a right angle ; but since 
 CK and HI are parallel, the angle CKA is equal to HIE 
 (I. 22.), and therefore is also a right angle, or CK is per- 
 pendicular to AB. 
 
 PART 11. 
 
 Geometrical Prohlems resolved by means of Compas- 
 ses, or by the mere description of Circles* 
 
 PROP. I. PROB. 
 
 To repeat a given distance in the same direc- 
 tion. 
 
 Let A and B be two given points; it is required io find, 
 by means of compasses only, a series of equidistant points 
 in the same extended line. 
 
 From B as a centre, with the given distance BA, de- 
 scribe a portion of a circle, in which inflect that distance 
 three times to C ; from C, with the same radius, describe 
 
PART II. 205 
 
 another circle, and InseK 
 the triple chords to D ; re- 
 peat that process from D, 
 E, &c. : The equidistant 
 
 points A, B, C, D, E, &c. will all lie in the same straight 
 line. 
 
 For, by this construction, three equilateral triangles are 
 formed about the point B, and consequently (I. SO. cor. 1.) 
 the whole angle ABC, made by the opposite distances BA 
 and BC, is equal to two right angles, or ABC is a straight 
 line. The same reason applies to the successive points, D, 
 E, &c. 
 
 PROP. 11. PROB. 
 
 To find the direction of a perpendicular from a 
 given point to the straight line joining it with 
 another given point. 
 
 Given the points A and B : to find a third point, sucl> 
 that the straight line connecting it with B shall be at right 
 angles to BA. 
 
 From A and B, with any conve- 
 
 , M . NT) 
 
 nient distance, describe two arcs in- 
 tersecting in C, from which, with /' 
 the same radius, describe a portion ,/S 
 of a circle passing through the points 
 
 A and B, and insert that radius three x'c 
 
 times from A to D : BD is perpen- — ' 
 
 dicular to BA. 
 
206 APPLNDIX. 
 
 For it is evident, from the last Proposition, that the arc 
 ABD is a semicircumference, and consequently (III. 19.) 
 the angle ABD contained in it is a right angle. 
 
 Scholium, The construction would be somewhat simpli- 
 fied, by taking the distance AB for the radius. 
 
 PROP. III. PROB. 
 
 To find the direction of a perpendicular let fall 
 from a given point upon the straight line which 
 connects two given points. 
 
 Let C be a point, from which a perpendicular is to be 
 
 let fall upon the straight line joining A and B. 
 
 From A as a centre, with the 
 
 C 
 distance AC, describe an arc, ; 
 
 and from B as a centre, with the | 
 
 distance BC, describe another ' | 
 
 arc, intersecting the former in ''^ ; ^ 
 
 the point D : CD is perpendicu- i 
 
 lar to AB. % 
 
 For CAD and CBD are evi- 
 dently isosceles triangles, and consequently (I. 7.) their 
 vertices must lie in a straight line AB which bisects their 
 base CD at right angles. 
 
 Scholium, It will be perceived that this construction dif- 
 fers not in any respect from the mode employed in Prop. 6. 
 Book I. of the Elements. 
 
PART II. 
 
 209 
 
 PROP. IV. PROB. 
 
 To bisect a given distance. 
 
 Let A and B be two given points ; it is required to find 
 the middle point in the same direction. 
 
 From B as a centre, with the radius BA, describe a se- 
 micircle, by inserting that distance successively from A to 
 C, D, and E ; from A as a centre, with the distance AE, de- 
 scribe a portion of a circle FEG, in which, from the point 
 E, inflect the chords EF and EG equal to EC; and from the 
 points F and G, with the same 
 radius EC describe arcs intersect- 
 ing in H : This point bisects 
 the distance AB. 
 
 For, by the first Proposition, 
 the points A, B, and E extend 
 in a straight line ; but the trian- 
 gles FAG, FHG, and FEG, be- 
 ing evidently isosceles, their ver- 
 tices A, H, and E (I. 7.) must 
 lie in a straight line ; whence 
 the point H lies in the direction 
 
 AB. Again, because EFH is an isosceles triangle, 
 AF*— HF^ = EA.AH (II. 20.); that is, AE*— EC% or 
 (III. 19. and II. 10.) AC* or AB* = EA.AH. Wherefore, 
 since EA is double of AB, the segment AH must be its 
 half. 
 
21© 
 
 AlPPENDlX. 
 
 pAop. v. prob. 
 
 To trisect a given distance. 
 
 Let it be required to find two intermediate points that 
 are situate at equal intervals in the line of communication 
 AB. 
 
 Repeat (App. II. 1.) the distance AB on both sides to 
 C and D ; from these points, with the radius CD, describe 
 the arcs EDF and GCH, from D and C inflect the chords 
 DE and DF, CG and 
 
 CH, all equal to DB, E/ \G: 
 
 and, with the same dis- 
 tance and from the 
 points E and F, G and 
 H, describe arcs inter- 
 secting in I and K: 
 The distance AB is tri- 
 sected by the points I 
 and K. 
 
 For it may. be de- 
 monstrated, as in the 
 last proposition, that 
 the points I and K lie 
 in the same direction 
 
 AB. In like manner, it appears (II. 20») that DG* 
 
 KG^ = CD.DK, or 9AB^-.4AB% or 5AB* = 3AB.DK ; 
 and consequently 5AB=3DK, or2AB=3AK, and AB=: 
 3BK. But, for the same reason, AB = 3AI. 
 
 
 ■■'•'-]C 
 
PART II. 
 
 211 
 
 T 13 C X) E F 
 
 I 
 
 PROP. VI. PROB. 
 To cut off any aliquot part of a given distance. 
 
 Suppose It were required to cut off the fifth part of the 
 distance between the points A and B. 
 
 Repeat (App. II. 1.) the distance AB four times, to F ^ 
 from F, with the radius FA, describe the arc GAH ; in- 
 flect the chords AG 
 and AH equal to AB, '^- 
 
 and, with that radius 
 and from the points G -A 
 and H, describe^ arcs 
 intersecting in I : AI 
 is the fifth part of the line of communication AB. 
 
 For, as before, the point I is situate in AB. But since 
 AGI is evidently an isosceles triangle, and AF is equal to 
 FG, it follows (II. 23. cor.) that AG* = AF.AI, and con- 
 sequently AB^=5AB.AI; whence AB=5AI. 
 
 PROP. VII. PROB. 
 To divide a given distance by medial section. 
 
 Let it be required to cut the distance AB, such that 
 BH* = BA.AH. 
 
 From B describe a circle with the radius BA, which 
 insert successively from A to D, E, C, and F ; from 
 the extremities of the diameter AC and with the chord 
 AE, describe two arcs intersecting in G ; and, from the 
 
212 
 
 APPENDIX. 
 
 ^ 
 
 points E and F with the distance BG, describe other two 
 arcs intersecting in H : This is the point of medial section. 
 For it is evident that this point H lies in the straight line 
 AB. And because the trian- 
 gles AGB, CGB have their 
 sides respectively equal, the 
 angle ABG (I. 2.) is a right 
 angle, and consequently (II. 
 10.) AG* = AB^ + BG^; but 
 AG=AE, and AE^ = 3AB* 
 (IV. 17. cor. 2.); wherefore 
 SAB* = AB* + BG% and 
 BG* = 2AB*. Now since 
 BE = EC, it follows, (II. 20.) 
 that HE*— BE* = CH.HB; but HE*— BE* = BG*— 
 BE* = AB*, and therefore AB* = CH.HB. Whence CH 
 is cut by a medial section at B, and consequently (II. 19. 
 cor. 1.) its greater segment BC or AB is likewise divided 
 medially at H by the remaining portion BH. 
 
 PROP. VIII. PROB. 
 To bisect a given arc of a circle. 
 
 Let it be required to bisect the arc AB of a cifcle whose 
 centre is C. 
 
 - From the extremities A and B with the radius AC, de- 
 scribe opposite arcs, and from the centre C inflect the 
 chord AB to D and E ; from these points, with the dis- 
 tance DB describe arcs intersecting in F; and from D or 
 E, with the distance CF, cut the given arc AB in G : AB 
 is bisected in that point. 
 
PART II. 213 
 
 For the figures ABCD and ABEC being evidently 
 rhomboids, DC and CE are parallel to AB, and hence con- 
 stitute one straight 
 
 'line; consequently y<(^ 
 
 the triangles DEC 
 and EEC having 
 their correspond- 
 ing sides equal, the 
 angle DCF is a 
 right angle, and 
 (II. 10.) DF^ = 
 DC^ + CE^ But, 
 
 in the rhomboid ABCD, DB* + CA^=2DC^+2CB* 
 (II. 22.), or BD^ = 2DC^ + CB^; and since DB = DF, 
 2DC* + CB^ = DC^ + CF% whence DC* + CB^ = CF% 
 or DC* + CG* = DG*, and therefore (II. 11.) DCG is a 
 right angle. And because CG is perpendicular to DC, it 
 is likewise (I. 22.) perpendicular to AB, and the triangles 
 CAP and CBP are equal (I. 21.), and the angle ACG 
 equal to BCG ; whence (III. 12.) the arc AG = BG. 
 
 PROP. IX. PROB. 
 
 To find the centre of a circle. 
 
 Assume an arc AB greater than a quadrant, and from 
 one extremity B, with the distance BA, describe a semi- 
 circle ADC, cutting the given circumference in D ; from 
 the points B and C, with the distance CD, describe arcs 
 intersecting in E, and, from that point with the same dis- 
 tance, describe an arc cutting ADC in F ; and lastly, fron^ 
 
214 
 
 APPENDIX. 
 
 the points A and B, with the distance AF, describe arcs 
 intersecting in G : This point is the centre of the circle 
 ADB. 
 
 For the isosceles triangles BEC, BEF, being evidently 
 equal, the angle FBC is equal to both the angles at the 
 base ; but FBC is (I. S2. El.) equal to the interior angles 
 BAF and BFA of the iso- 
 sceles triangle ABF, and ^^^ 
 hence that triangle is simi- ,-' /\ 
 lar to BEF. Wherefore >''' •' \ 
 BE : BF : : BA : AF, or 
 CD : BD : : BA : AG; 
 consequently the isosceles 
 triangles CBD and BGA 
 (VI. 12. cor.) are similar, 
 and the angle BCD is e- 
 qual to GBA ; BG is, therefore, parallel to CD, and hence 
 (I. 30. p.) the angle BDC, or BCD, is equal to GBD. 
 The triangles BGA and BGD, having thus the side BA 
 equal to BD, BG common, and equal contained angles 
 GBA and GBD, are (I. 3. El.) equal, and therefore the 
 side GA is equal to GD. The point G being thus equi- 
 distant from three points. A, D, and B in the circumfe- 
 rence, is hence (III. 8. cor.) the centre of the circle. 
 
 PROP. X. PROB. 
 
 To divide the circumference of a given circle 
 successively into four, eight, twelve, and twenty- 
 four equal parts. 
 
PART II. 
 
 215 
 
 1. Insert the radius AB three times from A to D, E, 
 and C ; from the extremities of the diameter AC, and with 
 a distance equal to the chord AE, describe arcs in- 
 tersecting in the point F ; and from A, with the distance 
 BF, cut the circumference on opposite sides at G and H : 
 AG, GC, CH, and HA are quadrants. 
 
 For, as before, AF*==AE^ = 3AB*; and the triangle 
 ABF being right-angled, 3AB^ = AF* = AB^ + BFS and 
 therefore BF' = AG* = 2AB* j whence (11. 12.) ABG is 
 a right angle, and AG a quadrant. 
 
 2. From the point F with -p 
 the radius AB, cut the circle in ,-^^ 
 I and K, and from A and C in- 
 ject the chord AI to L, H and 
 M; the circumference is divided 
 into eight equal portions by the 
 points A, I, G, K, C, M, H, 
 and L. 
 
 For BF*, being equal to 
 2AB*, is equal to the squares 
 of BI and IF, and consequently PIF is a right angle ; but 
 the triangle BIF is also isosceles, and therefore the angle 
 IBF at the base is half a right angle ; whence the arc IG 
 is an octant, 
 
 3. The arc DG, on being repeated, will form twelve e- 
 qual sections of the circumference. 
 
 For the arc AD is the sixth or two-twelfth parts of the 
 circumference, and AG is the fourth or three-twelfths ; 
 consequently the difference DG is one-twelfth. 
 
 4. The arc ID is the twenty-fourth part of the circum- 
 ference. 
 
 For the octant AI is equal to three twenty-fourths, and 
 
216 APPENDIX. 
 
 tbe sextant AD is equal to four twenty-fourths ; their dif- 
 ference ID is hence one twenty-fourth part of the circum- 
 ference. 
 
 PROP. XI. PR OB. 
 
 To divide the circumference of a given circle 
 successively into five, ten, and twenty equal parts. 
 
 Mark out the semicircumference ADEC, by the triple 
 insertion of the radius, from A and C, with the double 
 chord AE, describe arcs intersecting in F, from A, with the 
 distance BF, cut the circle in G and K, inflect the chords 
 GH and GI equal to the radius AB, and, from the points 
 H and I, with the distance \^ 
 
 BF or AG, describe arcs in- ^ 
 
 tersecting in L. ^Dx'' ^"^^"^^iE 
 
 It is evident from App. II. pry \t 
 
 7, that BL is the greater 7 \ 
 
 segment of the radius BH a[ i _q 
 
 divided by a medial section ; \ I 
 
 wherefore(;iV.23.cor.2. Eh) \ X / 
 
 AL is equal to the side of ^s^ ^^ 
 
 the inscribed pentagon, and ^ J^ 
 
 BL to that of the decagon inscribed in the given circle. 
 Hence AL may be inflected five times in the circumfe- 
 rence, and BL ten times ; and consequently the arc MK, 
 or the excess of the fourth above the fifth, is equal to the 
 twentieth part of the whole circumference. 
 
 Scholium. This proposition, and the preceding, include 
 the happiest application of the circle to the solution of 
 such problems. 
 
PART II. 
 
 217 
 
 PROP. XII. PROB. 
 
 From a given side to trace out a square. 
 
 Let the points A and B terminate the side of a square, 
 which it is required to trace. 
 
 From B as a centre describe 
 the semicircle ADEC, from A 
 and C, with the distance AE, 
 describe arcs intersecting in F, 
 from A, with the distance BF, 
 cut the circumference in G, 
 and from A and G, with the 
 
 radius AB, describe arcs intersecting in H : The points 
 H and G are corners of the required square. 
 
 For (App. II. 10.) the angle ABG is a right angle, and 
 the distances AB, AH, HG, and GB, are, by construction, 
 all equal. 
 
 PROP. XIII. PROB. 
 
 Given the side of a regular pentagon, to find 
 the traces of the figure. 
 
 From B describe through A the circle ADECF, in 
 which the radius is inflected four times, from A and C 
 with the double chord A E describe arcs intersecting in G, 
 from E and F, with the distance BG, describe arcs inter- 
 secting in H, from A, with the radius AB, describe a por- 
 tion of a circle, inflect BH thrice from B to L and from 
 
218 
 
 APPENDIX. 
 
 .t 
 
 A to O, and lastly from L and O, with the radius AB, 
 describe arcs intersecting in P : The points A, L, P, O, 
 B mark out the polygon. 
 
 For, from App. II. 7, it is evident that BH is the greater 
 segment of the distance 
 AB divided by a medial X. 
 
 section. Consequently ^y 
 
 (IV. 3. El.) the isosceles 
 triangles BAI, lAK, 
 KAL, ABM, MBN, 
 and NBO, have each of 
 the angles at the base 
 double their vertical an- 
 gle. Wherefore the an- 
 gles BAL and ABO are 
 each of them six-fifths of a right angle (IV. 4. cor.), and 
 hence (I. S3, cor.) the points L and O are corners of the 
 pentagon ; but P is evidently the vertex of the pentagon, 
 since the sides LP and OP are each equal to AB. 
 
 Scholium. The pentagon might also have been traced, as 
 in Book IV. Prop. 5, by describing arcs from A and B 
 with the distance HC, and again, from their intersection 
 P, and with the radius AB, cutting those arcs in L and O. 
 It is likewise evident, from Book IV. Prop. 8, that the 
 same previous construction would serve for describing a 
 decagon, P being made the centre of a circle in which AB 
 is inflected ten times. 
 
PART IT. 
 
 219 
 
 PROP. XIV. PROB. 
 
 The side of a regular octagon being given, to 
 mark out the figure. 
 
 Let the side of an octagon terminate in the points A and 
 B ; to find the remaining corners of the figure. 
 
 From the centres A and B, with the radius AB, de- 
 scribe the two semicircles AEFC and BEGD ; with 
 the double chord AF, and from A, C and B, D de- 
 scribe arcs intersecting in H, I ; from these points, with 
 the radius AB, cut the semicircles in K, L : on HI de- 
 scribe the square HMNI, by making the diagonals HN, 
 IM equal to BH, and 
 the sides equal to AB ; 
 and, on MH and NI, 
 describe the rhombus- 
 ses MOKH and 
 NPLI : The points A, 
 B,K,0,M,N,P,and I- 
 L, are the several cor- 
 ners of the octagon. 
 
 For (by App. II. 
 Prop. 10.) BH, AI are both of them perpendicular to 
 BA, and BKH, ALI are right angled isosceles triangles ; 
 HI is therefore parallel to BA, and HMNI, consisting of 
 triangles equal to BKH, is a square ; whence all the sides 
 AB, BK, KO, OM, MN, NP, PL, and LA of the octa- 
 gon are equal : But they likewise contain equal angles ; for 
 ABK, composed of ABH and HBK, is equal to three 
 half right angles, and BKO, by reason of the parallels BH 
 and KO, being the supplement of HBK, is also equal to 
 
220 APPENDIX. 
 
 three half right angles. In the same manner, the other 
 angles of the figure may be proved to be equal. 
 
 PROP. XV. PROB. 
 On a given diagonal to describe a square. 
 
 Let the points A and B be the opposite corners of a 
 square which it is required to trace. 
 
 From B as a centre describe the semicircle ADEC, 
 from A and C with the double chord AE describe arcs in- 
 tersecting in F, from C with the distance BF describe an 
 arc and cut this from A with the radius AD in G, and 
 lastly from B- and A with the 
 distance BG describe arcs inter- >< 
 
 secting in H and I : ABHI is ^ 
 
 the required square. /rrV^ \ 
 
 For, in the triangle AGC, the / .^, \ 
 
 straight line GB bisects the base, //' '\ \ 
 
 and consequently (II. 22.) AG* -^\ /^ ^ 
 
 + CG* = 2AB* + 2BG* ; but, *y ' 
 
 (by App. II. Prop. 10.) CG*= ^ 
 
 BF* = 2AB*; whence AG* = AB* = 2BG% and (II. 11.) 
 AHB is a right angle ; and the sides AH, HB, BI, and 
 lA being all equal, the figure is therefore a square. 
 
 PROP. XVI. PROB. 
 
 Two distances being given, to find a third pro- 
 portional. 
 
PART II. 221 
 
 Let it be required to find a third proportional to the dis- 
 tances AB and CD. 
 
 From any point E, K ^j j3 
 
 and with the distance '•.*". Ci OO 
 
 AB, describe a portion 
 
 of a circle, in which / ' \H 
 
 inflect FG equal to CD, 
 
 and from G, with that ,^r^i.^^-_L^.i^^, - t- 
 
 distance, describe the j^\ 133 
 
 semicirde FHI ; HI ^^^.^^-.......,B 
 
 is the third propor- 
 tional required. / /£' 
 For-the atigfes GEH 
 
 and IGH are each of ^^-- ^G i 
 
 them double the angle GFH or IFH at the circumference 
 (III. 17. EL); whence the triangles GEH and IGH must 
 ako have the angles at the base equal, and are consequent- 
 ly similar : Wherefore (VI. 12. El.) EG : GH : : GH : HI. 
 If the first term AB be less than half the second term 
 CD, this construction, without some help, would evident- 
 ly not succeed. But AB may be previously doubled, or 
 assumed 4, 8, or 16 times greater, so that the circle FGH 
 shall always cut FHI ; and in that case, HI, being like- 
 wise doubled, or taken 4, 8, or 16 times greater, will give 
 the true result. 
 
 PROP. XVII. PROB. 
 
 To find a fourth proportional to three givea 
 distances. 
 
222 
 
 APPEI9D1X. 
 
 Let it be required to find a fourth proportional to the 
 distances AB, CD, and EF. 
 
 From any point G, de- 
 scribe two concentric circles 
 HI and KL with the distan- 
 ces AB and EF; in the cir- 
 cumference of the first inflect 
 HI equal to CD, assume any 
 point K in the second 'cir- 
 cumference, and cut this in 
 L by an arc described from I 
 with the distance HK ; the 
 chord LK is the fourth pro- 
 portional required. 
 
 For the triangles ILG and HKG are equal, since their 
 corresponding sides are evidently equal ; whence the an- 
 gle IGL is equal to HGK, and taking away HGL, ther 
 angle IGH remains equal to LGK ; consequently the iso- 
 celes triangles GIH and GLK are similar, and GI : IH : : 
 GL : LK, that is, AB : CD : : EF : LK. 
 
 If the third term EF be more than double the first AB, 
 this construction, it is obvious, will not answer without 
 some modification. It may, however, be made to suit all 
 the variety of cases, by multiplying equally AB and the 
 chord LK, as in the last proposition. 
 
 PROP. XVIII. PROB. 
 
 To find the linear expressions for the square 
 roots of the natural numbers, from one to ten in- 
 clusive. 
 
PART II. 
 
 228 
 
 This problem is evidently the same as, to find the sides 
 of squares which are equivalent to the successive multiples 
 of the square constructed on the straight line representing 
 the unit. Let AB, therefore, be that measure : And from 
 B as a centre, describe a circle, in which inflect the radius 
 four times, from A to C, D, E, and F ; from the opposite 
 points A and E, with the double chord AD, describe arcs 
 intersecting in G and H,— with the same distance, and 
 from the points D, F, describe arcs intersecting in 1, — 
 and, with still the same distance and from E, cut the cir- 
 cumference in K ; 
 and from A and K, 
 with the radius AB, 
 describe arcs inter- 
 secting in L ; Then 
 will AK^ = 2AB% 
 AD* = 3AB*, AE* 
 =4AB% IK^ = 
 5AB%IG*=6AB*j 
 IC* = 7AB^GH* = 
 8AB% IA* = 9AB% 
 andIL* = 10AB^ 
 
 For, in the isosceles triangles ACB and BDE, the per- 
 pendiculars CO and DP must bisect the bases AB^'and 
 BE ; and the triangle ADI being likewise isosceles, IP = 
 AP, and consequently IB = AE = 2AB. But, from what 
 has been formerly shown, it is evident that AK* = 2AB* 
 and AD* = 3AB*; and since AE = 2AB, AE*=4.AB^ 
 In the right-angled triangles IBK and IBG, IK* = IB*-f- 
 BK*=4EB* + BK* = 5AB%IG* = IB* + BG* = 4AB* + 
 2AB* = 6AB*5 but (II. 23.) IC* = IB*-}-BC* + IB.2BO 
 = 4AB*+AB*^-2AB* = 7AB^ Again, GH being double 
 
 --■XT 
 
 ^ 
 
224- APPENDIX. 
 
 of BG, GH* = 4j . 2AB* =8AB% and AI being the triple 
 of AE, AI*=9AB* ; and lastly, lAL being a right-an- 
 gled triangle, IL* =IA* + AL* =9AB» + AB* = lOAB*. 
 If AB, therefore, denote the unit of any scale, it will 
 follow, that AK=V2, AD=V3, AE=V4, IK=V5, 
 IG= V6, IC= V7, GH= VS, IA= V9, and IL= VlO. 
 
ELEMENTS 
 
 OP 
 
 PLANE TRIGONOMETRY. 
 
 Trigonometry is the science of calculating the 
 sides or angles of a triangle. It grounds its con- 
 clusions on the application of the principles of 
 Geometry and Arithmetic. 
 
 The sides of a triangle are measured, by refer- 
 ring them to some definite portion of linear ex- 
 tent, which is fixed by convention. The mensu- 
 ration of angles is effected, by means of that uni- 
 versal standard derived from the partition of a 
 circuit. Since angles were shown to be propor- 
 tional to the intercepted arcs of a circle described 
 from their vertex, the subdivision of the circum- 
 ference therefore determines their magnitude. A 
 quadrant, or the fourth-part of the circumference, 
 as it corresponds to a right angle, hence forms 
 the basis of angular measures. But these mea- 
 sures depend on the relation of certain orders of 
 lines connected with the circle, and which it is 
 necessary previously to investigate. 
 
226 ELEMENTS OF 
 
 DEFINITIONS. 
 
 1 . The complement of an arc is its defect from a quadrant ; 
 its supplement is its defect from a semicircumference ; and 
 its explement is its defect from the whole circumference. 
 
 2. The sine of an arc is a perpendicular let fall from 
 one of its extremities upon a diameter passing through the 
 other. 
 
 3. The versed sine of an arc is that portion of a diame- 
 ter intercepted between its sine and the circumference. 
 
 4. The iarigent of an arc is a perpendicular drawn at 
 one extremity to a diameter, and limited by a diameter 
 extending through the other. 
 
 5. The secant of an arc is a straight line which joins 
 the centre with the termination of the tangent. 
 
 In naming the sine^ tangent, or secant, of the complement of an 
 arc, it is usual to employ the abbreviated terras of cosine, cotan- 
 gent and cosecant, A farther contraction is frequently made in 
 noting the radius and other lines connected with the circle, by 
 retaining only the first syllable of the word, or even the mere 
 initial letter. 
 
 Let ACFE be a circle, of which the diameters AF and CE 
 are at right angles ; having taken any arc AB, produce the 
 radius OB, and draw BD, AH perpendicular to AF, and BG, 
 
TRIGONOMETEY. 
 
 227 
 
 CI perpendicular to CE. Of this assumed arc AB, the com- 
 plement is BC, and the supplement 
 BCF ; the sine is BD, the cosine 
 BG or OD, the versed sine AD, 
 the coversed sine CG, and the sup- 
 plementary versed sine FD ; the 
 tangent of AB is AH, and its co- 
 tangent CI ; and the secant of the 
 same arc is OH, and its cosecant 
 01. 
 
 Several obvious consequences flow from these defini- 
 tions : — 
 
 1. Since the diameter which bisects an arc bisects also 
 the chord at right angles, it follows that half the chord of 
 any arc is equal to the sine of half that arc. 
 
 2. In the right-angled-triangle ODB, BD* + Or)* = 
 OB* ; and hence the squares of the sine and cosine of an 
 arc are together equal to the square of the radius. 
 
 3. The triangle ODB being evidently similar to OAH, 
 OD : DB ; OA : AH ; that is, the cosine of an arc is to 
 the sine, as the radius to the tangent. 
 
 4. From the similar triangles ODB and OAH, OD : OB 
 ; : OA : OH ; wherefore the radius is a mean proportional 
 between the cosine and the secant of an arc. 
 
 5. Since BD* = AD.FD, it is evident that the sine of an 
 arc is a mean proportional between the versed sine and the 
 
228 ELEMENTS OF 
 
 supplementary versed sine, or between the sum and differ- 
 ence of the radius and the cosine. 
 
 6. Hence also the chord of an arc is a mean proportional 
 between the versed sine and the diameter 5 for AB* = 
 AD.AF. 
 
 7. The triangles OAH and ICO being similar, AH : OA 
 : : OC : CI j and hence the radius is a mean proportional 
 between the tangent of an arc and its cotangent. 
 
 8. Since OD* = BG=^ = CG.CE, it follows that the cosine 
 of an arc is a mean proportional between the sum and the 
 difference of the radius and the sine. 
 
 The circumference of the circle is commonly divided 
 into 360 equal parts, called degrees, each of them being 
 subdivided into 60 minutes, and these again being each 
 distinguished into 60 seconds. It very seldom is required 
 to carry this subdivision any farther. Degrees, minutes, 
 seconds, or thirds, are conveniently noted by these marks, 
 
 o / // ttt 
 
 Thus, 23° 27' 43'^ 42''', signifies 23 degrees, 27 minutes, 
 43 seconds, and 42 thirds. 
 
 Scholium, To discern more clearly the connection of the 
 lines derived from the circle, it will be proper to trace their 
 successive values, while the corresponding arc is supposed 
 to increase. Let the arc AB', on the opposite side, be made 
 equal to AB, draw the diameter FOA, extend the diame- 
 ters 5'OB and 60B', join BB' and 66', and at A apply the 
 
TRIGONOMETRY. 
 
 229 
 
 double tangent HAH'. It is evident that BE=zbe, or that 
 the sine of the arc AB is equal to the sine of its supple- 
 ment ABb. But B'E and 6V, or the sines of ABF6' and 
 ABFZj'B' which lie on the opposite side of the diameter, 
 are likewise equal to BE ; that 
 is, the inverted sine of an arc 
 is equal to the sine of that arc 
 or of its supplement, augment- 
 ed, each by a semicircumfe- 
 rence. The arc AB, and its 
 defect A^FB' from a whole 
 circumference, have both the 
 same cosine OE ; and the sup- 
 plemental arc ABb, and its de- 
 fect from a whole circumference, have likewise the same 
 cosine, although with an inverted position. AH and OH 
 are respectively the tangent and secant not only of AB, but 
 of the arc ABbFb\ which is compounded of the original 
 arc and a semicircumference ; and the similar lines AH' 
 and OH', on the opposite side, are at once the tangent 
 and secant of the supplementary arc AB^, and of AB6F6'B', 
 likewise compounded of that arc and a semicircumference. 
 As the prolonged diameter Z>'OBH, therefore, turns a- 
 bout the centre, the sine and tangent both increase, till the 
 arc attains 90°, when the sine becomes equal to the radius, 
 and the tangent vanishes into unlimited extent. Between 
 90^ and 180^, the sine again diminishes, and the tangent, 
 re-appearing in the opposite direction, likewise contracts 
 by successive diminutions. In the third quadrant, the sine 
 emerges with a contrary position, and increases till it be- 
 comes equal to the radius j while the tangent, resuming 
 
230 ELEMENTS OF 
 
 its first position, stretches out till it vanishes away. Be- 
 tween 270^^ and 360", the opposite sine again contracts, 
 and the tangent, re- appearing on the same side, shrinks 
 also by degrees to a point. In the firs^t and fourth qua- 
 drants, the cosine lies on the same side of the centre, while 
 the secant stretches from it in the direction of the extre- 
 mity of the arc ; but, in the second and third quadrants, 
 the cosine shifts to the opposite side, and the secant shoots 
 from the centre in a direction opposite to the termination 
 of the arc. 
 
 The same phases are thus repeated at each succeeding re- 
 volution. Hence, if m denote any integral number, the sine 
 of an arc a is equal to the sine of the arc {^m — i) 180*^ — a, 
 and to opposite sines of (2/w-i) 180^ + « and of 2»2.180° — a, 
 the cosine and secant of an arc a are equal to the cosine 
 and secant of 2w.l80® — a^ and to the opposite cosines and 
 secants of (2?w — i) 180°— « and of (2w — i) 180° + «; and 
 the tangent or cotangent of an arc a is equal to the tangent 
 or cotangent of the arc (2w — i) 180° -J- «, and to the op- 
 posite tangents or cotangents of the arcs (27w — i) 180 — a 
 and 2»z.l80— «. 
 
 An arc may, by a simple extension of analogy, be con- 
 ceived to comprehend innumerable other arcs. Thus, the 
 arc AB, in fact, represents all the arcs which have their 
 origin at A and their termination at B j it therefore in- 
 cludes not only the small arc AB, but that arc as aug- 
 mented by successive revolutions, or the repeated addition 
 of entire circumferences. Hence the sine or tangent of an 
 arc a are the same with the sine or tangent of any arc 
 «.360«'+ff. 
 
TRIGONOMETRY. 
 
 231 
 
 PROP. I. THEOR. 
 
 The rectangle under the radius and the sine of 
 the sum of two arcs, is equal to the sum of the 
 rectangles under their alternate sines and cosines. 
 
 Let A and B denote two arcs, of which A is the great- 
 er; then, R.sin{A-{''B)=:sinA.cos'B-{-cosA.sinB. 
 
 For it is evident that AC will represent the sum of the 
 arcs AB and BC ; make BC equal to BC, and join OB 
 and CC, and draw HFH' parallel, and CE, FG, BD, 
 and HC'E' perpendicular, to the radius OA. 
 
 The triangles COF and C'OF, having the side CO equal 
 to CO, OF common, and the contained angles FOC and 
 FOC measured by the equal arcs 
 BC and BC, are equal ; wherefore 
 OF bisects CC at right angles. But 
 the triangles OBD and OFG being 
 similar, OB : BD : : OF : FG, or 
 HE, and consequently OB.HE = 
 BD.OF. The triangles OBD and 
 CFH are likewise similar, for the 
 right angle CFO being equal to HFG, if HFO be taken from 
 both, the remaining angle CFH is equal to OFG or OBD ; 
 whence OB : OD : : CF : CH, and OB.CH = OD.CF. 
 Wherefore OB.HE + OB.CH, or OB.CE = BD.OF+ 
 OD.CF. But BD and OD are the sine and cosine of the 
 arc AB, CF and OF the sine and cosine of BC, and CE 
 is the sine of the compound arc AC. Consequently, 
 B smAC=sinA'B cosBC + cos AB sinBC, 
 
 E'A 
 
2S2 ELEMENTS OF 
 
 Cor, 1. Hence, likewise, the rectangle under the radius 
 and the sine of the difference of two arcs, is equal to the 
 difference of the rectangles under their alternate sines and 
 cosines \ or R sin AU= sin AB cosBC-^cos AB sinBC. 
 
 Cor, 2. If the two arcs A and B be equal, it is obvious 
 thsit R sin2A=: sin A 2cos A, 
 
 Cor. 3. Let the arc A contain 45° ; then 
 R 5m(45<?=fcrB)=sm45°(co5B=±:5mB)= V lR''{cosB=t:sinB) 
 or RMn{^5'^z±zB)z:2RVl{cosB=±z$inB), 
 
 Cor. 4. Let 2A=C, and, by the second corollary, 
 
 RsinC=isin^C 2cos^C. 
 
 PROP. IL THEOR. 
 
 The rectangle under the radius and the cosine 
 of the sum of two arcs, is equal to the difference 
 of the rectangles under their respective cosines 
 and sines. 
 
 Let A and B denote two arcs, of which A is the great- 
 er ', then R cos{A + B) = cos A cosB — sin A si?iB. 
 
 For, in the preceding figure, the triangles OBD and 
 OFG being similar, OB : OD : : OF : OG, and OB.OG = 
 OD.OF, and the triangles OBD and CFH being likewise 
 similar, OB : BD : : CF : FH, or GE, and consequently 
 OB.GE = BD.CF. Wherefore OB.OG — OB.GE= 
 OB.OE = OD.OF— BD.CF ; that is, 
 
 R cosACzz cos AB cosBC — sin AB sinBC, 
 
TRIGONOMETRY. 333 
 
 Cor. 1. Hence, likewise, the rectangle under the radius 
 and the cosine of the difference of two arcs is equal to the 
 sum of the rectangles under their respective cosines and 
 sines ; or R.cos AC = cos AB cosBC+ sin AB sinBC, 
 
 Cor. ^. If A and B represent two equal arcs, it will follow, 
 that B'Cos2A = cosA^ — sinA^ = {cos A + smA){cosA — sin A) ; 
 or, since cosA^:=:R^ — sinA^^ 
 
 B cos2A=i^*-:-25^wA*=2co5A*—i2^ 
 
 Cor. 3. Since, sinA^=^R{R — cos2A)^ and 
 sinB^=l;R{R — cos2B); therefore 
 sfwA*— 5mB^ =4i2(co52B— C052A). 
 
 Cor, 4. Let the arc A be equal to 45^, and 
 R co5(45?=±=B)=5m4.5°(co5B=:p5/wB). 
 
 Cor. 5, Let 2A = C, and by the second corollary, 
 R cosC=:R^-'2sinlC*=z2cosiC^^R\ 
 
 PROP. in. THEOR. 
 
 Of the equidifferent arcs, the rectangle under 
 the radius and the sum of the sines of the ex- 
 tremes, is equal to twice the rectangle under the 
 cosine of the c.ommon difference and the sine of 
 the mean arc. 
 
 Let A — B, A, and A+B represent three arcs increasing 
 by the difference B ; then 
 
 R{^sin{A+B)+sin{A^B)\=z2cosB sitiA. 
 
234? 
 
 ELEMENTS OF 
 
 The property is easily deduced by combining the preced- 
 ing theorems; but it 
 will be more easily 
 perceived, by refer- 
 ring immediately to 
 the original figure. 
 The triangles OBD 
 and OFG being si- 
 milar, OB : BD : : 
 OF : FG, or OB : 
 BD:: 20F : 2FG 
 or CE+C'E', and 
 OB (CE+C'EO = 
 20F.BD; that is, E{sinAC'{-smAC)=:2cosBC sinAB. 
 
 Cor, 1 . Hence, likewise, of three equidifferent arcs, the 
 rectangle under the radius and the difference of the sines 
 of the extremes, is equal to twice the rectangle under the 
 sine of the common difference and the cosine of the mean 
 arc ; or Rlsin^A + B) — sin{A — B)\ = 2«»B cos A, 
 
 Cor, 2, Hence R(cos{A'-B)+cos{A-^'B)\=2cosB cos Af 
 
 and R(cos{A — B)— co5(A + B)'\ z=2sznB shiA. 
 
 For OB : OD : : OF : OG :t20F :20Gor OE'+OE, 
 and OB(OE'+OE) = 20F.OD; that is, 
 BicosAC + cos AC) = 2cosBC cos AB, 
 
 Again, OB : BD : : CF : FH : : 2CF : 2FH, or 
 OE'-— OE, and OB(OE'— OE)=2CF.BD ; that is, 
 
 Ji{cosAC^cosAC) = 2sinBC sin AB, 
 
TRIGONOMETRY. 235 
 
 Cor, 3. Let the radius be expressed by unit, and the arcs 
 B and A, denoted by a and na ; then collectively 
 2sm a.cos ?ia = si7i(n+ \)a — sin{n — 1)«, 
 ^cos a . sin na = sm{n + 1 )« + sin{ «— 1 )a, 
 2sin a.sin naz=:cos{n — \)a — cos[n-\-\)af and 
 2cos a.cos 7ia = cos{7j — l)a-\-cos{ii-{-l)a. 
 
 Cor. 4. Since versB = R — cosB, it follows that 
 R (sin{A + B)+sm{A — B)\ = 222 sin A — 2vei'sB sin A, 
 
 and consequently R sin{A + B) =2jB sin A — R sin{A — B)- 
 2versB sin A, oi R{sin{A+B) — sinA)=:R[sinA-sin{A'-B)\ 
 ^"^versB sin A, 
 
 In the same way, it may be shown that R{cos{A — B) — cos A) 
 z=R(cosA — cos{A + B)\ — 2t;^5B cos A, 
 
 Cor, 5. If the mean arc contain 60°, then R(sin(60''-\'B) 
 — S2w(60°— B)) = 2sinB cos60% or sinB 2sinS0°, But twice 
 the sine of 30° being (cor. 1. def.) equal to the chord of 
 60° or the radius, it is evident that sin{60^ + B)— 
 si7i{60° — B)=5/wB, or 
 sm(60o+B)= sin{60^'^B)+sinB. 
 
 Cor. 6. Produce CE to the circumference, join CI 
 meeting the production of FG in K, and join OK. Since 
 FK is parallel to CI and bisects CC, it likewise bisects 
 IC; and hence OK is perpendicular to KC, which is, 
 therefore, the sine of half the arc lAC, or of half the sum of 
 the arcs AC and AC, as CF is the sineof half their differ- 
 ence. But(II.21.El.)IC*-CC'^ = IC.2CE',orCK^-CF* 
 = CE.C'E' ; consequently siii^AB-siti^ BC = sin AC si7iAC', 
 or, employing the general notation, 
 
236 ELEMENTS OF 
 
 j/tzA* — smB* = sm(A + B) sin{\ — B) = (2. cor. 3.) 
 l7?(cos2B— C05 2A.) ■ ' 
 
 Scholium. By help of this proposition, the sines and co- 
 sines of multiple arcs are easily determined ; but the ex- 
 pressions for them will become simpler, if, as in cOr. 2. 
 the radius be supposed equal to unit. For A, 2A and 3 A 
 being three equidifferent arcs, 
 sinA-\-sin3Az=2cosA sin2A=:2cosA 2casA sin A, or 
 sinSAc=:4!C0sA^.sinA — sin A ; and 
 eosA + cosSA = 2cosA.cos2A = 2cosA (2cos A^-j) = 
 4cosA' — 2cosA, or 
 cos3 A = 4:cosA ' — 3cosA, 
 
 Again, since 2A, SA, and 4A are equidifferent arcs, 
 5m2A+s/w4A=2cosA sin3A:=ScosA^ sin A — 2cosA sinA^ 
 or sini!A=ScosA^ sin A — 4co5A sin A ; 
 cos2A + cos^A = 2co5 A.e:o53 A = 2cosA{^cosA ^ — 3co5 A), 
 or cos^A = ScosA* — ScosA^ + l* In like manner, as- 
 suming the equidifferent arcs 3 A, 4 A, 5 A, the sine and 
 cosine of 5 A are found ; and this mode of procedure 
 may be continually repeated. To abridge the notation, 
 however, it will be proper to express the sine and the co- 
 sine of the arc a, by s and c. The results are thus ex- 
 pressed in a tabular form : 
 
 Sin 2a = 2cs. 
 Sin %a = . 4c*5 — s.^ 
 Sin \a = 8c^5 — 4cs. 
 (1.) Sin 5a = I6c^s—I2c*s + s. 
 Sin 6a = 32c^s^S2c^s + 6cs, 
 Sin la = 64c<^5 — 80c*5 -f 24 ess, 
 &c. &c. &c. 
 
TRIGONOMETUY. 257 
 
 Cos 2« = 2c* — 1. 
 Cos 3« = 4cJ ■— 3c. 
 (2.)' Co5 4fl = 8c^ — 8c* + 1. 
 Co5 5a = 16c*— 20c5 +5c. 
 Co5 6a = S^c'^— 48c^+18c»— 1. 
 &c. &c. &c. 
 
 If in these expressions, i — s* be substituted for c*, in 
 the sines of the odd multiples of a, and in the cosines of 
 the even multiples, — the sines and cosines of such multiple 
 arcs will be represented merely by the powers of the sine On 
 
 Sin 2,a = 35— 45^. 
 (3.) Sin 5a = 55—2055 + 165. 
 
 Sin 7a = 75— 56s' +1125*— 645^ 
 &c. &c. &c. 
 
 Co5 2a rr + 1 — 25*. 
 (4,) Co5 4« = + 1 — 85* +85*. 
 
 Co5 6a = + 1 — 185^ + 485*— 32s*. 
 &c. &c. &c. 
 
 If the terras of the first table be repeatedly multiplied 
 by 2s, and those of the second by 2c, observing the sub- 
 stitutions of cor. 2, there will result expressions for the 
 sines and cosines. Thus, 2 sin a^ =z2s.s = — cos 2a + 1, 
 4 sin a' = — 2s.co5 2a + 2s = — sin 3a + sin a + 2s =r 
 — sin 3a + 3s, and 8 sin a* = — 2s.sm 3a + 2s.3s=r + cos 4a 
 — cos 2a — 3cos 2a + 3 = cos 4a — 4 cos 2a + 3 , Again, 2 cos a* 
 =2c.c = cos 2a +1,4 cosa^ = 2c,cos 2a+2c^ cos 2a + 
 
23S ELEMENTS OF 
 
 COS a •{• 2c -=. COS Sa + 3 cos a, and 8 cos a^ = ^Cxos So. + 
 2c.3cosazzcos^a'{-cos2a+Scos2a + S=:cos4!a+^cos2a-^S, 
 In this manner, the following tables are formed. 
 
 Sifi a =z s, 
 2 Sin «* = — cos 2a + 1 . 
 4 Sin a' = — 52/« S« 4- 35. — 
 
 (5.) 8 Sin fl* = + cos 4a5 — - 4 cos 2« + 3. 
 16 /Sm «* = 4- s/w 5a — 5 sm 3a + 10s. 
 82 Sin a^ z=i — cos 6a + 6 cos 4a — 15 cos 2a + 10. 
 64 Sin a' = — sin 7a + 7 si?i 5a — 2 Ism 3a + 35s. 
 &c. &c. &c. 
 
 ' Cos a =r c. 
 
 2 Cos a^ = cos 2a +1. 
 4 Cos a' = cos 3a + 3c, 
 l(6.) 8 Cos a* = cos 4a + 4 cos2a + 3. 
 16 Cos a' = cos 5a + 5 cosSa + 10c. 
 32 Cos a^ = cos 6a + 6 cos4a + 15 cos2a + 10. 
 64 Cos a' = cos 7a + 7 cos5a + 21 cos 3a + 35c. 
 . &c. &c. &c. 
 
 PROP. IV. THEOR. 
 
 The sum of the sines of two arcs is to their dif- 
 ference, as the tangent of half the sum of those 
 arcs to the tangent of half the difference. 
 
 If A and B denote two arcs; s?wA+smB : si?iA — si?iB 
 
 ^ A+B , A-B 
 
 : : tkin — t — : tan — - — 
 
TRIGONOMETRY. 
 
 231^ 
 
 For, let AC and AC be the sum 'aad difference of the 
 arcsABandBCjOrBC; 
 draw the perpendiculars 
 CE, and C'E', extend the 
 chord CC^, and apply at 
 B the parallel tangent 
 HBL, meeting in K and 
 L the diameter produced, 
 and draw OCH, OFB 
 and OC'H^ Because CE 
 
 is parallel to CE', and CK to HL, CE : CE' : : CK ; C'K 
 (VI. 2. El.) HL : HX ; and consequently CE+C'E' : 
 CE-C'E' : : HL + HX : HL—HL', that is, 2BL : 
 2BH, or BL : BH. But CE and CE' are the sines of 
 the arcs AC and AC, and BL and BH are the tangents 
 of AB and BC, or of half the sum and half the differ- 
 ence of those arcs. Wherefore sm AC + sin AC : sin AC — 
 
 sinAC : : ta7i 
 
 AC+AC 
 
 ta?i 
 
 AC—AC 
 
 2 2 
 
 Cor. 1. The sines of the sum and difference of two 
 arcs are proportional to the sum and difference of their 
 tangents. For CE : CE' : : HL, or BL+13H : H'L, 
 or BL — BH ; that is, resuming the general notation, 
 5/w(A + B) : sin{A — B) : : tanA-\-ta?il^ : tan A — ta7i^. 
 
 Cor. 2. Let the greater arc be equal to a quadrant; and 
 R+sinB : R-^sinB : : /«7?(45^+|B) : ^a!w(4.5«»— 4.B), or 
 co/(4;5°4-4.B). But, the radius being a mean proportional 
 between the tangent and cotangent of any arc, and the 
 cosine of an arc being a mean proportional between the 
 sum and difference of the radius and the sine, it follows 
 that R+sinB : cosB :: R: if«w(45^— ^-B), and 
 R-^sinB : cosB, or cosB : R+sinB :: R: ^«w(45<'+iB). 
 
240 ELEMENTS OF 
 
 Or, if instead of B, there be substituted its complement, 
 these analogies will become R+cosB : smB : : JR : tan^Bf 
 and R — cosB : sinB : : R : cot^B. 
 
 Cor, 3. Since cosB : R f: R^sinB : i?«»(45^ — |B), and 
 cosB :R:: R+sinB : ifa7?(45<' +I.B), therefore (VI. 19. El.) 
 cosB : R : : 2R : tan{4!5°-~i.B) + tan{4!5° +i.B) ; that is, 
 supposing B to be the complement of 2C, sin2C : 2R : : 
 R : tanQ + cotC, But (Prop. 1. cor. l,)R.sin2C — 2cosC 
 ^inC, and consequently cosC.sinC : R^ : : R : tanC+colC. 
 
 Cor, 4. Since (4< cor. def.) cosB : R : : R : secB, and 
 (3. cor. def) cosB : sin B • : R : ia?iB, therefore 
 cosB : R-\'SinB : : R : tanB-\'SecB, and consequently 
 (2. cor. def)^aw(45°+^B) = i^«wB + 5£?cB.— This also ap- 
 pears clearly from the figure, on supposing OH'=HX', 
 or the angle LOH' equal to OLH', and consequently the 
 arc AC equal to the complement of AB. 
 
 PROP. V. THEOR. 
 
 As the difference of the square of the radius 
 and the rectangle under the tangents of two arcs, 
 is to the square of the radius, — so is the sum of 
 their tangents, to the tangent of the sum of the 
 arcs. 
 
 Let A and B denote any two arcs ; then, 
 R*-^tanK,tanB : R'' : : tanA + taiiB : tan{ A+B.) 
 In reference to a diagram, let AB and BC^be the two 
 arcs, AD and BE their tangents, and AF consequently 
 the tangent of their sum HC. From the centre O, draw 
 
TRIGONOMETRY. 
 
 241 
 
 to 
 
 meet the extension of this tangent, draw OH perpen- 
 dicular to OD and OG making the 
 angle AOG equal to BOC; and 
 from D draw DI parallel to BE, 
 or (I. 23. El.) OH. 
 
 The triangle AOG is evidently 
 equal (I. 20. El.) to BOE, and 
 therefore AG equal to BE. Be- 
 cause the parallels BE and DI 
 (VI. 2. El.) cut the diverging 
 lines OD and OJ, BE or AG : DI 
 : : OB or OA : OD ; but the 
 right angled triangle DOH being 
 (VI. 15. El.) divided by th^ per- 
 pendicular OA into similar tri- 
 angles, OA : OD : : AH : OH, and 
 consequently AG : DI : : AH : OH, 
 or by alternation AG : AH : : 
 DI : OH. Again, since the parallels DI and OH are 
 intercepted by the diverging lines FH and FO, (VI. 2.) 
 DI: OH::FD:FH; wherefore AG : AH: : FD : FH, 
 and (V. 10. El.) GH : AH : : DH : FH : : (V. 19. 1. cor. 
 El.) DG : AF. Consequently (V. 25. cor. 2. El.) GH.AD : 
 AH.AD::DG:AF; but (VI. 15. cor. El.) AH.AD=OA% 
 and hence GH.AD = OA* — AD. AG ; wherefore 
 OA=^— AD.BE : 0A» : ; DG : AF. Now OA is the 
 radius, AD and BE the tangents of the arcs AB and BC, 
 DG their sum, and AF the tangent of the compound arc 
 AC ; consequently the proposition is manifest. 
 
 Cor, 1. Hence it follows, by changing the position of the 
 figure ; — That as the sum of the square of the radius, an4 
 
 n 
 
242 ELEMENTS OP 
 
 the rectangle under the tangents of two arcs, is to the square 
 of the radius, so is the difference of their tangents to the tan- 
 gent of the difference of the arcs. If A and B denote the two 
 arcs, then R^+tanA tan B : R* : : tan A-^tan B : tan{A-B,) 
 
 Cor, 2, Let the two arcs be equal; and 
 E^'-'tanA^ iR"- :: ^tanA : tan2A, 
 
 Cor, 3. Let the greater arc contain 45®, whose tan- 
 gent is equal to the radius, then R^ r+= R.tan^ : iJ* 
 : : RdtztanB : ^a5w(45^=fc:B), or R=^tanB : Rz±ztanB : : 
 R : i^«w(45°=l=B). 
 
 Scholium. Assuming the radius equal to unit, expressions 
 are hence easily derived for the tangents of multiple arcs. 
 Let t denote the tangent of an arc a ; then i — t^ : i : : 2t: 
 
 tan^azzj-j^ and i—t-^;;^ : i : : ^+7Z^-^^"^^=7Il3^- 
 In like manner, it will be found that 
 
 Tan 4«= 
 
 
 6Y—-20; 5+6^5 
 &c. &c. &c. 
 
