‘ee? ®) hed 1A Mob sh ty eh Sel We via Mie ded A tat oe RD aie rather oid ahs te ie Ur ee wv A (i La ( oy, ie wi ty pays) ibe ear ae ay Shih et aan aha ? Hot 9g “i ie ) yi * ve (i, “ilk is ow 13 | |o3 S 14 Wj = 15 J|S ) Cor.to AS 1.8 MATHEMATICAL WORKS OF THE LATE THOMAS SIMPSON, F.R.S. Printed for J. COLLINGWOOD, in the Strand. THE ELEMENTS OF GEOMETRY, with their Application to the Mensura- ‘tion of Superfices and Solids; to the Determinaticn of the Maxima and Minima of Geometrical Quantities, and-to the-Construetion of a great Variety of Geome- trical Problems, 8vo, a-new Edition corrected ; with additional Notes, and ian APPENDIX, containing a Description of the Analytical and Synthetical Modes of reasoning made use of by Mathematicians; an Account of the lost Analytical Works of EUCLID and APOLLONIUS, and of the several Attempts of the Moderns to restore them. TRIGONOMETRY, Plane and Spherical, with the Construction and Application of Logarithms, 8vo, 5th Edition, 3s. A TREATISE OF ALGEBRA, wherein the Principles are demonstrated and applied in many useful and interesting Inquiries, and the Resolution of a great Variety of Problems of different kinds. To which is added, the-Geometrical Con- struction of a great number of Linear and Plane Problems, with the Method of re- solviag thesame numerically, in 8vo, a new-Edition, carefully revised, from a Cor- rected Copy by the ‘Author, with an APpPENDIx containing a PRAXIS, or Collection of Questions, without Solutions ; applicable to the most useful purposes of Science, and calculated for the improvement of Students in Algebra. SELECT EXERCISES for Young Proficients in the Mathematics, contain- ing a great variety of Algebraical and: Geometrical Problems, with their Solu- tions; the Theory of Gunnery ;.a New and Comprehensive Method for finding the Roots of Equations in Numbers; and a Short Account of the Nature and First Prin- ciples of Fluxions, New Edition, 8vo. 7s. A TREATISE ON .THE NATURE AND LAWS. OF CHANCE, 8vo. The whole after a new, general, and conspicuous manner, and illustrated with a great variety of Examples. 4s. A SUPPLEMENT TO THE DOCTRINE OF ANNUITIES AND -REVER- SIONS, deduced from General and Evident Principles; containing the Valuation of Annuities and Reversions, for single and joint Lives; with the necesasry Tables. 8vo. 3s. 6d. : THE DOCTRINE AND APPLICATION OF FLUXIONS; containing (besides what is common on the subject) a number of new Improvements in the Theory and the Solution of a variety of new and very Interesting Problems in different branches of the Mathematics, in 2 vols, 8vo. A new Edition with Corrections, and an ApPENDIX, containing much new and important matter, by a Graduate of Trinity Coll. Cambridge. ESSAYS ON SEVERAL CURIOUS and USEFUL SUBJECTS, in Speculative and Mixed Mathematics; in which are explained the most difficult Problems of the First and Second Books of Sir ISAAC -NEWTON’S PRINCIPIA;- being a useful Intro- duction to Learners, for the understanding that illustrious Author. 4to. a new Edition, in the Press, under the superintendence of an eminent Mathematician. MATHEMATICAL DISSERTATIONS ona variety of PHYSICAL and ANALY- TICAL SUBJECTS ; the whole in a general and perspicuous manner. 4to, a new Edition in the Press. MISCELLANEOUS TRACTS on some curious and very interesting Subjects in © MECHANICS, PHYSICAL ASTRONOMY, .and SPECULATIVE MATHEMA- TICS, wherein the Precession of the EQUINOX, the Nutation of the EARTH’S AXIS, and the Motion of the MOON in her ORBIT, are determined. 4to. anew Edi« tion in the Press. ; ELEMENTS | ELEMENTS GEOMETRY. rs BOOK IL. DEFINITIONS, 1. GEoMETRY is that science, by which we compare such quantities together as have extension. Extension is distinguished into length, breadth, and thickness. 2. A Line is that which has length without breadth. The terms, bounds, or extremes of a Line are points, 3. A Surface is that which has fe length and breadth only, as C. C The bounds of a surface are lines. 4. A Solid is that, which has length, breadth, and thickness, as VL). The bounds of a Solid are surfaces. 5. A Right (or straight) line is that, which lies even! between its extremes, or which every where tends the same way, as AB. A B 6. A Plane-surface is that, which is every where perfectly flat and even, or which touches, in every part, any right-line extended between points any where taken in that surface. B ELEMENTS OF GEOMETRY. 7. An Angle is the inclination, or opening, of two right-lines meeting in D a point, as D. Cc 8. When one right line DC, standing upon another AB, makes the angles on both sides | equal, those angles are called right angles; aud that line CD is said to be perpendicu- lar to the other, AB, on which —— 1 it insists. A o 9, An Acute-angle is that, which is less than a right-angle, as E. E 10. An Obtuse-angle is that, which is greater than a right-angle, as F. EF 11. The distance of two points, is the right-line reaching from the one to the other. 12. The distance of a point from a line, isa right- line drawn from that point, perpendicular to, and ter- minating in, the line given. 13. Parallel (or egut- e - 2 distant) right-lines AB, CD are such, which being in the same plane-surface, if infinitely produced, would never meet, A B | a _ 14. A Figure, in Geometry, is a bounded sp Ce, and is either a surface, or a solid. 15. A right-lined plane Figure is that, formed in a bipenet ics tole whose terms, or bounds, are right- ines. u BOOK THE FIRST. 16. All plane Figures bounded by three right-lines, are called Triangles. 17. An Equilateral Triangle is that, whose bounds or sides are all equal, as A. > 18. An Lsosceles Triangle is, when two sides are equal, as B. a 19. A Scalene Triangle is, when all the three sides are unequal, as C. 20. A Right-angled Triangle is that which has one right-angle, as ACB; whereof the side AB opposite to the right-angle, is called the Hy- pothenuse. INA 21. An Obtuse-angled Triangle is that which has one obtuse angle. 22. An Acute-angled Triangle is that which has all its angles acute. 23. Every plane figure bounded by four right lines, is called a Quadrangle or Quadrilateral. 24. Any Quadrangle, whose opposite sides are parallel, is called a Parallelogram, as D. i 25. A Parallelogram, whose angles are all right-ones, is called a Rect- angle, as K. La B2 ELEMENTS OF GEOMETRY. 26. A Parallelogram whose sides are all equal, and its angles all right ones, is called a Square, as F. 27. A Parallelogram whose sides are all equal, but its angles not right, is called a Rhombus as G. 28. All other four-sided figures, besides these, are called Trapeziums. 29. A right-line joining any two opposite angles of a four-sided figure, is called a Diagonal, as CB. 30. That side AB upon c E which any parallelogram ; ACEB, or triangle ACB is supposed to stand, is called | the base; and the perpendi- KO Talenkadab cular CD falling thereon from ° i the opposite angle C, is called the altitude of the pa- rallelogram, or triangle. 31. All plane figures contained under more ‘than four sides, are called Polygons; those having jive sides, are called. Pentagons; those having six sides, Hexagons ; and so on. 32. A Regular Polygon is one whose angles, as well as sides, are all equal. | | | 33. A Circle isa plane figure, bounded by one curve-line. APCD, called its circumfer- ence, every where equally dis- D tant from a point KE within the ~ \ circle, which point is called the ©. | center. . OO ee BOOK THE FIRST, 34. The Radius of a circle, is the distance of the center from the circumference, or a right-line EA drawn from the center to the circumference. AXIOMS, oR SELF-EVIDENT TRUTHS. 1. Things, which are equal to one and the same thing, are also equal to each other. 2. Every whole is greater than its part. 3. Every whole is equal to all its parts taken to- gether. | A. If to equal things, equal things be added, the wholes will be equal. 5. If from equal things, equal things be taken away, the remainders will be equal. 6. If to, or from unequal things, equal things be added, or taken away, the sums, or remainders, will have the same difference, as the unequal things first proposed. 7. All right-angles are equal to one another. 8. More than one right-line cannot be drawn from one given point A to another given A B point B. 9. If two points D, | F, in a_ right-line Se ne Bohn MN, are posited at unequaldistancesDC, | SS ESOS LS ES ee FE, from another, G hy ) Cc Me right-line AB in the “in ee same plane-surface ; those two lines being produced on the side of the least distance EF, will meet each other. JO. If two C FE right-lines CA, CB, making an angle C, be res- pectively equal to two other right-lines FD, FE, making an ELEMENTS OF GEOMETRY. angle F, and the angles which they make C and F, be likewise equal; the right-lines AB, DE joining their extremes will be equal, and the two triangles ACB, DFE equal in all respects. If this should not appear sufficiently evident for an axiom, conceive the triangle DFE to be removed, and so applied to the triangle ABC, that the point F may coincide with C, and the side FD fall upon the side CA; then, because FD is supposed equal to CA, the point D will also fall upon A. And the angle F being equal to the angle C, the side FE will fall upon CB; and consequently the point E upon the point B, because FE is supposed equal to CB. Therefore, seeing all the bounds of the two triangles coincide, it is manifest, that not only the bases AB, DE, but the angles opposite to the equal sides, are also equal. When all the four lines CA, CB, FD, FE, are equal; the triangle DFE, being contrariwise applied to ACD, so that FE may coincide with CA, will, also, agree with the triangle ACB (as is manifest from the reasoning above:) and so, the angle EK (as D did be- fore) now coinciding with the angle A, the two angles K and D must necessarily be equal to each other, in. this case, where the triangle DFE is an isosceles one. POSTULATES, OR PETITIONS. 1. That, from any given point, to any other given point, a right-line may be drawn. 2, That a right-line may be produced, or continued out, at pleasure. 3. That, from any point as a center, with a radius equal to any right-line assigned, a circle may be des- cribed. Peery 4. That a right-line may be drawn perpendicular to another, at any point assigned ; and that it is also pos- sible to make a right-line, or a right-lined angle, equal to the whole, or to the half, of any right-line, or right- lined angle assigned. BOOK THE FIRST. This fourth Postulate is added, more for the sake of making the proper references, than through absolute necessity ; since, what is here barely assumed as pos- sible, is effected, and actually demonstrated, in the beginning of the Fifth Book, entirely independent of every thing but Axioms and the other Postulates, above laid down. It may also be proper to note here, that, though these Postulates are not always quoted, it will be easy to perceive where, and in what sense, they are to be understood. Notes and OBSERVATIONS, with the significations of Terms and Signs, used in this Work. A PRoposiTion is when something is either pro- posed to be done, or to be demonstrated, and is either a problem or a theorem. 2 A PROBLEM is a practical proposition, in which something is proposed to be done. A THEOREM is a speculative proposition, in which something is proposed to be demonstrated. A LEMMA is when some premise is demonstrated, in order to render the thing in hand the more easy. A CoROLLARY is a consequent truth, gained from some preceding truth or demonstration. A ScHOLIUM is when remarks and observations are made upon something going before. An HYPOTHESIS is a supposition or assumption, made either in the enunciation of a proposition, or in the course of a demonstration. The signification of SIGns. The sign =, denotes that the quantities betwixt which it stands are equal. The sign (, denotes that the quantity preceding it, is greater than that which comes after it. ‘The sign =, denotes that the quantity preceding it is less than that which comes after it. ; The sign +, denotes that the quantity which it pre- cedes is to be added. | | The sign —, denotes, that the quantity which it precedes, is to be taken away, or subtracted. A figure, or number, prefixed to any quantity, shows ® Def. 8. 6 Post, 4. ELEMENTS OF GEOMETRY. how often that quantity is to be taken, or repeated ; as*5 A shows, that the quantity represented by A, is to be taken 5 times. | When several angles are formed C D about the same point (as at B,) each particular angle is described by three letters, of which the middle one shows the angular igre point, and the other two, the lines“ B E that form the angle: thus CBD or DBC signifies the angle formed by the lines CB and DB. When, in any demonstration, you meet with several quantities joined the one to the other continually by the mark of equality (=,) the conclusion drawn from thence, is always gathered from the first and last of them; which are equal to each other, by virtue of the first axiom. Thus if A=>B=C=D, then will the first (A) and the last (D) be equal to each other. Also, when in the quotations you meet with two numbers, the first shows the proposition, and the se- cond the book. Moreover, Ax. denotes axiom; Post. postulatum; Def. definition; Hyp. hypothesis. Note also, that, whenever the word dine occurs, without the addition of. either right, or curved, a right-line is always understood: and that, when a line is said to be drawn to, or from an angle, the angular point is meant. THEOREM I. A line (as) standing upon another line (cp) makés with it two angles (ABc, ABD) which, taken together, are equal to two right-angles. If the angles ABC, ABD are equal, it is plain they make two right angles*; if unequal, let BE. be perpendicular to CD °, dividing the greater of them (ABC) into the parts EBC, EBA; then the former part B D EBC being a right-angle*, and By, A BOOK THE FIRST. , the remaining part EBA together with the whole less angle ABD, equal to another right-angle EBD‘; the «Ax. 3. whole, of both the proposed angles, taken together, must necessarily be equal to two right-angles 4. 4Ax.4. - COROLLARY. Hence all the angles at the same point (B) on the same side of a right-line (CD) are egal to two right-angles*. ¢ Ax. 3. THEOREM II. If one line (as) meeting two others (Bc, BD) mm the same point (B), makes two ungles with them (ABC, ABD) which together are equal to two right-angles; these lines (Bc, BD) will form one continued right-line. For, tf possible, let BH, and not BD, be the continua- tion of the right line CB; then the angles ABC and ABH TI being=two _ right-angles*¢= «1. 1. ABCandABD-/; if from these _ J Hyp. equal quantities, ABC com-C B Dp mon to both, be taken away, there will remain ABH =—ABDeE; which is impossible *. A ey THEOREM III. The opposite angles (DEB, AEC), made by two lines (DC, BA) méersecting each other, are equal, For DEB + DEA=two right- angles‘ = AEC+DEA: whence, D B by taking away DEA, common, there remains DEB=AEC*, 10 l Post. 4. ™ Post. 1. " Def. 8. ° Ax. 10. P Hyp. Cp ae 8 r Ax, 8, 5 Ax. 9. and Def. 13. t Post. 4. ELEMENTS OF GEOMETRY. THEOREM IV. Two right-lines (AB, CD) perpendicular to one and the same right-line (EF), are parallel to each other. If you say, they are not parallel; then let them, when produced out, meet in some point, as G. In EA, pro- C F D duced (if neces- sary) let there be taken EH= EG‘, andlet theright- H A 10) B G line FH be drawn”. The'triangles EHF and EGF, having EH=EG, the angle HEF=GEF”, and EF common, are therefore equal in all respects®: and so, the angle EF H being=EFG (EFD)=a right-angle?, HF DG (as well as HEG) must be one continued right-line?: which ts impossible’. Therefore AB and CD are parallels. SCHOLIUM. In this theorem, the posszbilzty of parallel lines (or such, which being infinitely produced, in the same plane, can never meet) is demonstrated: for KF may be drawn perpendicular to AB’; and CFD, again perpendicular to EF’; which last, it is demonstrated, will be parallel to AB. THEOREM V. Perpendiculars (uF, Gu) to one (AB) of two pa- rallel lines (AB, CD) terminated by those lines, are equal to each other; and also perpendi- cular to the other of the two parallels (cp). For, AB and CD being parallel to each other, GH can neither be greater nor less than EF*; and therefore must be equal to EF. If you say, that KF’ is not perpendicular to CD; then let FM be perpendicular to EF‘, meeting GH produced (if BOOK THE FIRST... ll) necessary) in M: so shall FM MIME: be parallel to AB”; and con- ¢ F }. to each other %. i 44.1, 12 5 Re ts m Ax, 4, ELEMENTS OF GEOMETRY. THEOREM VII. A line (AB) intersecting two parallel lines (sr, ap) makes the alternate angles (spc, Pcp) equal to each other. Let CF and DE be perpen- dicular to QP, and SR°; then these lines FC and DE are likewise parallels’; and so the triangles CFD and CDE, having the side CF=DE¥%, FD=CE*’, and the angle A F= kes, they will also have the angle FDC=ECD *. COROLLARY I. Hence, a line intersecting two parallel lines, makes the angles (BDR, BCP) on the same side, equal to each other ; for BDR (=CDS‘)=BCP*, COROLLARY II. Hence, also, a line falling upon two parallel lines, makes the sum of the two internal angles (SDC + QCD) onthe same side of tt, equaltotworight-angles: for the angle SDC being=PCD, and PCD+ QCD =two right angles’; thence is SDC +-QCD=also to two right-angles ™. THEOREM VIII. If a lane (AB) intersecting two other lines (Pa, rs), makes the alternate angles (pcr, cDs) equal to each other ; then are those two hnes parallel. For, if possible, let some rp other line DT, and not DS, be parallel to PQ"; then § must CDT = DC Pe= CDS?: which is impossible’, BOOK THE FIRST. 13 COROLLARY. Hence, if a line falling on two others, makes the angles (BDR, BCP) above them on the same side, equal to each other; then those two lines are pa- rallel: because SDC=BDR". "3.1. THEOREM IX. If one side (as) of a iriangle (asc) be produced, the external angle (cBD) will be equal to both the internal opposite angles (a, c) taken together. For, let BE be pa- rallel to AC*; then will C, Eeah'schs to the angle C=CBE‘, vgey and theangle A= DBE"; ee 7 therefore C+A=CBE Sy 7.1. = ; — 4 * Ax. 4, Msebivekihi: 1 Ws) nok B Ds axe. - COROLLARY. Hence the external angle of a triangle is greater than either of the internal, opposite angles. THEOREM xX. | The three angles of any plane triangle (asc) taken together, are equal to two right-angles. For, if AB be produced to D, C then C+A=CBD+%, to which 6%: equal quantities let the angle - CBA be added, then will C+A | +CBA = CBD+CBA‘=two ) | Dice right-angles °. A BD COROLLARIES. | 1. If two angles in one triangle, be equal to two 14 ELEMENTS OF GEOMETRY. angles in another triangle, the remaining angles will e Ax.5. also be equal’. 2. If one angle in one triangle, be equal to one angle tn another, the sums of the remaining angles” - will be equal’. | 3. If one angle of a aa be right, the other two taken together, will be equal to a risht-angle. i 4. The two least angles, of every triangle, are acute. | . PH.EO REM, X11. { The four inward angles of a quadrangle (ABO 5) taken together, are equal to four reght-angles. : Let the diagonal AC be AC drawn; then’ the three angles ery of the triangle ABC being = 410.1... two right-angles *, and those of the triangle ACD equal also to two right-angles“; it fol- lows that the sum of all the angles of both triangles, which A B make the four angles of the quadrangle, must be equal ‘Ax.4. to four right-angles °. COROLLARY I. Hence, tf three of the angles be right ones, the fourth will also be a right-angle. ] i | j | f 4 j i COROLLARY II. Moreover, if two of the four angles, be equal to two right-angles, the remaining two together will like- wise be equal to two right-angles. ee SCHOLIUM. If from any point P, within a polygon ABCDE, lines be drawn to all the angles, so as to divide the! BOOK THE FIRST. 15 whole into as many triangles APB, BPC, CPD, DPE, EPA, as the polygon has sides ; the sum of all the angles of these triangles, B D (which together make up, or - compose the angles of the poly- gon, over and above those about _ the point P) will be equal totwice A. EB as many right-angles as the polygon has sides (by 10 1.) Therefore, seeing all the angles about the point P, whereby the angles of all the triangles ex- ceed those of the polygon, are equal to four right- angles, it is manifest, that all the angles of the polygon, taken together, will be equal to twice as many right- angles, wanting four, as the polygon has sides. THEOREM XII. The angles (a,8,) at the base of an isosceles triangle (ABC) are equal to each other. For, let the line CD bisect, C or divide the angle ACB into two equal parts ACD, BCD, and meet AB in D: then the triangles ACD, BCD, having AC = BC¥*, CD common, and . F Def. 18. the angle ACD = BCD, will . also have the angle A = Be. A dy B : Agee 3 X. A COROLLARY I. Hence, the line which bisects the vertical angle of an isosceles triangle, bisects the base, and is also perpendicular to it". COROLLARY Ii. Hence it appears also, that every equilateral triangle is likewise equiangular, 16 ELEMENTS OF GEOMETRY. THEOREM XIiIl. In any triangle (ase) the greatest side subtends the greatest angle. Let AB be greater than AC ; Cc in which let there be taken AD = AC; and draw CD. The triangle ADC being isosceles, the angles ACD and ADC are iN D B ‘12.1. therefore equal’; whence ACB, * which exceeds the former of them, must also exceed kAx.2, the latter ADC*, and consequently, much more 'Cor.to9, exceed B, which is less than ADC ‘. 1. COROLLARY. Hence, 72 any triangle, the side that subtends the greatest angle, is the greatest ; because ACB can- not be greater than B, unless AB is greater than ™ 13,1. AG a THEOREM XIV. If the three sides (AB, AC, CB) of one triangle, be equal to the three sides (DE, DF, FE) of another triangle, each to each respectively ; then the angles opposed to the equal sides will and DEF beequal also be equal. . K B ta \, : .f D E ™Ax.10. in all respects”: *Hyp. therefore, AG being = DF = AC*, and BG = EF °32,.1. = BC”, the angle ACG is also = AGC?, and BCG PAx.4. = BGC’; and consequently ACB = AGB? = DFE: or, therefore the triangles ABC, DEF are equal in all respects”. Let the angle BAG = D;,AG om DF, and let GB and GC be drawn; so shall the triangles ABG BOOK THE FIRST. 17 SCHOLIUM. The demonstration of the last theorem, in obtuse- angled triangles, may admit of another case; which, however, is not necessary; because, if the triangle AGB (equal to DEF) be conceived to be formed on the longest side of ABC; then, all the angles CAB, CBA, GAB, GBA being acute’, the line CG will, 2 Cor. 4. always fail within the figure ACBG’, as in the pre-, (1)! sent case. Ba, THEOREM XV. If two triangles (ABC, DEF) mutually equ- angular, have two corresponding sides (AB, DE) equal to each other, the other corres- ponding sides will also be equal. If you say Cc Ik BC is great- er than EF; G from BC let a part BG be taken = E EFs,andlet A B D * Post, 4, AG be drawn. The triangles ABG, DEF having AB = DE, BG = EF, and B = EK (by hypothesis, ) will also have BAG = D‘; but D = BAC"; there-¢ Ax.10, fore BAG = BAC”; which is impossible. ipl w Some Ie COROLLARY. Pa Hence, equiangular triangles, having any two cor- responding sides equal, are equal to each other*, * Ax. 10.1. THEOREM XVI. Lf two right-angled triangles (anc, DEF) having equal hypothenuses (Ac, DF,) have two other sedes (BC, EF) likewise equal; the remaemng sides (AB, DE) will be equal, and the two trrangles equal in all respects. c 1s ELEMENTS OF GEOMETRY. In AB produced, an F take BG = ED, and ‘ let GC be drawn: then; the triangles . BCG, and DEF, ‘ ¥ Hyp. = Ax, @ Ax, 10. vn ae ¢ Cor. 1, LOMO 1, 415. 1, e Ax. 7. FS Hyp. EAD As h16.1. i Cor. 1. to 10, having BG = ED, A B G D E BC = EF», and the angle CBG = E*, will also have the angle G = D, and CG = DF4 = ACy: whence, the triangle ACG being isosceles, the angle’ G, or D, will be = A?%; and consequently F also = ACB°; therefore the triangles ABC and DEF, being mutually equiangular, and having AC = DF, they are equal in all respects“. THEOREM «XVII. If two triangles (asc, pEF) having two_ sides (ac, BC) of the one equal to two sides (D¥, EF) of the other respectively, have also the angles (a, D) subtended by two of the equal sides (Bc, EF) equal to each other; and if the an- gles (B, &) subtended by the other equal sides, be erther, both acute or both obtuse; then will the two triangles be equal in all respects. Let CG and FH be perpendicular to AB and Cc if a B B Asi Abia Ge dD ¢ H DE: then, the angle AGC being = DHF , A = D, and the side AC = DFY, CG will also be = FHs; whence, CB being=FE/, the angles GBC and HEF are likewise equal”, and so, the triangles ABC and DEF, being mutually equi-angular‘, and having the sides AC and DF equal, are equal in all respects ®. . | ) BOOK THE FIRST. 19 The demonstration is the same, when both the angles are obtuse, asin the triangles A6C, Dek : for, if Cb (=CB=FE)=Fe, the angles GbC and HeF¥ being equal (as before) the angles AOC and DeF will likewise be equal. rs Sire THEOREM XVIII. If two angles (a, 8) of a triangle (anc) be equal, the sides (Bc, Ac) subtending them will lke- wise be equal. {) Let CD bisect the angle ACB, and meet AB in D: then the tri- angles ACD, BCD being equi- angular*, and having CD com- k Cor. 1, to mon to both, they will also have AU AC=BC4 : 7 615.1. : A D B . COROLLARY. Hence, every equiangular triangle, is also equilateral. THEOREM XIX. Any two sides (Ac, BC) of a triangle (a BC) y taken together, are greater than the third side (AB.) In BC produced, let there D be taken CD=CA, and let AD bedrawn. The angles C ~D and DAC are equal”; q m 12,1. therefore BAD, which ex- ; ceeds the latter” must also n Ax, 2. exceed the former D; and R consequently BD (or BC+ A @ Cor. t AC) must exceed AB?. 13, 1, c2 20 P Hyp. ¥ Cor. 4, to 10.1, * Cor. to13. 1, °Cor. to 9, 1, ¢ Ax. 10. ELEMENTS OF GEOMETRY. | THEOREM Xx. Of all the right-lines (PA, PB, PC) falling from a given pornt (Pp) upon an infimte right-line (rs,) that (pA) zs the least which rs perpen- dicular to it ; and, of the rest, that (PB) which ws the nearest the perpendicular 2s less than any other (pc) at a greater distance. For BAP being a right- angle?, ABP will be acute, and therefore AP 4 BP’. . Afr P Jf Again; when PB and PC / | are both on the same side of the perpendicular PA; R Yi hve Cat Bes Acti then is CBP cright-angle* ct BCP%, and consequently P@c PB: If PB be on the contrary side of the perpendicular to PC; from AC, let AP be taken=AB;; then the two lines PB, PB will be also equal‘; and therefore PC, which exceeds the one (by the preceding case) will also exceed the other. THEOREM XXI. Of two triangles (ABC, DEF) having two sides (AB, BC) of the one, equal to two sides (DE, EF) of the other, each to each respectively, the base of that (asc) well be the greatest which is subtended under the greatest angle, Let the angle ABG=E, BG=EF (=BC) also let BOOK THE FIRST. 1 AG and CG be drawn, upon the last of which, pro- duced, let fall the perpendiculars BH and AI". “4.1. Since BG=BC”, and, consequently, GH=HC’, it is » Byp. evident, that GI (whether the point I be considered as * !®- !- falling between G and K, or between G and H) will be less than CI*; and therefore AG, or its equal DF“, * Ax. 2. also less than AC®. ee THEOREM XXII. Of two triangles (anc, DEF,) having one angle (BAC) in the one equal to one angie (EDF) wn the other, and the sides (Bc, EF) opposed to thematlso equal, that(apc)willhavethe greatest base, of which the opposite angle (acs) differs ihe least from a righi-angle. Let BG and EH be perpendicular to AC and DF, in which produced, take HK=HE, GI=GB, and BM=EH; also let MN be parallel to GA, meeting AB, produced if necessary, in N; aud let Cl and KF be drawn. N The angle ICG being=BCG4, and the latter of these greater than EF H‘, (or KFH4,) thence is [CB «Hyp. 'KFE; and consequently BICEK ; whence, also, eee BG GBDCEH (tEK) or its equal BM“; and there- ~~ | fore BAC BN, because AG and MN being parailels, both the points M and N will fall on the same side of AG. But BN, (as the triangles NBM, DEH are equiangular, and have BM=EH/) is=DEs: there-/ Hyp. and fore BA is also greater than DE. 71. ee SiG. 15 hAQ.1: i Ax. 6. k Ax. 3. 'Cor. 9.1. Lm ELEMENTS OF GEOMETRY. THEOREM XXIII. If, of two triangles (aBc, ABD) standing upon the same base (aB,) the one be wholly ncluded within the other, the two sides (av, BD). of the: encluded one taken together, will be less, and the angle (D) contained by them greater, res- pectively, than the two sedes (Ac, Bc,) and the contained angle (c) of the other. CasE 1. Jf the vertex of the contained triangle be in one side of the other: Then, ra AD isaAC+CD"; whence, by adding BD common, AD+BD will also be a= AC+CD+BD¥i or than its equal AC+BC#. But the angle ADB isaACB’ A B D CasE II. Jf the vertex be within the other triangle. Let AD be produced to meet BC in E: then (by case 1.) AD +BDaAE+BE, but AE+ BEZAC +BC, therefore AD +BDaAC+BC. Moreover, the angle ADBC- BEDcCC. A B THEOREM XXIV. The opposite sides (AB, DC) of any parallelo- gram (ABCD) are equal, as are also the oppo- site angles (8, D;) and the diagonal (ac) divides the parallelogram into two equal parts. BOOK THE FIRST... . 93 For AB; DC, and AD, BC being parallels”, the angle BAC is=DCA2, and BCA=DAC‘, therefore the equiangular trian- gles ABC, ADC”, having AC 7 common, are equal in all re-— A B spects !. 715.1, Def, 24, a’ fae P Cor. 1, to 10.-1. COROLLARY. Hence, if one angle (B) of a parallelogram be a right-angle, all the other three will be right ones: for D, being= B, is a right-angle ; and BCD is= 5, and DAB = D, by Theor, 'V. | THEOREM XXV. Every quadrilateral (asep) whose opposite stdes are equal, 1s a parallelogram. (See th* ~ preceding scheme.) Let the diagonal AC be drawn; then the triangles ABC, ADC being mutually equilateral’, they will also » Hyp. be mutually equiangular*; consequently AB willbe pa- :14. 1. rallel to DC, and AD to BC’. ‘8.1. THEOREM XXVI. The lines (AD, BC) yoineng the corresponding extremes of two equal, and parallel lines (AaB, Dc) are themselves equal and parallel. Let the diagonal BD be drawn. Beeause AB and DC are parallel", the angle ABD is = CDB”; there- « typ. fore, BA being = DC*, YS hocentheoe ital ok and BD common, the re- lat at maining sides and angles | will likewise be respec- tively equal’ ; and conse- ¥ Ax. 10. quently AD parallel to 7. B if ele 78,1, 24 @ 3. 1. dis Fi) ¢ 24, 1, 4 Hyp. e15,1. ELEMENTS OF GEOMETRY. THEOREM XXVIII. If, in one side (Ax) of a triangle (aBc,) from three points (D, F, 11) at equal distances (DF, FH,) dines: (DEM, FG, Ht) be drawn parallel to the base, the parts (EG, G1) of the other side (Ac) intercepted by them, will also be equal to each other. . Let NGM be parallel to AB, intersecting, HI and DE in N and M. Then, the triangles IGN, MGE, having the angle IGN = EGM’, ING = M+, and GN (= FH° = FD¢) = GM, will also have GI=GE*% c COROLLARY I. Hence it appears, that, 7f one side of a triangle be divided into any number of equal parts, and from the points of division, lines be drawn parallel to the base, cutting the other side, they will also di- vide tt into the same number of equal parts. COROLLARY II. Hence, also, z7/ two lines FG, HI, cutting the sides of a triangle, be parallel to each other, and another line DE be so drawn as to cut off FD=FH and GE=GI, this line DE will be parallel to the two former. THEOREM XXVIII. If in the sides of a square (aBcn,) equally dis- tant from the four angular points, there be taken four other posnts (E, ¥, G, U,) the figure (eFGH) formed by joing those points, shall also be a square, BOOK THE FIRST. 25 For the wholes AD, DC, E 7 CB, BA being equal’, and A D /Def, 26, also the parts AE, DF, CG, BH, the remaining parts BR Hyp. ED, FC, GB, HA must consequently be equal’; parle whence, all the angles D, fy C, B, A being equal’; the i Ax, 7, sides EF, FG, GH, HE will be equal likewise", c “Ax. 10. and DEF=AHE*, There- B G fore, because DEH is=A+AHE 4 if from these, the ig |. equal angles DEF, AHE be taken away, there will remain Hi F=A‘*=a right-angle’. By the same ar- gument (or by Theor. 25th, and the Corol. to the 24th) the other three angles will be right-angles. THEOREM XXIX. If all the sides of any quadilateral (ancn) be bisected, the figure (nEGu) formed by joining the points of bisection, will be a parallelogram. Draw the diagonals AC and BD. Because EF and HG are D parallel to AC”, they are also parallel to each other’. After the same manner is FG parallel to KH; therefore EFGH isa parallelogram ”, : THE END OF THE FIRST BOOK. ELEMENTS OF GEOMETRY. arama emer BOR Rot: DEFINITIONS. A, In a parallelogram ABCD, if two right-lines EF; HI, parallel to the sides, intersecting the diagonal] in the same point G, be drawn, dividing the parallelo-_ gram into four other pa- C K B | rallelograms; those two GD, GB through which the diagonal does not pass, are called Complements ; and the other two, HE, FI, parallelograms, about the diagonal. Dy iS te A Cc 2. EVERY rectangle is said to be contained under the two right- lines AB,BC that form its base and altitude. | A B The rectangle contained under two right-lines AB and BC is often, for brevity’s sake, denoted by AB x BC. But when the figure is a square, tt is usually repre-- sented by placing the number 2 over the letter, or letters, expressing the side of the square: thus AB*— denotes the square made upon-the line AB. | BOOK THE SECOND. 27 THEOREM I. The rectangles (BD, FH) contained under equal lines, are equal. For let the di- D C #H G agonals AC, EG be drawn: then, because AB=EF, BC=FG, and B= F*, the triangles ABC, EFG are A Biss eds FE equal’. And, in the very same manner will ADC ° Ax. 10. and EHG appear to be equal. Therefore the whole rectangle ABCD is also equal to the whole rectangle EFGH*. ¢ Ax. 4.1. “Hyp. THEOREM II. Parallelograms (ABCD, BCFE) standing upon the same base (Bc) and between the same parallel (BC, AF) are equal. For, since (in Fig. |.) the angle F=BEA%, and 4. Cor. 1. CDF=A 4, the triangles FDC, EAB are equiangular®; , {°7.): they are also equal’, because CF=BES: therefore, 45 40.1. if each be taken from the whole figure ABCF, there f15. 1. will remain ABCD= EBCF *. rote i . COROLLARY If. Hence, triangles, BAC, BFC, (fig. 2.) standing upon the same base, and between the same parallels, are also equal, being the halves of their respective parallelograms ‘. 24,1, 28 k 24.1, and Def. 1. Ax. 5. ks ” Constr. © Cor 2: to Z. 2. p.24.1. 9 Ax. 4. ELEMENTS OF GEOMETRY. COROLLARY Il. Hence all parallelograms, or triangles, whose bases and altitudes are equal, are equal among them- selves ; because all such parallelograms are equal to rectangles standing on the same bases, and between the same parallels; and these last are equal, by the preceding proposition. THEOREM Il. The complements (Ec, EA) of any parallelo- gram (AC) are equal. For, the whole triangle BCD being equal to the D T ¢ whole triangle DAB *. and the parts DIE, EFB respectively equal to the H parts DHE, EGB*, the remaining parts EC, EA, = must likewise be equal ’. A G 6B THEOREM IV. A trapezium (ABcp) of which two sides (AD, BC) are parallel, is equal to half a parallelogram, whose base is the sum of those two sides, and — tts altitude the perpendicular distance between them. | For, in AD produced, RB ‘. WT E take DF=BC; and let CG, DH and FE beall parallel to AB, meeting AF and BC produced, in ‘G, Hand E. Then AE A G D ary isa parallelogram of the same altitude with ABCD, having its base AF equal to the sum of AD and BC”: © but this parallelogram, because BG=HF . and CGD = CHD», is equally divided by the line CD‘; and so ABCD is the half of it. BOOK THE SECOND. | 99 THEOREM V. The sum of all the rectangles contained under a given line (Ap,) and all the parts (An, HG, GB) of another (aB,) any how divided, is equal to the rectangle contained under the two whole lines. Let ABCD be the rectangle contained under the two whole bo Cc lines, and let HF, GE be pa- rallel to AD, meeting DC in F and E. Then will AF, HE, GC be rectangles” of the same ” Cor, to altitude with AC‘; therefore G mar BN ASAD MAT AHSIAD Cis ae Ley HG, and GC=ADxBG‘*; consequently FADx {12 AB(=AC=AF+HE+GC)=AD x AH+ADxHG wot +ADxBG% -. THEOREM VI. Tf a right line (AB) be, any-wise, divided into two parts (AC, BC,) the square of the whole line will be equal to the squares of both the parts, together with two rectangles under the same parts. _ Let ABGI be the square of AB, and CBEF that of BC, and let EF and CF be pro- duced to meet the sides of the N B E square ABGI in M and N. From the equal quantities CM,EN ” take the equal quan- w 24,1, tities CF and EF, and there and De- remains FM=FN?; therefore A © R fin, 26. 7 = Ax. 5, 1. all the angles of the figure being right ones’, NM is a» Cot é square* upon FN(=AC;) and AF, FG are equal 24.1. to two rectangles under BC and AC*: but AG=BF aie ai +FI+AF+FG, or AB? =BC?+AC?242AC x BCS + Ay 3.1. 30 © Cor. 24. @ Ax, 24.1. © Def, 24. J Hyp. & Ax. 3. ah ie ELEMENTS OF GEOMETRY... COROLLARY I. Hence, the square of any line is equal to four times | the square of half that line. COROLLARY II. Hence, also, if two squares be equal, their sides must be equal; because unequal lines BA, BC have not equal squares. THEOREM VIL. The difference of the squares (ABEH, ACIK) of any two unequal lines (AB, AC,) ts equal to a@ rectangle under the sum and difference of the same lines. Ms In EB, produced, take BF= H Doo Ea AC; let FG be drawn parallel to EH, and let CI be produced both K ways, to meet EH and FG in D and G. It is evident that DF is a rectangle *, whose base GF(= CB") =the difference of the given lines A AB,AC; and whose altitude FE (because BE= BA‘, and BF = AC‘) is = the sum of the same G EF lines: but this rectangle DF is=DB+GBs=DB+ DK (because DK *=GB)=square AE—-square AI. THEOREM VIII. The square made upon the side (ac) subtending - the right-angle of a plane triangle (axBc,) ts equal to both the squares (BE, BG) made upon the sides (ac, Bc) contaiming that angle. BOOK THE SECOND. 31 Let the sides of the squares ft. AU D BE, BG be produced to meet - each other in L and D;in _. which take KL and IG each equal to AE (or AB;) and let CI,IK, and KA be drawn. K Since ABH and FBC (which are continued right- me lines’) are equal to each other", A f- ‘21. of Peay prey ED, and LG will ie Pee dies) CAs be all equal among themselves! ; and so the angles K, !24. 1. D, G and L being all right ones", EDGL will be a ™Hyp. & square, and consequently ACIK a square likewise". .9, )7 Now, if from the square DL, the four equal? trian- ° Ax. 10. gles ADC, CGI, ILK, and KEA be taken away, there will remain the square AI: and, if from the same DL, the two equal? parallelograms DB, BL ?1.2. (which are equal to the said four triangles, because DB=two of them!) be taken away; then there will remain the two squares BE and BG. Consequently the square AJ is=the two squares BE and BG’. 9 Ax. 5. The same, demonstrated otherwise. Let AD be the square on the hypothenuse AC, and BG, BI the two squares on the sides AB and BC: let MBH be parallel to AE, meeting GF (produced) in H ; and let EA be produced to meet GH in N. If from the equal” angles GAB, e tet nae CAN, the angle NAB, commonto yon is, both be taken away, there will re-G -? ; main NAG=BAC*; whence, as ¢ | the angle G is also=ABC’, and the side AG=AB*, the sides AN ind AC (= Al) are likewise equal” und therefore the parallelogram AM = the parallelogram AH”; which last, and consequently the ormer, is equal to the square BG* itanding on the same base AB, and ) | setween the same parallels.~ By the.same argument, he parallelogram CM is=the square BI: and, conse- juently, the square AD (=AM-+CM) = both the’ 4x4. squares BG and BI’. ‘ oh 2 Ax, 5. > Ax. 5. © Ax. 5. og, 2. ELEMENTS OF GEOMETRY. COROLLARY. Hence, the square upon either of the sides including the right angle, is equal to the difference of the sguares of the hypothenuse and the other side: ; or, equal to a rectangle contained under the sum and difference of the hypothenuse and the other side’. THEOREM IX. The difference of the squares 4 the two sides — (ac, BC) of any triangle (ABC) ts equal to the difference of 8: ve squares of the two lines, or distances (AD,BD) included between the ex- éremes of the base (AB) and the perpendicular (cD) of the triangle ;* 1.¢.4c’—BC*=AD*?—BD’. For, since AC? = DCz+4+AD2and BC2=DC?+BD, _, (by the precedent,) it is evident that the difference | of AC* and BC’ will be equal to the difference between ie + AD? and ’ DC?+BD%, or between | \ Fo Gee d) AD: and « BD?, by rie away yt Piers common, eof both. COROLLARY I. Since the rectangle under the sum and difference of | any two unequal lines, is equal to the difference of their squares 4, it follows, that, che difference of: the ~ * From Pappus, Math. Coll. Book VII. Prop, 120.—See also | D’Omerique, Anal. Geom. Introd, prop. 11., and R. Simson’s j Apollonius on Plane Loci, B. U1. Lemma 1 —Epiror. fe ) BOOK THE SECOND. a sguares (or the rectangle under the sum and dif- Jerence) of the two sides of any triangle, is equal to the rectangle under the sum and difference of the distances included between the perpendicular and the two extremes of the base; that is, AC2—BC?2 (=AC+ BC x AC—BC)=AD+ DBxAD— DB (=AD*—BD’?)=ABx AE. COROLLARY If. It follows, moreover, that, the difference of the squares (or the rectangle under the sum and difference) of the two sides of a triangle, is equal to twice a rectangle under the whole base, and the distance of the perpendicular from the middle of the base; that is, AC°’-—BC?=ABx2DE., For, let EK be the middle of the base, and make EF =ED; then AF being = BD‘, the excess of¢ Ax.5, AD above BD (or AF) will (in Fig. 1.) be=DF= 2Di.; therefore AD+BDxAD— BD (=sAC?2—°9.2. BC’*)=AB x2DE. Again (in Fig. 2.) AD + BD being = AD+AF 8" 4x4: =FD=2ED, and AD—BD=AB, we have, also, in this cases, AC?——-BC*=ABx2DE, THEOREM X. The square of one side (ac) of a triangle (aBc) ts greater, or less than the sum of the squares of the base (a8) and of the other side (Bc), by a double rectangle under the whole base (a8) and the distance (Bp) of the perpendicular from the angle (8) opposite to the side first mentioned; that is, greater, when the per- pendicular falls beyond the said angle (as in _ Fig. 1.); but less, when it falls on the contrary side of that angle (as in Fig. 2, and 3.) D oA 32 Car. 2. to 9.2, m Ax, 3. * Ay, 5, oe Ax. 4. ELEMENTS OF GEOMETRY. Let the square ABHF, on the base AB, be di- — vided into two equal‘ rectangles EF and EH by C C C 1 2 \ 3 ei E |B [A ESB A Bo pGs ow EeG) Abid: Get ne weenie the line EG, bisecting AB in E; and let the per- ; pendicular CD be continued to meet FH (pro-— duced) in I. In Fig. t. AC?—BC?=2EI'=2EH + 2BI”=AH © i (AB?)+2BI (2AB x BD); therefore, if from the first” Se eS ‘ and last of these equal quantities, AB* be taken away, ~ then AC 2less both BC *and AB?=2AB x BD”. In Fig. 2, and 3. BC?-AC?=2E1'=2B1—2BG" —2ABxBD—AB?; and so, by adding AB’ to the AB?+.BC?—AC?=2AB x BD *. “ COROLLARY I. “ Hence, in obtuse-angled triangles (ABC, fig. 1.) the Square (AC) of the side opposite the obtuse-angle (B) zs equal to the sum of the squares of the other sides (AB, BC,) together with twice a rectangle contained by one of those sides (AB,) and tts con=— tinuation (BD) to a perpendicular from the extre-\ i! first and last of these equal quantities, we have here — > | “ Hence also, in triangles (figs. 2. and 3.) the squares, mity of the other side: that is, AC?=AB*+ BC’+ i 2AB x BD.” ¢ e “ COROLLARY II. q of the base (AB) and either side, are equal to twice a rectangle contained by the base (AB) and the adjacent segment (BD) made by a perpendicular, ' together with the square of the other side (AC); “ that is, AB?+BC*=2ABxBD+AC2” >| | | | BOOK THE SECOND. THEOREM XI. The double of the square of a line (cr) drawn Jrom the vertex to the middle of the base of any triangle (asc), together with double of the square of the semi-base (Ax,) is equal to the squares of both the sedes (ac, Bc) taken togcether,® 2. €. 2CE*+2QAE*=AC?+BC?. For, let CD be perpendicular to AB: then, be- cause? AC? exceeds the sum of AK? and CE? (or C Cc K B Dish: ® Diu BE’ and CE*) by the double rectangle 2AEx ED (or 2BEKx ED); and because BC? is less than the Same sum by the same double rectangle: it is ma- nifest that both AC? and BC? together, must be equal to that sum twice taken; the excess on the one part making up the defect on the other. THEOREM XII. - The two diagonals (anc, BED) of a parallelo- gram (ABCD) bisect each other; and the sum of their squares is equal to the sum of the squares of all the four sides of the parallelo-. gram.t | _.™ Rappus’s Math. Coll. VII, 122; R. Simson’s Plane Loci, Book 11, Lem. 6. Schooten’s Exercit. Math, p. 62. + Gregory St. Vincent, de Quad. Circ., p. 33. Garnier, Reciproques de Geom. p. 22, D2 35 ° 10. Z. ELEMENTS OF GEOMETRY. For, the triangles D AEB, DEC. being equiangular ”, and hav- ing AB=DC%,° will also have AEK=CE, and BE=DE”. More- £ B over, because ZA EH’+ A 2ED?=* AD?+ CD’, by taking the double of these, we have 4AE? (‘ AC?) +4ED? (DB’)="AD? + BC? wo -+CD?+ AB’. THEOREM UAXIll. _ Tf from any point (r), to the four angles of a rectangle (apcp) four lines be drawn; the sums of the squares of those drawn to the opposite angles will be equal (I say that FA ?4+FC°=FB°+FD ’).* For, let the diagonals AC +, C and BD be drawn, bisecting 2 each other in E+, and let E, F be joined; then the tri- angles ABC,BAD being equal in all respects”, thence will AE (:AC)=DE(4DB). But FA?2+ FC?=’ 2A? (2DE’)+ 2EF*=’ FB?+ FD’. VX * Nov. Act. Acad. Petropol. Tom. 1. 1750. END OF BOOK THE SECOND. ELEMENTS OF GEOMETRY. a BO.0O K LU, DEFINITIONS, i. Any right-line FD, iw passing through E the cen- ter of a circle, and terminat- ing in the circumference at A B both ends, is called a ~ E Diameter. E D 2. An arch of a circle, is any portion of the peri- phery, or circumference, as ACB. | G 3. The chord or subtense of an arch ACB, is a right-line AB joining the two extremes of that arch. 4, A semi-circle is a figure contained under any liameter and either part of the circumference cut off oy that diameter. 5. A segment of a circle is a figure contained under in arch ACB and its chord AB. 6. A Sector of a circle is a figure contained under wo right-lines EF, EG, drawn from the center to the sircumference, and the arch FG included betwixt hem. When the two lines EF,EG, stand perpendi- sular to each other, then the Sector is called a ruadrant. 38 ELEMENTS OF GEOMETRY. 7. An angle ABC is said to be in a segment of a circle ABC, when, being in the periphery of the circle, the right-lines, BA, BC by which it is formed, pass through the extremes of the chord AC bounding that segment. B A. D 8. An angle ABC in the periphery, comprehended by two right-lines BA,BC, including an arch of the rs 9. A right-line AB is said to touch a circle, when, passing through a point (C) in the cir- cumference it cutteth off no part of the circle. each other, when the circumferences of both pass a %3 S through one point (C) and yet do not cut each other. _ circle, ADC, is said to stand upon that arch. A ral B. BOOK THE THIRD. 11. Two circles, in the same plane, are said to cut one another, when they fall partly within, and partly without each other; or, when their circumferences cut each other. 12. A &ight-line is said to be applied to, or in- scribed tn a circle, when both its extremes are in: the periphery of the circle. 13. A Right-lined figure is said to be inscribed in acircle, when all its angles are in the circumference CO, &c. being equal -each eee of the circle. 14. A Circle is said to be described about a right- lined figure, when the periphery of the circle passes through all the angles of that figure. 15. Aright-lined figure is said to be described about a circle, when all its sides touch the circle. 16. A circle is said to be tmscribed in a right-lined Jigure, when it is touched by all the sides of the right- lined figure. 17. A right-lined figure is said to be inscribed in a right-lined figure, when all the angles of the former are situated in the sides of the latter. THEOREM I. If the sides ( an, BC, CD, &e.) of a polygon in- scribed in a circle, be equal, the angles (aox, Boc, cop, &c.) at the center of the circle, subtended by them, will likewise be equat. ~ For,, AO; BO, to each other’, as well as AB, BC, CD, &c. are mutu- ally equilateral; and therefore have all the angles AOB, BOC, &c. equal to 39 $14.1, 40 © Def. 33, 1. #20. 1. ELEMENTS OF GEOMETRY. SCHOLIUM. On this proposition depends the division of ma- thematical instruments for taking and measur- ing of angles. For if, by repeated trials, or any other means, the circumference of a circle described about a center O, be divided into any number of parts AB, BC, CD, &c. so that the chords be equal; then it is evident, from hence, that all the angles AOB, BOC, COD, &c. which make up the four right-angles AOD, DOG, GOK, KOA at the center, will also be equal to each other, let the radius OA of the instru- ment be what it will.—In the division of the circle for practical uses, the number of parts into which the circumference is thus divided, or the number of equal angles at the center, is 360; which equal angles are called degrees ; so that a right-angle, consisting of 90 of these equal angles, is said to be an angle of 90 de- grees; every angle being denominated, from the de- grees and parts of a degree, which it contains; each degree being conceived to be subdivided into 60 equal parts, called minutes ; each minute again into 60 equal parts, called seconds; and so on to thirds, fourths, fifths, &c. at pleasure. THEOREM II. Any chord (AB) of a circle, falls wholly within the same: and a perpendicular (cD) let fall on any chord, from the center of the circle, will divide vt into two equal parts. Let C, A, and C, B be joined: and through any point E in the chord AB, let the right -line CEF be drawn, meeting the cir- cumference in F. It is evident, because CA= EN a CB*, that these equal lines are on different sides of the perpendicu- 3 lar CD 4; and so, CE being a CA or CF %, the point BOOK THE THIRD. Al ¥& (take it where you will in the line AB) and conse- quently the line AB itself, will fall within the circle* « Ax. 2. Moreover, because the triangles ACD, BCD have CA=CB and CD common, thence will AD be also Bent) fs: $16.1, COROLLARY. Hence a line bisecting any chord at right-angles, passes through the center of the circie, THEOREM III. Any two chords (aB, DE) equally distant from the center (0) of a circle, are equal to each other. Let the perpendiculars OF, B i OC be drawn, and let O, D and O, A be joined. Because OF=OC* » OD=OAY, and F ¢ Hyp. and C are both right-angles®, / Ber 33. therefore is DF = AC, and con- of 1. sequently DE = ODF i= 2AC# es = AB‘ i2, 3. A D k Ax. 4, 1. THEOREM IV, IV. In a circle (AEFB) the greatest line (AB) ts the diameter ; and of all others terminating in the circumference, that (cD) which is nearest the center (0,) ws greater than any other (EF) further from it. 1. Draw OC and OD; R shen it will appear that AB | ‘or OC+O0OD)c- CD” 2. Let OP be the distance »f CD from the center, and IQ that of EF, both taken n the same radius OR; Draw OF and OF; be- vause the triangles DOC, OR, have two sides equal 42 ; ELEMENTS OF GEOMETRY. »Def.33.1,each to each”, and have the contained angle DOC o Ax) 2 P21. 1, 73. 2. ry Ax. 4, 5419, 1. ¢ Ax, 2. “21.1, c“the contained angle FOE°; therefore, also, will the base DC be & the base FH”; and consequently tc any other chord, at the same distance, with IF %. COROLLARY. Hence, a right-line greater than the diameter, drawn Jrom any point within a circle, will cut the cir- cum/ference. THEOREM V. If to the circumference of a circle (AFEB), from any point (D) which ts not the center, right- lines (DA, DF, DE) be drawn, the greatest of all (pA) shall be that which passes through the, center (c;) and, of the rest, that (pF) whose other extreme (F) ts placed nearest, im the circumference, to the ext¥eme (a) of the greatest, will exceed any other (DE) whose eas treme (£) ts at a greater distance. D From the center C, let CE and CF be drawn. Then, 1. AD (=DC+CF’)c DF* | And, 2. Since DC is common, CF=CE, and DCF Cc DCE, therefore is DF — DE™. a | COROLLARY Tf. Because no two lines, DE, DF, drawn from D, on the same side of the diameter AB, can be equal t BOOK THE THIRD. © 43 each other”, three equal right-lines cannot possibly ws, 3. be drawn from the periphery to any point, besides _ the center of the circle: and, therefore, 2f from a potnt in any circle, three equal right-lines can be drawn to the periphery, that point is the center of the circle. COROLLARY II. ‘Hence it also follows, that no circle can be described to cut another FBG in more points than two: for, if it were possibie to cut it in three points G, E, F, then right-lines drawn from the center Q, to those points, would be all equal*, which is shown = Def.33.1, to be impossible ¥, unless when the center Q coin- Cor. 1, to _ cides with C; and then the circles themselves will 4-3: neither cut, nor touch, but coincide, and become one circle *. THEOREM VI. A right-line (rp) drawn through any point (a) en the circumference of a circle, at right-angles to the radius (wa) terminating in that point, will touch the circle. From any point in FD, F A B D to the center HE, let theright- ~~ >= line BE be drawn; which being greater than AE4% ociouet the point B must, necessa- ; b Def, 33. rily, fall ont of the circle ®; aepar and therefore, as the same i of L. argument holds good with regard to every other point: in the line FD (except A) itis manifest that this line cuts off no part of the circle, but touches it in one point only. THEOREM VII. If the distance (AB) of the centers of two circles, be equal to the sum of the two semt-diameters 44 © Constr. and Ax. 5. ad Def, 33, 1. let DCE be drawn perpendicu- lar to AB: then, BC being also =BN*, the circumferences of both circles will pass through the point C“: but the right-line DE (by the precedent) falls wholly above the one, and wholly below the other; there- fore the circles themselves fall wholly without each other, and touch in one point C only. ELEMENTS OF GEOMETRY. (am, BN,) the circles will touch each other, outwardly; and the right-line (ap) joining their centers, will pass through the point of. contact. In AB, take AC=AM, and COROLLARY. Hence, tf the centers of two circles be placed at a I } § distance, from one another, less than the sum of the | two semt-diameters, a part, at least, of the one wilt be contained within the other: but, if the distance | be greater than that sum, the two circles will then newther touch, nor cut each other. THEOREM Vill. circles touch inwardly ; and that radius (ca) } If the distance (cv) of the centers of two circles (car, DAB) be equal to the difference of the | two semi-diameters (CA, DE,) then well those | | of the greater, which 1s drawn through the center (D) of the less, will meet the two peri-— pheries in the point of contact. BOOK THE THIRD. From any point E in the cir- cumference of the less, to the two centers, let EC and ED be drawn. Because CA exceeds DE by the line DC’, or because DE+DC =CA‘= DA+DCzg, therefore is DA=DE"; and so che circumference of the circle D likewise passes through A ; but CA c CE: therefore every point in the periphery of the circle D (except A only) falls within the circle C: which was to be demonstrated. COROLLARY I. Hence, 7f the centers of two circles be placed at a distance from each other, greater than the differ- ence of the two semi-diameters, a part, at least, of the one will fall without the other ; but, if the dis- tance be less than that difference, the lesser circle will then be contained wholly in the greater, but _ without touching tt. COROLLARY II. dence, and from the precedent, it likewise appears, _ that 7f two circles touch, either inwardly or out- | ewardly, a right-line, drawn through their two cen- _ ters, will also pass through the point of contact: _ because they can only touch, when the distance of their centers is equal to the sum, or to the difference of their semi-diameters *. THEOREM IX. Uf the distance of the centers (¥, G) of two circles (pL, mH) be less than the sum, and greater than the difference of the two semi-diameters (FL, GM,) those circles will cut each other. 45 ® Cor. of 7. and Cor, 1, of 8, 46 'Cor, to 7, 3. ” Cor. to 8, 3. *Def. 11. of 3. The angle (ppc) at the center of a circle, is _. double to the angle (Bac) at. the circumfer= ence, when both angles stand upon the same arch (Bc). | 09, 1. P12. 4. q Ax. 4. t AX, 5. ELEMENTS OF GEOMETRY. For, sinee the distance of the twocentersissup- An ; ! posed less than p LG FO the sum of the se- mi- diameters, a part of the one | circle MH, falls within the other DL’; and since that distance is’ greater than the difference of those semi-diameters, a part of the same circle MH also falls without the eircie DL”: which was to be proved*. 3 THEOREM. X. Let the diameter ADE be drawn. In the first case (where AB passes through the center) BDC=A+Co=2A?, _ An the second case, BDE=2BAE, (by case 1.);} to which adding CDE=2CAE, we have BDC= 2BAC 4, Ln the third case, CDE=2CAE (dy case 1.) from] whence subtracting BDE=2B AE, there remains BDC=2BAC’, BOOK THE THIRD. THEOREM XI. All angles (wa¥, EBF) in the same segment, (EABF) of a circle, are equal to each other. CasE 1. Ifthe segment be greater ‘han a semi-circle ; from the center C draw CE and CF; then EAF md EBF being each of them = | io +ECF s, they must necessarily be \y qual to each other. Case II. Jf the segment be less than a semi-circle; let H pe the intersection of EB and AF: then the triangles AEH and 7 and BFH, having the ae AHE= BHF ¢, and AEH=BFH {by case 1.) they will also have EAH=FBH*. F THEOREM Angles (p, G) 2m the cir ‘cumferences, standing upon equal subtenses (AB, EF) of circles hav- eng equal diameters, are equal to cach other. And the subtenses of equal angles, in the cir- — cumferences of circles having equal diameters, are also equal. St ae A, XII. Q2 47 «Cor. 1. ta 107 3: 48 « Hyp. * Def. 33 of 1. ¥14. I 710. 3. @ Ax. 10. of 1. $10. 3. cAx. 4.1 er, joined ELEMENTS OF GEOMETRY. From the centers P and Q, let PA, PB, QE, QF. be drawn. : é 1. Hyp. Since AB=EF¥», and AP=BP*=EQ) =FQ”; therefore is P = Q», and consequently D_ (=1 P*=7Q)=G. | 2. Hyp. Because D=G, therefore P=Q*; whence | ig ae and PB=QF”, AB will also be) COROLLARY. Hence angles tn the circumference, standing upo , equal chords of the same circle, are equal. THEOREM XIII. The angle (AcB) ina semi-circle, is a right-angle, Let the diameter CDE be drawn. Because ACD = half ADE, and BCD = half BDE®, therefore is ACD+ A Bo BCD (=ACB)=half of _ ADE and BDE*‘=half two right-angles? = one right- angle. B THEOREM XIV. The angle (cas) included by a tangent to @ circle,and a chord (ac) drawn from the point of contact (a,) is equal to the angle (AEC) m the alternate segment. a Let the diameter AOF be drawn, and E, F be BOOK THE THIRD. 49 * The line DB falling wholly A Pp above the circles, OA is the Ta —————_ ¢ Def, 9. least line that can be drawn to 3. it from the center OF; and an 7 OAB is thereforea right-angles ; ee but FEA is alsoa right-angle?: £20. 1, therefore, if from these equal C *13. 3. angles, the equal‘ angles FAC, ‘11. 3. FEC (standing on the same arch FC) be taken away, there will remain BAC= AEC *. k Ax.5, THEOREM XV. The angle (DEC) made by two lines (DEB, CEA) ¢ntersecting each other within, or without a circle, is, in the former case, equal to the sum, and i the latter, equal to the difference, of two angles in the circumference, standing on the two arcs (DC, AB) intercepted by those lines. E elt Let the chord CB be drawn. © Then DEC=DBC+ACB,, in the first case. 9, 1. And DEC=DBC—ACB,, in the second case. 9.1. and = Ax. 5, \ COROLLARY. Tence, an angle (E) formed below, or above the cir- cumference of a circle, is greater, or less than an angle in the circumference, standing on the same arch, . E SO ELEMENTS OF GEOMETRY. THEOREM XVI. The vertical angle (Abc) of any oblique-angled triangle (AcB) inscribed in a circle (ABCD) 18 greater, or less than a right-angle, by the angle (cap) comprehended under the base (ac) and the diameter (AD) drawn from the extremity of the base. 7 i 13.3. For, BD being drawn, ABD will be a right-angle’, «11.3. and CAD=CBD*; therefore, in the first case, Aba 1ax.4. —right-angle+ CAD’; and in the second, ABC=! ™Ax. 5, right-angle—CAD™”™. THEOREM °XVH. ’ a | If any side (Bc) of a quadrilateral (aBcn) in scribed in a cirele, be produced out of th circle, the external angle (ECD) will be equa to the opposite, internal angle (Bav).* ‘I - *® From Commandine’s Commentary on Prop. 26; BUViITY Pappus’s Mathematical Collections. | BOOK THE THIRD. Let the diameter BF be drawn, and jet AF and CF be joined: then the angle « BAF being a right-angle (= BCF)= ECF”, and. DAF also=DCF° standing both on the samearch DF ; thence will the remainders BADand B ECD be also equal ?. COROLLARY. Hence the opposite angles BAD, BCD of any qua- drilateral inscribed in a circle, are together, equal to two right-angles. For, since BAD=ECD, therefore is BAD + BCD = ECD + BCD’=two right-angles:. THEOREM XVIII. ‘Through any three points (A, B, C) not situated am the same right-line, the circumference of a carcle may be described. ! Draw AB and BC, which let be bisected by the perpendiculars DG and EH, infinitely produced on that side of AB or BC, on which 'the angle ABC is formed. ji These perpendicu- lars, I say, will inter- 3ect éach other; and ‘the point of intersec- jon O; will be the center of the circle. 51 r Ax.4, ‘11. 52 t Ax. 2. “ Cor. 2. to 7.1. » Def. 13. of }. = Constr. ¥y Ax. 7. = Ax. 10. ® Ax. 1. > Def, 33. ]. ¢ Cor. to eB 2d Cor. to 17. 3. * Hyp- J As. Os £23. 1. ELEMENTS OF GEOMETRY. For, if DE be drawn, it is plain, that the angles” GDE, HED are less than two right-angles*; there- fore DG, EH, not being parallels", they will meet each other”. Hence, if from the point of intersection O, the right-lines OA, OB, OC be drawn, the triangles ADO, BDO, having two sides equal, each to each z and the angles ADO, BDO, contained by them, equal ¥, will likewise have AO=BO*. After the very same manner is CO=BO; therefore AO=BO=CO": whence the circumference of a circle described from the center O, at the distance of AO, will also pass through B and C°. SCHOLIUM. Hence the method of describing the circumference . of a circle through three given points, is manifest. THEOREM XIX; If the opposite angles (BAD, BCD) of a quadri- lateral (ABcp) be equal to two right-angles, @ circle may be described about that quadrila-. teral. | For the circumference of a cir- cle may be described through any three points B, C, D, (oy the precedent.) But,if you deny that it passes through A; then, through the center O, let OAF be drawn, and let it (if possible) pass through some other point F in the line OAF (for it must cut this line somewhere‘;) also let BF and | | DF be drawn. Because BFD+BCD=two right- angles‘= BAD+BCD*; thereforemust BF D= BAD, which is impossible&. Therefore the circumference: of the circle described through B,C and D, must also pass through A. | oo - = BOOK THE THIRD. THEOREM A. If Ala one extremity (A) of a diameter (az) of a circle (arB) a right-line (av) be drawn to meet the circumference (in E) and any perpendicular (cD) to that diameter, (ether within or without the circle) the rect- angle under the segments (DA, Ax) of the line so drawn from the extremity of the dia- meter, will be equal to the rectangle under the diameter, and the distance of the same extre- mity from the perpendicular, that is, DA x AE =BAXAC.* ' For EB being joined, the angles DCB, DEB are equal, because they are right-angles ; therefore a circle may be described through the four points D, C, E, B, . ‘by the last prop.) and consequently the rectangle DA x AE is equal to the rectangle BA x AC. } THEOREM Xx. of from two pownts (B, Cc) in the same diameter | rr (AD) equally distant from the center (0) of a circle, right-lines (BE, CE; BF, CF) be drawn ‘to meet, two by two, in the circumference ; the sum of the squares of any twa correspond- ang ones, will be equal to the sum of the squares of ray other two, meeting in like man- ner. * R. Simson’s Apollonius on Plane Loci. Prop. 8 and 9, 33> o4 k11, 2. 32.3. k Ax, 5. Cor. 1. to 9, 2. ELEMENTS OF GEOMETRY. an 9 . For, if OF and OF be (D2 drawn; then will BE ?+ (/ CE? =" 2BO” + 20E? 4 als D (2OF *)="BF 2+ CF *. THEOREM XXI. If two lines (AB, cp,) terminated by the pert- phery on both sides, cut each other within a circle, the rectangle (AP x BP) contained under the parts of the one, will be equal to the rect- angle (ce x DP) contained under the parts of the other. CasE 1. If one, of, the, two lines, (AB) passes through the center O.; then let OQ be drawn perpen- dicular to the other CD, and let OC be joined. It is plain, because QD=QC‘*, that DP is equal to the difference of the segments CQ and PQ*: but the ree- tangle under the sum and-difference of the two sides OC, OP, of any triangle. COP, is. equal to. the, ree- tangle under the whole base CP, and the difference of its two segments!; therefore, the sum of the two sides OC, OP. heing (=QA+OP)=AP, and their BOOK THE THIRD. difference (=OB—OP)= BP, thence is the rectangle contained under AP and BP equal to the rectangle contained under CP and DP. _ Case Il. If neither of the two lines’ pass through thecenter ; let the diameter EPF be drawn ; then, (dy case 1,) AP xBP= FP xEP=CP x DP. THEOREM XXII. If from two points (A, Cc) in the circumference | of a circle, two lines(aP, cp) be drawn, to pass through and meet without the circle; the rectangle (APxBP) contained under the other. Through the center’ O, et PE be. drawn, meeting che circumference in E and F; let OQ be perpendicular Lo. AP, and: let. A,.O be joined, Then, (by Cor. 1. to 9. 2.) the rectangle contained ander PF (=PO+0A) ud PE (=PO—OA) is= he» rectangle contained under AP. and PB. After whole and Rehe external. part of the one,. wall be equal to the rectangle (ce x DP) contained. under the whole and eae external part a the EF he very same manner. PF x PE= OP x DP.: there- lore AP x BP=CP’x DP”. t } / pe PF x PE°-=PC.x PD. > Ax. it, COROLLARY. dence, if PS bea tangent at S, and the radius OS _ be drawn; then, PF being= the sum of PO and OS, ‘and PE=their difference ; it follows, that PS’= * Cor. te 8, ae 56 ELEMENTS OF GEOMETRY. THEOREM XXIII. If from the center (c) of a circle, to a point (A) wn any chord (Bp), a line(ca) be drawn; the” square of that line, together with the rectangle » contained under the two parts of the chord, wll be equal to a square made upon the radius of the circle. Let EAF be another chord, RB perpendicular to CA, and let C, Vv Bas E, be joined. ECS 2 Since AF=AE4, thence will Da AE* =AExAF*= AB x AD:; to which equal quantities adding AC’, we have CE?=ABxAD +AC ?*, ; COROLLARY. Hence the square of a line(AC) drawn from any point in the base of an isosceles triangle (BCD) to the opposite angle, together with the rectangle of the parts of the base, is equal toa square made upon one of the equal sides of the triangle. 2 THEOREM XXIV. ‘ The rectangles contained under the correspond- ong sides of the equangular triangles (ABC, DEF) taken alternately, are equal. 4 I say, if A=Ty B=E and C=F, then will AB x DF=AC x DE. as BOOK THE THIRD. In BA_ produced, let AG be taken=DF; let GCB be the circumference of a circle passing through the three points B, C, G %, meeting CA produced in H; and let GH be joined. _ Because the angle H= By=K*?, HAG=*BAC= ‘D+, and AG= DF, thence ‘ss AH=DE?; and there- ‘ore AC x DE=ACxAH°=ABx AG? | THEOREM XxV. The rectangle under the two sides (ac, Bc) of any triangle (ABC) 2s equal to the rectangle under the perpendicular (cp) to, the base, and the diameter (ce) of the circumcribing circle ; For B, E being joined, | che angles A, E will be equal‘, and ADC, EBC ooth right-angles¢; and, consequently the triangles ACD, ECB equiangular’: herefore AC, EC; CD, CB being corresponding sides, opposed to equal ingles, the rectangle AC x 2B, contained under the | } } THEOREM XXVI. The square of a lne (cp) bisecting any angle (c) of atriangle (anc) and terminating in the opposite side (AB), together with the rectangle irst and last of them, will be equal to the rectangle EC x CD contained under the other two ¢. ¥11.3. 524,3. 57) 58: / E'18.'3. $11. 3., sg Ee + Constr. * Cor. 1, to 10. 1. t 24. 3. % Ax, 4. I, ELEMENTS OF GEOMETRY. (AD x BD) under the twosegments of that side, ws equal to the rectangle of the two sides (ca, CB) including the proposed angle. 0 Let CD be produced to meet the circumference of a circle, described * through the points A, C, B,in E; and let AE be drawn. The angles E and B, standing upon the same segment: AC, are equal’; and ACE is- equal DCB. (by hypothesis) ;. therefore the triangles AKC, DCB Ez are equiangular™; whereof: AC, CD; CE, CB are ‘ corresponding sides,, opposed to equal angles: there-— fore AC x CB=CD x CE*= CD*+CD'xDE°=CD *%& +AD'x DB’. THEOREM XXVIL. The rectangle of- the two: diagonals, (AC, BD): oft any quadrilateral (anon) inscribed: in a circle, 1s equal to the sum-of the two rectangles (ABX DC, AD xX BC) contained under the oppo- site Stdes. Let BF be drawn, making Ay the angle CBF=ABD, and : meeting AC in F: Because the angle BCF= ABD, and CBF=ABD’, the triangles CBF, DBA are equi- angular*; and therefore, BC, \: BD; CF, AD, being corres- . ponding ‘sides, the reetangles BC x. AD, and: BD x CE: will be: equal! Again, the angle: ABF) being CBD", A “WN Fe AD ty “ = ‘ aa oe > BOOK THE THIRD. ) 59) and BAF=BDC"; the triangles. ABF and BDC, are, ~ 11.3. ikewise, equiangular; and consequently, AB, BD ; AF, DC being corresponding sides, AB x DO=BD x AF; to which adding BCxAD=BDxCF (so sroved above, ) we have also AB x DC+BCx AD= BD x AF+BD x CF= BD xAC*, = §,2. HEOREM XXVIII. y the radius of a circle (OADF) be so divided nto two parts, that the rectangle under the whole and the one part shall be equal to the square of the other part; then will this last part be equal to the side (cv) of a regular decagon (ABCcDEF, &c.) mscribed m the circle; and that line whose square is equal to the two squares, of the whole and of the same part, will be equal to the side (ac) of a regular pentagon inscribed in the same crrcle. Draw the radii OA, oc, OD, OF ; also draw AD, cutting OC in G, and let AH be perpendi- cular to OG. The triangle ODG, having the angle COD A\ =3 DOF? = OAD*)= ODA‘, is isosceles? ‘moreover the Sie AOG, having AGO (= GDO + DOG*s = 2DO0C * ) = AOC, is likewise ‘isos seles¢ ; as is also the triangle CDG, because, ¢ 18. 1. CG being = AGO’, and CDG (CDA) = =FAD/, (3.1 ‘the triangles AOG, CDG are equiangular. Therefore, ‘Sait i CD, AO; CG, GO being corresponding sides, we have CGx AO (CG x CO)= CD xGOs=GO2, be- ¢ 24, 3. cause GO=GD=DC": whence the former part of "18. 1. the proposition is manifest. ® 60: ELEMENTS OF GEOMETRY. * 16.1. Again, because AG=AO, HG will be=HO « an q so GC being the difference of the segments HO and HC, we have (by Cor. 1. to 9. 2.) AC *—AO *=CO x CG=O0G? (as Pate te and Cone Tee REY AC * om AO #+0G?. 4 THE END OF THE THIRD BOOK. (Ba ELEMENTS OF GEOMETRY. & Rh at BOOK IV. DEFINITIONS. ) ema W HEN one magnitude contains another a ** certain a of times, exactly, the former is said ‘to be a multiple of the latter, and the latter a part _* of the former. B. “ When several magnitudes aie multiples of as “many others, and each contains its part the same “number of times, the former magnitudes are said to ** be eguz-multiples of the latter, and the latter magni- “tudes are called like parts of the former. 1. Ratio is the proportion which one magnitude bears to another magnitude of. the same kind, with ‘respect to quantity. The measure, or quantity of a ratio, is conceived by considering what part, or parts, the magnitude referred, called the antecedent, is of the other, to which at is referred, called the consequent.* | | } * 1. Between any two finite magnitudes of the same kind, there subsists a certain relation in respect to quantity, which is _ called their Ratio: The two magnitudes compared together are _ called the terms of the ratio, the first the antecedent, and the second the consequent. », 2, Of four magnitudes, the ratio of the first to the second is said | to be the sume as; or equal to, the ratio of the third to the fourth, _-when the first contains any part, whatever, of the second, just so * ' 62 ELEMENTS OF GEOMETRY. A, B, C. 2. Three quantities, or magnitudes A, B, C, are 2.4.8. said to be proportional, when the ratio of the first A to the second B, is the same.as the ratio of the second B, to the third C. ’ A,B,C,D. 3. Four quantities A, 'B, C, 1D, are said to be ?.4.5.10. mroportional, when the ratio of the first A to the second B, is the same as the ratio of the third C to the fourth D. : Lo denote that four quantities A, B, C, D, are pro- portional, they are usually written thus, A: B::C:D; and read thus, as A is to B, soisC to D. But when three quantities A, B, C are proportional, the middle one is repeated, and they are written thus, A:B::B:C.; ‘ Sometimes, however, they are “ written thus, A: B=C: D, and then they are read thus, the ratio of “ A to B is equal to the ratio of OF oye Bas 4, Of three proportional sinutiting ie middle one is said to be a mean-proportional between the other two; and the last, a third-proportional to the first and second. a 5. Of four proportional quantities, the ast is said to be a fourth-proportional to the other three taken ) in order, >» A, B, C, 6. Quantities are said to be continually pro- a g portional (or in continual proportion) when the first 6. | isto the second, as the second to the third, ag the third to the fourth, as the fourth to the fifth, and soon; (or, which is just the same thing, when the ratios of the first to the second, of the second to the third, of the third to the fourth, &c., are all. equal.) : of often as the third contains the like part of the fourth.—But the ratio of the first to the second is said to be greater than the ratio. of the third to the fourth, when the first contains a part of the second oftener than the third contains a like part of the fourth; and less when not so often, : , fs 8. When two ratios are equal, their terms are said to be proportional. BOOK THE FOURTH. 7. In a series, or rank of quantities continually soportional, the ratio of the first and third, is said to i duplicate to that of the first and second ; ; and the itio of the first and fourth, triplicate to that of the lest and second; “also ie ratio of the first to the ‘second, is called the sub-duplicate of the first to the ‘third. The swb-triplicate of the first to the ' fourth, &c. 8. Any number of quantities, A, B, C, D being iven, or propounded, the ratio of the first (A) to the ast (D) is said to be compounded of the ratios of the rst to the second, of the second to the third, and so n to the last. | 9. Ratio of equality, is that which equal quantities ear to each other. _ It may be observed here, that ratio of equality, and quality of ratios, are, by no means. synonymous erms: since two or more ratios may be equal, though he quantities compared are all unequal. Thus the atio of 2 to 1, is equal to the ratio of 6 to 3 (2 being he double of l, and 6 the double of 3); yet none of he Jour numbers are equal. _ 10. Jnverse ratio is, when the antecedent, is made he ae ee ‘and the consequent the antecedent. Thus of2: 76: 33 then, inversely, 1:2::3: 6. 1]. Alternate ratio is, when antecedent is com- yared with antecedent, and consequent with con- jequent, As of 2: W6-g1 3s pie ek alternation (or yermutation) i will be 2 : 6: pea 12. Compound ratio is, when the antecedent and consequent, taken as one quantity, are compared sither with the ear a, or with the antecedent. | Bie’, v 2: : 6 : 3; then, by composition, +d ; ‘aide $, and 241 : 2:3 643: 6. i 13. Divided ratio is, when the difference of the antecedent and consequent is compared, either with the consequent, or with the antecedent. - \ 63 64 7™ ‘ELEMENTS OF GEOMETRY. oie U S:slansy lB: Agsathen, aby division, 3- ie : 12- ra 4, and 3—1 : 38: : 12-4: 12. These four last definitions, which explain th names given by Geometers to the different ways ¢ managing and diversifying of proportions, are pu down here for the sake of order: but are not tod used, or referred to, in any shape, till those propertie and relations are demonstrated; which is cae u the three first Theorems of this book. 14. Similar (or like) right lined figures are such as have all their angles equal, one to anothe respectively, aud also the sides about the equal angle proportional. C fb ya Spee remind wee Thus if the angle A=D, B=E, os F; als PRs VANS ee ed Si BA: BC:: ED: EF §c., then the figures ABC, DEF are said to bi similar. é AXIOMS. j 1. The same quantity being compared with ever so many equal quantities, successively, will have th same ratio to them all. 2. Equal quantities, have to one and the | quantity, the same ratio. a 3. Quantities, having the same ratio to one and the same quantity, or to equal quantities, are equal. ...@ 4. Quantities, to which one and the same quantity has the same ratio, are equal. 