 Ihe^e/ormulcc might also be derived from expressions for 
 the sine and cosine of the multiple arc which involve the 
 powers of the tangent. Thus, from ( 1 ), sin 2a = 2cs =r 
 
 c* (2-j=c*.2^, and 5m3« = 4c*5— 5 = 3c*5— (i--c*)5 = 
 ^3 /s-— ^^=cH3^— J^'); again, from (2), c$s2a=2c*—i= 
 
TRIGpNOMETRY,- 243 
 
 c?— s»=c^ A— O =^'*(i — ^*)> and cosSasi4}C^ -—Scsz 
 c5-.3c(i-.c*)=c5(l— 3-il)=J^Ui— 3^*)- In this way, the 
 following tables are formed : 
 
 Sin 2a = c^.2L 
 SmSa=: c^{3t--'t^), 
 (8.) Sin ^a = c*(4j?— 4ifJ). 
 
 ;S/w 6« = c\6t^20t^ +6^0' 
 &c. &c. &c. 
 
 Co5 3a = c5(l— 3^»). 
 (9.) Co5 4«=^;*(l--6/^»+2f*). 
 
 Cos 5« = cs( 1—10^*4-5^*). 
 Co5 6a = c^{\-^l5t'- + \6t^-'t^). 
 &c. &c. &c. 
 
 The first set of expressions being divided by the second, 
 will evidently give the same results for the tangent of the 
 multiple arc. 
 
 PROP. VI. THEOR. 
 
 The supplemental chord of half an arc, is a 
 mean proportional between the radius, and the 
 sum of the diameter and the supplemental chord 
 of the whole arc. 
 
24 4. 
 
 ELEMENTS OF 
 
 This property, which is only a modification of cor. 2. to 
 Pr. 2. will admit of a more direct demonstration. For draw 
 the chord AB, the semichords AE and BE, and the supple- 
 mental chords CB and 
 CE, and the radius 
 OE. The isosceles 
 triangles AEB and 
 COE are similar, for 
 the angles OCE and 
 EAB at the base 
 stand on equal arcs AE and EB; consequently AE : AB 
 : : CO : CE. But, AC BE being a quadrilateral figure 
 contained in a circle, CE.AB = AE.CB + EB.CA = AE 
 (CA + CB), or AE : AB : : CE : CA + CB; wherefore 
 
 CO: CE:: CE: CA + CB, or CE» = CA(^^±^). 
 
 Cor, Hence, in small arcs, the ratio of the sine to the 
 arc approaches that of equality. For, let the semiarcs AE 
 and EB be again bisected in the points F and G ; and, 
 continuing their subdivision indefinitely, let the successive 
 intrrmediate chords be drawn. The ratio of the sine BD 
 to the arc AB may be viewed as compounded of the ratio 
 of BD to the chord AB, of that of AB to the two chords 
 AE and EB, of that of AE and EB to the four chords 
 AF, FE, EG, and GB, and so forth. But thpse ratios, 
 it has been shown, are the same respectively as those of 
 the supplemental chords CB, CE, CF, &c. to the diame- 
 ter CA. And since each of the ratios CB : CA, CE : CA, 
 CF : CA, 3cc. approaches to equality, it is evident that 
 their compounded ratio, or that of the sine to its corre- 
 sponding arc, must also approach to equality. 
 
TRIGONOMETRY. 24-5 
 
 Scholium, Hence the ratio of the sine BD to the arc 
 AB 15 expressed numerically, by the ratio of th« continued 
 product of the series of supplemental chords CB, CE, CF, 
 &c. to the relative continued power of the diameter CA* 
 The ratio may, therefore, be determined to any degree of 
 exactness, by the repeated application of the proposition 
 in computing those derivative chords. But a very con- 
 venient approximation is more readily assigned. Make 
 CD to ei as CB to CA, CI to CK as CE to CA, CK to 
 CL as CF to CA, and so forth, tending always towards 
 the limit Z ; then the ratio of CD to CZ, being com- 
 pounded of these ratios, must express the ratio of the sine 
 BD to its corresponding arc AB. Now CD : CB : : 
 CB : CA; consequently CI = CB, and CD : CI : : 
 CI : CA, or the point I nearly bisects DA. Again, 
 
 CE* = CA (^5A±^\ and therefore CE differs from 
 
 CA, by nearly the fourth part of the diiferenee between CB 
 and CA. These differences being small in comparison of 
 the quantities themselves, the series of supplemental chords 
 may be considered as forming a regular progression, each 
 succeeding term of which approaches four times nearer 
 to the length of the diameter. Wherefore IK=|DI, 
 KL=iIK, and so continually. But (V. 21. El.) as the 
 difference between the first and second term, is to the first, 
 so is the difference between the first and last tern>, or DI 
 itself, to the sum of all the terms, or the extreme limit 
 DZ ; that is, 3 : 4 : : DI : DZ ; and consequently 
 DZ=|DA. The i^tio of the sine BD to the arc AB is, 
 therefore, nearly that of CD to CD-f |DA, or of 3 CD to 
 CD+2CA. 
 
24;6 ELEMENTS OF 
 
 This approximation may be differently modified. Since 
 SCD=60A— 3DA, and CD+2AC=60A— DA, it fol- 
 lows that BD is to AB, as 60A— 3DA to 60A— DA. 
 But this ratio, which approaches to equality, will not be 
 sensibly affected, by annexing or taking away equal small 
 diflPerences. Whence the sine is to the arc, as 60A — 6DA 
 to 60A—4DA, or 30D to 0A+20D. But OD is to 
 OA, as the sine of AB is to its tangent ; and consequently 
 the triple of that arc is equal to its tangent together with 
 twice its sine. 
 
 Again, both terms of the ratio increased by the minute 
 difference DA become 60A — 2D A, and 60A ; wherefore 
 BD is to AB, as 30A— DA to 30A, or as 20C+0D to 
 SCO. Hence, if CP be 
 made equal to the radius 
 CO, and PBH be drawn 
 to meet the tangent, — 
 the arc AB will be near- 
 
 O DA 
 
 ly equal to the intercepted portion AH. For BD : AH : : 
 PD : PA, or 20C-f OD : 30C ; that is, as the sine BD 
 is to its arc AB. 
 
 Another approximation, of much higher importance, 
 may be hence derived ; for PD : PA : : BD : AH, or as 
 the sine to its arc nearly. But (V. 3. El.) PD.CD is to 
 PA.CD in the same ratio, and PA.CD= PD.CD + 
 AD.CD = (III. 26. cor. 1.) PD.CD + BD^ ; whence 
 PD.CD is to PD.CD + BD*, as the sine to its arc nearly. 
 If the arc be small, it is evident that OD will be very 
 nearly equal to AO, and consequently PD may be as- 
 sumed equal to 3AO, and CD equal to 2AO. Where- 
 fore 6A0^ : 6AO* + BD* : : BD : AB nearly; or, the ra- 
 dius being unit, and a and s denoting a small arc and its 
 
TRIGONOMETRY. 247 
 
 sine, 6 : 6+s^ : :s : a, and hence a = s +r— nearly. But 
 
 since a and s are very small, a^ will approach extremely 
 near to 5', and it may, therefore, be inferred conversely^ 
 
 that s = a — — . 
 6 
 
 A convenient approximation for the versed sine of an 
 
 arc is easily derived from the fundamental property of the 
 
 lines themselves; for 2AO.AD = AB* -, BD' + AD*, or 
 
 employing v to denote the versed sine, 2t;=5*4-^% and 
 
 «^=~ +^* If> therefore, the arc be small, it maybe suffi- 
 
 ciently near the truth to assume v=r— > but should great- 
 er accuracy be required, substitute this value of v in the 
 second term of the complete expression, andv=2r-+^> 
 which will form a very close approximation. 
 
 Calculation of the Trigonometric Lines. 
 
 The preceding theorems contain all the principles re- 
 quired in constructing Trigonometric Tables. The ra- 
 dius being denoted by unit, the several lines connected 
 with the circle are referred to that standard, and are ge- 
 nerally computed to seven decimal places. 
 
 The first object is to compute the Sines for every arc 
 of the quadrant. 
 
 Since the semicircumference of a circle whose radius is 
 unit was found, by the scholium to Prop. SO. Book VI. of 
 the Elements, to be 3.1 4 15926536, the length of the arc of 
 
248 ELEMENTS OF 
 
 one minute is .0002909, which, in so small an arc, may be' 
 assumed as equal to the sine, and consequently the versed 
 sine of a minute = |(.0002909)* = .000,000,042,308. 
 Whence, by cor. 3. to Prop. 3. sin{K + 1') = 2sinK — 
 2^mA X .000,000,042,308— 5m( A— 1') •, and therefore, by 
 a series of repeated operations, the intermediate arc being 
 successively 1', 2', 3', 4', &c. the sines of 2', 3', 4', 5', &c. 
 in their order will be calculated. 
 
 The numbers thus obtained will at first scarcely differ 
 from ^n uniform progression, the versed sine of 1', which 
 forms the multiplier of deviation, being so extremely small. 
 It is hence superfluous, to compute rigidly all those mi- 
 nute variations. The labour may be greatly shortened, 
 by calculating the sines for each degree only, and employ- 
 ing some abridged process for filling up the sines, corre- 
 sponding to the subdivision in minutes. 
 
 The arc of one degree being equal to ,0174533, it fol- 
 lows from the scholium to Prop. 6., that the sine of 
 1°=:.0174533—|.(.0174533)' =.0174524, and hence the 
 versed sine of 1°=4-(.0174524)^=:.0001523. Wherefore 
 5m(A-f- l°)=25/wA— 25fwA X .0001523— 5zw( A— 1°); or, if 
 from tmce the sine of an arc, diminished by its 6566*** part, 
 the sine of an arc one degree lower be subtracted, the re^ 
 mainder will exhibit the sine of an arc, which is one degree 
 higher. Thus, 
 
 Sin2^ =z2sinl--2sinl'' X .0001523=.0349048--.0000053 
 
 =.0348995. 
 Sin3°z:z2sin2°^2sin2° X .0001523— 5ml °=:.0697990— 
 
 .0000 1 pe—.0 1 74524= .0523360. 
 6'm4°=2sm3°— 25z>z3«^ X .0001523— .5^2°=. 1046720— 
 .0000160— .0348995.=:0697565. 
 
TRIGONOMETRY. <J49 
 
 After this manner, the sine for each degree is computed 
 in succession. 
 
 But the sines may be found, independently of the pre- 
 vious quadrature of the circle. Assuming an arc whose 
 chord is aheady known, it is easy, from Prop. 6. to deter- 
 mine the successive chords and supplemental chords of its 
 continued bisection. Let that arc be 60° j its chord is 
 equal to the radius, and (IV. 17. cor. 2.) its supplemental 
 chord ^v'Sz: 1.7320508076. Whence the supplemental 
 chordof30° = V(2 + 1.7320508076)=1.93185I6525. In 
 this way, by continued extractions, the supplemental chords 
 of 15°, 7° 30', 3° 45', and 1° 52% are succcjsively compu- 
 ted, the last one being equal to 1,9997322758. Again, the 
 chords themselves are deduced by a series of analogies ; 
 for 1.9318516525 : 1 : : 1 : .5l763809004.=chord of 30°, 
 and so repeatedly, till the chord of 1° 52'i., which is 
 .0327234633. Hence, taking the halves of those num- 
 bers, the sine of 56'i- = .0163617317 and the cosine of 
 56':^=9998661379, and therefore (cor. 3. defin.) the tan- 
 gent of that arc is .0163639215 j consequently the arc itself 
 4- (2 X. 0163617317 + .0163639215)=.01636246]6, and 
 thence the length of the arc of a minute is .0002908832086. 
 Wherefore the sine of 1'=.0002908882— .^(.0002908832>» 
 = .0002908882604^6, and the versed sine of 1'= 
 
 4.(.U0029088826046)* = .000000042308. 
 Employing these data, therefore, 
 
 Sifi2' = '2sini'-2sinl' X .000000042308=r.00Q58 17763845 ; 
 SinS' = 2si7i2'-'2sm2' X .000000042308—5^1'= 
 .0008726645152; and so forth. 
 But it is very seldom requisite to push the estimation 
 to such extreme nicety. The sines being calculated for 
 each degree as before, those corresponding to the subdivi- 
 
250 ELEMENTS OF 
 
 sion in minutes, may be found by a mere expeditious me- 
 thod, though founded on ulterior considerations. If the sines 
 increased uniformly, the sine of A^'+w' would exceed that 
 
 n 
 
 of A by the quantity — -(5mA+l° — sin\ — 1°) = B. But 
 
 the rate of this augmentation, being continually retarded, 
 occasions a defect, equal to w* X5W A X .000,000,04^2308 = C. 
 Again, since the retardation itself gradually relaxes, it re- 
 quires a small compensation, which may be estimated at 
 (60-~7^')B X .00000 1 3 =D. The sine of A° + /j' is then 
 very nearly =s?ViA + B — C + D. Thus, the sines of 31°, 32°, 
 and 33° being respectively .5150381, .5299193, and 
 .5446390, let it be required to find the sine of 32° 40'. 
 
 40 
 Here B = — -(5^w33°~sm31«^ = .0098670, 
 
 C = 1600 X sm32 X .0000000423=r.0000359, 
 and D=20 X .0098670 X .00000 1 3 = .0000003. 
 Whence sm32° 40' =i= .5299193 + .0098670 — 0000359 + 
 .0000003 = .5397507. 
 
 After the sines are calculated up to 60°, the rest are de- 
 duced from cor. 4. Prop. 3. by simple addition. Thus, 
 5m61°=52w59°+5ml°=.8571673 + .0l74524=.8746197. 
 
 The Versed Sines and supplementary versed sines are 
 only the difference and sum of the radius and the sines. 
 
 The Tangents are easily derived from the sines, by help 
 of the analogy given in the third corollary to the definition^. 
 Thus, C(?s32°: 52W32 .-.R: tan32°, or, .8480481 : .5299193 
 : : 1 : ,62^869^— tanS2°, Beyond 45°, the calculation is 
 simphfied, the radius being (cor. 7. defin.) a mean propor- 
 tional between the tangent and cotangent, or the cotan- 
 gent is the reciprocal of the tangent. 
 
TRIGONOMETRY. . 251 
 
 The Secants are deduced by cor. 4. to the definitions, 
 since they are the reciprocals of the cosines^ 
 
 From the lower tangents and secants, the tangents of arcs 
 that exceed 45® are most easily derived; for (cor. 4. Prop. 4.) 
 tan{^5°+a)=sec2a+taji2a. Thus, tan4!6°=sec2''+ta?t2°, 
 or 1.0355303 = 1.0006095 + .0349208. 
 
 PROP. VII. THEOR. 
 
 In a right angled triangle, the radius is to the 
 sine of an oblique angle, as the hypotenuse to tlie 
 opposite side. 
 
 Let the triangle ABC be right angled at B ; then 
 R : sinCAB : : AC : CB. 
 
 For assume AR equal to the given radius, describe the arc 
 RD, and draw the perpendicu- 
 lar RS. The triangles ARS 
 and ACB are evidently simi- ^^^ 
 
 lar, and therefore AR ; RS 
 : : AC : CB. But, AR being 
 the radius, RS is the sine of 
 the arc RD which measures 
 the angle RAD or CAB; and consequently^ : sin A 
 : : AC : CB. 
 
 Cor. Hence the radius is to the cosine of an angle, as 
 the hypotenuse to the adjacent side ; for R : sinC or cos A 
 : : AC : AB, 
 
25^ ELEMENTS OF 
 
 PROP. VIII. THEOR. 
 
 . In a right angled triangle, the radius is to the 
 tangent of an oblique angle, as the adjacent side 
 to the. opposite side. 
 
 Let the triangle ABC be right angled at B ; then 
 R:tanBAC::AB:BC. 
 
 For, assuming AR equal to the given radius, describe 
 the arc RD, and draw the per- 
 pendicular RT. The triangles C 
 Art and ABC being similar, 
 AR : RT : : AB : BC. But, 
 AR being the radius, RT is 
 the tangent of the arc RD 
 which measures the angle at 
 A ; and therefore R : tan A : : 
 AB:BC. 
 
 Cor. Hence the radius is to the secant of an angle, as 
 the adjacent side to the hypotenuse. For AT is the se- 
 cant of the arc RD, or of the angle at A ; and, from simi- 
 lar triangles, AR : AT : : AB ; AC. 
 
 PROP. IX. THEOR. 
 
 The sides of any triangle are as the sines of 
 their opposite angles. 
 
 In the triangle ABC, the side AB is to BC, as the sine 
 of the angle at C to the sine of that at A. 
 
TRIGONOMETRY. 253 
 
 For let a circle be described about the triangle ; and the 
 sides AB and BC, being chords 
 of the intercepted arcs or of the 
 angles at the centre, are (cor. def.) 
 equal to twice the sines of the 
 halves of those angles, or the 
 angles ACB and CAB at the 
 circumference. But, of the same 
 angles, the chords or sines (VI. 
 
 11. cor El.) are proportional to the radius; and conse- 
 quently AB : BC : : sinC : sinK, 
 
 Cor. Since the straight lines AB and BC are chords, 
 not only of the arcs AB and BC, but of the arcs ACB and 
 BAC, or the defects of the former from the circumference, 
 it follows that the sides of the triangle are proportional also 
 to the sines of half these compound arcs, or to the sines of 
 the supplements of their opposite angles. — A like inference 
 results from the definition, for the sine of an arc and that 
 of its supplement are the same. 
 
 PROP. X. THEOR. 
 
 In any triangle, the sum of two sides, is to the 
 difference, as the tangent of half the sum of the 
 angles at the base, to the tangent of half their difr 
 ference. 
 
 In the triangle ABC, 
 
 AB+ AC : AB— AC : ; tan^^ : tarP"^. 
 
 2 % 
 
254t ELEMENTS OF 
 
 From the vertex A, and with a distance equal to the 
 greater side AB, describe the semicircle FBD, meeting the 
 other side AC extended both ways to F and D, join BD 
 and BF, which produce to meet a straight line DE drawn 
 parallel to CB. 
 
 Because the isosceles 
 
 triangle DAB, has the /^ \B. 
 
 same vertical angle with 
 
 the triangle CAB, each 
 
 of its remaining angles 
 
 ADB and ABD is (1. 30. "^ 
 
 El.) equal to half the sum of the angles ACB and ABC ; 
 
 and therefore the defect of ABC from that mean, that is 
 
 the angle CBD, or its alternate angle BDE, must be equa} 
 
 to half the difference of those angles. Now FBD being 
 
 (III. 19. El.) a right angle, BF and BE are tangents of 
 
 the angles BDF and BDE, to the radius DB, and hence 
 
 are proportional to the tangents of those angles with any 
 
 other radius. But since CB and DE are parallel, CF, or 
 
 AB + AC : CD, or AB— AC : : BF: BE; consequently 
 
 ATI. An Ajy An . ACB+ABC , ACB— ABC 
 
 AB + AC: AB — AC:: tan i : tan > 
 
 2 '^ 
 
 or AB+AC : AB— AC : : cotlA : cot(B+hA), or 
 
 — co/(C+iA). 
 
 Cor. Suppose another triangle abc to have the sides eib^ 
 and ac equal to AB and AC, but containing a 
 
 right angle : It is obvious that tan-^ : tan-^ 
 
 ^ ACB+ABC ^ ACB— ABC 
 : : tan ^r : tan -, or 
 
 h 
 
 
TRIGONOMETRY. 
 
 255 
 
 that is, 
 
 R : tan{4i5-^b) : : cotl A : coi(B+\A), or--co2f(C+.lA). 
 Now, in the right angled triangle abc, ah or AB, is to «r, 
 or AC, as the radius, to the tangent of the angle at h. 
 
 PROP. XI. THEOR. 
 
 In any triangle, as twice the rectangle under 
 two sides, is to the difference between their 
 squares and the square of the base, so is the ra- 
 dius to the cosine of the contained angle. 
 
 In the triangle ABC, 2AB.AC : AB^ + AC^— BC^ : : 
 R : C05BAC ; the angle BAC being acqte or obtuse, 
 according as BC* is less or 
 greater than AB* + AC^. 
 
 For let fall the perpendicu- 
 lar BD. In the rigjit angled 
 triangle ADB, AB : AD : : 
 R : s2>iABD or C05BAC ; con- 
 sequently 2AB.AC : 2AD. AC 
 : : R : cosBAC. But (II. 23. 
 El.) twice the rectangle under 
 AD and AC is equal to the 
 difference of the squares AB _, . 
 
 and AC from the square of BC. Whence 
 2AB.AC : AB* + AC*— -BC* : : R : C05BAC. 
 
 Cor. The radius being denoted by unit, it follows (V. 6. 
 El.) that AB* + AC^— BC* = 2AB.ACco5BAC, and con- 
 sequently BC^=AB* + AC*— 2AB.ACC05BAC, or BC= 
 V(AB»+AC;i-~2AB.AC C05BAC). 
 
256 
 
 ELEMENTS OF 
 
 PROP. XII. THEOR. 
 
 In any triangle, the rectangle under the semipe- 
 rimeter and its excess above the base, is to the 
 rectangle under its excesses abovje the two sides, 
 as the square of the radius, to the square of the 
 tangent of half the contained angle. 
 
 In the triangle ABC, the perimeter being denoted by P, 
 |P(|P— AC : (iP--AB) (iP-.BC) : : R* : tan^BK 
 
 For, employing the construction of Prop* 29., Book VJ. 
 of the Elements; since the triangles BIE and BGD 
 are right angled, BI : IE : : R : tanlBE, or tan^B, 
 
 and BG : GD : : R : tanGBB, or tanlB ; 
 whence (V. 22. El.) BI.BG : lE.GD : : R* : tanhB\ 
 
 But it was proved that -q 
 
 IE.GD = AI. AG ; where- 
 fore BI.BG : AI.AG : : 
 R* : tanlB'', Now BI is 
 equal to the semiperime- 
 ter, BG is its excess above 
 the base AC, and AI, AG 
 are its excesses above the 
 sides AB and BC ; conse- 
 quently the proportion is 
 established. 
 
TRIGois^OMETEY. 257 
 
 PROP. XIIL THEOR. 
 
 In any triangle, the rectangle under two sides, 
 is to the rectangle under the semiperimeter, and 
 its excess above the base, as the square of the i a- 
 dius, to the square of the cosine of half the con- 
 tained angle. 
 
 In the triangle ABC, the perimeter being denoted by P, 
 AB.BC : 4P(,P— AC) : : RM cos.B^. 
 
 For, the same construction remaining ; in the right- 
 angled triangles BIE and BGD, 
 
 BE : BI : : R : sz>BEl, or cos B, 
 and BD : BG : : R : smBDG, or cos B ; 
 whence BE.BD : BI.BG : : R* : co5 B*. 
 
 But the quadrilateral figure EADC, being right angled 
 at A and C, is (III. 17. El.) contained in a circle, and 
 consequently (III. 16. El.) the angle AED or AEB is 
 equal to ACD or to DCB ; wherefore, ^ince by construc- 
 tion the angle ABE. is equal to DBC, the triangles BAE 
 and BDC are similar, and BE : AB : : BC : BD, or 
 BE.BD = ABBC. Hence AB.BC : BI.BG : : 
 "R^'.coslW. Now BI is the semiperimeter, and BG its 
 excess above IG or AC ; wherefore the proposition is de- 
 monstrated. 
 
258 
 
 ELEMENTS OF 
 
 PROP. XIV. THEOR. 
 
 In any triangle, as the rectangle under two side^ 
 is to the rectangle under the excesses of the semi* 
 perimeter above those sides, so is the square of 
 the radius, to the square of the sine of half their 
 contained angle* 
 
 In the triangle ABC, the perimeter being stili denoted 
 by P, AB.BC : (|P— AB) (iP— BC) ; : R' : sinlB', 
 
 For, the same construction being retained, in the right- 
 angled triangles BIEand BGD, BE : IE : : R : sin^B, 
 
 and BD : GD : : R : siiiiB ; 
 whence BE.BD : lE.GD : : R* : sinkB"^. 
 
 But it has been proved 
 that BE.BD= AB.BC, or 
 the rectangle under thecon- 
 taining sides of the trian- 
 gle; and IE. GD=AI. AG, 
 or the rectangle under the 
 excesses of the semiperi- 
 meter above the sides AB 
 and BC. Wherefore the 
 proposition is ' establish- 
 ed. 
 
 Scholium, > The three last propositions are demonstrated 
 here by an independent process; but they are only modi- 
 fications of the same principle, and might consequently be 
 derived from a comparison with the first of the train. 
 
 The eight preceding theorems contain the grounds of 
 trigonometrical calculation. A triangle has only five va- 
 
TRIGONOMETRY. 259 
 
 riable parts— the three sides and two angles, the remain- 
 ing angle being merely supplemental. Now, it is a gene- 
 ral principle, that, three of those parts being given, the 
 rest may be thence determined. But the right-angled tri- 
 angle has necessarily one known angle ; and, in conse- 
 quence of this, the opposite side is deducible from the con- 
 taining sides. In right-angled triangles, therefore, the 
 number of parts is reduced to four, any two of which being 
 the assigned, the others may be found* 
 
 JPROP. Xy. PROB. 
 
 Two variable parts of a right-angled triangle 
 being given, to find the rest. 
 
 This problem divides itself into four distinct cases, ac* 
 cording to the different combination of the data. 
 
 1. When the hypotenuse and a side are given* 
 
 2. When the two sides containing the right angle are given, 
 ^. When the hypotenuse and an angle are giveni 
 
 4. When either of the sides and an dhgle are giverii 
 
 The first and third cases are solved by the application 
 bf Proposition 7, and the second and fourth cases receive 
 their solution from Proposition 8. It may be proper, 
 however, to exhibit the several analogies in a tabulaif 
 form* 
 
260 
 
 ELEMENTS OF 
 
 J 
 
 1 
 
 ,a 
 
 SOLUTION. 
 
 I. 
 
 AC, 
 A B 
 
 A, or C, 
 BC 
 
 AC : AB : : R : sinC, or cos.\. 
 R : sin A : : AC : BC. 
 
 11. 
 
 AB, 
 BC 
 
 A, or C 
 AC. 
 
 AB : BC : : R : tanA, or cotC 
 cosX : R : : AB : AC, or 
 R ; secA : : AB : AC 
 
 III. 
 
 AC 
 
 A 
 
 AB 
 BC 
 
 R : cos\ : : AC : AB. 
 R : sin A : : AC : BC. 
 
 IV. 
 
 AB, 
 A 
 
 BC 
 AC 
 
 R : tank : ; AB : BC. 
 cosA : R : : AB : AC, or 
 R'.secA : : AB : AC. 
 
 In the first and second cases, BC or AC might also be 
 deduced, by the mere application of Prop. 11. Book II. 
 of the Elements : 
 
 For AC^ = AB* + BC% or AC= V (AB* + BC*) 
 and BC^ = AC^-AB* = (AC + AB) (AC-AB), 
 or BC= V((AC+ AB) (AC— AB)). 
 
 Cor. Hence the first case admits of a simple approxi- 
 mation* For, by the scholium to Proposition 6, it appears, 
 that, AC being made the' radius, 2AC + AB is to 3 AC, as 
 the side BC is to the arc which measures its opposite angle 
 CAB, or alternately 2AC+ AB is to BC, as 3 AC to the 
 
TRIGONOMETRY. 261 
 
 arc corresponding to BC. But the radius is equal to an 
 
 0/1/ III o 
 
 arc of 57 17 44? 48, or 57| nearly ; wherefore 3 AC is to 
 
 o 
 
 the arc which corresponds to BC, as 3 >C 57 a, or 172°, to the 
 number of degrees contained in the angle CAB, and con- 
 sequently 2AC + AB : BC: :172° ; the expression of the 
 angle at A, or AC+^AB : BC : : 86° : number of degrees 
 in the angle at A. 
 
 This approximation will be the more correct, when the 
 side opposite to the required angle becomes small in com- 
 parison with the hypotenuse ; but the quantity of error 
 can never amount to 4? minutes. 
 
 PROP. XVI. PROB. 
 
 Three variable parts of an oblique angled tri- 
 angle being given, to find the other two. 
 
 This general problem includes three distinct cases, one 
 of which again is branched into two subordinate divisions. 
 
 1 . Whtn all the three sides are given, 
 
 2. When two sides and ati angle are given ,• which angle 
 may either (1.) he contained by these sidesy or {2,) subtended 
 hy one of them, 
 
 3. When a side and two of the angles are given. 
 
 The first case admits of four different solutions, derived 
 from Propositions 11, 12, 13, and 14, and which have 
 their several advantages. The second case, consisting of 
 
262 
 
 ELEMENTS OF 
 
 two branches, is resolved by the application of propositions 
 9 and 10 ; and >he solution of the third case flows imme- 
 diately from the former of these propositions. 
 
 -T? 
 
 6 
 
 Given. 
 Sought. 
 
 SOLUTIOl^. 
 
 
 I. 
 
 
 AB, 
 BC, 
 
 ind 
 
 AC. 
 
 B. 
 
 AB . BC : (iP-AB) (^P-BC) ::R^: sin\B\ 
 iP(iP-AC) : (iP-AB) (iP-BC) ::7^*:/flwi'B^ 
 AB . BC : iP (iP-AC) ::R^: cos\B\ 
 ?AB . BC : AB' + BC^ ~AC^ ::R : cosB. 
 
 1 
 9 
 3 
 4 
 
 il. 
 
 1 
 
 AB, 
 BC, 
 
 and 
 
 c. 
 
 A, 
 
 and 
 
 AC. 
 
 AB ; BC : : ^iwC : sin A ; whence B, and 
 sinC'.sinB:: AB : AC 
 
 5 
 6 
 
 ^2 
 
 AB, 
 BC, 
 
 and 
 
 B. 
 
 A, 
 
 c, 
 
 Hnd 
 
 AC. 
 
 AB + BC : AB-BC : : cot\B : : co^(A+iB). 
 or - co/(C+iB). 
 rAB: BC: ;R: itanb; and 
 I R : tan(^5°'d) : : cot^B : eo^( A+^B), 
 or-co^(C+iB). 
 sinA : sinB : : BC : x\C, or 
 AC=V( AB^^-BC^— 2AB.BC CQ*B.) 
 
 7 
 
 8 
 
 9 
 10 
 
 rii. 
 
 
 AB, 
 A, 
 B, 
 
 and 
 thence 
 
 c. 
 
 BC, 
 AC. 
 
 sinC : sin A : : AB : BC, 
 sinC : sinB : : AB : AC. 
 
 11 
 12 
 
TRIGONOMETRY. 263 
 
 For the resolution of the first Case, the analogy set 
 down first, is on the whole the most convenient, particu- 
 larly if the angle sought do not approach to two right 
 angles. The second analogy may be applied with obvious 
 advantage through the entire extent of angles. The third 
 and fourth analogies, especially the latter, are not adapted 
 for the calculation of very acute angles ; they will, how- 
 ever, answer the best when the angle sought is obtuse. It 
 is to be observed, that the cosines of an angle and of its 
 supplement are the same, only placed in opposite direc- 
 tions ; and hence the second term of the analogy, or the 
 difference of AB^ + BC* from AC^, is in excess or defect, 
 acc6rding as the angle at B is acute or obtuse. — These re- 
 marks are founded on the unequal variation of tKe sine and 
 tangent, corresponding to the uniform increase of an arc. 
 
 The first part of Case II. is ambiguous, for an arc and 
 its supplement have the same sine. This ambiguity, how- 
 ever, is removed if the character of the triangle, as acute 
 or obtuse, be previously known. 
 
 For the solution of the second part of Case II. the first 
 analogy is the most usual, but the double analogy is the 
 best adapted for logarithms. In astronomy, this mode of 
 calculation is particularly commodious. The direct ex- 
 pression for the sid9 subtending the given angle is very 
 convenient, where logarithms are not employed. 
 
264* ELEMENTS OF 
 
 PROP. XVIi. PROB. 
 
 Given the horizontal distance of an object and 
 its angle of elevation, to find its height and abso- 
 lute distance, 
 
 Let the angle ABC, which an object A makes at the 
 station B, with an horizontal line, and also the distance BC 
 of a perpendicular AC, to find 
 that perpendicular, and the hy- 
 potenusal or aerial distance BA. 
 
 In the' right-angled triangle 
 BCA, the radius is to the tan- 
 gent of the angle at B, as BC to AC ; and the radius is 
 to the secant of die angle at B, or the cosine of the angle 
 at B is to the radius, as BC to AB. 
 
 PROP. XVIII. PROB. 
 
 Given the acclivity of a line, to find its corre- 
 sponding vertical and horizontal length. 
 
 In the preceding figure, the angle CBA and the hypo- 
 tenusal distance BA being given to find the height and 
 the horiz{;ntal distance of the extremity A. 
 
 The triangle BCA bein;. rigiit angled, the radius is to 
 the sine of the angle CBA as BA to AC, and the ra- 
 dius is to the cosine of CBA as BA to BC. 
 
TRIGONOMETRY. 
 
 S65 
 
 Scholium* If the acclivity be small, and A denote the 
 
 measure of that angle in minutes ; then AC=BA X 
 
 nearly. But the expression for AC, will be rendered 
 more accurate, by subtracting from it, as thus found, the 
 
 ^. ACJ 
 
 quantity ^^^. 
 
 In most cases when CBA is a small angle, the horizontal 
 distance may be computed with sufficient exactness, by de- 
 
 AC* 
 
 ducting ~jj^, or BA X A* X .000^000,0423, from the hy- 
 
 potenusal distance. 
 
 PROP. XIX. PROB. 
 
 Given the interval between two stations, and 
 the direction of an object viewed from them, to 
 find its distance from each. 
 
 Let BC be given, with the angles ABC and ACB, to 
 calculate AB and AC. 
 
 In the triangle CBA, the angles ABC and 
 ACB being given, the remaining or supple- 
 mental angle BAC is thence given; and 
 consequently, swBAC : 5/wACB : : BC: AB, 
 and smBAC : smABC : : BC : AC. 
 
 Cor, if the observed angles ABC and 
 ACB be each of them 60^, the triangle will 
 be evidently equilateral ; and if the angle at 
 the station B be right, and that at C half a right angle, 
 the distance AB will be equal to the base BC. 
 
266 
 
 ELEMENTS OF 
 
 PROP. XX. PROB. 
 
 Given the distances of two objects from any 
 station and the angle which they subtend, to find 
 their mutual distance. 
 
 Let AC, BC, and the angle ACB be 
 given, to determine AB. 
 
 In the triangle ABC, since two 
 sides and their contained angle are gi- 
 ven, therefore, by corollary to Propo- 
 sition 10. AC + BC : AC — BC ; : 
 cot\C : cot{A'\-\C)f then sin A : sinC : : 
 BC : AB; or (from the cor. to Prop. 11.) 
 AB= >/(AC*-hBC*— 2AC.BC cosC.) 
 
 Cor, By combining this with the preceding proposition, 
 the distance of an object may be found from two stations, 
 between which the communication is in- 
 terrupted. Thus let A be visible from 
 B and C, though the straight line BC 
 cannot be traced. Assume a third sta- 
 tion D, from which B and C are both 
 seen. Measure DB and DC, and ob- 
 serve the angles BDC, ABC and ACB. 
 In the triangle BDC, the base BC is 
 found as above ; and thence, by the preceding proposition, 
 $he sides AB and AC of the triangle ABC are determined 
 
TRIGONOMETRY. g67 
 
 PROP. XXI. PROB. 
 
 Given the interval between two stations, and 
 the directions of two remote objects viewed from 
 them in the same plane, to find the mutual dis^ 
 t^nce, and relative position of those objects. 
 
 Let the points A, B represent the two objects, and C, 
 D the two stations from which these are observed ; the in- 
 terval or base CD being measured, and also the angles 
 CD A, CDB at the first station, and DCA, DCB at the 
 second ; it is thence required to determine the transverse 
 distance AB, and its direction. 
 
 It is obvious that each of the points A and B would be 
 assigned geometrically by the intersection of two straight 
 lines, and consequently that the position of the objects will 
 not be determined, unless each of them appears in a diffe- 
 rent direction at the successive stations. 
 
 1. Suppose one of the stations C to lie in the direction of 
 the two objects A and B, 
 
 At C observe the angle BCD, and at 
 D the angles CDA and BDC. Then 
 by Prop. 9.slnCAD : sinCDA : : CD : 
 CA, and sinCBD : sinCDB : : CD : 
 CB ; the difference or sum of CA and 
 (pB is AB, the distance sought. 
 
 2. When neither station lies in the direction of the t'w^ 
 objects^ and the base Cf) has a transverse position. 
 
268 ELEMENTS OF 
 
 Find by Prop. 19. the distances AC and BC of both ob- 
 jects from one of the stations ^ 
 C J then the contained angle 
 ACB, or the excess of DCA 
 above DCB, being likewise gi- 
 ven, the angles at the base AB 
 of the triangle BCA, and the 
 base itself, may be calculated, 
 from the analogies exhibited 
 for the solution of the second 
 branch of Case second. For AC+BC : AC— BC : : 
 cotiACB : cot{\ACB'\-CAB), and thus the angle CAB is 
 found. Or more conveniently by two successive opera- 
 tions, AC : BC : : R : fan b, and R : tan{i5° — b) : : 
 cotkACB : co/(|ACB + CAB. Now, sinCAB : sinACB : : 
 BC : AB, or AB= V(AC^ + BC*— 2AC.BC cos ACB). 
 
 The inclination of AB to CD' in the first case is given 
 by observation, and in the second case it is evidently the 
 supplement of the interior angles CAB and DCA. A pa- 
 rallel to AB may hence be drawn from either station. 
 
 Cor. Hence the converse of this problem is readily sol- 
 ved. Suppose two remote objects A and B, of which the 
 mutual distance is already known, are observed from the 
 stations C and D, and it were thence required to deter- 
 mine the interval CD. Assume unit to denote CD, and 
 calculate AB according to the same scale of measures; 
 the actual distance AB being then divided by that result, 
 will give CD r For the several triangles which combine to 
 form the quadrilateral figure CABD, are evidently given 
 in species. 
 
TRIGONOMETRY. 269 
 
 PROP. XXII. PROB. 
 
 Given the directions of two inaccessible objects 
 viewed in the same plane from two given stations, 
 to trace the extension of the straight line con- 
 necting them. 
 
 Let the angles ACD, BCD be observed at C, and 
 ADC, BDC at D, with the base CD -, to find a point E 
 in the straight line ABF produced through A and B. 
 
 By the last proposition, find 
 AD and the angle DAB, and 
 assume any angle ADE. In 
 the triangle DAE, the angles 
 at the base AD, and conse- 
 quently the vertical angle 
 AED, being known, it fol- 
 lows, by Prop. 9., that 5m AED : sinE AD : : AD : DE. 
 Wherefore, measure out DE on the ground, and its ex- 
 tremity E will mark the extension of AB. 
 
 PROP. ;XXIII. PROB. 
 
 Given on the same plane the direction of two 
 remote objects separately seen from two stations 
 and their direction as viewed at once from an in- 
 termediate station, with the distances of those 
 stations, from the middle station, — to find the 
 mutual distance of the objects. 
 
S70 
 
 ELEMENTS OF 
 
 Let object A be visible from the station D, and B from 
 E, and both of them be seen at once from the station C j 
 the compound base DC, C£ be- 
 ing measured, and the angle 
 DCA, ACB and BCE, with 
 ADC and BEC, observed,— to 
 determine AB. 
 
 In the triangles DAC, CBE, 
 the sides AC and BC are found 
 by Prop. 19., and in the triangle 
 ACB, the base AB is thence found 
 by the application of Prop. 20. 
 
 It is evident that the mode of investigation will not be 
 altered, if the three stations D, C and E should lie in the 
 game straight line* 
 
 PROP. XXIV. PROB. 
 
 Given four stations, with the direction of a re-* 
 mote object viewed from the first and second sta- 
 tions, and the direction of another remote object 
 viewed from the third and fourth stations, all in 
 the same plane, — tb find the distance between 
 the objects- 
 Let the bases EC, CD, and DF be given, with the an- 
 gles ECD and CDF, and suppose that at the stations E 
 and C the angles CEA and EGA are observed, and the 
 angles BDF and BFD at D and F ; to find the transverse 
 distance AB. 
 
TRIGOifC^METRY. 
 
 211 
 
 In the triangles EAC and 
 DBF, find by Prop. 19. the 
 sides AC and BD ; and in the 
 triangle CAD, the sides AC, 
 CD, with their contained an- 
 gle ACD, being given, the base 
 DA and the angle CDA are 
 found by Case II. But the 
 distances DA, DB being now given, with their contained 
 angle ADB, the base AB is found by Prop. 20. ' 
 
 PROP. XXV. PROB- 
 
 The mutual distances of three remote objects 
 being given, with the angles which they subtend 
 at a station in the same plane, to find the relative 
 place of that station. 
 
 Let the three points A, B, and C, and the angles ADB 
 and BDC which they form at a fourth point D, be given j 
 to determine the position of that point. 
 
 1. Suppose the station D to he situate in the direction of 
 two of the objects A, C* 
 
 All the sides AB, AC and BC of the triangle ABC be- 
 ing given, the angle BAC is 
 found by Case I. ; and in the 
 triangle ABD, the side AB with 
 the angles at A and D being 
 given, the side AD is found by 
 Case III. and consequently the ^ 
 position of the point D is determined^ 
 
272 ELEMENTS OP 
 
 2. Suppose the three objects Ay B and C to lie in the same 
 direction. 
 
 Describe a circle about the extreme objects A, C and 
 the station D, join DA, DB and DC, produce DB to meet 
 the circumfe;rence in E, and join AE and CE. 
 
 In the triangle AEC, the side AC is given, and the an- 
 gles EAC and ECA, being equal (III. 16. El.) to CDE and 
 ADE, are consequently given ; wherefore the side AE is 
 found b}^ Case III. The triangle AEB, having thus the 
 sides AE, AB, and their contained angle EAB or BDC 
 given, the angle ABE, and its 
 supplement ABD are found by 
 Case II. Lastly, in the triangle 
 ABD, the angles ABD and 
 ADB, wth the side AB, are 
 given ; whence BD is found by 
 Case III. But since the angle 
 ABD and the distance BD are 
 assigned, the position of the station D is evidently deter- 
 mined. 
 
 3. Let the three objects form a triangle , and the station 
 jy lie either without or "within it. 
 
 Through D and the extreme points A and C describe a 
 circle, draw DB cutting the circumference in E, and join 
 AE and CE. 
 
 1. In the triangle AEC, the side AC, and the angles 
 ACE and CAE, which are equal (III. 16. El.) to ADB 
 and BDC, being given, the side AE is found by Case III. 
 
TRIGONOMETRY. 273 
 
 2. All the sides of the trian- 
 gle ABC being given, the angle 
 CAB is found by Case I. 
 
 3. In the triangle BAE, the 
 sides AB and AE are given, and 
 their contained angle EAB, or 
 thediiferenceofCAEandCAB, 
 are given, whence, by Case II., 
 the angle A BE or ABD is found. 
 
 4. Lastly, in the triangle 
 DAB, the side AB and the angles ABD and ADB being 
 given, the side AD or BD is found by Case III., and con- 
 sequently the position of the point D, with respect to A and 
 B is determined. By a like process, the relative position 
 of D and C is deduced ; or CD may be calculated by Case 
 XL from the sides AC, AD, and the angle ADC, which 
 are given in the triangle CAD. 
 
 It is obvious that the calculation will fail, if the points B 
 and E should happen to coincide. In fact, the circle then 
 passing through B, any point D whatever in the opposite 
 arc ADC will answer the conditions required, since the 
 angles ADB and BDC, being now in the same segment, 
 must remain unaltered. 
 
 PROP. XXVI. THEOR. 
 
 The mutual distances of three remote objects, 
 two of which only are seen at once from the same 
 station, being given, with the angles observed at 
 
274 ELEMENTS OF 
 
 two stations in the same plane, and the interme- 
 diate direction of these stations^ — ^to find their re- 
 lative places. 
 
 Suppose the three points A, B 
 and C are given, with the angle 
 AEB which A and B subtend at 
 E, and BFC, which B and C 
 subtend at F, and likewise the 
 angles AEF and EFC ; to find 
 the relative situation of each of \ / 
 
 those stations E and F. '^ ^ 
 
 Produce AE and CF to meet in D, and join BD. The 
 angle EDF, being equal to AEF+CFE— 180<^, is given. 
 Now in the triangle EBF, sinBFE : sinEBF : : EB : EF; 
 and in the triangle EDF, si7iET>F : siriBFE : : EF : ED ; 
 whence, (V. 23. El.) sinBFE.shiEBF : siiiEBF. sinBFE 
 : : EB : ED, and consequently the ratio of EB to ED 
 is found. Again the angle BED, being the supple- 
 ment of AEB, is given, (Prop. 10. cor.) si7iBFE.sinET>F 
 : sinEBF.sinDFE : : R : tan b, and R : tan (45 — b) : : 
 co^i:BED :— coi:(iBED+EBD), or co/(180<?-~i.BED-- 
 EBD), whence the angle EDB is given. The angles which 
 all the three objects A, B, and C subtend at the point D 
 are therefore all given, and hence the position of D is de- 
 termined by the preceding proposition. But BD, being 
 found, the several distances BE, ED, and BF, FD are 
 thence obtained, and consequently the position of each ox 
 the stations E and F is determined. 
 
TRIGONOMETRY. 
 
 ^75 
 
 PROP. XXVII. 
 
 Given the angles of elevation at which an ob- 
 ject is seen from three known points in a horizon- 
 tal plane, to find its position and altitude. 
 
 Let A, B, and C be the three points of observation, and 
 D the foot of the perpendicular from the given object to 
 the horizontal plane. It is evident from Proposition 17, 
 that the horizontal distances AD, BD and CD are pro- 
 portional to the co-tangents of the vertical angles at the sta- 
 tions A, B, C •, let these co-tangents be respectively denot- 
 ed by the lines L, M, and N. Divide AB, the base of the 
 triangle ADB, 
 internally and 
 externally at 
 the points E 
 and F, in the 
 ratio of L to 
 M, and the 
 lines DE and 
 DF joining the 
 vertex D must 
 (VI. 10. cor. 
 El.) bisect in- 
 ternally and externally the angle ; whence EDF is a right 
 angle, and (III. 19. El.) contained in a semicircle. In 
 the same manner, divide CB internally and externally at 
 G and H in the ratio of M to N, and on GH describe a 
 semicircle. The point D, being common to both semicir- 
 cles, must occur in their intersection. 
 
576 
 
 ELEMENTS OF 
 
 From this construction, the trigonometrical calculation, 
 is readily deduced. For L + M : M : : AB : BE, and 
 L— M : M : : AB : BF ; whence EF and its half DE, or 
 the radius KE, is found. In like manner, N + M : M : : 
 CB : BG, and N— M : M: : CB : BH; consequently DI = 
 
 — . In the triangle IBK, the sides BI and BK, 
 
 with their included angle, are given, and therefore (Prop. 
 10.) the angle BKI and the base IK are found. Again, 
 all the sides of the triangle IDK being given, the angle 
 IKD (Prop. 14.) is found. Hence, in the triangle ADK, 
 the whole angle AKD and its containing sides are given, 
 and therefore the base AD, or the horizontal distance of 
 the object from the station A is found, and consequently 
 its altitude. — The opposite semicircles will, likewise, by 
 their intersection, give, on the other side, a second posi- 
 tion for that point. In practice, the ambiguity would 
 easily be removed. 
 
 If the object be seen at the same elevation from all the 
 three points, the arcs of the circles will evidently pass in- 
 to tangents which bisect at right angles the sides of the 
 triangle ABC. The projection D of the object on the ho- 
 rizontal plane will then be the centre of the circle circum- 
 scribing that triangle, and therefore the radius, or the dis- 
 tance AD, will be found by Prop. 18. Book VI. of the 
 Elements. 
 
 If the three points of observation should lie in the same 
 straight line, the centres of the determining circles will 
 likewise occur in that line or its extension, and hence the 
 process of calculation will be greatly abridged. 
 
 General Scholium, In all the foregoing problems, the an- 
 gles on the ground are supposed to be taken by means of 
 
TRIGONOMETRY. 277 
 
 a theodolite ; which, being adjusted by spirit-levels, mea- 
 sures only horizontal and vertical angles, or decomposes 
 other angles into these elements. If the sextant be em- 
 ployed for the same purpose, such angles, when oblique, 
 must be reduced by calculation to their projections on the 
 horizontal plane. 
 
 In surveying an extensive country, a base is first care- 
 fully measured ; and the directions of the prominent dis- 
 tant objects being observed from both of its extremities, 
 they are all connected with it by a series of triangles. To 
 avoid, in practice, the multiplication of errors, these trian- 
 gles should be chosen, as nearly as possible, equilateral. — 
 After a similar method, large estates are the most cor- 
 rectly planned and measured ; the ordinary practice of 
 carrying the theodolite with a chain round the boundary 
 being subject to much inaccuracy. 
 
 If the inequality of the surface of the ground will not 
 admit of the measurement of a base of a sufficient length, 
 a smaller one may be selected at first, and another base 
 derived from this, by combining with it one or more tri- 
 angles. These triangles, to preclude the multiplication of 
 errors, should be as nearly as possible right-angled, and si- 
 milar, having their sides increasing in a continued propor- 
 tion. When this rate of increase is not less than the ra- 
 tio of the radius to the side of an inscribed equilateral tri- 
 angle, the number of intermediate triangles between the 
 measured and the computed base will be rather favourable 
 to the accuracy of the result. 
 
 The vertical angles employed in the mensuration of 
 heights, since they are estimated from the varying direction 
 of the level or the plummet, must evidently, when the sta- 
 tions are distant, require some correction. I^et the points A 
 
278 
 
 ELEMENTS OP 
 
 and B represent two remote objects, and C their centre of 
 gravitation ; with the radius CA describe a circle, draw 
 CB cutting the circumferer.ee in D and E, and join EA 
 and AD. The converging lines AC and BC will indicate 
 the direction of the plummet at A and B, the intercepted 
 arc AD, will trace the contour of a quiescent fluid, and 
 the tangent AZ, being applied 
 to A, will mark the line of the 
 horizon from that station. 
 Wherefore the vertical angle 
 observed at A is only ZAB, 
 which is less than the true an- 
 gle DAB, by the exterior an- 
 gle DAZ. But (III. 2J. El.) 
 DAZ being equal to the angle 
 AED in the alternate segment, 
 is (III. 15. El.) equal to half the angle ACD at the centre. 
 Hence the true vertical angle at any station will be found, 
 by adding to the observed angle half the measure of the 
 intercepted arc j and this measure depending on the cur- 
 vature of the earth, which is neither unifoim nOr quite re- 
 gular, must be deduced, for each particular place, from 
 the length to the corresponding degree of latitude. 
 
 Such nicety, however, is very seldom required. It will 
 be sufficiently accurate in practice to assume the mean 
 quantities, and to consider the earth as a globe, whose cir- 
 cumference is 24',856 miles, and diameter 7,912. The arc 
 of a minute on the meridian being, therefore, equal to 6076 
 feet, the correction to be added to the observed vertical 
 angle must amount to one second, for every 69 yards con- 
 tained in the intervening distance. 
 
 The quantity of depression ZD below the horizon is 
 
TRIGONOMETRY. 279 
 
 hence easily computed; for (III. 26. El.) AZ* = EZ.ZD, 
 or very nearly ED.ZD; and consequently the visual de- 
 pression of an object is proportional to the square of its dis- 
 tance AZ from the observer. In the space of one mile, 
 this depression will amount to y|g parts of a foot ; and 
 generally, therefore, it may be expressed in feet, by two- 
 thirds of the square of the distance in miles. Thus, at 
 twenty miles, the depression is 266| feet ; and at the dis- 
 tance of fifty miles, it amounts to 1666|, or nearly the 
 third of a mile. 
 