5. If two’ quantities be referred to a third, that which is the greatest will have the greatest ratio. } Wy BOOK THE FOURTH. | 6. If two quantities be each referred to a third, that is the greatest which has the greatest ratio. | 7. Ratios, equal to one and the same ratio, are also equal, one to the other. |' 8. If two quantities be divided into, or composed of parts, that are equal among themselves, or all of the same magnitude; then will the whole of the one, have the same ratio to the whole of the other, as the number of the parts in the one has to the number of equal darts in the other. | 9. If the double, treble, or quadruple, &c. of every dart of any quantity be taken, the aggregate will be he double, treble, or quadruple, &c. of the whole {uantity propounded. THEOREM I. Squmultiples of any two quantities (AB, CD) are in the same ratio as the quantities them- selves, Aisa fo poe Bk a y | | | ) Let the ratio of AB to CD be that of any one amber M (3) to any other number WN (4), or, which | the same, let AB contain M (3) such equal parts 65 ‘Aa, ab 6B*,) whereof CD contains the number *4~ ®: ¢: | (4). Let there be taken Ef, fe, oF any equi- ultiples of Aa, ad, 5B, respectively,; and let Gp ib gr, TH be the same multiples of Ce, cd, de, eD: * shall the whole EF be the same assigned multiple ( the whole AB, and the whole GH of the whole \D, as each part in the one, is of its correspon- ent in the other’. And, since the parts Aa, ab,» Ax.9.4. 3, Cc, cd, &c. are all equal‘, their equimultiples « py, i, t8> §F, Gp, pg, &c. will also be equal“, There- «ay 4.1. ire EF is to GH, as the number of the parts in EF | F 66 ELEMENTS OF GEOMETRY. is to the number of equal parts in GH+, or (which is the same) as the number of parts in AB, to the number of parts in CD, that is, as AB to CD% which was to be demonstrated. COROLLARY. Hence, like parts of quantities, have the same rati as the wholes ; because the wholes‘are equimultiples of the like parts. | THEOREM II. The two antecedents (AB, DE) of four propor tonal quantities of the same kind (AaB, BC DE, EF) are in the same ratio with the tw consequents (Bc, EF.) See Def. 11. Aa Be? Cc hl aR a a Dac, Be 77 Fen remem me WAT te Let the common ratio of AB to BC, and of DE ti EF, be that of any one number iM (5) to any othe number WN (3); then will AB contain M (5) sue) ee ia equal parts (Aa) whereof BC contains eN (3) ani ee DE will, in like manner, contain M (5) such equa parts (Dd), whereof EF contains N (3.) And st AB and DE, as well as BC and .EF, being eqi. multiples of Aa, and Dd, thence will AB: DE:: A fi.4, (Bb) : Dd(Ee) : : BC: EF* COROLLARY. The proportionality will subsist, when the const quents are taken as antecedents, and the antecedent as consequents. For BC : AB : : number of part in BC (or EF): number of parts in AB @ DE):: EF : DE* See Def. 10. a BOOK THE FOURTH. - THEOREM III. Of four proportional quantities (AB, BC, DE, EF) the sum, or difference, of the first antecedent and consequent (AB+ BC), ts to the first | antecedent, or consequent, as the sum, or ' difference, of the second antecedent and conse- quent (DE+EF) zs to the second antecedent, or consequent. See Def. 12 and 13. | Let what was premised in the demonstration of the receding theorem, be retained here: then will C (AB+BC) be in proportion to AB, as the umber of parts in AC is to the number of equal arts in AB%, or as the number of parts in DF DE-+EF) to the number of equal parts in DE, that ,as DF (DE+EF)isto DEs. Again, if from AB id DE, be taken away Bec=BC, and E/=EF, then \ill the difference Ac be in proportion to AB, as the ‘umber of parts in Ac (or Df) is to the number of arts in AB (or DE®,) that is, as Df isto DE. In je same manner it will appear, that AB+ IC,: BC :: DE+EF : EF; and AB—BC : jC :: DE—EF : EF. COROLLARY. | will appear from hence, that the swm of the greatest and least (AB+ EF) of four proportional quantities ‘(of the same kind) will exceed the sum (BC+DE) of the two means: because, AB being supposed © DE, Ac will be — Df, in the same proportion*: *3.4. and, if to these there be added BC+ EF (common 3) then will the sum Ac+BC+EF (AB+EF) be also C the sum Df/+BC+EF‘(DE+ BC.) ‘Ax. 6 1, F 2 67 & Ax. 8. 4, 68 ELEMENTS OF GEOMETRY. SCHOLIUM. In the demonstration of this and the preceding theorems, the antecedent and consequent are sup- posed to be divided into parts, all mutually equal) among themselves. But it is known to Mathe- maticians, that there are certain quantities or mag; nitudes that cannot possibly be divided in that manner, by means of a common measure. The theorems themselves are, nevertheless, true, wher applied to these incommensurables : since no twe quantities of the same kind, can possibly be assigned whose ratio cannot be expressed by that of two num bers, so near, that the difference shall be less than thi least thing that can be named. But if the matter viewed in this light should not appear sufficienth scientific, and you will not (Cn the precedin; theorem) allow the ratio of AC to BC, to b exactly the same with that of DF to EF, when AE BC, and DF, EF, are incommensurables ; then let i if possible, be as some quantity aC (less than AC is to BC, so is DE to EF. Let Bb be a part ¢ measure of BC less than the difference (Aa) tween AB and aB; Let Bp be that multiple of B which least exceeds Ba, and let gk be to EF, as p to BC. It is evident, that A , Bz Ya pB is 2 AB (because Lemans aoe cpm ‘Hyp. ap—Boa* Aa); and that gE is alsomDE, PZ Eo # because the ratio of gE to EF, being=that of pB i Ax.5.4. BC*, it must necessarily be athat of AB to BC’, or m Ax.6.4, DE to EF, and so gE a DE”. | Now if (as is supposed) the ratio of aC to BC be=the ratio of DF to EF, it must, of consequent nAx.h. bet? the ratio of gF to EF *, or (which is the? sam _inl?-* than the ratio of pC to BC which is impossible. at In like manner it will appear, that no quantity, th isc-AC, can possibly be to BC, as DF is to E Therefore AC: BC :: DF: EF. And by 7? same kind of argumentation (authorized and adopt! by Euclid himself, in his twelfth book) any di’ F BOOK THE FOURTH. * 69 culties, or scruples that may be elsewhere brought, rom the incommensurability of quantities, may be »bviated and remov>d. ' | | THEOREM IV. Uf, of four proportronals (aB, BC, PQ, QR) equi- ' multiples of the antecedents (AB, BQ) be taken | and compared with any equimultiples of their | respective consequents (BC, QR), the ratios _ well be the same, and the four quantities | proporttonals. _ Let the common ratio of AB, to BC, and of PQ to QR, be that of any one number M to any other number NV; so shall AB contain M such equal oarts whereof BC contains? N, and PQ, in like? Ax.8.4. manner, M/ such equal parts whereof QR contains N. _ Let CD, DE be taken each=BC, and RS, ST sach=QR, so that BE and QT may be equimultiples pf BC and QR; and let CD, DE; RS, ST be con- veived to be divided, each into the same nnmber of | 2a A. B fy D E | P ie BOR HSMG Wa ete yarts with BC, or QR. In like manner, let a@B and 2Q be taken as equimultiples of AB and PQ, &ec. Then will the number of parts in BD=number of darts in QS?, and thenumber of parts in BE=number of parts in QT'?: and so likewise with respect to aB 1 Ax. 4.1. ind pQ. Therefore aB is to BE, as the number of arts in @B to the number of (equal) parts in BE ?. or which is the same thing, as the number of parts in »Q to the number of parts in QT, that is, as pQ is to RT, which was to be demonstrated. 70 ELEMENTS OF GEOMETRY. : THEOREM V. ff of two ranks of quantitees (AB, BC, CD ; PQ aR, RS,) the ratio of the first and ate ee an the one, be equal to the ratio of the first and second wn the other, and the ratio of the second and therd, in the one, equal likewise to the ratio of the second and thard in the other; then, also, shall the ratio of the first to the third, be the same in the one rank as in the other. | Let the common ratio of AB to BC, and of PQ to QR, be still expressed as in the preceding demonstra. tions. Let, moreover, CD and RS be conceived to’ be divided, each into the same number of parts with BC and QR. LANE B & On D E PP } ; BK yn «qt ca ea MERC NES uv a EN SN | Because the quantities BC, CD, QR, RS, are pro- ” Hyp. portional’, their like parts Bb, Cc, Qg, Rr (being in| ‘Cor.to the same ratio with the wholes °) will also be propog ae: tionals; or, because Aa= Bd and Pp=Qg’, it will be! ‘“™ Aa:Ce:: Pp: Rr. But AB and PQ are equimul tiples of the antecedents Aa and Pp”; and CD, RS are equimultiples of the consequents Ce and Rr; 4, 4, therefore" AB: CD:: PQ: RS;* which was to demonstrated. * When in two ranks of quantities, the proportions are inordina asAB: BC: : QR: RS,andBC:CD: : PQ: QR; the same thing may be demonstrated ; 3 and that in the very ‘y same manner, €k- cept only that QR must here be divided into the same number of parts with AB, and PQ. BOOK THE FOURTH. COROLLARY I. {f other quantities DE, ST be taken, still proportional to the two next preceding them, so that CD: DE :: RS: ST; then, by the same argument (regard being had to AB, CD, DE in the one rank, and PQ; RS, ST in the other) it is evident that AB: DE::PQ:ST*. And thus we may go on,=-5.4, ' and the ratio of the first and last, will always be the same in one rank, asin the other. Therefore Lax ratios ¥ compounded of the same number of like, or » Def. 8. i still assuming other quantities, as many as we please; t | | equal ratios, are equal. COROLLARY If. .t is also evident from hence, that if any two quan- . tities be taken, proportional to the two consequents _ of an assigned proportion, they will also be propor- _ tionals when compared with the antecedents ; and vice versd. For, the two quantities CD and RS, when compared, successively, with the consequents and | antecedents of the given proportion AB: BC:: PQ : QR, appear to be proportional, in the one case, as well as in the other”. | THEOREM VI. Tf, to the two consequents (Bc, Kt) of four pro- | porteonals (AB, BC, IK, KL,) any two quanti- ties (CD, Lm) that have the same ratio to the respective antecedents be added; these sums and the antecedents will still be proportionals (LI say, if AB: BC::1K:KL,and AB: CD:: 1K - 7 >LM; then shall AB: BD::1K: KM.) ' For CD and LM foe A B eS dD being proportional to 7 ‘he antecedents AB und IK*, they and J the consequents (BC, = j}—-—— —— ” E 4 of 4. 25, 4, K. L M_M N “ Hyp. red J 72 °6. 4. ELEMENTS OF GEOMETRY. KL) will also be proportionals (Corol. 2 of the last): whence’ (6y comp.) BC : BD::KL:KM., And ‘so again (same Corol.) AB: BD: SIK: KM. COROLLARY. From this theorem it will appear, that, if the ratios of the corresponding quantities of two ranks with respect to the two first, are the same in both ranks. (AB: BC::IK: KL, AB:CD:: IK: LM, &c.;). then the ratio of all the quantities to the first, will also be the same in the one rank, as in the other. For, by adding DE and MN to the last conse-. quents (BD, KM) there will be had* AB: BE:: IK: KN (and so on, to any number of anon whatever.) Then (oy comp.) AB: AK::1K:IN._ When the quantities in both ranks, are of the same | kind, it will appear (by alternation) that the ratio. of the two sums, and that of every two correspond- ing terms, will be the same. The six theorems here delivered, on proportions of magnitudes in general, comprehend all that is most’ useful in that subject. — What relates to the proportions of extended magnitudes, under dif- ferent limitations, and figures, as far as regards right-lines and surfaces, will be the subject of the remaining part of this book. Note, whenever you Lapis with several ratios connected bi y hs sign :: (like theses A:B:: @ ap Di Vek Oa CLs § a the concluston to be drawn, ts always from the first and last of the two ie | rattos. THEOREM VIL. Triangles (acD, BCD,) and also parallelograms (ADCQ, BDcP,) having the same altitude, are to one another in the same ratio as their bases (AD, BD). BOOK THE FOURTH. 73 ' Let the base AD be to the base D in the ratio of any one num- P___© Q er m(3) to any other number * | (2,) or which is the same‘, let D contain m (3) such equal arts whereof BD contains the umber 2 (2.) Then, the trian- les ACp, pCq, BCr, &c. made B y drawing lines from the points rPDGPA f division to the vertex C, being all equal*; the tri-¢ Cor 2. agle ACD will be to the triangle BCD, as the num- ‘?.?. er of equal parts in the former, to the number of yual parts in the latter, or as the number of parts 1 AD to the number of parts in BD, that is, as AD » BD», whence, also, the parallelograms ADCQ, DCP, being the doubles of their respective triangles %, re likewise in the same ratio as their bases AD and , Dé ° 4,4, ¢ Ax. 8. 4. SCHOLIUM. If the bases AD and BD are incommensurable to ach other, the ratio of the triangles cannot be other van that of their bases. _ For, if possible, let the triangle C iCD be to the triangle ACD, ot as BD to AD, but as some ther line ED, greater than BD, ‘to AD. _ Let AN be a part, or measure ff AD, BE‘, and let DF be E yd Lem. 1. vat multiple of AN which least FB D Noagrs sceeds DB; also let CE and CF be drawn. It is ianifest, that the point F falls between B and EK, be- ause (by Hyp.) BF a AN, and AN aBE. More- ver, the ratio of FCD to ACD is the same as that of 'D to AD (by the precedent.) But the ratio of BCD » ACD (or of ED to ADS) isc the ratio of FD to/ Hyp. Ds, or of FCD to ACD; and consequently BCD pace "CD"; which is impossible‘; by the same argument, he 6 ' will appear, that the triangle BCD cannot be to the of 4. iangle ACD, as a line, less than BD is to AD. "Ax, 2. herefore BCD: ACD:: BD: AD. ie 74 & Cor. 2. to 2. 2. 1], 4, — eo _ ELEMENTS OF GEOMETRY. If this Scholium should appear difficult to the learner, 1t may not be amiss to omit wt entirely; since it is only put down for the sake of ‘those who may be scrupulous about the business of incommensurables , to whom tt may not be improper to observe, that no- thing more ts taken for granted here, than what i effected by means of the first Lemma in the 8th book; which, being demonstrated from axioms, and one sim gle theorem in the jirst book, is referred to, here, though not given till hereafter, for reasons already hinted at, in this note. THEOREM VIII. Triangles (ABC, DEF) standing upon equal bases (AB, DE) are to one another, in the same rate as their altitudes (cu, FI.) Let BP be perpendicular to AB, and equal to CH in which let there be taken BQ= FI, and let AP and AQ be drawn. The triangle ABP=ABC*, and ABQ= DEF® but ABP (ABC): ABQ (DEF)::’BP (HC): B@ (FI,) || THEOREM IX. : i’ Tf, parallel to the bases of any two parallelo’ grams (AC, EG,) two lines (ra, MN) be drawn, so as to cut the sides proportionally (AP: AD BOOK FHE FOURTH. ag P tem: EH,) then will those parallelograms | ha ther corresponding parts (AQ, EN) be also proportionals. | Por, AQ: AGw C APeAD*:: EM: 7. Gry EEE ENE GH ee Q He Ind therefore, by ~M N "2: alternation, AQ : BN::AC ;HG?: AL 1 ate THEOREM X. If four lines are proportional (AB:cCD:: DE: BF,) the rectangle (ar) under the two ex- tremes, will be equal to the rectangle (cr) under the two means. And, if the rectangles under the extremes and means of four given lines (AB, CD, DE, BF) be equal, then are those four lines proportional. In DE let DG be taken=BF, and let GH, parallel o DC, be drawn. } 1. Hyp. AF: CG se ODP DE IVE Lp yagi SP or: DE: DG": Hu >E: CG?; therefore, ae consequents of the irst and last of these A 8 Sogn i D qual ratios being the same quantity CG, the two an- acedents AF and CE must be equal ’*. * Ax. 3. } 2 Hyp. AB: CD: ag CG?:: CE: CG':: DE: ¢ax,2. 4, IG? (BE.) SCHOLIUM. | From the same demonstration and scheme, it will | ppear, that the two antecedents of four proportional wnes (AB, CD, DE, BF) are in the same ratio to ‘ 76 aks 6 Ax, 4. 4. © Cor. 2. of 6, 2. ELEMENTS OF GEOMETRY. each other, as the two consequents: for, if in DC there be taken DP=BF, and PQ be drawn parallel to DE; then AB : DE:: AF: PE‘:: CE: PE:: CD: PD (PF.) | THEOREM Xl. The rectangles under the corresponding lines, of two ranks of proportionals, are themselves proportionals. (I say, if AB:BC::CD: DE, and BF: BG::DH: DI, then will he rect, AF - recl. BM :: rect. CH: rect. DQ.) For, in BG and DI (produced if necessary ) let there be taken BF =BF, DH=DH, and let FP be paralle| i ¢ Q ‘eo D Dek to BC, ie ue to tes then AF: BP:: AB: BC’: CD: DE:::CH: DN*; whence (alternately) AF. CH::BP: DN, and so likewise is BM to DQ*% whence (again by alternation) AF : BM:: CH: DQ. COROLLARY I. Hence, the squares of four proportional lines, are themselves proportional. COROLLARY II. Hence also, the sides of four proportional squares (AB?, BC2, CD*, DE’) will be proportional. For. let the line RS be taken such, that AB: BC:: CD) RS ; then, since AB*?: BC?:: CD?: RS? (by Coral i.) and AB?: BC?:: CD?: DE? (by supposition, thence will’ RS? —DE?; therefore‘ RS= DE, ang consequently AB: BC: : CD: DE4 ad Ax. 1.4, BOOK THE FOURTH. 79 THEOREM XIL. 1 line (Bx) drawn parallel to one side (cD) of a triangle (Acp) divides the other two sedes pro- portionally. (I say, AB: AC:: AE: AD, AB :AE!ED, and AC: BC:: AD: ED.) | Let AB be to AC, as any one number 7 (3) is to any other number 2 (5), or, which is the same, let AB hontain m (3) such equal parts whereof AC contains 2(5.) Then, if from the voints of division, lines be FY jrawn parallel to the side CD, they will also divide © AE and AD into the like number of equal parts‘, Cor. 1. therefore AE is to AD, as the number of equal parts ‘27! in AE to the number of equal parts in AD, or as the aumber of equal parts in AB to the number of equal parts of AC, that is, as AB to ACY. In the same/Ax.8. manner, AK is to ED, as the number of parts in AK to the number of parts in ED, or as the number of partsin AB to the numberof parts in BC, that is, as AB to BC. Also, in the same manner, AC: BC:: AD: ED. The same otherwise. Draw CE andDB. Then will the triangles BEC ‘and EBD be equal to each other’; whence, by .add-/ Cor. 1. ing BEA to both, AEC will be also=AB Dé. wae pei « Ax.4. 1. vty _ But, AB: AC: : triang. AnD : : triangle AEC +* (ABD) : > AD*; and AB : BC : eco AEB : triangle CEB? (DEB) : “AE : ED. WTS 4 B 78 a ro and 5. 4, * Hyp. “Cor. of 12. 4. m Cor. 2. to 27,1. ELEMENTS OF GEOMETRY. COROLLARY. Hence a right line, which divides two sides of a triangle proportionally, as parallel to the remain- ing side: because AD is divided in the same ratio with AC, when BE is parallel to CD; but not else +, SCHOLIUM. From this last theorem, whatever relates to the composition and division of ratios, when these re- spect the comparison of right-lines, will appear ex~- ceedingly obvious. For, let AB, B BG. AD, and @"S; 7, at ae CO DE, be. pro- EN RA portionals: and “A from any point D A, let. two in- definite right- lines AP, AQ be drawn; - in A é D roe which take AB=AB, BC=BC, AD=AD, and DE =DE; also take Be= BC, De=DE, and let BD, CE, and ce be drawn. Since AB : BC :: AD: DE*, thence is EC pa- rallel to DB’; and so, Be being=BC, and De = DE, ec will also be parallel to DB™. Therefore, AC (AB+BC): AB:: AE (AD+DE): AD; AC (AB+BC): BC :: AE (AD+DE): DE; Ac (AB—BC): AB:: Ae (AD—DE): AD; Ac (AB—BC): BC: : Ae (AD—DE): DE; And, AC (AB+BC): Ac (AB—BC): : AE (AD +DE) : Ae (AD—DE). | P THEOREM XIII. The parts (DE, FG) of the two sides of a triangle, entercepted by right-lines (DF, EG) drawn parallel to the base (Bc), are in the same ratio with the wholes. (DE: FG:: AB: AC). BOOK THE FOURTH. For, DF, and EG being A arallel to each other, thence wil DE: AE: FG: AG°; 012. 4, gerefore, by (alternation) D E DE: FG :: AE: AG?, P2, 4, m the same manner, AF : iG:: AB: AC® Conse- FF G juently> DE : FG ':’AB : iC 4, B C | COROLLARY. Tence, if ever so many lines be drawn parallel to the _ base, cutting the sides of a triangle, every two ‘ corresponding segments will have the same ratio4. THEOREM XIV. n triangles (asc, abc) mutually equiangular, _ the corresponding sides (AB, ab, AC, ac) contaming the equal angles (a, a) are pro- | portional. _In AB and AC (produced if necessary) take AD = ‘b, and AE=ace, and join D, E. | C Ba b A D | The triangles abc and ADE having ad=AD, ac= eo .E, and the angle a=A, have also the angle ADE’ ’ Ax. 10. =abe=ABC*; whence DE will be parallel to BC; Hyp. jad therefore AB : AD (ab): : AC: AE“(ac). Cor. to The same otherwise. | “12.4, Because AB x ac=ab x AC”, therefore is AB : ab Rate i AC: ac*. sins ! COROLLARY. Dy | 8.1. of 4 [ence equiangular triangles are similar to eachother’. » Def.14. 80 ELEMENTS OF GEOMETRY. THEOREM XV. If two triangles (asc, abe) have one angl (aBc) ix the one, equal to.one angle (bac) in th other, and the sides (AB, ab, Ac, ac) about thos angles proportional; then are the triangl equiangular. | In AB and AC, take AD=ab, and AE=ac, an let DE be drawn. D Bad > Hyp. Since AB : ab (AD) :: AC : ac (AE®), therefor ‘Cor.to is DE parallel to BC*; whence the angle B= ADE 12.4. —b)* and the angle C=AED=c. | me THEOREM XVI. If two triangles (asc, bc) have one angle (A in the one, equal to one angle (a) zn the other and the sides (AB, ab, cB, cb) about ether o the other angles proportional ; then will th triangles be equangular, provided these las angles (B, b) be, either, both less, or bot greater, than right-angles. | In AB, let AD be taken=ab, and let DE be draw parallel to BC, meeting AC in E. C A D Be 6 @ Then will the triangles ABC, and ADE, b BOOK THE FOURTH. 81 quiangular/; therefore, CB : ED :::AB :)ADe¢/Cor.1. : AB :.ab*:: CB: cbhi;.and consequently ED= SamagE b* : whence the triangles abc and ADE (having abn ay. 44, =AD, ch=ED, and a= A) will be equal in all re- carp: oects', provided the angles abc and ABC (=ADE), are re either both less or both greater than right-angles, ° ‘herefore, since the latter of these equal triangles . abc, ADE) is equiangular to ABC, the proposition is ianifest. THEOREM XVII. f two triangles (asc, abe) have all: their sides, respectively proportional (ACM aM : AB::ab::¢B: cb) then are those triangles equeangular. |In AC and AB, take AE=ac, and AD=ab, and jin KE, D. LOS i ON. mom D Ba b Since AC: AE (ac):: AB: AD (ab)", the triangles * Hyp. 4BC, ADE are equiangular™; hence CB : EJ) - 215.4, JB: AD? (ab): : CB: cb™; ‘and consequently °14. 4, ID=cbe: therefore the triangles abe, ADE, being pax. 4.4, Hutually equilateral, they must also be mutually equi- ‘gular’; and consequently abc, as well as ADE, «14.1. éuiangular to ABC. | THEOREM XVIII. « right-line (cv) bisecting any angle (acB) of a triangle (axBc) divides the opposite side (AB) nto two segments (Ap, BD) having the same ratio with the sides (AC, cB) containing. that angle, 7 G 82 ELEMENTS OF GEOMETRY. Let AE.and BF be perpen- dicular to CDE. Then the triangles ACE, CBF, and ADE, BDF being, re- "Hyp and spectively, equiangular’, it nd 2, &, 2239 Ast: Wis THEOREM XIX. si44, Willbe AD: BD :: AE: BFs 1g tp bealde ye) a F D E A perpendicular (ev) let fall from the righi angle (c). upon the hypothenuse (Av) of | right-angled triangle (ABc) will be a mea proportional between the two segments (AD, BD of the hypothenuse ; and each of the sed containing the right-angle, will be a mea proportional between its adjacent segment, an the whole hypothenuse. For since the angle BDC Cc tAx.7. is=BCA‘, and B common, the triangles BDC, BCA, are “ Cor. 1.to equiangular”: after the same 10.1. manner ADC and ABC ap- ,& pear to be equiangular. DOT aD “13, 3. 710, 4. Because the angle in a semi-circle is a right-angle Therefore, by Theor. XIV. BD: CD :: CD.» AD AB: BG: : BC: BD AB : AGC :: AC: AD COROLLARY. it follows, that, if from any point C, 2 | periphery of a semi-circle ACB, a perpendicul CD be let fall upon the diameter, AB, and from i same point C, to the extremities of that diameti two chords CA, CB be drawn; the square of th perpendicular will we equal to a rectangle una the two segments of the diameter; and the squé of each chord, equal to a rectangle under the wh diameter and its adjacent segment: for because) .. the above proportions, we have CD?=BD x A) BC?=AB x BD, and AC*=zABx AD. BOOK THE FOURTH. THEOREM XX. Tf, in similar triangles (axzc, EFG) from any two equal angles (ACB, EGF) to the opposite sides, _ two right-lines (cD, Gu) be drawn making | equal angles with the homologous sides (cB, GP); those right-lines will have the same _ ratto as the stdes (AB, EF) on which they fall, _ and will also divide those sides proportionally. C G ) VANS fee | BE : A D H - | For, the triangles ADC, EHG, and BDC, FHG | ‘as well as the wholes ABC, EFG) being equiangular ”. ’ Hyp. Thence is* AB: EE. (:: AC: EG):: CD: GH; 4:4. 83 2 and Ax. and AD: EH (::DC:HG):: BD: FH. 7. THEOREM XXI. tf, m two triangles (anc, ABD) having one side _ (AB) common to both, from any point u in that side, two lines (ur, uG). respectively, pa- raltel to two contiguous sides (Bc, BD) be drawn to terminate in the two remaining sides (Ac, AD); those lines (uF, 1G) will have the same ratio as the sides (Bc, BD) to which they are parallel. ; For, AB : AH :: BC HF’, and AB: AH:: ’D : HG; therefore, ly equality, BC : HF \: BD : HG; whence, ilternately, BC : BD: : IF, HG?. = C «14,4, COROLLARY. fence, if BC=BD, then also will HEF=HG. ae G 2 r, / 84 a 20. 4, ’Schol. 12. 2.4. « Hyp. 6 Schol, e ELEMENTS OF GEOMETRY. | | “THEOREM A,’ The side (nr) or (EH) of a square (EFIH) in- | scribed in a triangle (ABc) 2s to the base of © the triangle (aw) as the perpendicular (cp) 2 to the sum of the base, and perpendicular | (AB+CD). c | For EF (PD) : CP: AB: CD+. Therefore, by composition, EF : CD: : AB : AB+CD*, and con- sequently EF : AB:: CD: AB+CD*.” a. A es as «THEOREM B.” ; Tf a right line, (an), be unequally divided wn (c), andin it, produced, there be taken (co) to(Ac), as (Bc) to(Ac—cB) ; and upon (0), as a center, at the distance of (oc), @ circle (cpp) be described, and two lines (AvP, BP), be drawn from (A, and B), to meet any where, in the pe- riphery of that circle, then will those lines (AP, Bp,) be every where in the constant ratio of (actia BOs Draw the radius OP ; then since CO: AC:: BC : AC—CB* there- fore, by composition CO: AOFe BCRTACs. whence, by alternation : CO + BCR AO. AG; therefore, by division CO: BO:: AO: CO’, or, (by Ax. 2 and 3) AIS a * R. Simson’s Apollonius, or Plane Loci, prop. 2. Book I —See also Galileo’s Discoursi e Demonstraziont Matematiche, to, Leyden 1638, p. 39....46, or the English Translation by Weston, Lond. 1730.—Ludlam’s Rudiments of Mathematics, App. to Re= aead on Euclid ; and Playfair s Geometry at the end of Book Vij —EDpITorR. BOOK THE FOURTH. |» 85 PO: BO: : AO: PO; wherefore seeing the sides about the common angle, O, are proportional, the 15.4. _ triangles BOP, POA, are equiangular °, and therefore _ the other sides are proportional, that is PO (CO) : -AO:: PB: AP“; whence by equality BP : AP + : 414.2. Pat A Cy * Q, E. D. | THEOREM XXIL. Lf, at any two points (F, G) in two lines (AB, Ac) meeting each other, two perpendiculars (FD, GD) be erected, so as to meet each other: the distance (Ab) of their concourse Jrom that of the proposed lines, will be to the distance (FG) of the two points themselves, in the ratio of one of the sard lines (ac) to a perpendicular (ce) falling from the extreme of (Ac) upon the other (AB). Ed Let FD be produced to meet AC in H.” | _ Since the angles AFD and AGD are right- ones‘, the circumference ~ of a circle will pass * Hyp. through all the four | points A, F, D, Gi: age and so the angles GFD, - GAD, standing on the same subtense GD, will oe equal’; and conse- A’ e315. | as jnently the triangles AHD, FHG equiangular’: there- f Cor. 1. to fore AD) FG: : AH + Hs :: AC: CE, 10. 1. A 6 14; 4, 86 7, and 3. of |. 414.4. k 11.4. 67.4, m 20, 4. "10. 4, Tf through any point (P) in a triangle (ABO) | ‘Hence, if AD=BD, then also will AF x CE=CF x ELEMENTS OF GEOMETRY. THEOREM XXII. three right-lines (Ax, BF, CD) be drawn, from the angular pots to cut the opposite sides, the segments (AD, BD) of any one side (AB) will be to each other, as the rectangles (A¥ | X CE, BEX CF) under the segments of the other sides taken alternately.* itin H and. G.. It is manifest, that, the Therefore AF: CF::AB: CG, CE: BE:: CH: AB*. Let GCH be parallel G C H to AB, and let AE and : | BF be produced to meet triangles FBA, FCG; | EAB, EHC; APB, : GPH, are equiangular®. A Pub. Whence AF x CE: CF x BE:: CHx AB: CGX AB*:: CH: CG::'; but CH: CG:: AD: BD"; therefore, by equality, AFxCE:CFxBE:: AD: BD. AN COROLLARY. BE, and therefore AF : CF :: BE: CE”. iF |. THEOREM XXIV. Equiangular triangles (asc, EFG) are i pro portion to one another, as. the squares (AK, em) of their homologous sides. ; * Prise Quest. Ladies’ Diary, 1735. BOOK THE FOURTH. | C \ Upon AB and ar ye 4F let fall the per- E UN vendiculars CD 4 BE E nd GH, and let 7 = he diagonals BI, "L be drawn. ] K OF M. Because ABC: ABI:: CD: AI (AB)°:: GH? :°8.4. JF (EL) :: EFG : EFL°; therefore, alternately, ” 29. 4. iBC: EFG:: ABI: EFL1:: AK: EM”. rang | to 2, 2, THEOREM XXV. reangles (ABC, DEF) having one angle (a) i the one, equal to one angle (p) in the other, are in the ratio of the rectangles (Ac X AB, DF x DE)-contained under the sides including the equal angles. | AF FE | B NS, | x p D Q 1D /Upon- AB and DE, let fall. the perpendiculars CP ed- FQ. Then ACxAB:CPXxAB:: ACH: CP# 23°7. 4, IF: FQt:: DF x DE: FQ x DE; whence, alternately, ¢ 14. 4. iCxAB: DFxDE:: CPx AB: FQxDE: : tangle ABC : triangle DEF «, * Cor. 1, 4. COROLLARY. tone, f the rectangles of the sides contain equal angles, be equal, or the sides themselves recs. srocally proportional”, the criangles will be equal.» 10, 4, The same also holds in parallelograms, being the doubles of'such trian gles. ing the 87 89 ELEMENTS OF GEOMETRY. THEOREM XXVI. a | All semalar right-lined figures (ABCDE, FGHIK) | are in proportion to one another as the squares” of their homologous sides (AB, FG.) : D A ee meee | t | £ ee a! es are Ne / \ | A B ihe ; H faideeh ye 4 Or Draw the right-lines BE, BD, GK, GI. epef.14. Because A=F, and AB: AE:: PG: FK +, ti of4. triangles BAEK, GFK are equiangular®; therefore, if ¥15.4, from AED=FKI®, there be taken AEB=FKG, the Ax.5. vemainders BED, GKI will also be equal. Where: ofl. fore, since ED: KIG: FA: KF?*):: EB: KG”, the "14.4. triangles EBD, KGI are likewise equiangular ’. In the same manner it will appear, that DBC, IGH are also | equiangular. #)| 424, 4, Therefore, because ABE: GFK ( : “BE’: GK?) > BED VGKliGs Gb 3 Sr hee “BDC : GIB, it is evident, that the sum of all the antecedents (ABCDE) is to the sum of all the consequents (FGHIK) as the first antecedent ABE is to the first + Corto consequent GFK ®, or as AB? to* FG?; which was 6.4. be demonstrated. ad * H ‘ n, THEOREM XXVIl.- If three right-lines (AB, DE, PQ) are prop | tional, the right-lined figure (aBc) upon the first, will be in proportion to the similar, and css BOOK THE FOURTH. 89 _ sumilarly described, Jigure (DEF) on the se- | cond, as the Jirst line (AaB) to the third (PQ.) | See Def. 7. [Yd | \ ec a " | A Bid wR PN For, AB : PQ :: AB? : ABxPQ* (DE): :e7.4, ABC: DEF« +4 Hyp. - | and 10 | of 4, | 26, 4. COROLLARY, . pits dence, similar right-lined Jigures, are in the dupli- _ cate ratio of their homologous sides! wed | i. imrrpeoetens? 2 Se NT as . ! THEOREM XXVIII. 'f four right-lines (AB, CD, EF, Gu) be propor- | tional, the right-lined figures described upon them ; being similar, and similarly situated, | shall also be proportional (ABI: CDK :: EM : GO.) | ean SR ) | ‘ 7 N O Ko BC DE EG H For, ABI: CDK (: : #AB?: CD?:; EF’. GH?) £26... \. EM : GOs. | ' Cor. 1. to 11. 4, +26, 4.. ™§,. 2. ” Ax. 3. of 4. ELEMENTS OF GEOMETRY. THEOREM XXIX. If upon the three sides of a right-angled triangle (apc) as many right-lned figures (cD, BE, BF) like, and alike situate, be described, that (cv) upon the hypothenuse (Ac) will be equal to both the other two (BE, BF) taken together, | Cc Er For, BE: BF :: AB’: | -'BC?; therefore (by com- position) BE+BF : BE: : AB?+ BC?(=" AC’) : AB’ -: CD: BE®*;. and con- D sequently BE+BF=CD". Sana Gt END OF BOOK THE FOURTH. «) | | “4 t i ELEMENTS OF GEOMETRY. ic OF Bhi 6 PROBLEM I. Mp OM the greater(s8) Ff two unequal lines C D \B, CD) to cut off; or ike away a part (ax) / cual to the less (cv.) A E B ‘From A asa center, with ‘radius equal to CD, let \ te circumference of a cir- (2 be described ¢, cutting ° @ Post. 3 4B in E; and the thing is done. Ax, 2 a ee a PROBLEM Ul. 3. € Post. 1, other: and, from the point, B, of their intersect h Ax, 2. + Hyp. ' Def. 11. 3. ™ Constr. and Def. 33. of 1. " Obs. on Ax. 10. Constr. P Ax. 10. 9 Post. 3. * Cor. to - pounded, take two then the angle EAB being in the semicircle’, is a rig ELEMENTS OF GEOMETRY. PROBLEM III. At a given point (A) in an mfinite reght-li (PQ) to erect a perpendicular. al In the line pro- equal. distances AC, and AD*; from the centersC’ and D&, with any equal radii greater than AC(orAD,) let two circles EMR and FNS be described; which will cut efi draw BA, ap the thing is Hohe: For, let the points R, E, and F, S, be those in wt the infinite line PQ intersects the ‘circumferences! the two circles EMR and FNS”: then AF being | FD* (or CR‘) 2 AR’; and AS DS? (or Cl c AE", the point F falls within, and the point without the circle EMR; and so the two circles (¢ each other’. If, therefore, from the point of int section, BC and BD. be drawn; then the trian} CBD being isosceles™, the angles BCD, BDC at { base will be equal"; and so, CA being = Al j @)! and CB=BD”, the angle CAB is also=: DAB? . Otherwise. > - | From any point D above the line PQ, as a center, through the given point A, let the cir- cumference of a cir- cle be described ! intersecting PQ in 4 E’; draw the diameter EDB, and also BA’; ‘thi angle, and consequently BA is perpendicular to] | which was to be done. | ‘ BOOK THE FIFTH. 93 * COROLLARY: fe the former of these constructions. it appears, that, from any two points, with two equal radii, ‘greater, each, than half the distance of those points two circles be described : those circles will cut each other.. | PROBLEM IV. a. : ‘ ‘ i ‘wor rom & given point (A) upon an infinite right- tne (pQ) to let fall a perpendicular (aB.) From the given point 4 as a center, let an arch A c acircle be described, me s as to pass below PQ : ad intersect it in M and }; from which points, wth any equal radii, pater than half MN, l¢ two other arches be o described, and from tl point of their inter- sition C, let the right-line CBA be drawn; which wil be perpendicular to PQ. |For, let AM, AN, CM and CN be drawn ; then Al being= AN*, and MC=NC», the angle A MB. is « Def. 33, =ANB*, and CMB=CNB’; and consequently AMC wee 1. =ANC¥#: whencé (as AM=AN, and MG=NG)s 8 per. tl, triangles AMC, ANC are equal in all respects*; 33.1, ad so, the angle MAB being= NAB, the angle MBA * ir Se isikewise=NBA®*. _ | | vane ¥y Ax, 4. * Ax. 10, | | oF | | | | * Post. 3. > Constr, e1.5. # Post. 3. € Post. 1. F Constr. ELEMENTS OF GEOMETRY. The same otherwise. From any point C in the line PQ, asa cen- ter, let the circumfer- ence of a circle be des- cribed through the given p_ point A‘, intersecting Se PQ in D°; and from | the center D, with a TN , “4 radius equal to the dis- “ ¥ | S tance of the points A g and D, let another circle mEn be also described, cut ting the former ADE in E; then draw ABE for th perpendicular required. | For, conceiving right-lines to be drawn from C am D, to A and K, the triangles ACE, ADE will be bot of them isosceles’; and so the demonstration is th same with that of the preceding method. Bb PROBLEM V. r To bisect, or divide inio two equal parts, an given right-lined angle (PAQ) In the lines contain- ing the given angle, take AC=AD*; and upon the centers C and D, with any equal radii, let two circles be described *, so as to intersect each other; | a | , and from the point of in- Pl tersection E draw EA, 4 and the thing 1s done. 4 For let, CD, CE, and DFE be drawn‘; then, # triangles ACD and ECD being both isosceles’, | and Def. angle ACD will be=ADC, and ECD=EDCs; a 33. of 1 g Ax. 10. b Ax, 4. ‘ consequently the whole angle ACE=the whole ang ADE*: whence (AC being=AD, and EC=ED the angle CAE is also=the angle DAES. | | | | BOOK THE FIFTH. | 95 PROBLEM VI. To besect a given right-line (aB.) From the extremes A, B, of the given line, with equal radii, describe two circles, so as to cut each other‘; and between the “Wo points of intersection jraw CD, cutting AB in 5 and the thing is done. D | For, if AC, AD, BC and BD be drawn, the tri- hngles ACB, ADB being isosceles*, thence is Be eet ingle CAB=CBA4, and DAB=DBA!?; and conse- et ce juently CAD=CBD™: whence the triangles ACD: Obs. on ind BCD are equal in all respects"; and so the angle Ax. 10. ACE being=BCE, AC=BC, and CE common,” 4* * ‘Post. 3. and Cor. to 3, of 5, : 2™ Ax, 10, hence is AE also= BE», ; COROLLARY. dence, it is manifest, that CD not only bisects AB, but is also perpedicular to it”, ” Ax. 10. | and Def. 8. of I, | PROBLEM VI. "rom a given point (A) in a given right-line (AQ) to draw a hie (AK) whech shall make | with the former an angle, equal to an angle | given (HBG,) In BG and AQ take two qual distances BC, AD°: and 2.5 t © and D erect the two per- endiculars CI and DF to BG nd AQ’; and in DF take DE 3.5, qual to the part CD of the for- er, intercepted by the lines ontaining the given angle HBG?; hs nen through E draw AEK*, ” Post. 3. nd the thing is done: "Ax. 10, ¢ Post. 3. ®14, 1, -AD, BF, let two arcs of‘cir- ELEMENTS OF GEOMETRY. SCHOLIUM. Having, in the seven preceding problems, effected and demonstrated, by means of the axioms only, what-_ ever was assumed in the fourth postulate, as barely possible: we are now authorized, by the most rigid. laws of geometrical reasoning, to. make use of any theorem or conclusion, whatsoever, derived in the preceding books, in virtue of those assumptions, by which the process and result can be rendered the most obvious and eligible-—Accordingly, by having re- course here to Theorem XIV, of the first book, we shall be able to arrive at a construction of the last problem, better adapted to practice than that above” laid down. From the centers A and B, at any equal distances cles, FC, DR, be described ‘, intersecting the given lines in D, F, and C; also from D, with a radius equal to the distance of the points F, C, let another circular arch pEr be described, cutting the former DR, in E; then draw AEK, and the thing is done. For, conceiving right-lines A LQ. to be drawn from F to C, and from D to E, the tri- angles BFC and ADE will be equilateral to each other, by construction, and therefore equiangular also*. a PROBLEM VII. ~ i To describe a triangle, whose three sides shall be equal to three given lines (A, B, C ;) provided any two of them, taken together be greater than the third. 7 | | BOOK THE FIFTH. 97 Arm | ee eG K L Make FG=B”, and from the centers F and G, with ~2. 5, the intervals, or distances A and C, let two circles DKL, HKL be described”; which will cut each = post, 3. other; and if, from the point of intersection K, the ines KF and KG be drawn, then will FKG be the riangle required. _ For, the distance FG of the two centers, is less than he sum of the radii A and C*; and greater than their 4x. 6.1. |” lifference (because B+A being CC *: thence is BO: pyp. 2—A¥;) therefore the two circles cut each other“: 9.3. consequently FK=A, FG=B, and GK=C*, ox 33. PROBLEM 1X. Pious a given point (a,) to draw a right-line - (rs) parallel to a given right-line (eQ.) | At any point in Hie m iven line, make 1 yual to the distance of R_A 7S ye points A and B°: and 62. 