 But the effect of the earth's curvature is modified by an- 
 other cause, arising from optical deception. An object is 
 never seen by us in its true position, but in the direction of 
 the ray of light which conveys the impression. Now the 
 luminous particles, in traversing the atmosphere, are, by 
 the force of superior attraction, refracted or bent continu- 
 ally towards the perpendicular, as they penetrate the lower 
 and denser strata ,• and consequently they describe a cur- 
 ved track, of which the last portion, or its tangent, indi- 
 cates the apparent elevated situation of a remote point. 
 This trajectory, suffering almost a regular inflexure, may 
 be considered as very nearly an arc of a circle, which has 
 for its radius six times the radius of our globe. Hence, to 
 correct the error occasioned by refraction, it will not only 
 be requisite to diminish the effects of the earth's curvature 
 by one-sixth part, or to deduct, from the vertical angles, 
 the twelfth part of the measure of the intervening terres- 
 trial arc. The quantity of horizontal refraction, however, 
 as it depends on the density of the air at the surface, is ex* 
 tremely variable, especially in our unsteady climate. 
 
NOTES, 
 
 AND 
 
 ILLUSTRATIONS. 
 
NOTES 
 
 AND 
 
 ILLUSTRATIONS. 
 
 NOTES TO BOOK I. 
 
 DEFINITIONS. 
 
 1. 1 HE primary objects which Geometry contemplates are, 
 from their nature, incapable of decomposition. No wonder 
 that ingenuity has only wasted its efforts to define such ele- 
 mentary notions. It appears more philosophical to invert the 
 usual procedure, and endeavour to trace the successive steps 
 by which the mind arrives at the principles of the science. 
 Though no words can paint a simple sound, this may yet be 
 rendered intelligible, by describing the mode of its articula- 
 tion. 
 
 The founders of mathematical learning among the Greeks 
 were in general tinctured with a portion of mysticism, trans- 
 mitted from Pythagoras, and cherished in the school of Plato. 
 Geometry became thus infected at its source. By the later 
 Platonists, who flourished in the Museum of Alexandria, it 
 was regarded as a pure intellectual science, far sublimed above 
 the grossness of material contact. Such visionary metaphy- 
 sics could not impair the solidity of the superstructure, but did 
 contribute to perpetuate some misconceptions, and to give a 
 wrong turn to philosophical speculation. It is full time to re- 
 store the sobriety of reason. Geometry, like the other scien- 
 ces which are not concerned about the operations of mind, 
 must .ultimately rest on external observation. But those uiti- 
 
284? NOTES AND ILLUSTRATIONS, 
 
 /■ 
 
 mate facts are so few, so distinct, and obvious, that the subse- 
 quent train of reasoning is safely pursued to unlimited extent, 
 without ever appealing again to the evidence of the senses. 
 The science of Geometry, therefore, owes its perfection to 
 the extreme simplicity of its basis, and derives no visible adr 
 vantage from the artificial mode of its construction. The 
 axioms are here rejected, as being totally useless, and rather 
 apt to produce obscurity. 
 
 2. The term Surface^ in Latin superficies^ and in Greek ixi^»' 
 vetxj conveys a very just idea, as marking the niere expansion 
 which a body presents to our sense of sight. Line, or ^^uf^fAx, 
 signifies a stroke ; and, in reference to the operation of writ- 
 ing, it expresses the boundary or contour of a figure. A 
 straight line has two radical properties, which are distinctly 
 marked in different languages. It holds the same undeviating 
 course — and it traces the shortest distance between its ex- 
 treme points. The first property is expressed by the epithet 
 recta in Latin, and droite in French ; and the last seems in- 
 timated by the English term straight, which is evidently de- 
 rived from the verb to stretch. Accordingly Proclus defines 
 a straight line as stretched between its extremities — k stt xk^a/v 
 
 S. The word Point in every language signifies a mark, thus 
 indicating its essential character, of denoting position. In 
 Greek, the term e-ny^cc was first used : but, this being de- 
 graded in its application, the diminutive a-n^eiov, formed from 
 e-7ifi»,a signal, came afterwards to be preferred. The neatest 
 and most comprehensive description of a point was given by 
 Pythagoras, who defined it to be *< a monad having position." 
 Plato represents the hj/postasisy or constitution of a point, as 
 adamantine ; finely alluding to the opinion which then prevail- 
 ed, that the diamond is absolutely indivisible, the art of cut- 
 ting this refractory substance being the discovery of modern 
 ages, 
 
 4. The conception of an Angle is one of the most difficult 
 
NOTES AND ILLUSTRATIONS. 285 
 
 perhaps in the whole compass of Geometry. The term cor- 
 responds, in most languages, to corner^ and therefore exhibits 
 a most imperfect picture of the object intimated. Apollonius 
 defined it to be *^ the collection of space abont a point.'* Eu- 
 clid makes an angle to consist in " the mutual inclination, or 
 xA^o-K, of its containing lines," — a definition which is obscure 
 and altogether defective. In strictness, this can apply only, 
 to acute angles, nor does it give any idea of angular magni- 
 tude ; though this really is as capable of augmentation as the 
 magnitude of lines themselves* It is curious to observe the 
 shifts to which the author of the Elements is hence obliged to 
 have recourse. This remark is particularly exemplified in the 
 20th and 21st Propositions of his Third Book. Had Euclid 
 been acquainted with Trigonometry, which was only begun to 
 be cultivated in his time, he would certainly have taken a 
 more enlarged view of the nature of an angles 
 
 5. In the definition of Reverse Angle, I find that I have been 
 anticipated by the famous mechanician Stevin of Bruges, who 
 flourished about the end of the sixteenth century. It is sa- 
 tisfactory to have the countenance of an authority so highly 
 respectable. 
 
 6. A Square is commonly described as having all its angles 
 right. This definition errs however by excess, for it contains 
 more than what is necessary. The original Greek, and even 
 the Latin version, by employing the general terms «^6oyavicvt 
 and rectanglum, dexterously, avoided that objection. The 
 word Rhombus comes from pifi/ietv, to sling, as the figure repre- 
 sents only a quadrangular frame disjointed. The Lozetig^^^jsi.. 
 heraldry and commerce, is that species of rhombus which is 
 composed of two equilateral triangles placed on opposite sides 
 of the same base. 
 
 7. It scarcely deserves notice, but I will anticipate the ob- 
 jection which may be brought against me, for having changed 
 the definition o{ Trapezium, The fact is, that I have only re- 
 stricted the word to its appropriate meaning, from which 
 
286 NOTES ANI) ILLUSTRATIONS. 
 
 * 
 
 Euclid had, according to Proclus, taken the liberty to depart. 
 In the original, it signifies a table ; and hence we learn the 
 prevailing form of the tables used among the Greeks. In- 
 deed the ancients would api.tar to have had some predilection 
 for the figure of the trapezium, since the doors now seen in 
 the ruins of the temples at Athens are not exactly oblong, 
 but wider below than above. 
 
 8. Language is capable of more precision, in proportion as 
 it becomes copious. As I have confined the epithet right to 
 angles, and straight to lines, I have likewise appropriated the 
 word diagonal to rectilineal figures, and diameter to the circle. 
 In like manner, I have restricted the term arc to a portion of 
 the circumference, its synonym arch being assigned to the use 
 of architecture. For the same reason, I have adopted the term 
 eqitivalent, from the celebrated Legendre, whose Elemens de 
 Geometric is one of the ablest works that has appeared in our 
 times. These distinctions evidently tend to promote perspi- 
 cuity, which is the great object of an elementary treatise- 
 Euclid and all his successors define an isosceles triangle to 
 have only two equal sides, which would absolutely exclude 
 the. equilateral triangle. Yet the equilateral triangle is after- 
 wards assumed by them to be a species of isosceles triangle, 
 since the equality of its angles is inferred at once as a co- 
 rollary from the equality of the angles at the base of an 
 isosceles triangle. This inadvertency, slight as it may appear, 
 is now avoided. 
 
 PROPOSITIONS. 
 
 9. The tenth Proposition may be very simply demonstra- 
 ted, in the same manner as the next or its 
 
 converse, by a direct appeal to superposi- 
 tion or mental experiment. For, suppose 
 a copy of the triangle ABC were inverted 
 and applied to it, the sides BA and BC 
 being equal, if B A be laid on BC, the side 
 BC again will evidently lie on BA, and 
 the base AC coincide with CA, Consequently the angle 
 
NOTES AND ILLUSTRATIONS. 287 
 
 BAC, occupying now the place of BCA, must be equal t© 
 this angle. 
 
 It may be worth while to remark, that Euclid's demonstra- 
 tion of this Proposition, which, being placed near the com- 
 mencement of the Elements, has from its intricacy been styled 
 the Pons Asinoruniy is in fact essentially the same with what 
 has now been given. This will readily appear on a review of 
 the several steps of his reasoning : — 
 
 The sides BA and BC of the isosceles triangle being pro- 
 duced, the equal segments AD and CE are assumed, and AE, 
 CD joined.— 1. The complex triangles ABE and CBD are 
 compared : The sides AB and BC are equal, 
 and likewise BE and BD, which consist of 
 equal parts, and the contained angles EBA 
 and DBC are the same with DBE ; whence 
 (I. 3.) these triangles are equivalent, and the 
 base AE equal to CD, the angle BAE equal 
 to BCD, and the angle BE A to BDC — 2. The 
 additive triangles CAE and ACD are next 
 compared: The sides EC and EA being equal to DA and 
 DC, and the contained angle CEA equal to ADC, the tri- 
 angles are (I. 3.) equivalent, and therefore the angle CAE is 
 equal to ACD. — 3. Lastly, since the whole angle BAE is 
 equal to BCD, and the part CAE to ACD, the remainder 
 BAC must be equal to BCA. 
 
 Now this process of reasoning is at best involved and cir- 
 cuitous. The compound triangles ABE and CBD consist of 
 the isosceles triangle ABC joined to each of the appended 
 triangles ACE and CAD ; when, therefore, as the demon- 
 stration implies, ABE is laid on CBD, the common part ABC 
 is reversed, or it is applied to CBA, and the other part ACE 
 is laid on CAD. But the superposition of ABC or CBA is 
 easily perceived by itself; nor is the conception of that invert- 
 ed application anywise aided by having recourse to the super- 
 position, first of the enlarged triangles ABE and CBD, and 
 then contracting these by the superposition of the subsidiary 
 
288 NOTES AND ILLUSTRATIONS. 
 
 triangles ACE and CAD. In this, as in some other instances, 
 Euclid has deceived himself, in attempting a greater than 
 usual strictness of reasoning. 
 
 10. The fourteenth Proposition may be demonstrated other- 
 wise. 
 
 Draw (1. 5. El.) BE bisecting the angle ABC. The angle 
 BE A (I. 8. El.) is greater than the inte- 
 rior angle EBC or EBA, and therefore 
 (1. 13. El.) the side AB is greater than 
 AE. In like manner, the angle BEC is 
 greater than the interior angle EBA or 
 EBC, and consequently (I. 13. El.) the 
 side CB is greater that CE. Wherefore the two sides AB and 
 CB, being each of them greater than the adjacent segments 
 AE and CE, are together greater than the whole base AC. 
 
 11. The fifteenth Proposition might also be demonstrated 
 otherwise. For join BE (I. 12,) the exterior angle BEC af 
 the triangle BAE is (I. 12.) greater than 
 
 *the interior ABE or (I. 10.) AEB, which ^ 
 
 again is the exterior angle of the triangle 
 ECB, and therefore (I. 12.) greater than 
 GBE. Whence (I. 13.) the side BC oppo- ^ -^ ^ 
 
 site to the greater angle is greater than CE, or CE the differ- 
 ence between the sides AB and AG is less than the third side 
 BC. 
 
 12. From the property that two sides of a triangle are to- 
 gether greater than the third side, may be derived the gene- 
 ric character of a straight line : 
 
 The .shortest line that can be draton betiveen tuoa points, is a 
 straight line. 
 
 Let the points A and B be connected by straight lines 
 joining an intermediate point C ; and the two sides AC and 
 BC of the triangle ACB are greater than AB (I. 15.). Now 
 
NOTES AND ILLUSTRATIONS. 289 
 
 let a third point D be interposed between A and C ; and be- 
 cause AD and DC are together 
 greater than AC, add BC to both, 
 and the three lines AD, DC, and 
 CB are greater than AC and BC, 
 and consequently still greater than 
 AB. Again, suppose a fourth 
 point E to connect B with C ; 
 and the sides BE and CE of the triangle BCE being greater 
 than BC, the four straight lines AD, DC, CE, and EB are 
 together, by a still farther access, greater than AB. By thus 
 repeatedly multiplying the interjacent points, two sides of a 
 triangle will at each successive step come in place of a third 
 side, and consequently the aggregate polygonal or crooked 
 line AFDGCHEIB will acquire continually some farther ex- 
 tension. Nay, since there is no limit to the possible number 
 of those connecting points, they may approach each other 
 nearer than any assignable interval ; and consequently the 
 proposition is also true in that extreme case where the boun- 
 dary is a curve line, or of which no portion can be deemed 
 rectilineal. 
 
 The proposition now demonstrated is commonly assumed as 
 an axiom. It is indeed forced uj.on our earliest observation, 
 being suggested by the stretching of a cord, and other fami- 
 liar occurrences in life. But thus to multiply principles, ap- 
 pears quite unphilosophical. The two radical properties of a 
 straight line — the congruity of its parts — and its shortness of 
 trace — are distinct, though connected. The latter is shown 
 to be the necessary consequence of the former ; but it would 
 be impossible, by any direct process, to infer the uniformity 
 of straight lines, from their marking out the nearest routes. 
 
 In the demonstration, I could not avoid introducing the 
 consideration of limits. This will occasion, I presume, no mate- 
 rial difficulty, since the reasoning is actually the same as that 
 by which our most familiar conceptions are gradually ex- 
 panded. 
 
 Mr Schwab, author of a small tract, entitled Elemens de 
 Geometrief and published at Nancy in 1813, has endeavoured 
 
290 NOTES AND ILLUSTRATIONS'. 
 
 to define a straigiit line as that which, being turned like an 
 axis about its two extremities, all its intermediate points will con- 
 stantly preserve the same position. This ingenious idea I have 
 adopted, in distinguishing the character of a straight line. 
 
 The same intelligent writer has, I find, referred the gene- 
 ration of angles to a revolving motion. He considers the right 
 angle as derived from the quartering of a whole revolution ; 
 and he likewise views, as I have done, the angle which a portion 
 of a straight line makes with its opposite portion, as formed by 
 g semi-revolution. 
 
 13. In reference to the eighteenth Proposition, the ingenious 
 MrT. Simpson has very justly 
 
 remarked, in his Elements of 
 Geometry, that the demon- 
 stration which Euclid gives 
 of this proposition is defec- 
 tive, since it assumes that 
 the point G must lie below 
 the base AC. He has there- 
 fore legitimately supplied the 
 
 deficiency of the proof; and it is surprising that so rigorous 
 a geometer as Dr Robert Simson should have so far yielded 
 to his prejudices, as to resist such a decided improvement. 
 The demonstration inserted in the text appears to be rather 
 simpler and more natural than that of Mr T. Simpson, 
 
 14. The nineteenth Proposition is capable of being demon- 
 strated directly. 
 
 Let the triangles ABC and DEF have the sides AB and 
 BC equal to DE and EF, but 
 the base AC greater than DF ; 
 the vertical angle ABC is great- 
 er than DEF. 
 
 From the greater base AC cut 
 oflt' AG equal to DF, construct 
 (I. 1.) the triangle AHG ha- 
 ying tjie sides AH and GH e- 
 
NOTES AND ILLUSTRATIONS. 291 
 
 qual to AB and BC or DE and EF, join HB, and produce HG 
 to meet BC in I. 
 
 Because HI is greater than HG, it is greater than the equal 
 side BC, and therefore much greater than BI. Consequently 
 the opposite angle IBH of the triangle BIH is (I. 13.) greater 
 than BHI. But AB being equal to AH, the angle HBA is 
 (I. 10.) equal to BHA, and therefore the two angles IBH 
 and HBA are greater than IHB and BHA, that is, the whole 
 angle CBA is greater than IHA or GHA. And since the 
 sides of the triangle AGH are by construction equal to those 
 of EDF, the corresponding angle AHG is equal to DEF 
 (I. 2,) ; and hence the angle ABC, which is greater than 
 AHG, is likewise greater than DEF. — In like manner, this 
 may be demonstrated, if BH should f^ll without the base, 
 
 15. It is not difficult to perceive that the whole structure of 
 geometry is grounded on the mqtual comparison of triangles, 
 the simplest of all the rectilineal figures. The conditions which 
 iix the equality of those elementary portions of surface, are all 
 contained in the 2d, 3d, 20th and 21st propositions of this Book. 
 Such original theorems derive their evidence from the super- 
 position of the triangles themselves ; which, in reality, is no- 
 thing but an ultimate appeal, though of the easiest and most 
 familiar kind, to external observation, The same conclusions, 
 however, might be deduced more concisely,- from the circum- 
 stances required to determine the constitution of an indivi- 
 dual triangle. Suppose AB, BC, and AC, any one of which is 
 shorter than the other two conjoined in a straight line, to be 
 three inflexible rods moveable at pleasure. — (1.) Place them 
 with their ends meeting each other, and they will evidently 
 rest in the same position, and contain a distinct triangle,— 
 which corresponds to Proposition 2. — 
 (2.) Having joined the rods AB and BC 
 at B, continue to open them at that point, 
 till they form a given vertical angle ABC ; 
 their position then becomes fixed, and con- 
 sequently determines the rod AC which 
 connects their extremities and completes the 
 triangle. This inference evidently agrees 
 
292 KOTES AND ILLUSTRATIONS. 
 
 with Proposition 3.--(3.) While the rod AC retains its place, 
 let two rods AB and CB of unlimited length, and applied at 
 the ends A and C, be opened gradually till the one forms with 
 AC a given angle CAB, and the other a given angle ACB ; 
 it is evident that AB and BC will then rest crossing each 
 other in those positions, and containing a determinate triangle, 
 of which the vertex B is their point of mutual intersection. 
 This property corresponds with Proposition 20. — (1.) Let the 
 rod AB of a given length make a given angle with the unli- 
 mited rod AC, and applying at the end B another given rod, 
 turn this gradually round till it meets AC. If BC exceeds 
 the distance of B from AC, it will evidently, after stretching 
 beyond AC, again come to meet that boundary. With such 
 conditions, therefore, the rods might contain two determinate 
 triangles, the one acute and the other obtuse, and which are 
 hence distinguished from each other by those obvious charac- 
 ters. This qualified property, omitted in most elementary works, 
 Is yet of extensive application, and was requisite to complete 
 the conditions of the equality of triangles. It corresponds 
 with Proposition 21, 
 
 The four preceding theorems are reducible, however, to a 
 single property, which includes all the different requisites to 
 the equality of triangles. The sides of a triangle are obvious- 
 ly independent of each other, being subject to this condition 
 only, that any one of them shall be less than the remaining 
 two sides. But since all the angles of a triangle are together 
 equal to two right angles, the third angle must, in every case, 
 be the necessary result of the other two angles. A triangle 
 has, therefore, only five original and variable parts — the three 
 sides and two of its angles. Any three of these jiarts being as- 
 certained^ the triangle is absolutely determined. Thus — when 
 (1.) all the three sides aii«e given, — when (2.) two sides and 
 their contained angle are given, — when (4.) two sides and 
 an opposite angle are given, with the affection of the triangle, 
 or when (3.) one side and two angles, and thence the third 
 angle are given, — the triangle is completely marked out. 
 
 M. Legendre, in a very elaborate note to his Elemens dc 
 
NOTES AND ILLUSTRATIONS. 293 
 
 Geometrie, has sought, with much ingenuity, to deduce a pri^ 
 ori the radical properties of triangles i'rom the theory of 
 Junctions. But, like other similar attempts, his investigation 
 actually involves in it a latent assumption. This subtle logi- 
 cian sets out with the principle which would seem almost in- 
 tuitive, that a triangle is determined when the base and its ad- 
 jacent angles are given. The vertical angle, therefore, de- 
 pends wholly on these data,^— the base and its adjacent angles. 
 Call the base c, its adjacent angles A, B, and the vertical or op- 
 posite angle C. This third angle, being derived from the quan- 
 tities A, B and c, must be a determinate function of them, or 
 formed from their combination. Whence, adopting his nota- 
 tion, Cz=.(p : (A, B, c). But the line c is of a nature hetero- 
 geneous to the angles A and Bj and therefore cannot be com- 
 pounded with these quantities. Consequently C=r^ : (A, B), 
 or the vertical angle C is a function merely of the angles A and 
 B at the base ; and hence the third angle of a triangle must 
 depend wholly on the other two. 
 
 To a speculative mathematician this argument is very se- 
 ductive, though it will not bear a rigid examination. Many 
 quantities in fact appear to result from the combined relation 
 of other quantities that are altogether heterogeneous. Thus, 
 the space which a moving body describes, depends on the 
 joint elements of time and velocity, things entirely distinct in 
 their nature ; and thus, the length of an arc of a circle is com- 
 pounded of the radius, and of the angle it subtends at the 
 centre, which are obviously heterogeneous magnitudes. For 
 aught we previously knew to the contrary, the base c might, 
 by its combination with the angles A and B, modify their re- 
 lation, and thence affect the value of the vertical angle C. In 
 another parallel case, the force of this remark is easily per- 
 ceived. Thus, when the sides a, b and their contained angle C 
 are given, the triangle is determined, as the simplest observa- 
 tion shows. Wherefore the base c is derived solely from these 
 data, or c=(p : («, 5, C). But the angle C, being heterogene- 
 ous to the sides a and by cannot coalesce with them into an 
 equation, and consequently the base c is simply a function of 
 a and 6, or it is the necessary result merely of the other two 
 
29* NOTES AND ILLUSTRATIONS. 
 
 sides. In other words, as the third angle of a triangle depends 
 on the other two angles, so the base of a triangle must have its 
 magnitude determined by the lengths of the two incumbent 
 sides. Such is the extreme absurdity to which this sort of 
 reasoning would lead ! In both of these instances, indeed, the 
 conclusion is admitted by implication, only in the one it is 
 consistent with truth, while in the other it is palpably false. 
 — That such an acute philosopher could overlook the fallacy 
 of his argument, can only be ascribed to the influence which 
 peculiar trains of thought acquire over the mind, and to the 
 extreme facility with which elementary principles insinuate 
 and blend themselves with almost every process of reasoning. 
 
 The objections here directed against the celebrated abstruse 
 attempt to demonstrate, a priori^ the equality of triangles from 
 the nature of equations, and the properties of homogeneous 
 quantities, have, generally, I believe, been deemed conclu- 
 sive. I have scarcely heard, indeed, of a geometer of any emi- 
 nence in the island, (except the learned writer of a critique 
 which appeared in the Edinburgh Review), who is not per- 
 fectly convinced of the fallacy lurking in the argument advan- 
 ced by its very ingenious inventor. On this occasion, I shall 
 take the liberty of introducing an extract from a letter to me, 
 dated October 20. 1816, from an old friend and fellow-student, 
 who now stands decidedly at the head of our mathematicians. 
 
 ** With regard to Legendre*s demonstration, I am of opi- 
 nion, that there is involved in the mise en equation^ (reduced 
 to an equation,) a principle which is equivalent to Euclid's 
 12th axiom, (If a straight line meets two straight lines, so as 
 to make the two interior angles on the same side of it taken 
 together less than two right angles, these straight lines, being 
 continually produced, will at length meet on that side on which 
 are the angles which are less than two right angles.) Using 
 the notation of your book, his assumption is, that C = 
 (p : (A, B, c) : Now, this means that we shall get the angle C, 
 by combining the angles A and B with the line c, in a certain 
 way ; and it is implied, that this is true, whatever value the line 
 c may have ; or, in other Words, it is true for all values of c. 
 Suppose then an individual triangle, of which c is the base, 
 
NOTES AND ILLUSTRATIONS. 295 
 
 and A, B, the angles at Its extremities ; conceive an indefinite 
 number of lines, of any lengths, c', c", d", &c. and at the ends of 
 each of these lines, angles to be made equal to A and B, — will 
 a triangle be thus formed upon each of the lines c', c", c'", &c* 
 or not ? If you say that you cannot allow the existence of such 
 triangles without proof, you agree with the Greek geometer, 
 but then you must deny the legitimacy of Legendre's equa- 
 tion, C z= (p: (A, B, c) ; for it supposes the possibility of such 
 triangles, since it is a determination of the third angle of each 
 of them from knowing the base and the other two angles. If 
 you grant the possibility of the triangles, then Legendre's 
 equation will be established ; but you also admit Euclid's 12th 
 axiom : For you assume, that two lines drawn at the extremi- 
 ties of any third line, so as fo make with it two angles equal 
 to any two angleis of a triangle, do meet one another when pro- 
 duced. On examination you will find, that the only relation 
 generally true of two angles of a triangle is this, that they are 
 together less than two right angles. I cannot, therefore, ad- 
 mit, that Legendre's demonstration contributes in any degree 
 to remove the difficulty in geometry. The intrinsic evidence 
 of a principle, or proposition, is the same whether it be ex- 
 pressed in common language, or translated into the language 
 of functions. Grant to the geometer the same assumption 
 which is implied in the functional equation of the analyst, and 
 he will be no longer embarrassed with the theory of parallel 
 lines. Legendre endeavours to justify his equation, by sa)'- 
 ing that two triangles are identical when they have their bases 
 equal, and likewise the angles adjacent to their bases equal, 
 each to each. But this does not prove, that of all the infinite 
 number of triangles which can be formed upon a line greater 
 or less than the base of a given triangle, there is always one 
 that has the angles at its base equal to the angles at the base 
 of the given triangle. If this be thought a more self-evident 
 principle than those that geometers have employed, let it be 
 transferred to geometry, and that science will no longer have 
 need to borrow aid from the theory of functions." 
 
 To these acute and judicious remarks, I think it unneces- 
 sary to subjoin any farther observations ; but, in justice to the 
 
296 KOTES AND ILLUSTRATIONS. 
 
 illustrious author of the argument drawn from the higher ana- 
 lysis, I must state, that he still remains persuaded of its legiti- 
 macj'. In- a very flattering letter, which he did me the honour 
 to write, bearing date, Paris, 5th February 1816, he thus ad- 
 verts to the subject in dispute. ** Ayant un tres grande idee 
 de la superiorite de vos lumieres, Monsieur, j'eprouve un re- 
 gret d'autant plus vif de voir que vous n'approuviez pas, ou 
 meme que vous regardiez comme illusoire la demonstration 
 que j*ai donnee dans mes notes du principe sur les trois angles 
 du triangle. J'ai cependant la conviction intime que cette 
 demonstration est parfaitement rigoureuse, et j'ose vous prier 
 d'y donner encore quelqu' attention, persuade que vous recon- 
 noitrez son exactitude. La loi de I'homogenite est une loi 
 generale, qui n'est jamais en defaut, et qui doit 6tre range6 
 parmi les principes elementaires les plus generaux et les plus 
 simples. L'angle est un quantite que je mesure toujours, par 
 son rapport avec I'angle droit, car I'angle droit est I'unite na- 
 turelle des angles : Dans cette notion tres simple, une angle 
 est toujours un nombre. 11 n'en est pas de meme des lignes : 
 une ligne ne peut entrer dans le calcul, dans une equation 
 quelconque, qu'avec une autre ligne que sera prise pour unite, 
 ou qui aura un rapport connu avec la ligne unite. 
 
 «* Ainsi Inequation C = (f>: (A, B, c) rapportee, pag. 403, ou 
 A, B, C, sont des angles, et par consequent des nombres, ne 
 sauroit subsister, a moins que c ne disparoisse. Car si c ne dis- 
 paroit pas, il faudra qu'une longueur absolue c soit determinee 
 par des nombres, sans que Punite de longueur soit connue, ce 
 > qu? est une absurdite. L'objection faite, pag. suiv, sur Tequa- 
 tion c = ^ : («, ^, C) se resout tres facilement. Rien n'em- 
 peche que C, qui est un nombre, (par rapport a Pangle droit 
 pris pour unite), ne soit une fonction de a, b, C, pourvu que 
 cette fonction soit de nulle dimension, c'est-a-dire, pourvu que 
 le fonction de a, b, C se reduise a une fonction de deux rapports, 
 
 tels que — , — . Et en effet, c'est ce qui a lieu d'apres I'equa- 
 
 1+ ~—S- 
 
 fion trigonometrique, cos C = ^ "\^ — r— -^ = a * 
 
 2 ae 2 _ 
 
NOTES AND ILLUSTRATIONS. 297 
 
 Ajouteral-je ti ces raisons, une idee qui m'est venue plu- 
 sieurs fois. Suppose que le meme triangle, dont ^^ 
 
 yous vous occupez, soit mis sous les yeux d'un 
 etre intelligent, dont la stature et celle des objets 
 qui l*environnent soient cent fois plus grandes 
 que celle des objets environnans — mon raisonne- 
 ment sera toujours le meme, et ne perdra rien de 
 la force. Croiez-vous, cependant, qu'il fut possible que c restat 
 dans Inequation, C = <p : (A, B, c) ? Et si c restoit, les geans 
 dont je parle deduiroient-iis de cette equation la meme valeur 
 que vous ? II faudroit que cela fut, car Pobjet a les memes 
 dimensions dans les deux cas." 
 
 I am sorry that, on reconsidering the subject maturely, I 
 cannot assent to the force of this reasoning, however clearly and 
 neatly it is here developed. The whole stress of the argu- 
 ment, it may be perceived, lies in the distinction which M. 
 Legendre endeavours to establish between angles and lines, 
 — a distinction which I hold at bottom to be merely arbitrary.' 
 Angles and lines are both equally real quantities, though of 
 different kinds ; they are capable of being measured, and con- 
 sequently represented by numbers, by referring each of them 
 to some determinate measure or unit of its own denomination. 
 Angles are measured or expressed numerically by angles, 
 and lines by lines. It is true, that the mensuration of angles is 
 facilitated by a reference to the subdivision of the circuit or 
 entire revolution ; yet even this mode of denoting angular 
 magnitude is evidently only conventional. As standards for 
 measuring straight lines, nature has furnished the limbs of 
 the human body, and the extent of our globe itself. Such 
 units of mensuration are not indeed very definite or readily 
 attainable ; but they are not therefore the less real or pro- 
 minent. Nor is there any' essential difference in principle be- 
 tween the expressing of an angle by degrees, of which 360 or 
 400 are contained in a complete revolution, and the denoting 
 of a straight line on the French system, for instance, by the 
 number of metres it includes, each of which is the forty mil- 
 lionth part of the entire circumference of the earth. Angles 
 and lines hence present to the mind no radical or absolute 
 
1298 NOTES AND ILLUSTRATIONS- 
 
 discrimination, and therefore the argument grounded on such 
 a distinction must lose all its efficacy. 
 
 Admitting, however, what the slightest inspection readi- 
 ly confirms, that the third angle is merely derived from the 
 other two, M. Legendre demonstrates with great elegance, 
 the property that the three angles of a triangle are equal to 
 two right angles. Letting fall from the right angle a perpen- 
 dicular on the hypotenuse, he divides any right-angled tri- 
 angle into two subordinate triangles, which have each of them 
 two angles equal to those of the original triangle ; whence the 
 acute angles of that triangle are alternately equal to the angles 
 which compose the right angle. But every triangle may be 
 divided into two right-angled triangles, by letting fall a per- 
 pendicular from the vertex on the base, and consequently the 
 acute angles of both these triangles, and which form the angles 
 at the base, and the vertical angle of the primary triangle, — 
 are together equal to two right angles. 
 
 This theorem may be proved somewhat more directly. In 
 the triangle ABC, let the angle CBA be greater than ACB, 
 and draw BD, and then DE, making the angles ABD and 
 BDE each equal to ACB. The 
 triangles ABC and ADB having ^J? 
 
 the common angle BAC and the ^^^ I \ 
 
 angle ACB equal to ABD, their ^^ I ^^\ 
 
 third angles ABC and ADB ^ ^ ^ 
 
 must be equal. But the tri- 
 angles BCD and BDE have also a common angle CBD, and 
 equal angles DCB and BDE : whence the third angle BDC 
 is equal to BED, and therefore the supplementary angle ADB, 
 equal to ABC, is equal to DEC. Again, the triangles ABC 
 and DEC having two common or equal angles, their third 
 angles BAC and EDC are equal ; wherefore the three angles 
 ABC, BC A and BAC of the original triangle, are respective- 
 ly equal to BDA, BDE and EDC, and hence equal to two 
 right angles. — If the triangle ABC be equiangular, divide it 
 into two scalene triangles ABD and CBD, the angles of 
 which, or the angles of the original triangle, together with the 
 
KOTES AND ILLUSTRATIONS. 299 
 
 adjoining angles ADB and BDC, must be equal to four right 
 angles, and consequently the angles of that triangle are equal 
 to two right angles. 
 
 But the proposition is easily derived from another view of 
 the subject. If we suppose a ruler turning 
 about the point A, to change its direction 
 AC into AB, then opening at B till it gains 
 the direction BC, and finally wearing about 
 the point C till it acquires the opposite po- 
 sition C A ; thus changing its direction with 
 respect to a remote object, by three successive openings all 
 to the same side, the ruler, being now reversed, must have 
 performed half a circuit ; that is, the three angles of a tri- 
 angle, which constitute those openings, are equal to two right 
 anij 
 
 The profound geometer already quoted, pursuing his re* 
 fined argument, has, from the consideration of homogene- 
 ous quantities, likewise attempted to deduce the proportiona* 
 lity of the sides of equiangular triangles. But in this ab- 
 struse research, assumptions are still disguised and mixed up 
 in the progress of induction. Such indeed must be the case 
 with every kind of reasoning on mathematical or physical ob- 
 jects, which proceeds a priori, without appealing, at least in 
 the first instance, to external observation. Of this kind are 
 some of those ingenious analytical investigations respecting 
 the laws of motion and the composition of forces. In fact, 
 no elementary physical truth can ever be discovered by any 
 process of calculation, which merely combines or embodies 
 the various assumptions that have been tacitly made into a ge- 
 neral result. The principle o'i sufficient reason, introduced by 
 Leibnitz, appears to be nothing but an artificial mode of dress- 
 ing out an hypothesis, which the celebrated Boscovich has well 
 exposed in his excellent notes to a didactic poem by Stay, 
 entitled Philosophia Recentior. 
 
soo 
 
 NOTES AND ILLUSTRATIONS. 
 
 14. Proposition twenty-second. The subject of parallel 
 lines has exercised the ingenuity of modern geometers ; for 
 Euclid had only endeavoured to evade the difficulty, by styl- 
 ing the fundamental proposition an axiom. The investigation 
 now given seems to be one of the best adapted to the natural 
 progress of discovery. It is almost ridiculous to scruple 
 about admitting the idea of motion, which I have employed 
 for the sake of clearness. But even that futile objection 
 might be obviated, by considering merely the successive 
 positions of the straight line extending through the given 
 point. . 
 
 15. Proposition thirtieth. That invaluable instrument, 
 Hadley's quadrant, is founded on the 
 second corollary, annexed as an ob- 
 viousconsequenceoftheproposition. 
 A ray of light S A, from the sun, imr 
 pinging against the mirror at A, is 
 reflected at an angle equal to its in* 
 cidence ; and now striking the half- 
 silvered glass at C, it is again re- 
 flected to E, where the eye like- 
 wise receives, through the transpa- 
 rent part of that glass, a direct ray 
 from the boundary of the horizon. 
 Hence the triangle AEC has its ex- 
 terior angle ECD and one of its in- 
 terior angles CAE, respectively double of the exterior angle 
 BCD and the interior angle CAB, of the triangle ABC ; 
 wherefore the remaining interior angle AEC, or SEZ, is 
 double of ABC ; that is, the altitude of the sun above the 
 horizon is double of the inclination of the two mirrors. But 
 the glass at C remaining fixed, the mirror at A is attached 
 to a moveable index, which marks their inclination. 
 
 The same instrument, in its most improved state, and fitted 
 with a telescope, forms the sextant, which, being admirably 
 calculated for measuring angles in general, has rendered the 
 most important services to geography and navigation. 
 
NOTES AND ILLUSTRATIONS. 301 
 
 16. Proposition thirty-fourth. This problem is generally 
 constructed somewhat differently. 
 
 In AB take any point C, and on BC (I. 1. cor.) describe 
 an equilateral triangle CDB, on 
 its side DB, another DEB; and 
 on DE the side of this, a third 
 equilateral triangle DFE ; join the 
 last vertex F with the point B ; 
 and BF is the perpendicular re- 
 quired. 
 
 Because the triangles CDB and 
 DBE are equilateral, the angles CBD and DBE are each of 
 them equal to two-third parts of a right angle (I. 30. cor. 1.) ; 
 and the triangles BDF, BEF, having the sides BD, DF equal 
 to BE, EF, and the side BF common, are (I. 2.) equal, and 
 consequently the angles FBD and FBE are equal, and each 
 of them the half of DBE. The angle FBD, being therefore 
 one-third part of a right angle, and the angle DBA two-third 
 parts, the whole angle FBC must be an entire right angle, ox* 
 the straight line BF is perpendicular to AB. 
 
 BOOK II. 
 
 1. A simple proposition might be here introduced. 
 
 A straight line bisecting ttvo sides of a triangle^ is parallel t9 
 the base. 
 
 The straight line DE which joins the middle points of the 
 sides AB and BC, is parallel to the base AC of the triangle 
 ABC. 
 
 For join: AE and CD. Because the triangles ADC, BCD 
 stand on equal bases AD, DB, and have the same vertex or 
 altitude, they are (II. 2.) equivalent, and therefore ADC is 
 
S02 NOTES AND ILLUSTRATIONS. 
 
 half of the whole triangle ABC. For the 
 
 same reason, since CE is equal to EB, the 
 
 triangle AEG is equivalent to AEB, and 
 
 is consequently half of the whole triangle 
 
 ABC Whence the triangles ADC and 
 
 AEC are etjuivalent ; and they stand on a. G C 
 
 the same base AC, and have therefore the same altitude 
 
 (11. 3.), or DE is parallel to AC. 
 
 Cor. Hence the triangle DBE cut off by the line DE, is 
 the fourth part of the original triangle. For bisect AC in G, 
 and join DG, which is therefore parallel to BC. The triangle 
 ADG is equivalent to GDC (II. 2.), and GDC, being the half 
 of the rhomboid CE, is equivalent to DEC, which again is 
 (II. 2.) equivalent to DEB. The triangle ABC is thus divided 
 into four equivalent triangles, of which DBE is one. Hence 
 also the rhomboid GDEC is half of the original triangle. 
 
 2. From the preceding proposition the following theoreni is 
 easily derived: 
 
 Straight lines Joining the successive middle points of the sides 
 of a quadrilateral Jigure,Jbrm a rhomboid. 
 
 If the sides of the quadrilateral figure ABCD be bisected, 
 and the points of section joined in their order ; EFGH is a 
 rhomboid. 
 
 For draw AC, BD. And because FG bisects AB, BC, it 
 is parallel to AC ; and for the same 
 reason, EH, as it bisects AD and 
 DC, is parallel to AC. Wherefore 
 FG is parallel to EH (I. 28.). In 
 like manner, it is proved that EF is 
 parallel to HG ; and consequently 
 the figure EFGH is a rhomboid or 
 parallelogram. 
 
 It is likewise evident, that the inscribed rhomboid is half of 
 the quadrilateral figure ; for IG is half of the triangle ABC 
 ^nd IH is half of the triangle ADC. 
 
NOTES AND ILLUSTRATIONS. 303 
 
 3. Proposition fourth. This problem is of great use in prac- 
 tical geometry. The plan, for instance, of any grounds, how- 
 ever irregular, is divided into a number of triangles, which 
 are successively reduced to a simple triangle, and this again 
 is converted (by II. 6.) into a rectangle. Instead of comput- 
 ing, therefore, each component triangle, it may be sufficient 
 to calculate the area of the final triangle or rectangle. 
 
 4. Proposition ninth. On this proposition is founded the 
 method of offsets, which enters so largely into the practice 
 of land-surveying. In measuring a field of a very irregular 
 shape, the principal points only are connected by straight 
 lines forming sides of the component triangles, and the dis- 
 tance of each remarkable flexure of the extreme boundary is 
 taken from these rectilineal traces. The exterior border of 
 the polygon is therefore considered as a collection of trape- 
 zoids, which are measured by multiplying the mean of each 
 pair of offsets or perpendiculars into their base or intermediate 
 distance. 
 
 5. Proposition tenth. This beautiful property is easily de- 
 rived from Propositions fifteenth and sixteenth of Book II. 
 
 1 . Let ABC be a triangle right-angled at B ; produce the base 
 AB till AD be equal to the perpendicular BC ; on the com- 
 pound line BD describe the square BDEF, and make DG and 
 EH equal to AB, and join AG, GH and HC. 
 
 The triangles ABC and GD A, having the sides AB, BC evi- 
 dently equal to DG, AD, and the right angle at B equal to 
 that at D, are (I. 3.) equal. In 
 the same manner, the triangles HEG 
 and CHF are proved to be equal to 
 ABC. But (I. 30.) the exterior an- 
 gle GAB is equal to the interior 
 angles ADG and AGD, from which 
 take away the equal angles CAB 
 and AGD, and there remains GAC 
 equal to ADG, and consequently a 
 right angle. Wherefore the quadrilateral figure AGHC, havln* 
 
304 
 
 NOTES AND ILLUSTRATIONS. 
 
 likewise all its sides equal, is a square. But by Prop. 15. Book 
 II. the square BDEF, described on the sum of the sides AB and 
 BC, is equivalent to the squares of those sides, together with 
 twice their rectangle. Now (cor. 5. Book U.) the rectangle 
 under AB and BC is double of the triangle ABC ; and conse- 
 quently the square BDEF is equivalent to the squares of AB 
 and BC, and the four triangles CBA, ADG, GEH and HFC ; 
 but the same square is equivalent to the interior square 
 AGHC, with those four triangles ; "wherefore the squares of 
 the base AB and of the perpendicular BC, are equivalent to 
 the single square described on the hypotenuse AC. 
 
 2. From the base AB, cut off a part AD equal to the per- 
 pendicular BC, and on the remaining portion BD construct 
 the square BDEF ; produce DE and EF, till EG and FH be 
 equal to AD, and join AG, GH, and HC The triangles 
 CBA, ADG, GEH, and HFC are proved to be equal as before. 
 Again, the angle CAG being equal to the angles CAB and 
 DAG or BCA, the acute angles of the right-angled triangle 
 ABC, is consequently a right angle.' 
 Wherefore the quadrilateral figure 
 ACHG is a square. But, by Prop. 
 16. Book 11. the square BDEF, de- 
 scribed on BD the difference be- 
 tween the base AB and BC the 
 perpendicular is equivalent, to the 
 squares of AB and BC, diminished 
 by twice their rectangle, or by the 
 four triangles CBA, ADG, GEH, and HFC. But the square 
 BDEF is evidently equivalent to the square ACHG described 
 on the hypotenuse AC, diminished by those triangles, and 
 therefore equivalent to the squares of the base AB and of the 
 perpendicular BC. 
 
 This famous proposition appears to have been brought from 
 the East by Pythagoras. Both the demonstrations now given 
 are common in Persia and India. The second mode, however, 
 
 1 
 
NOTES AND ILLUSTRATIONS. 
 
 305 
 
 would seem to be the favourite, since the figure used is in 
 Hindustan styled the bridal chair or conchy in allusion, no doubt, 
 to its prolific virtues. This figure, and the preceding one, are 
 well adapted for exhibiting the result, by the dissection and 
 transposition of their several parts. The very meagre treatises 
 of geometry written in the ancient Sanscrit language, and the 
 versions of Euclid's Elements by Persian or Arabian commen- 
 tators, display some variety of such dissections. The method 
 generally adopted is ascribed to the Persian astronomer Nassir 
 Eddin, who flourished in the thirteenth century of our aera, 
 under the munificent patronage of the conqueror Zingis Khan. 
 It may gratify the young student in geometry to see the 
 mode of performing this dissection. Having drawn AL pa- 
 rallel to BF, and IC and GO pa- 
 rallel to DB, place the triangle CKA y^ X. II l^" 
 on CFH, invert the triangle GO A 
 on ADG, place the triangle GOM 
 on AKN, and transfer the small tri- 
 angle GIN to HLM. In this way, ,7._>^! j-— '4q 
 
 the square AGHC is transformed 
 into the two squares CKLF and 
 ADIK. By reversing the process, 
 the squares of the side^ of the right- 
 angled triangle may be compounded into the single square of 
 the hypotenuse. 
 
 X 
 
 o 
 
 \ 
 
 \:jsi 
 
 \ 
 
 K 
 
 / 
 
 6. It was a favourite speculation with the Greek geome- 
 ters, to express numerically the sides of a right-angled tri- 
 angle. The rules which they delivered for that purpose are 
 equally simple and ingenious. For the sake of conciseness, 
 it will be convenient, however, to adopt the language of sym- 
 bols. Let 71 denote any odd number ; then, 
 
 according to Pythagoras, w, 
 
 n*— I 
 
 and^I+i, 
 2 ' 
 
 or 
 
 according to Plato, 2 w, w* — 1 and w*-|-l, will repre- 
 sent the perpendicular, the base, and hypotenuse, of a right- 
 angled triangle.— Thus, n being supposed equal to 3, the num- 
 bers thence resulting are 3, 4, and 5, or 6, 8, and 10. These 
 
306 NOTES AND ILLUSTRATIONS. 
 
 analytical expressions are fundamentally the same, and are 
 easily derived from Proposition 17. Book II. : For 
 
 2^^^X2 = (2«)^ — Or without having recourse to algebraical 
 notation, since the square of the perpendicular is equivalent 
 to the difference between the squares of the hypotenuse and 
 of the base, it must, by Prop. 17. Book II. be equivalent to 
 the rectangle under the sum and difference of the hypotenuse 
 and base. Wherefore, if the perpendicular be an odd num- 
 ber, its square may be divided into two contiguous factors, 
 the one even and the other odd. Thus, assuming the perpen- 
 dicular equal to 3, its square 9 gives, by division, 4 and 5, for 
 the base and hypotenuse ; if the perpendicular be 5, the square 
 25 is parted into 12 and 13, for the corresponding base and 
 hypotenuse ; or if this perpendicular be denoted by 7, whose 
 square is 49, the base and perpendicular must, by partition, be 
 24 and 25. Again, if the perpendicular be supposed to be an 
 even number, its square may be divided into two adjacent 
 factors, whose sum is the half and their difference 2. Thus, the 
 perpendicular being 4, the half of its square, or 8, is split into 
 -3 and 5, for the base and hypotenuse ; if 6 be the perpendicu- 
 lar, the half of its square, or 18, is divided into 8 and 10, for 
 the base and hypotenuse ; and were 8 to represent the perpen- 
 dicular, the half of its square, or 32, gives \5 and 17, for the 
 corresponding base and perpendicular. 
 
 7. We may here introduce, from the Mathematical Collec- 
 tions of Pappus, an elegant extension of the famous Tenth 
 Proposition. 
 
 In any triangle, rhomboids described on the ttvo sides, are to- 
 gether equivalent to a rhomboid described on the base, and limit- 
 ed by these and by parallels to the line which joins the vertex tvith 
 their point of concourse. 
 
 Let ADEB and BGFC be rhomboids described on the two 
 sides AB and BC of the triangle ABC ; produce the summits 
 DE and FG to meet in H, join this point with the vertex B, 
 to BH draw the parallels AK, CL, and join KL. It is obvi- 
 ous that AK and CL, being equal and parallel to BH, are 
 
NOTES AND ILLUSTRATIONS. 307 
 
 likewise equal and parallel to each other, and that the figure 
 AKLC is a parallelogram or rhomboid — This rhomboid is 
 equivalent to the two rhomboids BD and BF. 
 
 For produce HB to meet the base AC in I. And because 
 the rhomboids KI and AH 
 stand on the same base AK 
 and between the same 
 parallels, they are equiva- 
 lent (II. I. cor.) ; but 
 the rhomboids AH and 
 BD, standing on the same 
 base AB and between the 
 same parallels, are also equivalent. Whence KI is equiva- 
 lent to BD. And in the same manner, it may be proved that 
 LI is equivalent to BF. Consequently the whole rhomboid 
 K!C is equivalent to the two rhomboids BD and BF, 
 
 If the triangle ABC be right-angled at B, this theorem will 
 pass into a case of the twenty-sixth of Book VI. ; the rhom- 
 boid, described on the hypotenuse, being equivalent to the 
 similar rhomboids described on the two sides. When these 
 rhomboids become squares, the proposition becomes the same 
 as the tenth ; the only difference in the construction being, 
 that a square AKOC (p. 52.) is constructed above the hypote- 
 nuse AC, instead of the square ADEC constructed below it. 
 
 8. Ffoto the proposition in the last article, an important 
 theorem may be derived, which deserves a place in an ele* 
 mentary work : 
 
 In any triangle^ the square described on the base is equivalent 
 to the rectangles contained by the two sides and their segments in- 
 terceptedjrom the base by perpendiculars let Jail upon them from 
 its opposite extremities. 
 
 Let the perpendiculars AP, CN be let fall from the points 
 A, C upon the opposite sides BC and AB of the triangle 
 ABC ; the square of AC is equivalent to the rectangles con- 
 tained by AB, AN, and by BC, CP. 
 
 For complete the rhomboids ADHB and CFHB, and let 
 fall the perpendiculars BR and BS upon DH and FH. 
 
308 
 
 NOTES AND ILLUSTRATIONS. 
 
 It is manifest, that the rhomboids AH and CH are equiva- 
 lent to the square of AC. But the rhomboid AH is equivalent 
 to the rectangle contained by AB and BIl 
 II. 1 . cor.). Comparing the triangles BH R 
 and ACN; the angle BRH, being a 
 right angle, is equal to ANC ; and the 
 two acute angles BHR and RBH, being 
 together equal to a right angle, are e- 
 qual to DAN and NAC ; but DAB is 
 equal to DHB (1. 26.), whence the angle 
 RBH is equal to NAC. These triangles 
 BHR and ACN, having thus two angles 
 respectively equal, and the correspond- 
 ing side BH in the one equal to AD or 
 AC in the other, are therefore equal 
 (I. 20.), and consequently the side BR 
 is equal to AN. The rectangle AB and 
 BR, which is equivalent to the rhom- 
 boid AH, is hence equivalent to the 
 rectangle contained by AB and AN (IL 
 1. cor.). 
 
 In the same manner, it may be demonstrated, by compa- 
 ring the triangles BHS and PAC, that the rectangle under BC 
 and BS, which is equivalent to the rhomboid CH, is equivalent 
 to the rectangle contained by BC and CP. Wherefore the 
 two rectangles of AB, AN and BC, CP are together equiva- 
 lent to the square described on AC. 
 
 If the triangle ABC be right-angled at the vertex B, the 
 perpendiculars CN and AP will evidently meet at the vertex, 
 and consequently the rectangles AB, AN and BC, CP will 
 become the squares of AB and BC. And hence the beauti- 
 ful Proposition II. 10- is derived, being only a remarkable 
 case of a much more general property. 
 
 9. Proposition tenth. It may be proper to notice likewise 
 an extension of this beautiful proposition, which is easily de- 
 monstrated, after a similar mode, from the decomposition of 
 tjie figure. 
 
NOTES AND ILLUSTRATIONS. 
 
 309 
 
 Equilateral triangles described on the sides of a right-angled 
 triangley are together equivalent to an equilateral triangle de- 
 scribed on the hypotenuse. 
 
 Let ABC be a right-angled triangle, around which are con- 
 structed the equilateral triangles ADB, BEC and CFA ; the 
 triangles ADB and BEC are equivalent to CFA. 
 