8. ‘om the centers A and , with an interval rales | CB, let 2 circles be de- gS @ then through BP B C Q *Post. 3. leir intersection D, let the line RS be drawn, and ie thing is done. Let AB and AC be drawn’. It is plain, that the * Post. 1. tro circles will cut each other’, because the sum of/9.3. H 98 € Constr. k19,1, ‘Ax. 1, k 25,1. ' Post. 1. m7. 5, "8. 1. ° 3; 5: and Teo. P Post. 3. 99,3. FOO. * Constr. * Cor. to 24.1, “ Def. 26. - other?in D; from which point ELEMENTS OF GEOMETRY. their semi-diameters (=AB+BCs) is greater thar AC": therefore, if ADS and CD be also drawn, ther will AB=BC=CD=DA }, and therefore RS will by parallel to PQ *. The same otherwise. From A, to any point in PQ, draw AB’; make th angle SAB=PBA™: and then AS will be parallel t PQ”. eerie meme PROBLEM X. Upon a given line (aw) to describe a squat (ABCD.) | Make AC perpendicular, and C equal to AB’; and from the centers B and C, let two circles, with the radius AB or AC, be described ?, intersecting each draw DB and DC, and the thing ts done. Wi For, all the four sides being b * equal, ‘by construction, the figure is a parallelogram! and therefore the angle A being a right-angles, th other three will be all right-angles‘’, and ACDB square", SCHOLIUM. a By the same method a rectangle may be describe the sides being gwen. PROBLEM XI. e | To divide a given line (azn) into any propos number of equal parts. 7 BOOK THE FIFTH. 96 _ From the extremes of the given line AB, lraw two _ indefinite ines AP, BQ parallel o each other’; in each f these lines let there 'e taken as many equal listances AM, MN, YO, OC; Bo, on, nm, va (of any length at leasure) as you would ave AB divided into” 1en draw Mm, Nn, Oo, itersecting AB in EK, F, G, and the thing is done. , | For MN and man being equal and parallel", FN will * Constr. 2 parallel to EM’; ‘and in the same manner will GO 728. 1. 2 parallel to FN: therefore, AM, MN, NO, &c: sing all equalt, Ak, EF, FG, will likewise be equal, toe, ' 0 ode The same otherwise. (In any right-line JP, drawn from A, tke as many equal (stances (AM MN, 10) wanting one, as yu would have AB qvided into; then, ving drawn the in- ¢finite line OBQ, in itake an equal num- Q tr of parts or distances OB, BC, CD, each of the laigth of OB, and let DN be drawn, cutting AB in (; make GF, FE, each equal to BG, and the thing 2 done, ; For, if AD and BN be drawn, they will be parallel, «Cor. to (ecause OA: ON:: OD: OB?;) and so, the trian- psi gs BNG, ADG, being equiangular ‘, it will be BG : <3 arq" fz::BN:AD?4:: ON: OA“ Therefore BG is 7.1. t2 same part of AG, as ON is of OA. | Teer hat } ‘ H 2 100 bY2, 4. ™ 12, 4. ELEMENTS OF GEOMETRY. PROBLEM XII. To two given lines (ap, BC) to find a third pro- por ‘ronal. | From any point A, draw two inde- finite lines AP, AQ, in which take Ad —- AB Ac = BC and 6D = BC*; draw bc, and paral- lel to 6c, draw DK cutting AQ in Bs 4 then cE will be the third proportional required : For Ab (AB) Ac (BC) :: 8D (BC): ck*. PROBLEM XIII. To three given lines (AB, ac, BD) to Jing fourth proportional. Having drawn AP and AQ, as in the pre- ceding problem, take therein Ab=AB, Ac =AC,andbD=BDi; draw bc, and parallel toit,draw DE, inter- secting AQ in E; then is cE the fourth proportional required. _ | For, Ab (AB): Ac (AC). -6D (BR); cE.™. ai PROBLEM XIV... 9) Between two given lines (AB, BC) to find a met proportional. BOOK THE FIFTH. 101 | In the indefinite line AP, take AO= AB, and A 228 eR 09C=BC"; bisect AC mt oo E°, and from the B Brin renter E, at the dis- af lance of EKA, or EC, et a semi-circle ADC Senet ve described”; erect p »D,. perpendicular to AC 1, cutting the cir- C E is Fe Bits umference in D; then will 6D be the mean propor- ional required. _ For, Ab (AB) : BD : : 6D : BC (BC)’. r19. 4. | and it’s Corol. PROBLEM XV. ‘0 divide a given line (Ac) into two parts (aB, BC) having the same proportion as two given | lines (AM, AN). | From A draw AD, making passat alts Sali M ay angle with AB; in which ike Am=AM, and mn= M N IN‘; draw nC, and mB pa- A eC £2. Se illel thereto’, meeting AC in ‘9. 5. Lo enen wit AD : Bos ¢ im (AM) : mn* (MN). "12. 4, 7hich was to be done. PROBLEM XVI. ‘9 add a line (wc) to a line given (ap) so that the whole compounded line (ac) shall be in proportion to the part added, as one given tine (AN) is to another (mn). 192 agi 2 * 9.95. 12.4 214.5. 76,5, 69.5, * Hyp. @ Cor. to 19. 4. « Hyp. and 10, 4, ELEMENTS OF GEOMETRY. From A draw AD, making any angle with BA; in which take An=ANY”, and 2m= NM; draw mB and 7C pa- rallel thereto*, meeting AB - produced, in C; then will AC: BCO:: An( AN): mn# (MN). Which was to be done PROBLEM XVII. oS To divide a given line (AaB) into two such part. (ac, BC) that the rectangle contained unde them, shall be equal to the rectangle unde two given lines (pM, MN); provided that th given rectangle is not greater than the squar of half the line (as) to be divided. E 1 ita Q A.C 0: G Bip N Between PM and MN take a mean-proportion MQ*; make BD perpendicular to AB, and equal MQ;; bisect AB in O%, from which, as a center, let semi-circle be described; and draw DE parallel 1 BA?®, which (because BD is less than the radius‘) w, meet the circle in some point EK; from which, upc AB let fall the perpendicular EC: so shall ACxB =" C*= DB?(=QM*)=PMxMN*’ Which wi | to be done. PROBLEM XVIII. 7 To a given line (AB) to add another line (Be such that the rectangle, under the: who BOOK THE FIFTH. 103 compounded line (ac) and the part added, shall be equal to a rectangle under two given lines (PM, MN). Between PM and MN ake a mean-proportional 4S’; make BD perpendi- 14.5, ular to AB’, and equal had o RS*; bisect AB in O;, ages raw OD, and take OC ida =OD": so shall ACx A O BMGLG 3;C=PMxMN, as was TE A BR TS, 0 be done. Ries ' For, if through A and B, from the center O, the lircumference of a circle be described, cutting DO in ), and KE, C be joined ; then, the triangles OC kK, ODB having OK=OB, OC=OD*, and the angle EOB Const. ommon) will be equal in all respects’; and so, EC apt eing a” tangent to the circle in KE, we have" AC x 76.3 }C—CEH’*=°BD*?=RS*= PM. x MN». noise ° 1.2, P 18 4, PROBLEM XIX. "o divide a given line (Bu) into two such parts, _ that the square of the one (Bc) shall be equal | to the rectangle under the other (cu) and a _ second given line (AB). | Taking BA in the same ‘raight line with BH, be- ween them let a mean roportional BD be found‘; & 714.5, ‘isectAB in O"; Draw OD, 6.5. od make OC=OD;3 so aallBC?-=CH x AB,aswas A O Bio js » be done. : _ For by the demonstration of the precedent, AC x iC (=CE*=BD’*) =AB x BH;; from each of which aking away ABxBC, there remains BC’=AB kCHs, +5. 2. and Ax. 5, I. 104 * Post. 3. * 923, * Def, 33. of I, ELEMENTS OF GEOMETRY. COROLLARY. If AB=BH, then will BC*=BH x CH; in which case the line BH is said to be divided according to extreme and mean-proportion. | i PROBLEM XX. | In a given circle, to apply y, or imscribe a line (AB) equal to a given line (cE,) less than the diameter of the cirele. SE Es ES, oy E From any point A in the, B | circumference, with the ra- dius CE, let a circular arch mn be described *, cutting the given circle in B“; then A draw AB, and the ‘thing as done”. Boh TOE ee ot PROBLEM XxXI. To draw a tangent to a given circle (c) rough a given point (a). B A Diane EK D “| \ a ay ail BOOK THE FIFTH. 105 CASE I. Lf the given point be in the circumference; 1en, to the center C, draw AC, and perpendicular to .C¥ draw BAD; which will touch the circle at A*. pas | CASE II. Jf the point A be without the circumfer- vce; then drawn AC, which bisect in P®; and from °*6.5. ie center P, at the distance of AP, or CP, leta semi- ircle AEC be deseribed ‘*, cutting the given circle in ¢ Post. 3, ¢; then draw AED, which will be the tangent 29.3. :quired ©; because (CE being drawn) AEC is a right- <6. 3. igle/. £13. 3. ; | PROBLEM XXII. [pon a given line (PQ) to describe a segment of a circle (pzQ) to contain an angle (x) equal to a given angle (BAC). Make AD perpendi- qlar to AB; also e3.5 iake PQO, and QPO, «ech, equal to DAC# h7.5 (he difference between te given angle and a izht one); then upon te point of intersection : (, asa center, at the Xe cstance of OP (or OQ) 1; a circle be described ; and the thing is done. For the angle H=right-angle+QPO ‘= DAB+ ‘16,3. JAC*= BAC. k Constr. SCHOLIUM. In the same manner the problem may be con- svucted, when the given angle is acute; only the lies PO, QO must then be drawn on the other side 0 (PQ) as is manifest from the 16th theorem of the 31 book. : 106 kG. 5, ! Proof of ISic: mo. 5. °4, 5. AB and AC, be bisect- ELEMENTS OF GEOMETRY. PROBLEM XXIII. | Let any two sides, ) About a given triangle (anc) to describe a cirele ed by two perpendicu- jars DF and EEF*; A | which will intersect each other in the center (F) of the required circle’; from whence the circle may be de- scribed. | j SCHOLIUM. By the same method, the circumference of a cirel may be described through any three given points, nm situated in the same right-line: | Li Also from hence the center of a circle may be founs. by having the segment of a circle given. | io i PROBLEM XXIV. | To inscribe a circle in a given triangle (asc). Bisect any two of the angles, A and B, by the lines AD and BD”, meeting each other in D; make DE perpendicular to AB’; then, if from the center D, at . te distance of \ DE, a circle be A ty | described, it will touch all the sides of the triangle. | BOOK THE FIFTH. 107 For, let DG and DF be perpendicular to AC and C?’; then the triangles ADE, ADG, having two ’*->- agles equal, each to each (by construction) and AD ommon, will not only be equiangular %, but also have * nora )E=DG". By the same argument DE=DF; ee 1. ih ,erefore the circumference of the circle also passes jrough G and F*; but it touches the sides of the’ Heke 385 iangle in those points’, because G and F are right- 16,3," agles “. “ Constr. fo PROBLEM. XXY. in a given circle (ars) to describe a triangle, equiangular to a given triangle (PQR). From the center C, m caw the radii CA, Y ‘EK, CB, making the BR agles ACE and BCE C qual, each, to the agle R”; join A, B, A B ad make the angle p Q JSBF=Q”. and from x ie point F, where BF cuts the circle, draw FA; so all AFB be the triangle required. For, ABF=Q?, F (=ACE"%=R*;3 and conse- = Constr. AY Sat gently BAF =P *. y ar Mie MeN i Se? © Lk tw Constr, | PROBLEM XXVI. * Cor Ibout a given circle (0,) to describe a triangle, | equangular to a given triangle (ac). QO? i eg B ry 108 | ELEMENTS OF GEOMETRY. Produce out the side AB both ways; and dray : the radii OP, OR, OQ, so as to make the angle PO} 7,5, = EBC, and POQ=DAC*; then draw three right $21.5. lines to touch. the circle in the points P, Q and. a and the thing 7s done. For, if PQ be drawn, the angles SQP and SPQ, NA Ax ?.1. be less than the two right-angles SQO and SPO« ¢Cor.2, and so PS and QS, not being parallels’, they wil to7.1. meet each other’; therefore, as the like may be in and ~— ferred with regard to PT and RT, é&c. it is manifes) 5.1. that the three tangents form a triangle STH. Now Cor.20 POR+T being =two right-angles © = ABC+EBC. toll. 1. and POR=EBC8; thence will T=ABC: and, yy th same argument, S=BAC; whence alsoH=C. PROBLEM XXVIL. 4 In a circle given (ABcpd) to wnscribe a square, € Constr. Draw two diameters AC and BD perpendicular ‘| 3.5, each other"; then draw AB, BC, CD and DA; 8 shall ABCD be a square inscribed in the circle. For, the angles AOB, P| BOC, DOC and DOA (as well as the sides OA, OB, OC, OD, containing *Ax.7.1. them) being equal‘, the and Def. opposite sides AB, BC, ; CD, DA will likewise be *Ax.10. equal*: and the angles ofl. ABC,BCD,CDA, DAB, are all of them right- 13.3, angles’, and therefore are equal. SCHOLIUM. ‘=| If two other diameters ac, bd be drawn (by pr 5.) to bisect the angles AOB, BOC, a regular octagor Aa Bb Ce Dd may be inscribed in the circle. And if all the angles at'the center O, he again bisected, a regular polygon of sixteen sides, may in like mannet be determined ; and so on, at pleasure. BOOK THE FIFTH, 109 | | PROBLEM XXVIII. h a cercle given (ABGE) to inscribe a regular pentagon ‘At the center O, upon B fe diameter FG, erect te perpendicular OB ™, wk leeting the ¢ircumfer- nn acein B; divide OG in C l (6y prob. 19.) so that Fe G (H?=GH x OG; that i take OR=:0F, and JH=dist. RB: Then ‘aw BH; which will. ~ 1: equal to the side of x te pentagon”; from whence the figure itself may be “28. 3. escribed. and 8, 2. = SCHOLIUM. Hence a regular decagon may be inscribed in the ircle; the side being =OH °. 28,3. | PROBLEM XXIX. a circle given, to inscribe a regular hexagon _(ABCDEF.) From the extremes of B C ay diameter AD, apply 1B, AF, DC, and DE (ual, each, to the radius .O?; then join B,C,and 4 1, F; and the thing is one. | For, if the radii OB, .C, OE, OF be drawn; ye triangles AOB and ‘OC, being equilateral’, will also have. the angle ¢ Constr. /AB=DOC'; whence AB is parallel (as well as*14. 1. qual) to OC*; and consequently BC and AO are:Cor. to | 8. 1. 110 +26. * Ax, 4,1, € 3. 5. “ Proof of 96. he “ Hyp. ELEMENTS OF GEOMETRY. likewise equal and parallel‘: Therefore, seeing the triangles AOB, BOC, COD, &c. are equilateral, and alike in all respects; not only the sides, but also. the angles ABC, BCD, &c. of the hexagon, will be | among themselves". bi COROLLARY. Hence it appears, that the side of a regular hexagon, inscribed in a circle, is egual to the semi-diameter, or radius. ie SCHOLIUM. Besides the figures constructed in the preceding problems, and those arising from thence by continual bisections, or taking the differences, no other regular polygon can be described, from any known method, purely geometrical, that is, by means of right-lines and circles only. "4 PROBLEM XXX. ci About a given circle to describe a regular poly. gon, of the same number of sides with a regular polygon (ABCDEF) mscribed in the circle. From the center O, to the angles of the in- scribed polygon, draw OA, OB, OC, &c. and perpendicular thereto draw PAQ, QBR, RCS‘, &c. intersecting” inky Q, Ro “SVT AN so shall PQRSTV be the polygon that was to be described. . al For, by taking away glia D : | the equal” angles OAF, OAB, OBA, OBC, &c. from the equal (right) angles OAP, OAQ, OBQ, OBR, &e. the remainders FAP, BAQ, ABQ, CBR, &c. will BOOK THE FIFTH. . ill jso appear to be equal*: therefore the triangles FAP, * Ax. 5. 1. 4BQ, BRC, &c. (having also FA=”"AB=BC, &c.) ge equal in all respects’; and so the angles P, Q, R, 115.1. (ce. as well as the sides PQ, QR, RS, &c. are equal znong themselves *. *Ax.4. 1, PROBLEM XXXI. sny two circles (ACE, ace) being given, to de- _seribe a polygon wm, or about the one (ace) that shall be similar to any polygon described \2n, or about the other (AcE.) First, having drawn i e radii OA, OB, &. = B C6 Le { the angles of the ¢ven inscribed polygon JBCDEF; make, at A te center o, the angle \ cb = AOB, boc = ¥ } . | J)C*, &c. Then, the chords ab, bc, ed, &e. being «7, 5, cawn, I say, the polygon abcdef will be similar tthe given one ABCDEF. For, the triangles AOB, 5; BOC, boc; &c. being equiangular*, the angles Cor. 1. 4BC, abe must also be equal’; and¢ AB: ab(:: to 10.1. (B:0b)::BC:bc. Inthe same manner, the other Seer, tresponding angles are equal, and the sides contain- 414.4. iz them proportional: Therefore the two polygons =="E a2 similar °. ¢ Def, 14. Again, having 4. dawn the radii OA, / x % : 3, OC, &c. to the A Lae of contact of Se t> given circum- P \ s:ibing polygon yp FRSTU; draw © l=, lewise the radiioa, © 0, oc, &c. making the angle aa&b=AOB, boc=BOCY,/7.5. «>. perpendicular to whichs draw pq, gr, rs, &c. so shall ¢3. 5, » t2 polygon pgrstv be similar to the given one IRSTU. For, the angles OAQ, OBQ, cag, odg, 112 RAx. 7.1. 2 Constr. K1}.1. and Ax. hal £16, 1. ™ 14, 4, ® Def, 14. 4. °14.4, and Cor. to 11. 4, P26, 4, ELEMENTS OF GEOMETRY. being all equal”, and AOB also=aob'; the remainn angles AQB, agb of the two quadilaterals AOBt aobg must be equal*; as must likewise their haly OQB, ogd (for the right-angled triangles OAQ, OB having OA=OB, and OQ common, have also OQ —OQB!.) In the same manner is ORB=o7b, & whence, the triangles POQ, pog: QOR, gor, & being equiangular, it follows” that PQ: pg (:: OG og):: QR : gr : And so of the rest. Therefo PQRSTU and pqrstv are similar ”. | COROLLARY. It appears from hence, that the similar inscribed p lygons, as well as the circumscribing ones, are | proportion, as the squares of the radii of their re pective circles. For, in the former case AO*: a ::° AB?: ab?::? ABCDEEF : abcdef; and, in tl latter,° AO? + ao* : : PO*® : po* :':. PQ’: pom PQRSTU : parstv. THE END OF THE FIFTH BOOK. | 7 — _ i . : <-> ae ae eee . ‘ ELEMENTS OF GEOMETRY. BOOK VI. b | till PROBLEM If. 4 make a square equal to a given rectangle (ABCD.) ly one side AB of the E rctangle, produced, take Et=the other side BC; oect AE in O*; and from “6.5. I: center O, at the distance E OA, or OF, let a semi- A sicle AFE be described; ul let CB be produced P Cc cmeet the circumference in F; then a square licribed on BF (by 10. 5.) will be equal to the given etangle ABCD, yeaa a a i rn | f- - PROBLEM IL. [ make a square equal to the sum of two given ‘quares. | | 1 114 ©3.5. a8, 2. e3.5. f Cor. to 8.2. ELEMENTS OF GEOMETRY. Let AB and BC be the sides of the two given squares. Draw two indefinite lines BP, BQ, at right-an- gles to each other’; in which take BA=BA, BC = BC, and join A,C; then asquare described on AC (by 10. 5.) will be equal to the sum of the two squares de- scribed upon AB and BC“. SCHOLIUM. In the same manner a square may be made equ to the sum of three, or more, given squares: for AB, BC, CE be taken as the sides of the given square then, by making BH=AC, BE=CE, and drawi EH, it is evident, that a square upon EH, will equal to the sum of the three squares upon AB, B and CE; or that, EH?=BH?* (AC?) + BE*=AB BC?+ CE*. PROBLEM II. To make a square equal to the difference of et given squares. 4 i iG Let AB and BC (taken in 7 the same strait line) be equal E z to the sides of the two given ee | squares, 4 Upon the center B, with the radius BA, let a circle be described, and make CE per- Be pendicular to BC *, meeting the circumferé in E: so shall a square described on CE (by 10. be equal to BE? (BA*)—/ BC’. Big BOOK THE SIXTH. 115 PROBLEM IV. 'o make a triangle equal to a given quadrila- teral (ABCD.) | Draw the diagonal AC, Iso draw DE parallel to D Cc \Cs, meeting BA produced wetay €9- 5, 1 Kk, and join CE; then N vill the triangle BCE=the iven quadrilateral ABCD. _ For, the triangles ACE, «CD, being, upon the same £ B ase AC, and between the ame parallels AC and ED, are equal"; therefore, if * Cor. 1. «BC be added to each, then also will BCE=, '?.2. iBCDi. Ax. 4. 1. PROBLEM V. ‘o make a triangle equal to a given pentagon | (ABCDE.) ‘Draw DA and DB,,. D ad also EH and CF arallel to them’*, ,eeting AB produced 1 H and F; then raw DH and DF; » shall the triangle 'HF=the pentagon © 'BCDE. Ht —-r B ‘For the triangle DHA is=DEA!’, and DFB= 'Oor.1. ‘CB’; therefore DHF (=DHA+DAB+DFB= ‘2-2. 'EA+DAB+DCB)=ABCDE”. m Ax. 4.1, | 12 | : kQ, oe 116 29, D. °6,5, P 3, 5. ELEMENTS OF GEOMETRY PROBLEM VI. Upon a given line (e¥), to make a rectangl equal to a given triangle (ABC). Through C, the RF vertex of the tri- angle, draw KN parallel to the base AB"; and _bisect AB with the per- pendicular LQ®, meeting KN in K; iv B also draw BP perpendicular to AB”, intersecting K in I; then in AB, produced, take BM= EF, ai draw MIQ ‘cutting LQ in Q; draw QO and Mt parallel to AM and LQ’, meeting each other in € then will INOP be the rectangle Tequired. i For, it is evident, that LI, IO and LO are | rectangles ?: therefore IN=BM’=EF ‘, and 10° ye ABC*, The same otherwise. From the vertex C, C upon the base AB, I let fall the perpendi- i | cular CD”; make EH perpendicular to 4 EF *, and equal toa A D i fourth proportional to 2EF, AB, and iy ¥: then | rectangle EG contained under EF and EH will ¢ equal to the triangle ABC. ! For since, by construction, 2EF : AB: CD: Fl, therefore is2EF x EH=AB x CD +, and consequet} EF x EH=1AB x CD=ABC~ BOOK THE SIXTH. SCHOLIUM. _ By either of the two preceding methods, a paralle- ‘gram having a given angle may be described upon a vven line, equal to a given triangle; if, instead of BP, MLQ, or BDC, FEH being right-angles, you jake them all equal to the angle given: the rest’of ‘e construction being the same. PROBLEM VIL. ‘pon @ given line (AB) to describe a rectangle equal to a given right-lined figure (PQRS). ‘Let the given FF ; fiure be divided aR S iyo triangles PQR, Cc ! D * IRS: and -upon ~ | ti; given line AB (7 the precedent) ‘A/ °°" BY O™ a a lca rectangle ABDC, equal to the triangle PQR, eens also upon CD make the rectangle JFE equal to the triangle PRS: go shall ABEF, much is a rectangle (because both ACK, and BDE 4» continued right lines*) be= PQR + PRS°=PQRS : wich was to be done. ca SCHOLIUM. wy £4) 117, 69. 1. CRA Nae and 4. 1. _ When the figure given has not more than five sides, - | construction will be more easy, by first finding a tngle equal to it (6y prob. 4. or 5.), and then nking a rectangle equal to that triangle. But if the igtre be a rectangle, the easiest way of all, will be to ‘ae a fourth proportional BF to the given line AB ul the two sides PQ and PS of the given ‘rectangle @ 13. 5.); which fourth proportional will be the uitude of the rectangle required. For, since Ay): PQ:: PS: BF (6y Constr.) therefore (by C4.) ABx BF=PQxPS 118 e Ax. 4. "Or Oy Ls £9; 4. ELEMENTS OF GEOMETRY. PROBLEM VIII. To describe a rectangle equal to the sum 0 difference of two given right-lined figures. Let the two given figures be ABN and PP. * By the precedent, let two rectangles AD and A respectively equal to ABN and P, be described, ¢ the same, or different sides of AB, according as t difference, or sum, of the two figures is required: th: will the rectangle CF be equal to that sum or diffe ence * COROLLARY. Hence two lines having the same ratio with two gw right-lined figures, are determined : | For AC: AK: : AD (ABN): AF’ (P). PROBLEM Ix. To make a square equal to any right-lined Fgh given (ABCDF). -. a BOOK THE SIXTH. 119 D Upon AB describe rectangle AE equal } » ABCDEF®; then Cc 7.6 iake a square BH B j qual to that rect- SA agle’, and the thing 1.6, : done. SCHOLIUM. | After the same manner (from prob. 8.) a square ‘ay be described equal to the sum, or difference, of iy two given right-lined figures. | , i PROBLEM X. ‘o describe a figure (rGuIK) equal, and similar to a given right-lined figure (ABCDE). D { G Draw AC and AD, and also FG equal to ABt;i9. 5. nike the angle GFH=BAC*, HFI=CAD, and: 7.5. (K=DAE *; likewise make FH=AC, FI=AD, 11 FK=AE!’; then draw GH, HI, and IK, and the: ax. 10. ‘ing is done. 1. For, since the triangle FGH=ABC!and FHI= VD‘, &c.; therefore is the whole polygon FGHIK, 1.0, equal to the whole polygon ABCDE”. ™ Ax. 4. 1. 120 ” Def. 14, 4. - by making the triangles FGH, FHI, &c. respectively ELEMENTS OF GEOMETRY. Moreover, these equal triangles being also equi- angular’, it is manifest, that G=B, GHI=BCD, HIK=CDE, and so on; therefore, FG being also —AB, GH=BC, HI=CD', &c. the two polygons ABCDE, FGHIK are similar to each other”. a SCHOLIUM. The figure FGHIK may be otherwise constructed, equilateral to ABC, ACD, &c. as is evident from 14. 1. and Ax. 4. | PROBLEM XI. Upon a given line (as) to describe a figure {(ABCDE) stmilar to a given right-lined figun (PQrst.) P 4X B Draw PR and PS, and in PQ (produced, if needful take Pg=AB; draw qr parallel to QR °, meeting PI inv; also draw rs and st, parallel to RS and ST intersecting PS and PT in sand ¢: then upon AB, (b the precedent), describe a polygon equal and similart Parst; and the thing is done. 3 For, since any angle BCD (qrs) of the polygo ABCDE, is equal to its correspondent QRS? and also CB (gr) : CD. (7s): : RQ: RS, therefoi jets two polygons ABCDE, PQRST, are like to ea other’. F BOOK THE SIXTH. 12); SCHOLIUM. | ‘This last problem may be otherwise constructed making the triangles ABC, ACD, ADE equi- agular to the triangles PQR, PRS, PST, re- vectively. For, then the angle BCD being=QRS, CDE= IST, &c.; and alsoBC : QR (2: AC: PR): :*Ax.4.1. (iD: RS‘, &c. the two polygons must therefore be ‘14. 4. ‘milar to each other - " pe 14, ot 4. j PROBLEM XII. 0 describe a figure similar to a right-lined SJigure gwen (parst), which shall be to it in a given ratio of one right-line to another. In PQ (produced if ne- essary) take Pn to Pg in te given ratio of the figure t be described to the figure even”; and in the same line Prd: JQ, take Pg equal to a mean- }oportional between Pz and f2*; upon which (by the * 14. 5, zecedent) let Pgrst, similar t PQRST, be described, and t2 thing is done. iFor, since Pn : Py: : Pg : PQ (by Constr.) ; ‘ y terefore Pn : PQ: : Pgrst : PQRST ». 7,4, PROBLEM XIII. - 4 describe a figure that shall be equal to one right-lined figure given (re), and similar to ‘another (ABCD). 122 His «14, 5. °11. 6, °7.4. 427.4, ¢ Ax, 4.4. ELEMENTS OF GEOMETRY. Upomw AB make the rectangle ABFG=ABCD4, and upon AG make the rectangle AGNE=P*; in AB take AI equal to a mean proportional between AB) and AK“; and upon AI let AIKH be described simi- lar, and alike situated, to ABCD, and the thing is done. . For ABCD (AF): P (AN):: AB: AE®s:) ABCD : AIKH®; and therefore AIKH=P °. * PROBLEM XIV. To describe a figure similar to a given right- | lined figure (azcn). which.shall be to another given right-lined figure (p) in a given ratio of one right-line (s) to another (R). BOOK THE SIXTH. 123 _ Make the rectangle ABFG=ABCDzs, and the rect- «7. 6. angle AGNE=P; also in AF, produced, take AQ =a fourth-proportional to R,S,and AE’; then, (by the * 13. 5. precedent), make AIKH similar to ABCD, and equal to the rectangle AG x AQ: then will AIKH (AG x | fame P. CAG XAE t2,AQ.: AB? :: S : Rt; { Contr. which was to be done. ag THE END OF THE SIXTH BOOK. ELEMENTS OF GEOMETRY. | ae re BOOK VIL. | DEFINITIONS. 1. A RIGHT-LINE is said to be perpendicular t a plane, when it is perpendicular to all right lines that can be drawn in that plane, from the point or which it insists. 2. One plane is said to be perpendicular to anothe plane, when all right-lines drawn in the one, perpen dicular to the. common section, are perpendicular tc the other. 3. Parallel-planes are those which are every where equally distant, the one from the other. 4. A Solid is that, which has length, breadth, anc thickness. f 5. Similar solids are such as are bounded by ar equal number of similar planes. oe 6. A Prism is a solid, the planes ~of whose sides are parallelograms, and whose two ends, or opposite bases, are plane, rectilineal figures, parallel to each other. * BOOK THE SEVENTH. 7. A parallelepipedon is a lid bounded by six parallelo- , rams, of which the opposite ones re parallel, equal, and like to wch other. 8. An upright prism, or parallelepipedon, is that, ' which the planes of the sides are perpendicular to te plane of the base. 9. A rectangular paral- lepipedon is that whose ounding planes are all rect- agles, and which stand at ght-angles one to another, lov. When all the bounding planes re squares, the parallelepipedon is alled a ‘cube. Il. A Pyramid is a vlid, whose base is any — ‘ght-lined plane figure, aod whose sides are tri- L ogles, having all their €ttices united in a point, bove the base, called the ertex of the pyramid. B ‘hus ABCLE represents a 1, and base BCLE. C pyramid, whose vertex is © : Cc 12. A Cylinder (DCcd), is a olid generated by the rotation f a rectangle ACDB about one fits sides AB, supposed at rest ; vhich quiescent side AB ig called ae axis of the cylinder. 125 126 ELEMENTS OF GEOMETRY. A 13. A cone(ACc) isa solid generated by the rotation of a right-angled triangle ABC about its perpendicular AB, which is called the axis of the cone. c C 14. A Sphere is a solid generated by the rotation ( a semi-circle about its diameter. 15. The Frustum of a pyramid, or cone, is th: part which remains when any part next the verte: cut off by a plane parallel to the base, is take away. 16. The altitude of a pyramid, or prism, is the pe pendicular distance of the vertex, or upper plan from the plane of the base. | 17. Every rectangular parallelepipedon is said tol contained under the three right-lines that are + length, breadth, and altitude. ¢ 18. A Plane is said to be extended (or to pass). t a right-line, when every part of the latter is Plage in, or touched by the former. AN AXIOM. Upright prisms (AabcCBA, Ddef FED)of the san altitude, standing upon bases (ABC, DEF) equal ap like to each other, are themselves equal. To see the evidence of this Axiom in the strongest light, conceive co f a right-lined plane fi- gure PQR 7to be _JSormed, equal and like in all re- B S| spects to the bases ABC, ve 7* DEF of the two prisms: BOOK THE SEVENTH. upon which, conceive the prisms to be placed, one after another, so that their bases may coincide there- with. Then, because the planes of the sides stand, in both cases, perpendicular to the plane of the base, upon the same lines PQ, QR, PR, and are carried up to the same height, tt is manifest that the bounds of the two solids, when thus placed, have the very same position, and, consequently, that the solids themselves occupying (successively) the same identical space, are ‘equal the one to the other. | | That by any two right-lines (AB, AC) meeting in ja point, 2 plane may be extended. A POSTULATE. In order better to com- ¥E prehend the sense and de- D \sign of this Postulate, let a plane BDEC, extended by the right-line joining the ‘points B and C, be con- ceived to be revolved about ‘upon that line, till it meets 8 with, or takes in, the point A; then the plane in- cluding, in that position, all the three points B, C, ‘and A, it also includes, or is extended by, the right- lines AB, AC, BC, joining those points ; which are in the same plane with their extremes (by def. 6. 1.) _ Hence tt appears, that by any three points, a plane imay be extended; and that all the three sides of any i right-lined triangle, are in the same plane. THEOREM I. The common section of two planes (AB, CD) ts a right-line. 127 128 ELEMENTS OF GEOMETRY. For, between the two ex- treme points I, F of the com- * Post. 1. mon section leta right-line EF * Def.6.1. be drawn’; then, that line Se ae being in the plane AB®, and also in the plane CD °, it must of consequence, be the common section of them both THEOREM II. If a right-line (as) be perpendicular to tw other right-lines (CE, DF) cutting each other at the common section (A,) 2é will be perpen dicular to the plane (cDEF) passing by thost two lines. t Take AC, AD, AE, AF all equal to one another; and having joined CD, DE, EF, CF, let there be drawn through A, in the plane CDEF, any right-line GH, meeting CF and DE in Gand H; and let BC, BG, BF, BD, BH, and BE be also joined. 6 Constr. Because AC’=AE=AD=AF, and CAF=DAE* 3.1. therefore is CK = DE‘, and the angle FCA (or GCA, Ax. 10.1. —DEA or (HEA); and so, GAG being likewise= e 15. 1, ae and AC= AE °, thence will AG=AH Def. 2. 7, COROLLARY I. aM -ence it will appear, that the plane AB (according | to the sense of the definition) is perpendicular to the plane ED: Fora right-line QR drawn in the former, perpendicular to the common section C D> being also* perpendicular to PQ, ité (and conse ¢ Hyp. an’ quently the plane AB in which it is) will be per- , Def.1.' _pendicular to the plane ED*. Bie Meare Bee K 2 oy “ 132 # Cor. to e5. 7. h Def. 1. 7. ? Def. 6.1. k4, 1.and Def. 24, 124, 1, ™ Def. 3. ELEMENTS OF GEOMETRY. COROLLARY II. Hence it also appears, that a line standing at right- angles to one of two perpendicular planes at any point (1) in the common section (CD) must be tm the other plane: For the line IK, in the plane ED, is perpendicular to the plane AB; besides which line, another perpendicular to AB, from the same point I, cannot be drawn’, f THEOREM VII. Planes (nr, Gu) to which one and the same right-line (AB) is perpendicular, are parallel to each other. | From any point C, in the plane EF, let CD be drawn parallel to AB; which (as well as AB) will be perpendicular to both the planes’; and so the angles A, B, C, D (when AC and BD are joined) being all right-ones*, the figure ABDC (the sides of which AC, BD | are in the same plane with the parallels AB, CD’ will therefore be a rectangular parallelogram *; an consequently CD=AB4 By the very same argu ment, all other perpendiculars, terminated by the tw planes, are equal among themselves ; which was to b demonstrated”. . | COROLLARY. Hence, all right-lines perpendicular to one of tw parallel planes, are also perpendicular to the othe SCHOLIUM. From the two last Theorems, the sense, and pri priety of the two definitions of perpendicular, an parallel planes, appear manifest. a | BOOK THE SEVENTH. THEOREM VIII. Bighi-lines (As, cp) parallel to one and the | same right-line (er) though not:in the same | plane with at, are also parallel to each other. Let GH and GI be drawn per- pendicular to EF, in the planes ew Se) al AF and ED of the proposed ? parallels. Then" shall GF he perpendicular to the plane pass- ing by HGI; and HB, ID will also be perpendicular to the THEOREM IX. Tf two right-lines (aB, Ac) meeting each other, be respectivel y parallel to two other right-lines _ (DE, DF) also meeting each other, Dat not _ beng in the same plane with them; the an- gles (BAC, EDF) contained by those lines, will be equal. Take AB, AC, DE, DF all equal to ach other, and let BE, AD, CF, BC, “uF be drawn. Then AB and ED, as Vell as AC and DF, being equal and same plane®, and therefore parallel to each other”. ys z, P4.é. 133 harallel?, BE and CF will be both Y Hyp. and qual, and parallel, to AD’, and there- Constr. £26; 1. ore equal, and parallel, to each other °; Peay vhence BC is also equal to EF”; and and 8,7. o, the triangles ABC, DEF being mu- ally equilateral, the angles BAC, es “DF are likewise equal‘. Beh. li | THEOREM X. 0 two right-lines (AB, Ac) meeting cach other, . be respectively parallel to two other right-lines 134. «8, # ps » Constr. =§,1. OT. Constr, lit Keb 4 Constr. AI a $9, 7. § Def. 1. 7. h Def. 3. 7. io) Er HUF k 26.1, plane EFHG; also let EI, ELEMENTS OF GEOMETRY. (DE, DF) also meeting each other, and not being in the same plane with them, the planes (BAC, EDF) extended by those lines, will be. parallels, Let AG be perpendicular to the plane BAC, meeting the plane EDF in G; in which last plane, let GH and GI be drawn parallel to ED and DF; and they will also be parallel to AB and AC“; - whence, seeing the angles GAB and GAC are both right”, AGH and AGI must likes wise be right-angles*; and so AG being perpendicular to the plane EDF“ (as well as to BAC?,) the two- planes are parallel to eath other *. * Pinte 2255 THEOREM Xt. The sections (nF, Gu) made by a plane eral cutting two parallel planes (AB, CD,) are alse parallel, the one to the other. | $0 } Let EG and FH be drawn "a - parallel to each other, inthe A ~ CNuS t ‘ ? : FK be perpendicular to the plane CD, and let IG, KH be joined: Then, EG being parallel to FH4, and EI to FKe, the angle GEI is = i y | HFK*; but the angle EIG pig ei is also=FKH, being both right-angles¢; and El. is=FK*: Therefore EG will be equal? (as well as parallel) to FH; and consequently EF likewise Pe rallel to GH *. COROLLARY, it appears from hence, that parallel lines, terminated by the same parallel planes, are equal to each other. ~ BOOK THE SEVENTH. 135 | THEOREM XII. Tf, from the two extremes of a right-line (AB) _ cutting a plane (cv,) two perpendiculars (AF, BG) be drawn to the plane; the right-line (FG) joining the points where they meet the | plane, will pass through the point (&) in which _ the proposed line (AB) cuts the plane, so as to be divided by it into two parts (FE, EG,) hav- ing the same ratio to cach other as those two perpendiculars (AF, BG.) t | For, if AF be produced to f, the lines Af and BG (which are both perpendicular to the plane CD’) will be parallel to each other”; therefore AB and FG being both in the same plane with these parallels, (iD which their extremes are posited”, they must necessarily (as they are not themselves parallels) intersect each other ; And so the alternate angles FAKE, GBE being equal?, as 3B o7. ts well as the opposite ones FEA, GEB?, thence will ’3. 1. FE:EG:: AF : BG’; which was to be demon- 114.4. strated. | COROLEARY. | / Hence, if in the plane CD, the lines FC, GD be made _ parallel, the one to the other, and in them be taken | Fa=FA, and Gb=GB; then will the line (ab) ' joining the points a and b, cut FG in the very same point in which itis cut by AB. For, if e be taken as the intersection of ab and FG, the triangles aFe, | Geb will be equiangular”; whence Fe: eG:: Fa Aer (FA) :°Gb(GB)::FE:EG*. Therefore, seeing #14. 4. ¥G is divided in one and the same ratio, both by ‘12-7. _ eand E£, these points must necessarily coincide". — ” Axe 2 y and 95. 4. | f 156 ELEMENTS OF GEOMETRY. THEOREM XI. If two planes (aw, CD) cutting each other, be both perpendicular to a third plane (au,) their common section will ulso be perpendicular to the same plane (Gu.) For, from the extreme point F of the common section, let the right-line FE be erected perpendicular to the plane GH: which line being in both 'Cor.2,to the planes AB, CD’, it must 6.7. necessarily be their common | section. Therefore the common section is perpendis ™Constr. cular to the plane GH ~. | THEOREM XIV. Lf, from the angular pownts (A, a) of two equal angles (BAC, bac) two right-lines (av, ad) be drawn, or elevated on high, above the planes of the sard angles, so as to form equal angles with the lines first given, each to its correspon dent (pAB=dab, pac=dac,) and of, from any pownts (m m) in those elevated lines perpendt= culars (mN, mn) be let fall upon the planes (Bac, bac) of the first-mentioned angles ; these perpendiculars will be, in proportion, as the parts (AM, am) of the elevated lines included between them and the angular points (a, a) Jirst named. | Make AD and ad equal to each other; and in the planes ADB, ADC, adb, adc, draw DE, DF, de, GH perpendicular to AD and ad; and from their inter- sections with AB, AC, ab, ac, draw EF and ef, meet- — ing AN and an (produced) in G and g, and let D, G, and d, g, be joined. BOOK THE SEVENTH, 137 i The angles ADE, ADF being both right-ones”, not « Constr. nly the line AD, but the plane ADG extended by -2, 7. , is perpendicular to the plane EDF”. But the same 6.7, . lane ADG is also perpendicular to the plane HAF ?: ‘herefore the common section EF is likwise perpen- icular to the plane ADG*%; and consequently the 213. 7. ogle EGA a right-one’. By the very same argu- "Def. 1. 7. ent, ega is a right-angle. Now the triangles ADE, de; ADF, adf being equal in all respects‘, and the ‘Hyp, and ngle KAF=eaf‘, the triangles AEF, aef are also Hyp. qual and alike"; and so, the angle AEG being= «ay. 10.1, eg, EGA=ega, and AE=ae, thence is AG=ag”,* 15.1. ad the angle DAG (MAN)=* dag (man,) because *16.1. DG, adg are both right-angles°. Therefore MN: m:: AM :am?. 14, 4, COROLLARY. fence, the two perpendiculars MN, mn subtend equal angles at the points (A, a) from whence the two elevated (or inclining) lines are drawn. | | THEOREM XV. P. | f any solid (ac) having a .rectilinear base (aBcD,) whereof the planes (ab, Be, cd, aD) of the sides are parallelograms, be cut by a plane parallel to the base, the section (EFGH) will be equal, and sumilar to the base. 138 > Hyp. get te > Def. 24 €24,1, 49. 7, Gd and all parallel to each other’, AE 1 Def. 13. y & Hyp. 426.1. ‘8. 7. k25, 1, ELEMENTS OF GEOMETRY. For, the plane EF GH being parallel / to ABCD *, EF is therefore parallel to 4 AB“; and so, AF being a parallelo- @ gram >, EF is equal (as well as parallel) to AB*. Inthe same manner is FG equal, and parallel to FG, &c. Whence f& also the angle EFG is=the angle ABC; and so of the rest. Therefore EFGH is both equilateral and equi- angular to ABCD. ”~ D COROLLARY. Hence, the opposite bases of a prism are equal am similar (as well as parallel) to each other. y THEOREM XVI. If, from one of the angular points (A) of any parallelogram (ac) a right-line (ar) be ele vated above the plane of the parallelogram, so-as to make any angles (EAB, EAD) woth th éwo contiguous stdes (AB, AD) and there be als drawn, from the three remaining angular points, three other right-lines (BF, CG, DH parallel, and equal to the former (ax;) then. the extremes of those lines being joined, I say, the figure (AG) thus described, will be a pa- ralleleptpedon. | For AK, BF, CG, DH being paar and BF are in the same plane’, as are also AK and DH, &c. Therefore, all these lines being ¢ equal among themselves, AF, AH, DG, and BG are paral- lelograms*; and so, EF being ao D 5 | parallel to AB, parallel to DC, | parallel to HG‘, EF and HG are in the same plane: and EG is also a parallelogram *, equilateral to its op- BOOK THE SEVENTH. 139 iosite AG; but EG is equiangular, and parallel (as vell as equilateral) to its opposite AC; because, XF being parallel to AB, and EH to AD, the angle "EH is therefore=BAD 4, and the plane EG parallel ‘9.7. o the plane AC™ And in the same manner, the 10. 7. ther opposite parallelograms appear to be equiangular nd parallel (as well as equilateral.) Therefore the olid AG, bounded by them, is a parallelepipedon™. = Def. 7. COROLLARY. at f the angle A of the parallelogram AC be a right- one, and AE be erected perpendicular to the plane AC; then will the paralielepipedon be a rectan- gular one: for, all the three contiguous planes AC, AF, AH being rectangular °, their opposites will be °Hyp. and rectangular likewise’; and so, the angles HGF, ue HGC being right-ones, HG will be perpendicular to the plane GB?; and consequently both the planes ‘2.7. EG and DG likewise perpendicular to the plane BG”. And so of the rest. "6.7. SCHOLIUM. | In this Theorem, a way to describe a parallele- pepedon of any given dimensions, 18 indicated ; and he consistence of the 7th and 9th definitions evinced. r ‘THEOREM XVII. Rectangular parallelepipedons (AG, ag) standing upon equal bases (ac, ac), and having equal ultitudes (AE, ae) are equal. Let the rectangles OK nd KL, equal and like to he bases AC and ac of the wo solids, be so formed, hat NK may be in the lame straight line with KM; hen shall PK be also in he same straight line with Ax. 3. because PF (AB,) QG (BC) are equal, as well as ‘7-1. parallel*.) Therefore PFGQ being =plmg’, QGHR= ¢Constr. gmn 1°, &e. the said excess will’ consequently be== | and < ODpl+ipS!* a ODSIY 4 OMNI (4 T?) as Q25°5°5. 7 which was to be done.—This last construction 18 fax. 2. equally applicable to other curvilinear figures; ‘the * "yp. former is peculiar to the circle. | COROLLARY. It follows from henee, that a magnitude, which is _ greater than any polygon that can be described in, and less than any polygon that can be formed about, a given circle, must be equal to the circle ttself: _ seeing that a polygon may be inscribed, which (as well as that formed about the circle) shall exceed any quantity less than the circle itself, be the differ- ence ever so smail; and because a polygon may be formed about the circle, which (as well as that in the circle) shall be less than any quantity that ex- ceeds the circle. THEOREM IL. Every circle (AcE) ts equal to a rectangle (orsr) under its radius (on) and a right-line (oT) equal to half the circumference. | It is evident, in the first place, that the proposed rectangle ORST is greater than any polygon A BCDIctk 152 ELEMENTS OF GEOMETRY. mS 7 G yp that can be described in the circle: For, drawing OA OB, &c. And also Ov perpendicular to AB; iti ‘Cor. to 2. plain, that the triangle AOB (* Ov x FAB) will be les 2. thant OA x 1AB (or OR x4AB:) And, .in the sam 440 ue an’ manner, BOC 4 OR xiBC, &e. Consequently, thi b Ay. 2, 1, Whole poly gon ABCDEF is less than* OR xiAB4 15, 2, OR xiBC, &c. that is', less than a rectangle (Om under OR and Op= half the perimeter (AB+ BC4 CD, &c.) But this rectangle (Om) is, itself, less thay OS, because, Op (half the perimeter of the polygon ™Post.3, 1S less than OT m (half the circumference of the circle, Consequently the polygon ABCDEF is less than th rectangle OS. But, secondly, it will appear, that the same rectangll ORST is less than any polygon HIKLMN thatcanb described aboutthecircle: For, if OH OI, &c. be joined and the radius OP be drawn to the point where HI touche Cor. to 2. the circle ; then will the triangle HOI="OP xi 2. (=ORx-zHI.) In the very same manner lOK= ORx4IK, &c. and therefore the whole polygoi cAx.4.1, HIKLMN=°OR xtHI+ORxiIK, &c.=? a reet PS, 2. angle (On) under OR and Og= half the perimete (HI+IK+4KL, &c.;) which rectangle is manifest]! greater than OS, since Og (=half the perimeter of th 1Post.3. polygon) is ereater than OT, Seeing, therefore, that the rectangle OS is zrealel than any polygon that can be described in the circle, ani Cor. to 1, less than any polygon that can be described about th circle; it must be equal to the circle’. THEOREM III. : All circles (Ack, ace) are tn proportion to on another, as the squares of their radii (A0° ao’. BOOK THE EIGHTH. 153 Let Q: circle ace::.AO?: ao?; then I say, that 2Y=circle ACE. For, first it is evident that Q is sreater than any polygon ABCDEF that can be des- ribed in the circle ACE: Because, if another poly- gon abedef, similar thereto, be* described in the circle «31.5. wce; then will polyg. ABCDEF : polyg. adbcdef 1: :* AO*: ao’): : Q: “circle ace; where the first con+ ‘Cor. to peguent (polyg. abcde) being less than the second (or, ede |han the circle in which it is inscribed”) itis manifest, w 4x 2, jbat the first antecedent ABCDEF must also be less | ‘ban the second Q*. *2. 4. | In the same manner it will appear, that Q is less yhan any polygon HIKLMN that can possibly be de- icribed about the circle ACE: For, if about the other sircle ace, a similar polygon hiklinn be described! ; 731.5. then will HIKLMN : Adkimn (::* AO? : ao?) :: Q:+Cor to circle ace; where the first conseguent (hiklmn) being 3': > sreater than the second (ace,)°® the first antecedent Ses AIKLMN must therefore be also greater than the ‘<2. 4, second Q. | Therefore, seeing that Q is greater than any polygon hat can be described in the circle ACE, and less than my polygon that can be described about the circle; t must be equal to the circle”. 7 d oe to 1, | SCHOLIUM. ' After the same manner, other similar curvilinear figures are proved to be in proportion to one another, us the squares of their diameters, or other homologous limensions ; by means of the second construction of ‘he first proposition ; it being very easy to demonstrate ‘hat the polygons formed from thence, whether both ELEMENTS OF GEOMETRY. within, or both without two similar figures, will them- selves be similar. . THEOREM IV. The circumferences of all circles (Azcn, abed) are in the same proportion as theer radu (oB, ob.) ? Let OF, oe be RB KR squares on the radii OB, ob ; and let OG, ae: og be two rectangles contained under the “| Kal a same radii and right- ‘ ee ¢ lines OH, of, respec- tively equal to the semi - circumferences ABC, abc. Then, these rectangles being on to the circles themselves, it will therefore be, Ok: OG:; oe, og*. Andin this same ratio aref also the bases OC, OH; oc, of: whence (by equality and alterna- tion) OC (OB): o¢ (0b).:: OH: oh: : 20H (circumf. ABCD): 2oA (circumf. abcd.) . LEMMA Q. : If a solid (ac) generated by the revolution a any plane figure (EBCF) about a queescent axis (EF,) be cut by a plane perpendicular to the axis; the section will be a circle, having 2b center in the point (0) where it mects the axis. For, from _O, in the ia tee generating plane EBCF, 4 PeaGlOAae Sy aaah) draw OR perpendicular 7 / > to the axis EF, meeting BC in R. Then, since thislineOR, during the whole revolu- tion, every where pre- serves its perpendicularity a . "| to the axis EF, it is there- De he P| roy BOOK THE EIGHT H. 155 ‘e always in the plane passing through O perpendi- dlar to the said axis’: and consequently, as itS*Cor, to 2 hgth also continues the same in every position, 7: 2 line Rrrrr described, in that plane, by the extreme pint R, by which the section is bounded, must be the ‘cumference of a circle’, of which the point O is the ' Def.33.1 oter. COROLLARY. _ et ance, the bases of cylinders and bones, and all séc- tions parallel to them, are circles. LEMMAS, right-lhne (PQ) standing per; SUnHs cary to the plane of a cylinders base (and not exceeding ‘the axis cr) falls wholly within, or wholly ‘without the cylinder, according as the point |(P) on which it insists, ts entied wethen, or without the circumference of the base. From the center C,tothegiven BR F QDC@ pint P, draw CP; take, in CF [ id PQ, any two egual distances L, PN, and let LN be drawn, ne the surface of thecylinder M ‘Because CL and PN are pa- illel*, and therefore both in the ka. 7, ‘me plane, LN is parallel, and | Def, 13, qual to CP ™, Therefore, when ~ "26.1; (P is less than the radius CG, LN will be less than 24.1. (G, or than its equal LM”; aiid so the point N must » per. 12. ill ‘within the cylinder’. And the same is equally 7. jue with regard to any other point in the line PQ. ee Ts lut, when CP is greater than CG, LN will also be geater than HG (LM;;) and the point N will then fall at of the cylinder °. 156 ELEMENTS OF GEOMETRY. THEOREM V. | Every cylinder is equal to.a rectangular pa a lelepipedon of equal base and aie Oe I say, if the base ace of the cylinder aS be eqil to the base IKLM of the rectangular parallelepiped) IP, and the altitude OH of the former be also eqil to the altitude KO of the latter: then the two sol: will be equal. i For, first, it is evident, that. the cylinder exceg any paralielepipedon (Ip,) of the same given altitay, whose base Lidin is less than the base (ace) of the ( R im MOE linder. Because a polygon (abcde) may be descritl °1.8. in the circle ace, that shall exceed IKlmn?; up which, an upright prism (of the given altitude) m 718.7. be coustituted ‘ ; ; which will be less than the cylin as being wholly contained therein; since (by Lemut 3.) ail tight- -lines drawn perpendicular to the base,! "Cor. 2. to the planes of the sides’, from any points in ab, be, = 6.7. fall wholly within the cylinder, and consequently rb planes themselves, i in which they are. But this (oc 41.7. tained prism is greater than the parallelepipedon Ip: therefore the cylinder itself must, necessarily, ‘Ax.2. greater than Ip*. In like manner it will appear, that the cylinder 5 less than any parallelepipedon, Ir (of the same al- tude) whose base IKvé Ep ecoe that of the cylinde: BOOK THE EIGHTH. a polygon (ABCDEF) may be described about ¢3, nor greater than IP ; it must necessarily be equal qit. COROLLARY. i Ince, whatever is demonstrated (in the 2st, 22nd, ind 23rd Theorems of the preceding book,) with espect to the proportions of prisms, holds equally ‘rue in cylinders also; being equal to prisms of equal base and altitude. | SCHOLIUM. | rom the same demonstration, it will likewise ap- jar, that every regular solid, whose sections, by \ * mines perpendicular to the base, are all rectungles, sequal to a parallelepipedon of equal base and altt- ge; and consequently, that all solids of this kind ‘hich may be comprehended under the name of (linderoids) will be equal among themselves, when LLir altitudes, as well as bases, are equal. t j i LEMMA A. ; two solids (HAu, hah) of the same altitude, have their sections by planes parallel to the ‘bases, at all equal distances therefrom, equal \to each other; it ts proposed to demonstrate | (under certain restrictions specified hereafter ) that the solids themselves will be equal. fLet Il, KK, &c. 2, kk, &c. be sections of the two lids by planes parallel to the bases HH, hh, dividing te altitudes AB, ab into parts BC, CD, &c. bc, ed, -. all mutually equal to each other.» Then, every ‘o corresponding sections being equal” (HH=AA, | =i, &c.) the upright solids HNNH, hanh ; LOOT, 1 circle ace that sball be less than I, it is evi- be ADE 80} ent, that the content of the whole figure will be the um of all these, or 280 square inches. But, when ae given lines are expressed by fractions, or very wge numbers, the work will be somewhat shortened, y finding the content of every two triangles, having he same base (forming what is called a T'rapezium,) t one operation ; that is, first add the two perpendicu- ars together, and then multiply half their sum by the ommon base of the two triangles. Thus, in the last ex- imple, the half-sum of the two perpendiculars BF and JG being 10, if this number be, therefore, multiplied by (0, the measure of the common base AC, the product vhich is 200, will be the content of the trapezium \ECDA;; to which 80, the content of the triangle ADE reing added; the sum will be 280, the same as before. _ But, if the polygon proposed be a regular one, that is, me whose sides, and angles are all equal, the shortest vay of all, is, to multiply half the sum of all the sides 'y the length of the line drawn from the middle of any ‘ide to the center of the polygon. ‘Thus, suppose it vere required to find the area of a pentagon, whose lide is 250, and perpendicular distance of the center rom the middle of any side, 172 inches ;—Here 250 x5 = 1250 = the perimeter, or sum of the sides; hen +#5° x 172=625 x 172=107500 the superficia] content of the pentagon, in square inches. The reason if puch is obvious, from the demonstration to T’heor, » B. VIII. Having shown how the area of any right-lined igure may be computed, it will be proper here, to say omething with regard to the area, and periphery of he circle. | 17! 172 OF THE MENSURATION OF It is well known, that to determine the true area of a circle, and to find a right-line exactly equal to its circumference are looked upon, by mathematicians, as absolutely impossible: But, though neither the one nor the other can be accurately: known, yet several ways have been invented by which they may be approximated, to any assigned degree of exactness, That which I am now going to lay down, though less expeditious than some others, seems, nevertheless, t be the most proper for this place, as depending on the most simple and evident principles: I shall therefore begin with premising the following " LEMMA. a If av be a diameter, and Aas, Bc two equal are of the same circle, and if the chords vB, Dt be drawn; then pBp’=7ADXDC+FAD’. ‘ For, if in DA produced, there be taken AF = DC and BF, BA, BC and the radius BE be drawn; then the external angle FAB, of the trapezium ABCD being equal to the internal opposite angles DCB (6 17, 3.) also AF=DC, and AB=CB (by Hyp.) ; iti evident, that FB is also=DB, and consequently tht angle F=FDB=DBE: And so the isosceles triangle DEB, DBF being equiangular, it will be as DI (GAD): DB: : DB: DF (DC+AD); and conse quently DB’=4ADxDC+AD*% Q. LE. D. COROLLARY. | Hence, if the diameter AD be denoted by the numbe' SUPERFICIES AND SOLIDS. 2, the chord DB will be denoted by 7 DO+2: whence, it appears, that, if the measure of the sup- plemental-chord of any arch be increased by the number 2, the square-root of the sum will be the _ supplemental-chord of half that arch. Now, to apply this to the matter proposed, that is, o the finding of the area and circumference of the sircle; let the arch ABC be taken equal to 4 of the semi-periphery ACD; then will the chord AC be squal to the radius AE (by 29. 5.); and therefore, since ACD is a right angle (by 13. 3.) DC? (=AD*— AC, by 8. 2.) will be=4—1=3; and consequently DC = 1/3= 1,7320508075, &c. Wherefore, seeing the supplemental chord of + of the semi-periphery is 1,7320308075, we shall by the preceding corollary, A/ 24-1,7320508075==1 ,93185 16525 V 24-1,9318516525—=1 9828897227 V/ 241,9828897227=1,9957178465 / 24-1,9957178465=1 9989291743 V/ 2-+1,9989291743—=1 9997322757 A/ 24-1,9997322757 =1 9999330678 V 2-+-1,9999330678= 1/ 3,9999330678 have chord of leone of the periphery. for the supplemental de Ww © Now, therefore, since it is found that 3,9999330678 is the square of the supplemental-chord of 3%, of the gemi-periphery, let this number be subtracted from 4 the square of the diameter, and the remainder -0,0000669322 will be the square of the chord of the same arch; therefore the chord itself being= 4/0,0000669322=0,00818121, let this number be mul- tiplied by 768, or twice 384, and the product 6,28317 ‘will be the perimeter of a regular polygon of 768 “sides, inscribed in the circle; which, as the sides of the polygon very nearly coincide with the circum- ference of the circle, must also express the length of _ the circumference itself, very nearly. _—— 173 174 OF THE MENSURATION OF — | 7 But, in order to show how near this is to the truth, let AB represent one side of a regular polygon of 768: sides, inscribed in the circle (whose length, we have found above to be 0,00818121) and let ab be a side of another similar polygon, described about the circle; and from the center O, let ON be drawn, bisecting AB and ab in M and N: Then, since AM is=2AB = 0;0040906, and AO=1, it is plain, that OM? (AO? — AM”) will be=0,99998327, and consequently OM= 0,99999163 ; whence, because of the similar triangles AOB, a0, &c. we have 0,99999163(OM) : 1 (ON) >: AB: ab: : 6,28317.(the perimeter of the inscribed polygon) : 6,28322 the perimeter of the circumscribed polygon. But the circumference.of the circle being. greater than the perimeter of the inscribed polygon, and less than that of the circumscribed one, it must, | consequently, be greater than 6,28317, and less than. 6,28322 ; and must, therefore, be equal to 6,2832, very near, since this number exceeds the perimeter of the inscribed polygon by no more than 0,00003, | and is less than the perimeter of the circumscribed one by 0,00002, only. , 4 From the periphery thus found, the area of the circle will also be ‘known; being equal to the product of half the periphery into the radius (by 2. 8.) that is, =3,1416 x 1=3, 1416. Therefore, sinceitis proved (in Theor. 3, and 4, of 8.) that the peripheries of circles are in proportion as their | SUPERFICIES AND SOLIDS.) diameters, and the circles themselves as the squares of those diameters; it follows, that, as 2 is to 6,2832, or as 1 to 3,1416:: the diameter of any circle to its peri- phery; and as 4 to 3,1416, or as 1 to 0,7854 : : the square of the diameter to the area. But, if you would rather have the proportions in whole aumbers, and the case proposed does not require any great degree of accuracy, then, instead of the fore- going, those of rchimedes may be used, viz. 7: 22° :: diam. : circumf. and 14: 11 :: square diam. : area. Which proportions differ but little from those above, as will appear from the following example: wherein the diameter of a circle being given 28, its circum- ference and area are required. Here, according to the first proportions, I multiply 28 by 3,1416, for the ‘circumference, and the square of 28 (or 784) by 10,7854 for the area; and there results 87,964 and 615,75, respectively. But, according to the propor- ‘tions of Archimedes, the circumference will be found equal to 88, and the area 616; which differ very little from the former. | By knowing the proportion between the diameter of a circle and the circumference, and between the square of the diameter and the area, the convex su- perficies of solid bodies may be determined. Thus, | The convex superficies of a cylinder is found, by first finding the circumference of the base, and then ‘multiplying by the altitude of the solid. Thevefore, if to that product, the area of the two ‘circular ends be added, the sum will be the whole superficies of the ‘cylinder. _ To find the convex superficies of a cone, multiply half the length of the slant side by the circumference ‘of the base. _ The convex superficies of any frustum of .a.cone is found, by multiplying the sum of the peripheries of the:two ends into half the length of the slant side of the frustum. _ To find the 'superficies ofa sphere, multiply the pe- irtphery‘of ‘the wreatest, or generating, circle by tte 175 176 OF THE MENSURATION OF diameter: Or, multiply the square of the diameter by 3,1416. | The convex superficies of any segment of a sphere is found, by multiplying the periphery of the greatest circle of the sphere into the altitude of the segment. — The demonstration of these last rules; for finding the curve surfaces of solid bodies (which is not given in the Hlements, for reasons mentioned hereafter) is inserted at the end of this section. OF THE MENSURATION OF SOLIDS. As every superficies is measured by a square, whose side is unity (as one inch, one foot, one yard, &c.) every solid is measured by a cube, of which the side is also an unit. Thus, let the solid to be measured, E F 2 pa A! B hate ia ih be the rectangular parallelepipedon AF, and let th cube P, whose side is one inch, be the measurin unit; also let the length AB, of the base AC, be? inches, the breadth BC 2 inches, and the altitude Al of the solid 5 inches: Then, because the area of th base ABCD is 2 times 4 (or 8) square inches, it i SUPERFICIES AND SOLIDS. ‘easy to conceive, that, if the solid were to be only one inch high (instead of 5,) the content would be just the same number (8) of cubical inches; because then, upon each of the eight equal squares into which the whole base ABCD is divisible, a cube of one inch might be erected, so as to compose a parallelepipedon on that base, of one inch high. Therefore, seeing ‘hat the content of the solid, at one inch high, is 8 »ubical inches, the whole content at 5 inches high, must consequently be 5 times 8, or 40 cubical inches (since the whole solid AF may be considered, as com- yosed of 5 such heights of cubes, one ranged above mother. ) _ And, generally, whatever the dimensions may be, tis manifest (from 21 and 22. of 7.) that the paral- elepipedon will contain the cube P, as many times as he base ABCD contains the base of the cube, re- yeated as often as the altitude AH contains the alti- ude of the cube. | Therefore the content of any parallelepipedon will e found, by multiplying the area of the base by the wltitude of the parallelepipedon. Thus, for example, ‘the two dimensions of the base be 16 and 12 inches, nd the height of the solid 10 inches; then, the area f the base being 192, the content of the solid will be 920 cubical inches. | From the content of a parallelepipedon, thus known, tat of a prism, or a cylinder, will likewise be known; very such solid being (by 20. 7. or 5. 8.) equal to a ‘arallelepipedon of equal base, and altitude. There- ore, multiply the area of the base (found by the rules or superficies) ¢nto the height of the prism, or cylinder, nd the product will be the content. Hence the content of any pyramid, or cone, is also btained ; being (by Cor. 3. to.8. 8.) equal to 4 part fa prism, or cylinder, of the same base and altitude. ‘herefore, multiply the area of the base by + of the ttitude, and the product will be the answer. | Every sphere being (by 11. 8.) equal to 3 parts of cylinder of the same diameter and altitude’; the con- nt of any sphere will, therefore, be found, by multi- lying the area of its greatest, or generating,’ circle | N | ~!I ~t 178 of a whole pyramid, of the same OF THE MENSURATION OF into 2. of its diameter: Or (because the area of such circle is to the square of the diameter, in proportion as 0,7854 to 1,) let the cube of the diameter be multiplied by the fraction ,5236 (= of 0,7854,) and the product will be the content. Thus, if the measure of the dia- meter be 20, its cube will be 8000 ; which, multiplied | by ,0236, will give 4188,8 for. the measure of the sphere’ S solidity. The manner of finding the’ content of any frustums of the M solids above determined, is col- B lected from Fheor. 10. and 1}. is Bb. VIUIl. hi Let the frustum (MN), first proposed, be that of a pyramid ; \ then, having found the content given base and altitude; say, N ‘ as any side A of the lower end / _ or base, is to its correspondent B of the upper, so is the said content of the whole pyramid to a fourth ‘proportional. | And, as A is (again) to B, so is the quantity last found to another proportional : which two propor tionals, added to the content first determined, wi il give the true content of the frustum. But when the opposite bases of the frustum are squares, the rule will be more simple, and put on @ better form: For then the area of the base being A’, the content: of a whole pyramid thereon, of the same altitude with the frustum, will be equal to the paral- lelepipedon Cx A’, C being 4 of the given alta of the frustum. But A:B::CxA?: CxAxB (by 22. 7.) afl 1 ~A:B::CxAxB:CxB2, Therefore CxA?#O KAXB+C xB? (=CxA?+AxB+B%, by Schol, 10 20. 7.) is the true content, in this case. fi Hence, to jind the content of the frustum of any square yramid, add the product of the two sides af the lower and upper ends to the sum uf their suai and then inet the aggregate by + of the Srustuam height. SUPERFICIES AND SOLIDS. rian Galen ere [cys Q From the content here found, that of any coneca/ rustum (PQ) is readily obtained ; being in propor- on to the content (Cx A ?+AxB-+B *) of the frus- am of a square pyramid circumscribing it, as the ase of the former is to the base of the latter (oy Cor. . to 8. 8.), or as the fraction ,7854 is to unity: And », will be equal to the ,7854 part of CxA’+Ax B -B*=ExA’*+Ax8+8B?; by taking E=,7854 x C the ,2618 part of the whole given altitude. ‘There- are, to find the content of any frustuin of a cone, add ie product of the diameters of the two ends to the um of their squares ; then multiply the aggregate by efrustum’s height, and the product, again, by the action ,2618. 7 names / } ‘Hence, and from Theor. 11. B. VIII. a rule tor ding the content of any segment IAK of @ spiere, ay also be deduced. For, it appears, fram, thence, at the segment proposed, IAK, is equal to the differ- ‘ce between a conical frustum FCDH and a cylin- r ECDG of the same altitude, standing upon a base, M 2 179 180 OF THE MENSURATION OF whose radius CA is equal to that (AO) of the sphere itself. But the content of the frustum FCDH, if the twe diameters CD, FH be represented (as above) by 4 and B, and the ,2618 part of the altitude (D) by E will be=E x A? +A x B+B ® (that is, equal to a pa rallelepipedon whose altitude is E, and base=A ? + A xB-+B%): And the content of the cylinder EC DG will be=3E x A 2, or Ex 3A 2. Therefore the difference (or the content of the seg ment IAK) will be=Ex2A *—A x B—B 2 (Schol. 20. 7. and Az. 5. 1.) But 2A ?—A x B—B ? is composed of A?—AxE and A2—B?; the former part of which A?—AxF is=A—BxA (by 5. 2.)=2D x A (because A—B (01 CD—FH)=FE+HG=2AB or 2D); and the latte A?—B’?=A—Bx AB (by 7. 2.)=2D x ZA—2D. | Whence the sum of both will consequently be= 2D x3A—2D; and the content of the segment itsel! =E x 2D x3A—2D=,5236 x D?x3A—2D (because 2E=,5236D). | Therefore, fo find the content of any segment of a sphere, multiply the square of the segment’s heighé by the excess of thrice the spheres diameter above tht double of that height ; and then multiply by the frac tion ,5236. | The demonstration of the rules for determining the superficial content of the cylinder, cone and spheré and of their several segments, or frustums, is collected pi from the two Lemmas here subjoined. ur a LEMMA 1. The upper superficies, or area of all the sides of a regular pyramid, in which a cone may be inscribed, is equal to a rectangle under the perimeter of the base, and half the length of the cone’s slant side. | ») ey. ae | SUPERFICIES AND SOLIDS. _ For, let BCDE, &c. be the base of the pyramid, nd BPGM that of the inscribed cone ; and from the “ertex A to the point P where any side DF ofthe polygon puches the circle, let AP be drawn. Then, since the triangle ADE is=1AP x DE=4ABx DE; and as he like holds good with regard to every other side f the pyramid, it is evident, that the sum of all the ides, or the whole superficies of the pyramid (exclu- ive of the base) will be equal to AB x DE + EF +&c.; nat is, equal to a rectangle under }AB and the whole erimeter of the base. COROLLARY I. lence it will also appear, that all the sides of any _Srustum Bg of the pyramid, will be equal to a rect- _angle under half the length of each side and the _ sum of the perimeters of the two ends: For, the area of the side DEed being=iPp x Di +de, or -+Bb x DE+de (by 4. 2.), the area of all the sides _ will, therefore, be=1Bd x Dib +de+ EF +e/+«c, 4 181 182 OF THE MENSURATION OF COROLLARY II. Therefore, seeing that the foregoing conclusions hold universally, whatever the number of the sides may be; and as the pyramid, by increasing the number ofits sides, approaches nearer and nearer, continually, to the inscribed cone, which is its limit; thence will the upper or conve superficies of the cone (as) well as that of the pyramid) be equal to a rectangle, under half the length of its slant side, and the peri-| meter of its base.* | * The rule above given, applies only to a right cone, that is, to a cone, whose axis is perpendicular to its base ; no method has yet been discovered of determining, accurately, the curve surface of an oblique cone, though many attempts have been made, at different) times, by some of the most eminent mathematicians in Europe:| the following are among the principal which have come under the Editor’s observation : ist. In 1648, Richard White,-an Englishman, published a work at Rome, in 4to, entitled, “‘ Hemispherium disseetum, opus geome: Lricum waveee accessit appendix, de inscriptione in sphera coni scaleni, et de superficie ejus...+..” in which he shows that the curve surface of an oblique cone is nearly equal to the area of ¢ circle, the radius of which is a mean proportional between 4) and !\/ (h2-+r 2)+18-+453 where r denotes the radius of the cone’s base, h the perpendicular height, and S and s the greatest ant least slant sides. See Leyhourn’s Mathematical Repository, N.S Vol. 1. page 113. 2 4 - ond. In the Miscell. Berol. Tom. II. 1727, Peter Varignn published his “ Schediasma de superficie cont ad basin circularen| obliqui ope longituninis curve ss... with an “ additio...” by Leibnitz. Be. | 8rd. In the Nov. Comment, Acad. Petropol. Tom. I. ad ann. 1747 1748, is a memoir by the illustrious Euler, de superficie conorun sca'enorum aliorumque corporum conicorum ...+-- and in the Wov) Act, Acad. Petropol. Tom. II. 1785, is another memoir, by th same profound analyst, de superficie coni scalent, ubt imprimi ingentes difficultates, quee in hac investigatione occurrunt, perpend UNLUT wae é 4th. D’Alembert, in Tome I. of his Opuscules Mathematiques Paris, 1761, has a memoir, “ de la surface d'un cone, qui a po base une ellipse ... pa. 236. avec un addition au memotre precedent, pa. 244; and another, de la surface des cones obliques. 5th. In the Comment. Soc, R. Gotting, the celebrated. Abr. Goti Keestner has a memoir, de cont scaleni superficie et rett. ae 6th and lastly. In the Ladies Diary for 1753 (Q. 369), af ap) proximation is given by Mr. Thomas Simpson, the author of thes Elements ; and the subject is again brought under consideration, 1) the same Diary for 1806 (Prize Q.) q | 2 SUPERFICIES AND SOLIDS. And the convex superficies of any Jrustum of the { i | ' t | j i | \ H | ] | | . F ae er oe > cone will, also, be equal to a rectangle under half the length of its slant side, and the sum of the peri- pheries of its two ends, or bases. ‘Whence it likewise follows, that the convex surface of a cylinder will be equal to a rectangle under half its altitude, and twice the periphery of its base (or under the whole altitude, and once that periphery 5) because then the two ends are equal.—From this Corollary, the rules for finding the superfices of the cylinder and cone, are given. LEMMA 9. Tf a regular polygon azcpnr, &c. of an even number of sides, together with rts inscrebed circle Rasg, be supposed to revolve about the (produced) diumeter RS, as an axis; the su- perficies of the solid generated by the polygon, well be equal to a rectangle under its axis AF and a right-line equal to the circumference - nasg of the inscribed circle. | ou al | From the center O, to the point of contact Q, of _ any side BC, let the radius OQ be drawn; also draw | BBM, QPg, Cc, &c. perpendicular to AF, and BN _ perpendicular to CL. 183 184 OF THE MENSURATION OF Because the solid generated by the plane BbcC is the frustum of acone, the convex superficies of that frustum | generated by BC, is equal to a rectangle under 3BC | and the sum of the peripheries of the two circles des- cribed by Bb and Cc, (by Cor. 2, to the precedent): , But the sum of these two peripheries, as (QP is an | arithmetical mean between Bb and Ce,) is equal to. twice the periphery Qq; and therefore the convex su- | perficies of the said frustum equal to ; BC x 2 periph. Qg=BC x periph. Qg. But, because of the similar. triangles OPQ, BNC, we have BC: BN (bc): : OQ. : PQ:: periph. RQSq: periph. Qg (by 4. 8.); and consequently BC x periph. Qg=bc x periph. RQSg=. the superficies generated by BC. By the very same. argument, the superficies generated by any other side CD is=cd x periph. RQSqg: Whence it is mani-_ fest, that the superficies of the whole solid is=Ab+6c- +cd+&c. x periph. RQS7=AF x periph. RQSg. | COROLLARY I. Since the superficies of the solid is, universally, equal. to AF x periph. RQSg, let the number of sides of the’ generating polygon be what it will; and as the said: superficies, by increasing the number of sides, ap- proaches nearer and nearer, continually, to the su-: perficies of the inscribed sphere, which is its limit; therefore the superficies of the sphere, itself, is equal to a rectangle under its axis RS and periphery RQSgq ; and the convex superficies of any segment of a sphere vRw, is likewise equal to a rectangle under its axis (or height) Re and the same peri- phery RQSq; since it is proved, that the cor- responding superficies of CBAML, is universally — equal to Ac x periph. RQSg. | COROLLARY IT. Z Hence it also appears, that the superficies of every sphere%s equal to four times its generating circle; Because (by 2. 8.) the-circle RQSg=3RS x periph. RQSq=iRS x periph. RQS¢. SUPERFICIES AND SOLIDS. ‘In deriving these conclusions, as well as those de- ending upon the preceding Lemma, the Reader must ave observed, that something is assumed, which is >t demonstrated in any part of these Elements. But .is will not, I imagine, be considered as a fault, by ‘ose who know, that it is impossible to prove in a janner perfectly regular and geometrical, tbat a curve lurface, of any kind, is equal to a plane-one of an as- igned magnitude. Plane surfaces are compared ‘ith one another, in virtue of the 10¢h Axiom; in ‘hich, whatever relates to the equality of plane fi- ares, has its original. But no principles have been ee admitted into the common, or lower Geometry, by thich a curve-surface can be compared with a plane- ae; nor even by which the proportion of any one urve-line to a right-line can be known: Nor can it demonstrated by all the Geometry in Huclid’s ilements, that the periphery of a circle is less than ie perimeter of its circumscribing square—We can etermine the proportion of solids bounded by curve- urfaces, by describing other solids in, and about them, pas to differ less from them, than by any assigned art, however small. But, in comparing the sur- aces, this method fails; because, let the number of ides of the inscribed, or circumscribed solid, be ever o great, or let the solid itself approach ever so near » the proposed one ; the two surfaces, after all, will ave no part in common on which a demonstration an be formed, but will still be distinct things. Be- bra such a. comparison can possibly be made, in a egular and scientific manner, new principles must be ‘aid down: But these belong to, and are best supplied athe Modern Geometry, or Method of Elusions. t i 185 OF THE MAXIMA AND MINIMA OF 4 ® GEOMETRICAL QUANTITIES. canbusnaiiehakoakcsaiaae vm THEOREM L. 7 | i If from two given points A, B, on the same sid of an indefinite ne ra (im the same plan with them) two lines ak, BE be drawn to me on, and make equal angles AEQ, BEP with th said line pa; the lines so drawn, taken toge ther, shall be less than any other two aG, BE drawn from the same points to meet on th same line PQ, For, let BNM be per- pendicular to PNQ, and let AE be produced to meet it in M, alsolet MG be drawn. Then the triangles at MNE, BNE, having the tHe angle MEN (=AEQ*) ©Coutr, =BEN*, MNE=BNE¢ and NE common; have also MN=BN, and ME M | 415.1, ==BE“%; whence also MG=BGe: But AM (AE4 rox tot BE) a AG + MGY, or, than its equal AG+BG 719,1, Q. E.. OF THE MAXIMA AND MINIMA, &c. 187 THEOREM IL. a right-lnes AP, BP; AQ, BQ, that can be _ drawn from two given points A, B, to meet, two by two, on the convexity of a given circle rPaR: those two av, Be taken together, shalt be the least, which make equal angles with the tangent MeN (or with the radius DB) at the point of concourse P. SX Ry, _ For, if to any point m in the part of the tangent in- tercepted by AQ and BQ, there be drawn Az and Bn; M then AP+BP 3 An+ Bas, and An+ Bn JAQ+ BQ’: Consequently AP+BP 3 AQ+BQ. QF. D. & Theor. 1. hQ3, 1. D _ This demonstration holds equally true, when the curve RPR is supposed of any other kind; provided all tangents to it, fall entirely without the curve. ; THEOREM III. Uf, in a given triangle Bc, a point is to be de- termined, so that the sum of all the three lines drawn from thence to the three angles, shalt be the least possible; I say, the position of that point must be such, that all the angles formed about it by those lines, shall be equal among themselves, 188 * Theor. 