 For let fall the perpendiculars DG, 
 BH and FI, and join CD, BF, CG, BI 
 and HF. It is evident (I. 21.) that 
 the^ perpendiculars DG and FI bi- 
 sect the bases AB and AC, and di- 
 vide the triangles ADB and CFA 
 into two equal triangles. But the 
 angle DAB is equal to CAF, being 
 angles of an equilateral triangle : 
 add BAC to each, and the whole 
 angle DAC is equal to BAF. But 
 the containing sides DA and AC are respectively equal to 
 BA and AF, and consequently (I. 3.) the triangle ADC is 
 equal td ABF. Now the triangle ADC is composed of the 
 three triangles ACG, ADG, and DCG, and the triangle ABF 
 is composed of ABI, AFI, and FBI ; but, since AB and AC 
 are bisected in G and H, the triangles ACG and ABI are 
 (II. 2.) halves of the original triangle ABC, and consequently 
 equivalent to each other. Wherefore the remaining triangles 
 ADG and DCG are together equivalent to AFI and FBI. 
 But DG and CB being both perpendicular to AB, are (1. 22.) 
 parallel ; and, for the same reason, BH is parallel to FI. 
 Whence (II. 1.) the triangle DCG is equivalent to DBG, and 
 the triangle FBI equivalent to FHI; and therefore the tri- 
 angles ADG and DBG, or the whole triangle ADB, must be 
 equivalent to AFI and FHI, or the whole triangle AFH. — in 
 like manner, it may be shown that the triangle BEC is equi- 
 valent to the triangle CFH ; and consequently the equilateral 
 triangles ADB and BEC are equivalent to AFH and CFH, 
 which make up the whole triangle AFC. 
 
 This demonstration is the second of those given by the cele* 
 brated Italian geometer Torricelli, the favourite disciple of 
 Galileo, and inventor of the barometer. 
 

 T> n 
 
 
 T 
 
 
 G — ^ 
 
 
 
 
 S(L 
 
 V <. 
 
 ; u : 
 
 13 
 
 310 NOTES AND ILLUSTRATIONS. 
 
 10. A useful proposition may be introduced here : 
 
 The square described on a straight line, is equivalent to the 
 squares of the segments into tvhich it is divided, and tmce the 
 rectangles contained hy each pair of these segments. 
 
 The square of AB is equivalent to the squares of AC, of 
 CD and of DB, with twice the rectangles of AC, CD, of AC, 
 DB, andofCD, DB. 
 
 For make AE and EF equal to AC and CD draw EM, FL 
 parallel to AB, and CH, Dl parallel to AG. 
 
 It is manifest that AO is the square 
 of AC, OQ the square of CD, and QK Q_II_XJf 
 the square of DB. Nor is it less obvi- 
 ous that the two rectangles CN and 
 EP are contained by AC, CD, that the 
 two rectangles NL and PI are contain- 
 ed by CD, DB, and that the two rect- 
 angles DM and FH are contained by 
 AC, DB. But those squares and those double rectangles 
 qomplete the whole square of AB. Wherefore the truth of 
 the Proposition is established. # 
 
 Co)'. Hence if a straight line be divided into three portions, 
 the squares of the double segments AD, BC, together with 
 twice the rectangle under the extreme segments AC, BD, 
 are equivalent to the squares of the whole line AB and of the 
 intermediate segment CD. For the squares FD, HM, toge- 
 ther with the equal rectangles GP, NB, evidently fill up the 
 whole square AB, with the repetition of the internal square 
 OQ ; that is, the squares of AD and BC, with twice the rect- 
 angle AC, DB, are equivalent to the squares of AB and CD. 
 
 11. Since rectangles correspond to numerical products, the 
 properties of the sections of lines ar'e easily derived from sym- 
 bolical arithmetic or algebra : 
 
 1. In Prop. 14. let AC be denoted by a, and the segments 
 ©f AB by hf c and d; then a{b-{-c-\-'d')z=ab-{-ac-\'ad. 
 
 2. In Prop. 15. let the two lines be denoted by a and b j 
 then ia + byz::a' + b^+2ab. 
 
NOTES AND ILLUSTRATIONS. Sll 
 
 a. In Prop. 16. let the two lines be denoted by a and b ; 
 then («— 6)*=«^+^^— 2«6. 
 
 4. In Prop. 17. let the two lines be denoted by a and b ; 
 then (a + Z»)(a— 6)=a^— ^^ 
 
 5. In the Proposition contained in the last paragraph of the 
 notes on this Book, let the segments of the compound line be 
 denoted by a, b and c ; then 
 
 {a + b + cy=a^-{-b^+c'^ + ^ab + ^ac + 2bc. 
 
 6. In Prop. 18. let the two lines be denoted by a and ^/ 
 
 /a4-b\^ /a — b\^, 
 then a^-h^^ = i(«+^)*+i(«--*)'=2(^-2~J '^\~2~ ) 
 
 7. In Prop. 19. let the whole line be denominated by a, and 
 its greater segment byx; then x^z=aia—x), andx^+ax:=a\ 
 
 whence xzzdtz^-^ -^ ==±ifl(V| — \). Hence, if unit 
 
 represent the whole line, the greater segment is .618033984'28, 
 &c. and the smaller segment .38196601572, &c. 
 
 From Cor. 1. an extremely neat approximation is likewise 
 obtained. Assuming the segments of the divided line as at 
 first equal, and each denoted by 1, the following successive 
 numbers will result from a continued summation ; 
 
 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, &c. 
 which are thus composed, 
 l+,2=z3, 2+3 = 5, 3+5 = 8, 5+ 8= 13, 8+ 13 = 21, &c. 
 
 These numbers form, therefore, a simple recurring series, 
 a kind of approximation which was first noticed in this actual 
 case early in the seventeenth century, by Gerard, an ingeni- 
 ous Flemish mathematician. 
 
 Hence, if the original line contained 144 equal parts, its 
 greater segment would include 89, and its smaller segment 55 
 of these parts, very nearly ; but 55 X 144 = 7920, being only 
 one less than 7921, the square of 89. 
 
 12. Proposition nineteenth, cor. 2. This problem may, how- 
 ever, be constructed somewhat differently, without employing 
 the collateral properties. 
 
312 NOTES AND ILLUSTRATIONS. 
 
 For bisect AB in C (I. 7.)> draw (I. 5. cor.) the perpen- 
 dicular BD equal to BC, join AD and 
 continue it until DE be equal to DB 
 or BC, and on AB produced take AF 
 equal to AE : The line AF is the re- 
 quired extension of AB. For make DG 
 equal to DB or BC ; and because 
 
 (11. 17. cor. 2.) the rectangle EA, AG, together with the square 
 of DG or DB, is equivalent to the square of DA, or to the 
 squares of AB and DB ; the rectangle EA, AG, or FA, FB, 
 is equivalent to the square of AB. 
 
 13. Proposition twenty-third. This proposition is of great 
 use in practical geometry, since it enables us to divide a tri- 
 angle, of which all the sides are given, into two right-angled 
 triangles, by determining the position, and consequently the 
 length, of the perpendicular. 
 
 Thus, suppose the base of the triangle to be 15, and the 
 two sides 13 and 14: Then 15* + 13' — 14' = 225 + 
 169 — 196= 198, which shows that the perpendicular falls 
 
 198 
 within the triangle ; and — ^ =: 6.6) the segment adjacent to 
 
 the short side, whence the perpendicular rr/y^ ((13/ — {6.6 j^) = 
 
 ^(169— 43.56)= 11.2. The area is therefore 15 X 5.6 = 84. 
 
 Again, let the base be 9, and the two sides 17 and 10: 
 
 Then 17* — 9' ■— 10' = 289 — 81 — 100 = 108, indicating that 
 
 108 
 
 the perpendicular falls without the base. Wherefore, = 
 
 18 
 
 6, the external segment, and a/( 10»— -6') = ^(100—36) = 
 
 9x8 
 ^^64 = 8, the perpendicular; which gives —Q — = 36, for 
 
 the area of the triangle. 
 
 Lastly, if the base were 10, and the sides 21 and 17 : Then 
 2P— 17'— 10' =441 --289.— 100= 52, which shows that 
 the perpendicular falls somewhat beyond the base. Whence 
 
 — ■=■ 2.6, the external segment; and v^ ( 17' — 2.6') = 
 
NOTES AND ILLUSTRATIONS. SIS 
 
 V^(289 — 6.76) = v^282.24^=: 16.8, which gives 84 for the 
 area, as in the first example. 
 
 The same results are obtained by applying the Twenty-First 
 Proposition. Thus, in the first example, the distance of the 
 
 perpendicular from the middle of the base is — = .9, 
 
 and therefore the segments of the base are 8.4, and 6.6» In 
 the second example, the distance of the perpendicular from 
 
 the middle of the base is = 10.5, and consequently 
 
 the segments of the base are 15 and 6. In the last example, 
 the distance of the perpendicular fronf the middle part of the 
 
 21*— 17* 
 base is - — = 76, and the segments of that base are hence 
 
 12.6 and 2.6. — The length of the perpendicular and the area 
 of the triangle are, in each case, therefore, easily deduced 
 from these data. 
 
 M. From the corollary to the last proposition is derived a 
 very simple construction of the problem, " to find a square 
 equivalent to a given rectangle." 
 
 Let ABCD be the given rectangle, 
 of which the side AD is greater than 
 AB. In AB or its production, take 
 AE equal to the half of AD and place 
 it from E to F ; then AF being joined, 
 is the side of the equivalent square. 
 For (II. 23. cor. El.) since the sides '^-^^ 1 / 
 
 AE and EF of the triangle AEF are ' '^4''* 
 
 equal, the square of AF is equivalent '^ 
 
 to the rectangle under twice AE and AB, that is, from the 
 construction, the rectangle under AD and AB. 
 
 The same construction might likewise be deduced from 
 the demonstration of the celebrated property of the right- 
 angled triangle. For, in the figure of page 52, suppose BO 
 were drawn to the hypotenuse AC, making an angle ABO 
 
314' NOTES AND ILLUSTRATIONS. 
 
 equal to BAO or BAG ; since the two acute angles are to«. 
 gether equal to a right angle, the angle BCA is equal to the 
 remaining portion CBO of the right angle at B, and conse- 
 quently the triangles AOB and COB are isosceles, and the 
 sides OA, OB and OC all equal. Wherefore AB, the side of 
 a square equivalent to the rectangle ADMN or that under 
 AK and AN, is determined by making AO equal to the half of 
 AK or AC and inserting it from O to B. — The inspection of 
 the same figure also points put the mode of dissecting the 
 rectangle, and thence compounding the square ; for a perpen- 
 dicular let fall from K on AB is evidently equal to GB or AB. 
 Hence, on AF, in the original construction, let fall the per- 
 pendicular DG, transpose the triangle FBA in the situation 
 DHI, and slide the quadrilateral portion into the place of 
 KAHI ; the rectangle ABCD is now transformed into the 
 square KGDI. — A slight modification will be required when 
 AB is less than the half of AD. 
 
 In this construction of the problem, the application of the 
 circle which (III. 37. EL) is indispensably required, is only 
 not brought into view — When the side AD is double of AB, 
 the point G coincides with F, and the rectangle is resolved 
 into three triangles, which combine to form a square. 
 
 15. To this Book some neat propositions may be subjoined. 
 
 PROP. I. THEOR. 
 
 If, from the hypotenuse of a right-angled triangle, portions be 
 cut off equal to the adjacent sides ; the square of the middle seg- 
 7nent thus formed, is equivalent to twice the rectangle contairied 
 hy the extreme segments* 
 
 Let ABC be a triangle which is right-angled at B ; from 
 the hypotenuse AC, cut off AE equal to AB, and CD equal 
 
NOTES AND ILLUSTRATIONS. 315 
 
 to CB : Twice the rectangle under AD and CE is equivalent 
 to the square of DE. 
 
 For the straight line AC being divided 
 into three portions, the squares of AE and 
 
 CD, together with twice the rectangle AD, 
 CE are equivalent to the squares of AC j^ jO E~C 
 and DE (art. 10.). But the squares of 
 
 AB and BC, or those of AE and CD, are equivalent to the 
 square of AC (II. 10.). There consequently remains twice 
 the rectangle AD, CE equivalent to the square of DE. 
 
 By an inverse process of reasoning it will appear, that if 
 twice the rectangle AD, CE be equal to the square of DE, 
 the straight line AC, so composed, is the hypotenuse of a 
 right-angled triangle, of which AB and BC are the sides. 
 
 This proposition will furnish another convenient method of 
 discovering the numbers which represent the sides of a right- 
 angled triangle : For since DE^=:2AD.CE, it is evident that 
 iDE*=:AD.CE ; and consequently, expressing DE by an even 
 whole number, and resolving ^DE* into the factors AD and 
 
 CE, AD-f-DE and CE + DE will represent the two sides, and 
 AD + CE-f DE the hypotenuse. Thus, if 2 be taken, the fac- 
 tors of half its square are 1 and 2, which produce the numbers 
 3, 4, and 5. Again, if 4 be assumed, the factors are 2 and 4, 
 or 1 and 8 ; whence result these numbers, 6, 8, and 10, or 5, 
 12, and 13. In this way, a very great variety of numbers can 
 be found, to express the sides of a right-angled triangle. 
 
 PROP. II. THEOR. 
 
 The squares of lines draiunjrom any point to the opposite cor- 
 ners of a rectangle are together equivalent, 
 
 Iffromapoint E, either within or without the rectangle 
 ABCD, straight lines be drawn to the four corners, the 
 squares of AE, EC are together equivalent to the squares of 
 BE, ED, 
 
31^ 
 
 NOTES AND ILLUSTRATIONS. 
 
 For join E with F, the intersection of the diagonals AC, 
 BD. Because it follows readily from Prop. 27- Book I. that these 
 diagonals are equal, and bisect each 
 other, the lines AF, BF, CF, and 
 DF are all equal. Wherefore the 
 squares of AE, EC are equivalent 
 to twice the square of AF, and 
 twice the square of EF (II. 22.); " 
 and the squares of BE, ED are like- 
 wise equivalent to twice the square 
 of BF and twice the same square 
 of EF ; consequently, the squares of 
 AF and BF being equal, the squares 
 of AE, EC, are together equivalent 
 to the squares of BE, ED. 
 
 PROP. HI. THEOR. 
 
 If straight lines be draxionjrom the angular points of a triangle 
 to Used thb opposite sides, thrice the squares of these sides are 
 together equivalent to Jour times the squares of the bisecting 
 lines. 
 
 Let the sides of the triangle ABC be bisected in D, E, and 
 F, and straight lines drawn from these points to the opposite 
 vertices ; thrice the squares of the sides AB, BC, and AC 
 are together equivalent to four times the squares of BD, CE 
 and AF. 
 
 For, by Proposition II. 22. the squares of AB, BC are equi- 
 valent to twice the square of BD and 
 twice the square of AD, that is, half the 
 square of AC ; the squares of BC, AC 
 are equivalent to twice the squares of 
 CE and half the square of AB ; and the 
 squares of AC, AB are equivalent to 
 twice the square of AF and half the 
 square of BC. Whence the squares of the sides of the tri- 
 
NOTES AND ILLUSTRATIONS. 317 
 
 angle, repeated twice, are equivalent to twice the squares of 
 BD, CE, and AF, with half the squares of the sides of the tri- 
 angle. Consequently four times the squares of AB, BC, and 
 AC are equivalent to four times the squares of BD, CE, and 
 AF, with once the squares of AB, BC, and AC ; wlierefore 
 thrice the squares of the sides AB, BC, and AC are together 
 equivalent to four times the squares of the bisecting lines BD, 
 CE, and AF. 
 
 PROP. IV. THEOR. 
 
 The squares of the sides of a quadrilateral figure are together 
 equivalent to the squares of its diagonals ^ together tmth Jour times 
 the square of the straight line joining their middle points. 
 
 Let ABCD be a quadrilateral figure, in which the straight 
 lines AC, BD, drawn to the opposite corners, are bisected at 
 the points E, F ; the squares of AB, BC, CD, and DE, are 
 together equivalent to the squares of AC, BD, togjBther with 
 four times the square of EF. 
 
 For join EF. And because AC is bisected in F, the squares 
 of AB and BC are equivalent to twice the square of AF and 
 twice the square of BF'(II. 22.) ; .^ 
 
 and, for the same reason, the squares 
 of CD and DA are equivalent to 
 twice the square of AF and twice 
 the square of DF. Consequently 
 the squares of all the sides AB, BC, 
 CD, and DA, are equivalent to four 
 times the square of AF— or the 
 square of AC — with twice the squares of BF and of DF. But 
 twice these squares of BF and DF is equivalent (II. 22.) to 
 four times the square of BE, or the square of BD, with four 
 times the square of EF; whence the squares of all the sides 
 of the quadrilateral figure are together equivalent to the 
 squares of its diagonals AC, BD, with four times the square 
 of the straight line EF which joins their points of equal sec- 
 tion. 
 
818 NOTES AND ILLUSTRATION*. 
 
 This general theorem seems to have been first given by 
 the illustrious Leonard Euler in the Petersburg Memoirs. It 
 evidently comprehends the twenty-fourth Proposition of 
 this Book ; for when the quadrilateral figure becomes a 
 rhomboid, the diagonals bisect each other, and the middle 
 points E and F coincide ; whence the squares of all the sides 
 are equivalent simply to the squares of those diagonals. — If 
 this rhomboid again becomes a rectangle, it will have equal 
 diagonals, and consequently, as in the 10th Proposition of the 
 Second Book, the squares of the sides of a right-angled tri- 
 angle are equivalent to the square of the hypotenuse. 
 
 BOOK III. 
 
 1. Proposition fifteenth. Hence angles are sometimes mea- 
 sured by a circular instrument, from a point in the circum- 
 ference, as well as from the centre. 
 
 2. Proposition eighteenth. On this proposition depends the 
 construction of amphitheatres ; for the visual magnitude of an 
 object is measured by the angle which it subtends at the eye, 
 and consequently the whole extent of the stage, the interme- 
 diate objects being purposely darkened or obscured, will be 
 seen with equal advantage by every spectator seated in the 
 same arc of a circle. 
 
 3. Proposition twenty-second. To erect a perpendicular, any 
 point D is taken, as in Prop. S^. Book T,, 
 and from it a circle is described passing 
 through C and B ; the diameter CDF, by d. 
 its intersection at the point B, determines 
 the position of the perpendicular BF. To r— 
 let Jail a perpendicular, draw to AB any 
 
 straight line FC, which bisect in D, and from this poiirt as a 
 
NOTES AND ILLUSTRATIONS, 
 
 319 
 
 centre describe a circle through the points C, B and F ; FJB 
 is the perpendicular required. 
 
 4-. To this Book may be subjoined some useful propositions. 
 
 PROP. I. THE OR. 
 
 The inclination of tiuo straight lines is equal to the angle tcT' 
 minated at the circumference by the sum or difference of the arcs 
 which they intercept^ according as their vertex is xoithin or xvith' 
 out the circle. 
 
 If the two straight lines AB and CD intersect each other 
 in the point E within a circle ; the angle AED which they 
 form, is equal to an angle at the circumference and standing 
 on the sum of the intercepted arcs AD and BC. 
 
 For draw the chord BF parallel to CD. Because ED and 
 BF are parallel, the angle AED (I.' 22.) 
 is equal to the interior angle ABF, 
 which stands on the arc AF ; but since 
 the chords BF and CD are parallel, the 
 arc BC is equal to DF (III. 1 8.) and 
 consequently the arc AF, which termi- 
 nates at the circumference an angle 
 equal to AED, is the sum of the two 
 intercepted arcs AD and BC. 
 
 Again, if the straight lines AB and CD meet at E, without 
 the circle, their inclination AED is e- 
 qual to an angle at the circumference, 
 having for its base the excess of the 
 arc AD above BC^ 
 
 For BF being drawn parallel to CD, 
 the arc BC is equal to FD, and conse- 
 quently the arc AF is the excess of 
 AD above BC ; but the angle ABF 
 which stands on AF, is equal to the 
 interior angle AED. 
 
320 NOTES AND ILLUSTRATIONS. 
 
 Cor, Hence if two chords intersect each other at right 
 angles within a circle, the opposite intercepted arcs are equal 
 to the semicircumference. 
 
 This proposition is of some utility in practice, for an angle 
 may be hence measured by help of a circular protractor, with- 
 out the trouble of applying the centre to its vertex or the 
 point of concourse of the sides. The same principle is likewise 
 applicable to the construction of some optical instruments, 
 adapted to measure lateral angles by the intersection of micro- 
 meter wires. 
 
 PROP. II. THEOR- 
 
 If a circle he described on the radius of another circle, any 
 straight line dratvn from the point where they meet to the outer 
 circumference, is bisected hy the interior one, ^ 
 
 Let AEC be a circle described on the radius AC of the 
 circle ADB, and AD a straight line 
 drawn from A to terminate in the ex- 
 terior circumference ; the part AE in 
 the smaller circle is equal to the part 
 ED intercepted between the two cir- 
 cumferences. 
 
 For join CE. And because AEC is 
 a semicircle, the angle contained in it 
 
 is a right angle (III. 19.) ; consequently the straight line CE, 
 drawn from the centre C, is perpendicular to the chord AD, 
 and therefore (III. 4.) bisects it. 
 
 PROP. III. THEOR. 
 
 If, on each side of any point in the circumference of a circle^ 
 equal arcs be repeated; the chords which join the opposite points 
 of section will be together equal to the last chord extended till it 
 meets a straight line drawn through the middle point a?id either 
 extremity qfthefrst chord. 
 
NOTES AND ILLUSTRATIONS. 321 
 
 Let BAG be the circumference of a circle, in which the 
 arcs AB, BC, CD on the one side of a point A, and the cor- 
 responding arcs AE, EF, FG on the other side, are all assum- 
 ed equal ; the chords BE, CF, and DG, are together equal to 
 the line GH, formed by extending GD till it meets the pro- 
 duction of AB. 
 
 For join FD and CE, and produce this to meet GH in the 
 point I. 
 
 Because the arcs BC 
 and CD are equal to 
 EF and FG, the chords 
 BE, CF, and DG are 
 parallel ; but, for the 
 same reason, since the 
 arcs BC and CD are e- 
 
 qual to AE and EF, the chords BA, CE and DF are likewise 
 parallel. Hence the figures HBEI and ICFD are rhomboids, 
 and therefore the extended chord GH, being composed of the 
 segments HI, ID, and DG, is equal to the sum of their op- 
 posite chords BE, CF and DG. — It is obvious that the same 
 train of reasoning may be pursued to any number of equal 
 arcs. 
 
 PROP. IV. THEOR. 
 
 Ifjrom any 'point in the diameter of a circle or its extension, 
 straight lines be drawn to the ends of a parallel chord / the. 
 squares of these lines are together equivalent to thi squares of the 
 segments into xuhich the diameter is divided^ 
 
 Let BEFD be a circle, and from the point A in its extend- 
 ed diameter the straight lines AE and AF be drawn to the 
 ends of the parallel chord EF . the squares of AE and AF 
 are together equivalent to the squares of AB and AD. 
 
 For, from the centre C, let fall the perpendicular CG upon 
 EF (I. 6.), and join AG and CE. 
 
 Y 
 
322 
 
 NOTES AND ILLUSTRATIONlS. 
 
 Because CG cuts the chord EF 
 at right angles, GE is equal to GF 
 (ill. 4.) ; wherefore the squares of 
 AE and AF are equivalent to twice 
 the squares of AG and GE (11. 22.) 
 But ACG being a right-angled tri- 
 angle, the square of AG is equiva- 
 lent to the squares of AC and CG 
 (II. 10.), or twice the square of AG 
 is equivalent to twice the squares of 
 AC and CG. Wherefore the squares 
 of AE and AF are equivalent to 
 twice the three squares of AC, CG, 
 and GE. Of these, the two squares 
 of CG and GE are equivalent to the square of CE or CB, for 
 the triangle CGE is right-angled. Consequently the squares 
 of AE and AF are equivalent to twice the squares of AC and 
 CB. But the straight line BD being cut equally at C and un- 
 equally at A, the squares of the unequal segments AB and 
 AD are together equivalent to twice the squares of AC and 
 CB (II. 18. cor.); whence the squares of AE and AF are toge- 
 ther equivalent to the squares of AB and AD. 
 
 u\-I5 
 
 PROP. V. THEOR. 
 
 The rectangle under the segments of a chord is greater or less 
 than the rectangle under the segments into tvhich a perpendicular 
 from the point &f section divides a diameter, by the square of that 
 perpendicular — according as it lies without or xiithin the circle. 
 
 Let the perpendicular CF be let fall from a point C in the 
 chord ACB upon a diameter DE ; the rectangle BC, CA, is 
 greater or less than the rectangle EF, FD, by the square of 
 the perpendicular CF, according as this lies without or with- 
 in the circle. 
 
 First, let the perpendicular CF lie without the circle, and 
 join CE and DG. 
 
NOTES AND ILLUSTRATIONS^ 
 
 S23 
 
 The square of the hypotenuse 
 CE is equivalent to the squared 
 of FE and CF (II. 10.). But 
 the square of CE is composed of 
 the rectangles CE, EG, and CE, 
 CG (II. 14.) ; and the square of 
 FE is composed of the rectangles 
 FE, ED, and FE, FD : Where- 
 fore the rectangles CE, EG and 
 CE, CG are equivalent to the rectangles FE, ED and FE, 
 
 FD, together with the square of CF. And since EGD, stand- 
 ing in a semicircle, is a right angle (III. 19.), its adjacent 
 angle CGD is also right, and the angle opposite to this at F 
 is right; consequently (III. 17. cor. 1.) a circle might be de- 
 scribed through the four points C, G, D, F. Whence (III. 26.) 
 the rectangle CE, EG is equivalent to FE, ED : and taking 
 these from the terras of the former equality, there remains the 
 rectangle CE, CG, that is, (III. 26.) AC, CB, equivalent to 
 the rectangle FE, FD, together with the square of CF. 
 
 Next, let the perpendicular CF lie within the circle. 
 
 The same construction being made, 
 the rectangle CE, EG is still equiva- 
 lent to the rectangle FE, ED. But 
 the rectangle CE, EG is (II. 14-.) e- 
 quivalent to the rectangle CE, CG, 
 and the square of CE, or the squares 
 of FE and CF; and the rectangle FE, 
 ED is equivalent to the rectangle FE, 
 FD and the square of FE. From 
 
 these equal quantities, therefore, take away the common 
 square of FE, and there remains the rectangle CE, CG, or 
 AC, CB, with the square of CF, equivalent to the rectangle 
 
 FE, FD. 
 
 Lastly, if the perpendicular CF lie partly without and part- 
 ly within the circle, the Proposition must be slightly modi- 
 fied. 
 
 The former construction being retained; Because the 
 square of CE is equivalent to the squares of CF and 
 
324. 
 
 NOTES AND ILLUSTRATIONS. 
 
 FE, the rectangles CE, EG and CE, 
 CG are together equivalent to the 
 square of CF and the dift'erence be- 
 tween the rectangle FE, ED and FE, 
 FD ; but the rectangle CE, EG is e- 
 quivalent to the rectangle FE, ED, 
 and consequently the rectangle CE, 
 CG, or the rectangle AC, CB, is 
 equivalent to the difference between 
 the square of CF and the rectangle 
 FE, FD. 
 
 In the first case, if the square of FH be equivalent to the 
 rectangle FD, FE, the square of CH will be likewise equiva- 
 lent to the rectangle CG, CE ; for the rectangle AC, CB, be- 
 ing equivalent to the rectangle FD, FE, or the square of FH, 
 together with the square of CF, must (II. 10. El.) be equiva- 
 lent to the square of CH. 
 
 PROP. VI. THEOR. 
 
 A straight line draxunjrom the vertex of a triangle through the 
 intersection of two perpendiculars Jrom the extremities of the base 
 to the opposite sides, is likemse perpendiculiir to the base. 
 
 In the triangle ABC, the straight line BFG drawn from the 
 vertex B through F, the intersec- 
 tion of the perpendiculars AE and 
 CD from A and C upon the oppa- 
 site sides CB and ABis perpendi- 
 cular to the base AC. 
 
 For join DE. Because BDF and 
 BEF are right angles, the quadrila- 
 teral figure ADEC (III. 17. cor. 1.) is 
 contained in a circle; and for the 
 
 same reason, the quadrilateral ADEC is contained in a circle. 
 Wherefore the exterior angle BDE (III. 17. cor. 2.) is equal to 
 ACE ; but ( III. 1 6. ) BDE is equal to the angle BEE in the same 
 Segment, which is therefore equal to ACE or GCE, and con- 
 
NOTES AND ILLUSTRATIONS. 
 
 325 
 
 sequently the quadrilateral CEFG is also contained in a cir- 
 cle. Whence (III. 17.) the opposite angles CEF and CGF 
 are equal to two right angles, and CEF being a right angle 
 by hypothesis, EGF must likewise be right ; or the straight 
 Ixne BFG is perpendicular to the base AC. 
 
 PROP. VII. PROB. 
 
 Through a given point , hetvueen two diverging straight lines^ to 
 dratu a straight line that shall have equal segments terminated by 
 them. 
 
 Let AB and AC be two diverging straight lines given in 
 a position, and F an intermediate point, through which it is re* 
 quired to draw GFH, such that the intercepted segments FG 
 and FH shall be equal. 
 
 This may be easily effected, by drawing a parallel from F to 
 AB, and doubling the portion so cut off, from A to G, to mark 
 the position of GFH. But the problem may be constructed in 
 another way, which, though more complex, is important in its 
 application to the Theory of Lines of the Second Order. 
 
 Draw AD bisecting the angle BAC, and upon it let fall the 
 perpendicular FE, which pro- 
 duce both ways to B and C ; 
 from B erect BD perpendicu- 
 lar to AB, join DF ; and EFH, 
 being drawn perpendicular to 
 it, is the line required*- 
 
 For join DC, DGand DH. 
 The right-angled triangles 
 ABD and ACD are (I. 20.) 
 equal, and consequently BDC 
 is isosceles. But GBD and 
 
 GFD being right angles, and therefore equal, the quadrilateral 
 figure GB, FD (III. 16.) is contained in a circle, and hence 
 the angle DGF is equal to DBF ; for the same reason, since 
 DCH and DFH are right angles, the quadrilateral figure 
 PCHF is likewise contained in a circle, and hence the angle 
 
326 NOTES AND ILLUSTRATIONS. 
 
 DHF is equal to DCF. Consequently the angle DGF is equal 
 to DHF, and the right-angled triangles DFG and DFH are 
 equal, and the base FG equal to FH. 
 
 If the point F were taken in the extension of the line EB, 
 the perpendicular to DF may then be shown to have equal 
 segments intercepted by the sides of the exterior angle form- 
 ed by AG and the production of C A beyond the vertical point 
 A. 
 
 BOOK IV. 
 
 r. The equilateral triangle, the square, the pentagon, tho 
 hexagon, and other polygons derived from these, were the 
 only regular figures known to the Greeks. The inscription 
 of all the rest has for ages been supposed absolutely to trans- 
 cend the powers of elementary geometry. But a carious 
 and most unexpected discovery was lately made by Mr 
 Gauss, now Professor in the University of Gottingen, who 
 has demonstrated, in a work entitled Disquisitiones Arith- 
 meticce, and published at Brunswick in 1801, that certain very 
 complex polygons can yet be described merely by help of 
 circles. Thus, a regular polygon containing 17, 257, 65537, 
 &c. sides, is capable of being inscribed, by the application of 
 elementary geometry ; and in general, when the number of 
 sides may be denoted by 2"+ 1, and is at the same time a 
 prime number. The investigation of this principle is rathei: 
 intricate, being founded on the arithmetic of sines and the 
 theory of equations ; and the constructions to which it would 
 lead are hence, in every case, unavoidably and most excessive- 
 ly complicated. Thus the cosine of the several arcs arising 
 from the division of the circumference of a circle into seven- 
 teen equal parts, are all contained in this very involved ex- 
 pression : 
 
 l-ZCn+Sv"^?— V'(34-.2-/17)— 2^/(34.t.2V17)) 
 
NOTES AND ILLUSTRATIONS. 327 
 
 As the radicals may be taken either positive or negative, their 
 various combinations, rightly disposed, will produce eight dis- 
 tinct results. 
 
 Let 5r denote the circumference ; then 
 
 cos — = oos^ == .9324722294, cos ±L=cos^=:' 
 17 17 17 17 
 
 .7390089172, cos — = cos — =.H51SS3558 cos — = 
 17 17 17 
 
 eos ~ = .0922683595 cos ^ = cos ?^=— .27366229901, 
 cos ^^z=cos^^z=^ .6026346364, cos l^=::cos^=: 
 — .8502171357, and cos 1^ =r cos ^ = — .9829730997. 
 
 2. Pythagoras was the first who remarked tlie simple pro- 
 perty, that only three regular figures, — the square, the equi- 
 lateral triangle, and the hexagon, — can be constituted about 
 a point. Here the mystic philosopher might again admire 
 the union of the monad with the triad. — It may not be super- 
 fluous perhaps to observe, that on this property is founded 
 the adaptation of patchwork, and the construction of tessellat- 
 ed pavement. 
 
 S. Several interesting propositions may be annexed to this 
 Book. 
 
 PROP. I. THEOR. 
 
 The square of the side of a regtdar octagon inscribed in a cir- 
 clCf is equivalent to the rectangle contained by the radius and the 
 difference between the diameter and the side of the inscribed 
 square* 
 
 Let ABCD be a square inscribed in a circle, and 
 
328 
 
 NOTES AND ILLUSTRATIONS. 
 
 AEBFCGDH an octagon, which is formed evidently by the 
 bisection of the quadrants AB, BC, CD, and DA : The square 
 of AE is equivalent to the rectangle under AO and the dif- 
 ference between AB and AC. 
 
 For draw the diameter EG. It 
 is manifest, that the triangles 
 AIO and BIO are right-angled 
 and isosceles ; and because AO 
 is equal to EO, and AI perpendi- 
 cular to it, — the square of AE 
 (IF. 23. cor. E|.) is equivalent to 
 twice the rectangle under EO 
 and EI, or the rectangle under 
 AO and twice EI. But EI is 
 the difference of EO and 10, and 
 
 twice EI is, therefore, "equal to the difference of twice EO or 
 AC and twice 10 or AB. Whence the square of AE, the 
 side of the octagon, is equivalent to the rectangle under the 
 radius and the difference of the diameter and AB the side of 
 the inscribed square. 
 
 PROP. II. THEOR. 
 
 In and about a given circle, to inscribe and circumscribe an 
 equilateral triangle. 
 
 Let AEB be a circle, in which it is required to inscribe an 
 isosceles triangle. 
 
 Draw the diameter AB, describe (I. 1.) the equilateral tri- 
 angle ADB, join CD meeting the circumference in E, draw 
 {I. ^23.) EF, EG parallel to AD, BD, and join FG : The tri- 
 angle EFG is equilateral. 
 
 For the triangles ADC, BDC having the two sides DA, AC 
 equal to DB, BC, and the third side DC common to both, are 
 (I. 2.) equal, and the angle DC A is equal to DCB ; whence 
 the arc AE is (III. 12.) equal to BE. And the triangle ADB 
 (I. 10. cor.) being likewise equiangular, the angle DBA is 
 
NOTES AND ILLUSTRATIONS. 
 
 329 
 
 equal to DAB, and the arc AEM equal to BEL, and the re- 
 maining arc ME equal to LE. 
 But EF and EG being parallel 
 to LA and MB, the arcs AF and 
 BG are equal to LE and ME, 
 and to each other ; hence FG is 
 parallel to AB, and the inscribed 
 triangle FEG is (L 29.) equi- 
 angular, and consequently equi- 
 lateral. 
 
 Again, let it be required to 
 describe an equilateral triangle 
 about the circle AEB. 
 
 The same construction remaining ; at the points F, E, and 
 G, apply the tangents HI, HK, and KI, to form the circum- 
 scribing triangle IHK : This triangle is equilateral. 
 
 For because IH is a tangent and FG is inflected from the 
 point of contact, the angle IFG is equal to the angle FEG in 
 the alternate segment (IIL 21.), and therefore IH is parallel 
 to EG (L 22. cor.). In like manner it is proved, that HK, 
 KI are parallel to GF, FE, and consequently (I. 29.) the 
 angles of the triangle IHK are equal to those of FEG, and 
 therefore equal to each other. 
 
 Cor. Hence the circumscribing equilateral triangle con- 
 tains four times that which is inscribed ; for the figures EFIG, 
 EHFG, and EFGK are evidently equal rhombuses, and con- 
 tain equilateral triangles which are all equal. Hence also the 
 side of the circumscribing, is double of that of the inscribed, 
 equilateral triangle. 
 
 PROP. III. THEOll. 
 
 To inscrihe and circumscribe a circle in and about a given re^ 
 gidar pentagon. 
 
 Let ABCDE be a regular pentagon, in which it is required 
 to inscribe a circle. 
 
330 
 
 NOTES AND ILLUSTRATIONS. 
 
 Draw AO and EO to bisect the angles at A and E, let fall 
 the perpendicular OF, and from O as a centre, with the dis- 
 tance OF, describe a circle FGHIK : This circle will touch 
 the pentagon internally. 
 
 For, from the point O, let 
 fall perpendiculars on the op- 
 posite sides of the figure. The 
 angles EAO and AEO, being 
 the halves of the angles of the 
 pentagon, are equal, and con- 
 sequently the triangle AOE is 
 isosceles, and the perpendicu- 
 lar OF bisects the base. And 
 the triangles AOG and BOG, 
 having the angles OAG and 
 OGA equal to OBG and 
 
 0GB and the common side OG, are (1.20.) equal. Again, the 
 triangles BOG and BOH have now the angles OBG and 
 0GB equal to OBH and OHB, with the side BO common to 
 both, and are therefore equal. In like manner, all the tri- 
 angles about the centre O are proved to be equal ; conse- 
 quently the perpendiculars OF, OG, OH, 01, and OK are 
 equal, and the circle touches the pentagon in the points F^ 
 G, H, I, and K. 
 
 Next, let it be required to describe a circle about the pen- 
 tagon. 
 
 From the same centre O, with the distance OA, describe a 
 circle : It will pass through the points B, C, D, E ; for the 
 triangles about O being all equal, the straight lines OA, OB, 
 OC, OD, and OE must be likewise equal. 
 
 PROP. IV: THEOR. 
 
 In and about a regular hexagon to inscribe and circumscribe a 
 circle. 
 
 Let ABCDEF be a regular hexagon, in which it is requi- 
 red to inscribe a circle. 
 
NOTES AND ILLUSTRATIONS. 
 
 331 
 
 Draw AO and FO, bisecting the angles BAF and AFE 
 (1.5.) ; and from the point of intersection O, with its distance 
 from the side AF, describe a circle : This circle will touch 
 the hexagon internally. 
 
 For let fall perpendiculars 
 from O upon the sides of the 
 figure. It may be demonstra- 
 ted, as in the last proposi- ' 
 tion, that the triangles AOB, 
 BOC, COD, DOE, and EOF 
 are all equal to AOF; and, in 
 like manner, it will appear 
 that the intermediate bisect- 
 ed triangles are equal. Hence 
 the perpendiculars OG, OH, 
 
 OI, OK, OL, and OM, are all equal, and a circle must touch 
 these at the points, G, H, I, K, L, and M. 
 
 Again, let it be required to describe a circle about the hex- 
 agon. 
 
 From the same point O, as a centre, with the distance 
 OA, describe a circle, which must pass through the points B, 
 C, D, E, and F; for the straight lines O A, OB, OC, OD, 
 OE, and OF were proved to be equal. 
 
 Cor. Hence, in any regular polygon, the centre of the in- 
 scribing an4 circumscribing circle is the same, and may be de- 
 termined in general, by drawing lines to bisect the adjacent 
 angles of the figure. 
 
 BOOK V. 
 
 DEFINITIONS. 
 
 1. The words ;vey«? in Greek and ratio in Latin, signifying 
 reason or manner of thought^ indicate vaguely a philosophical 
 conception. The compound term haXoyisc comes nearer to 
 
332 NOTES AND ILLUSTRATIONS. 
 
 this idea ; but its correlative, proportio, marks very distinctly 
 a radical similarity of composition. 
 
 The doctrine of proportion has been a source of much con- 
 troversy. In their mode of treating that important subject, 
 authors differ widely ; some rejecting the procedure of Euclid 
 as circuitous and embarrassed, while others appear disposed 
 to extol it as one of the happiest and most elaborate monu- 
 ments of human ingenuity. But, to view the matter in its 
 true light, we should endeavour previously to dispel that mist 
 which has so long obscured our vision. The Fifth Book of 
 Euclid, in its original form, is not found to answer the pur- 
 pose of actual instruction ; and this remarkable and indisputed 
 fact might alone excite a suspicion of its intrinsic excellence. 
 The great object which the framer of the Elements had propo- 
 sed to himself, by adopting such an artificial definition of pro- 
 portion, was to obviate the difficulties arising from the consi- 
 deration of incommensurable quantities. Under the shelter of 
 a certain indefinitude of principle, he lias contrived rather to 
 evade those difficulties than fairly to meet them. Euclid seems 
 not indeed to grasp the subject with a steady and comprehen- 
 sive hold. In his Seventh Book, which treats of the properties 
 of number, he abandons his former definition of proportion, 
 for another that is more natural, though imperfectly developed. 
 Through the whole contexture of the Elements, we may dis- 
 cern the influence of that mj'sticism which prevailed in the 
 Platonic school. The language sometimes used in the Fifth 
 Book would imply, that ratios are not mere conceptions of the 
 mind, but have a real and substantial essence. 
 
 The obscurity that confessedly pervades the fifth book of 
 Euclid being thus occasioned solely by the attempt to extend 
 the definition of proportion to the case of incommensurables, 
 the theory of which is contained in his tenth book — the perti- 
 nacity of modern editors of the Elements in retaining such an 
 intricate definition, appears the more singular, since, omitting 
 all the books relating to the properties of numbers, they have 
 not given the slightest intimation respecting even the existence 
 of incommensurable quantities. 
 
 The notion of proportionality involves in it necessarily the 
 
NOTES AND ILLUSTRATIONS. 335 
 
 idea of number. The doctrine of proportion hence constitutes 
 a branch of universal arithmetic ; and had I not, on this occa- 
 sion, yielded to the prevalence of custom, I should, after the 
 example of M. Legendre, have rejected it from the Elements 
 of Geometry, and deferred the consideration of the subject till 
 I came to treat of Algebra, where it is sometimes indeed given, 
 but in a very contracted and insufficient form. The properties 
 themselves are extremely simple, and may be regarded as only 
 the exposition of the same principle under different aspects. 
 The various transformations of which analogies are susceptible, 
 resemble exactly the changes usually effected in the reduction 
 of equations. 
 
 According to Euclid, " The first of four magnitudes is said 
 to have the same ratio to the second which the third has to the 
 fourth, when any equimultiples whatsoever of the first and third 
 being taken, and any equimultiples whatsoever of the second 
 and fourth ; if the multiple of the first be less than that of the 
 second, the multiple of the third is also less than that of the 
 fourth; or, if the multiple of the first be equal to that of the 
 second, the multiple of the third is also equal to that of the 
 fourth ; or, if the multiple of the first be greater than that of 
 the second, the multiple of the third is also greater than that 
 of the fourth." This definition, however perplexed and ver- 
 bose, is yet easily derived from that which appears to furnish 
 the simplest and most natural criterion of proportionality ; 
 For, let A : B : : C : D ; it was stated as a fundamental prin- 
 ciple, that, if the mth part of A be contained n times in B, the 
 With part of C will likewise be contained n times in D. Whence 
 ?jA=?nB, and wC=wD ; which is the basis of Euclid's defini- 
 tion. But when the terms are incommensurable, such equali- 
 ty cannot absolutely subsist. In this case, no single trial would 
 be sufficient for ascertaining proportionality. It is required 
 that, every multiple whatever, twA, being greater or less than 
 ?iB, — the corresponding multiple, ?wC, shall likewise be con- 
 stantly greater or less than nD. Actually to apply the definition 
 is therefore impossible ; nor does it even assist us at all in di- 
 recting our search. In the natural mode of proceeding, by as- 
 suming successively a smaller divisor, we are, at each time, 
 brought nearer to th^ incommensurable limit* But Euclid's 
 
334 NOTES AND ILLUSTRATIONS. 
 
 famous definition leaves us to grope at random after its object, 
 and to seek our escape, by having recourse to some auxiliary 
 train of reasoning or induction. 
 
 The author of the Elements has likewise given what Dp 
 
 ,.^]^arrow calls a metaphysical definition of ratio : *' Ratio is a 
 mutual relation of two magnitudes of the same kind to one an- 
 other, in respect of quantity^ This sentence, as it now 
 stands, appears either tautological, or altogether devoid of 
 meaning; and Dr Simson, anxious for the credit of Euclid, 
 considers it, in his usual manner, as the interpolation of some 
 
 Omskilful editor. I am inclined to think, however, that the 
 passage will admit of a version which is not only intelligible, 
 but conveys a most correct idea of the nature of ratio. The 
 
 original runs thus : Aoyo^ sa-n ^vo yAyzSm o[Aoy%vm 4 x^lss Unhixol-^ei 
 
 TTgej »h.MKec Tirtnec, iryjitrni, Now the term ci-jjAix*?, on which the 
 whole evidence hinges, though commonly rendered quantus, 
 may be translated quotus^ as expressing either magnitude or 
 multitude. In its primitive sense, it probably denoted ivamher, 
 and came afterwards to signify quantity ^ as this word itself has, 
 in the French language, undergone the reverse process, \w 
 confirmation of this opinion, it maybe stated, that the relative 
 term «A<x<fl: properly denotes age^ and thence stature or size. 
 According to this interpretation, therefore, ** Ratio is a cer- 
 tain mutual habitude of two homogeneous magnitudes with re- 
 spect to quotityt or numerical composition." 
 
 It is very vmfortunate that, from the poverty of language, 
 and the slow progress of science, the terms used in com- 
 >^mon life, though unavoidably deficient in precision, were 
 V. adopted into Geometry. But the vagueness of expression is 
 nowhere more apparent than in what concerns Proportion.—. 
 Thus, the words denoting time are, in most dialects, blended 
 with those which signify number. To express how often a part 
 is contained In a whole, we intimate how many xmys It is to be 
 placed, how many /hidings are required, or how many ti7nes the 
 operation of admeasurement must be repeated. In the Greek 
 and Latin languages, the adverbs compounded from plica, a 
 fold, are very extensive. In English, the corresponding terms 
 are limited, and mark too obviously their composition : for 
 
NOTES AND ILLUSTRATIONS. 355 
 
 duplex^ tripleXf qiiadruplex, we have double, triple or quadruple^ 
 ivoqfold, threefold or fourfold. But our application of the 
 word ijoay is still more confined : we have only tvoice and thricci 
 or tiao laays and three ways. When we seek to go farther, we 
 are absolutely obliged to borrow the word time; thus, we say 
 that one number is four or five times greater than another ; or 
 that it would require the addition of the part so often, to form 
 the whole. The German language involves the same idea 
 without bringing it so prominently forward ; the termination 
 mal, the same originally with our word meal, referring to the 
 regular succession of the hours of refreshment. The French 
 is in this instance more happy, the term fois, derived from voye, 
 in the Latin and Italian via, a ivay, having been abridged from 
 ioutevoye or always, and converted into a general adverb. 
 
 2. Proposition fourteenth. This proposition is easily de- 
 rived from geometry ; for, since of 
 
 proportional lines the rectangle un- 
 der the extremes is equal to that of 
 the means, the segments AG and 
 AH of the diameter in the figure 
 are (III. 7- El.) the greatest and 
 least terms of an analogy, of which 
 AB and AD are the intermediate 
 terms, and consequently (111.6. El.) the diameter GH, or the 
 sum of AG and AH, is greater than the chord BD, or the sum 
 of AB and AD. 
 
 3. PS'oposition twenty-seventh. The numerical expression 
 of the ratio A : B, may be deduced indirectly, from the series 
 of quotients obtained in the operation for discovering their 
 common measure. 
 
 Let A contain B, m times, with a remainder C ; B contain 
 
 C, n times, with a remainder D ; and, lastly, suppose C to con- 
 tain D, p times, with a remainder E, and which is contained in 
 
 D, ^ times exactly. Then DrryE, Czr^^D+E, BrrwC-j-D, 
 and A=w2B-i-C ; whence the terms D, C, B, and A, are suc- 
 cessively computed; as multiples of E ; A and B will, there- 
 
336 NOTES AND ILLUSTRATIONS. 
 
 fore, be found to contain E their common measure K and L 
 times, or the numerical expression for the ratio of those quan- 
 tities is K : L. 
 
 It is more convenient, however, to derive the numerical ra- 
 tio, from the quotients of subdivision in their natural order ; 
 and this method has besides the peculiar advantage of exhi- 
 biting a succession of elegant approximations. 
 
 The quantities A, B, C, D, &c. are determined, as before, 
 by these conditions: ArrmB+C, B=wC+D, C=pD+E, 
 D=yE-f-F, &c. But other expressions will arise from substi- 
 tution: For, 
 
 1. A=mB-t-C=?w(«C4-D)-fC=r(w?«+l)C4.wD, or, put- 
 ting ?w.«-}-l=?w', Azzm'C^-wD. 
 
 2. A=m'C4-w«D=w2'(pD-f E)4-mD=(m'p+m) D+m'E, 
 or, putting m^p-^m^m", A=:?w"D-j-w'E. 
 
 or, putting ?n^^.q-\-m'=m"', A=zm'"E-\.m'^F, 
 
 Again, the successive values of B are developed in the same 
 manner : 
 
 l.B=«C+D=«(;9D+E)+D=(«;74-l)D+wE, or, putting 
 7i.p^ 1 = n', B= w'D+ wE. 
 
 2.B=n'D+wE=w'(yE4.F)+«E=(w'y4-w)E+?2'F,or,put. 
 ting n'.q-^nz=:n", B=w"E+w'F. 
 
 These results will be more apparent in a tabular form : 
 
 A=w2B+C, I Bz=:nC+D, 
 
 =:m'C4-W7D, 
 
 =w"D+w'E, 
 
 z=:nrE^m"F, 
 
 =«'D+wE, 
 
 =Z7l"E+7l% 
 &C. 
 
 The substitutions are thus arranged : 
 
 7W.W-}-l = 7?i', 
 
 m'.p-{-niz=:m"f 
 m".q-{-m'zzm"'i 
 &c. 
 
 n.p'\-\:=.n', 
 n* .q'\'nz=:7i" y 
 
 Whence, the law of the formation of the successive quanti- 
 ties, is easily perc5eived. 
 
KOTES AND ILLUSTKATIONS. S37 
 
 But, to find the ratio of A to.B, it is not requisite to know 
 the values of the remainders C, D, E, &c. Suppose the sub- 
 division to terminate at B; then A=mB, and consequently 
 A : B, as mB : B, or m : 1, If the subdivision extend to C, 
 then A=7?2'C, and B=nC ; whence A : B, as m' : n. In gene- 
 ral, therefore, the second terra, in the expressions (or A and 
 B, may be rejected, and the letter which precedes it consider- 
 ed as the ultimate measure, and corresponding to the arithme- 
 tical unit. Hence, resuming the substitutions, and combining 
 the whole in one view, it follows, that the ratio of A to B may 
 thus be successively represented ; 
 
 l.m:l, 
 
 2. mw-f- 1 : w, or jw' : n, 
 
 3. m'p^m : np-\- 1, or w" : n'. 
 
 4. m"q-{''m' : n'q-\-nj or vi'" : n", 
 &c. &c. &c. 
 
 The formation of these numbers will evidently stop, when 
 the corresponding subdivision terminates. But even though 
 the successive decomposition should never terminate, as in the 
 case of incommensurable quantities, — yet the expression thus 
 obtained must constantly approach to the ratio of A : B, since 
 they suppose only the omission of the remainder of the last 
 division, and which is perpetually diminishing. 
 
 4. Proposition twenty-pinth. The sanie conclusion is de- 
 rived from the division of surds. Thus — = 1 -f- 
 
 I \/2+l ^. a/2— 1 , . • ,, ^ 
 
 y^ , =" — ^=:2-\-- — -— , and then contmually the ex- 
 pansion of the same residue —.- — - , which therefore gives 2 
 
 as a repeated integral quotient. Hence m being 1 and n, p, 
 y, r, &c. all equal to 2, the successive approximations are, by 
 the last note, 1 : 1, 2 ; 3, 5 : 7, 12 : 17, 29 : 41, 70 ; 99, &c. 
 The ratios of the squares of these numbers are 4 : 9, 25 ; 49, 
 144 : 289, 841 : 1681, 4900 : 9801, thus approaching rapidly 
 to the ratio of one to ttvo, but alternately in excess and de- 
 fect. 
 
 z 
 
ss$ 
 
 NOTIS AND ILLUSTRATIONS. 
 
 BOOK VI. 
 
 ] . Proposition first. The consideration of diverging lines 
 furnishes the simplest and readiest means, for transferring the 
 doctrine of proportion to geometrical figures. The order which 
 Euclid has followed, beginning with parallelograms, and thence 
 passing from surfaces to lines, appears to be less natural. 
 
 2. Proposition fourth. It will be proper here to notice the 
 several methods adopted in practice, for the minute subdivi- 
 sion of lines. The earliest of these — the diagonal scale — de- 
 pending immediately on the proposition in the text, is of the 
 most extensive use, and constituted the first improvement on 
 astronomical instruments. 
 
 Thus, in the figure annexed, the extreme portion of the ho- 
 rizontal line is divided into ten equal parts, each of which 
 again is virtually subdivided into ten secondary parts. The 
 subdivision is efiected by means of diagonal lines, which de- 
 cline from the perpendicular by intervals equal to the primary 
 divisions, and which are cut transversely into ten equal seg- 
 ments by equidistant paral- 
 lels. Suppose, for example, 
 it were required to find the 
 length of 2 and 38—100 parts 
 of a division ; place one foot 
 of the compasses in the se- 
 cond vertical at the eight in- 
 terval which is marked with a dot, and extend the other foot, 
 along the parallel, to the dot on the third diagonal. The 
 distance between these dots may, however, express indiffe- 
 tently 2.38, 23.8, or 238, according to the assumed magnitude 
 of the primary unit. 
 
NOTES AND ILLUSTRATIONS. 3S9 
 
 Nunez, or Nonius, in a Treatise De CrepuscuUs, printed at 
 Lisbon in 154^2, proposed one more complicated. He placed 
 a number of parallel scales, or concentric circles, differently 
 divided, and forming a regular ascending gradation of 89, 88, 
 87, &c. equal parts, from 90 to 46 inclusive. An index laid 
 any where across these scales might, therefore, be presumed 
 to cut at least on.e of them at some of the divisions, and hence 
 the intercepted space would be expressed by a corresponding 
 fraction: 
 
 But the method of subdivision which was afterwards intro- 
 duced by Peter Vernier, a gentleman of Franche Comte, and 
 published by him in a small tract printed at Brussels in 1631, 
 being itself an improvement on the method used in the con- 
 struction of Tycho Brahe*s astronomical instruments, is much 
 simpler and far more ingenious. It is founded on the diffe- 
 rence of two approximating scales, one of which is move- 
 able. Thus, if a space equal to n — i parts on the limb of 
 the instrument be divided into n parts, these evidently will 
 each of them be smaller than the former, by the nth part of a 
 division. Wherefore, on shifting forward this parasite or 
 Vernier scale, the quantity of aberration will diminish at each 
 successive division, till a new coincidence obtains, and then 
 the number of those divisions on that scale will mark the frac- 
 tional value of the displacement. 
 
 Thus in the annexed figure, nine divisions of the primary 
 
 scale, forming ten equal parts in the attached or sliding scale, 
 
 the moveable 
 
 2:ero stands 1 f f ? ffi'l?i ? 
 
 beyond the " ■ Vi '^ ^TH-Vf?T- | ■ J^ i M i | i i i i j^, 
 
 first interval 
 
 between the third and fourth division. To find this minute 
 difference, observe where the opposite sections of the scales 
 come to coincide, which occurs under the fourth division of 
 the sliding scale, and therefore indicates the quaAtity 1.34. 
 
 3. Proposition fifth. This problem could be otherwise sol- 
 ved. Through B draw the inclined straight line CBG extend- 
 ed both ways, in this take any point C, and make BD, DE, 
 
34d 
 
 NOTES AND ILLUSTRATIONS. 
 
 EF, FG, &c. each equal to BC, complete the parallelograoi 
 ABCI, and join ID, IE, IF, IG, &c. cutting AB in the point 
 K, L, M, N, &c. ; then is the segment AK the half of AB, 
 AL the third, AM the fourth, and AN the fifth part of the 
 same given line. 
 
 For the segments of the straight line AB must be propof- 
 tional to the segments of the parallels AI and BG, intercepted 
 by the diverging lines ID, IE, IF, IG, &c. Thus, AK : KB. 
 : : Al : BD; but, by con- 
 struction, BC or AlrrBD, 
 whence (V. 4.) AK=KB, 
 and therefore AK is the 
 half of AB. Again, AL : 
 LB : : AI : BE ; and since 
 BE=2AI, it follows that 
 LB=2AL, or AL is tho 
 third part of AB. In the 
 same manner, AM : MB : : 
 AI : BF; but BF=3AI, 
 whence MB=3AM, or AM 
 is the fourth part of AB. 
 And, by a like process, it may be showrn that AN is the fifth 
 part of AB. 
 
 4. Proposition seventeenth. The solution of this impor- 
 tant problem now inserted in the text, was suggested to me 
 by Mr Thomas Carlyle, an ingenious young mathematician, 
 formerly my pupil. But I here subjoin likewise the original 
 construction given by P»ppus, which, though rather more 
 complex, has yet some peculiar advantages. 
 
 Let AB be a straight line, which it is required to cut, so 
 that the rectangle under its segments shall be equivalent to a 
 given rectangle. 
 
 On AB describe the semicircle AFB, at A and B apply tan- 
 gents AD and BE equal to the sides of the given rectangle, 
 and both in the same or in opposite directions, according as 
 the line is to be cut internally or externally ; join DE, and 
 from tlie point F where it meets the circumference, draw the 
 
NOTJES AND ILLUSTRATIONS. 
 
 34,1 
 
 perpetidiciilar FC ; this will divide the given line AB into A^* 
 and BC, the segments required. 
 
 For the right angle DFC is 
 equal (III. 19.) to the angle 
 AFB contained in the semi- 
 circle, and consequently their 
 difference from AFC or the 
 angles DFA and CFB are 
 equal. For the same rea- 
 son, the angle AFB being 
 likewise equal to CFE, add 
 or take away CFB, and the 
 angle BFE will be equal to 
 AFC. But AD being a tan- 
 gent, and AF a straight line 
 inflected to the circumfe- 
 rence, the exterior angle 
 DAF is equal (III. 21.) to 
 the angle in the alternate seg- 
 ment AF or the angle CBF (III. 17. cor. 2.). Again, BE 
 being a tangent and BF an inflected line, the exterior angle 
 EBF is equal to BAF. Wherefore the triangles DAF and 
 AFC are similar to BFC and BFE ; and hence AD : AF : : 
 CB : BF, and AF : AC : : BF : BE ; consequently (V. 16.) 
 AD : AC : : CB : BE, and (V. 6.) AD.BE=AC.CB. 
 
 Cor. If the sides of the given rectangle be equal, the con- 
 struction of the problem will become materially simplified. 
 
 First, in the case of internal section: The tangents AD 
 BE being equal, it is evident that DE 
 must be parallel to AB and the per- 
 pendicular FC parallel to EB. Whence, 
 employing this construction, or erect- 
 ing the perpendicular BE equal to the 
 sides of the given square, and drawing 
 the parallel EF to meet the circumference F, from which is 
 let fall on AB the perpendicular FC, the rectangle under the 
 segments AC and CB is equivalent to the square of BE ; which 
 also follows from Prop. 26. cor. 1. Book III, 
 
 A- 
 
 \- 
 
 ^\. 
 
 CK 
 
312 
 
 NOTES AND ILLUSTRATION S. 
 
 Next, in the case of external section ; The opposite tan. 
 gents AD, BE being equal, 
 the triangles AGD and BGE 
 are evidently equal, and 
 therefore DE passes through 
 the centre. Hence the tri- 
 angles BGE and FGC are 
 also equal, and GC equal to 
 GE. The modified construc- 
 tion is therefore to erect the 
 
 perpendicular BE equal to the side of the given square, join 
 GE, and where this cuts the circumference apply the tangent 
 FC to meet AB produced : Then AC and CB are the required 
 external segments of the 'given line AB. For it is evident that 
 the rectangle AC, CB will be equal to the square of BE ; 
 which is also deduced from Prop, 26. cor. 2. Book III., since 
 CF is now a tangent and AC.CB=CF* or BE». 
 
 If AB be equal to BE, the construction will exactly corre- 
 spond with what was before given. 
 
 In applying this problem to the construction of quadratic 
 equations, it is necessary previously to ascertain the precise 
 import of the ordinary signs used in Algebra, when extended to 
 geometrical quantities The signs -f- and — intimate, in general, 
 nothing more than that the number, or the magnitude express- 
 ed by number, to which they are respectively prefixed, is to 
 be added to, or taken away from, any other number, with which 
 it comes to be combined. It would be more correct language, 
 therefore, to call the quantities carrying such signs additive 
 and subtractive, implying merely a casual and mutable rela- 
 tion ; instead of the usual appellations of positive and negative, 
 which seem to bestow a distinct and absolute character, and 
 have hence led incautious reasoners into mystery and paradox. 
 A similar degree of reserve is indispensable in Geometry. 
 Following the European mode of writing from left to right, we 
 might fancy it almost natural to draw a line in the same direction : 
 When we want to extend a line, we apply an additional line to 
 the right; but when we seek to contract it. we retrace a deft- 
 
NOTES AND ILLUSTRATIONS. 
 
 843 
 
 cient line to the left. Thus, if NO be annexed to the right of 
 
 MN, there results f__ l_ I i 
 
 MO; or if NO' be M o' N O 
 
 taken to the left of the extremity N, there will remain MO'c 
 The position of NO or NO^ to the right or left, will, therefore, 
 in reference to a combination with any line MN, have the 
 same effect as the signs of addition or subtraction produce in 
 Algebra. Following out the same analogy, while lines drawn 
 upwards may correspond to additive quantities, lines drawn 
 downwards must express subtractive quantities. 
 Quadratic equations are reducible to these four forms : 
 
 1. x^ •}- ax=z •{• be 
 
 2. x^-—ax=z ^ be 
 
 3. a:* 4" ^^ = — ^^ 
 
 4. X* — ax = — be. 
 
 The two first may be constructed from the second case of 
 Proposition seventeenth ; and the two last will receive their 
 construction from the first case of that problem. We shall 
 resume the equations in their order : 
 
 1. x^^ax= +bc, then j:= — -|db:i/^+ 3c, these beingt 
 
 2 '4 
 
 roots, the greater subtractive, and the less additive. 
 
 Employing the construction of the second case of the pro 
 blem, let AB = a, AD = b, 
 and BE = — c, since it 
 stretches below AB ; if BC 
 represent — x, then CA, in 
 the reverse position, will 
 be denoted by — a — x. 
 Wherefore BC x CA = 
 ( — a — x)x=r — ax — x\ 
 and consequently AD.BE 
 = — be = — ax — x*, or, by 
 inversion, a:' + ar = -j- be. 
 The roots' are, consequent- 
 ly, the shorter segment BC which is additive, and the longer 
 segment BC which is subtractive. 
 
 wo 
 
3i4' NOTES AND ILLtSTRATIONS; 
 
 2. x'^ax=-\-bc, then j:=+ 1=1=^-1 +^c; there being now 
 
 likewise two roots, but the greater additive, and the less sub- 
 tractivp. 
 
 Here AB, AD and BE being denoted by a, by and — c, as 
 before ; if AC represent x, C'B in a reverse position will be 
 expressed by a^x. Consequently AC'.OB =r (a— ar)x = 
 
 ax — x^i and therefore AD.BE= — hcz=.ax x% or x^ ax 
 
 =;:+ be. The roots are hence the greater segment AC, which is 
 additive, and the less segment AC, which is subtractive. 
 
 In this case, the quadratic equation will alwdys admit of a 
 double solution, since the radical part of the root is both ad- 
 ditive and subtractive, while the circle crossing AB must ne- 
 cessarily cut it in two parts. 
 
 The third and fourth forms of the equation are constructed 
 by the application of the first case of the problem. 
 
 3. x'-\-ax-=:-—bcy then xz=. — %r±i\^^-r -^bc ; the two roots 
 
 having the same character, and both of them subtractive. 
 
 Let AB = a, AD = h, and BE 
 =rc; if BC denote— X, AC or 
 AB — BC, will be expressed by 
 a Jr X. Whence AC.BC = 
 (fl-{-ar)-^=-«^-^% andAD.BE 
 = 5c = — ax — x"". By transpo- 
 sition, therefore, :r* 4- ax = — be. 
 The values x are consequently 
 BC and BC, both of ihem sub- 
 tractive. 
 
 4. x^—ax^z-^bc, then a=z-\- ^ =±: y - — bcj both rootshaving 
 
 likewise the same character, but additive. 
 
 Let AB, AD, and BE be expressed as before by a, b and c ; 
 if AC represent x, CB will be denoted by a — x. Wherefore, 
 AC.CB=(a — x)x=ax — j;% and AD.BE=bc=:ax — j;\ Con- 
 sequently by transposition x* — ax= — be. The roots of this 
 equation are, therefore, expressed by AC and AC, both of 
 them additive. 
 
KOTES AND ILLUSTRATION^. S4^ 
 
 When the rectangle undef the perpendicular AD and BE, 
 becomes equivalent to the square of half of AB, the circle 
 touches AB, and the two points C and C merge in a single 
 
 point. At this limit, too, the radical part z±z /_ — l>c of the 
 
 Value of :r vanishes, and there results a single root, which is ad- 
 ditive or subtractive according to the sign of the second term 
 of the quadratic equation. If it were sought that the rectan- 
 gle linder AD, BE, or under the segments AC, CB, should 
 exceed the square of the half of AB, the circle would not meet 
 this straight line, while the radical would evidently become 
 impossible, and thus betray the same incongruity of hypo- 
 thesis. 
 
 It may be observed, that the algebraical solution of these 
 quadratic equations flows from the geometrical construction. 
 For, suppose AB were bisected in O ; it is evident that 
 AD.BE=AC.CB=:AO^-.OC% or ,OC^--AO», or OC'= 
 AO*— AD.BE, of AD.BE-f-AO% according as the intersec- 
 tion takes place within or without AB. Wherefore OC always 
 
 represents the radical part z±:^ — z^bc of the expression for 
 
 the values of or, which are formed by its combination with OA. 
 If the construction of Pappus be used, while the perpendi- 
 culars AD, BE, and the transverse line DE remain the same 
 as before, the intersection of this with a circle described on 
 AB determines the position of a perpendicular to it, dividing 
 the diameter internally or externally into the required seg- 
 ments. 
 
 i?. 
 
 Proposition eighteenth. To this proposition might be 
 added a corollary : That four times the area of a triangle is to 
 the rectangle under any ttvo sides, as the base to the radius of the 
 circumscribing circle. 
 
 For the area of the triangle ABC is (Prop. 5. 11.) equivalent 
 to half the rectangle contained by the base AC and the per- 
 pendicular BD, and consequently four times this area is equi- 
 valent to twice the rectangle AC, BD. But (VI. 18.) the 
 
84-6 NOTES AND ILLUSTRATIONS. 
 
 rectangle under the sides AB and BC 
 is equivalent to the rectangle under the 
 perpendicular BD and BE, the diame- 
 ter of the circumscribing circle, or to 
 twice the rectangle under BD and the 
 radius of that circle. Whence four 
 times the area of the triangle is to the 
 rectangle under the sides AB and BC, 
 as twice the rectangle under BD and AC to twice the rectan- 
 gle under BD and the radius of the circumscribing circle, or 
 as the base AC to that radius. 
 
 Let a, b and c denote the three sides of a triangle, and S 
 half their sum or the semiperimeter ; then, combining Prop. 29. 
 Book VI. with this corollary, the radius of the circumscribing 
 
 circle will be expressed by ^ ,,^ ^ j^ — j— rr- — .. Thus*, 
 
 if the sides of the triangle be 13, 14, 15, the radius of the cir- 
 
 6. Proposition nineteenth. This well-known proposition is 
 now rendered more general, by its extension to the case 
 of the exterior angle of the triangle. The two cases com- 
 bined afford an easy demonstration of the corollary to Propo- 
 sition 7. Book Vi. ; for the straight lines bisecting the vertical 
 and its adjacent angle form a right-angled triangle, of which 
 the hypotenuse is the distance on the base between the points 
 of internal and external section. 
 
 7. Proposition twenty-third. The latter part of the scho- 
 lium was added to this proposition, with a view to ex- 
 plain the principle of the construction of the pantagraphy a 
 very useful instrument contrived for copying, reducing, or 
 even enlarging plans. It consists of a jointed rhombus DBFE, 
 framed of wood or brass, and having the two sides BD and BF 
 extended to double their length ; the side DE and the branch 
 DA are marked from D with successive divisions, DO being 
 made to BO always in the ratio of DP to BC j small sliding 
 
NOTES AND ILLUSTRATIONS. 
 
 347 
 
 boxes for holding a pencil or tracing point are brought to the 
 corresponding graduations, 
 and secured in their positions 
 by screws ; the point O is 
 made the centre of motion, 
 and rests on a fulcrum or 
 support of lead; and the 
 tracer is generally fixed at 
 C, while the crayon or draw- 
 ing point is lodged at P, 
 From the property of diver- 
 ging lines intersectLpg paral- 
 lels, the three pomfs O, P 
 and C must evidently range 
 in the same straight line, and which is divided at P in the de^ 
 terminate ratio. While the point C, therefore, is carried along 
 the boundaries of any figure, the intermediate point P will, 
 by the scholium, trace out a similar figure, reduced in the 
 proportion of OC to OP or of OB to OD, and which, in the 
 present instance, is that of three to one. 
 
 But the point P may be placed in the fulcrum, the tracer 
 inserted at O, and the crayon held at C ; in which case, C 
 would delineate a figure which is enlarged in the ratio of OP 
 to PC or of OD to DB. If the points O and P were now 
 brought to coincide with A and E, the distances AE and EC 
 being equal, the original figure would be transferred into a 
 copy exactly of the same dimensions. 
 
 In reducing small figures, however, artists commonly pre- 
 fer another method, which is partly mechanical. The origi- 
 nal is divided into a number of^mall squares, by means of 
 equidistant and intersecting parallels. Other reduced squares 
 are drawn for the copy, which is then filled up, by observing 
 the same relative position and form of the boundaries. — One 
 material advantage results from this practice ; for if oblongs 
 be used in the copy instead of squares, the original figure will 
 be more reduced in one dimension than another, which is of- 
 ten very convenient where height and distance are represent- 
 ed on different scales. 
 
3*5 NOTES AND ILLUSTRATIONS. 
 
 8. Proposition twenty-eight. The curious properties of 
 the crescents, or lu7iu/ce, contained in the first corollary, were 
 discovered by Hippocrates of Chios, in his attempts to square 
 the circle. But a beautiful extension of them was briefly 
 suggested by the Reverend Mr Lawson, and afterwards ex- 
 plained and demonstrated by Dr Hutton of Woolwich, in 
 whose ingenious Mathematical Tracts it now appears. It is 
 a mode of dividing a given circle into equal portions, and 
 contained within equal circular 
 boundaries. For example, let 
 it be required to cut the circle 
 APBQ into five equal spaces. 
 Divide the diameter AB into 
 five equal parts at the points 
 C, D, E and F ; on AC, AD, 
 AE, and AF describe the se- 
 micircles AGC, AID, ALE, 
 and ANF, and on BC, BD, 
 BE, and BF, towards the op- 
 posite side, describe the semicircles BHC, BKD, BME, and 
 BOF ; the circle APBQ will be divided into five equal por- 
 tions, by the equal compound semicircumferences AGCHB, 
 AIDKB, ALEMB, and ANFOB. 
 
 For the diameter AB is to the diameter AD, as the circum- 
 ference of AB to the circumference of AD, or (V. 3.), as the 
 semicircumference APB to the semicircumference AID ; and 
 AB is to BD, as the semicircumference APB to the semicir- 
 cumference BKD. Wherefore (V. 20.) AB is to AD and BD 
 together as the semicircumference APB to the compound 
 boundary AIDKB ; and consequently these interior bounda- 
 ries AGCHB, AIDKB, ALEMB, and ANFOB, are all equal 
 to the semicircumference of the original circle. 
 
 Again, the circle on AB is to the circles on AE and AF, as 
 the square of AB to the squares of AE and AF ; and conse- 
 quently (V. 20.) the circle on AB is to the difference between 
 the circles on AE and AF, as the square of AB to the diffe- 
 rence between the squares of AE and AF, that is (II. 17.), 
 the rectangle under the sum and difference of AE and AF, or 
 
NOTES AND ILLUSTRATIONS. 349 
 
 twice the rectangle under EF and AS, the distance of A from 
 the middle point of EF. Whence the circle APBQ is to the 
 difference of the semicircles ALE and ANF, or the space 
 ALEFN, as the square of AB to the rectangle under AS and 
 EF; and, for the same reason, the circle APBQ is to the 
 space FOBME, as the square of AB is to the rectangle under 
 BS and EF; consequently (V. 20.) the circle APBQ is to the 
 compound space ALEMBOFN, as the square of AB to the 
 rectangles under AS and EFand BSand EF, or the rectangle 
 under AB and EF ; but the square of AB is to the rectangle 
 under AB and EF, (V. 25. cor. 2.) as AB to EF, which is the 
 fifth part of AB; wherefore (V. 5.) any of the intermediate 
 spaces, such as ALEMBOFN, is the fifth part of the whole 
 circle. 
 
 9. Proposition twenty-ninth. This elegant theorem admits 
 of an algebraical investigation. Put AC=«, AB=^, BC=:c, 
 and let s denote the scmipgrimeter, and T the 
 area of the triangle ; then, by Prop. 23. Book II., 
 2AC.CD = a^4-c^-~^% consequently CD=: 
 
 *±f!=:£, and BD^=BC^— CD^ = 
 
 2a .A. D C 
 
 c*— 'v^^^^t|^-^)^ and, therefore, by Prop. 5. Book IL, T'= 
 
 AC^BP' __ 4aV— (g^ 4- c' — b^y 
 
 4 ~ 16 
 
 I^ut this expression, consisting of the difference oftwo squares, 
 may be decomposed, by Prop. 17. Book II. ; whence T^z= 
 2ac + a'- + c^^b'' 2ac-^a''—c'^ -f. b^ _ (a + cf — 6^ b^—{a—cf 
 4 * 4 ~ 4 " i ' 
 
 and, decomposing these factors again, 
 _a±b^ a — b + c a-^-b — c — a-^-b+c 
 ~ 2 * 2 * 2 2 "' 
 
 Now, ^±1+^ = s, ^-^ + ^= s--b, ^ + ^-' = 
 ^^^~^=5 — o-'y wherefore we obtain, by substitution, 
 
 T = >V/(5(5— «) (5— W (5-C) ). 
 
 Suppose the sides of the triangle to be 13, 14, and \5 ; then 
 
350 
 
 NOXES AND ILLUSTRATIONS. 
 
 the area is = v'( 21. 8.7.6) = ^^7056 = 84-. If the sides were 
 21, 17 and 10, the area would be the same, for >v/(24?.3.7.14)= 
 ^^7056 = 8^. 
 
 This mosl^ useful proposition was known to the Arabians, 
 but seems to have been re-invented in Europe about the latter 
 part of the fifteenth century. 
 
 Another corollary might be subjoined to this proposition : 
 As the scmiperimeler of a triangle is to its excess above the base, 
 so is iJie rectangle under its excesses above the tivo sides to tJte 
 square of the radius of the inscribed circle, 
 
 ForBIrBG:: EI : DQ, 
 and consequently (V. 25. 
 cor. 2.)BI:BG:: EI.DG : 
 DG* ; but it was proved 
 that EI.DG is equivalent to 
 AG. AI, and hence BI : BG 
 : : AG.AI : DG\ Now 
 BI has been shown to be 
 the semiperimeter, and BG, 
 AG and AI its excesses 
 above the base and the other 
 two sides of the triangle, 
 of which DG is the radius 
 of the inscribed circle. 
 
 Hence let the sides of the triangle be denoted by a, b and 
 c, and the semiperimeter by S ; the square of the radius of the 
 
 inscribed circle will then be expressed by — ^^^-^ — ^ • 
 
 Suppose, for example, the sides of the triangle were 13, 14- 
 and 15, the radius of the inscribed circle would be the square 
 
 O >T /? ■■ ~ 
 
 root of ~j^ — , or of 16, that is 4. 
 
 Employing the same notation, it is not diflScult to perceive 
 that the continued product of all the sides of a triangle must 
 be equivalent to the product of twice their sum into the ra- 
 dii of the inscribed and circumscribing circles. Thus, 
 
 13.14.15=: 2730=84^.4.81. 
 
NOTES AND ILLUSTRATIONS. 351 
 
 Recurring to the last figure, it is evident that BG : BI : : 
 DG : EI : : DG.EI : EI% or, since DG.EI = AG.AI, 
 BG : BI : : AG.AI : EP ; that is, As the excess of the peri- 
 meter above the base is to the scmiperimeter itself, so is the rect- 
 angle under its excesses above the other two sides of the triangle to 
 the square of the radius of the circle of external contact below the 
 base. Thus, in the triangle taken for illustration, 6:21 : : 8.7 : 196, 
 and consequently the radius of the circle under the base is 14. 
 Again, 7 : 21 : : 8.6 : 144, and the radius of the circle touch- 
 ing externally the side 14 is therefore 12. And, in the same 
 manner, 8 : 21 : : 7.6 : llOJ; which gives 10^ for the radius 
 of the circle applied beyond the shortest side 13. 
 
 10. Proposition thirtieth. A similar and very important 
 problem, which formerly occupied a place in the text, must 
 not be omitted. It likewise furnishes an ingenious and concise 
 approximation to the quadrature of the circle, first published 
 at Padua in the year 1668, by James Gregory, my illustrious 
 predecessor in the mathematical chair of the University of 
 Edinburgh ; and seems the more deserving of attention, as it 
 probably led that original author to the investigation of the 
 Method of Series. 
 
 Given the area of an inscribed, and that of a circumscribed, re- 
 gular polygon ; to find the areas of inscribed and circumscribed 
 regular polygons, having double the number of sides. 
 
 Let TKNQ and HBDF be given similar inscribed and cir- 
 cumscribed rectilineal figures ; it is required thence to deter- 
 mine the surfaces of the corresponding inscribed and circum- 
 scribed polygons AKCNEQGT and VILMOPRS, which have 
 twice the number of sides. 
 
 From the centre of the circle, draw radiating lines to all the 
 angular points. It is evident that the triangles ZXK and ZAB 
 are like portions of the given inscribed and circumscribed fi- 
 gures TKNQ and HBDF; and that the triangle ZAK, and 
 the quadrilateral figure ZAIK are also like portions of the de- 
 rivative polygons AKCNEQGT and VILMOPRS. And since 
 
352 
 
 NOTES AND ILLUSTRATIONS. 
 
 XK is parallel to AB, ZX : ZA : : ZK : ZB (VI.2.); butZX 
 is to ZA as the triangle ZXK is to the triangle ZAK (V. 25. 
 cor. 2.), and, for the same reason, ZK is to ZB as the triangle 
 ZAK is to the triangle Z AB ; whence ZXK : ZAK : ; ZAK : 
 ZAB, and consequeDtly the derivative inscribed polygon 
 AKCNEQGT is a mean proportional between the ins/^ribed 
 and circumscribed figures TKNQ and HBDF. 
 Again, because ZI bisects 
 
 the angle AZB, Z A is to ZB, 
 t>r ZX is to ZK, as At to IB 
 (VI. 10.), and consequently 
 (V. 25. cor. 2.) the triangle 
 XZK is to the triangle AZK, 
 as the triangle AZI to the "l\ 
 triangle IZB. Hence the in- , \\ 
 scribed figure TKNQ is to ^^ 
 its derivative incribed figure 
 AKCNEQGT as the trian- H 
 gle AZI to the triangle IZB ; 
 wherefore (V. 11. and 13.) TKNQ and AKCNEQGT toge- 
 ther are to twice TKNQ, as the triangles AZI and IZB, or 
 a\ZB, to twice the triangle AZI, or the space AII^Z, — that 
 is, as HBDF to VILMOPRS. And thus the two inscribed 
 polygons are to twice the simple inscribed polygon, as the 
 surface of the circumscribing polygon to the surface of the de- 
 rivative circumscribing polygon with double the number of 
 sides. 
 
 Cor. Hence the area of a circle is equivalent to the rectan- 
 gle under its radius and a straight line equal to half its cir- 
 cumference. For the surface of any regular circumscribing 
 polygon, such as VILMOPRS, being composed of a number 
 of triangles AZI, which have all the same altitude ZA, is equi* 
 yalent (II. 6.) to the rectangle under ZA and half the sum of 
 their bases, or the semiperimeter of the polygon. But the 
 circle itself, as it forms the ultimate limit of the polygon, 
 must have its area, therefore, equivalent to the rectangle un- 
 der the radius ZA, and the semicircumference ACE. 
 
 Scholium, This solution, it was obseryed, affords one of the 
 
 1 
 
NOTES AND ILLUSTRATIONS. 
 
 353 
 
 best elementary methods of approximating to the numerical ex- 
 pression for the area of a circle. Supposing the radius of a circle 
 to be denoted by unit ; the surface of the circumscribing square 
 will be expressed by 4, and consequently (IV. 15. cor.) that of its 
 inscribed square by 2. Wherefore the surface of the inscribed 
 octagon is =-v/2^4=2,8284;27 1247; and the surface of the cir- 
 cumscribing octagon is found by the analogy, 2-f- 2.8284271247: 
 2 X 2 : : 4 ; 3.3137084990. Again, ^^ (2. 828427 1247 X 
 3.3137084990)= 3.0614674589, which expresses the area of 
 the inscribed polygon of 16 sides ; and 2.8284271247-h 
 3.0614674589 : 2x2,8284271247, or 5,8898945836 : 
 5.6568542494 : : 3.313708499 : 3.1825979781, which de- 
 notes the area of the circumscribing polygon of 16 sides. Pur- 
 suing this mode of calculation, by alternately extracting a 
 square root and finding a fourth proportional, the following 
 Table will be formed, in which the numbers expressing the 
 surfaces of the inscribed and circumscribed polygons conti- 
 nually approach to each other, and consequently to the mea- 
 sure of their intermediate circle. 
 
 Number of 
 
 Area of the in- 
 
 Area of the circum- 
 
 Sides. 
 
 scribed Polygon. 
 
 scribing Polygon. 
 
 4 
 
 2.0000000000 
 
 4.0000000000 
 
 8 
 
 2.8284271247 
 
 3.313708*990 
 
 16 
 
 3.0614674589 
 
 3.1825979781 
 
 32 
 
 3.1214451523 
 
 3.1517249074 
 
 64 
 
 3.1365484905 
 
 3.1441184852 
 
 128 
 
 3.1403311570 
 
 3.1422236917 
 
 256 
 
 3.1412772509 
 
 3.1417503692 
 
 512 
 
 3.1415138011 
 
 3.1416321807 
 
 1014 
 
 3.1415729037 
 
 3.1416025026 
 
 2048 
 
 3.1415877253 
 
 3.1415951177 
 
 4096 
 
 3.1415914215 
 
 3.1415932696 
 
 8192 
 
 3.1415923456 
 
 3.1415928076 
 
 16384 
 
 3.1415925766 
 
 3.1415926921 
 
 32768 
 
 3.1415926344 
 
 3.1415926632 
 
 65536 
 
 3.1415926488 
 
 3.1415926560 
 
 131072 
 
 3.1415926524 
 
 3.1415926542 
 
 1262144 
 
 3.1415926533 
 
 3.1415926537 
 
 J52428S 
 
 3.1415926535 
 
 3.1415926536 
 
S54« NOTES AND ILLUSTRATIONS. 
 
 The computation of this table might be greatly abridged, 
 by attending to the successive formation of the numbers. Let 
 a and b denote the area of an inscribed and circumscribing po- 
 lygon of the same number of sides, and a' and b' the areas 
 of correspo^ing polygons having double the number of sides. 
 Since «' =\/ a,b, when a and b approach to equality, it is ob- 
 vious that a' = — "J- nearly, or a^ — «= -^ : Wherefore, af- 
 
 ter the sides of the polygon are multiplied, the numbers of the 
 first column v/ill be formed, by constantly adding half their 
 difference from those of the second column. Again, because 
 