2, * Ax, 6.1, ' Cor. to 8. 1. ™ Hyp. OF THE MAXIMA AND MINIMA If you deny it, then let A some point EK, at which the angles BEA, CEA are unequal, be the required one. Upon the center A, through KE, let the circum- ference of a circle RER be described ; and let D be that point in it, where ~ the angles ADB and ADC B are equal. | Because BD+ CD 3 BE+ CE‘, therefore is AD+ BD+CD3AE+BE+CE*; which is repugnant Therefore no point at which the angles are unequal can be the required one. Q. EL. D. The same otherwise. Let the point P A be that, at which all the angles APB, APC, BPC, are equal; * and from f UV any other point Q, Gi, i upon the lines form- P S 4 ing them, let fall the \bwt | three perpendiculars | A> Qa, Qb, Qe. I say, a TQ) first, that the sum of 7 = the three distances B Cc Aa, Bb, Ce, intercepted by those perpendiculars, an¢ the three given points A, B, C, will be equal to thi sum of the three first distances AP + BP + CP. a} For, if through the intersection M of Pc and bQ rv be drawn parallel to Aa, meeting Bb (produced. in v, and PS, parallel to aQ, in S; it is evident (be cause the angle v=! BPa=vPM=2 of a right-angle™ that the triangle PyM is equilateral; and that thi Se ee EL. | * The determination of the position of a point, at which, line, drawn from three given points, shall form any given angles, is giver among the Geometrical Constructions, in the next section. | OF GEOMETRICAL QUANTITIES. 189 ght-angled triangles QMc, QMr, having QMe (= -Mb=vMdb)=QMr, have also cM=7rM ; to which it MP=Mv be added; so shall cP=rv=aP+vS= P+P0. | And, if to the first and last of these, AP+ BP+Cc 2 (again) added; then will AP+BP+CP=Aa+ b+Cc, as was asserted. | Whence the Theorem itself is exceedingly obvious : or seeing that the sum AP+BP+4CP is but equal to je sum of the three bases Aa, Bb, Cc, it must neces- irily be less than that of the three hypothenuses AQ, iQand CQ. Q. #. D. \ N THEOREM IV. ‘The greatest triangle anv that can possibly be contained under two right-lines, given mm length, and any other right-line joining thear ex- tremes, will be when the two given lines AB, BD make right-angles with each. other. } For, let BC be - D qual to BD, and the C Cc jugle ABC either ‘reater, or less than ae right-angle ABD; ‘et also CF be drawn ‘arallel to AB, meet- B he BD (produced, if A | ‘ecessary) in F, and let A, F, and A, C be joined. | Then the angle BFC being a right-one", it is evi- *5. 1. vent, that BC (BD) is c BF°; and therefore the 0. 1. riangle ABD, being ~ ABF”, is also greater than» Ax. 2. ts equal? ABC. Q. E. D. | * Cor. to ; sme i ii THEOREM V. ; : IF all triangles aBc, ABD, having the same base AB, and the sum of their other stdes the same, - the isosceles one ACB, ts the greatest. 190 ¥ 22. 5. and 11, 1. *16, 1, Cor. to 2. 3 090. 1. b Cor. 2, 2. and Ax. 2, bisects AB*, but also OF THE MAXIMA AND MINIMA. Let CH be perpendicu- lar to AB, and DEF paral- lel to AB, intersecting HC (produced if need be) in E; likewise let AE and BE be drawn. ' It is manifest, that the angles AEF, BED are . | equal’; therefore AE+ BE A. H B 3AD+BDs5; or than its equal AC+BC?; and so ' the triangle AEB, falling within the triangle ACB« must be a ACB”; and therefore ADB (=AEB *) = "ACB. Q. £. D. + THEOREM VI. Of ali triangles anc, ABD standing upon the same base AB, and having equal vertical an- gles ACB, ADB, the isosceles one ACB is the greatest. Let ACDB bea seg- ment of a circle, in which the equal angles ACB, ADB are con- tained’; make CEG perpendicular, and DE parallel, to AB; from the center O draw OD, and let A, E and B, E be joined. It is evi- dent, that CG not only passes through the cen- ter O*. Therefore, OD (OC) being > OE, the trie angle ACB will also bec AEB, or than its equal ADB , Q. Op Dd. “phe “ | OF GEOMETRICAL QUANTITIES. 191 FHEOREM VII. If all right-lines DE, FG that can be drawn to cut off equal areas ADE, AFG from a given triangle asc, that DE is the least, which makes the triangle ADE, cut off, an rsosceles one. Let AFG be the circumference of a circle passing ‘brough the three points A, F, G; also let PH be verpendicular to FG, at the middle point P, meeting he circumference in H, and let FH and GH be lrawn. The triangle FHG, being isosceles*, is there- «Ax. 10. ore’ FAG 4, or than its equale ADE: Whence, as ¢Theor. 6. he triangles FHG, ADE are equiangular/’, the base 7 Hye. *G of the greater, must consequently exceed*the base’ g}y, Ar JE of the less. Q. H. D. Pett. THEOREM VIII. Of all right-lines EF, GH, Gu that can be drawn | through a given point v, between two right- lines BA, BC given in position ; that EF which ts bisected by the given point p, forms with them the least triangle (EBF). 192 3. and 7. L. *Hyp. ke 15, ti tAx, 2. ™ Ax. 6. °15. 1. P24, 1, 9 Cor. to 2. 2, OF THE MAXIMA AND MINIMA For, if EI, parallel to BC, be drawn, meeting GH in I; the* equiangular triangles: DFH and DET having DF=DEi, will be equal*; and DFH wil) therefore be less, or greater than DEG 4, according aj BG is greater or less than BE. In the former casi let DEBH common, be added to both ; so shall FEE be a HGB*. And if, in the latter case, DGBF b added, then will HGB be c~ FEB”; and conse quently FEB (in this case also) 3 HGB. Q. £. D. COROLLARY. If DM and DN be drawn parallel to BC and BA the two equal°® triangles DEM, DEN, taken toge ther (since EM=DN °=MB?) will be equal to thi parallelogram DMBN7?; and consequently to thi parallelogram DMBN=;BEF 23 BGH. Whence it is manifest, that a parallelogram 1s always les than half a triangle in which it is inscribed, excep when the base of the one 2s half the base of the other in which case the parallelogram is just half th triangle. OF GEOMETRICAL QUANTITIES, SCHOLIUM. | From the preceding Corollary alone, it may be iGM which can possibly be described about, and the reatest parallelogram EF Bz that can be described in, ny curve ABCD, concave to its axis AE, will be rhen the sub-tangent FG is equal to half the base EG f the triangle, or to the whole base EF of the paral- tlogram; and that the two figures will be in the atio of two to one. _For let HN be aside of any other circumscribing ‘langle (EHN) touching the curve in C, and meeting .Brinr: ‘Then, the curve being concave to its axis, ae point 7 will fall above B ; whence, if rm be drawn arallel to Bn, then will EGM=2BE—32rE GEN. gain, if IC, parallel to EM, be produced to meet iM in p, and CK and pq be drawn parallel to AE; sen, also, will BE=;EGMcpEc CRB, as wasto be Lown. | THEOREM IX. 'f all right-lined figures, contained under the same number of. sides, and inscribed in the same circle, that rs the greatest whose sides are : , ? all equal. 1] ery easily made to appear, that the least triangle’ 193 194 OF THE MAXIMA AND MINIMA For, if possible, let some polygon ABCFE whose sides CF, EF are unequal, be the greatest. - ie | Let CDE be an isosceles triangle described in the Theor, 6. Same segment with CFE; which beingc-CFE’, the and 11.3. whole polygon ABCDE will also bet~the whol polygon ABCFE; which is repugnant. Therefor the polygon is the greatest when the sides are al equal. Q. E. D. 4 G THEOREM xX. Of all right-lined figures, contained under th, same perimeter, and number of sides, th, greatest is, when the sides are all equal. __ B A ci For, if ABCDE be the greatest possible, the tri angle CDE must, manifestly, be — any other triangl OF GEOMETRICAL QUANTITIES. (FE upon the same base, of which the sum of the cher sides is also the same. But, (by Theorem V.) te greatest triangle, when the base, and the sum of {2 sides are given, is that whose sides are equal: ‘nerefore DC and ED are equal. In the same 1anner it appears, that BC=CD, &c. Q. E. D, ee a THEOREM XI. J all the sides of a polygon, except one, be igeven in length, and their posttton be required, so as to make the polygon itself the greatest possible ; I say, their position must be such, that two lines drawn from the extremes of the unknown sede to any angle of the polygon, shall form a right angle. ; P ‘ot, if you would have the polygon ABCDEF to. e.he.greatest possible, and yet ADF, subtended by h pars aay side AF nota right-angle:': Then let >) be a right-angle, contained under PS=AD, and 02 19. 196 OF THE MAXIMA AND MINIMA OS=FD; and upon PS and OS, let the figures PS RQ and OST be described, equilateral, and equal, to ‘10.6. ADCB and FDE‘. ‘Theor.4. The triangle PSOC- ADF‘; therefore, PSRQ “Constr. being=ADCB, and OST=FDE ‘, the whole polygor ’ Ax.6.1. PQRSTO is. alsocthe whole polygon ABCDEF” which is repugnant. : o | COROLLARY. Hence, because the angle in a semi-circle is a ‘right #18. 3. angle; it appears, that the greatest polygon thatea be contained under any proposed number of give sides, and one other side any how taken, will be when it may be inscribed in a semicircle, of whic, the indetermined line will be the diameter. | } / } THEOREM XIl. A polygon ABCDEA in a circle, ts greater tha any other polygon pQrstP, whatever, whos sides are the same both in length and numbe: Let AF be the diameter of the circle, and join] F; also make the angle PTO=AEF, TO=EF, an let PO be drawn. , | Op S Fp ( Because AB=PQ, BC=QR, CD=RS, DE =£ viyp. and EF=TO¥», the polygon ABCDEF, being i: scribed in a_ semi-circle, will bec the polygé *Ther. PQRSTO*; and, if from these, the equal triang! il. AEF, PTO be taken away ‘there will rema ABCDEAC”PQRSTP. @. #. D. - OF GEOMETRICAL QUANTITIES. | SCHOLIUM. _ That the magnitude of the greatest polygon, which van be contained under any number of unequal sides, joes not at all depend upon the order in which those a - A. E lines are connected to each other, will appear, thus, Let ABCDE be the greatest, one way, or according to one order of the sides; and upon BD leta triangle BDF be constituted whose sides DF and BF are, respectively, equal to BC and DC;; then, the triangles BCD, BFD being equal, the whole polygons ABCDE and ABFDE will likewise be equal, notwithstanding their equal sides BC, DF, &c. are placed according to different orders. THEOREM XIII. Of all polygons, contained under the same perimeter, and number of sides ; that whose sides, and angles, are equal, is the greatest. For, the greatest polygon that can be contained vunder a given perimeter, is one whose sides are all equals, But of all the polygons of this sort, that is the greatest which may be inscribed in a circle’: ‘Therefore, the greatest of all, is that whose sides are 197 a Theor. ‘all equal, and which may be inscribed in a circle, © ‘or whose angles, as well as the sides, are all equal. QQ. E. D. | 198 © Theor. 4. ¢ Hyp. € Ax, 6. f13.3 | OF THE MAXIMA AND MINIMA THEOREM XIV. The greatest area that can possibly be contain by one right-line, any how taken, and any other line or lines, whatever, of which the sum as given: will be, when two right-lines drawn from the extremes of the unknown line Sfirsi mentioned, to meet any where in the grven boundary, make right-angles with each other. D A 5 © Q For, if you would have the area ACDEBA, con tained by some right-line AB, and ACDEB whos length is given, to be the greatest possible, anc ADB, at the same time, not a right-angle: Then, le PSQ be a right-angle, contained under PS=AD, anc QS=BD; and, having joined PQ, upon PS and Qs‘ conceive two figures PRS and QST to be formed equal and alike in all respects to ACD and DBE. | Since the area PSQC- ADB®; itis manifest, tha’ the area PRSTQP, contained by the right-line (PQ. and PRSTQ (=ACDEB*) will also be the ane ACDEBA ®*; which ts repugnant. Therefore the area ACDEBA cannot be the creates possible, unless the angle ADB bearight one. Q. £. 4 COROLLARY. Hence, because the angle in a semi-circle is a right angle%, it is evident, that che area will be the greates possible, when the given length or boundary, form the arch of a semi-circle; of which the wedelenias right-line proposed is the diameter. OF GEOMETRICAL QUANTITIES. Le THEOREM XV. af all plane figures ABCD, EFGH, contained ' under equal perimeters (or limuts ), the circle | (aBcp) zs the greatest. ¥ | D | For, if the diameter AC be drawn, and EFG be aken equal to the arch ABC; then the area ABCA will (by the precedent) be — the area EFGE, contained oy EFG and the right-line EG ; and ADCA will also vet EHGE: Therefore ABCD TC EFGH. QaETD. Vk COROLLARY. dence it appears, that the greatest area that can possibly be contained by a right-line AB, and a curve-line AeB, both given in length; will be, when | the latter is an arch of a circle. For, let AnB be any other curve line, equal to AeB, ind let the whole circle AeBCD be completed ; which will (it is proved) be greater than the mixed figure AnBCD; and consequently, by taking away the sommon segment ABCD, there will remain AeBA AnBA. —— THEOREM XVI. The greatest parallelepipedon that can be con- ' tained under the three paris of a given line ; , _ AB, any how taken, will be when all the parts are equal to cach other. 400 OF THE MAXIMA AND MINIMA For, ¢f possible, let Pap Dd 1 two parts AE, ED be 4.7—_—_*#-"—_+ BS unequal. Bisect AD in C; then will the rectangle’ under AE (AC+CE) and ED (AC—CE) be 3 AC? 7.2, (or ACxCD) by the square of CE’. Therefore the’ solid AE x ED x DB will also bethe solid ACx 421.77 CDxDB*; which ts contrary to hypothesis. } { COROLLARY. | | Hence, of all rectangular parallelepipedons having’ the sum of their three dimensions the same, the cube. is the greatest. THEOREM XVII The greatest parallelepipedon ac?xcp that can possibly be contained under the square of one part Ac, of a given line an, and the other; part BC, any how taken; will be, when the: former part is the double of the latter. | _ For, let Ac and ~C | Be be any other Ks oD : | Bo parts into which the given line AB may be divided; and let AC and Ac be bisected in D and d. So shall ‘Cor. 06. AC?x CB=4AD x DC x CBic4Ad x de x cB *(AC* 2.and7. cB‘) by the precedent. Q. E. D. ® Cor. to 6, 2, : { THEOREM XVII. «of The hypothenuse aB of @ right-angled triangle azc being given; the solid Bc x ac* contained under one leg sc and the square of the other Ac, will be the greatest possible, when the square of the latter leg ac ts double to that of the former BC. OF GEOMETRICAL QUANTITIES. 201 For, if CD be con- REINS yeived perpendicular to AB, and DE to AC; it E willbe AC? (ABxAD'): : Leb B*::AD:AB"™:: DE: m O17, 3C"; and consequently, n 14. 4. AC*?x BC=AB’?xDE°; A D B °23.7. which (as AB* is given) will, evidently, be the zreatest possible, when DE, or its square? (DE”) is? Cor. 2. to he greatest possible. But DE?: AD?::9BC*(BDx , &-?. 4B‘): AB?:: BD: AB”; and therefore DE* x AB 4j. ir =AD’*xBD?°; which (and consequently DE’) will ye the greatest possible, when AD is the double of BD’; that is, when AC? (AD x AB) is the double of’ hyd BC?(BDxAB). Q. £. D. ; THEOREM XIX. ; The altitude wc of the greatest cylinder uc that ‘can possibly be inscribed in any cone ADE, 2s one third part of the altetude ax of the cone, and the cylinder itself $ parts of the cone. For, let gh be any A. other cylinder inscribed in the cone; and it will be, AC?xBC : CG*x Peer? AC* ss CG? 3: Mer. cot: : Ac*xBe : 'eg* x BC; whence, by Sit alternation, AC*xBC : Ac?x Be: : CG*xBC: °g*xBe : and so like- DE - Wise is the cylinder HG ~ Sowes *Ry ae “3. and 5. fo the cylinder*Ag; but AC*xBCc "AC? x BGs: 0° therefore HGc~Ag. Again, since AC=2AB*, and 4,°°" therefore CG=3BE; we also have, cylinder HG: = Ax. 4. cone ADE (or cylinder? DN): : CG?(gBE?) :* Cor.3. fees: 331. .Q. £. D. neh Be Slr £ W : Cor, 11, 4. 202 OF THE MAXIMA AND MINIMA SCHOLIUM. From this proposition, by reasoning as in the Scholium to Theorem VIII, it will appear, that the) least cone that can be described about, and the greatest cylinder that can possibly be described tm, any solid generated by the rotation of a curve, con- cave to its axis, will be, when the sub-tangent is two-thirds of the altitude of the cone, or twice the altitude of the cylinder ; and that the two figures wall be in the ratio of nine to four. From whence the dimensions of the greatest and least cylinders and cones, that can be described in, and about solids generated by curves, to which the method of drawing tangents is known, may be readily determined.” " “¢ The first ideas of Maxima and Minima are to be found in the Elements of Euclid, or flow immediately from them: thus it ap. pears by Prop. 5, B. II, that the greatest rectangle that can be made of the two parts of a given line, any how divided, is wher the line is divided equally in the middle: Prop. 7, B. III, shows’ that the greatest line that can be drawn from a given point withit a circle, to its circumference, is that which passes through the center ; and, that the least line that can be so drawn, is the con: tinuation of the same to the other side of. the circle :—Prop. 8, ib shows the same for lines drawn from a point without the circle and many more instances of a similar nature might be pointed out in the Elements. Other writers on the Maxima and Minima art Apollonius, in the whole 5th Book of his Conic Sections i= Archimedes, in Prop. 9. of his treatise on the Sphere and Cylinder where he demonstrates, that of all spherical segments under equa superficies, the hemisphere is the greatest: Serenus, in his 2nc Book, or that’on the Conic Sections :—Pappus, in many parts of hi Mathematical Collections ; as in Book U1, Prop. 28, &c., B. Vi Prop. 31, &c., where he treats of some curious cases of variabl geometrical quantities ; showing how some increase and decrease both ways, to infinity, while others proceed only one way, by in crease or decrease to infinity, and the other way to a certain magni tude, giving a Maximum and a Minimum; also B. VII, Props. 1g). 14, 165, 166, &c.' These are the principal writers on th geometrical Maxima and Minima among the ancients, to whic) may be added some others of a more recent date, as Richari White in his Hemispherium Dissectum, 4to. Rom. 1647. Viviani_ de Maimis et Minimis, &c. Fol. Flor. 1659.—Simon Lhuillier ¢ Geneva; Bossut; Le-Gendre; Dr. Hutton; Dr. Olinthe Gregory, &c. &e.” a, THE CONSTRUCTION OF GEOMETRICAL PROBLEMS. PROBLEM I. a ere ih a gwen triangle ABC, to wnscribe a square DEFN.* CONSTRUCTION. i | ‘From any point M, in aither side, upon the base IAB, let fall the perpendicular MG; make MR perpendicu- dar, and equal, to it, and é let ARE be drawn, meeting Ly the other side of the triangle in KE; thendraw ED parallel, A GNSS FB pod EF and DN perpendicular, to sina and the thing is done. ‘ DEMONSTRATION. Let RS be drawn parallel to EF: Then (by dies triangles) RS (MG): EF CG: AR: AE): 7 DE: Therefore, as MG and MR are equal, srg con- _ struction, EF and DE will likewise be equal. .| By the same method a rectangle may be inscribed ina triangle, whose sides shall be ina given ratio ; if MR and MG (instead of being equal) be taken in ‘the given ratio; the rest of the construction being | exactly the same. * prreidue Anal, Geom. B. II, Prop. ols 204 THE CONSTRUCTION OF The same constructed otherwise. Upon the base AB, from the vertex C, let fall the perpendicular CP, and make AM parallel to CP, and equal to AB; also draw PM, cutting the side AC in D, and draw DE parallel to AB, meet- ing BC in E; then upon AB let fall the perpendi- culars DG, EF, and the A thing zs done. DEMONSTRATION. It is evident, from the construction, that the figure DEFG, is a rectangle. And, by similar triangles AP: AM (AB):: PG (QD): DG; also (by 20. 4) AP: AB:: DQ: DE; therefore DG=DHE, and con- sequently DEFG isa square. Q. #. D. l Ss PROBLEM II. Inu given triangle asc, to inscribe a rectangle EFGH, equal to any given right-lined figure Q, not exceeding half the triangle. | CONSTRUCTION. On the base AB (by 7, 6.) leta rectangle ABPL be constituted =Q ; and let LP meet the perpendicular CD of the triangle (pro- duced) in K. Then (by 17. 5.) let CD be divided in I, so that ClIx DI=CD A xDK (that is, let two semi-circles be described on CD and CK; drawing MN and NI parallel to CD and AB): Soshall DI be the altitude of the required rectangle. | | ; | ; GEOMETRICAL PROBLEMS. DEMONSTRATION. : Since (by Constr.) C1x DI (=NI?=MD*)=CD x DK, thence will DI: DK :: CD: CI:: AB: EF ‘by 20. 5.); and consequently DI x EF (by 10, 4.)= ABxDK=Q. Q. E. D. fia kit Limitation. That the Problem will be impossible, when Q is greater than half the triangle, is evident from the Construction, as well as from the Theorem on p. 192. It may also be observed, that there is another way, besides that used above, for dividing ‘CD in the manner proposed ; which (though not more obvious) is in point of conciseness, rather preferable ; and is thus. | _, Having (as before) found a mean proportional DM between CD and DK, and bisected CD in O; from ‘M to CD apply MR=OD, and take OI= RD. So jshall Clx DI~OD’—OI? (Oy 7. 2)=MR’*—RD* (by Hyp.)= DM=CD x DK (as before). te PROBLEM IIL. i} . | In a given circle APBQ, to describe a rectangle equal to a given right-lined figure, RSTU, noe exceeding half the square of the diameter. : | CONSTRUCTION. Pp J ON V 1 ZN. B hg Tax R s Upon the diameter AB describe the - rectangle, | ABKI=RSTU (ay 7. 6.); and from the point C, 205 206 THE CONSTRUCTION OF where the side KI intersects the periphery of the circle, draw CA and CB, parallel to which draw BD and AD; then will. ACBD be the rectangle that was | to be constructed. DEMONSTRATION. The lines AC, BD, and AD, RC being parallel (by | Constr.) and the angle ACB a right one (by 13. 3.) | the figure ACBD isa rectangle (by Cor. to 24. 1.) and D is also in the circumference of the circle. — But ACBD=2ACB=ABKI=RSTU. | Limitation. ‘That the Problem will be impossible, when BK isc~1 AB, or when BI (RT) isc 3 AB?, is manifest from hence: because KI will then fall entirely | above the circle. | a PROBLEM IV. : To draw a line xt parallel to a gwen line AG, { which shall terminate in two other lines as, — AC, given by position, so as to form with’ them a triangle Ax, equal to a given rectangle ADEF. i CONSTRUCTION. ! 4 Let FE, produced, meet AG and AC, inG and H; and, in AB, take a mean propor- tional AK between GH and 2EF; then draw KL parallel to AG, and the thing is done. A D iK DEMONSTRATION. The triangles AKL, HGA being equiangular, it will — be AKL: HGA : : AK *(=GH x? EF, by Constr.) . : GH? 3): EF 4 i GH (by 7.4): :EFx AF: GHx AF (= HGA): * ‘Therefore, the consequents being | : equal, the antecedents AKL and EF x AF must also be equal, Q. £. D. A GEOMETRICAL PROBLEMS. | | PROBLEM V. “ . ele Jetween two lines aB, ac, given by position, to apply a line xx, equal to a gwen line MN, so that the triangle ax formed from thence, j ° . | shall be of a given magnitude.* | CONSTRUCTION. | re, A k : L M0 N/K S . Having bisected MN in D on MD describe a rect- nogle MDEF (dy 7. 6.)=the magnitude given: also nm MN let a segment of a circle be described (by 22. .) to contain an angle=A; and from its intersection vith EF, draw HM and HN; then make AK=HM, 1L=HN, and join K,L. So shall the base KL be Iso=the base MN, and the triangle AKL equal and ke in all respects, to HMN; which last (dy Cor. to . 2.) is manifestly, equal to the magnitude given IDEF. Q. £. D. PROBLEM VI. | . ° Through a given point Pp, to draw a line EPD to | meet two henes aB, ac, given by position, so | that the triangle ave formed from thence, | shall be of a given magnitude.t | _* D’Omerique, Anal. Geom. III, 36. Saunderson’s Algebra, [ rt. B09. Appendix to the Author's Algebra, prob. 33. _F Newton’s Arith. Universalis. Prob. 20 j~Leslie’s Geometry, age 305, 207 208 THE CONSTRUCTION OF CONSTRUCTION. : Draw FPH parallel to AB, intersecting AC in F; and on AF, let a pa- rallelogram AFHI be formed, equal to the given area of the triangle: Make IK _ perpendicular to AI, and equal to FP; and from kK, to AB, ) | apply KD=PH; then draw DPE, and the thing; : | done. DEMONSTRATION. The triangles PHM, PFE and MDI, by reason: the parallel lines, are similar; and therefore, since tt three homologous sides PH (KD), FP (IK) and L are such, as to form a right-angled triangle (dy Constr the triangle PHM on the first of them, is equal to bot the other two FPE and MDI (by 29. 4.): and, if{ these equal quantities, AFPMI be added; then wi AFHI bealso ADE. j Limitation. This problem will be impossible, whe KD (PH) is less that KI (PF); that is, when th area given is less than a parallelogram under AF an 2FP PROBLEM VIL. From a given polygon aBcpErn, to cut off part AIKFH, equal to a given rectangle MN bya line (1x), either parallel to a given lin AQ, or passing through a given point P.* ~~ | r | ax * Newton, Arith. Universalis, Cor. to Prob. 20. GEOMETRICAL PROBLEMS. | CONSTRUCTION. bon ON let a rectangle OQ be constituted (dy 7. 6.) (ual to AGFH; then, by Prob. the 4th, or 6th, ac- te triangle GKI= MQ, and the thing is done. The Demonstration is manifest from the Con- suction. | ‘And, in the same manner, the polygon may be «vided according to any given ratio; because, the \hole being given, each part will be given. PROBLEM VIII. } ) divide a given triangle ABC ito any proposed number of parts (AKM, KN, LC) so as to have any given proportion to each other ; ‘by means of lines drawn parallel to one of the sides Bc of the triangle. P bet BA, and EF be produced to meet in G; and (rding to the case proposed, draw IK, so as to make 209 210 THE CONSTRUCTION OF CONSTRUCTION, Let AB be divided into parts, AE, EF, ¥ B, hav- ing the same given pro- portion to each other, as the parts of the triangle are to have. Upon AB let a semi-circle AHIB be described; and per- pendicular to AB, draw EH, FI, meeting the cir- cumference in H and I: From the center A, | through H and I, describe the ares HK, IL, meeting | AB in K and L; then draw KM and LN parallel to. BC, and the thing is done, DEMONSTRATION. 4 The triangles AKM, ALN and ABC, are in pro- . portion, to one another, as AK? (ABx AE), AL? (AB x AF) and AB? (6y 19. and 24. 4.); that is, as AE, AF, and AB (6y 7. 4.) Whence (by divisitn) the proposition is manifest. | 2 ied | ee eee ‘" PROBLEM IX. : he To divide a given hne Pa into any proposed number of parts, so that similar. right-lined Jigures pMm, ML/, aq, described upon them, shall have the same given ratio among them- selves, as an equal number of right-lines AB, | AC, AD assigned. ‘ | CONSTRUCTION. Upon the greatest AD of the given lines AD, AC, Z a GEOMETRICAL PROBLEMS. AB, describe a semi-circle AEFD; and perpendicu- ar to AD, draw BE and CF, meeting the circumfer- snce in E and F; and, having drawn PR, at plea- ure, in it take PH=dist. AE, HI=dist. AF, and IK =AD; draw KQ, and parallel thereto draw HM, .L; which will divide PQ, as required. DEMONSTRATION. | PMm: LQg:: PM?: LQ? (dy 26. 4.) :: PH? (= AR*?=ABxAD, by 19. 4.): IK? (AD’*):: AB: AD by 7.6.) Inthe very same manner it appears, that MLI: LQg::AC: AD. Q. £. D. | - PROBLEM X. fo determine the position of a point P, so that | lines drawn from thence to the extremes of | three right-lines AB, CD, EF, given in length _ and position, shall form three triangles app, €PD, EPF, mutually equal to each other. CONSTRUCTION. _ Let the given lines be produced to meet in G and 1; in which take Gm=AB, Hs=EF, and Gn, Hz, ja " | fi esi] Vim G aa ator ie ee H jual each to CD: Complete the parallelograms impn, Hrts; and let the diagonals Gp and Hié be roduced till they meet in P, and the thing is done. P2 THE CONSTRUCTION OF DEMONSTRATION. Let PA, PB, Pm, &c. be drawn. The triangles GPn, GPm, having the same base GP, and equal al- titudes (because GMpn is a parallelogram) are there-_ fore equal to each other: But CPD=GPn, and APB =GPm, (by Cor. to 2. 2.); whence CPD and APB! are likewise equal. And, from the very same reason- ing, it appears, that CPD and FPE are equal. { PROBLEM XTr ae From two given points a, 8, to draw two lines) AC, BC, to meet in a line pu of any kind, given by position: so that the difference o their squares shall be equal to a square given (MtN*). | a) CONSTRUCTION. Make AF perpendicular to =p ™M the line AB, and equal to re MN; draw BF, which bi- sect with the perpendicular GH; and from its intersec- tion with AB, draw HC per- pendicular to AB, meeting DE in C; draw AC and “Af BC, and the thing is done. For BC? — AC*?= BH?— AH? (by 9, 2.)=FH’?—AH? =AP’=MN*%, . Ni yj . PROBLEM XIU. | From two given points a, B, to draw two ln AC, BC, to meet in a Line pK of any kine given by position; so that the sum of the squares shall be equal to a given square, MN» GEOMETRICAL PROBLEMS. CONSTRUCTION. | Bisect AB with the »yerpendicular FG, in which take —FP=FA, ind draw APQ=MN; im AB (produced, if ne- ‘essary ) let fall the per- yvendicular QR; from \ to FG apply AH= AR, and about the cen- er F, through H, let he circumference of a circle be described ; and from ts intersection © with the given line DE, draw CA und CB, and the thing ts done. DEMONSTRATION. Let BH be drawn; which being=AH=AR=RQ \because FP=AF); thence will, AC ?+BC ?=AH ” -BH? (by 20. 3.)=AR?+RQ?= AQ*=MN* d. EL. D. : | PROBLEM XIII. From two given potnts A, B, to draw two lines AC, BC, meeting in a line DE of any kind, _ given by position ; so as to obtain the ratio of two unequal right-lines Mm, Nn assigned.* CONSTRUCTION, ‘Having joined the civen points, divide AB . n F (6y 15. 5.) so that AF:BF::Mm: Na; nake Af and Fé parallel o each other, taking the ormer = AF, and the atter =F B; and through ' * See Pappus, Math. Coll. VII, 155 ;—D’Omerique, Anal. Geom. \II, 35 ;—the Author's Select Exercises, Prob. 19, the Appendix ohis Algebra, Lemma, page 334;—and Theo. B, on page 84 of his volume. 213 Q14 THE CONSTRUCTION OF their extremes draw fbO, meeting AB, produced, in O; from whence, with the radius OF, let the semi. circle FCG be described, cutting DE in C; then draw : AC and BC, and the thing is done. J DEMONSTRATION. fp | Because OA: Af (AF): : OF : Fd (FB), we have (by division) OA: OF : : OF: OB; or OA: OC:; OC: OB. ‘And so, the triangles OAC, OCB, having one angle FOC common, and the sides about it pro- portional, must, therefore, be similar (by 15. 4); whence the other sides will also be proportional, ov AC: BE ::0A:,0C, COP) .2 Af 12 to A (by Constr.) Q. E. D. Note, When, in either of the two preceding proba the circle described, neither cuts nor touches the given line DE, the thing proposed to be done, will be im- possible ; as no two lines drawn from A and B, to meet above the circumference, can possibly have their ratio, or the sum of their squares, the same as two lines meeting in the circumference. PROBLEM: XAILV: From two given pots A, B, to draw two lines) AC, BC, meeting in a right-line DE, grven by position; so as to make therewith two angle ACD, BCE, whose difference shall be equal to an angle given, bch. | CONSTRUCTION. : Make AFG perpendicu- f lar to DE, and FG=AF; ¢ Paee. and, having drawn GB, on it let a segment of a cir- cle GCB be described (by 22. 5.) to contain an angle > equal to the supplement (beg ) of the given one beh; A “B and from its intersection (C) with DE, draw CA sl CB; and the thing is done. , GEOMETRICAL PROBLEMS. _ For, if BC and GCH be drawn, then will ACD= aCD+=ECH=BCE-—BCH (bch). boFn the same manner, the Problem will be con- . structed, when, instead of the difference of ACD and BCE, that of ABC and BAC is given: Because, when DE is parallel to AB, the latter difference is equal to the former ; and, in all other cases, differs from it by wide the given angle GCI, expressing the inclination of the said lines. —When the sum of the angles ACD and BCE is given, the angle ACB is also given: And here, nothing more is necessary, than barely to des- cribe, upon AB, a segment of a circle to contain the said given angle ACB. | LEMMA. Jf, of any three proportional lines, AB, DB, FB, _ the difference av of the two extremes be bi- | sected wn G; and uf on the greatest AB, a@sS @ base, a triangle, asc be so formed, that vis less side xc shall be to the distance MG of the perpendicular from the bisecting pont G, in the given ratio of AB to vB; then shall the | greater side we exceed the less ac by the given line DB. DEMONSTRATION. ‘Because FM _ exceeds _ /AM by 2GM, BM wih © exceed it by 2GM-+BF; and the rectangle under ‘this excess and the whole . ‘base AB (=2MGxAB+ A MG D ER “BF x AB) will therefore (by 9. 2.) be=BC?—AC’. “But (by hypothesis and 10. 4.) MG x AB=AC x BD and BF x AB=BD?: Therefore 2AC x BD+BD *= “BC#—AC®*; and, by adding AC *, common, AC *+ -2ACxBD+BD? (or the square of AC+BD, by 6. | 2.) will be= BC *; and, consequently, AC+ BD=BC. (QQ. ELD. 215 216 THE CONSTRUCTION OF This Lemma is not only of use in the Problem next following, but will be found a ready instrument in the Solution of many others ; for which reason it is here put down. PROBLEM XV. From two given points a, B, to draw two lines AC, BC to meet ina right-line DE, given by position ; so that their difference shall be equal to a given right-line Bd.* CONSTRUCTION. In AB, take a third- DK C proportional BF to BA and Bd; and, having bi- sected AF in G, take GI =Bd; make GH and IK perpendicular to AB, meeting DE in H and K; and draw HAL, to which from K, apply KL=AB; and parallel thereto draw & AC, meeting DE in C; join B, C; and the thing ts done. DEMONSTRATION. Let CM be perpendicular to AB. Then, because of the parallel lines, it will be AC: KL (AB):: HC: HK::GM: GI (Bd.) Whence (by alternation) AC being to GM, as BA to Bd; and AB, dB, FB being also proportionals; the whole construction is manifest from the Lemma premised. — lf the sum, instead of the difference, of the two lines (AC, BC) be given, the method of construction will be exactly the same, without the least alteration of any one step; provided that Bd be first of all taken Cin BA, produced) equal to the given sum, instead o! the difference. MA ie ese MRP VN 3 Lf) SON: NL * Gregory St. Vincent, de Quad. Circ, prop. 82 ;—D’Omerique, Anal. Geom. III, 26. GEOMETRICAL PROBLEMS. PROBLEM XVI. “rom two given points, A, B, to draw two lines AC, BC, so as to meet in a right-line DE paral- | lel to that (AB) jouneng the said points, and | that the rectangle (ac x BC) contained by them, | shall be equai to a rectangle given, ADEF*. CONSTRUCTION. _ Bisect AB and AF, in x and H; and, having rawn GI perpendicular to .B, to it, from A, apply .O=AH; and from the enter O, through A and B, >t the circumference of a ircle be described inter- < ecting DE in C3; from A GH K BFE vyhence draw CA and CB, and the thing is done. DEMONSTRATION. For, if CK be drawn perpendicular to AB, it is \vident (by 25. 3.) that AC x BC=CK x 2AO0=AD KAF. Q. £. D. PROBLEM XVII. er es Through a given point P, so to draw a line ¥PE _ that the parts P¥, PE, ttercepted by that / point and two lines aB, CD, given in po- sition, shall obtain a given ratro.t | * D’Omerique, Anal. Geom, I, 31. Simp. Ezer. Prob. 21. + Apollonius, de Sectione, Rationis ;—Newton’s Principia, B. II, Lemma 7. CONSTRUCTION. Through P, to any | P point in AB, draw PG, in which (by 13. 5.) take PH to PG, in the given ratioof PF toPE; draw ¢@ HF parallel to AB, meet- jng CD in F; then draw FPE, and the thing is done. POLE | For, the triangles PGE, PFH being equiangulai (by 3. and 7. 2.) thence is PE: PE :: PH: PG : and so PF and PE (as wellas PH and PG) are in the rage given. n In this construction, it is necessary, that one of the two given lines should bea right-one: The other may, it is manifest, be a line of any kind whatever. 218 THE CONSTRUCTION OF . : PROBLEM XVIII. Through a given pornt Pp, to draw a line Gu te minating wna right-line aB, and in a line tp of any kind, given both by position; so that the rectangle under the two paris PG, PH, shall be of « a given magnitude.* es 4 CONSTRUCTION, Having drawn EPF per- pendicular to AB, take therein PF (dy 7, 6.) so that PE x PF =the magnitude given: Upon PF, as a dia- meter, let a circle be des- cribed, intersecting CD in G: then draw GPH, and the thing is done. AK Hoa A, Problem of Apollonius, de Seatione Spaiii. See Hal / Restitution of that work, Svo. Oxon. 170%, neg «| | GEOMETRICAL PROBLEMS, 219. DEMONSTRATION. If FG be drawn, the triangles PFG and PHE, aaving FPG=HPE, and FGP (=a right-angle, by 13. 3.)=PEH, will be equiangular: And, consequently ‘by 24. 3.) PGxPH=PF x PE = the magnitude given. Q. L. D. | PROBLEM XIX. Through a given point P, between two right-lines | ap, AC, given by position, so to draw a line ED, that the sum of the segments (Apd+ AE) cut off by it, from the two former, shall be equal to a given line rs.* CONSTRUCTION. | Draw PF and PG parallel to AB and \AC; in BA _ pro- duced take ~AM= AF, Ane MN AP : ‘then divide GN in I ‘ 'D (by 17. 5.) so that R—~— . id a > GDxND=AMxAG; draw DPE, and the thing ts done, m4 DEMONSTRATION. By similar triangles, GD : GP (=AF=AM): : FP (=AG): FE; and consequently GD x FE=AM xAG=GDxND (by Constr.): Whence FE=ND, and therefore FE+ AF+AD(AE+AD)=ND+AM +tAD=MN=RS. Q. #£. D. Limitation. It appears from the construction, that the problem will be impossible, when (GM) the excess ‘of RS above PF and PG, is less than the double of a _mean-proportional between these two quantities. * Select Exercises, probs 43. 220 THE CONSTRUCTION OF PROBLEM Xx. | Through a given point P, between two right-lines AB, AG, given by position, so to draw a line DE, that the difference (AD—ar) of the segments cut off by it from the two former, shall be equal to a given line rs. | CONSTRUCTION. Draw PF and PG pa- rallel to AB and AC; in * R AB take AM=AF, and . MN=RS; then (by 18. 5.) let a line ND be added to GN, so that GDx ND =PGx PF: draw DPE, and the thing is done. A. MG De DEMONSTRATION. | Because of the similar triangles PGD, EFP, we have GDxFE=PGxPF=GDxND (by Constr.) and consequently FE=ND; whence AD—AE=AN' —AF(AM)=MN=RS. Q. E. D. | PROBLEM XXI. Between two right-lines NBN, MBM, given by posvtion, to apply a line DE of a given length, : and which (produced, tf necessary,) shall pass through a given point a, equally distant from the sad given lines.* ; * Algebra, Appendix, prob. 72. GEOMETRICAL PROBLEMS. CONSTRUCTION. Let FG be the length ak iven: and, having ‘} rawn AB, make the ingles F, G equal each » ABD (or ABE); nd in AB produced ake AC (by 18. 5.) so gaat AC x BC=HG?: rom CG to NN apply SD=HG: then draw JAE (DEA), and the M hing is done. DEMONSTRATION. Because AC : CD:: CD: BC (by Constr. and 10. L.) the triangles CAD and CDB (having one angle sommon, and the sides about it proportional) will be squiangular (by 15. 4.): And therefore, since CDA= DBC=ABE (by hypothesis), the circumference of a circle may be described through all the four points, C, D, B, E’ (by 11. 