 b^z=—, — , by substitution b' = - — -—.. and hence b — i' =i: 
 ~~~r-j = (b — a) r> — 7~T ; but, since a and b come to differ little, 
 
 the fraction - — — , may be reckoned to jr. or b — b^z= - — -— 
 Ha'^b •' * 4- 
 
 very nearly. Consequently the higher numbers in the second 
 column may be filled up, by subtracting one-fourth of the com- 
 mon difference. It follows likewise, from combining this result 
 with what has been shown before, that a number in the second 
 column, diminished by the third^2LVi of the common difference, 
 must give very nearly the final result. Thus, the areas of the 
 inscribed and circumscribing polygon of 2048 sides, being 
 3,1415877253 and 3.1415951177, thdir difference is 73924, and 
 the third of this, or 24641 , taken away from the greater, leaves 
 3.1415926536, for the ultimate value, or the area of the circle 
 itself. 
 
 Of the two modes of approximating to the mensuration of 
 the circle, the one contained in the text, though not so direct^ 
 is on the whole simpler than the other. In the course of my 
 geometrical lectures, I generally mentioned, that the first pro- 
 position of the fourth book, by enabling us to discover a series 
 of regular polygons with the same sides continually doubled, 
 admitted of an easy application. But not having pursued the 
 calculation to any length, I neglected the obvious advantage 
 which results from reducing the perimeter at each step to the 
 same extent, till I was led to reconsider the subject, in con- 
 sequence of meeting with the small work of Schwab, before 
 quoted. It somehow had escaped my notice, that M. Legen- 
 dre, in the additions to his Geometry, has cursorily treated the 
 subject in the same way. 
 
NOTES AND ILLUSTRATIONS. 
 
 355 
 
 The numbers contained in the last table were copied and 
 interpolated from the tract of James Gregory, entitled Vera 
 Circuli el Hyperbolce Quadraturat as reprinted in the Opera 
 Fana of Huy gens. For the calculation of the table contain- 
 ed in the text, and of other two tables which will be annexed 
 to this note, accompanied by several acute remarks concern- 
 ing the formation of the successive numbers, I am indebted to 
 the very obliging assiduity of a young friend, Mr G. A. Wal- 
 ker Arnott, whose solid talents and unwearied application pro- 
 mise the happiest fruits. 
 
 Let the same mode of computation be applied to the suc- 
 cessive polygons derived from the hexagon. The radius of 
 the circle being unit, the perpendicular from the centre to the 
 base of each component triangle of the inscribed hexagon will 
 be = v^l, and consequently the area of the figure = 4-V'^ =^ 
 2.5980762114. Again, each side of the circumscribing hexa- 
 gon is = ^4 =: 2-v/j, and therefore its area, or that of the six 
 contained triangles, is = 6-/^= 2^3=-/ 12 = 3464-1016151, 
 or one-third more than the former. Hence the following table 
 is constructed. 
 
 Number of 
 
 Area of the 
 
 Area of the 
 
 Sides. 
 
 Inscribed Polygon. 
 
 Circumscribing Polygon. 
 
 6 
 
 2.5980762114 
 
 3.4641016151 
 
 12 
 
 3.0000000000 
 
 3.2153903092 
 
 24 
 
 3.1058285412 
 
 3.1596599421 
 
 48 
 
 3.1326286134 
 
 3.1460862150 
 
 96 
 
 3.1393502030 
 
 3.1427145996 
 
 192 
 
 3.1410319509 
 
 3.1418730499 
 
 384 
 
 3.1414524723 
 
 3.1416627470 
 
 768 
 
 3.1415576079 
 
 3.1416101766 
 
 1536 
 
 3.1415838921 
 
 3.1415970343 
 
 3072 
 
 3.1415904632 
 
 3.1415937487 
 
 6144 
 
 3.1415921060 
 
 3.1415929274 
 
 12288 
 
 3.1415925167 
 
 3.1415927220 
 
 24576 
 
 3.1415926194 
 
 3.1415926707 
 
 49152 
 
 3.1415926450 
 
 3.1415926579 
 
 98304 
 
 3,1415926514 
 
 3.1415926547 
 
 196608 
 
 3.1415926531 
 
 3.1415926539 
 
 393216 
 
 3.1415926535 
 
 3.1415926537 
 
 786432 
 
 3.1415926536 
 
 3.1415926536 
 
35G 
 
 NOTES AND ILLUSTRATIONS. 
 
 If the method employed in the text for discovering the ra- 
 dius of the circle, which has twice the number of sides under 
 the same extent of perimeter, be applied to the hexagon or its 
 elementary equilateral triangle, the numbers will stand as be- 
 low. 
 
 Number of 
 
 Radius of Inscrib- 
 
 Radius of Circum- 
 
 Sides. 
 
 ed Circle. 
 
 scribing Circle. 
 
 6 
 
 .8660254038 
 
 1.0000000000 
 
 12 
 
 .9330127019 
 
 .9659258263 
 
 24 
 
 .9494692641 
 
 .9576621969 
 
 48 
 
 .9535657305 
 
 .9556117687 
 
 96 
 
 .9545887496 
 
 .9551001222 
 
 192 
 
 .9548444359 
 
 .9549722705 
 
 384 
 
 .9549083532 
 
 .9549403113 
 
 768 
 
 .9549243322 
 
 .9549323217 
 
 1536 
 
 .9549283270 
 
 .9549303243 
 
 3072 
 
 .9549293257 
 
 .9549298250 
 
 6144 
 
 .9549295753 
 
 .9549297002 
 
 12288 
 
 .9549296378 
 
 .9549296690 " 
 
 24576 
 
 .9549296534 
 
 .9549296612 
 
 49152 
 
 .9549296573 
 
 .9549296592 
 
 98304 
 
 .9549296582 
 
 .9549296587 
 
 196608 
 
 .9549296585 
 
 .9549296586 
 
 393216 
 
 .9549296586 
 
 .9549296586 
 
 Wherefore, .95492965855 : 1 : : 3 : 3.1415926536; and hence 
 3.1415926536 is the nearest expression, consisting often de- 
 cimal places, for the area of the circle whose radius is 1. But 
 the semicircumference in this case denoting also the surface, 
 the same number must represent the circumference of a cir- 
 cle whose diameter is 1. Consequently, if D denote the dia- 
 meter of any circle, the circumference will be expressed ap- 
 proximately, by 3.1415926536 xD; whence the area will be 
 ^D^ X 3.1415926536, or D^ X .7853981634. 
 
 By help of the note to Prop. 27. Book V. lower numbers may 
 be found, approximating to the same results. For in this case 
 
NOTES AND ILLUSTRATIONS. 357 
 
 5K=S, K='7, 7>=16, and ^=11: whence, remounting from these 
 conditional equahties, the ratio of the diameter to the circmn- 
 ference of a circle is denoted progressively, by 1 : 3 — by 7 : 
 22— by 113: 355— and by 1250 : 3927- Tjie ratio of 1 to 3 
 is the rudest approximation, being the same as that of the dia- 
 meter of the circle to the perimeter of its inscribed hexagon ; 
 the ratio of 7 to 22 is what was discovered by Archimedes ; the 
 ratio of 113 to 355, in which the three first odd numbers ap- 
 pear in pairs, was first proposed by Adrian Metius of Alkmaer, 
 Professor of Mathematics and Medicine at Franeker, who died 
 in 1636 ; and the ratio of 1250 to 3927, the same as 1 to 
 3.1416, is that generally adopted by the Hindus. Hence also 
 the circle is to its circumscribing square nearly — as 1 1 to 1 4, 
 or, still more nearly — as 355 to 452. 
 
 To this Book may be added the following Propositions. 
 PROP. I. THEOR. 
 
 JfJroTii any point in the circumferejice of a circle, straight lines 
 he drawn to the extremities of a chord, and meeting the perpendi- 
 cular diameter, they mil divide that diameter, internally and ex- 
 ternally, in the same ratio. 
 
 Let the chord EF be perpendicular to the diameter AB of 
 a circle, and from its extremities F and E straight lines FG 
 and EG be inflected to a point G in the circumference, and 
 cutting the diameter internally and externally in C and D ; 
 then will AC : CB : : AD : : DB. 
 
 For join AG and BG, and draw HBI parallel to AG. 
 
 Because AEGB is a semicircle, the angle AGB is a right 
 angle (III. 19.); wherefore AG and HI being parallel, the alter- 
 nate angle GBI is right (I. 22.), and likewise its adjacent angle 
 GBH. But the diameter AB, being perpendicular to the chord 
 
358 
 
 NOTES AND ILLUSTRATIONS. 
 
 EF, must (III. 4. and 13.) bisect the arc FAE, and therefore 
 the angle EGA is equal to AGF (III. 12. cor.) or (lil. 17.), 
 its supplement. And since AG is parallel to HI, the angle 
 EGA is equal to the an- ■• 
 
 gle GIB or its supple- '^- 
 
 ment (1. 22.); and for 
 the same reason, the an- 
 gle AGF is equal to the 
 alternate angle GHB. 
 Whence the angle GIB 
 is equal to GHB ; but 
 theanglesGBIandGBH 
 being both right angles, 
 are equal, and the side 
 GB is common to the 
 two triangles BIG and 
 BHG, which are, there- 
 fore, equal (I. 20.), and 
 consequently BH is e- 
 qual to BI, and AG : BH : : AG : BI. Now, because the pa- 
 rallels AG and BH are intercepted by the diverging lines AB 
 and GH, AG : BH : : AC : CB (VI. 2.) ; and since the pa- 
 rallels AG and BI are intercepted by the diverging lines GD 
 and AD, AG : BI : : AD : DB. Wherefore, by identity of 
 ratios, AC : CB : : AD : DB, that is, the straight line AB is 
 cut in the same ratio, internally and externally, or the whole 
 line AD is divided harmonically in the points C and B. 
 
 Cor, 1. As the points E and G 
 come nearer each other, it is ob- 
 vious that the straight line EGD 
 will approach continually to the 
 position of the tangent, which is 
 its ultimate limit. Hence the tan-, 
 gent and the perpendicular, from the point of contact or mu- 
 tual coincidence, cut the diameter proportionally, or AC : 
 CB : : AD : DB/ It is, therefore, evident (VI. 7.) that, O be- 
 ing the centre, OC : OB : : OB : OD. 
 
NOTES AND ILLUSTRATIONS. 
 
 359 
 
 Cor. 2. Since OC : OB : : OB : OD, it follows (V. 19. cor. 2.) 
 that OC : OD : : OB^— OC^ or AC.CB : OD^ —OB' or 
 AD.DB ; whence, by division, CD : OD : : AD.DB— AC.CB, 
 or VI. 7. COB.) CD^ : AD.DB. 
 
 PROP. 11. THEOR. 
 
 If two straight lines he inflected J'rom the extremities of the 
 base of a triangle to cut the opposite sides proportionally, another 
 straight line, dravonjrom the vertex through their point of con - 
 course, will bisect the base. 
 
 In the triangle ABC, let AE and CD, drawn from the ex- 
 tremities of the base to cut the opposite sides proportionally, 
 intersect each other in F, join BF, which produce if necessary 
 to meet the base in the point G ; AG will be equal to (jC, 
 
 For join DE. And because the sides AB and BC are cut 
 proportionally, DE is parallel to AC 
 (VI. 1. cor,), whence BD : BA : : 
 BH : BG (VI. 1.) ; but BD : BA : : 
 DE : AC (VL 2.), and therefore 
 BH : BG : : DE : AC, Again, the 
 parallels DE and AC being cut by 
 the diverging lines AE and CD, 
 DE ; AC : : DF : FC (VI. 2.) and 
 DF : FC : : FH ; FG (VI. 1.) ; where- 
 fore BH : BG : : FH : FG, or BF is 
 
 cut internally and externally in the same ratio. But DH be- 
 ing parallel to AG, BH : BG : : DH ; AG ; and since DH is 
 also parallel to GC, HF : FG : : DH : GC ; whence DH ; 
 AG : : DH : GC, and consequently AG is equal to GC. 
 
 PROP. III. THEOR. 
 
 If a semicircle be described on the side of a rectangle, and 
 through its extremities two straight lines he drawn from any point 
 
360 
 
 NOTES AND ILLUSTllATIONS. 
 
 in the circumfej-ence to meet the opposite side produced both xvat/s ; 
 the altitude of the rectangle xuill he a mean proportional hetxueen 
 the segments thus intercepted. 
 
 Let ABED be a rectangle, which has a semicircle ACB de- 
 scribed on the side AB, and the straight lines CA and CB 
 drawn from a point C in the circumference to meet the exten- 
 sion of the opposite side DE ; the altitude AD of the rectan- 
 gle will be a mean proportional between the exterior segments 
 FDandEG. 
 
 For, the angle ADF, being evidently a right angle, is equal 
 to the angle ACB, which stands in a semicircle (III. 19.). and 
 
 the angle DFA is equal to the exterior angle BAG (I. 22.); 
 wherefore (VI. 11.) thp triangle FAD is similar to ABC. In 
 the same manner, it is proved that the triangle BGE is simi- 
 lar to ABC ; whence the triangles D AF and BGE are similar 
 to each other, and consequently (VI. 11.) FD : AD ; : BE or 
 AD: EG. 
 
 If the straight lines CD and CE be drawn, they will(VI. 2.) 
 divide the diameter AB into segments AH, HI, and IB, which 
 are respectively proportional to the segments FD, DE, and EG 
 of the extended side DE. Consequently when ABED is a 
 square, and therefore DE a mean proportional between FD 
 and EG, it must follow that HI is likewise a mean propor- 
 tional between AH and IB. 
 
 If the rectangle ABED have its altitude AD equal tp the 
 side of a square inscribed within the circle, the square of the 
 
NOTES AND ILLUSTRATIONS. 
 
 361 
 
 diameter AB is equivalent to the squares of the two segments 
 AI and BH. For FD : AD : : AD : EG, whence (V. 6.) 
 FD.EG = AD.% or sFD.EG = 2AD* ; but (IV. 15. cor.) 
 2Ap= = AB^ or DE% and consequently sFD.EGnDE* ; 
 wherefore (VI. 2.) 2AH.IB = HP, and hence, by the first ad- 
 ditional proposition to Book II., the segments AI, BH are the 
 sides of a right-angled triangle, of which AB is the hypote- 
 nuse, or AB*=AP+BH=. 
 
 PROP. IV. THEOR, 
 
 A chord of a circle is divided vi continued proportion, hy 
 straight lines iriflected to any point in the opposite circumference 
 from the extremities of a parallel tangent, xiohich is limited by 
 another tangent applied at the origin of the chord. 
 
 Let AB, AC be two tangents applied to a circle, CD a 
 chord drawn parallel to AB, and AE, BE straight lines in- 
 flected to a point E in the opposite circumference ; then will 
 the chord CD be cut in continued proportion at the points F 
 and G, or CF : CG : : CG : CD, 
 
 For join BD, BC, and CE. Because the tangent AB is 
 equal to AC (III. 22. cor.), 
 the angle ABC is equal to 
 ACB (I. 10.); but ABC 
 is equal to the angle BCD 
 (I. 22.), and to the angle 
 BDC (III. 21.); whence 
 (VI. 1 1 .) the triangles B AC 
 and BDC are similar, and 
 AB : BC : : BC : CD, and 
 
 consequently (V. 6.) BC*r=AB.CD. Again, the triangles 
 CBG and CBE are similar, for they have a common angle 
 CBE, and the angle BCG or BCD is equal to BDC or BEC 
 (III. 16.): Wherefore BG : BC : : BC : BE, and BC^=BG.BE. 
 Hence AB.CD=BG.BE, and AB : BE : : BG : CD ; but FG 
 being parallel to AB, AB : BE : : FG : GE (VI. 2.), and 
 
362 
 
 NOTES AND ILLUSTRATIONS. 
 
 consequently FG : GE : : BG : CD ; therefore (V. 6.) FG.CDr- 
 BG.GE; antLsince (III. 26.) BG.GEnCG.GD, it follows 
 that CG.GD=FG.CD, and FG : CG : : GD : CD, and hence 
 (V. 10.)CF: CG::CG:CD, 
 
 PROP. V. THEOR. 
 
 If, from the vertex of a triangle, two straight lines he dratm, 
 making equal angles tvith the sides and cutting the base; the 
 squares of the sides are proportional to the rectangles under the 
 adjacent segments of the base, * 
 
 In the triangle ABC, let the 
 straight lines BD and BE make the 
 angle ABD equal to CBE ; then 
 AB^ : BC : : DA.AE ; ECCD. 
 
 For (III. 9. cor.) through the 
 points B, D, and E describe a cir- 
 cle, meeting the sides AB and BC 
 of the triangle in F and G, and 
 join FG. 
 
 Because the angles DBF and EBG are equal, they stand 
 (III. J 6. cor.) on equal 
 arcs DF and EG, and con- 
 sequently (III. 18. cor,) 
 FG is parallel to DE. 
 Whence (VI. 1.) AB : 
 BC:: AF:CG, and there- 
 fore (V. 13.) AB^:BC^ :: 
 AB.AF:BC.CG;but(III. 
 26.) AB.AF = DA.AE, 
 and BC.CG = ECCD. 
 Wherefore AB^ : CD^ : : 
 DA.AE : ECCD. 
 
 If the triangle ABC be right-angled at C, and the vertical 
 
NOTES AND ILLUSTRATIONS. 
 
 36^ 
 
 lines BD and BE cut the base internal- 
 ly ; then BC^ + AC.CE : BC* : : AE : CD. 
 Tor make AH equal to EC. Because 
 AB^ : BC" : : DA.AE : EC.CD, and (II. 
 10.) AB^=AC^ + BC% therefore AC^ + 
 BC^ : BC" : : DA.AE : EC.CD, and, by 
 division, AC^ : BC: : DA.DE*— EC.CD : 
 EC.CD. But,by successive decomposition, D A. AE-^EC.CD=: 
 DA.AC—DA.EC— EC.CD = DA.AC~-EC.AC=AC.HD; 
 whence AC* : BC* : : AC.HD : EC.CD, and (V. 13. and cor.) 
 AC.EC : BC* ; : EC.HD : EC.CD, or (V. 3.) : : HD : CD ; con- 
 sequently (V. 9.) BC* + AC-EC : BC^- : : HC : CD ; but, AH 
 being equal to EC, HC is equal to AE ; wherefore BC*+ 
 AC.EC : BC* : : AE : CD. 
 
 If the vertical lines BD, BE cut the base AC of a right- 
 angled triangle ACB 
 ejfternally ; then will 
 BC^~AC.EC : BC^ :: 
 AE : CD. For make 
 AH = EC. It is de- 
 monstrated as before, 
 that AC^- : BC^ : : 
 DA.AE-EC.CD : EC.CD ; but DA.AE-EC.CDrrDA.AC-f- 
 DA.EC— EC.CD = DA.AC— EC.AC = AC.HD : wherefore 
 AC* : BC* : : AC.HD : EC.CD, and AC.EC : BC* : ; EC.HD : 
 EC.CD : : HD : CD, and consequently BC*— AC.EC : BC* : : 
 HC or AE : CD. 
 
 PROP. VI. THEOR. 
 
 The perpendiculqr tvithin a circle, is a mean proportional to 
 the segments Jbrmed on it hy straight lineSf draximjrom the ex- 
 tremities of the diameter ^ through any point in the circumference. 
 
 Cet the straight lines AEC and BCG, drawn from the ex- 
 
36* 
 
 KOTtS AND iLLUSTR4TION.V' 
 
 tremitics of the diameter of a circle 
 through a point C in the circumfe- 
 rence, cut the perpendicular to AB j 
 the part DF within the circle is o 
 mean proportional between the seg- 
 ments DE and DG. 
 
 For the angle ACB, being in a se- 
 micircle, is a right angle (IIL 19.), 
 and the angle ABG is common to 
 the two triangles ABC and GBD, 
 which q^re, therefore, similar (Vf. 
 11.}. Hence the remaining angle 
 BAG is equal to BGD, and conse- 
 quently the triangles ADE and GDB 
 are similar; wherefore AD : DE : ; 
 DG:DB, and (V. 6.) AD.DB=: 
 DE.DG. But (III. 26. cor.), the 
 rectangle under AD and DB is equi- 
 valent to the square of DF ; whence 
 DE.DG=DF% and (V. 6.) DE : 
 DF :DF:DG. 
 
 The Appendix to the books of Geometry cannot fail, by its 
 novelty and singular beauty, to prove highly interesting. The 
 first part is taken from a scarce tract of Schooten, who was 
 Professor of Mathematics af Leyden, early in the seventeenth 
 century. But the second and most important part is chiefly 
 selected from a most ingenious work of Mascheroni, a cele- 
 brated Italian mathematician; which in 1798 was translated 
 into French, under the title of Geomeirie du Compas. It will 
 be perceived, however, that I have adapted the arrangement 
 to my own views, and have demonstrated the propositions 
 more strictly in the spirit of the ancient geometry. 
 
NOTES AND ILLUSTRATIONS. 365 
 
 NOTES TO TRIGONOMETRY. 
 
 1. The French philosophers have, at the instance ofBor- 
 da, lately proposed and adopted the centesimal division of 
 the quadrant, as easier, more consistent, and better adapted to 
 our scale of arithmetic. On that basis, they have also con- 
 structed their ingenious system of measures. The distance of 
 the Pole from the Equator vi^as determined with the most scru- 
 pulous accuracy, by a chain of triangles extending from Calais 
 to Barcelona, and since prolonged to the Balearic Isles. Of 
 this quadrantal arc, the ten millionth part, or the tenth part of 
 a second, and equal to 39.371 English inches, constitutes the 
 metre, or unit of linear extension. From the metre again, are 
 derived the several measures of surface and of capacity ; and 
 water, at its greatest degree of contraction, furnishes the stan- 
 dard of weights. 
 
 It would be most desirable, if this elegant and universal sys- 
 tem were adopted, at least in books of science. Whether, 
 with all its advantages, it be ever destined to obtain a general 
 currency in the ordinary affairs of life, seems extremely ques- 
 tionable. At all events, its reception must necessarily be very 
 slow and gradual ; and, in the meantime, this innovation is pro- 
 ductive of much inconvenience, since it not only deranges our 
 habits, but lessens the utility of our delicate instruments and 
 elaborate tables. The fate of the centesimal division may fi- 
 nally depend on the continued merit of the works framed aftei' 
 Ihat model. 
 
 2. The remarks contained in the preliminary scliolium, will 
 obviate an objection which may be made against the succeed- 
 ing demonstrations, that they are not strictly applicable, ex- 
 cept when the arcs themselves are each less than a quadrant. 
 But this in fact is the only case absolutely wanted, all the de- 
 rivative arcs being at once comprehended under the definition 
 of the sine or tangent. To follow out the various combina- 
 tions, would require a fatiguing multiplicity of diagrams ; and 
 
36S NOTES AND ILLUSTRATION*. 
 
 such labour would still be quite superfluous, because the mo Je 
 of extending or accommodating the results from the general 
 principle is so easily perceived. 
 
 3. The general properties of the sines of compound arcs 
 may be derived with great facility from Prop. 20. of Book VL 
 of the Elements. For, since AB.CD + BC.AD=AC.BD, it is 
 evident that ^AB.iCD-|-iBC.iAD=:^AC.|BD; but (cor. 1. 
 def. Trig.) the semichord of an arc 
 is the same as the sine of half the 
 arc, and consequently, by substitu- 
 tion, sin \Pi.^sin\CT> + sin\\^Q sin 
 iABCD ±= sin\k^C x sin^J^QT>. 
 Let iAB=±L, iBC=M, and iCD 
 i=N ; wherefore iABCI> = L+ 
 M + N, iABC = L + M, and 
 iBCD=M + N, and hence the ge- 
 neral result ; sinljsin^ + siriMsiniJu 4. M -{- N)=:^/w(L-|-IVI)' 
 5in(M-f-N}, in which L, M and N are any arcs whatever. 
 This expression, variously transformed, will exhibit all the 
 theorems respecting sines. For the sake of conciseness, let 
 the radius be denoted as usual by 1, and the semicireurafe- 
 rence by tt. 
 
 I. Put A=M, B=N, and let L be the complement of A. 
 
 Then, cosK sin^ -f sin A sin ( A 4- B -f - — A ) = sin ( - — A -|- A ) 
 
 s?n(A4-B); that is, since the sijie of an arc increased by a 
 quadrant is the same as its cosine, sinh. C05B + cosK sinBzz 
 5^»(A+B). 
 
 2. Let the arc B be taken on the opposite side, or substitute 
 — B for it in the last expression, and sinAcosB — cosAsinBzz- 
 sin{A—B). 
 
 3. In art. I, for A substitute its coniplement ; theft 
 
 sin{A-^B)=zsinQ-A+B)=sin[^ + A—B)=cos{A^B)yand 
 
 hence cosAcosB-^sinAsiuB=cos(A — B). 
 
 4. In art. 2, likewise substitute for A its complement, and 
 the result will become cos AcosB^-^siuAsinBz^ cos (A-^B). 
 
NOTES AND ILLUSTRATIONS. 367 
 
 5. In art. 1, let ArrB, and 2sinAcosA=zsin2A, 
 
 6. In art. 4, let A=B, and cosA''-sinA''=zcos2A. 
 
 7. In art. 3, let A=B, and cosA'^-\-sinA'^:=il. 
 
 8. Add the Jbrtnulce in art. 1. and 2, and 2sinAcosB = 5i» 
 (A+B)4-5f«(A— B). 
 
 9. Subtract the Jormidce of art. 2. from that of art. 1, and 
 2cosAsinB=s?n{A'\-B) — sin{A — B). 
 
 10. Conjoin the Jbrnmlce of art. 3. and 4/, and 2cosAcosB=!: 
 cos{A+ B ) + cos ( A— B ) . 
 
 11. Take the Jormu/cu of art. 4. from that of art. 3, and 
 2sinAsi7i'B=cos{A — B) — co5(A+B). 
 
 12. In art. 8, let B be the complement of A, and 2««A*= 
 
 sin ( A + A)-{-sin ( A — - -f A)= 1 — cos^Az^versQA. 
 
 J 3. In art. 9, let B be the complement of A, and 2cosA'z:: 
 5/w(A4- -— A) — {sin A \'A)=zl-\-cos2A=suvers2A. 
 
 14. In art. 5, instead of A substitute its half, and 2sin\A X 
 cos^AzrsinA^ 
 
 15. In art 6, likewise substitute the half of A for A, and 
 {cos^AY — {sin\Ay:=:cosA. 
 
 16. In art. 12, for A substitute its half, and 2(«n^A)* = 
 1 — cosA, or sinlAzz \/{\{ i — cosA))z=\/\versA. 
 
 17. Make the same substitution in art. 13, and 2(co5^A)'= 
 i-f-co^A, or cos\A=:x/{{{\-^co6A))=:^{suversA. 
 
 18. In art 8, transform A and B into A-[-B and A — B, and 
 consequently, for A-}-B and A — B, substitute 2A and 2B ; 
 then 2sin{A -{- B)c<^5(A — B) = sin2A-{~sin2B, or ^m(A4-B) 
 cos{A^B)=^si7i'2A+sin2B). 
 
 19. Make the same transformation in art. 9, and 2co5( A-f- B) 
 cos{A — B) = si7i2A — «'«2B, or cos(A + B)sin(A — B) = 
 \{sin2A—sin2B). 
 
 20. Repeat this transformation in art. 10, and 2cos{A^B) 
 ^;o5(A--B)= cos2A + cos2B, or cos{Aj^B) cos{A ^ B) = 
 ^cos(2A+cos2B). 
 
 2 ! . The same transformation being still made in art. l ] , 
 2sin{A+ B)5f;< A_B)=co62B— C052 A , or sin(A -}- B)sin{A--B)z= 
 i{cos2B—coi,2A). 
 
 22. Suppose L=:N=:B, and M=A-— B; then the general 
 
3G8 NOTES AND ILLUSTIIATIONS. 
 
 expression becomes sinB'^siu{A~B)sin{A-{-B)=zsinA\ or 
 sin{A~^.B)si7i{A—B ) =:sinA'-—smB\ 
 
 23. Instead of A in the last article, take its complement, and 
 
 sin{'^- A^B)si7i(^- A— B) = cosA^—sinB\ or cos(A — B) 
 cos(A-{- B)=zcosA' — sinB'» 
 
 24. Compare art. 21. with 22, and l(cos2B^cos2A)=:smA^-^ 
 sinB\ 
 
 25. Comparing likewise art. 20. with 23, and l{cos2A+ 
 cos2B)=:cosA'- — sinB^. 
 
 26. Resolve the difference of the squares in art. 22. into its 
 factors, and 5m( A-|. B)sin(A~^B)=:(si7iA^sinB)(sinA—siiiB). 
 
 27. Make a similar decomposition in art. 23, and co*(A4-B) 
 cos(A — B)= ( C05 A+ smB){cosA — sinB ) . 
 
 24-. In art. 18, instead of A and B take their halves, and 
 sinA+sinB=2sinl{A-{.B)cosl{A'^B). 
 
 25. Make the same change in art. 19, and sinA-^sinBz= 
 25fw^( A— •B)co5i ( A-f B). 
 
 26. Change likewise art. 20, and cosB-{.co$A=z2cosl(A-\.B) 
 co4(A~-B). 
 
 27. Do the same thing in art. 21, and cosB — cos A = 2sin^ 
 {A--B)sin\(A+By 
 
 From the third additional proposition to Book III., a very 
 simple expression may be derived for the sum of the sines of 
 progressive arcs. Suppose the diameter AO were drawn ; then 
 BE + CF+ DG = HG = HO + DO, or 2si7iAB-{. 25m AC + 
 2si7iAD = HO + sinADy and siTiAB + si7iAC -}- si7jAD = 
 1H0+ Isin AD = iAO.taiiBAO + ^siriAD, Wherefore, in 
 
 general, si7i a -j- sin2a -\- shiSa sItv 7ia =: \vers 7ia.cot^a-^ 
 
 \sin Tia. Hence the sum of the sines in the whole semicircle 
 is=:c(H\a. Thus, if the sines for each degree up to 180°, the 
 radius being unit, were added together, the amount would be 
 114,58866. 
 
 4. On examining the formation of the successive terms of the 
 first and second tables, it will appear that the coefficients are 
 certain multiples of the powers of 2, whose exponents likewise 
 at every step decrease by two. It is farther manifest, that if 1 , 
 
NOTES AND ILLUSTRATIONS. 369 
 
 A, B, C, &c. 1, A', B', C, &c. and 1, A\ B\ C\ &c. denote the 
 multiples corresponding to the arcs 7i.a, n + i««» ^^^ n—~i.a ; 
 then A + 1=A', B+ A'=B', C+B'r: C, &c. Whence the va- 
 lues of A, B, C, &c. are determined, either by the method of 
 finite differences, adopting the appropriate notation, or from 
 the theory of functions. Thus in the first table, AA=1, and 
 
 A = « — 2 ; AB=A'=»— 3, and 6.=^=^^; AC = B'= 
 
 , and C = ^ — T Wherefore m general 
 
 (1.) Sin na= 2^^^c"~^^— ;z^2.2"-^c"-^^+ ^ ^'~^^^""^ ,2>^V-^^~ 
 n — ^,n — 5.W — 6^„ 7 „ ? . ^ 
 
 — 2""*^C'^^5+ &c. 
 
 ( 2. ) Cos nazz 2»-i.c"— «.2"-^.c«-2^ !!:!!^ .2«-^c"-!r~ 
 
 ,2?^^.c"-^+ &c. 
 
 The third and fourth tables are evidently formed by multi- 
 plying constantly by 2cos 2a or 2 — 4s*, and subtracting the 
 term preceding ; or the multiplication by 4s* produces the se- 
 cond differences of the successive quantities. Hence in the 
 former, AaA=4w", AAB=4A", &c. ; 
 
 wherefore AArrw-f-i.w-f-i, and A := ^ ^o • — 5 
 
 AU-2( -g -^J-, ^ , 
 
 _ _, n.n — i.w+T.w — 3.W+3 ^ . . _ .1 . , , 
 
 and B = J^ . ^ -^--. But m the fourth table, 
 
 2.3.4.5 
 
 AAA+4, AAB+4A", AAC'=4B''; and consequently A A =r 
 2w4-2, andA=| ; AB=S(2.«+2.w-t-2)==~i-^^^^±^, and 
 
 ^_«\«--2£2+_2 Wherefore in general, 
 2.«3.4 
 
 w* — I 3 , w* — I w* — 9 5 
 
 2.3 2,3 4-5 
 
 (3.) Sm nazzn.s^n s^-^n — ^s 
 
 2.3 4.5 6.7 ^ 
 
 B 
 
3/0 NOTES AND ILLUSTRATIONS. 
 
 (4.) Cos «a=l «*+- . 5* . -^6^4-. &c 
 
 In the fifth and sixth tables, the coefficients are evidently 
 the same as those of the power of a binomial, only proceeding 
 from both extremes to the middle terms. Hence, according 
 as n is odd or even, 
 
 (5.) 2«-i sin u''=z±=sin na=fin.sin{n—2)a:=i=:nJ^—sin(n-4!)az:f: 
 
 n — I n — 2 . ^^ „ 
 
 n ._, — sin(^n — 6)adb:«&c. ; and 
 
 2«— 1 sin a" =r=t:co5 tw, =+= n.cos(n — 2)az±zn!^ — cos{n — ^jaqR 
 n. . ■co5(«— 6)«, &c. 
 
 2 3 
 
 Again, 
 
 (6.) 2»-i cos a"=:cos na^n,cos(n — 2)aJ^n, ~-—^cos{n — 4a)4- 
 
 n — I n — 2 , 
 71. . — -; — »cos{n — 6}a, &c. 
 
 In these three expressions, half the last term, which corre- 
 sponds to the middle in the expansion of the binomial, is to be 
 taken, when n is an even number. 
 
 It will be satisfactory likewise to subjoin an investigation of 
 the sine of the multiple arc, as derived from the Theory of 
 Functions. 
 
 It appears from inspecting the successive formation of the 
 sines of the multiple arcs, 1. that the odd powers only of j oc- 
 cur ; 2. that the coefficient of the first term is only n, and the 
 other coefficients are itR functions of third, fifth, &c. orders ; 
 and 3. that since, in the case when wrzi, the rest of the coef- 
 ficients evidently vanish, those coefficients in general, as af- 
 fected by opposite signs, must in each term produce a mutual 
 balance. 
 
 Let therefore sin jiazzn.s-^-n.s^J^n^s^ S:c. ; where s denotes 
 
NOTES AND ILLUSTRATIONS. 371 
 
 / in mil 
 
 the sine of the arc cr, and n, n, n, &c. the successive odd or- 
 ders of the functions of w. It is evident, from (Prop. 3. cor. 2. 
 Trig.) that, by substitution 
 
 (« + i)+(«--i))5+^(« + i) + («-i))53+((n + i) + (n-i)J5S 
 
 / /// ///// 
 4- &c. —^^/{^s^) sin na=(2—s^—^s\ &c.) {ns+ns^-^-ns^, &c.) 
 
 =2«5+(2« — 7^)$3_|.(27^_?^__J-w)s■', &c. Now, equating corre- 
 sponding terms, and rejecting the powers of 5, we obtain these 
 general results : 
 
 /// /// /// / «/■// //y/ ///// /// / 
 27l'ez=2n^; (w-}-l)-J-(?2-f.I)=2W—W; (72 -f.I)-|-(w — l)=27t — 71 ^W. 
 
 It remains hence to discover the several orders of the func- 
 tions of w. 
 
 1. The equation 2n'=^2n' contains a mere identical proposi- 
 tion ; but other considerations indicate that n must always de- 
 note the first term, or that the first function of w is w itself. 
 
 Ill in III I 
 
 2. The equation (w -J- 1) -j-(w — 1)=2« — w fixes the conditions 
 of the third function of w, which, from the nature of the rela- 
 tion, is obviously imperfect, and wants the second term. Put 
 therefore, ?i"'=«?z^-f./3w ; and, by substitution, 2«w^-{-6««w-f- 
 2/3w=2fitw^-f-2/3w— w. Equating now the corresponding terms, 
 and 6«=— I, or «= — ^; but <«-f-/3=o, and therefore /3=-}-|.. 
 
 /// w' — I 
 
 Whence n =.=— |.?z3-|- ^n =— m.-^^ • 
 
 . iiiii mil 'III! m 4 
 
 3. Again, m the third equation, (w+ 1) -t-(w— i)=2w--w — Jw, 
 
 mil 
 
 substitute ?z=«n5-{-/3«3-t-yn, and the conditions of the fifth or- 
 der of the function of n will be determined by this compound 
 expression : 2xn^ -f (20« -f- 2^)n3+ ( 1 0^ -f 6/3-}- 2y)» = 2«»^-f- 
 (2|3-t-|)n3-t- (2y — I- — t)w. Equate the corresponding terms, 
 
 and 20«-t- 2/3= 2/34- 1., or «=—=:—__ In like manner, 
 10^-1-6^4-27=27—^—;^, and ^= — y^— ^— ^V= —^ = 
 
 1Q 9 
 
 ^^^^^ ; but ^+/3+ 7 = 0, whence 7=53^5 . . : . . ; .. 
 
372 NOTES AND ILLUSTRATIONS. 
 
 Collectively, therefore, r=^^-^Q^^+g^^=;,;!^^.^!=zL^ 
 ^ 2.3.4..5 • 2.3 4.5 
 
 Whence, resuming all the terms, sin na=:ns — w. - ""-js ^ 
 
 «*— 1 w*i-9 . ,' 
 
 «• a g — . rv-s — &c. as before. 
 
 From the expression for the sine of a multiple arc, may be 
 deduced the series for the sine of any arc, in terms of the arc 
 
 itself, and conversely. Let na=A, and therefore ar=— ; if « 
 
 n 
 
 be supposed indefinitely great, then a must be indefinitely 
 
 sniall, and consequently in a ratio of equality to s. Whence, 
 
 A 
 
 substituting A for na, and — for s in the general expression, 
 
 there results, 5f«A==A---^!=^A*+^!=L^!^!=I?.:^_&c. 
 
 .^,3 n*^ 2,3 4.5 n* 
 
 But n being indefinitely great, the composite fractions — ^— 
 &c. are each in effect equal to unit, which forms their 
 
 .a ' 
 
 extreme limit. Consequently, assuming that modification. 
 
 Again, putting a=r A and 5=8, suppose n to be indefinitely 
 small, and sinnaz=:naz=:nA ; whence, by substitution, 
 
 «A = »S-«.'il=i S3+ «. — .'-^ S=_, &c. and 
 A=S - !^ S^+ n."-!^'!^^^ S5-&C. 
 
 2.3 ' 2.3 4.5 
 
 But, if 72 vanish from all the terms, the series will pass into 
 a simpler form. 
 
 A= S'+ ^-S'+ll. S5+ -il^-?^ S7 + , &c. 
 
 ' 2.3 ^ 2.3.4.5 ^ 2.3.4.5.6.7 ' 
 
 By a similar investigation, the series for the cosine of an arc 
 is likewise found. 
 
 ^ 1/2^2.3.4 2.3.4.5.6 "* 
 
NOTES AND ILLUSTRATIONS. 
 
 373 
 
 These series' are very commodious for the calculation 6f 
 sines, since they converge with sufficient rapidity when the 
 arc is not a large portion of the quadrant. Though the method 
 explained in the text is on the whole much simpler, yet as the 
 errors of computation are thereby unavoidably accumulated, 
 it would be proper at intervals to calculate certain of the sines 
 by an independent process. 
 
 The series' now given furnish also various modes for the rec- 
 tification of the circle. Thus, assuming an arc equal to the 
 
 radius, its sine is, 1- 
 
 1 
 
 1 
 
 ^— &c. =.84<147l, and its 
 2.S ' 2.3.4^.5 
 
 cosine is, 1 — - i -i—&c.= 440302. But that arc evidently 
 2 '2.3.4 
 
 approaches to 60*, of which the sine is \/l=.866025, and the 
 cosine .500000. Wherefore (Pr. 1. Trig.) the sine of the dif- 
 ference of these two arcs is .866045 X -540302— .81.1471 X 
 .500000= .047 1 8, and consequently, by the series, that inter- 
 val itself is .0472. Hence the length of the arc of 60° is 
 1.0472, and the circumference of a circle which has unit for 
 its diameter is 3 X 1.0472=3.1416; an approximation extreme- 
 ly convenient. 
 
 5. The Fifth Proposition may be otherwise demonstrated from 
 the corollaries at p, 363. 
 
 Let AB and BC, or BC, be two 
 arcs, of which AB is the greater ; 
 make AD, or AD', equal to BC, 
 and apply the respective tangents. 
 Because OAE is a right-angled tri- 
 angle, and OG', OF, are drawn, 
 making equal angles with OA and 
 OE, it follows, that OA^— AE.AG': 
 O A^ : : EG' : AF, and consequently 
 R^^tanAB . tanBC : R^ : : 
 tanAB + tanBC : tan{AB -|- BC). 
 Again, since OG and OF' make 
 equal angles with OA and OE, it is 
 
37i NOTES AND ILLUSTRATIONS. 
 
 evident that OA^+AE.AG ; OA* : : EG : AF\ and hence 
 R*+ianABtanBC :1V:'. tanAB-^tanBC : ^an(AB— -BC). 
 
 6. Tlie radius being expressed by unit, the sum of the tan- 
 gents of the angles of any triangle is equal to the number 
 arising from their continued product. For, let A, B, and C, 
 denote the several angles of the triangle ; and since two of 
 these, such as A and B, are supplementary to the remaining 
 one C, the tangent of A4-B is the same (schol. def. Trig.) as 
 that of the third angle in an opposite direction. Whence 
 
 - — ^ "v. ^ — n = — tanC, and therefore ian A 4- ta7i B = 
 
 — tan C -\- tan A tan B tan C, or tan A -|- tan B 4- i<^^ C = 
 tan A tan B tan C. 
 
 7« The properties of the tangents are easily derived fram 
 those of the sine&, 
 
 -. m » . T> sin A , siiiB sinAcosB-X-cosAsinB 
 
 1. TanA-^-tanBzz ^^4- — 5= x- — ^ = 
 
 ' cos A* cosB cosAcosB 
 
 (art. 1. N0.3.)*-^<4±^'. 
 
 ' cosAcoso 
 
 2. Change the sign of B in the last article, and tanA — tanBzz 
 sin{A—B) 
 
 cosAcosB 
 
 3. Instead of A and B in art. I. substitute their complements, 
 
 andco«A+.o/B = £i'?i^. 
 
 * smAsmp 
 
 4. Make the same substitution in art. 2, and cot B — cot A-=: 
 sin {A — B) 
 
 sinAsinB 
 
 5. Tan(A + B) = ^1(1^ '=^^''' ^' "^^ *' ^^^ ^'^ 
 
 sinAcosB+cosAsinB ^ ^^. ^.^.^^^ ^ ^^^ ^ ^^^g ^^, 
 
 cosAcosB — smAstnB » o 
 
 , . ^ . , A . T.% tanA-^tanB cotB-\-cotA 
 
 dnAsznB.giyes tan(A+B)= i^^^^A^anB^co^ co^A-l* 
 
 6. Change the sign of B in the last article, and /o«(A— B)=: 
 
 fan A — tanB _ cotB — cotA 
 IJ^tanAtanB "" cotB cot A + V 
 
NOTES AND ILLUSTRATIONS. 375 
 
 7. Divide the expression in the first article by that in the se- 
 
 , , sin( A+B) tanA-^-tanB cotB + cotA 
 
 cond, and . . ' p {= ^ a ^ — 15= ^i> — rx- 
 
 sin{A — B) tan A — tanB cutB — cotA 
 
 8. In the last article, change the sign of B, and instead of A 
 
 • , . . , ^ , cos{A+ B) cot B — tan A 
 
 take Its complement, and — /a pv = . t. . . r> = 
 ^ ' cos{ A-^B) cot B 4- ^aw B 
 
 cotA — ta7iB 
 
 cot A-\- tanB' 
 
 9. Divide the expression of art. 12. NO. 3. by that of art. 5., 
 
 1 — co*2A__ 2s2mA* ___sinA__ 
 sin2A "^^sinAcosA cos A "" 
 
 10. Divide the expression of art. 5. in the same number, by 
 
 ^u . c .10 J sin2A 2sinAcosA sin A ^ ^ 
 that of art. 13. and =-~ — .rA = o a*— = — r=^«wA. 
 1 -|- C052A 2co5 A* cos A 
 
 11. Multiply the expressions of the tviro preceding articles, 
 
 - 1 — cos2 A , A 2 ^ A / 1 — C052A 
 
 12. Decompose the expression in art. 9., and tanAzz . — 
 
 ^.J^ = cosec2A— co^2A. 
 stn2A 
 
 13. In the last article, change A into its complement, and 
 co^ A= coscc2A + co/2 A. 
 
 14. Subtract the last expression from the one preceding it, 
 and tanA — cotA= — 2co^2A, or tanAzzcotA — 2co^2A. 
 
 15. In art. 9, 10, and 11, for 2A and A, take A and ^A, and 
 
 i4 — 1 — cosA __ sin A _ / i — cos A 
 tan^A^ siwA ""l+cosA""V i-f-cosA* 
 
 16. Multiply the expressions of art. 1. and 2., and(fawA+ifawB) 
 (tanA^tanB):^tanA^^tanB^= ^^L^^±^^^^, 
 
 17. Multiply the expressions of art. 3. and 4., and vco^B+co^A) 
 
 .A, sinlA—B) sinAJfB) 
 {cotB^cotA) = cotB^^otA'= -^IfH^SB^^^- 
 
 18. Divide art. 28. of NO. 3. by art. 29., and^f]^|±^ = 
 
 2sin\ ( A + B ) cos\ ( A—B ) _ tanl{ A+B) 
 2coi\[ A 4- B ) sin\ ( A—B ) "tan^ ( A—B f 
 
376 NOTES AND ILLUSTRATION'S. 
 
 19. Divide art. 30. of the same NO. by art. 31 ., and ^?^2±£^ - 
 
 cosB — cos A. 
 
 2cosl{A + B) cq4( A~- B)_ cotl ( A+ B ) 
 
 2sinl{A + B) ;s2«i( A—B)"" ^a«i( A— B ) * 
 
 Since by art. 14. cot A — 2cot2A=ztanA, if the arc A and its 
 compound expression be continually bisected, there will arise : 
 \dot\A — cot Azz^tan^A 
 fcotfA-^cot^A=z-^tanfA 
 ^cot^ A— ~co/;^A=|^awf A 
 &c. &c. &c. 
 Wherefore, collecting these successive terms, and observing the 
 effects of the opposite signs, the general result will come out, 
 
 1 A 1 A 
 
 ^^J^ot—^^cotAzzlianlA+^tanfA+^Jan^,A . . . + -^ian~. 
 
 If n be supposed to become indefinitely large, then 
 
 1 ,A 1 1 . „. . , 1 1 2» 1 
 
 — • cot ^^zz 
 
 ¥"=¥'— li '' "^^'"^"'^^^ 2^- A =T«- A' ^'^ A ' 
 
 tan 
 
 2" 2" 
 
 and consequently j- zzcotA-^-ltanlA-i-ftan^A-i- Ifan^A ^ 
 
 ^ya«^A+&c. 
 
 This neat and very simple investigation is given in the se- 
 cond French edition of Cagnoli's Trigonometry, printed at 
 Paris in 1800, and forming the completest treatise which has 
 yet appeared on the subject. It was also, and nearly about 
 the same time, communicated by my friend Mr Wallace of the 
 Royal Military College at Sandhurst, a geometer of the first 
 order, to the Royal Society of Edinburgh ; another instance 
 of that accidental coincidence which has occurred so frequent- 
 ly in the history of mathematical discovery. 
 
 8. It is obvious that the terms of the series for the tangent 
 of the multiple arc are formed out of the coefficients of the 
 powers of a binomial. Wherefore, 
 
NOTES AND ILLUSTRATIONS. 377 
 
 {1.)Tanna-- ^ ^ 
 
 n^—i , n — I n — 2 n — 5^, 
 i—n 1^ J- n, -7:- •— — -/*— &c. 
 
 2 2 3 4 
 
 Hence also, 
 
 „ V „. , . w — I n — 2,„ , ?i — I w— 2?i — 8 n — 4«, 
 (8.) Szw na=:cnfnt-^7i.-- t^-{.Ji,—~ ;: — . /o — 
 
 ^ ^ 2 3 2 o 4 • 5 
 
 &c.) and 
 
 (9.) Cos na = c«(l — n. /* + w. . . /4— 
 
 ^ — ^ J^T^. ^ — 3 ^ 7?— 4 « — 5 .C o N 
 
 "• 2 3*4*5^6 *^ 
 
 9. The series for the tangent in terms of the arc, is easily de- 
 rived, by the theory of functions, from the expression of the 
 
 tangent of the double arc. Since ^ffw2a=: ^ =2^4- 2/3 j. 
 
 I t 
 
 n^ 4- &c. Put tzza-{- Aa3 ^ Ba^ + &c., and, by substitu- 
 tion, tan2a zz 2a -^ sAa^ + 32Ba^ -f &c.= 2a + (2A + 2)a3^ 
 (2B + 6A -J- 2)a^-}-, &c. Equating, therefore, the corre- 
 sponding terms, we obtain, sA = 2A + 2, or A = i, and 
 52B = 2B + 6A + 2, or soB = 4, and B = ^. Whence, 
 in general, /«»a=a-J-§-a3-|,^a^ &c. Again, revert this se- 
 ries, and az=t--^i^+j.i^-^^t^+&c. 
 
 The last series affords the most expeditious mode for the 
 rectification of the circle. Assume an arc a, whose tangent t 
 
 is one-fifth part of the radius, and tmiazz — J^ — 1~? • 
 ^ 1—6/^+^* ~" 119* 
 
 consequently (Prop. 5. Trig.) tajt (4a — 45°) = -i-=r 
 
 .004,184,100,418. Wherefore, computing the terms of the 
 series, ^=.197,395,559,850, and 4a=.789,582,239,400. In 
 like manner, we find 4«--45°=. 004, 184,076,000, and hence 
 the difference between these values, or .785,398,1634 exhibits 
 the length of the octant ; which number, multiplied by 4, 
 gives 3,1415926536 for the circumference of a circle whose 
 diameter is 1 . 
 
378 NOTES AN1> ILLUSTItATIONS. 
 
 10. Proposition sixth, with its corollaries, would furnish a 
 simple quadrature of the circle. The sine of a semiarc being 
 equal to half the chord, it follows that the ratio of an arc to 
 its chord is compounded of the successive ratios of the radius 
 to the cosines of the continued bisections of half that arc. 
 Assuming therefore the arc of 60°, whose chord is equal to the 
 radius, the logarithm of the ratio of the circumference of a 
 circle to its diameter will be thus computed : 
 
 Arith. comp. log. Cos 15° = .0150562219 
 
 Co* 7* 30' = .0037314339, 
 
 Cos 3° 45' = .0009308547 
 
 Cos 1° 52' SO" = .0002325891 
 
 Cos 0° 56' 15" = .0000581395 
 
 Cos 0° 28' 7i" = .0000145344 
 
 One-iliird of the last term, ~ .0000048448 
 
 Logarithm of 3 . = .477] 212547 
 
 .4971493730, which 
 exceeds only by 3 in the last place the logarithm of 
 3,l4l59i654. As the successive terms come to form very 
 nearly a progression that descends by quotients of 4, the third 
 of the last one is, for the reason stated in page 245, consider- 
 ed as equal to the result of the continued addition. 
 
 » 11. An elegant mode of forming the approximate sines cor- 
 responding to any division of the quadrant, may be derived 
 from the principles stated in the calculation of trigonome- 
 tric lines : For the successive differences of the sines for the 
 arcs A — B, A, and A+ B, are sin A. — sin(A — B,) and 
 5fw (A 4. B) — sin A; and consequently the difference be- 
 tween these again, or the second diflference of the sines, is 
 sin(A 4. B) 4- 5m( A — B)~2«V/A=(Prop. 3. cor. 3. Trig.) -* 
 2versBsinA, The second differences of the progressive sines 
 are hence subtractive, and always proportional to the sines 
 themselves. Wherefore the sines may be deduced from their 
 second differences, by reversing the usual process, and recom- 
 pounding their separate elements. Thus, the sines of A~E 
 
NOTES AND ILLUSTRATIONS. 379 
 
 and A being already known, their second and descending dif- 
 ference, as it is thus derived from the sine of A, will combine 
 to form the succeeding sine of A+ E, which ia—^versBsinA^ 
 (sinA — sin(A — B) ) -f- sinA. It only remains then, to deter- 
 mine, in any trigonometrical system, the constant multiplier 
 of the sine, or twice the versed sine of the component arc. 
 Suppose the quadrant to be divided into 24 equal parts, each 
 containing 3° 45' or 225'. The length of this arc is nearly 
 
 OO 111 .11 
 
 ^ •4o='|^» ^"^ consequently twice its versed sine =(t^)' — 
 ( __) in approximate terms. If the successive sines, corre- 
 
 Zoo 
 
 sponding to the division of the quadrant into 24? equal parts, 
 
 be therefore continually multiplied by the fraction —-, or di- 
 
 vided by the number 233, the quotients thence arising will 
 represent their second differences. But, since 233 is nearly 
 equal to 225, or the length in minutes of the primary or com- 
 ponent arc, and which differs not sensibly from its sine, — this 
 last may be assumed as the divisor, the small aberration so 
 produced being corrected by deferring the integral quotients. 
 In this way the following Table is constructed. 
 
 It will be seen that the number 225, which expresses the 
 length of the component arc, and therefore represents very near- 
 ly its sine, is here employed as the constant divisor. Thus, 225, 
 divided by 225, gives a quotient 1 ; and this, subtracted from 225 
 leaves 224, which, being joined to 225, forms 449, the sine of 
 the second arc. Again, 449 divided by 225, gives 2 for its in- 
 tegral quotient, which taken from 224, leaves 222 ; and this, 
 added to 449, makes 671, the sine of the third arc. In this 
 way, the sines are successively formed, till the quadrant is 
 completed. The integral quotients, howev«er, are deferred ; 
 that is, the nearest whole number in advance is not always ta- 
 
 38 
 ken. Thus the quotient of 1315 by 225, is 5 — , which ap- 
 
 preaches nearer to 6, and yet 5 is still retained. These ef- 
 forts to redress the errors of computation are marked with 
 ■ asterisks. 
 
580 
 
 NOTES AND ILLUSTRATIONS. 
 
 Parts of the 
 quadrant. 
 
 Arcs. 
 
 Sines. 
 
 IstDiff. 
 
 2d Diff. 
 
 Arcs. 
 
 1 
 
 225' 
 
 225 
 
 224 
 
 1 
 
 3" 
 
 45' 
 
 r» 
 
 450 
 
 449 
 
 222 
 
 2 
 
 7 
 
 30 
 
 3 
 
 675 
 
 671 
 
 219 
 
 3 
 
 11 
 
 \5 
 
 4 
 
 900 
 
 890 
 
 -215 
 
 4 
 
 15 
 
 
 
 5 
 
 1125 
 
 1105 
 
 210 
 
 5 
 
 18 
 
 45 
 
 0^ 
 
 1350 
 
 1315 
 
 205 
 
 ♦ 5 
 
 22 
 
 30 
 
 7 
 
 1575 
 
 1520 
 
 199 
 
 6 
 
 26 
 
 15 
 
 8 
 
 1 800 
 
 1719 
 
 191 
 
 * 7 
 
 30 
 
 
 
 9 
 
 2025 
 
 1910 
 
 183 
 
 8 
 
 33 
 
 45 
 
 JO 
 
 2250 
 
 2093 
 
 174 
 
 9 
 
 37 
 
 30 
 
 1 1 
 
 2475 
 
 2267 
 
 164 
 
 10 
 
 41 
 
 15 
 
 12 
 
 2700 
 
 2431 
 
 154 
 
 * 10 
 
 45 
 
 
 
 13 
 
 2925 
 
 2585 
 
 143 
 
 11 
 
 48 
 
 45 
 
 14 
 
 3150 
 
 2728 
 
 131 
 
 12 
 
 52 
 
 30 
 
 15 
 
 3375 
 
 2859 
 
 119 
 
 * 12 
 
 56 
 
 15 
 
 ]6 
 
 3600 
 
 2978 
 
 106 
 
 13 
 
 60 
 
 
 
 17 
 
 3825 
 
 3084 
 
 93 
 
 13 
 
 63 
 
 45 
 
 18 
 
 4050 
 
 3177 
 
 79 
 
 14 
 
 67 
 
 30 
 
 T9 
 
 4275 
 
 3256 
 
 65 
 
 14 
 
 71 
 
 \5 
 
 20 
 
 4500 
 
 3321 
 
 51 
 
 14 
 
 75 
 
 
 
 21 
 
 4725 
 
 3372 
 
 37 
 
 * 14 
 
 78 
 
 45 
 
 22 
 
 4950 
 
 3409 
 
 22 
 
 15 
 
 82 
 
 30 
 
 23 
 
 5175 
 
 3431 
 
 7 
 
 13 
 
 86 
 
 15 
 
 1 - 
 
 5400 
 
 3438 
 
 
 
 15 
 
 ( 
 
 90 
 
 
 
 Each of the three composite columns, we may observe, 
 really forms a recurring series. In the second quadrant, the 
 first differences become subtractive, and the same numbers 
 for the sines are repeated in an inverted order. By continuing 
 the process, these sines are reproduced in the third and fourth 
 quadrants, only on the opposite side. 
 
 Such is the detailed explication of that very ingenious mode, 
 which, in certain cases, the Hindu astronomers employ, for 
 constructing the table of approximate sines. But, ignorant 
 totally of the principles of the operation, those humble calcu- 
 lators are content to follow blindly a slavish routine. The 
 Brahmins must, therefore, have derived such information from 
 people farther advanced than themselves in science, and of a 
 
NOTES AND ILLUSTKATIONS. 861 
 
 bolder and more inventive genius. Whatever may be the pre- 
 tensions of that passive race, their knowledge of trigonometri- 
 cal computation has no solid claim to any high antiquity. It 
 was probably, before the revival of letters in Europe, carried 
 to the East, by the tide of victory. The natives of Hindustan 
 Baight receive instruction from the Persian astronomers, who 
 were themselves taught by the Greeks of Constantinople, and 
 stimulated to those scientific pursuits by the skill and liberali- 
 ty of their Arabian conquerors — This opinion seems to derive 
 strong confirmation from the Lilawati, a very meagre and de- 
 fective practical treatise of arithmetic and geometry, which I 
 had some time since an opportunity of examining, with the 
 kind assistance of the learned Dr Wilkins, at the library of the 
 India House. Of that singular performance, a translation from 
 the original Sanscrit by Dr John Taylor, printed at the ex- 
 pence of the Literary Society at Bpmba}', has just reached us, 
 and will enable the European mathematicians, who are ac- 
 quainted with the state of science at the revival of letters in 
 Italy, to reduce the lofty pretensions of the Brahmins to their 
 just level. They will perceive the utter nakedness of a sys- 
 tem, which, in the language of ignorance and oriental exagge- 
 ration, the Hindus represented as endued with a sort of magical 
 virtue, that would enable the person who understands it " to 
 tell, in the twinkling of an eye, the number of leaves on a tree, 
 or of blades of grass in a meadow, or the number of grains of 
 sand on the sea shore." 
 
 The principles before stated lead to an elegant construction 
 of the approximate sines, entirely adapted to the decimal scale 
 pf numeration, and the nautical division of the circle. Suppose 
 a quadrant to contain 16 equal parts, or half points ; the length 
 
 22 111 
 
 of each arc is nearly— ^•^:;—= — -, and consequently twice its 
 ver&ed sine is( — )*, or, in round numbers, — , It will be 
 
 1 1-^ 105 
 
 sufficiently accurate, therefore, to employ 100 for the constant 
 divisor. The sine of the first being likewise expressed by 100,^ 
 let the nearer integral quotients be always retained, and the 
 sine of the whole quadrant, or the radius itself, will come out 
 
S82 
 
 NOTES AND ILLUSTRATIONS. 
 
 exactly 1000. The first term being divided by 100 gives 1 for 
 the second difference, which, subtracted from 100, leaves 99 
 for the first diflference, and this joined to 100, fornns the second 
 term. Again, dividing 199 by 100, the quotient 2 is the se- 
 cond difference, which, taken from 99, leaves 97 for the first 
 difference, and this added to 199, gives the third term. In 
 like manner, the rest of thq terms are found. 
 
 Half 
 
 Arcs. 
 
 Sines. 
 
 IstDiff. 
 
 2d Diff. 
 
 Excess. 
 
 Correct 
 
 points. 
 
 
 
 
 
 
 Sines. 
 
 1 
 
 5« 3?i' 
 
 100 
 
 99 
 
 1 
 
 3 
 
 97 
 
 2 
 
 11 15 
 
 199 
 
 97 
 
 2 
 
 4 
 
 195 
 
 3 
 
 16 521 
 
 296 
 
 94 
 
 3 
 
 5 
 
 291 
 
 4 
 
 22 30 
 
 390 
 
 90 
 
 4 
 
 6 
 
 384 
 
 5 
 
 28 7k 
 
 480 
 
 85 
 
 5 
 
 7 
 
 473 
 
 6 
 
 33 45 
 
 565 
 
 79 
 
 6 
 
 8 
 
 557 
 
 7 
 
 39 22i 
 
 644 
 
 73 
 
 6 
 
 9 
 
 635 ^ 
 
 8 
 
 45 00 
 
 717 
 
 66 
 
 7 
 
 10 
 
 707 
 
 9 
 
 50 37 1 
 
 783 
 
 58 
 
 8 
 
 9 
 
 774. 
 
 10 
 
 56 15 
 
 841 
 
 50 
 
 8 
 
 8 
 
 833 
 
 11 
 
 61 52| 
 
 891 
 
 41 
 
 9 
 
 7 
 
 884 
 
 12 
 
 67 30 
 
 9^^ 
 
 32 
 
 9 
 
 6 
 
 9'^6 
 
 13 
 
 73 7h 
 
 964 
 
 22 
 
 10 
 
 5 
 
 959 
 
 14 
 
 78 45 
 
 986 
 
 12 
 
 10 
 
 4 
 
 982 
 
 15 
 
 84 22| 
 
 998 
 
 22 
 
 
 3 
 
 995 
 
 16 
 
 90 00 
 
 1000 
 
 
 
 
 
 The errors occasioned by neglecting the fractions accumu- 
 late at first, but afterwards gradually diminish, from the effect 
 of compensation. The greatest deviation takes place, as might 
 be expected, at the middle arc, whose sine is 707 instead of 
 7l7. Reckoning the error in excess as limited by lO, and de- 
 clining uniformly on each side, the correct sines are finally de- 
 duced. The numbers thus obtained seldom differ, by the 
 thousandth part, from the truth, and are hence far more accu- 
 rate than the practice of navigation ever requires. This sim- 
 ple and expeditious mode of forming the sines is not merely 
 an object of curiosity, but may be deemed of very consider- 
 
NOTES AND ILLUSTRATIONS, 383 
 
 able importance, as it will enable the mariner, altogether in* 
 dependent of the aid of books, to the loss of which he is often 
 exposed by the hazards of the sea, to construct a table of de-- 
 parture and difference of latitude, sufficiently accurate for every 
 real purpose. 
 
 12. In trigonometrical investigations, it is often requisite to 
 determine the proportion which the difference of an arc bears 
 to that of its related lines. With this view, let A denote the 
 increment or finite difference of the quantity to which it is 
 prefixed. 
 
 1. In art. 29. of NO. S. change A into A+AA, and B into 
 A ; then will 
 
 AsinA=9,si?i\6,Acos{^PsL-\-\di.Ay 
 
 2. Make the same change in art. 31. of that number, and 
 AcosA= — 25mJ^AA.5m(A4-iAA), 
 
 3. In art. 2. of NO. 7. let a similar change be made, and 
 
 .. . sinAA 
 
 AtanAzz . 
 
 cosAcos{A-{-AA) 
 
 4. Do the same thing in art. 4>. and 
 
 . . sin A A 
 
 AcotA= 
 
 sinAsin{A-\. aA) 
 
 5. In art 22. of NO. 3. make a like substitution, and 
 AsinA^ = sinAAsin(2A + a A). 
 
 6. Let the same change be made in art 23., and 
 Aco5A*= — sinAAsin{2A-\- A A). 
 
 7. Do the same thing in art. 16. of NO. 7. and 
 ^ j__5i«A A(5m2 A-f A A^ 
 
 '~ cos A^ cos A + aA)* 
 
 8. Lastly, let a similar change be made in art. 17. of that 
 
 number, and 
 
 sin A A( sin 2 A 4-^ A) 
 AcofA^ = : ^^ = • 
 
 stnA^sin{A~\-AAy- 
 
 If the differences be conceived to diminish indefinitely and 
 pass into differentials, these expressions, in coming to denote 
 only limiting ratios, will drop their excrescences and acquire 
 a much simpler form. Thus, adopting the characteristic dj 
 
33i NOTES AND ILLUSTEATIONS. 
 
 t 
 ^Ince the ratio of an arc to its sine is ultimately that of equa- 
 lity, and the sine of A+g?A may be considered as the same 
 with the sine of A ; it follows, that 
 
 1. d smA—-\~cosAdA, 
 
 t2. d cosA=z — sin Ad A. 
 
 3. dtanA=+.^^ . 
 
 cosA^ 
 
 4. d cot A =— 
 
 StnA^ .; , ; 
 
 5. d, sinA^ — -f- 2sinAcosAdA^ 
 
 6. dcosA^zz — 2sinAcosAdA. 
 2tanAdA 
 
 7. dtanA^=z^ 
 
 S. dcotA^zz^. 
 
 cosA"- 
 
 2cotAdA 
 
 sinA^ 
 
 ■J 3. Since, by NO. 12. d si?iA=cosAdA, or the variation of 
 the sine of an arc is proportional to its cosine ; it follows that, 
 near the termination of the quadrant, the slightest alteration 
 in the value of a sine would occasion a material change in the 
 
 arc itself. Again, from the same Note, d tanA= — —-, or the 
 
 cosA"- 
 
 Tariation of the tangept is inversely as the square of the cosine, 
 
 and must therefore increase with extreme rapidity as the are 
 
 approaches to a quadrant. 
 
 14.. It is convenient to reduce the solution of triangles to al- 
 gebraic Jbrmulce. Let a, h and c denote the sides of any plane 
 triangle, and A, B, and C their opposite angles. The various 
 relations which connect these quantities may all be derived 
 from the application of Prop. 11. 
 
 o. But, since (art. 16. NO. 3.) sin\A.'-=i\{i—cosA\ it fol- 
 lows, by substitution, that «n|A'=f^£Z^!:z£!±^ 
 
 - — }-. — ^=:L_L 11 x_/ and therefore, s denoting the 
 
NOTES AND ILLUSTRATIONS. 385 
 
 setniperimeter, Sm|A* = ( f— M^— g) . ^j^jch corresponds to 
 
 Prop. 14. 
 
 3. Again, because (art. 17. Note 3.) cos\A^z=\{l+cosA), 
 
 by substitution, cos^A' = ^ ^^ -=^— ^^^ = 
 
 V. \ -r ^''^^^ ^ ' ^ % and consequently 
 
 Co5|A'=: ^-T — ^; which agrees with Prop. 13. 
 
 4. The second expression being now divided by the third, 
 gives tan^A"^ =4 ^-^ — ', corresponding to Prop. 12. • 
 
 These are the Jbrmulce wanted for the solution of the first 
 case of oblique-angled triangles. To obtain the rest, another 
 transformation is required. 
 
 5. It is manifest that sinA'zz 1~cq5A'= ^^'<^'— (^'+g'— «')^ 
 
 4T* 
 
 and consequently, by Note 5. Book VI., sinA'^ = -—-^ or 
 
 o c , 
 
 2T 2T 
 
 sin A z= -7- • Fcg: the same reason, sinB = — , and hence 
 be ac 
 
 ^?^=-? ; which corresponds to Prop. 9. 
 sino b 
 
 6. Again, by composition. '^^^= |=|. and therefore, 
 
 by art. 18. Note 7- 
 
 a^ tani{A-^ B) ^^^^^^ ^ ^j^j^ p ^ 
 
 a+b tan^{A+B)' ^ * 
 
 7. Lastly, transforming the first expression, there results, 
 a = ^(b^+c^-^2bc co5A)=>/( (^— c)*4-2^c ver^A) 
 
 = >/ ( (b+cy^2bc{l-\-cosA)). 
 
 The iprecedmg Jbrmulce will solve all the cases in plane tri- 
 gonometry ; but, by certain modifications, they may be some- 
 times better adapted for logarithmic calculation. 
 
 2c 
 
836 NOTES AND ILLUSTRATIONS. 
 
 8. Divide the terms of art. 6. by a, and ?^i(^~^)= ± . 
 
 !etA=: tanx, andf^-t^^l^j-f ^ ^^ = (art. 6. Ko, 7.) 
 
 ^aw (45°-^^). Wherefore, — zz tan x, and tan(^5° — x) = 
 
 ianlC tanl(A—B)z=tanlC cot(\C-\-B)= 
 ta7i\C(—cot(\C+A)). 
 
 9. Again, from art. 7. « = >/ ( (^^-tc)*+2^c versA)zs 
 
 2bc 
 (h — <?)'V^(1 + /T — w. wr^A) ; consequently find tan xzs 
 
 X^ — -y' vers A-zz^-r — dn\Ay and a •=. (h — c) secxzz^^^^^. 
 
 C C COS X 
 
 10. But the expression in art. 1., bya different decomposition, 
 
 2bc 
 ^ivesa=:-\/((^+c'-26c5Mver5A))=:(5+0V'(^""7AX~T^ ^"*'^^*^) i 
 
 wherefore find sin x — X-, — VsmersA =r 2 ~- cos\Ai and 
 
 11. Other expressions are likewise occasionally used. Thus, 
 by art. 1., 2bc.cosA-=zb^-\-c^ — a*, or c' — 2bc»cosAzza^ — 5*, and, 
 solving this quadratic, we obtain czz b cos A z±z \/ (a* — 6^4- 
 b^cosA^) zz b cosAz±ZA/(a''—b'- sinA^, or c=^ cosAztz\/{{a+b 
 sinA){a-'bsinA) ). When two sides and an angle opposite to 
 one of them are given, the third side is thus found by a direct 
 process. 
 
 12. From art. 5., c=za-^-x ; but C being a supplementary 
 
 angle, its sine is the same as that of A-f-B, and consequently 
 
 .sinAcosB^cosAsinB. t, • -i . z* 
 c = a{ — J-^ ^ ). By a similar transformation, 
 
 __ sinC __ , sinC v__ a 
 
 "" A7«(B+C)"" ^sinBcosC-{-cosBsinC''^cosB'^-smBcotC' 
 
 ;S. Lastly, from art. 3. of Note 7, cotA+coiC = -J~^t§^^ 
 
» NOTES AND ILLUSTRATIONS. 387 
 
 -, . ^f ^ ■=: -L-y and therefore cotX = -^—^01 C = 
 
 b — a cosO . m asinQ, , 
 
 —7^', or tan A =r , r-^^i 
 
 a smC — a un\j 
 
 If the angle A be assumed equal to 90°, the preceding^or- 
 wm/<^ will become restricted to the solution of right-angled 
 triangles. 
 
 14. From art. 1., co*A= 0= -"H^'"~^'; whence, a* =i*+c% 
 
 which expresses the radical property of the right-angled tri- 
 angle. 
 
 15. From art. 5.,-S^ = — , and consequently sin B = — , 
 
 which corresponds with Prop. 7. 
 
 ,_ . . ^ ^, . , b sinB sinB , 
 
 16. Agam, from the same article, — = . ^ = — 5, ana 
 
 ° c stnC cosB 
 
 therefore tan B =: — = cotC, 
 c 
 
 For the convenience of computing with logarithms, other 
 expressions may be produced. 
 
 17. Thus, from art. 14., i*=a'— c*, and hence 
 bz=^/((a+c){a^c)). 
 
 c* c 
 
 18. Since a^=zb* (1— -^), put j- zztan x, and a=.b(sec c)=: 
 
 b 
 
 cosx' 
 
 19: Lastly, because 5* = a* (I4. — j-), put — = sin x, and 
 
 6 =: a.cos x. 
 
 Besides the regular cases in the solution of triangles, other 
 combinations of a more intricate kind sometimes occur in 
 practice. It will suffice here to notice the most remarkable 
 of these varieties. 
 
 20. Thus, suppose a side, with its opposite angle and the 
 sum or difference of the containing sides, were given, to de« 
 
388 NOTES AND ILLUSTRATIONS. 
 
 termine the triangle. By art. 5., a=^4^ = ^^, '^and 
 
 therefore ^=-^0^^^^ = ^ sM+sinC = ^'''' ^' ^"^ 
 
 18 Notes ) (^ + 02.^'^MB+C)^q4(B+C) _ (^+c)cq4(B+C) 
 '' 2sin^{B + C)cosl{B--.C] "" co^^B— C) ' 
 
 "But co5i(B+ C)=sinlAy and hence co4(B— C)='i^+£M^ ; 
 
 and the difference of the supplementary angles B and C be- 
 ing known, these angles themselves are hence found. 
 
 In like manner, it will be found that sinl{B-^C)=: {lf—c)cos\A^ 
 
 a 
 21. Let a side with its adjacent angle and the sum of the 
 other sides be given, to determine the triangle. By art. 4. 
 
 ianlA'' = ^ -'~~^— and tan |B*= n ""^~^ . whence ^a«|A* 
 s.s — a s,s — 6 
 
 tanlB^ = £- — _^l ^ > and consequently faw|A^«w|B= 
 
 Again by art. 1., 2^cco5A=: 5^ + c* — a% or a* — b^ — c*= 
 — 2bc.cosA, and adding 2a6 + 26^ to both sides, a^ + 2ab + 
 g=— c*=2tti4-25'— 25c.co5A, or {a-\-b)*—c^z;::2b(a-\-b—c.cosA) ; 
 whence ( {«+ ^) -j- c)((a + b) — c) = 26(a-f-i — c.cosA), and 
 
 "" * {a-{-b)—c.cosA 
 If the sign of b be changed, and the supplement of its adja- 
 cent angle therefore assumed, we shall obtain 
 
 * * c — {a — b) c.cosA — [a — b) 
 
 The relation of the sides and angles of a triangle might also 
 be in some cases conveniently expressed by a converging se- 
 
 ries, 
 
 ^, b smB sinB sinB 
 
 Thus — =-7— r =r 
 
 a sin A sin{B-\-C) sinBcosC-{- cosBsinC 
 and consequently b sinB cosC -f- b cos B sin C = a sin B, or 
 
 bstnC ^^J!!^^tanB, Wherefore, by actual division, ^««B= 
 
 'b cosC cosB' 
 
NOTES AND ILLUSTRAf IONS. 389 
 
 —siiiC -i — rSinC cosC H — -smCcosC* + —TsinCcosC^-\-&c.; 
 and, in substituting the powers of this expression for those of 
 the tangent in the series of Note 9., we obtain B= — smC-{- 
 
 ^sinC cosC +5^3(4co5C'— 1 ) sinC + -^{ 2 cos C -^ I) sin C 
 
 h b^ h'^ h* 
 
 cosC +&€.; or — sinC 4- -— sin2Q-\--—sinSC •{■-—. sin^Q + 
 
 &c. 
 
 In certain extrenje cases, approximations can likewise be 
 employed with advantage. Thus, suppose the angles A and 
 B to be exceedingly small ; then, by the last paragraph of 
 page 247, their versed sines are very nearly equal to lialf the 
 squares of the sines. Wherefore, sinQ, or sin ( A + B ) =r (art. 1 . 
 Note 3.), sinA{\ — \sinB^)-\'sinE{l'—\sinA.^) nearly, and con- 
 sequently, by art. 5., c=(a + ^) (I — \sinK sinB) ; or, the arcs 
 being nearly equal to their sines, substitute c for a-\- bin the 
 second or differential term, and c~a^b — |cAB. Again, put 
 C = TT — 6, or d = A + B, and {a-\-b){\sinAsinB)=i\siii\sin^ 
 
 l^zTzf^LQz nearly, orc=a4-i— 1^* — —7-. 
 
 15. Propositiipn twenty-fifth, which is employed with great ad- 
 vantage in maritime surveying, admits likewise of a convenient 
 analytical solution. Let the given distances AB, BC and AC be 
 denoted by a, b and c, and the observed angles ADB and CDB 
 
 by m and n; then (art. 5. Note 3.) BD = 5^"?^_?- 
 
 *' * Sin m 
 
 b sinBCD b sin m sinB AD , b sin m — a sin n 
 
 ^ , or — -. — = ■ T^.,.^ and - — : : — ~ 
 
 Sinn asmn smBLu b stn m-\- a sin 7t 
 
 sinBAD-^sinBCD _ . ^ lo xr f 7 ^ /a«i(BAD~BCD) 
 5mBAD4-5f«BCD - ^^''^' ^^' ^"^^^ ^'^ 7awi(BAD + BCD)- 
 But the angles ABC and ADC of the quadrilateral figure 
 DABC being evidently given, the sum of the remaining an- 
 gles BAD and BCD is given, and each of them is consequent- 
 ly found. Hence the triangles ABD and CBD are imme- 
 diately determined. 
 
390 NOTES AND ILLUSTRATIONS. 
 
 This most useful problem was first proposed by Mr Town- 
 ley, and solved in its various cases by Mr John Collins, in the 
 Philosophical Transactions for the year 1671. The second so- 
 lution"'given in the text is borrowed from Legendre. 
 
 16. The reduction of oblique angles to their projection on a 
 horizontal plane, is commonly solved by the help of sphe- 
 rical trigonometry. It admits, however, of a simple and ele- 
 gant general solution, derived from the arithmetic of sines. 
 Let a and b denote the two vertical angles, or the acclivities 
 of the diverging lines, A the oblique angle which these con- 
 tain, and A' the reduced or horizontal angle. Since the mag- 
 nitude of an angle depends not on the length of its sides, as- 
 sume each of them equal to the radius or unit, and it is evi- 
 dent that the base of the isosceles triangle thus limited will be 
 the chord of the oblique angle A, the perpendiculars from its 
 extremities to the horizontal plane, the sines, — and the hori- 
 zontal traces or projections, the cosines, of the vertical angles 
 a and b. The base of the isosceles triangle forms the hypotenuse 
 of a right-angled vertical triangle, of which the perpendicular 
 is the difference between the vertical lines. Consequently the 
 square of the reduced base is equal to the excess of the square 
 of the chord of A above the square of the difference of the 
 sines of a and ^, or 
 
 (cor. 6. def. Trig.) 2 — ^cosA. — (sin a — sinby^: 
 (II. 16. El.) 2 — 2cosA — sin a* — si?i b* •\-'2sin a sin bz=i 
 (2. cor. def. Trig.) cos a^'\-CQS b^ -^^sin a sin b — 2cosA: 
 Wherefore (Prop. 11. Trig.) in the triangle now traced on the 
 horizontal plane, 2cos a cos b cos A.' zz2cosA — 2sin a sin b ; and 
 multiplying by \ sec a sec b, there results (cor. 4. def. Trig.), 
 
 1 . Cos A' =r sec a sec 5 cos A — tan a tan b. 
 
 This expression appears concise and commodious, but it 
 may be still variously transformed. 
 
 For vers A' = 1 — co5 A' = 1 + tan a tanb — sec a sec b cos A 
 = sec a sec b {cos a cosb-{'Sin a sin b — cosA)=z 
 (Prop. 2. Trig.) seca sec b{cos{a — b) — cos A) : whence 
 
 2. VersA'zzsec asecb{ versA—vers(a — b).) 
 
NOTES AND ILLtJSTRATIONS. 391 
 
 Again, because (2. cor. l.andS. cor. o,Trig.)versA'zz2sm\A'*' 
 and versA—vers(a^b)z^2sm^:±i±Z^, sin A-(g--&)^ ^e ob. 
 
 tain, by substitution, 
 
 3. 5i;ziA* = sec a secb(isin^i±J^^Z^. sin A-(| -f ^)), 
 
 Of these JbrmuicBf the first, I presume, is new, and appears 
 distinguished by its simplicity and elegance. The last one 
 however, is, on the whole, the best adapted for logarithmic 
 calculation* 
 
 When the vertical angles are skftall, the problem will admit 
 of a very convenient approximation. For, assumiDg the arcs 
 a, b as equal to their tangents, it follows, by substitution, that 
 cosA'=cosA^(l~^a'')>^{l+b^)-<ib^cosA((l+ld'Xl+lb^))—ab 
 =co5A(l+^rt^4-^6*— )a3, nearly. Whence, by Note 13, the 
 decrement of the cosine of that oblique angle is 
 ab—cosA{ia^-^\b^); but 
 
 (II. 17. El.) «^=(«_+i).^(«JZ^)S and . 
 
 (II. 18. El.) Xa^+xi« = (fL±i)*+(?Lri)- 
 wherefore the decrement of co^A'zs 
 
 Consequently the increment of the oblique angle itself is, hf 
 Note 13, 
 
 Such is the theorem which the celebrated Legendre has gi- 
 ven, for reducing an oblique angle to its projection on the ho- 
 rizental plane. It is very neat, and extremely useful in prac« 
 
392 NOTES AND ILLUSTRATIONS. 
 
 tice. But to connect it with our division of the quadrant, re- 
 quires some adaptation. Let a and b express the vertical aftigles 
 
 in minutes; then will ^(l±i)z;«,^iA_(l=i)^ cot\A.^^ 
 
 denote, likewise in minutes, the quantity of reduction to be 
 applied to the oblique angle. 
 
 17. In computing very extensive surveys, it becomes necessary 
 to allow for the minute derangements occasioned by the con- 
 vexity of the surface of our globe. The sides of the triangles 
 which connect the successive stations, though reduced to the 
 same horizontal plane, may be considered as formed by arcs of 
 great circles, and their solution hence belongs to Spherical 
 Trigonometry. But, avoiding such laborious calculations, for 
 which indeed our Tables are not fitted, it seems far better to 
 estimate merely the deviation of those incurved triangles from 
 triangles with rectilineal sides. For effecting that correction 
 two ingenious methods have lately been proposed on the Con- 
 tinent. The first is that employed by M. Delambre, who sub- 
 stitutes the chords for their arcs, and thus converts the small 
 spherical, into a plane, triangle. This conversion requires two 
 distinct steps. 1. Each spherical angle, or that formed by tan- 
 gents at the surface of the globe, is changed into its corre- 
 sponding plane angle contained by the chords. Let a and /S 
 express the sides or arcs in miles ; and the angles of elevation, 
 or those made by the tangents and the respective chords, will 
 
 , , 21600^ , 21600 
 
 be (IlL 29. El.) denoted by 94^5 6'^^ ^ 24856 ^^ ^" minutes, 
 
 1350' ^ 1350' , t. u r . , . 
 
 or oj7y7-«* and o .q^ 7 ^* insert these, thereiore, m place of a 
 
 and b in the fonrmla of the preceding note, and the quantity of 
 reduction of the angle A, contained by the small arcs ct and/3, 
 
 will be (U -f- fiYianlA — (u — (ifcot\K) Y^xi^^ seconds. 
 
 2. Each arc is concerted into its chord : But, by the Scholium 
 to Proposition VL of the Trigonometry, an arc » is to its chord, 
 
NOTES AND ILLUSTRATIONS. 393 
 
 «* 
 
 as 1 to 1— "FjSi; wherefore the diminution of that arc in pass- 
 
 ing into its chord, amounts to the 375~600000 ^^^^ ^^ ^^® 
 whole. 
 
 These reductions bestow great accuracy, and are sufficient- 
 ly commodious in practice. But the second method of^cor-' 
 recting the effects of the earth's convexity, and which was 
 given by M. Legendre, is distinguished by its conciseness and 
 peculiar elegance. That profound geometer viewed the sphe- 
 rical triangle as having its curved sides stretched out on a 
 plane, and sought to determine the variation which its angles 
 would thence undergo. Analysis led him, through a compli- 
 cated process, to the discovery of a theorem of singular beau- 
 ty. But the following investigation, grounded on other prin- 
 ciples, appears to be much simpler. 
 
 Let A and B denote any two angles in the small spherical 
 triangle, and a and /3 express in miles the opposite sides, or 
 those or its extension upon a plane. Since (Prop. 9. Trig.) 
 u : /i : : sinA : sinB, there must exist some minute arc &j such 
 that sin» : sinfi : : sin{A -f- 6) : sin(B + 6.). But (art. 1. Note 3.) 
 sin {A + 6) zzsinA-^ 6 cos A, and (Schol. Prop. VI. Trig.) 
 
 sinec = ec — -^i whence u — ^ : /3 — g- : : sin A -^ 6 cos A: sinB 
 
 -|- 6 cosB. Now fi'.x:: sinB : sinAy and therefore, (V. 9. El.) 
 
 1— -g- : i — -g- : : sin A sinB -f 6 cos A sinB : sinAsinB + 
 
 6sinA cosB. But the first and second terms being very nearly 
 equal, and likewise the third and fourth, — it is obvious that the 
 analogy will not be disturbed, if each of those pairs be increa- 
 
 sed equally. Hence 1 : 1 + — g — : : sin A sinB : sin A sinB 4- 
 
 6(sinAcosB — cosA sitiB) ; and since (Prop. I. Trig.) sin A 
 cos B — cos A sin B = sin (A— B), therefore (V. 10. El.) 
 
 1 : — p — : : sin A sinB : 6 sin{A — B). Consequently, since x and 
 3 are proportional to 677? A and sinB, $ (sin A^B) =: sin A sinB 
 
39 4( NOTES AND ILLUSTRATIONS- 
 
 ^ ,-, ^ (^m A*— 52«B*; =: (Proposition III. cor. 5. Tri- 
 6 ' 6 • 
 
 gonometry,)-g (5m(A+B)5m(A— B)), or 6 = —sin (A + B). 
 
 But the sine of the sum of A and B is the same as that of their 
 supplement C, or of the angle contained by the sides cc and /3, and 
 
 consequently & is the third part of-^^mC, the area of the tri- 
 angle, or the third part of the excess of the angles of the sphe- 
 rical, above those of the plane triangle. Wherefore the sines 
 of the sides, which, in the spherical triangle, are as the sines 
 of their opposite angles, are likewise proportioned, in the plane 
 triangle, to the sines of those angles, increasing each by the 
 common excess. It is hence evident, that the angles of the 
 plane triangle are obtained from those of the spherical, by de- 
 ducting from each the third part of the excess above two right 
 angles, as indicated by the area of the triangle. 
 
 The whole surface of the globe being proportioned to 720'', 
 
 720^ 
 that of a square mile will correspond to 24856 x 7912 » °^ 
 
 the >yKOQ part <» a second. Hence each angle of the small 
 spherical triangle requires to be diminished by «,4 sinC 
 
 455:28 ^^ ^^^°"^'- 
 
 18. Another problem of great use in the practice of delicate 
 surveying, is to reduce angles to the centre of the station. In- 
 stead of planting moveable signals at each point of observa- 
 tion, it will often be found more convenient to select the more 
 notable spires, towers, or other prominent objects which oc- 
 cur interspersed over the face of the country. In such cases, 
 it is evidently impossible for the theodolite or circular instru- 
 ment, although brought within the cover of the building, to 
 be placed immediately under the vane. The observer ap- 
 proaches the centre of the station as near, therefore, as he can 
 with advantage, and calculates the quantity of error which the 
 
NOTES AND ILLUSTRATIONS. 
 
 395 
 
 minute displacement may occasion. Thus, suppose it were 
 required to determine the 
 angle AOB which the re- 
 mote object A and B sub- 
 tend at O, the centre of a 
 permanent station : The 
 instrument is placed in the 
 immediate vicinity at the 
 point C, and the distance 
 CO, with the angle of devia- 
 tion OCA, are noted, while 
 the principal angle ADCB 
 is observed. The central 
 angle AOB may kence be 
 computed from the rules 
 of trigonometry; but the 
 calculation is effected by simpler and more expeditious me- 
 thods. Since (I. 30. E\.) the exterior angle ADB is equal 
 both to AOB with OAC, and to ACB wJth OBC ; it is evi- 
 dent that AOB = ACB + OBC— OAC. But the angles OBC 
 and OAC, being extremely small, may be considered as equal 
 
 CO 
 
 to their sines, and (art. 5. Note 14.) &in OBC = /jg sinBCO, 
 
 CO 
 and sinOAC == q^ sinACO ; wherefore the angle AOB at. 
 
 f sinBCO sinACO^ 
 the centre, is nearly equal to ACB + C0\^ 
 
 = ACB + C0 r^^7>H^^^''^ - 
 
 -)• 
 
 OA ; 
 
 Call the dis- 
 
 OB 
 
 sinACO 
 
 OB OA" 
 
 tances AC and BC of the point of observation, a and b, the 
 distances AO and BO of the centre a^ and &' ; the displace- 
 ment CO, and the angle ACO of deviation m and ^, while the 
 subtended angles ACB and AOB are denoted by C and C, 
 and the opposite angles ABO and OAB by A and B ; then C 
 
 ~r) 3438'. If the centre O lies oh 
 AC, the correction of the observed angle, expressed in mi- 
 nutes, will be merely \-r^sinC } 3438', 
 
 ^c + ,»(^^2i^ 
 
396 NOtES AND ILLUSTRATIONS. 
 
 But the problem admits of a simpler approximation. Let 
 a circle circumscribe the points A, O, ami B , and cut AC in 
 E. The angle AOB = (IlL 16. El.) AEB = ACB -f CBE ; 
 
 CE 
 but sinCBE = jg^ sinACB, and sinOEC = sinAEO or ABO 
 
 CO 
 is equal to ^ sinCOE or AEO— ACO, and hence by com- 
 
 ,. . . r.r.T^ CO ^iwACB 5m(AB0->AC0) ^. 
 bmation sm CBE = ^ SaBO ^^^^^^ 
 
 therefore, EB is nearly equal to OB, and the small angle CBE 
 
 may be regarded as equal to its sine, the correction to be add- 
 
 m 
 ed to the observed angle is denoted in minutes by -r? 
 
 sinCsin(A — <P)^,^„ ^, . ... ., .„ . 
 ^^^ 34<38. Ihis quantity, it is evident, will entirely 
 
 vanish when (p becomes equal to A, or the angle ABO equals 
 ACO ; in which case, the point of observation C coincides 
 with E, or lies in the circumference of a circle that passes 
 through the two remote points A and B and centre of the sta- 
 tion. To place the instrument at E, therefore, would only re- 
 quire to move it along CA, till the angle AEO be equal to 
 ABO. 
 
 Both these methods for the reduction of an angle to the 
 centre are given by Delambre ; but, in his calculations, he ge- 
 nerally preferred the last one, as being simpler and sufficient- 
 ly accurate for practice. The investigation, however, will be 
 found to be now considerably shortened. 
 
 19. The accuracy of trigonometrical operations must depend 
 on the proper selection of the connecting triangles. It is very 
 important, therefore, in practice, to estimate the variations 
 which are produced among the several parts of a triangle, by 
 any change of their mutual relations. Suppose two of the 
 three determining parts of a triangle to remain constant, while 
 the rest undergo some partial change ; and let, as before, the 
 small letters a, b and c denote the sides of the triangle, and 
 the capitals A, B and C their opposite angles. 
 
NOTES AND ILLUSTRATIONS. S9*7 
 
 Case I. — When two sides a and b are constant. 
 
 Since the angles A and B, after passing into A -J- AA and 
 B + AB, must have their sines still proportional to the oppo- 
 
 sinA. sin B 
 
 site sides, it is evident that ,-(a+aA) = liM^+'E^y ^^^ 
 
 sin(A 4- AA) — sin A. sin{B-\-AB)—sinB 
 consequently ^(a+ AA) + ;s//zA = sin{B+AB)+^inB * 
 wherefore, by alternation and art. 7. Note 12., 
 ^ tanl^ AA ^aw(A4-^,AA) 
 * taniAB^ tan{B^^ABy 
 Next, in the incremental triangle formed by the sides Cp 
 c+Ac, and the contained angle A A, (art. 1. Note 12.) 
 
 J^ ^^^(B + JAB) ^^^ j^^^^^ reciprocally, 
 
 C + ^Ac-" coth AA ir j» 
 
 '^aw^AA"" co^(B-J-6AB) 
 
 In like manner, from the incremental triangle contained by- 
 the sides c, c-{- Ac and the angle aB, it follows that 
 iAc __ C'\-lAc 
 
 ' tanh aB co^(A-f-. aA) 
 
 Again, the base of the incremental isosceles triangle con- 
 tained by the equal sides b, b, and the vertical angle aC, is 
 (art. 15. Note 12.) ^bsin^AC; wherefore, in the incremental 
 triangle formed with the same base and the sides c and c+Ac, 
 
 bv art. 20. Note 12., co5(A+iAA) = ^ ^+|^^'"^^^^ ; 
 
 whence 
 
 sin^ AB __ __ a co5( a 4- J a A) 
 *5m|AC ^+:;Ac 
 
 After the same manner, it will be found that 
 
 sinlAA __ acoA(B4 -|AB( 
 
 *m.^AC f-f-. Ac 
 
 Muhiply the expressions of art. 4-. into those of art 3. and 
 
 ^Ac __ bsin{A-\- ^aA) 
 ' sinh^C cosh AB * 
 
398 NOTES AND ILLUSTRATIONS. 
 
 7. 
 
 Multiply likewise the expressions of art. 2, and 5., and 
 §Ac _ asin{B+hAB) 
 
 sin t AC cosh A A 
 
 If, in all the preceding formula^ the increments annexed to 
 the varying quantities be omitted, there will arise much sim- 
 pler expressions for the differentials. 
 
 # ] ^^ — . ^«wA 
 
 *o dc ^ c 
 
 ' dK cotB" 
 
 * 3. ^ = ^— 
 
 ' dB cotA 
 
 *4. ^=-,A.o5A. 
 dC c 
 
 *5.^=:^±cosB. 
 dC c 
 
 *6. ~=: bsinA. 
 dC 
 
 *7. ^=a«nB. 
 
 Case 11 — When one side a, and its opposite angle A, are con" 
 stant. 
 
 Since (art. 5/Note 12.) -^= A, it is evident that 
 smA stnB 
 
 a sinBzzb sinA, and taking the differences by art. 1. of Note 
 10. Ab sin A = 2a sinlABcos{B -f i aB), whence !J:l\^ -_ 
 
 gCQg(B+|AB) * ^"^ consequently, by art. 5. of Note 12. 
 g^ f^iAB _ ^^V^^AC_ ^z/iB 
 
 lhb~ iA<^ ~^C05(B+|aB)* 
 
 In like manner, it will be found that 
 
 ^m^AB ^sin\AC _ sinC 
 
 i^C — ^AC "" "~CC05(C-f.iAC)' 
 
 Combine the two last expressions, and 
 
10. 
 
 NOTES AND ILLUSTRATIONS. 399 
 
 Ac ~" cos{C^^AC) 
 The differentials are discovered, by rejecting the modifica- 
 tions of the variable quantities, 
 ^ dB __ sinB __ tanB 
 
 ' db "^ b cosB "" b, ' 
 ^ c?B __ __^ jinC __ tanC 
 dc "~ ccosC "" C~~" 
 
 #10 ^^ — ^^^^ 
 
 dc cosC 
 
 Case III — When one side a, and Us adjacent angle B, are con* 
 stant. 
 
 In the incremental triangle contained by the sides ^, 5 4- A5 
 and Ac, it is evident, (art. 5. Note 12.), that 
 2j Ac -^^ _ ^ ^+A 
 
 '^mAC"" 52wAA~"«w(A-|-AA)"~ sinA ' 
 
 Again, in the same incremental triangle, (art. 6. Note 
 12.) 
 ig^ ^A^ _.__ l^l> _ b + \Ab 
 
 ' tan^AQ tan\AA. ^aw(A-f-iAA)' 
 
 Or, transforming the preceding expression, 
 
 \Ab tan\AK , , 
 
 7-rv— r= — • r-~TirT~mr:* ^"^ consequently 
 b^\Ab ^aw(A-i-iAA) ^ ^ 
 
 \^b tan\AA / . , XT . ».v 
 
 '-T-= ~ r .HA+^aAA)+^a.|AA = ^^^'^ ^- ^^^^ 7-) 
 
 52w(A+AA) ^ * V««(A+AA;) 
 
 wherefore, 
 13 |A& __ jA^ _ ^/ co5(A4-^AA )x 
 
 * 5W^aC 5iw^AA ^5m(A-|-AA) / 
 
 Again, in the same incremental triangle, by art. 20. Note 
 ]2. 
 
 co5(A+|AA)= — (— co4aC)= — cos\aA ; whence 
 
 Ac Ac 
 
 14, ^^ _ CQ^(A4-|AA) 
 *Ac"~" co^^aA *■ 
 
400 NOTES AND ILLUSTRATIONS. 
 
 The differentials are found as before, by the omission of the 
 
 minute excrescences. 
 
 dc dc b 
 
 * 11. -77:: :r: 
 
 dC dA smA 
 
 12 —- —-^ * 
 
 dC dA tanA 
 
 ^^^ db db , (cosA\ , , . 
 
 * 14'. -y-zzcosA, 
 dc 
 
 To compute the values of the finite differences, when these 
 differences themselves are involved in their compound expres- 
 sion, the easiest method is to proceed by repeated approxima- 
 
 tions. Thus, from art. 3» Acr=——^^^^7jY^-TT (2c+Ac) ; as- 
 
 taizkAB 
 sume, therefore, first, Ac = — c otiA-L ^aA\ ^^ ' ^"^ then, ^c 
 
 tanlAB tan^AB^ 
 
 = - coiiA + iAA) (2^- cot{A+lAA) ^')' ^"^ '^ ^^" '^^' 
 dom be requisite to advance beyond two steps ; though the 
 process, if continued, would evidently form an infinite converr 
 ging series. 
 
 When only one part of a triangle remains constant, the ex- 
 pressions for the finite differences will often become extremely 
 complicated. It may be sufficient in general to discover the 
 relations of the differentials merely. To do this, let each in- 
 determinate part be supposed to vary separately, and find, by 
 the precedingy<>nwM/<^, the effect produced ; these distinct ele- 
 ments of variation being collected together, will exhibit the 
 entire differential. 
 
 The materials of this intricate Note appear in Cagnoli, but 
 the subject was first started by our countryman Mr Cotes, a 
 mathematician of profound and original genius, in a brief tract, 
 entitled Estimatio errorum in mixta Mathesi, It is unfortunate 
 that I have not room for explaining the application of those 
 JbrmulcB to the selection and proper combination of triangles in 
 nice surveys. 
 
NOTES AND ILLUSTRATIONS. 401 
 
 'iO. Having in some of the preceding notes briefly pointed 
 out the several corrections employed in the more delicate geo- 
 desiacal operations, I shall subjoin a few general remarks on the 
 application of trigonometry to practice. The art of surveying 
 consists in determining the boundaries of an extended surface. 
 When performed in the completest manner, it ascertains the 
 positions of all the prominent objects within the scope of ob- 
 servation, measures their mutual distances and relative heights, 
 and consequently defines the various contours which ;iiark the 
 surface. -But the land-surveyor seldom aims at such minute 
 and scrupulous accuracy ; his main object is to trace expedi- 
 tiously the chief boundaries, and to compute the superficial 
 contents of each field. In hilly grounds, however, it is not the 
 absolute surface that is measured, but the diminished quantity 
 which would result, had the whole been reduced to a horizon- 
 tal plane. This distinction is founded on the obvious princi- 
 ple, that, since plants shoot up vertically, the vegetable pro- 
 duce of a swelling eminence can never exceed what would 
 have grown from its levelled base. All the sloping or hypo- 
 tenusal distances are, therefore, reduced invariably to their 
 horizontal lengths, before the calculation is begun. 
 
 Land is surveyed either by means of the chain simply, or by 
 combining it with a theodolite or some other angular instru- 
 ment. The several fields are divided into large triangles, of 
 which the sides are measured by the chain ; and if the exterior 
 boundary happens to be irregular, the perpendi<jular distance 
 or offset is taken at each bending. The surface of the com- 
 ponent triangles is then computed from Prop. 29. Book VI. of 
 the Elements of Geometry, and that of the accrescent space by 
 Note 4. to Prop. 9. Book 11. In this method the triangles 
 should be chosen as nearly equilateral as possible ; for if they 
 be very oblique, the smallest error in the length of their sides 
 will occasion a wide difference in the estimate of the surface. 
 The calculation is much simpler from the application of Prop. 5. 
 Book 11. of the Elements, the base and altitude of each tri- 
 angle only being measured ; but that slovenly practice appears 
 liable to great inaccuracy. The perpendicular may indeed be 
 ti-aced by help of the surveying cross, or more correctly by 
 
 2d 
 
402 NOTES AND ILLUSTRATIONS. 
 
 the box sextant, or the optical square, which is only the same 
 instrument in a reduced and limited form ; yet such repeated 
 and unavoidable interruption to the progress of the work will 
 probably more than counterbalance any advantage that might 
 thence be gained. 
 
 The usual mode of surveying a large estate, is to measure 
 round it with the chain, and observe the angles at each turn 
 by means of the theodolite. But these observations would re- 
 quire to be made with great care. If the boundaries of the 
 estate be tolerably regular, it may be considered as a polygon, 
 of which the angles, being necessarily very oblique, are there- 
 fore apt to affect the accuracy of the results. It would serve 
 to rectify the conclusions, were such angles at each station 
 conveniently divided, and the more distant signals observed. 
 The best method of surveying, if not always the most expedi- 
 tious, undoubtedly is to cover the ground with a series of con- 
 nected triangles, planting the theodolite at each angular point, 
 and computing from some base of considerable extent, which 
 has been selected and measured with nice attention. The la- 
 bour of transporting the instrument might also in many cases 
 be abridged, by observing at any station the bearings at once 
 of several signals. Angles can be measured more accurately 
 than lines, and it might therefore be desirable that surveyors 
 would generally employ theodolites of a better construction, 
 and trust less to the aid of the chain. 
 
 The quantity of surface marked out in this way is easily 
 computed from trigonometry Adopting the general nota- 
 tion, the area of a triangle which has two sides, and their in- 
 
 ab 
 eluded angle known, it is evident, will be denoted by -^'sinC, 
 
 and the area of a triangle of which there are given q}\ the 
 
 a' sinB sinC» 
 angles and a side, is ■J'-^Ji^^A" 
 
 From the same principles may be determined the area of a 
 quadrilateral figure inscribed in a circle. Let the sides a and b 
 contain an acute angle A, and the opposite sides c and ^ must 
 contain the obtuse supplementary angle. The common base of 
 these triangles, or diagonal of the quadrilateral figure, is hence 
 
NOTES AND ILLUSTRATIOKS. 40S 
 
 expressed by -/(«'+ ^'-2aico5 A), and by -v/(c'+a?'+ 2^^ ^05A); 
 and consequently a^-f 5'— c* — d^=2ab cosK-{- 2cd cos A, 
 
 <'--A=^^Sw^"- Wherefore l+co.A= 
 
 2a^+ 'icd "~ ^ab^ ''led 
 
 a^^^^^lab^^i^^^ consequently 
 
 '2a6 + 2cc? 2ab-\-cd ^ ^ 
 
 (I4-C05A) (1 — cosA) = l — co5A*=5iwA^ = 
 la^J>?-±=d_Y M->>j'-(oY)\ But the area of the qua- 
 
 drilateral figure, or that of its two component triangles, is 
 sinhi-^- — \z=i\sinK'^ah-\^'icd)y and therefore its square is= 
 4^inh} ( 2a^+ 2cf^)S or ^^.{a + hf-^c-^f.ia^by-^c^dY = 
 
 4< * 4 "" 
 
 a^bJ{-c — d a-\-b^c+d a — b^c-{-d — g^hJ^cJf-d 
 2 ' 2 * 2 * 2 • 
 
 Or, if 5 denote the semiperimeter, the square of the area will 
 be expressed by 5 — a,s — b.s — c,s — d. If one of the sides d were 
 supposed to vanish, the quadrilateral figure would pass into a 
 triangle, whose area would be s.s~a,s — b.s — c, — the same as 
 was before investigated. 
 
 The English chain is 22 yards, or 66 feet in length, and equi- 
 valent to four poles ; it is hence the tenth part of a furlong, or 
 the eightieth part of a mile. The chain is divided into a hun- 
 dred links, each occupying 7.92 inches. An acre contains ten 
 square chains or 100,000 links. A square mile, therefore, in- 
 cludes 640 acres ; and this large measure is deemed sufficient, 
 in certain rude and savage countries, as the Back Settlements 
 of America, where vast tracts of new land are allotted merely 
 by running lines north and south, and intersecting these by 
 perpendiculars, at each interval of a mile. 
 
 The Scotch chain consists of 24 ells, each containing 37.069 
 inches, and ought therefore to have 74-. 138 feet for its correct 
 length. The English acre is hence to the Scotch, in round 
 numbers, as 11 to 14, or very nearly as the circle to its cir- 
 cumscribing square. But this provincial measure is gradual- 
 
404' KOTES AND ILLUSTRATIONS. 
 
 ly wearing into disuse, and already the statute acre seems to 
 be generally adopted in the counties south of the Forth. 
 
 21. Levelling is a delicate and important branch of general 
 surveying. It may be performed very expeditiously by help 
 of a large theodolite, capable of measuring with precision the 
 vertical angle subtended by a remote object, the distance 
 being calculated, and allowance made for the effect of the 
 earth's convexity and the influence of refraction. But the 
 raore usual and preferable method is to employ an instrument 
 designed for the purpose, and termed a spirit-levelj which is 
 accompanied by a pair of square staves, each composed of two 
 parts that slide out into a rod of ten feet in length, every foot 
 being divided centesimally. Levelling is distinguished into 
 two kinds, the simple and the compound; the former, which 
 rarely admits of application, assigns the difference of altitude 
 by a single observation; but the latter discovers it from, a 
 combined series of observations carried along an irregular sur- 
 face, the aggregate of the several descents being deducted 
 from that of the ascents. The staves are therefore placed 
 successively along the line of survey, at suitable intervals ac- 
 cording to the nature of the ground and not exceeding 400 
 yards, the levelling instrument being always planted nearly in 
 the middle between them, and directed backwards to the first 
 staff, and then forwards to the second. The difference between 
 the heights intercepted by the back and the fore observation, 
 must evidently give at each station the quantity of ascent or 
 descent, and the error occasioned by the curvature of the globe 
 may be safely overlooked, as on such short distances it will 
 not amount at each station to the hundredth part of a foot. To 
 discover the final result of a series of operations, or the diffe- 
 rence o^ altitude between the extreme stations, the measures 
 of the back and fore observations are all collected severally, 
 and the excess of the latter above the former indicates the ep- 
 tire quantity of descent. 
 
 As an example of levelling, I shall take the concluding part 
 of a survey, which my friend Mr Jardine, civil engineer, has 
 recently made for the Town-Council of Edinburgh, with a de- 
 gree of accuracy seldom attempted, in tracing the descent 
 from the Black and Crawley springs, near the summits of the 
 
NOTES AND ILLUSTRATIONS. 
 
 40^ 
 
 PentUnd chain, to the Reservoir on the Castlehill, with a view 
 to the conducting of a fresh supply of water from those heights. 
 To avoid unnecessary conaplication, however, i shall only no- 
 tice the principal stations. The figure annexed represents a 
 profile or vertical section of the ground, LV is the level of the 
 Black spring, and the several perpendiculars from it denote 
 the varying depth of the surface, referred to the base assu- 
 med 700 feet below. The stations marked are as follow : 
 
 L Lowest point in the Meadow. 
 
 M Cleansing cocks on the north side of the Meadow. 
 
 N Sunk fence in Lord Wemyss's garden. 
 
 O Air cock in Archibald's nursery. 
 
 P South side of Lauriston road. 
 
 Q Bottom of Heriot's Green Reservoir. 
 
 R Head of Hamilton's close. 
 
 S Strand on south side of Grassmarket. 
 
 T Cleansing cock on north side of Grassmarket. 
 
 U Gaelic Chapel. 
 
 V Upper side of the belt of Castlehill Reservoir. 
 
 
 
 Back Ob- 
 
 Fore Ob- 
 
 
 Stations. 
 
 Distance. 
 
 servation. 
 
 servation. 
 
 Ascent. 
 
 
 Feet. 
 
 Feet 
 
 Feet. 
 
 Feet. 
 
 L 
 M 
 
 370 
 
 4.59 
 
 2.04 
 
 
 2.55 
 
 N 
 
 640 
 
 8.68 
 
 3.05 
 
 8.18 
 
 O 
 
 905 
 
 9-12 
 
 2.22 
 
 15.08 
 
 P 
 
 1236 
 
 29.43 
 
 2.11 
 
 42.40 
 
 Q 
 
 1493 
 
 16.24 
 
 1.40 
 
 57.24 
 
 R 
 
 1925 
 
 2.54 
 
 26.98 
 
 32.80 
 
 s 
 
 2260 
 
 4.69 
 
 53.28 
 
 —15.79 
 
 T 
 
 2352 
 
 4.22 
 
 4.42 
 
 —15.99 
 
 U 
 
 2540 
 
 32.40 
 
 1,25 
 
 15.15 
 
 V 
 
 2705 
 
 94.77 
 
 997 
 
 99.95 
 
406 NOTES AND ILLUSTRATIONS. 
 
 Black spring, being 620.05 feet above the level of the Mea- 
 dow, is therefore 520.1 feet higher thiin the belt of the reser- 
 voir. The numbers exhibited in the last column, are obtained 
 by taking the differences of the aggregates of the two preced- 
 ing columns. Where the ground either sinks or rises suddenly, 
 some intermediate observations are here grouped together in- 
 to a single amount. Thus, three observations were made be- 
 tween O and P, two between P and Q, three between Q and 
 R, five between R and S, three between T and U, and no fewer 
 than nine between U and V. The slight sketch between 
 the perpendiculars from Q and R, shows the mode of planting 
 and directing the instrument. 
 
 The mode of levelling on a grand scale, or determining the 
 heights of distant mountains, will receive illustration from the 
 third volume of the Trigonometrical Survey, which Colonel 
 Mudge has been kindly pleased to communicate to me before 
 its publication. I shall select the largest triangle in the series, 
 being one that connects the North of England with the Borders 
 of Scotland. The distance of the station on Cross Fell to that 
 on Wisp Hill, is computed at 235018,6 feet, or 44.511 miles, 
 which, reckoning 6094?^ feet for the length of a minute near 
 that parallel, corresponds, on the surface of the globe, to an 
 arc of 38' 33".?. Wisp Hill was seen depressed 30' 4-8" from 
 Cross Fell, which again had a depression of 2' 31" when view- 
 ed from Wisp Hill. The sum of these depressions is 33' 19", 
 which, taken from 38' 33".7, the measure of the intercepted 
 arc, or the angle at the centre, leaves 5' 14".7, for the joint 
 effect of refraction at both stations. The deflection of the 
 visual ray produced by that cause, which the French philo- 
 sophers estimate in general at .079, had therefore amounted 
 only to .06805, or a very little more than the Jifteenth part of 
 the intercepted arc. Hence, the true depression of Wisp Hill 
 was 30'48"— 16'39".5=14'8''.5; and consequently, estimating 
 from the given distance, it is 967 feet lower than Cross Fell. 
 
 From Wisp Hill, the top of Cheviot appeared exactly on 
 the same level, at the distance of 185023.9 feet, or 35.0424 
 miles. Wherefore, two-thirds of the square of this last num- 
 ber, or 819, would, from the scholium at page 276, express in 
 feet the approximate height of Cheviot above Wisp Hill. But 
 refraction gave the mountain a more towering elevation than 
 
NOTES AND ILLUSTRATIONS. 407 
 
 it really had ; and the measure being reduced in the former 
 ratio of 38' 33".7 to 33' 19", is hence brought down to 708 feet. 
 
 Again, the distance 292012.7 feet, or 55.3054? miles, of 
 Cross Fell from Cheviot, corresponds to an arc of 4-7' 54?'/.8, 
 which, reduced by the effect of refraction, would leave 
 41' 23".8 for the sum of the depressions at both stations. Con- 
 sequently, Cheviot had, from Cross Fell, a true depression of 
 only 23/ 4<4"— 20' 4r'.9 or 3> 2".l, and is therefore lower than 
 that mountain by 258 feet. 
 
 These results agree very nearly with each other. The height 
 of Cross Fell above the level of the sea being 2901, that of 
 Wisp Hill is 1934, and that of Cheviot 2642 or 2643. In the 
 Trigonometrical Survey, the latter heights are stated at 1940 
 and 2658 ; a difference of small moment, owing to a balance 
 of errors, or perhaps to the adoption of some other data with 
 respect to horizontal refraction, and which do not appear on 
 record. 
 
 From the same valuable work, I am tempted to borrow ano- 
 ther example, which *has more local interest. From Lums- 
 dane Hill, the north top of Largo Law, at the distance of 
 189240.1 feet, or 35.84 miles, appeared sunk 9' 32" below the 
 horizon. Here the intercepted arc is 31' 3", and the effect of 
 the earth's curvature, modified by refraction, is 13' 24". 8 ; 
 whence the true elevation of Largo Law was 1 3' 24".8 — 9' 32", 
 or 3' 52//.8, which makes it 213 feet higher than Lumsdane 
 Hill, or 938 feet above the level of the sea. In the Trigono- 
 metrical Survey, this height is stated at 952 ; but I am in- 
 clined to prefer the former number, having once found it by 
 a barometrical measurement, in weather not indeed the most 
 favourable, to be only 935 feet. 
 
 Through the kindness of Captain Colby of the Royal Engi- 
 neers, who has for several years so ably conducted the survey 
 under the direction of Colonel Mudge, I am enabled to subjoin 
 some more examples, from the observations made last season. 
 From Dunrich Hill the station on Cross Fell appeared de- 
 pressed 19' 21", at the distance of 349,343 feet or 66.1634 
 miles. This corresponds on the same parallel to an intercepted 
 arc of 57' 19"; the half of which, diminished by one-twelfth of 
 the whole, gives 23' 53, for the effect of curvature modified by 
 
40S NOTES AND ILLUSTRATIONS. 
 
 refraction. Cross Fell had therefore an elevation of 4' S2^', 
 the excess of 23' 53'' above 19' 21'', which, at the given dis- 
 tance, makes it to be 461 feet higher than Dunrich Hill. Con- 
 .sequently, the altitude of Dunrich Hill above the level of the 
 sea is 2901 — 4-61^ or 2440 feet. This altitude^ detern>ined 
 IVom nearer bases, was only 2421 feet. 
 
 Again, from Cairnsmuir upon Deugh, at the height of 2597 
 feet above the sea, the top of Ben-Lomond appeared with a 
 depression of 18' 24", the distance being nearly 352,004 feet, 
 or 66.6673 miles. The intercepted arc on the earth's surface 
 was hence 57' 45^", and the effect of curvature, as modified 
 by refraction, 24' 4". Wherefore, R : ^a«6'40", the real ele- 
 vation : : 352,004 : 580, which, added to 2597, gives 3177 for 
 the altitude of Ben-Lomond. 
 
 We shall select another example, which affords an approxi- 
 mation to the diameter of our globe. From the station at the 
 observatory on the Calton-hill, at the altitude of 350 feet, the 
 horizon of the sea was found depressed 18' 12" But refrac- 
 tion being supposed to have diminished the effect by one- 
 twelfth part, if the eleventh part be added of this remaining 
 quantity, there will result 19' 43" for the true measure of de- 
 pression. The angle at the centre is consequently the half 
 of 1 9' 43" or 9' 51 1" ; wherefore, tan 9' 51.^" : 11 : : 350 : 122,048 
 feet, or 23.1 152 miles, the distance at which the extreme visual 
 ray grazes the sea. Again, tan 9 51|" : R : : 23.1152 : 4030 
 miles, the radius of the earth, a near approximation to the 
 real measure, or 3956. It should be noticed, that the state of 
 the tide would have some effect in modifying the angle of de- 
 pression. Thus, on the 12th May 1816, at 7f p. ni. the de- 
 pression towards the mouth of the Firth of Forth, between 
 the Isle of May and the Bass Rock, was found to be 18' 14" ; 
 but it was 18/ 16" in a direction more to the north and near 
 the Fife coast, because the sea had ebbed nearly five hours, 
 the current outwards running first along the northern shore. 
 On the following day, at three quarters after twelve o'clock, 
 and therefore two hours and a half before high water, the de* 
 pression about the middle of the Firth was 18' 9", and only 
 IS' 6" on the northern shore, the tide then flowing up princi- 
 pally in the middle of the channel. 
 
ROTES AND ILLUSTIIATIONS. 409 
 
 ^22. MARITIME Surveying is of a mixed nature : It not only 
 determines the positions of the remarkable headlands, and 
 other conspicuous objects that present themselves along the 
 vicinity of a coast, but likewise ascertains the situation of the 
 various inlets, rocks, shallows and soundings which occur in 
 approaching the shore. To survey a new or inaccessible coast, 
 two boats are moored at a proper interval, which is carefully 
 measured on the surface o( the water ; and from each boat the 
 bearings of all the prominent points of land are taken by means 
 of an azimuth compass, or the angles subtended by these points 
 and the other boat are measured by a Hadley's sextant Ha- 
 ving now on paper drawn the base to any scale, straight lines 
 radiating from each end at the observed angles, as in Prop. 
 21. of the Trigonometry, will by their intersections give the 
 positions of the several points from which the coast may be 
 sketched. — But a chart is more accurately constructed, by 
 combining a survey made on land, with observations taken on 
 the water. A smooth level piece of ground is chosen, on 
 which a base of considerable length is measured out, and sta- 
 tion staves are fixed at its extremities. If no such place can 
 be found, the mutual distance and position of two points con- 
 veniently situate for planting the staves, though divided by a 
 broken surface, are determined from one or more triangles, 
 which connect with a shorter and temporary base assumed 
 near the beach. A boat then explores the offing, and at every 
 rock, shallow, or remarkable sounding, the bearings of the sta- 
 tion staves are noticed. These observations furnish so many 
 triangles, from which the situation of the several points are 
 easily ascertained.— When a correct map of the coast can be 
 procured, the labour of executing a maritime survey is mate- 
 rially shortened. From each notable point of the surface of 