3. or 19.3.) And so the angle DEC, standing on the same subtense (DC) with DBC, will be equal to it; and, consequently, equal to EDC. Therefore the isosceles triangles EDC, F GH being equiangular, and having CD=HG, their bases ED and FG will also be equal. Q. #. D. | PROBLEM XXIL ‘In a given circle aBpc, to apply a chord AB equal to a given line rs (less than the diameter) and which shall pass through a given point P. | 222 THE CONSTRUCTION. OF CONSTRUCTION. Inscribe CD=RS, upon which, from the center O, A let fall the perpendicular OF; also draw OP, upon which let a semi-circle be described, and in it apply OE=OF; lastly, through ,, P and E, let AB be drawn; Cc which will, manifestly be * , (es equal to CD (=RS, by 3.3.) RR as being equally distant from the center, by con: struction. When the point given is placed without the citble PROBLEM XXIII, ' the construction will be no ways different. { ; Through a given point P, so to draw a line AB. | that the parts Ap, BP, intercepted by tha pownt and the circumference of a given corel shall have a given difference, DE. 4 CONSTRUCTION. ag: From the center C, draw CP, upon which let a semi- circle PQC be deseribed, and A in it apply PQ, equal to half DE, producing the same, both | ways, to meet the circumfer- ence in A and B: so shall AQ = BQ (by 2. 3.); and therefore PB (=BQ+ PQ=AQ+PQ= AP+2PQ)=AP+DE, which was to be done. GEOMETRICAL PROBLEMS. PROBLEM XXIV. Lom a given point Pp, to the circumference of a given circle c, to draw a right-line PBA, so as to be divided in a given ratio by a line RBS of any kind given in position. | CONSTRUCTION. 'To any point D in the ecumference, draw PD, » viich divide at E in the rio given; and, having cawn PF and the radius (D, parallel to the latter, caw EF, meeting the frmer in F; from whence, t| RS, apply -FB=FE; S\ D ten through B draw PA, and the thing ts done. } ‘Let CA be drawn. Then, because of the pa- ilel lines CD, EF, it will be, CD (AC) : EF CB) :: PC: PF; and so the triangles PAC, PBF vill likewise be equiangular (by 16.4.) | And there- fre PB: BA :: PF: FC:: PE: ED (by 13.4.) De Sy ep v B , DEMONSTRATION. PROBLEM XXV. through a given pomt Pp, to draw a lhne GH, terminating in the circumference of 4a given circle BGA, and in a line MN of any kind, given by position; so that the rectangle under the two parts PG, pH, shall be of a given | magnilude. | 2! 3 224 THE CONSTRUCTION OF CONSTRUCTION. Through P draw the diameter AB, in which let PD be so taken (by 7. 6.) that PAx PD= the magnitude given. ; From any point E in the circumference, draw EP, and the radius EC ; and, having joined BE, draw DF parallel to BE, and from its intersection with PE, draw FO pa- rallel to EC, meeting » PB in O; from whence, to MN, apply OH=OF; | and through P draw HG for the line required. | “4 { DEMONSTRATION, | | Let PH be produced (if necessary) to meet the circumference of the circle in 1; and let C, I be. joined. | The lines OF, CE being parallel, thence will | PO: PC:: OF (OH): CE (CI); and therefore OH | and CI will likewise be parallels (6y 16. 4.) Therefore _ PH. :. Pl a7 Bower Get oP top hye eee Eas PBa whence, alternately, PH : PD :: PI: PB:: PA: PG (6y 21. 3. and LO. 4.) and consequently PH x PG=PD xPA. Q. £..D. : bd ¥ ¥ } | PROBLEM XXVI. From the circumference of a given circle c, toa line MN of any kind, given by position, so to | draw a right-line nr, as to be both equal and | parallel to a given right-line PQ. | GEOMETRICAL PROBLEMS. CONSTRUCTION. « From the center C, draw CD equal and Y parallel to PQ; and, / i from D to MN, apply ,; DF=the radius CB; | & draw CE equal and parallel to DF; then, N EF being drawn, it MEG 26.11) be «Ro eae aud parallel to CD ; which was to be done. | PROBLEM XXVII. From the point of intersection Pp of two given circles 0 and c, so to draw a line pr, that the part ar intercepted by the two peripheries, shall be equal to a given line aB.* | CONSTRUCTION. | Upon the line OC join- ng the two centers, leta emi-circle ODC be de- ‘leribed, in which apply )D=7AB; and parallel aereto draw PR, and the hing is done. DEMONSTRATION. : | Let CD and OF parallel to CD, be drawn, meet- ag PR in E and F. Then, the angle ODC being a tere oe gue. TS a EW Be : * A problem of Apollonius, de Inclinationibus ; See Horsley’s : ad Burrow’s Restitutions of that Treatise, 4to, 1770, and 1779. | Q 225 126 THE CONSTRUCTION OF right one (by 13. 3.) and PR parallel to DO, EK, and’ F will also be right-angles, and EF=DO (dy 24. 1.): And so, PE—PF being (=DO)=+AB, it is manifest, that 2PE (PR)—2PF (PQ)=AB, or that QR = AB. Q. HE. rey Rb 1 } PROBLEM XXVIII. t 4, From a given point vp, im the lne passing through the centers of two given circles c and o; so to draw a line pd, that the parts DE, FG, intercepted by those circles, shall be ma given ratio, (VIZ. as p as to q.)* = CONSTRUCTION. CH eee. ole eonae o P Take CH to OR,as p tog; and CI to OR, as PC to PO: Upon HI let a semi-circle be described, in tersecting the circle CDE in K ; through which poin draw IKL, and make CL perpendicular thereto ; ~@ which distance, from the center C, let a circle MLb described: Then, if from P a line PD be drawn t touch this circle, the thing zis done. a | :| DEMONSTRATION. i Let CD, CK, HK, FO be drawn, and also CM A ON, perpendicular to PD, and HQ to CL. ) The right-angled triangles CDM, CKL havin CD=CK, and CM=CL, have also DM=LK=Q) (since, by 13. 3. the angle HKI, as well as L, is, right-one). Moreover (by Construction) CI: 0) ay "RF | | t v * See Burrow’s Apollonius, Prob. 3. GEOMETRICAL PROBLEMS, YOF):: PC.: PO:: CM (CL): ON; and so h the triangles CIL, OFN (having their sides propor- tional) must be similar; and consequently OF N also similar to CHQ : whence, as QH (or DM): FN: : CH: FO (OR) :: p: q (by Construction.). Q. E. D. PROBLEM XXIX. f linemN passing through the common center 0, of two given circles MEN, KDF, so that the parts CD,:ED, ¢ntercepted by that line and the two peripheries, shall obiain a given ratio. CONSTRUCTION. _ Let QOM be the given angle, (0 Which ECM shall be equal : in OM produced, let KA be so ‘aken, as to be to the radius OK n the given ratio of ED to 2D; upon which let a segment of a circle ABK be described 0 contain an angle equal to 20M ; and from its intersection vith the circle MEN, draw 3A ; parallel to which, draw the adius OD; and then, through J; draw EC parallel to QO. DEMONSTRATION. Let BK, BO,. EO be drawn, and also EP parallel .o DO, meeting OQ and OM in Q and P. | Then, the triangles AKB, POQ, having ABK= 20Q (by Construction), and KAB=OPQ (by 7. 1.) jave their external angles OKB, EQO also equal (by 9. 1.): Therefore, because EQ (=OD)= OK, and KO=BO, thence is OQ=KB (by 17. 1.); ind therefore, as the triangles POQ, ABK are equl- | qQ 2 To draw a line xc, to make given angles with a 227 228 THE CONSTRUCTION OF angular,’ PQ is also=AK: But (because of the pa-: rallel lines) CD : ED: : DO (OK): PQ (AK), Q. E. D. rte = a OEE PROBLEM Xxx. To determine a point Pp, so that three perpend- culars drawn from thence, to as many right- lines AB, AC, BC, given by posttion, shall ob- tain the ratio of three given hnes m, n, and p respectively.* CONSTRUCTION. Take AE and BF each=™m, and draw EM and F} parallel to AB; in which take EG=n, and FH=p, through G and H draw AP and BP, and the pointo concourse P, will be that which was to be de termined. ! DEMONSTRATION. | Upon the sides of the triangle ABC let fall the per pendiculars GL, GK, PR, PQ, PS; and let GI, pa rallel to AC, be drawn. | * See Sir Isaac Newton’s Universal Arithmetic, Prob. 21 Edit, 1720. i. GEOMETRICAL PROBLEMS. ! The angle LEG being=LAK=GIK (Cor. |. éo 7. ..) and L and K being both right-angles (by Constr.), he triangles EGL and GIK are similar; and there- lore IG (m): EG (m):: GK: GL:: PQ: PR by 21. 4.). And, in the very same manner, mem: : PQ: PS. QF D. After the same way, the Problem may be con- tructed, when the lines drawn from P are required to qake any given angles with the lines upon which hey fall. | PROBLEM XXXI. To determine a point p, so that three lines Pa, PB, Pc, draw from thence to three given | points a, 8, C, shall obtain the rairo of three _ given lines a, b, and c, respectively.* CONSTRUCTION. Cc ——»—- =r —<— | Having joined the given points, in AB take AF =a, : nd Al=c; make the angles AFG and AIK equal, } | b * See Sir Isaac Newton's Arithmetic, Prob, 28.—App. to impson’s Algebra, Prob. 53.—Simpson’s Exercises, Prob. 56. Vieta, 'p. Math. pe 345, | | | | 229 230 ~ then draw BP to make the angle ABP=AHF, and it THE CONSTRUCTION OF | each, to ACB; and from the centers F and G, with the radii 6b and AK, let two ares be described, inter- | secting in H; from which point draw HF and HA; > will meet AH (produced).in the point P, required. | AM DEMONSTRATION. a Jaen Let BP, CP, and GH be drawn. y triangles ACP, AHG being equiangular (by 1d. A.) it will be AP: CP:: AG: GH (AK):: FA (a): al Cee a ilo - a f a oe. 4 PROBLEM XXXII. | To determine the position of a point Pp, at which, *) lines drawn from three given points a, B, © shall make given angles, one with another.* CONSTRUCTION. Draw AB, upon which (6y 22. 5.) let a segment of a. circle APB be described, capable of containing the given angle which the lines drawn from A and B are to include; and let the whole circle be completed; make the angle ABD equal to that which the lines drawn from A and C are to include; and * Simpson’s Algebyg, Appendix. Prob. 48, 49, Vieta, Opera page 345. | jee > GEOMETRICAL PRO wid i from the point D, where BD meets the circu mference, through ©, let DP be drawn, meeting the circumfer- ence in P; which is the point required. — \\ Bein) jr DEMONSTRATION. x . \ 2 > be=the given angle ABD (dy 11. 3.) both standi the same arch AD: And APB is also of the give magnitude by construction. hits PROBLEM XXXIII. Any two unequal segments as, cv of a rght- line av being given, as well in position as | length ; to determine a point p ina line MN ia of any kind given by position ; at which the ‘two angles apB, CPD, subtended by those segments, shall (if possible) be equat to each other. 7 CONSTRUCTION. Make Ba and De ™! G N . parallel to each other, | taking the former= BA, jand the latter=DC; and through their ex- tremes draw caO, meet- ing DA produced, in oe O: Take OF a mean- proportional between ~ é OB and OC; and from the center O, with the interval ‘OF, let a circle FG be described; which (when the | Problem is possible) will cut (or touch) MN, and the point of intersection P, will be ‘hat required. ad DEMONSTRATION. Let PO, PA, PB, PF, PC, and PD be drawh. _ Because OD: De (DC):: OB: Ba (BA) it will be (by division) OD:0C::0B:0OA; and conse- quently OD x OA=OCxOB=OP? (by Constr.) | : } | } ; | For, AP and BP being drawn, the angle ARC will 231 232 THE CONSTRUCTION OF Therefore, seeing thatOA: OP ::OP:OD, and thatthe © angle O is common to both the triangles OAP, OPD, these triangles must be equiangular (dy 15. 4.) and consequently APF=OPF (OFP)—OPA (ODP)= | DPF (by 9.1.). In the very same manner, because OB: OP :: OP: OG, the angle BPF will be=CPF; and, consequently, APB also=CPD. Q. £#. D. Bh PROBLEM XXXIV. Two right-lbnes ap, ac beng given, both m length and position; from the point of theor | concourse A, so to draw another line a1, that - two perpendiculars, 1p, iN, falling from its extreme 1, upon the two given lines, shall. cut off alternate segments BP, CN wn a given ratio, each, to the line At, so drawn. CONSTRUCTION. | Preah Nie C | GQ ——__— ‘i / { A gt ag Deel ag 3 Let the given ratio of Al, BP and CN be that of - the lines p, g and 7, respectively. Take BE=g, and CL — =?r; making BD, EG, CD and LG perpendicular to | AB and AC, soas to meetin Dand G: draw DA | and DG, and from G to AD, apply GH=p, and pa-— rallel thereto draw AI, meeting DG produced (if needful) in I, and the thing is done. a GEOMETRICAL PROBLEMS. 933 DEMONSTRATION. h h | cause of the parallel lines, AI: GH (:: ID: e ):: BP: BE; whence, alternately, Al: BP:: : BE (gq). In the very same manner, Al : cp 2m, Q. L. D. Amtech oS, } } | | | Zetween two lines AG, BH, gzven both in position | and length, to draw a line MN, which shall | be in a given ratio to each of the segments | me, NH, cut off from the two given lines. \ PROBLEM XXXvV. CONSTRUCTION. Let the given ratio of IN, MG and NH be aat of r, m and 7, respec- wwely: In AG and BH ike Gg=m, and Hh=n;. od, having drawn GH, arallel to it draw Al ; to thich, from g, apply gK =r; draw GKN, meeting \H in N, and parallel to ig draw MN, and the wing is done. | _ DEMONSTRATION. Because of the parallel lines, it will therefore e MN: GM:: gK (r): Gg (m;) and likewise fie ek (r) : : GN: GK: : NH: HA (2) 5 rhich last (by alternation) is MN : NH: : r: 2. 2. £ | | 234 THE CONSTRUCTION OF PROBLEM XXXVI.- To draw a line vr, to cut three other lines av AG, BC, given by position; so, that the be parts DE, EF, entercepted by those lines shall be respectively equal to two given line: de, ef.* : : | : CONSTRUCTION. Upon the right-lines df, ef let two segments circles, daf, ecf; be described, to contain an respectively equal to BAG and BCG: Then (¢ Prob. 27.) draw fa, so that the part ac, intercept by the two peripheries, shall be equal to AC; join) d, and take AF=af, AD=ad; then draw DF, the thing is done. | | DEMONSTRATION. 7 The triangles FAD, fad, having AF =af, AD =a and A=a (by Construction), are equal in all respect and therefore (if ce be drawn) the triangles FCE, fe having F=f, FCK=fce, and CF=cf, will also || equal and alike: Therefore, seeing the wholes DF, ; and the parts FE, fe are equal, the remaining pat DE and de, must likewise be equal. Q. E. D. | ae. 4 ~ * e > sz ‘7 * Newton’s Universal Arithmetic, Prob. 32, Edit. 1720, or Pre 22 of the other Editions ; Principia, Cor. to Lemma 26, B. I. GEOMETRICAL PROBLEMS. PROBLEM XXXVII. hrough @ given point Pp, to draw a line pp¥} to cut three lines ac, cB, ABb, given by po- | sition; so that the parts DE, EF wntercepted | by those lines, shall obtain a givén ratio.* ie CONSTRUCTION. a In CA produced, take ATT to CA in the given ratio » f£DEto EF; and, having joined B, T, draw PQ arallel thereto; and from its intersection with BC, raw QN parallel to CT; also draw AR parallel to 3C: And in RB take RH (dy 18. 5.) so that NH x > bee TR: Then draw AH, and DPEF parallel oit, and the thing is done. DEMONSTRATION. | Because (by Constr. and 10. 4.) NH : ER 92: RH:: QE: AR (by 14.4.) it follows (6y uternation) that NH: QE::TR:AR:?: TB: BC by; 12. 4.) :: BN: BQ: Therefore EH (when jrawn) will be parallel to QN or AD (by Cor. to 12. 4.); and so, DEHA being a parallelogram, we wave again (by similar triangles) as DE (AH): EF::IH:1E::AT:AC. .Q. £. D. ; _ When the segments AD, BE, cut off from the given limes AC, BC, are required to bein a given ratio ‘instead of DE and EF), the construction will be the same; provided that CT be taken to BC in the ratio given. For, then AD (EH): BE:: CT: BC. * See David Gregory’s Astronomy, B. V Prob. 8. 236 THE CONSTRUCTION OF PROBLEM XXXVIII. To draw a hne ascn to cut four other lines MONP, NQ, ROT, SMT, given by position; so that the three parts AB, BC, CD, intercepted by those lines, may obtain the ratio of three given lines m, n, and Pp» respectively .* CONSTRUCTION. co) From any point / in NP draw fbg parallel to TMS intersecting NQ in 6; and, having taken bg to bf, in the given ratio of n to m, draw N gk: Moreover, having taken OH to MO in the given ratio of n to Pp, draw THG; and from (G) its intersection with NE, draw GI and GF parallel to MP and ST, respectively, cutting NQ and TR in B and C; through which two points draw ABCD, and the thing is done. . Gi DEMONSTRATION, 4 imilar §4B : BC :: BF : BG::df:bg:: m:n; P wn eae oe BC : CD::CG: CI::OH:MO::n:p. Q. EH. D. a * See Sir. I. Newton’s Principia, Cor. to Lem. 27, B. I, and Prop. 41, B. I1l.+David'Gregory’s Astronomy, B. V. Prop. 11.— Playfair on Porisms ; and Leslie’s Geom, Analysis, Prop. 25. triangles GEOMETRICAL PROBLEMS. PROBLEM XXXIX. 0 draw two lines cA, cB, from a given pont C, | to terminate in two other lines PM, PN, given by position, and which, together with the line ' aB, jorning ther extremes, shall form a triangle, ABC, similar to a given one, abe. CONSTRUCTION. | Upon ab let a segment of a circle apb be described, ‘o contain an angle equal to MPN, and let the whole sircle be completed ; draw PC, and also ae, making phe angle bae= CPN, and intersecting the periphery ne; and throtigh c draw ep, meeting the periphery n p; make the angles PCA and PCB respectively aqual to pea, and pcb; then join A, B, and the thing is done. | DEMONSTRATION. If pa and pb be drawn; then will the angle bae (CPB)=cpb, both standing on the same arch be; therefore, APB being also=apb (6y Constr.), the re- maining angles APC and ape must consequently be equal; whence, as PCA=pca, and PCB=pcb (by Constr.), the triangles APC, ape, and BPC, bpe are equiangular; and therefore AC : ac (2° PC 2 pe) »:CB:cb. Andso the triangles ABC, abc, having the sides about the equal angles ACB, acb, propor- tional, are similar to each other (6y 15. A.) QED. } } 237 238 THE CONSTRUCTION OF PROBLEM XL. | Lo describe a triangle par, equal and similar, to a given triangle pqr, which shall have its angular points” placed in three right lines. ABC, BD, AE, guven by positeon.* | CONSTRUCTION. Upon the two sides pq, pr, let two segments of cir. cles pbg, par be described, to contain angles respec tively equal to CBD and CAE: Then draw pa (by Prob. 27.) so that the part ba, intercepted by the two peripheries, shall be equal to BA; and, hay- ing joined 6, g, and a, r, make BP=bp, BQ=bg, AR=ar, and let PQ, PR, and QR be drawn: for the sides of the triangle required. .| DEMONSTRATION. a The triangles PAR, par; PBQ, pbg, being equal and alike in all respects (by Constr.), not only the Sides PR, pr; PQ, pq, but the angles QPR, gpr, will be equal ; and, consequently, the two triangles PQR, pqr also equal and like to each other... Q. E. D. te eee * Principia, Lemma 26, Book I. THE CONSTRUCTION OF PROBLEM XLI. To describe a trapezium similar to a given one . efgh, having wts angular points placed in four right-lines CN, BM, AL, CRAK gzven by posetion.* 7 CONSTRUCTION. Let three points p, g, r be found, from whence, as centers, segments of circles may be described, upon 7, e8, eh, to contain angles equal to the three given angles KAL, KBM, KCN, respectively ; draw pq, \n which, produced (if necessary) take gs to gp: in she proportion of BC to AB; and, having drawn rst, make ec perpendicular thereto, intersecting the three circles in a, 6b, c. Take AK: AB::ae:ab; and make the angles CKH, CHG, and CEF, respectively equal to ceh, ceg, and cef ; then let H, G, and G, F, be joined; and, I say, the trapezium EFGH will be similar to the given one ée/gh. | | DEMONSTRATION. | Let af, bg, ch, be drawn, and also pw, qu, perpen- dicular to ec. So shall ab=eb —ea=2ev—2ew=2vw ; and be=ec—eb=2et—2ev=2vt: And therefore ab: be::vw:ivt:: pg: gs (by 13. 4.) :: AB: BC (by * Principia, Lemma 27, Book I, Const.) Again, seeing the triangles EAF, eaf; EBG, 239 240 THE CONSTRUCTION OF ebe; EKCH, ech, are respectively, equiangular (by Constr.) it will be EG:eg:: EB:eb:: AE: ae (h Constr. and Composition) :: EF : ef (by 14. 4.); anc so the triangles GEF, gef (having the sides about the equal angles proportional) are similar. And, in the very same manner, the two remaining triangles EGH. egh (and consequently the whole trapeziums EF GH efgh) will appear to be similar. Q. £. D. | PROBLEM XLII. | To describe the circumference of a circle through two given points A,B, which shall touch a right-line cD, given by posrtion.* 7 CONSTRUCTION. Draw AB, which bisect in Ex, by the perpendicular EF, meeting GD in F; from any point H in EF, draw HG perpen- dicular to CD; and, having drawn BF, to the same apply HIl= HG, and parallel there- | to draw BK, meeting EF in K ; then from the center K, with the radius BK, let a circle be described, and th thing ts done. | DEMONSTRATION. | ee | Join K, A, and draw KL perpendicular to CD; Then, because of the parallel lines, HG : HI: : KL: KB (6y 21. 4.); whence, as HG and HI are equal, F ) nh SS an sey ae I er | CE * This, and the seven Problems which immediately follow it, be- long to the celebrated work of Apollonius, “ de Tactiontbus,” now lost ; other constructions may be seen in Vieta’s Apollonius Gallus Prob. 2.; and in Lawson’s Tangencies, Prob. 7.—See also Newtons Arith. Univ. Prob. 38.—Simpson’s Algebra, App. Prob. 43, ané Leslie’s Geometry. p a GEOMETRICAL PROBLEMS. .L and KB are equal likewise: But it is evident, from ne construction, that KA is=KB; therefore KB= hKue=KA. Q. £.D. _ Limitation. Because two equal lines HI, Hz may le applied from H to BF (except when DIH isa ight-angle) the problem will therefore admit of two ylutions. But when a perpendicular let fall from H pon BF is greater than HG, the problem will be npossible. And the like is to be understood in the onstruction of the subsequent problems. | PROBLEM XLIII. 0 describe the circumference of a circle through a given point A, which shall touch two right- lines BC, BD, given by posrtion.® CONSTRUCTION. Gh ‘From the point of concourse B of the two given aes, draw BA; and also BN, to bisect the angle BD; from any point E in BN, upon BC, let fall 'e perpendicular EF, and to BA apply EG=EF, rallel to which draw AH, meeting BN in H; then é * Vieta, Apoll. Gall. Prob. 4.—Lawson’s Tangencies, Prob. 8.— ewton, Arith. Univ. Prob. 44.—Simpson’s Alg. App. Prob, 42.— uslie’s Geom. page 338. R 241 242 THE CONSTRUCTION OF from the center H, with the interval HA, let a cirele be described, and the thing is done. DEMONSTRATION. Upon BC and BD let fall the perpendiculars HI and HK; which are manifestly equal, because (dy Constr.) the angle HBI=HBK: Moreover, as EF and EG are equal, HI and HA are also equal (6y 21. 4.) Q. E. Dz , PROBLEM XLIV. To describe a circle, which shall touch a given circle sam, and two right-lines BC, BD, given by positeon.® » CONSTRUCTION, a 5 wie Q ‘Draw PQ parallel to BC, at the distance of th radius Aa; and through the point of concourse Bo the two given lines, draw NBP, bisecting the angl CBD, and meeting PQ in P; moreover, from an) point EF in PN, upon PQ, let fall the perpendicula, ‘ Aceh * Vieta’s Apollonius Gallus, Prob. 5.—Lawson’s Tangencie, ns ydipt’ Isaac Newton’s Algebra, Prob, 45,—Leslie’s Geometry GEOMETRICAL PROBLEMS. EF, and from the same point, to PA, apply EG= EF; draw AH parallel to EG, meeting the periphery of the given circle in M, and the right-line PN in H, ‘rom which last point, as a cenier, through M, describe she circumference of a circle; and the thing is done. | DEMONSTRATION. _ Draw HS perpendicular to BD, and HK to PQ, ‘ntersecting BC in I. _ Then, because EG and EF are equal (by Constr.) dA and HK (Oy 21. 4.) are likewise equal; from vhich take away KI=AM (Aq) and the remainders dI, HM will be equal: But, it is evident, that HI is =HS, because BH bisects the angle IBS; therefore JI=HM=HS. Q. EL. D. PROBLEM XLV. To describe the circumference of a circle through | two given points a, B, and which shall also touch another circle Odb, given in position and magnitude* CONSTRUCTION. Bisect the given distance AB with the perpendicular ‘Ei, in which (6y Prob. 15.) let the point C be so SEE SS 2 AS RES SSE Ss aS iy Vieta, Prob. 8.—Lawson, Prob. 12.—D’Omerique, I, 47.— .¢. Mathematician, Prob. 27.—Newton’s Algebra, Prob. 46.— Islie’s Geom. page 341.— Math. Comp. 1811.— Leybourn’s i dies Diaries, vol. iv. page 265.—Cameret’s Apollonius, where {+ Problem is divided into eight cases. Re 43 244 THE CONSTRUCTION OF determined, that CO (when drawn) shall exceed CA by the radius Od of the given circle. Then that point, it is manifest, will be the center of the circle to be described, PROBLEM XLVI. Through a given point a, to describe the cir- cumference of a circle, whach shall touch a given circle 8, and also a right-line PQ, given by position.* | CONSTRUCTION. Make AH perpendicular to PQ, and BD to AH; and having drawn AB, in it take BF a third propor- tional to AB and_ the radius Bd, and let AF be bisected in G: then draw MN (by Prob. 35.) so that MN, NH, and MG may oe be in the same given ratio, among themselves, as BD, BA, and Bd: and at M and N let two perpendiculars be erected on AB and AG; which will intersect each other in the center C of the required circle. ! DEMONSTRATION. Let CA and CB be drawn, and also CK perpendi- cular to PQ. Because MN : NH: : BD: BA (Oy Constr.) :: MN : AC (by 22. 4.), thence, is NH (CK) —AC. And since (by Constr.) NH (AC): MG?: BA : Bd, it is also manifest, from the Lemma on p. 215. that BC=AC+Bd. Q. #. D. * Vieta, Prob. 6.—Lawson, Prob. 10.—The Mathematician, Prob. 26.—Leslie, page 342 ;—and Camerer’s Apollonius, where he, distinguishes 15 cases of this Problem. or GEOMETRICAL PROBLEMS. * | PROBLEM XLVII. To describe a circle, oamb, which shall touch two given circles Aka, BED, and also a right- dine CD, given by posrtion.* | | CONSTRUCTION. | | ‘avs From the radius BF of the greater circle, take away (FA equal to the radius AE of the less; and from the center B, with the interval Bh, describe the circle ‘Bhv ; also draw PQ parallel to CD, at the distance of Fhor AE: Then, by the last Problem, let the center O of a circle be found, whose circumference shall pass ‘through A, and touch PQ, and Bhv; and the same point O will, likewise, be the center of the required ‘circle Oamb. DEMONSTRATION. | Draw On perpendicular to PQ, cutting CD in m; ‘also let OA and OB be drawn intersecting the circles Aand Bin aand v: Then, since (by Consir.) AO=vO =n0, and Aa (AE)=bv (Fh)=nm, let these last be respectively taken from the former, and there will re- main Oa=Ob=Om. Q. LE. D. ; | * Vieta, Prob. 7.—Lawson, Prob. 11.—Newton, Arith. Univ. Prob. 40. 245 46 THE CONSTRUCTION OF PROBLEM XLVIIL. To describe the circumference of a circle through a given point A, which shall touch two other circles B and C, given in position and magni- tude.* CONSTRUCTION. To the centers of the given circles, draw AB and AC ; in which take Bb=a third-proportional to AB, and the radius BE; and Cc=athird-proportional to AC and the radius CF: bisect Ad and Ac in G and H, and let Fr be drawn parallel to AB, meeting BCinr. Then (6yProb. 34.) draw AO, so that OM ' and ON being drawn per- 4 pendicular to AB and AC, the three lines AO, GM and HN shall have the same given ratio among them-. selves,as AB, BE and Fr. Then shall the point 0. be the center of the required circle. > 3 DEMONSTRATION. For, since (6y Constr.) AO: GM :: AB: BE; and AO >.HNwG@ vAB.: Fr).::-AG os. O13 ee manifest from the Lemma on p. 215, that OB=AO+ BE; and OC=AO+CF. Q. E. D. * Vieta, Prob. 9.—Lawson, Prob. 13.—Newton, Arith. Univ. Prob. 41.—Rob. Simson, Opera Reliqua, Appendix.— Math, Comp. 1811.—Correspondance sur l’Ecole Polytechnique, Tom. 1. et 2.—, Leybourn’s LZ. Diartes, 1V, App.—Camerer’s Lemmas to Apollonius de Tactionibus, App. Prob. 5, where he shows that this Problem admits of 40 different cases, or varieties, according to the different. pesitions of the given point, with respect to the two given circles, GEOMETRICAL PROBLEMS. SC€HOLIUM. In the very same manner,.a@ cirede may be described to touch three given circles ;* the Problem amounting tono more than, Yo jind a point from whence lines, drawn to three given points, shall have given t * See Vieta, Prob. 10.—Lawson, Prob. 14.—Newton, Princiia, Lemma 16, Book I.—Arith, Universalis, Prob. 47.—L’Hospital’s Conic Sections, Book X, Ex. 4. Cor. I. “ Vieta, in a dispute which he had with Adrian Romanus, a Geometrician of the Low Countries, proposed to him this Problem. The solution which Romanus gave of it, though obvious, was very indifferent, viz. by determining the center of the circle sought, in the point of intersection of two hyperbolas: for, as the Problem is _aplane one, it may be solved by plane geometry; by this, Vieta solved it, and very elegantly, in his Apollonius Galius, printed at Paris, 1600: his solution is the same as that given in Newton’s Universal Arithmetic. Another solution may be seen in Lemma 16, Book I. of the Principia. (this question being there necessary for some determinations in Physical Astronomy), where Newton, by _aremarkable dexterity, reduces the two solid loci of Romanus to _ the intersection of two straight lines.—Moreover, Descartes attempted to solve this Problem by the Algebraic Analysis, but without success; for, of the two solutions which he derived from thence, he himself acknowledges (see his Lettres, tom. iii, lett. 80, $1), that one furnished him with so complicated an expression, that he would not undertake to construct it in a month; while the other, though somewhat Jess complicated, was not so very simple f ee ‘as to encourage him to set about the construction of it.—Lastly, the Princess Elizabeth of Bohemia, who, it is well known, honoured » Descartes with her correspondence, deigned to communicate a . solution to this Philosopher ; but, as it is deduced from the algebraic calculus,’ it labours under the same inconveniences as that of Descartes.” —Montucla, Hist. Math. Tom. II. See also Frisi, Op. Math. Tom. I, Prob. 20, page 64, Milan, 1782. _ —Nova Acta Acad. Petrop. Tom. 6, p. 95, et seq. where three solu- tions of this problem are given, namely, one by Euler, which pro- ceeds wholly on the principles of Trigonometry, and two by Fuss, . the first of which is purely algebraical, and the second is an _ analytical process, conducted on trigonometrical principles. The following Problem was proposed by Descartes to Fermat : _ it is given here, because ofits similarity to the Apollonian Problem. ' Suppose four things to be given in position, consisting of Points, ' Planes, and Spheres, given in Magnitude, which may be taken of "any one of these kinds, exclusively, or of any two of the kinds, ' indifferently, or lastly, of all the three kinds, promiscuously : it is required to describe a Sphere which shall pass through each of the given Points, and also touch each of the given Planes or Spheres. —— ——— = = a _ 247 248 THE CONSTRUCTION, &c. differences::* Since a point so found, will always be the center of the required circle, as well when the three given circles are to touch that circle inwardly, as when they are all required to touch it outwardly. * Two very elegant solutions of this Problem, one by Algebra, the other a geometrical construction, may be seen in the Author's Select Exercises for Young Proficients in the Mathematics, 8v0, which is an excellent practical performance, perhaps; indeed, the very best in the English language, which can be put into the hands of young men who wish to acquire a correct and elegant taste, by the study of a masterly specimen of analytical investi gation. THE END OF THE GEOMETRICAL CONSTRUCTIONS, i vs 2 | }) ity ° A | \ : NOTES, GEOMETRICAL AND CRITICAL. et er es [AXIOM 10. Book 1. What is here laid down as an Axiom, would more properly have been made a proposi- ion, had it admitted of such a demonstration, as is perfectly consistent with geometrical strictness and purity. But the aying of one figure upon another, whatever evidence it may afford, is a mechanical consideration, and depends on no Postulate. _ Fheor. 4 and 5. Book 1. There is scarcely any thing nore obvious to sense, and at the same time more difh- cult to demonstrate, than the first and most simple proper- cies of parallel lines. Even when we have (in Theor. 4.) >broved the possibility of the existence of such lines, we cannot from thence infer, that their distance from one ano- her is every where the same ; without having recourse to an Axiom, which, though very evident to sense, cannot be demonstrated. These difficulties only arise from our not javing any properties previously demonstrated, by which the orogress of a right-line, produced out, can be traced, with vespect to its distance from some other right-line assigned ; nothing less being required here, than the proportionality of the sides of equiangular triangles, which is not proved vefore the middle of the Fourth Book, and which depends ipon these very principles. | Schol. to Theor. 5. Book1. As there are several condi- ons requisite to make up the definitions of a rectangle and -quare, it was necessary to show here, that the several pro- verties ascribed to those figures, are not incompatible one vith another. Euclid is very strict in this particular, and never undertakes to demonstrate any thing relating to a agure, till he has proved the possibility of the existence of such figure by an actual construction. Theor. 22. Book 1. This proposition, which is not in Euclid, is of considerable use, being often wanted in de- 249 250 NOTES, GEOMETRICAL AND CRITICAL, termining the Maxima and Minima, in Mathematic inquiries. | Theor. 27. Book 1. This Theorem, though not in Eucli¢ is also very useful, at least, to our design: by it we are ne only enabled to divide a right-line into any number of equi parts, without the help of proportions, but also to demor strate that most important proposition, That the homologon sides of equiangular triangles are proportional. It is true the method pursued here, is not exactly conformable to th idea of proportions delivered in the 6th Def. of Euclid: Sth Book, But, even in that light, the demonstratio. will be equally easy, without inferring it from the pro portionality of triangular spaces, ; Theor..1. Book 11. This proposition, which is not men tioned by Euclid, otha thought unnecessary ; but it muy either be demonstrated, : or assumed, as it is B EC A’) wanted in almost every proposition of the se- cond book: By means of it, we also arrive at a A. D general, and very easy b CONE a demonstration of that / \ important Theorem,— That all parallelograms {AEFD, aefd) which stand upon equal bases Q ba and have equal altitudes, are themselves equal. For, when) is known that these parallelograms AEFD, aefd, are equal respectively, to rectangles ABCD, abcd, of the same bas! and altitude (which is proved in Prop, 2.) it is also manifes that they must be equal to one another, as their equal rect angles ABCD, abcd are shown (by Theor. 1) to be equal the one to the other. This determination is more genera than that given by Euclid, inthe 36th Prop. of his first book where he demonstrates the equality of parallelograms, whos equal bases are inthe same straightline: Which may, perhaps be thought sufficient for the whole ; because, if the basesar not in the same right-line, one of the two figures may b it also follows, Ava; ;A:a that A? : a3;: BCA: bca; or, that similar pa- 255 256 ‘advance on that subject. NOTES, GEOMETRICAL AND CRITICAL. rallelepipedons are to one another, as thecubes of their homo- logous sides. —The proportionality of similar parallelepipe. dons, described upon proportional lines, is also included in the same Theorem; being no other than that case of it, where the proposed ratios are all equal. si Theor. 8. Book Vill. The demonstration of this Theorem might have been delivered under a form somewhat different, by assuming two other solids (without regard to figure), the one less, and the other greater than the proposed parallele. pipedon IP, and proving that the cylinder must, also, be greater than the one, and less than the other: which is done by means of Lem. J. that is, by taking Pm, or Pt, so small a part of IP, as to be less than the difference between the given parallelepipedon, and either of the said solids: from whence the demonstration will proceed on, in the same man- ner in which we have given it. But, as these additional considerations would have increased the number of schemes, and lengthened the process, without adding one jot to the degree of evidence, 1t was thought proper, for the sake of the learner, to omit them. A Theor. 