 the water, the bearings of two known objects on the land are 
 taken, or the intermediate angles subtended by three such ob- 
 jects are observed. In the first case, those various points have 
 their situations ascertained by Prop. 21. and the second case 
 by Prop. 25. of the Trigonometry. To facilitate the last con- 
 struction, an instrument called the Station-Pointer has been 
 invented, consisting of three brass rulers, which open and set 
 at the given angles. 
 
410 NOTES AND ILLUSTRATIONS. 
 
 23. The nice art of observing has in its progress kept pace with 
 the improved skill displayed in the construction of instruments. 
 Surveys on a vast scale have lately been performed in Europe, 
 with that refined accuracy which seems to mark the perfection 
 of science. After the conclusion of the American war, a me- 
 moir of Count Cassini de Thury was transmitted by the French 
 Government to our Court, stating the important advantages 
 which would accrue to astronomy and navigation, if the dif- 
 ference between the meridians of the observations of Green- 
 wich and Paris were ascertained by actual measurement. A 
 spirit of accommodation and concert fortunately then prevail- 
 ed. Orders were speedily given for carrying the plan into 
 execution ; and General Koy, who was charged with the con- 
 duct of the business on this side of the Channel, proceeded 
 with activity and zeal. In the summer of 178'i, a fundamental 
 base, rather more than five miles in length, was traced on 
 Hounslow Heathj about 54< feet above the level of the sea, and 
 measured with every precaution, by means of deal rods, glass 
 tubes, and a steel chain, allowance being made for the effects 
 of the variable heat of the atmosphere in expanding those ma- 
 terials. The same line was, seven years afterwards, remea- 
 sured with an improved chain, which yet gave a difference on 
 the whole of only three inches. The mean result, or 27404'.2 
 feet, at the temperature of 62° by Fahrenheit's scale, is there- 
 fore assumed as the true length of the base. Connected with 
 this line, and commencing from Windsor Castle, a series of 
 thirty-two primary triangles was, in 1787 and 1788, extended 
 to Dover and Hastings, on the coast of Kent and Sussex. 
 Two triangles more stretched across the Channel. The hori- 
 zontal and vertical angles at each station were taken with sin- 
 gular accuracy by a theodolite, which the celebrated artist 
 Ramsden had, after much delay, constructed, of the largest 
 dimensions and the most exquisite workmanship. At the 
 same period, a new base of verification was measured on Rom- 
 ney Marsh, 15^ feet above the sea, and found, after various 
 reductions, to be 28535.6773 feet in length. This base, com- 
 puted from the nearest chain of triangles dependent on that 
 of Hounslow Heath, ought to have been 28533.3 ; differing 
 scarcely more than two feet on a distance of eighty miles. 
 The mean, or 28534'.5, is adopted for calculating the adjacent 
 
NOTES AND ILLUSTRATIONS. 
 
 411 
 
 and subsequent triangles. These triangles near the coast 
 were unavoidably confined and oblique ; but their sides are 
 generally deduced from larger and more regular triangles, ex- 
 panding over the interior of the country. The annexed figure 
 exhibits the most interesting portion of this memorable sur- 
 vey, and represents the various combination of triangles. At- 
 tached to it is a scale of English miles. 
 
 A Frant Church. 
 B Goodhurst Church. 
 C Hollingborn Hill. 
 D Tenterden Church. 
 E Fairlight Down. 
 F Allington Knoll. 
 G Lydd Church. 
 H Ruckinge. 
 I High Nook. 
 
 K Folkstone Turnpike. 
 
 L Padlesworth. 
 
 M Swingfield Church. 
 
 N Dover Castle. 
 
 O Church at Calais. 
 
 P Blancnez Signal. 
 
 R Fiennes Signal. 
 
 S Montlambert Signal. 
 
 KL The base of verification. 
 
 hm 
 
 CE± 
 
 Calculation of the sides of the Triangles, 
 
 ACE 
 A 70*» 23' 2" 
 C 52 II 2 * 
 E 48 25 55 
 
 A 27 
 B 136 
 C 16 
 
 ABC 
 
 4 36.13 
 27 35.87 
 27 48 * 
 
 141744.4 
 
 113926 
 107895.7 
 
 71298.5 
 44391.2 
 
 ABE 
 A 43* <S' 25.'/87 
 B 105 39 28.86 
 E 31 2 5.27 
 
 BCD 
 
 68 13 10.5 
 44 38 44.04 * 
 67 7 56.46 
 
 -JQ 
 
 93629.2 
 
 71887.5 
 54376.5 
 
412 
 
 NOTtS AND ILLUSTRATIONS. 
 
 DDE 
 
 B 49° 39' 2>i".77 7\637.'2 
 
 D ()4 59 25.81 93629:2 
 
 li 35 20 58.42 
 
 CDF 
 
 C 40 5'6.96 * 6]777-5 
 
 T> 91 34 22.01, .96039.8 
 
 r 48 24 39 
 
 DFG 
 
 1) 43 45 23.18 47850.9 
 
 ¥73 27 66169.2 
 
 G 63 14 9.82 * 
 
 DEG 
 
 D 62 32 32.51 71692.2 
 
 E 54 59 17-31 
 
 G 62 27 50.18* 71637.2 
 
 478509 
 
 
 EFG 
 
 E 21 18 
 
 37* 
 
 r 32 59 
 
 23 
 
 G 125 42 
 
 
 
 106926.2 
 
 FGI 
 
 F 33 8 46.1 31363.7 
 
 G 26 57 29.9* 23185.7 
 
 1 121 53 44 
 
 FHI 
 
 E 91 27 19-5 28534.5 
 
 H 54 19 18.5 
 
 I 34 13 22 16053 
 
 FGK 
 
 F 109 50 3935 84662.8 
 
 G 38 2 23.76 554631.6 
 
 K 33- 6 56.89 * 
 
 E 13 
 G 154 
 L 12 
 
 EGL 
 38 2.95 
 
 5 54 4 
 16 2.65 
 
 79536.1 
 14739.2 
 
 F;K 
 
 5170cS 
 
 1 79 41 0.5 . 
 
 K 24 17 6.25 
 
 IKL 
 
 I 14 48 25.5 * 14714.3 
 
 K 57 2 48305.2 
 
 L 108 9 345 
 
 KLM 
 
 K 60 27 39.5 170566 
 
 L 70 54 5.5 18525 8 
 
 M 48 38 15 
 
 30560.4 
 31555 7 
 
 42562.7 
 
 KMN 
 K 19 43 53.5 
 M 75 36 40 
 N 34 39 26.5 
 
 KLN 
 K 130 11 33 
 L 34 29 42.5 
 N 15 18 44.5 
 
 ELN 
 
 E 6 6 39.43 
 
 L 152 15 25.15 186119 
 
 N 21 37 55.42 * 
 
 ENP 
 
 E 25 33 55.02 ♦ II6660 
 
 N 110 55 29.83* 252505,6 
 
 P 43 30 35.15* 
 
 ENS 
 43 19 53.32 
 87 30 29.38 
 49 9 31.9 
 
 N 23 
 P 119 
 
 S 36 
 
 25 
 41 
 53 
 
 NPS 
 0.25 
 41.64 
 18 11 
 
 168827 
 245786 
 
 77237 2 
 
NOTES AND ILLUSTRATIONS, 413 
 
 .. In this register, each angle in the successive triangles is, for 
 the sake of conciseness, marked by the single letter affixed to 
 it, and the computed length of its opposite side in feet ranges 
 in the same line. Tlie addition of an asterisk denotes that an 
 angle was not actually observed, but only deduced from cal- 
 culation. The oblique triangles ABC and ABE have their 
 sides BC and BE derived from other larger triangles, which 
 were nearly equiangular. The triangles ELN and E-NP had 
 their angles discovered from conjoined observations. In ge- 
 neral the several angles, as affected by the spherical excess, 
 were corrected for computation by a sort of tentative process. 
 It results from a train of calculations, that Dover Castle lies 
 south 67° 44' 34" east, and at the distance of 328231 feet or 
 62.165 miles, from Greenwich Observatory. On their part, 
 the French astronomers, under the direction of Cassini, car- 
 ried forward the trigonometrical operations from Dunkirk to 
 Paris ; employing Borda*s repeating circle, an instrument much 
 smaller and less perfect than Ramsden's theodolite, but form- 
 ed on a principle which always procures the observer a near 
 compensation of errors. From a comparison of the whole, it 
 follows, that the meridian of the Observatory of Paris lies 2° 
 19' ^l" east from that of Greenwich, differing only nine se- 
 conds in defect from what the late Dr Maukelyne had pre- 
 viously determined from combined astronomical observations. 
 The success with which that great survey was attended, 
 gave occasion both in France and England to still more ex- 
 tensive projects. The National Assembly, amidst other es- 
 sential improvements which it meditated, having resolved to 
 adopt a general and consistent system of measures, the length 
 of a degree of the meridian at the middle point between the 
 pole and the equator was proposed as a permanent basis. But 
 to secure greater accuracy in determining the standard, it had 
 been decided to prolong the observations on both sides of the 
 mean latitude, and trace a chain of triangles over the whole 
 extent from Dunkirk to Barcelona. This bold plan was exe- 
 cuted in the course of the years 1792, 1793, 1794 and 1795, 
 with equal sagacity and resolution, by MM. Delambre and 
 Mechain, who, during all the horrors of revolutionary com- 
 
4H« 
 
 NOTES AND ILLUSTRATIONS. 
 
 motion, yet pressed forward their operations in spite of obsta- 
 cles and dangers of the most sickening kind. After the va- 
 rious triangles, amounting in total to 115, had been observed, 
 they were connected, in the neighbourhood of Paris, with a 
 base of more than seven miles in length, and measuring, at 
 the temperature of 16^° on the centigrade scale, or 6l^° by 
 Fahrenheit, 6075. 9 toises from Melun to Lieursaint. A base 
 of verification was likewise traced near the southern extremity 
 of the line of survey, extending 6006.25 toises along the road 
 from Perpignan to Narbonne. This base appeared not to differ 
 one foot from the calculation founded on the other, though 
 separated by a distance of 400 miles, — a convincing proof of 
 the accuracy with which the observations had been made. A 
 specimen of the French triangulation is given in the figure be- 
 low, where the vertical line represents the meridian of Dun- 
 kirk, with the distances expressed by intervals of 10,000 toises. 
 
 A St Martin du Tertre. 
 
 B Dammartin. 
 
 C Pantheon at Paris. 
 
 D Belle Assise. 
 
 E Brie, 
 
 F Montlheri. 
 
 G Lieursaint. 
 
 H Melun. 
 
 I Malvoisine, 
 
 K Torfou. 
 
 L Foret. 
 
 M Chapelle. 
 
 N Pithiviers. 
 
 O Bois Commun. 
 
 P Chatillon. 
 
 Q Chateau- neuf. 
 
 R Orleans. 
 
 GH The primary base. 
 
NOTES AND ILLUSTRATIONS. 
 
 415 
 
 Calculation of the sides of the Triangles, 
 
 ABC 
 A 76^ 2'30\66 17310.3013 
 B 57 20 17.82 150173211 
 C 46' 37 11.52 
 
 BCD 
 
 B 59 52 2.20 15756.8013 
 C 48 17 34.50 13601.3539 
 D 7150 23.30 
 
 CDE 
 C 37 1 40.59 9516.5896 
 D 57 21 1.87 13305.8528 
 E 85 37 17.54 
 
 CEF 
 C 61 13 47.94 13101.0845 
 E 55 51 48.75 12370.8194 
 F 62 54 23.31 
 
 EFI 
 
 E 40 32 37.60 
 F 45 18 40.41 
 1 74 8 41.99 
 
 8852.8293 
 12374.2130 
 
 FIG 
 
 F 49 34 22.32 8369.1673 
 I 76 47 42.98 10703.5616 
 G 53 37 54.70 
 
 IGH 
 
 I 40 36 56.68 
 G 75 39 29.67 
 H 63 43 33.65 
 
 6075.S993 
 9042.5510 
 
 FIK 
 
 F 55 10 1.03 7357.8627 
 
 I 43 52 3.25 6212.1595 
 
 K 80 57 55.72 
 
 IKL 
 53° 22' 24" 93 8349.1059 
 81 36 49.90 10^92.0814 
 45 45.17 
 
 I 70 51 
 L 62 47 
 M 46 20 
 
 ILM 
 
 37.77 13438.2345 
 29.54 12650.5655 
 52.69 
 
 LMN 
 
 L 68 35 59.16 14402.0625 
 M 51 5 13.26 12036.0949 
 N 60 IS 47.58 
 
 MNO 
 
 M 31 58 52.87 9190.1355 
 N 91 55 5.70 17341,8323 
 O 56 6 1.43 
 
 NOP 
 
 N 31 53 2.40 4877.2386 
 
 O 52 33 5,48 7330.6l66 
 
 P 95 33 52.12 • . 
 
 OPQ 
 
 O 62 31 30.34 10446 5520 
 P 93 17.27 11758.3955 
 Q 24 28 1239 
 
 PQR 
 
 P 50 28 6.42 12053.9075 
 Q 87 35 8.93 15614.7105 
 R 41 56 44.65 
 
 Through the whole process of their survey, the French 
 astrgnomers have certainly displayed superior science. In de- 
 
4-16 NOTES AND ILLUSTRATIONS. 
 
 ducing the correct results, they seem to exhaust all the re- 
 finements of calculation. The angles measured by the re- 
 peating circle, it was necessary to reduce, not only to the ho- 
 rizontal plane, but generally besides to the centre of observa- 
 tion. This would have required much nice and tedious com- 
 putation ; the labour of performing such reductions was how- 
 ever greatly simplified and abridged, by help of concise fov' 
 mulcB^ and the application of auxiliary tables. There is even 
 room to suspect that those ingenious philosophers have car- 
 ried the fondness for numerical operations to an excess, and 
 often pushed the decimal places to a much greater length in 
 their estimates than the nature of the observations themselves 
 could safely warrant. 
 
 In the spring of 1799, the registers of all these operations 
 . were referred to a commission, consisting of the ablest mem- 
 bers of the Institute, and some other learned men deputed 
 from the countries then at peace with France. The various 
 calculations were carefully examined and repeated ; and a 
 comparison of the celestial arc with that which had been mea- 
 sured in Peru having given — - for the oblateness of the earth, 
 
 the length of the quadrant of the meridian, or the distance of 
 the pole from the equator, was finally determined at 5130710 
 toises, the ten millionth part of which, or the space of 
 443.295936 lines forms the metre. This standard was after- 
 wards definitively decreed by the Legislative Body. 
 
 Mechain, however, still anxious to realize his early project 
 of extending the meridian as far as the Balearic Isles, again 
 repaired to Spain, and conducted with incredible exertions a 
 chain of triangles over the savage heights from Barcelona to 
 Tortosa, and was about to observe the altitude of the stars, 
 and measure the base of Oropesa, when, worn out by continued 
 fatigue, he caught an epidemic fever, which fatally closed his 
 meritorious labours, at Castellon de la Plana, in the kingdom 
 of Valentia, about the latter part of September 1805. — The 
 prosecution of the plan was subsequently committed to 
 MM. Biot and Arago, who brought it to a fortunate conclu- 
 sion. In the winter of 1806 and the spring of 1807, these 
 
•NOTES AND ILLUSTRATIONS. 4-17 
 
 philosophers contin»ed the series of triangles from Barce- 
 lona to the kingdom of Vaientia, and joined that coast with 
 the Balearic Isles, by an immense triangle, of which one of 
 the sides exceeded an hundred miles in length. At such 
 prodigious distances, the stations, however elevated, and not- 
 withstanding the fineness of the climate, could not be seen 
 during the day ; but they were rendered visible at night, by 
 combining Argand lamps with powerful reflectors. These ob- 
 servations give a result which agrees almost exactly with what 
 had been already found by Delambre and Mechain. If the 
 mean were adopted, it would yet scarcely affect the length of 
 the metre by the diminution of a four millionth part, making 
 this to be 443.322 lines of the toise brought by the Academi- 
 cians from Peru. The meridional arc extending from Dunkirk 
 to Formentera, measures 12° 22' 13".395 ; and from this ample 
 basis, the circumference of the earth is computed to be 24855.42 
 English miles, and the ratio of its axes that of 308 to 309. 
 
 The fourth volume of the Base Metrique, containing the ac- 
 count of the trigonometrical observations made by Biot and 
 Arago in Spain and the Balearic Isles, has been long promi- 
 sed ; and I was induced, for a considerable time, to defer the 
 publication of this edition, in the hope of being able to draw 
 come additional information from such a valuable source. In 
 the prosecution, however, of the French measurement, an ap- 
 plication from the Institute has been transmitted by Count 
 Laplace to Colonel Mudge, to have Ramsden's Zenith Sector 
 erected near Yarmouth, in order to connect the English arc 
 thence across the sea to near Dunkirk, with the meridional 
 measurement extending through France and Spain to For- 
 mentera, which would have the important advantage of being 
 nearly bisected by the parallel of 45^. This proposition, I am 
 happy to learn, will be carried into immediate effect. 
 
 In England, the prosecution of the trigonometrical survey, 
 without aiming at such splendid views, has, suitably to the ge- 
 nius of the people, been directed to objects of more domestic 
 interest, and perhaps real utility and importance. The per- 
 plexing inaccuracy of our best maps and charts had long been 
 the subject of most serious complaint. It was in consequence 
 resolved to extend the series of connected triangles over the 
 
 2E 
 
418 NOTES AND ILLUSTRATIONS. 
 
 wliole surface of the Island. But the death of General Roy, 
 happening so early as 1790, threatened to prove fatal to the 
 completion of his favourite scheme, for which the talents and 
 experience he possessed had so eminently fitted him. Af- 
 ter some interruption, however, an opportunity was embraced 
 of resuming that noble plan ; and it was, under the direction 
 of the Board of Ordnance, committed to the care of Colonel 
 Mudge, who, with equal ability and undiminished ardour, has, 
 during the space now of upwards of twenty years, been engaged 
 in carrying on the most extensive and varied system of ope- 
 rations ever attempted, and in a style of execution which re-: 
 fleets on him the highest credit. In 1793 and 1794^, the chain 
 of primary triangles was continued from Shooter's Hill to Dun- 
 nose in the Isle of Wight, including a great part of Surry, Sus- 
 sex, Hants, Wiltshire and Dorsetshire, and connecting with a 
 new base of verification measured on Salisbury Plain. This 
 base had, after correction, a length of 36571-4^ feet, or 6.92697 
 miles, having lost almost a whole foot in being reduced from 
 an elevation of .588 feet to the level of the sea. It differed 
 scarcely an inch from the computation founded on the base of 
 Hounslow Heath. In 1795, the triangles were carried into 
 Devonshire ; and they were continued in 1796 through Corn* 
 wall to the Sciliy Islands. The West of England became the 
 scene of repeated operations. In 1798, a third base was mea- 
 sured on King's Sedgeraoor near Somerton, and found, after 
 various corrections, to be 27680 feet, or 5.242425 miles, dif- 
 fering only about a foot from the result of the calculation de- 
 pendent on that of Salisbury Plain. The survey now advanced 
 to the centre of England, and was extended in 1803 to Clif- 
 ton in Yorkshire ; another base of verification, 2.6342.7 feet 
 in length, having been measured at Misterton Carr,on the north 
 of Lincolnshire. The triangles were next carried towards 
 Wales, and made to rest on a base of 24514.26 feet, stretch- 
 ing from the western borders of Flintshire to Llandulas in Den- 
 bighshire. From this last base, numerous triangles have been 
 extended in different directions ; one series bending through 
 Anglesea and by Cardigan Bay, to tlie Bristol Channel ; an- 
 other penetrating into the central parts of England ; while a 
 third series stretches northwards, through Lancashire, Cum- 
 
KOTES AND ILLUSTRATIONS. 419 
 
 berland and Westmoreland, into Scotland, and uniting with 
 the collateral chain of Misterton Carr from Yorkshire and 
 Northumberland, is prolonged to the heights immediately be- 
 yond the Firth of Forth. We look forward with anxiety to 
 the conclusions of this arduous undertaking. The mountains 
 and islands near the western coast of Scotland will furnish tri- 
 angles of vast extent. Colonel Mudge will not omit, we are 
 confident, the opportunities that such stations may afford to 
 determine the quantity of horizontal refraction, noting at the 
 same time the variable state of the atmosphere. The indica- 
 tions of the hygrometer would then require attention. We 
 have perfect reliance in the accuracy of his observations ; yet 
 it would be desirable in all cases, as in the French operations, 
 that the third angle of each triangle were actually measured. 
 It would likewise be satisfactory, in surveying the more moun- 
 tainous tracts, that the barometer should always accompany 
 the theodolite, that both modes of determining the altitudes 
 of the stations might be compared. 
 
 The triangulation has been extended along the east coast 
 of Scotland as far as the county of Banff and the borders 
 of Ross-shire. It has also been carried towards the same 
 points from Cumberland, through the heights of Galloway and 
 Dumfries-shire, to the summit of Ben-Lomond; and from Dum- 
 bartonshire and the vicinity of Glasgow in a north-easterly di- 
 rection, connecting all the remarkable mountains of Perth- 
 shire. The sands of Belhelvie, a few miles westward of Aber* 
 deen, the spot formerly pointed oiit by General Roy, is now 
 selected for a base of verification, which Colonel Mudge in- 
 tends to measure in person this summer. It would no doubt 
 Be very desirable to have another intermediate base determi- 
 ned nearer the west side of the island. For this purpose, the 
 plain between Kinniel and Carron, in the Carse of Falkirk, 
 might seem eligible. 
 
 Besides the principal triangles thus determined, a multitude 
 of subordinate ones were ascertained in the progress of the 
 survey, which serve to connect all the remarkable objects 
 that occurred over the face of the country. The capital points 
 were hence established for constructing: the most accurate 
 charts and provincial maps. A number of royal military sur- 
 
1-20 NOTES AND ILLUSTRATIONS. 
 
 veyors, of approved skill, have since been constantly employed 
 in filling up the secondary triangles, and embodying the ske- 
 leton plans. The various materials are collected at the draw- 
 ing-room of the Tower, and there adjusted, reduced and com- 
 bined. Under the same able direction, an extensive establish- 
 ment has been formed in those spacious apartments, where 
 a voluminous series of maps, on the largest scale, are not 
 only delineated but engraved. This truly national work ad- 
 vances with great activity, and has already proved highly ad- 
 vantageous to the public service. The Ordnance Maps, in 
 elaborate accuracy, and even beauty of execution, surpass 
 every thing hitherto designed. 
 
 The publication of these valuable geographical details, after 
 having been suspended for some years, is again free. Five 
 parts have already appeared, including Devonshire, Essex, 
 Sussex, Dorsetshire, Kent, the Isle of Wight, Hampshire and 
 Cornwall. Other maps are in a state of great forwardness, as 
 far northward as the parallel from Caernarvon through Shrews- 
 bury and Warwick to twenty miles beyond Boston in Lincoln- 
 shire. The completion of a work of such vast magnitude 
 must require proportional tinve and perseverance. The ma- 
 ritime counties will probably be first given to the public, and 
 the districts of the interior afterwards delivered. 
 
 For a concise and perspicuous exemplification of all the re- 
 finements adopted in the practice of-tf igonometrical surveying, 
 I have much satisfaction in referring to the late work of Baron 
 Zach sur P Attraction des Montagues ; nor can I omit this op- 
 portunity of testifying my respect and regard for that able 
 and very learned astronomer, in whose interesting society I 
 made a delightful excursion, in the month oi August 1814, 
 from Lyons by Orange to Vaucluse, and thence by Avignon 
 to Marseilles, where he was then residing, as chamberlain to 
 her Highness the Dowager Duchess of Saxe-Gotha. 
 
 22. To determine geometrically the altitude of a mountain re- 
 quires, it hence appears, a nice operation performed with some 
 large instrument. The barometrical mensuration of heights 
 is therefore, in most cases, preferred, as much easier and often 
 more exact. This curious application was early suggested, by 
 
NOTES AND ILLUSTRATIONS. 
 
 421 
 
 the objections themselves which ignorance opposed to Torri- 
 celli's iiipnortal discovery of the weight of our atmosphere. 
 But more than a century elapsed before the improvements in 
 mechanics had completely adapted the machine to that pur- 
 pose, and experiment combined with observation had ascer- 
 tained the proper corrections. Barometers of various con- 
 structions are now made quite portable, and which indicate 
 with the utmost precision the height of the mercurial column 
 supported by the pressure of the atmosphere. 
 
 The air which invests our globe, being a fluid extremely 
 compressible, must have its lower portions always rendered 
 denser by the weight of the incumbent mass. To discover the 
 law that connects the densities with the heights in the atmos- 
 phere, it is only requisite, therefore, to apply the fact which 
 experiment has established,-^that the elasticity counterbalan- 
 cing the pressure is exactly proportioned to the density. The 
 elasticity of the air at any point of elevation, is hence mea- 
 sured by a column possessing the same uniform density, with 
 a certain constant altitude. Let AB denote the height of this 
 equiponderant column, and the perpendicular Bl its density ; 
 and suppose the mass of air below to be distinguished into nu- 
 merous strata^ having each the same thickness BC. It is evi- 
 dent that the weight of the niinute stratum at B will be ex- 
 pressed by BC ; whence AB is to A,C, or BI to CK, as the. 
 pressure at B to the augmented pressure at C, and therefore 
 the density at C is denoted by CK, Again, having joined IC, 
 
 Be^Di5.FG:.lC 
 
 and drawn KD parallel, BI : CK : : BC : CD ; and conse- 
 quently CD will, on the same scale of density, express the 
 weight of the stratum at C. Hence, AC is to AD, as CK to 
 DL, or as the density at C is to that at D. It thus appears, 
 that, repeating this process, the densities BI, CK, DL, &c. of 
 the successive strata form a continued geometrical progression. 
 But the same relation will evidently obtain at equal though 
 sensible intervals. Thus, the density of the atmosphere is re- 
 
422 NOTES AND ILLUSTRATIONS. 
 
 duced nearly to one half, tor every % miles of perpendicular 
 ascent. At 7 miles in height, the corresponding density is 
 one-fourth ; at 10| miles, one-eighth ; and at 14 miles, one- 
 sixteenth. 
 
 The difference of altitude between two points in the atmos- 
 phere, is hence proportional to the difference of the logarithms 
 of the corresponding densities or vertical pressures. But the 
 heights of mountains may be computed from barometrical 
 measurement to any degree of exactness, by a simple nume* 
 rical approximation. Since AB, AC, AD, &c. are continued 
 proportionals, it follows that AB: BC : : AB+ AC+AD, &c. ; 
 BC-f-CD-fDE, &c. or BH. Let n denote the number of sec- 
 tions or strata contained in the mass of air, and — ( AB4- AH) 
 
 will nearly express the sum of the progression AB, AC, AD, 
 &c.; wherefore, AB+ AH : BH : : 2AB : wBC, or the absolute 
 difference of altitude. The height AB of the equiponderant 
 column, reduced to the temperature of freezing water, is near- 
 ly 'zG^O feet ; and hence this general rule, — As the sum of 
 the mercurial columns is to their difference, so is the constant num- 
 ber 52,000 to the approximate height. This number is the more 
 easily remembered, from the division of the year into weeks. 
 
 Two corrections depending on the variation of temperature 
 are besides required. 1. Mercury expands about the 5,000th 
 part of its bulk, for each degree of the centigrade scale ; and 
 hence the i,mall addition to the upper column mil he found, hy 
 removing the decimal point four places to the left, and multiplying 
 hy tvoice the difference hettveen the degrees of the attached ther mo ^ 
 meters. 2. But the correction afterwards applied to the prin- 
 cipal computation is of more consequence. Air has its vo- 
 lume increased by one 250th part, for each degree of heat 
 on the same scale. If therefore, the approximate height, ha- 
 ving its decimal point shifted back three places, be midtiplied by 
 ituice the sum of the degrees on the detached thermometers, the 
 product ivill give the addition to he made. If it were worth 
 while to allow for the effect of centrifugal force in diminishing 
 the pressure of the aerial column, this will be easily done be- 
 fore the last multiplication takes place, by adding to twice the 
 degrees on the detached thermometers the j(^M par^ of the 
 piean temperature corresponding to the latitude. 
 
NOTES AND ILLUSTRATIONS. 423 
 
 An example will elucidate the whole process. In August 
 1775, General Roy observed the barometer on Caernarvon 
 Quay at 30.091 inches, the attached thermometer being 15° .7, 
 and the detached 15°. 6 centigrade, while on the Peak of Snow- 
 don the barometer stood at 26.409, the attached thermometer 
 marking 10°.0, and the detached 8°.8. Here, twice the dif- 
 ference of the attached thermometers is 11°.4, which multi- 
 plied into .00264 gives .030, for the correction of the upper 
 barometer. Next, 30.091 + 26.439 : 30.091 — 26.439, or 
 56.530 : 3.652 : : 52000 : 3359. Again, twice the sum of the 
 degrees marked on the detached thermometers is 48.8, which 
 multiplied into 3.359 gives 164 ; wherefore, the true height of 
 Snowdon above the Quay of Caernarvon is 3359-}- 164, or 3533 
 feet. The correction for centrifugal force is only 7 feet more. 
 
 This mode of approximation may be deemed sufficiently 
 near, for any heights which occur in this island ; but greater 
 accuracy is attained by assuming intermediate measures. To 
 illustrate this, I shall select another example. At the very 
 period when General Roy was making his barometrical obser- 
 vations at home, Sir George Shuckburgh Evelyn found the 
 barometer to stand at 24.167 on the summit of the Mole, an 
 insulated mountain near Geneva, the attached and detached 
 thermometers indicating 14°.4 and 13°.4, while they marked 
 16°.3 and 17°.4 at a cabin below and only 672 feet above the 
 lake, the altitude of the barometer at this station being 28.132. 
 Now, 3.8x.0024=.009, and 24.167-^009=24.176; the arith- 
 metical mean between which and 28.1 32 is 26.154 ; and hence, 
 separately, 50.330 : 1.978 : : 52000 : 2044, and 54.286 : 1.978 : : 
 52000 : 1895. Wherefore, joining these two parts, 2044-}- 1895, 
 or 3939 expresses the approximate height. Tiie final correc- 
 tion is 61.6x3.939=243, or 254 feet, if allowance be made 
 for the effect of centrifugal force,^ and consequently the Mole 
 has its summit elevated 4865 feet above the lake of Geneva, 
 and 6063 above the level of the sea. 
 
 In general, let A and A -|- w6 denote the correct lengths of 
 the columns of mercury at the upper and the lower stations ; 
 the approximate height of the mountain will be expressed by 
 
 \2A:fb + 2A4.3^ + 2A + 5b '" + 2A + 2n^].b} ^^^^' 
 
424? NOTES AND ILLUSTRATIONS. 
 
 If w were assumed a large number, the result would approadb 
 to the accuracy of a logarithmic computation, though such an 
 extreme degree of precision will be scarcely ever wanted. 
 
 To expedite the calculation of heights from barometrical 
 observations, I have now caused Mr Gary, optician in London, 
 to make for sale a sliding-rule, of an easy and commodious 
 construction. That small instrument, which should be accom- 
 panied with a barometer of the lightest and most portable 
 kind, will be found very useful to mineralogical travellers who 
 have occasion to explore mountainous tracts. Nothing could 
 tend more to correct our ideas of physical geography, than to 
 have the principal heights in all countries measured, at least 
 with some tolerable degree of precision. But the elevation of 
 any place above the sea may be ascertained very nearly, from 
 the comparison of even very distant barometrical observations, 
 especially during the steadiness of the fine season in the hap- 
 pier climates. In the summer of 181 4, Engelhardt and Par- 
 rot, two Prussian travellers, by a series of fifty-one barometri- 
 cal observations, made along the distance of 711 miles, from 
 the Caspian to the Black Sea, ascertained the former to be 
 334< English feet below the level of the latter, which complete- 
 ly oversets the supposition of any subterranean communica- 
 tion existing between those seas. By the same mode may be 
 traced a profile or vertical section, that shall exhibit at one 
 glance the great features of a country. As a specimen, I have 
 combined and reduced the sections which the celebrated philo- 
 sophic traveller Humboldt has given of the continent of Ame- 
 rica, running in a twisted direction from Acapulco to Vera 
 Cruz, and connecting the Pacific with the Atlantic Ocean. 
 
 A Acapulco. 
 
 f Venta de Chalco. 
 
 a Peregrino, 
 
 g St Martin, 
 
 B Chilpansingo. 
 
 E La Puebla de los Angeles. 
 
 b Mescala. 
 
 h El Pinal 
 
 c Tepecuacuilco. 
 
 I Perote, 
 
 d Puente de Istla, 
 
 k Cruz Blanca. 
 
 C Cuernavaca. 
 
 F Xalafa. 
 
 e La Cruz del Marques, 
 
 G Vera Cruz. 
 
 D Mexico. 
 
 
TO^oOd 
 
 NOTES AND ILLUSTRATIONS. 
 
 A 
 
 42J 
 
 J01009 
 
 The divided scale expresses the horizontal distance in miles, 
 while the parallels, on a much larger scale, mark the elevation 
 in feet. This profile is really composed of four successive 
 sections, which are distinguished by opposite shadings. The 
 survey proceeded first along the road from Acapulco to Mex- 
 ico, thence to Puebla de los Angeles, next to Cruz Blanca, 
 and finally to Vera Cruz. These several directions and dis- 
 tances are expressed in the ground plan. 
 
 An attempt is likewise made in this profile, to convey some 
 idea of the geological structure of the external crust : 
 Limestone is represented by straight lines slightly inclined 
 
 from the horizontal position. 
 Basalt, by straight lines slightly reclined from the perpendi- 
 cular. 
 Porphyry, by waved lines somewhat reclined. 
 Granite, by confused hatches. 
 Amygdaloid, by confused points. 
 
 But the easiest way of estimating within moderate limits the 
 elevation of a country, is founded on the difference between 
 the standard and the actual mean temperature as indicated by 
 deep wells or copious aod shaded springs. Professor Mayer 
 
426^ 
 
 NOTES AND ILLUSTRATIONS. 
 
 ©f Gottingen, from a comparison of distant observations (m 
 the surface of the globe, proposed aformuhf which, with a 
 slight modification, appears to exhibit correctly the tempera- 
 ture of any place at the level of the sea. Let (p denote the la- 
 titude ; and 29 cos(p*, 
 er .14^ suvers 2<p, b 
 will express, in de- 
 grees of the centi- 
 grade scale, the me- 
 (diura heat on the 
 coast. But the gra- 
 dations of climate 
 are more easily con- 
 eeLved by help of a 
 geometrical diagram . 
 From the centre C, 
 draw straight lines 
 to the several de- 
 grees of the qua- 
 drant, and cutting 
 the interior semi- 
 circle; then the radius CA denoting 29% the perpendiculars 
 from the points of section will intercept segments proportional 
 to the mean temperature expressed on DE.^ 
 
 The higher regions are invariably colder than the plains ; 
 and I have been able, after a delicate and patient research, to 
 fix the law which connects the decrease of temperature with 
 the altitude. If B and b denote the barometric pressure at the 
 
 -7 ■^) Q5 express, 
 
 ©n the centigrade scale, the diminution of heat in ascent. 
 Hence, for any given latitude, that precise point of elevation 
 
 B 
 
 may be found, at which eternal frost prevails. Put x rr 
 
 and t = the standard temperature ; then / — -— x ) 25=^, or 
 
 V ^ 
 x^ -{• .04<txz=: 1, which quadratic equation being resolved, 
 
 gives the relative elasticity of the air at the limit of congela- 
 tion, whence the corresponding height is determined. From 
 these data the following table has been calculated. 
 
NOTES AND ILLUSTRATIONS. 
 
 42* 
 
 
 Mean temperature at 
 
 
 
 Mean temperature at 1 
 
 
 Lati. 
 tude. 
 
 the level of the Sea. 
 
 Height of Curve 
 of Congelation. 
 
 T f- 
 
 Latu 
 tude. 
 
 the level of the Sea. ' 
 
 Height of curve 
 
 of Congelation. 
 
 Feet. 
 
 7402 
 
 0° 
 
 Centigrade 
 29°.00 
 
 Fahrenheit. 
 
 46° 
 
 Centigrade 
 1S°.99 
 
 Fahrenheit. 
 570.2 
 
 84°.2 
 
 15207 
 
 1 
 
 28.99 
 
 84.2 
 
 15203 
 
 47 
 
 13.49 
 
 56.3 
 
 7133 
 
 2 
 
 28.96 
 
 84.1 
 
 15189 
 
 48 
 
 12.98 
 
 55.4 
 
 6,865 
 
 5 
 
 28.92 
 
 84.0 
 
 15167 
 
 49 
 
 12.43 
 
 54.5 
 
 6599 
 
 4 
 
 28.86 
 
 83.9 
 
 15135 
 
 
 
 
 
 5 
 
 28.78 
 
 83.8 
 
 15095 
 
 50 
 
 11.98 
 
 53.6 
 
 6334 
 
 6 
 
 28.68 
 
 83.6 
 
 15047 
 
 51 
 
 11.49 
 
 52.7 
 
 6070 
 
 7 
 
 28.57 
 
 83.4 
 
 14989 
 
 52 
 
 10.99 
 
 51.8 
 
 5808 
 
 8 
 
 28.44 
 
 83.2 
 
 14923 
 
 53 
 
 10.50 
 
 50 9 
 
 5548 
 
 9 
 
 28.29 
 
 82.9 
 
 14848 
 
 54 
 
 10-02 
 
 50.0 
 
 5290 
 
 
 
 
 
 53 
 
 9.54 
 
 49.2 
 
 5034 
 
 10 
 
 28.15 
 
 82.6 
 
 14764 
 
 56 
 
 9.07 
 
 48.3 
 
 4782 
 
 11 
 
 27.94 
 
 82.3 
 
 14672 
 
 57 
 
 8.60 
 
 47.5 
 
 4534 
 
 12 
 
 27.75 
 
 82.0 
 
 14571 
 
 58 
 
 8.14 
 
 46.6 
 
 4291 
 
 13 
 
 27.53 
 
 81.6 
 
 14463 
 
 59 
 
 7.69 
 
 45.8 
 
 4052 
 
 14 
 
 27.30 
 
 81.1 
 
 14345 
 
 
 
 
 
 15 
 
 27.06 
 
 80.7 
 
 14220 
 
 60 
 
 7.25 
 
 45.0 
 
 3818 
 
 16 
 
 26.80 
 
 80.2 
 
 14087 
 
 61 
 
 6.82 
 
 44.3 
 
 3589 
 
 17 
 
 26.52 
 
 79.7 
 
 13947 
 
 62 
 
 6.39 
 
 4.3.5 
 
 3365 
 
 18 
 
 26.2.3 
 
 79.2 
 
 1379S 
 
 63 
 
 5.98 
 
 42.8 
 
 3145 
 
 19 
 
 25.93 
 
 78. 
 
 13642 
 
 64 
 
 5.57 
 
 42.0 
 
 2930 
 
 
 
 
 
 65 
 
 5.18 
 
 41.3 
 
 2722 
 
 30 
 
 25.61 
 
 78.1 
 
 15478 
 
 66 
 
 4.80 
 
 40.6 
 
 2520 
 
 21 
 
 25.28 
 
 77.5 
 
 13308 
 
 67 
 
 4.43 
 
 40.0 
 
 2325 
 
 22 
 
 24.93 
 
 76.9 
 
 13131 
 
 68 
 
 4.07 
 
 39.3 
 
 2136 
 
 23 
 
 24.57 
 
 76.2 
 
 12946 
 
 69 
 
 3.72 
 
 38.7 
 
 1953 
 
 24 
 
 24.20 
 
 75.6 
 
 12755 
 
 
 
 
 
 25 
 
 23.82 
 
 74.9 
 
 12557 
 
 70 
 
 3.39 
 
 38.1 
 
 1778 
 
 26 
 
 23.43 
 
 74.2 
 
 12354 
 
 71 
 
 3.07 
 
 37.5 
 
 1611 
 
 27 
 
 23.02 
 
 73.6 
 
 12145 
 
 72 
 
 2.77 
 
 37.0 
 
 1451 
 
 28 
 
 22.61 
 
 72.7 
 
 11930 
 
 73 
 
 2.48 
 
 36.5 
 
 1298 
 
 29 
 
 22.18 
 
 71.9 
 
 11710 
 
 74 
 
 2,20 
 
 5;.o 
 
 1155 
 
 
 
 
 
 75 
 
 1.94 
 
 35.5 
 
 1016 
 
 50 
 
 21.75 
 
 71.1 
 
 11484 
 
 76 
 
 1.70 
 
 35.1 
 
 887 
 
 31 
 
 21.31 
 
 70.3 
 
 11253 
 
 77 
 
 1.47 
 
 34.6 
 
 767 
 
 32 
 
 20.88 
 
 69.5 
 
 11018 
 
 78 
 
 1.25 
 
 34.2 
 
 656 
 
 33 
 
 20.40 
 
 68.7 
 
 10778 
 
 79 
 
 1.06 
 
 33.9 
 
 552 
 
 34 
 
 19.93 
 
 67.9 
 
 10534 
 
 
 
 
 
 35 
 
 19.46 
 
 67.0 
 
 10287 
 
 80 
 
 .87 
 
 S3.6 
 
 457 
 
 36 
 
 18.98 
 
 66.2 
 
 10036 
 
 81 
 
 .71 
 
 33.3 
 
 571 
 
 87 
 
 18.50 
 
 65.3 
 
 9781 
 
 82 
 
 .56 
 
 33.1 
 
 294 
 
 38 
 
 18.01 
 
 64.4 
 
 9523 
 
 83 
 
 .43 
 
 32.8 
 
 226 
 
 39 
 
 77.51 
 
 65.5 
 
 9263 
 
 84 
 
 .32 
 
 32.6 
 
 167 
 
 
 
 
 
 85 
 
 .22 
 
 32.4 
 
 117 
 
 40 
 
 17.02 
 
 62.6 
 
 9001 
 
 86 
 
 .14 
 
 32.3 
 
 76 
 
 41 
 
 16.52 
 
 61.7 
 
 8738 
 
 87 
 
 .08 
 
 32.2 
 
 44 
 
 42 
 
 16.02 
 
 60.8 
 
 8473 
 
 88 
 
 .04 
 
 3-^.1 
 
 20 
 
 43 
 
 15.51 
 
 59.9 
 
 8206 
 
 89 
 
 .01 
 
 32.0 
 
 5 
 
 44 
 
 15.01 
 
 59.0 
 
 7939 
 
 1 90 
 
 .00 
 
 32.0 
 
 
 
 45 
 
 14.50 . 
 
 58.1 
 
 7671 
 
 1 
 
 
 
 
428 NOTES AND ILLUSTRATIONS. 
 
 This table will facilitate the approximation to the altitude 
 of any place, which is inferred either from its mean tempera- 
 ture or its depth below the boundary of perpetual congelation. 
 The decrements of heat at equal ascents are not altogether 
 uniform, but advance more rapidly in the higher regions of 
 the atmosphere. At moderate elevations, however, it will be 
 sufficiently near the truth, to assume the law of equable pro- 
 gression, allowing in this chmate one degree of cold by Fah- 
 renheit's scale for every ninety yards of ascent, and for every 
 hundred yards in the tropical regions. Thus, the tempe- 
 ratures of the Crawley and Black springs on the ridge of 
 the Pentland hills, were observed by Mr Jardine, where they 
 first issue from the ground, to be 46°.2 and 45° ; which, com- 
 pared with the standard temperature at the same parallel of 
 latitude, would give 567 and 891 feet of elevation above the 
 sea. The real heights found by levelling were respectively 
 5634 and 882; a coincidence most surpiising and satisfactory — - 
 This ready mode of estirnation claims especially the attention 
 of agriculturists. 
 
 Dr Francis Buchanan informs me, that he found the tempe- 
 rature of a spring at Chitlong, in the Lesser Valley of Nepal, 
 to be 14°.7 centigrade. But the mean temperature in the pa- 
 rallel of 27°38' being 22o.8, the density of the atmosphere cor- 
 responding to difference 8°.l, is .8510, which gives 4500 feet 
 for the corrected altitude. From other observations of the 
 same accurate traveller, we may conclude that Kathmandre, 
 the capital of Nepal, is elevated about 2780 feet above the 
 level of the sea. 1 found myself the temperature of the ce- 
 lebrated fountain of Vaucluse, which gushes with such vo- 
 lume as to form almost immediately a respectable river, to be 
 13° centigrade, or 2° less than what corresponds to its latitude 
 or 43° 55'. It may hence be inferred, that Vaucluse is 1080 feet 
 above the level of the Mediterranean. 
 
 The rule stated above for computing the measurements by 
 the barometer, seems to give results somewhat less, on the 
 wbple, thau tbose which are obtained from geometrical ob- 
 servations. It would ensure greater accuracy, perhaps, to 
 view the approximate height as answering to a temperature 
 one degree under the point of congelation ; and consequent* 
 ly, in applying the last correction, to add unit to the indi- 
 
JJOTBS AND ILLUSTRATIONS. 429 
 
 cations of the detached thermometers. But the whole sub- 
 ject demands a more thorough investigation. The elasticity 
 of air is affected by moisture as well as heat, although the 
 want of an exact instrument for measuring the former has 
 hitherto prevented its influence from being distinctly noticed. 
 
 When the hygrometer which I have invented shall become 
 better known to the public, it may not seem presumptuous 
 to expect, in due time, more correct data concerning the mo- 
 difications of the atmosphere. Yet, after all, in ascertaining, 
 the volume of a fluid subject to incessant fluctuation, it would 
 be preposterous to look for that consummate harmony which 
 belongs exclusively to astronomical science ; nor can I help 
 regarding the introduction of 5ome late refinements into the 
 formulce for measuring heights by the barometer, which would 
 embrace the minutest anomalies of atmospheric pressure, as 
 rather a waste of the powers of calculation. 
 
 I shall now subjoin a concise table of the most remarkable 
 heights in different parts of the world, expressed in English 
 feet. The altitudes measured by the barometer are marked 
 B, while those derived from geometrical operations, and taken 
 chiefly from the observations of Colonel Mudge, are distin- 
 guished by the letter G. 
 
 Snae Fiall Jokul, on the north-'west point qflcelandy 4558 G, 
 
 Hekla, volcanic mountain in Iceland^ - - 3950 G 
 
 Sulitelma, m Z«/?/aw^, - * . 5910 B 
 
 Snahatta, centre of the 'Norwegian mountains, - 8 1 20 B 
 
 Harebacke, Alpine ridge ofNortvaj/j - 4575 B 
 
 Pap of Caithness, - - - . 1929 
 
 Ben Nevis, Inverness-shire, ... 4380 B 
 
 Cairngorm, Inverness-shire, - - - 4080 B 
 
 Cairnsmuir upon Deugh, Galloxuay, - - 2597 G 
 
 Ben Lawers, Perthshire, - - « 4015 B 
 
 Ben More, Perthshire, - - . - 3870 B 
 
 Schihallien, Perthshire, - - - 3281 G 
 
 Ben Ledi, Perthshire, - - - - 3009 B 
 
 Ben Lomond, Stirlingshire, - - . 3240 B 
 
 Lomond Hills, east and west, Fifeshire, 1466 and 1721 G 
 
 -Soutra Hill, on the ridge of Lammermuir, - 1716 G 
 
 Coulter Fell, Lanarkshire, - _ . 2440 G 
 
 Carnethy, high point of the Pentland ridge, - 1700 B 
 
430 
 
 NOTES AND ILLUSTRATIONS. 
 
 Tintoc Hill, Lanarkshire, - - _ . 2306 (J 
 
 Leadhills, the house of the Director of the mines, 1280 B 
 
 Broad Law, near Crook Inn, Peebles-shire, - 2741 G 
 
 Queensbery Hill, Dumfries-shire, - - 2259 G 
 
 Caifnsmuir of Fleet, Galloxioay, - - 2329 G 
 
 Hert Fell, near Moffat, - - - ' 2635 G 
 
 Dunrich Hill, Roxburghshire, - - , - 24'21 G 
 
 Elden Hills, near Melrose, Roxburghshire, - 1364 G 
 
 Whitcomb Hill, Peebles- shire, - - 2685 G 
 
 Lother Hill, Dumfries- shire, - - 2396 G 
 
 Ailsa Rock, in the Firth of Clyde, - - 1 103 G 
 
 Crif Fell, near New Abbey, Kirkcudbright, - 1831 G 
 
 Kells Range, Galloway, - - - 2659 G 
 
 Goat Fell, in the Isle of Arran, - - 2865 G 
 Paps of Jura, south and north, in Argyleshire, 2359 and 2470 
 
 ' Snea Fell, in the Isle of Man, - - - 2004 G 
 
 South Beru\e, in Isle of Man, - - 1584G 
 
 Macgillicuddy's Reeks, County of Kerry, - 3404 
 Sliebh Donard, the highest of the Mourne Mountains, 2786 G 
 
 Helvellyn, Cumberland, - - - 3055 G 
 
 Skiddaw, Cumberland, - - - 3022 G 
 
 Saddleback, Cumberland, - - - 2787 G 
 
 Whernside, Yorkshire, - - - 2384 G 
 
 Ingleborough, Yorkshire, - - - 2361 G 
 
 Shunnor Fell, Yorkshire, - - - 2329 G 
 
 Snowdon, Caernarvonshire, - - . 3571 G 
 
 Cii^er IdiVis, Caernarvonshire, - ^ » - 2914 G 
 
 Beacons of Brecknock, - - . 2862 G 
 
 Plynlimmon, Cardiganshire, - - 2463 G 
 
 Penmaen Mawr, Caernarvonshire, • - 1540 G 
 
 Malvern Hills, Worcestershire, - - 1444 G 
 
 Cawsand Beacon, Devonshire, - - 1792 G 
 
 Rippin Tor, Devonshire, - - - 1549 G 
 Brocken, in the Hartz- forest, Hanover, - '- 3690 
 
 Priel, in Upper Austria, - - - 7000 B 
 
 Peak of Loranitz, in the Carpathian ridge, - 8870 B 
 
 Terglou, in Carniola, - - - - 10390 B 
 
 Mont Blanc, Switzerland, - - - 15646 G 
 
 Village of Chamouni, below Mont Blanc, - 3367 G 
 
 Jungfrauhorn, Sxvitzerland, - - 13730 
 
NOTES AND ILLUSTRATIONS* 
 
 4>U 
 
 ^t Goihardy Switzerland, - - 9075 
 
 Hospice of the Great St Bernard, - - 8040 B 
 
 Village of St Pierre, on the road to Great St Bernard, 5338 B 
 
 Passage of Mont Cenis, - - - 
 
 Gross-Glockner, between ike Tyrol andCorintUa, 
 
 Ortler Spitze, in the Tyrol, 
 
 lligiberg, abo-oe the lalce of Lucerne^ 
 
 Dole, the highest point of the chain of Jura, 
 
 Mont Perdu, in the Pyrenees, 
 
 Loneira, in the department of the high Alps, 
 
 6778 B 
 12780 B 
 15430 
 
 ^408 
 
 5412 B 
 11283 
 14451 
 
 Peak of Arbizon, in the department of the high Pyrenees, 8344 
 
 Puy de Dome, in Auvergne, - - ^ 4858 G 
 
 Montd'Or, . . - . 6202 G 
 
 Summit of Vaucluse, wear ^w^wow, - - 2150 
 
 Village on Mont Genevre, - - 5.945 B 
 
 St Pilon, near Marseilles, - - - 3295 G 
 
 Soracte, 7iear Rome, - - - 2271 G 
 
 Monte Velino, in the kingdom of Naples, - 8397 G 
 
 Mount Vesuvius, volcanic mountain beside Naples, 3978 
 
 iEtna, volcanic mountain in Sicily, - 10963 B 
 
 St Angelo, in the Lipari Islajids, - - 5260 
 
 Top of the Rock of Gibraltar, - - 1439 B 
 
 Mount Athos, in Rumelia, - - - 3353 
 
 Diana's Peak, in the Island of St Helena, - 2692 
 
 Peak of Teneriffe, one of the Canary Islands, - 12358 B 
 
 Ruivo Peak, the highest point of Madeira, - 5162 
 
 Table Mountain, near the Cape of Good Hope, 3520 
 
 Chain of Mount Ida, beyond the plain of Troy, 4960 
 
 Chain of Mount Olympus, in Anatolia, - 6500 
 
 Italitzkoi, in the Altaic chain, - - 10735 
 
 Awatsha, volcanic mountain in KamtchatJca, - 9600 
 
 The Volcano, in the Isle of Bourbon, - - 7680 
 
 Ophir, in the centre of the Island of Sumatra, - 13842 
 
 St Elias, on the western coast of North America, 12672 
 
 White Mountain, in the State of Massachusets, 6230 B 
 
 ^ Chimborazo, highest summit of the Andes, - 21440 B 
 
 Antisana, volcanic mountain in the kingdom of Q,uito, 19150 B 
 
 Shepherd station on that mountain, - - 13500 B 
 
 Cotopaxi, volcanic mountain in the kingdom of Qjuito, 18890 B 
 
 Tonguragua, volcanic mountain^ near Riobomba, 16579 B 
 
NOTES AND ILLUSTRATIONS; 
 
 Rucu de Pichincha, in the kingdom of Quito, - 15940 G 
 
 Heights of Assuay, the ancient Peruvian road, 15540 B 
 
 Peak of Orizaba, volcanic mountain east from Mexico, 17390 G 
 
 Lake of Toluca, in the kingdom of Mexico, - 12195 B 
 
 City of Quito, - - - 9560 B 
 
 City of Mexico, - - - 7476 B 
 
 Silla de Caraccas, part of the chain of Venezuela, 8640 B 
 
 Blue Mountains, in the Island of Jamaica, - 7431 
 
 Pelee, in the Island of Martinique, - - 5100 
 
 Morne Garou, in the Island of St Vincent* s, - 5050 
 
 In this list of altitudes, I have not ventured to insert the 
 Himalaya Mountains, or Great Central Chain of Upper Asia, 
 to which some recent accounts from India would assign the 
 stupendous elevation from 23,000 to 27,000 feet. Such at 
 least are the results of observations made with a small sextanC 
 and an artificial horizon, at the enormous distance of 226 or 232 
 miles, as computed indeed from very short bases. But even 
 with the best instruments, and under the most favourable cir- 
 cumstances, the determination of minute vertical angles is liable 
 to much uncertainty. The progress of accurate observation 
 has uniformly reduced the estimated altitudes of mountains. 
 
 I shall conclude with briefly stating the French measures^. 
 The Parisian foot was to the English, or the toise to the fa- 
 thom, as 1.065777 to 1, or nearly as 16 to 15. The metre, 
 or base of the new system, and equal to 39.371 English 
 inches, ascends decimally, forming the decametre or perch, the 
 hectometre, the kilometre or mile, and the myriametre or league, 
 equivalent to 6.213856 of our miles; and descending by the 
 same scale, it forms successively the decimetre or palm, the 
 centimetre or digit, and the millimetre or stroke. The square 
 of the decametre constitutes the are, and that of the hectametre, 
 the hectare or acre, and equal to 2.47117 English acres. The 
 cube of a metre, or 35.3171 feet, forms the unit of solid mea- 
 sure or the stere, that o^ a decimetre, or 61.028 inches forming 
 the litre or pint ; and the weight of this bulk of water at its 
 greatest contraction makes the kilogramme ov pound, equivalent 
 to 2.1133 pounds Troy, ih^ gramme answering to 15.444 grains, 
 
 FINIS. 
 
ERRATA. 
 
 P. 36. line 9. /or triangles read triangle 
 — 144. bottom, /or B : C : : C read C : D : : E : F 
 — 233. 5th line from bottom, /or Of the equidifferent arcs, read Oi three 
 equicUft'erent arcs, 
 
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