8. Book VIII. This Theorem is not so useful as the Corollaries that follow from it, which are all of very great importance: In the 3rd and 4th of them, the propor. tion of all kinds of prisms and pyramids is assigned, with- out the assistance of the usual demonstrations given for this purpose; which, though sufficiently evident in themselves, are often found a little perplexing to learners, on account of the schemes, wherein so great a number of lines is necessary. Observations on the Doctrine of Proportion, — Nothing, in the course of these notes, has been said rela- tive to the Theorems ou Proportions, though so nice and cri- tical a subject, and though the method I have therein pur: sued may stand in need of some apology. But, indeed, the whole of what I have to offer on this head, was too much to be comprised in the compass of one single note, and could not so properly be delivered in several detached ones: For which reason, I shall here throw together all that I have to There are two objections that may be brought against the method in which proportions are treated of in this work; the one, grounded on the impossibility of dividing every mag- nitude into equal parts; and the other, on the incommen- surability of two or nore magnitudes of the same kind, when compared with each other, The first of these objections ap- NOTES, GEOMETRICAL AND CRITICAL, ears, to me, to have very little weight. For, though a mag'ni- ‘udé may be so constituted, that the division of it into an ‘ssigned number of equal parts cannot be, actually, effected sy any geometrical construction; yet it is no less evident, or that reasons that every such magnitude has really its ‘bird, fourth, or other assigned part, though we are at a loss low to take it; or, in other words, it seems very clear to ‘onceive, that in every proposed magnitude, whatever its ‘igure may be, a less magnitude is contained, which, repeated ‘n assigned number of times, shall be equak to the magni- ude given. If, as the most rigid judges allow, every plane /gure is equal to some square, and every solid equal to some arallelepipedon ; then the parts of the square, or parallele- ‘ipedon, which are actually determimable by a geometrical onstruction, will also be lke parts of the figure first pro- ‘osed, and such as we conceive to be taken. : ' The other objection, depending on the incommerisurability f magnitudes, is a matter of real difficulty, which we have ken some pains to obviate in the Scholiato our 3rd and 7th: heorems. uclid, himself, seems to have been not a little ubarrassed with it, if we may be allowed to judge from the ifferent methods he has left usin his 5th and 7th books : he former whereof, which is suited to include the business “incommensurables, being nothing near so easy and na- iral as the latter. It has, it is true, the advantage of being eneral ; but, that the principles whereon it is grounded are either so simple, nor so evident as might be wished for, € many disputes about them, since Ewclid’s time, by someters of the first rank, will, im a great measure, evince. ud farther, it seems sufficiently plain, from Euclid’s own athority, that he himself was not entirely pleased with his va performance on this head; or that he was convinced, at ast, that it had not every advantage: For, otherwise, it ll be very difficult to account for bis having demonstrated jany things in his 7th book, by another method, whose de= puvestions had been actually given before, in the 5th, ader a different ferm. For these reasons, when I see the ‘travagant commendations that have been lavished on this ‘2 book of Euclid, I am no farther convinced by them, (an that great men may sometimes launch out too far in Lhalf of opinions which they have adopted. And I believe tat, whoever has read the notes on the 5th book, by that feat restorer of Kuclid, Professor Simson, will be apt to Caclude, that those high encomiums are a little misapplied. Iieed, if all that is advanced in those notes be allowed of, Think the author of them has proved too much; and this S 208 NOTES, GEOMETRICAL AND CRITICAL. superb fabric of proportions, reared with so much art, stand upon a tottering foundation. Itis not by choice that I ge out of my way to play the critic ; but as the writers agains the vulgar and indistinct notion of proportions (as they tern it) are very severe in their censures, and assume a grea superiority, from the boasted accuracy of their reasonings it may be necessary to show my reader, that, though wha he is here taught on proportions, is liable to some objections the method which some so greatly prefer, has also its diffi culties; and that there are other objections to it beside its obscurity. And this I shall make appear from th learned commentator’s own authority and concessions ; an ‘1 order thereto, shall first refer to his note on Prop. 1 which proceeds thus: “ It was necessary to give another de < monstration of this proposition, because that which is 7 << the Greek and Latin, or in other editions, is not legit «© mate. For the words greater, the same, or equal, lesse ‘shave a quite different meaning when applied to magn <¢ tudes and ratios, as is plain from the 5th and 7th defin <‘ tions of Book 53; by help of these, let us examine tl <¢ demonstration of the 10th Prop. which proceeds thus, &e, He then goes on in a long note, to show the insufficiency a demonstration, which had been received, by all, as pe fectly genuine and satisfactory ; and at last comes to th conclusion: ‘ Wherefore the 10th Proposition is not suf <¢ ciently demonstrated. And it seems that he who has giv « the demonstration of the 10th Proposition, as we now ha “it, instead of that which Euclid or Eudoxus had give << has been deceived in applying what is manifest, when wW << derstood of magnitudes, unto ratios, véz. that a magnitu «< cannot be both greater and less than another. That tho ‘‘ things which are equal to the same are equal to one an ‘“‘ ther, is a most evident Axiom when understood of ne “ tudes, yet Euclid does not make use of it to infer t ‘6 these ratios which are the same to the same ratio, are ¥ <¢ same to one another; but explicitly demonstrates this “ Prop. 11 of Book 5. The demonstration we have giv < of the 10th Prop. is no doubt the same with that of E << dovus or Euclid, as it is immediately and directly deriv “ from the definition of a greater ratio, viz. 7th of 5.” — Here the weight of the objection rests ‘on its not havi been proved, that, of three given magnitudes A, B, C, 4 ratio of A to € could not, at the same time, de both grea and less than that of B to C. But if, in the demonstrati¢ here rejected as insufficient, there 1s any real flaw, it chargeable on the definition of a greater and less’ rat NOTES, GEOMETRICAL AND CRITICAL. s the reasoning from it is clear, strong, and perfectly gientific. And I would seriously ask the contemners f the vulgar and confused notion of proportions, if a jefinition, by which it cannot be known, whether the atio of the first to the second of four given magnitudes, nay not, at the same time, be both greater and less than nat of the third to the fourth, is really calculated to afford jose very accurate ideas they pretended to? This com- hentator has too much penetration not to be aware of the orce of this objection, which he has attempted to obviate |. one particular case. But the new preposition given by um, for that purpose, ought to have preceded the 10th, and ( : : P have been demonstrated, independent of it. This he also’ t ems apprized of, when he says, that ‘it cannot be easily | demonstrated without the 10th, as he that tries to do it will ) find.” But, be that as it will, Iam not at all clear that ys $8 demonstration of the 10th, is the same with that of Ludoxus or Euclid.” Euclid, or (if you please) Eudoxus, yes never (that I know of) refer to any detinition, till it has ven proved, either by an actual construction, or by some de- jonstration previous to that in hand, that such definition ivolves no absurdity or conditions that are incompatible one ith another. If, therefore, it was conceived possible, that definition of a greater and less ratio could involve so eat an absurdity, as that, by it, the ratio of A to C might the same time be both greater and less than that of B to 5 this point, according to the method prescribed by Euclid, ght to have been cleared up, not by means of propositions rived in virtue of that very definition, but by others ante- lent thereto, and independent thereupon. And, to me, the 1 Prop. seems the proper place for the doing of this, where lmight be easily introduced, either in the Prop. itself, or I way of Corollary. It is there proved, that if, of three ‘gnitudes A, B, C, the first A is greater than the second then certain equimultiples of A and B may be taken sh, that being compared with some multiple of C, the mul- de of A shall be greater, and that of B less than the said ‘tiple of C. Whence, by the definition of a greater ratio, ratio of A to C is greater than that of Bto C. To which azht be added—And because A is greater than B, any Meee whatsoever of A must be greater than the same itiple of B; and, consequently, no equimultiples what- ee of A and B, can possibly be so taken, that the mul- ‘Le of A shal) be equal to, or less than some multiple of and that of B greater than the multiple of C: for, if the iltiple of B be greater than the multiple of C, the multiple g 9 ~~ 259 260 NOTES, GEOMETRICAL AND CRITICAL, of A, which is greater than that of B, must also be greater than the multiple of C. Wherefore the ratio of A to C cam not (by the definition) be less (as well as greater) than the ratio of B to C. “y But, notwithstanding all that has been proved on this” head, either here, or by that gentleman himself, the same objection occurs again in Prop, 13, where it remains in its full force. For, though it be allowed, that “* there are some. « equimultiples of C and E, and some of D and F, such, ‘ that the multiple of C is greater than the multiple of D, << but the multiple of E not greater than the multiple of Fy” yet it is not demonstrated, nor in any sort shown, that other equimultiples of those quantities cannot be taken such, that the very contrary shall happen.—If the demonstration of the 10th Prop. has been justly rejected by this gentleman him- self, as insufficient, because the impossibility of a contrary conclusion had not been shown; can it be thought that this: 18th Proposition is, at this day, sufficiently demonstrated, where the same objection occurs, and that in a much greater latitude? Ihave a much better opinion of this editor’s dis- cernment than to imagine, that his passing this matter over in silence proceeded from his not being aware of the diffi- culty ; but it seems to me, that his great dislike to the vulgar idea of proportion (so often testified in the course of; his notes) would not permit him to borrow any thing from thence, however evident, and though this objection, that strikes deep at the very root of proportions, might by meals, thereof be very easily removed. I say, the very root of pro- portions is deeply struck at in this objection ; because both the alternation and equality of ratios (ex equali sc. dist.) are grounded on the said 13th Prop. and which, therefore, ‘till the objection is removed, must be allowed to stand upon an uncertain foundation. es The principle hinted at above, whereby the difficulty might be obviated, is, that if a magnitude of any kind be given, or propounded, there may (or can) be another mag- nitude of the same kind which shall have to it any ratio assigned. This assumption Dr. Simson will by no mean: admit of (though Euclid himself, in Prop. 2. of his 12th book, has used it); and, in along note on Prop. 18. is angry with Clavius for having recourse to it ; affirming, ‘* That the ‘‘ demonstration (given by means thereof) is of no force ;” anc that “the thing itself cannot (as far as he can discern) bt «demonstrated by the preceding propositions, so far i817 ‘< from deserving to be reckoned an Axiom, as Clavius, afte ‘s other Commentators, would have it.” That the assump: NOTES, GEOMETRICAL AND CRITICAL. tion cannot be generally demonstrated by the preceding pro- positions (nor even by all the propositions in the Elements) I readily assent to; but then, because a thing, exceedingly obvious in its own nature, cannot be demonstrated, is it ‘therefore less proper for an Axiom? [I should rather take ‘the other side of the question, and maintain that nothing ought to be madean Axiom, which can be demonstrated. ‘But we are not, it seems, allowed to have any idea of pro- portion but what is contained in the 6th and 8th (or, as this author makes them, the 5th and 7th) definitions of Zuclid’s 5th book. And, in his note on the new Prop. marked A, he is again displeased with Clavius, for thinking it suf- ciently evident, from the nature of proportionals, that if, bf four proportional magnitudes, the first antecedent is rreater than its consequent, the second antecedent will also de greater than its consequent. <‘* As if there was (says he) “any nature of proportionals antecedent to that which is to be derived and understood from the definition of them.” Now I cannot help thinking, with Clavius, that there was a hature, or idea of proportion antecedent to that given in the 3th and. 8th definitions of Euclid’s 5th book: For, that mankind, long before the time of Euclid, bad some way to show or express in what degree one magnitude was greater or less than another, cannot be doubted: And this was the irst and natural idea of proportion: And I look upon those lefinitions, as refinements, only, on the simple and natural dea, in order to take in the business of incommensurables ; vhereby the original notion is so much obscured, that it ‘equires some skill, even to see that it is at all contained in hese definitions. I entirely agree with this gentleman, that ‘very demonstration ought to be strictly derived from prin- ‘iples before established : But then, whether is it more eligi- ale, to have recourse to an Axiom founded (as all other Axioms are) on the evidence of sense and reason, or to an sbscure and perplexed definition, which may, for any hing that has been proved to the contrary, involve an ‘bsurdity ? _ That there is something very ingenious and subtle in the loctrine of proportions, as delivered in Euclid’s Sth book, annot be denied. All that I contend for, is, that the prin- FR i ) iples on which it is built are obscure, and not so firmly stablished, as to authorize its partizans to assume that ‘reat superiority they lay claim to, in point of geometrical trictness. _ [have intimated above, that the principle is rejected, by hich the consistence of the definition of a greater and less itio might be established, without much difficulty: But I ) NOTES, GEOMETRICAL AND CRITICAL. would not be thought to mean, that the same thing cannot possibly be effected any other way, because I am satisfied that it may be done from the consideration of multiples alone: But a demonstration of this sort is not easy.—Were I to treat of proportions from the plan laid down in the 5th book of Euclid, I would entirely reject the 10th and 13th propositions, and every thing else founded on the definition of a greater and less ratio, as being of no other use in the Elements, than to open the way to those important Theorems onthe alternation and equality of ratios, which may be better demonstrated without them, from the definition of equal ratios alone; which, from the conditions of it, can admit of no absurdity ; and whose consistence is evinced in Prop. 16, and still more clearly in the first of the sixth. In the 14th of the 5th, whereon the alternation of ratios is grounded, itis necessary to demonstrate, ‘* That if, of four “* proportional magnitudes, of the same kind, A, B, C, D, “the first be greater than the third, the second shall be “* greater than the fourth; and if equal, equal; and if less, “less.” Which may be very easily done, independent both of the 10th and 13th, in the manner following : un First, let A be greater than C, bela yest 3 Vyabllng C Ba all ' LD) Boks Go.» ola Of A and C (by Prop. 8.) let such equimultiples be taken, that the multiple of A shall be greater, and. that of C less, than some multiple of B; let Eand G be any two such equimultiples of A and C, and F the multiple of B; so that E shall be greater than F, and G less than F; and let H be the same multiple of D, as F is of B, Therefore, because E and G are equimultiples of the first and third, and F and H also equimultiples of the second and fourth; and seeing that (by Hyp.) E is greater than F ; it is evident, from the definition of equal ratios, that G must likewise be greater than H: Therefore much more shall F (which ex- ceeds G) be greater than H; whence also B shall be greater than D (6y Ax. 4.) B and D being like parts of F andH, When A is less than C, it will be demonstrated in the Same manner, that B is also less than D. But when A is equal to C, no new demonstration is necessary ; since neither the 10th nor the 13th have any thing to do in this case, | Again, in the 20th Prop. (in which the 10th and 13th also enter) we are to prove, “ That if there be three mag- *‘nitudes (A, B, C) and other three (D, E, F) which taken NOTES, GEOMETRICAL AND CRITICAL, two and two have the same ratio (A: B: : D: E, B: C:: “E: F) if the first (A) be greater than the third (C), the «fourth (D) shall be greater than the sixth (Ff); and isif equal, equal; and if less, less.”? Which may likewise be done without the assistance of either the 10th or the 13th, in the same manner, above specified. _ For if A be greater than C, then of A and C (dy Prop. 8.) such equimultiples may be taken, that the multiple of A shall be greater, and that of C less, thansome multiple of B; let G and I be two such equimultiples of A and C, t = | ) G ' ish th ) A. B : Cc , | D E \ yd. q K L M and let H be the multiple of B, so that G shall be greater than H, and I less than H ; moreover take L the same mul- tiple of E, as H is of B; and K and M the same equimul- tiples of D and F, as Gand Lare of A and C. Therefore, since of the four proportionals A, B, D, E, equimultiples G, K of the first and third, and equimultiples H, L of the second and fourth, are here taken, it is manifest, from the definition of equal ratios, seeing G is greater than H (by Hyp.) that K must also be greater than L, And in the very same manner, because B, C, E, F are proportionals, and H is ‘greater than I, L will likewise be greater than M. Therefore much more shall K, which exceeds L, be greaterthan M. And ‘consequently (by Ax. 4.) D shall also be greater than F. ‘When A is less than C, the demonstration is the same : "The other case, when A is equal to C, does not require, nor indeed admit, of any improvement. q | ee Note to Problem XXXVIII. As the history of this problem is not a little curious, the Editor presumes that the following account of it may be ac- ceptable. . | 66 Sir Isaac Newton haying demonstrated that the trajec- ‘tory of a comet isa parabola, reduced the actual determi- nation of the orbit of any particular comet, to the solution of a geometrical problem, depending on the properties of the ‘parabola, but of such considerable difficulty, that it 1s ne~ ‘cessary to take the assistance of a more elementary problem, 263 264 NOTES, GEOMETRICAL AND CRITICAL. ain order to find, at least nearly, the distance of the comet from the earth, at the times when it was observed. The ex- pedient for this purpose, suggested by Newton himself, was to consider a small part of the comet’s path as rectilineal, and_ described with a uniform motion, so that four observations of the comet being made at moderate intervals of time from one another, four straight lines would be determined, viz, | the four lines joining the places of the earth and comet, at the times of observation, across which, if a straight line were drawn, so as to be cut by them into three parts, in the same ratios with the intervals of time above mentioned, the line so drawn would nearly represent the comet’s path, and | by its intersection with the given lines, would determine, at least, nearly the distances of the comet from the earth, at the | times of observation. _ The geometrical problem here employed, of drawing a. line to be divided by four other lines given in position, into~ parts having given ratios to one another, had been already | solved, by Dr. Wallis and Sir Christopher Wren, and to. their solutions Sir Isaac Newton added three others of his | own, in different parts of his works; yet none of these. geometers observed that peculiarity in the problem which | rendered it inapplicable to astronomy ; this was first done , by M. Boscovich, but not till after many trials, when, on its . application to the motion of comets, it had never Jed to any : satisfactory result : the errors it produced, in some instances, | were so considerable, that Zanotti, seeking to determine’ by | it, the orbit of the comet of 1739, found that his construce ! tion threw the comet on the side of the sun opposite to that 3 on which he had actually observed it... This gave occasion to Boscovich, some years afterwards, to examine the different — cases of the problem, and to remark, that, by a curious | coincidence, this happened in the only case which could be - supposed applicable to the astronomical problem above — mentioned; in other words, he found, that, in the state of the data which must there always take place, innumerable | lines might be drawn, that would be all cut in the same | ratio, by the four lines given in position. This he demon- strated ina dissertation published at Rome, in 1749, and since that time in the third volume of his Opuscula. A de- monstration of it, by the same author, is also inserted at the end of Castilleoneus’s Commentary of the Arithmetica - Universalis, where it is deduced froma construction of the _ general problem, above given in the text, by Mr. Simpson.— See the late Professor Playfuir’s ingenious paper “ On the | Origin and Investigation of Porisms,’” Edinb, Trans. Vol. HI. Epiror, 4 Rs Bs — ON MATHEMATICAL ANALYSIS, &c. | APPENDIX. On the Analytical and Synthetical Modes of Reasoning | made use of in the Mathematics. | | There are two methods of reasoning made use of in the ‘mathematical sciences—the synthetical and the analytical. ‘By referring back to the Greek origin of the words syn- thesis and analysis, we shall find that one signifies compo- sition, and the other resolution, or decomposition : nothing appears clearer, at the first glance, than these denominations, and we easily conceive, that the methods they denote, are the reverse, one of the other. | The Elements of Euclid are according to the synthetic method. That author having formed certain axioms, and made certain demands, advances his propositions, which he ‘proves by means of what precedes, and thus continually passes from simple to composite, which is an essential cha- racter of synthesis. | | The first usage of analysis, in geometrical researches, is attributed to Plato. By this method, we suppose the pro- »osed problem is already solved, from which it results, that »ne condition is fulfilled, or, what comes to the same, that in equality takes place among certain magnitudes given, and others sought. It is by finding out the consequences of the condition which we supposed fulfilled, or of the equa- ‘ity which is the consequence of it, that, in the end, we dis- cover the unknown quantity, or trace the proceeding which ‘tis necessary to follow, in order to perform what is de- ‘manded. - The following are the definitions which Vieta has given of analysis and synthesis, after Theon, a geometrician of Alex- indria, who, from his living so much nearer the times of the incients, was better able to judge of their opinions than we ire. . | Est veritatis inquirendz via queedam in mathematicis, ‘€ quam Plato primus invenisse dicitur, a Theone nominata ‘analysis, et ab eodem definita, adsumptio quesiti tam- ‘quam concessi per conséequentia ad verum concessum. “Ut contra synthesis, adsumptio concessi per conse- 26 w 2 266 ON MATHEMATICAL <‘ quentia ad quesiti:sinem et comprehensionem.”’—Vieta Opera, p. 1. There is likewise a definition of analysis and synthesis given | in the preface to the 7th Book to Pappus’s Mathematical. Collections, which is as follows :— | ‘«* Analysis is the way, which, proceeding from the thing «¢ demanded, arrives, by means of certain established con- <¢ sequences, to somewhat known before, or placed among «‘ the number of principles acknowledged as true. This «¢ method makes us go from a truth, or proposition, through. <‘ all its antecedents; and we call it analysis or resolution, “<* or an inverted solution, In synthesis, on the contrary, we. <¢ begin from the proposition last found in analysis, ordering’ ‘© properly the above antecedents, which now present them-, *¢ selves as consequents, and by combining them among. << themselves, we arrive at the conclusion sought, from which. “« we proceeded in the first case. a ‘© We distinguish two sorts of analysis: in one, which, <‘ may be named contemplative, it is proposed to discover. ‘‘ the truth or falsity of an advanced proposition ; the other, *< is related to the solution of problems, or the research of, <¢ unknown truths. 4 ‘* In the first, by assuming as true, or as formerly known, 6 the subject of the advanced proposition, we proceed, by, <‘ the consequences of the hypothesis, to something known; *‘ and, if the result is true, the proposition advanced is like- «« wise true. The direct demonstration is lastly formed, by. “‘ taking, in an inverse order, the different parts of the ana- ‘lysis. If the consequence to which we arrive at last, is << found to be false, we conclude that the proposition ana-. *¢ lysed is likewise false. : ‘© When we have to solve a problem, we suppose it to be “¢ already known, and we proceed upon this supposition till, << we arrive at something that is really known. If the last. ¢ result which we can obtain, is comprised in the number of “‘ what geometricians call known truths, the question pro- “© posed can be resolved. The demonstration (or more pro- © perly the construction) is then formed, by taking, in an «* inverse order, the different parts of the analysis. The im- ‘s¢ possibility of the last result of the analysis, will prove evi- ‘* dently, in this case, as in the preceding one, that of. the “ thing demanded. | ** There is likewise, in the solution of every problem,, “< what is called the Diorism or Determination, that is to, “‘ say, the reasonmg by which we show when, in what man- ANALYSIS AND SYNTHESIS. ‘se ner, and how many different ways, the problem can be * solved.” ' The demonstration of Theorems, in the manner called veductio ad absurdum, is, properly speaking, an analytical proceeding ; for we suppose the converse proposition is true, ‘and, by seeking consequences which are absurd, make it ‘appear that the hypothesis is so likewise. | The characteristics of synthesis and analysis, it is hoped, ‘are rendered tolerably clear by the preceding description. In the first method, the proposition enunciated is always the last consequence of the chain of reasonings which forms the ‘demonstration: it is a composition; for we add principle ‘continually to principle, until we come to this consequence. ' In analysis, on the contrary, by supposing the question lpesolved, we take in the whole of the subject, and, making it pass with different forms, or making divers traductions of the enunciations, we come to the svlutions sought. ' But, in the solution of problems, before we begin our ‘analysis, some previous preparation, or mental contrivance is commonly necessary, in order to form a connexion be- ‘tween the data and quesita, which cannot be brought within any regular rules, being various, according to the na- ‘ture of the problems proposed. Thus, right-lines must ‘sometimes be drawn in particular directions, or of parti- cular magnitudes, bisecting, perhaps, a given angle, or fall- ‘ing perpendicularly on a given line; or tangents must be ‘drawn from a given point to a given curve; or circles must ‘be described from a given center with a given radius, or ‘touching given lines, or given circles, and other similar ‘operations. And as these antecedent preparations cannot be distinctly pointed out for every particular. case, it is here that the learner will have an opportunity of showing ‘his ‘taste and acuteness in choosing the best effectéons on ‘which to found his analysis, and thence to obtain the most ‘elegant construction and demonstration the problem admits ‘of; for the solution of the same problem may often be de- ‘rived in a variety of ways; and however apt some may be ‘to ridicule the notion of taste, when applied to subjects of ‘this kind, all persons imbued with a genuine relish for the ‘mathematical sciences, well know, that as much real taste ‘may be shown, in being able to distinguish between the dif- ferent degrees of elegance in mathematical investigations, ‘as ‘a student in the Belles Lettres, or a professed critic can ‘show, by a due relish forthe sublime or beautiful in poetry or oratory, 267 268 ———S— ACCOUNT OF THE LOST WRITINGS OF j A Brief Account of the lost Writings of Eucuip, APOLLo- | Nius, and others, on the GEOMETRICAL ANALYSIS OF THE| ANCIENTS, and of the several attempts of the Moderns t restore them. HY Pappus* of ALEXANDRIA, in the preface to the seventh book of his very valuable Mathematical Collections, has enumerated those works which treated of Analysis, under the name of ‘* Resolution Books,” or ‘* Books of Solutions,” which were considered by the ancients as forming a kind of second elements, on which they appear to have set a high value. 4 These tracts are intituled De Sectione Rationis, de Sectione Spatii, de Sectione Determinatéd, de Tactionibus, de Incli- nationibus, and De Locis Planis,—each in two books, and all of them the productions of the elegant and _ prolific. genius of Apollonius; to which may be added Euelid’y Book of Data, and his three books De Porismatibus ;—none aE * Paprus was a very eminent mathematician of Alexandria in Egypt, towards the latter part of the fourth century: he is particularly men- tioned by Suidas, who says he flourished under the Emperor Theodo- sius the Great, who reigned from the year 379 to 395 of Christ: his: writings show him to have been a consummate mathematician Many of. his works are lost, or at least have not yet been discovered. The prin- cipal of these are, his Mathematicai Collections, in 8 books, the first of, which, and part of the second, are now lost. He wrote also 4 Com-. mentary on Ptolemy’s Almagest ; An Universal Chorography ; A De- scription of the Rivers of Lytia ; A Treatise of Military Engines ; Com- mentaries on Aristarchus of Samos, concerning the Magnitude and Dis- tance of the Sun and Moon, &c.; of these there have only been pub- lished, the Mathematical Collections, in a Latin translation, with a large commentary, by Frederic Commandine, in folio, at Pisa, in 1588; anda second edition of the same, also in folio (but less correct than the first) at Bologna in 1660. In 1644, Mersenne exhibited a kind of abridg-' ment of these books, in his Synopsis Mathematice, in 4to, but this contains-only such propositions as could be understood without, figures. In 1655 Meibomius gave some of the Lemmata of the 7th book, in his Dialogue upon Proportions. In 1688 Dr. Wallis printed the last 12 propositions of the second book, at the end of his Aristarchus of Samos. In 1703, Dr. David Gregory gave part of the Preface to the 7th book, in the Prolegomena to his Euclid. And in 1706 Dr. Halley. gave that Preface entire, in the beginning of his edition of Apollonius. on the Section of Ratio; detached portions of the preface have also been given by Snellius, Dr. Simson, and Camerer. ‘i An analysis of the contents of the Mathematical Collections, which’ are exceedingly curious, may be seen in Dr. IJutton’s Mathematical, Dictionary, art. Pappus ; and another, much more minute and extensive, in the Rev. Dr. Trail’s Life of Robert Simson. aa EUCLID, APOLLONIUS, &c. of these works, however, have come to our hands, except the reatise De Sectione Rationis, and Euclid’s Data. _ These books of Analysis contained complete investigations, poth by resolution and composition, in all their possible vases, of a great variety of propositions of a very general sature, and of frequent recurrence in geometrical inquiries. n the solutions of such problems, the geometers of antiquity hroceeded with the utmost caution, and were careful to re- mark every particular case, that is to say, every change in ‘he construction, which any change in the state of the Data could produce. The different conditions from which the so- utions were derived, were supposed to vary one by one, vhile the others remained the same; and all their possible combinations being thus enumerated, a separate solution was xiven wherever any considerable change was observed to have aken place. Nothing, it is evident, that was any way con- hected with the problem in hand, could escape a geometer vho proceeded with such minuteness of investigation. | The use of those general propositions was the more im- mediate resolution of any geometrical problem, which could »e reduced to a particular or known case of any one of them, is by such a reduction, the proposed problem was considered is fully resolved, because it then became only necessary to ipply the analysis, composition, and determination, of that varticular case of the general proposition, to this particular problem which was shown to be comprehended in it. This ipparatus of separate solutions, with the determinations of every possible case, which, to the advantageous use of these general problems, is essentially necessary, may seem for- aidding, and if regularly perused without examples of its ipplication, may sometimes appear tedious and uninteresting, ind this, perhaps, may have created some prejudice against hestudy ofthem. Dr. Halley, indeed, seems to think that hese books were intended merely for the instruction of oeginners in the study of the Geometrical Analysis; but it is ‘ustly observed by Dr. Simson, that, though they may be very advantageously employed for that purpose, yet, it is nanifest from their contents, and also from the manner in vhich they are referred to, in the small remains which we dossess of the ancient Geometry, that the chief object of ‘hem was what has now been stated. Indeed in Pappus vhere are some examples of problems being resolved by a eduction of them to cases of one or other of these general oroblems ; for instance, the 87th proposition of his seventh 200k (a problem useful in the treatise on Inclinations,) is vy analysis brought to a case of determinate section. 269 270 ACCOUNT OF THE LOST WRITINGS OF Another example is prob. 164 of the same book, the las Lemma belonging to the Porisms, which is a problem re solved by Pappus, by reducing it to a case of the section ¢ space, and the composition of the problem is considered ¢ completed, merely by a reference to that particular case This tract of Apollonius being lost, and no restoration of having been made at the time of Commandine, the case re ferred to is resolved by him, in his notes on Pappus, thoug in a very operose and tedious manner. Schooten also resolve this problem (Ewvercit. Math. p. 104, 105), and seems t blame Pappus for not solving it directly, but by a reference, to a case of the section of space, from which it appears tha Schooten was not aware of the real purpose which thes, ancient books were intended to answer. } ‘ I1.—On the Data of Euclid. {| Euclid’s Data is the first in order of the books writte by the ancient Geometers to facihtate and promote th. Method of Resolution, or Analysis: it consists of a numbe. of propositions, which are designed to show what things ca) be known, or determined, from those that, by hypothesis, ar. already known, and is one of the few which have been pre served in the original language. a «In general, a thing is said to be given or known, whiel is either actually exhibited, or can be found out; that is ‘¢ which is either known by hypothesis, or that can be demon “ strated to be known; and the propositions in the book o ‘© Kuclid’s Data show what things can be found out, or known ‘¢ from those which, by hy pothesis, are already known ; so that “in the analysis or investigation of a problem, from th “ things that are laid down as known, or given, by the help 0, << these propositions other things are demonstrated to be given) ‘s and, from these, other things are again shown to be given | “and so on, until that which was proposed to be investigated ‘¢ or found out in the problem, is demonstrated to be given, <‘ and when this is done, the problem is solved, and its com: «© position is made, and derived from the compositions of thi: “ Data which were made use of in the analysis. And thus thy ‘¢ Data of Euclid are of the most general and necessary use it <<‘ the solution of problems of every kind, and were constantly ‘‘ used by the ancients in their resolutions of problems: “ though it was not their practice to refer to the particular pro. << positions ; and this application of the Data is continued by * such as follow strictly the ancient method of analysis.” ; ri EUCLID, APOLLONIUS, &c. On the two Books of Apollonius, de Sectione Rationis, or, On the Section of Ratio. | Apollonius wrote a Treatise on the Section of Ratio, hich is still extant, and is comprised in two books; the ob- ct of this treatise is, the solution of a single problem, sub- f vided into a multitude of cases, and mar tked with various snitations ; ; it may be thus enunciated : * Through a given point to draw a straight line, so as to cut off segments from two straight lines given in post- , fton, which segments shall be adjacent to given points in the said straight lines, and shall also have a given ratio to éach other.’’ The two books on the Section of Ratio, were long sup- ysed to be lost, and accordingly Willebrordus Snellius, son Rodolphus Snellius, professor of Mathematics at Leyden, tempted, about the year 1607, at the early age of 17, to store them, in a small work published at Leyden in the dlowing year; and the attempt reflects great credit on its ivenile author, though it wants that purity, fulness, and egance, which distinguish the geometrical compositions of ancients. About a century afterwards, the famous Dr. wulley, with much sagacity, and incredible labour, reco- *red these two books from an Arabic manuscript, for rmerly 1 the possession of the learned Selden, but now in the odleian Library; this is the only complete work of Apol- ius now extant, and the only work in existence from hich a correct view of the mode of solving problems among fe ancients is to be obtained : a correct Latin translation of as valuable relic, was published by the Doctor in DET; at sford, in 1706. | | h Ni | i L—On the two Books of Apollonius, de Sectione Spatiz, | or, On the Section of Space. ‘The next in order, among the analytical writings of the cient Geometers, is a Treatise on the Section of Space, i in ‘0 books, and likewise by Apollonius. ‘The general. problem which forms the subject of this ieatise, j is somewhat similar to the last: it may be thus ex- lessed : ‘“ Through a given point to draw a straight line so as to ‘cut off segments from two straight lines given in position, ‘which segments shall be adjacent to given points in the said ‘straight lines, and shall contain a rectangle equal to a given 4 space, ” 271 72 ACCOUNT OF THE LOST WRITINGS OF IV.—On the two books of Apollonius, de Sectione Deter. minata, or On Determinate Section. : After the treatises on the Section of Ratio, and the Sectior of Space, comes the treatise on Determinate Section, by th author of the two last, and like them, comprised in twe books; the object was the solution of the following genera, proposition : bi To cut an indefinite straight line in one point, so, that 9, ** the segments contained between the point of section sought “and certain given points in the said line, either the squan “* on one of them—or, the rectangle contained by one of then “* and a given line—or, the rectangle contained by two of them «* —may have a given ratio—either, to the square on one oj *< the remaining lines—or, to the rectungle contained by one) * them and a given line—or, to the rectangle contained by “ two of them.” is.) This is Pappus’s enunciation of the general problem, as corrected by Dr. Simson. Montucla divides it into the three following cases : y | 1. Two points, A and B being given in position, ing straight line, it is required to find a third point, P, in the same line, such, that the square-PA may. be to the square ol PB; or the rectangle under PA, anda given line, D, to the square of PB; or the square of PB, to the rectangle unde PB, and a given line, D, in a given ratio. a | 2. Three points, A, B, and C, being given, in a straight line, it is required to find a fourth point, P, in the same line, such, that the square of PA may be to the rectangle under) PB and PC; or the square of PB to the rectangle unde PA and PC; or the rectangle under PA and PB, to the rectangle under PC, and a given line in a given ratio. : | . ; i" ) 3. Four points, A, B, C, and D, being given in @ straight line, it is required a fifth point, P, in the same line, such, that the rectangle under PA and PB may be to the rectangle under PC and PD; or, the rectangle under PA and, PC, to the rectangle under PB and PD, in« given ratio; where P may have either an internal or an external position, with respect to the given points. bh These two books on Determinate Section, long since lost, were restored by the younger Snellius,* before mentioned but m a very imperfect manner, without the necessary dis- tinctions of the various situations of the points (called by. Apollonius Epitagmata, which Snellius, it would seem, did not understand), and without a complete exposition of the determinations, which, as is well known, are necessary to the EUCLID, APOLLONIUS, &c. orfect solution of any problem .*—These books have again yen restored by Dr. Simson, who so passionately admired, d so thoroughly understood, the spirit of the ancient apery, in a style the most luminous and complete: the irned Professor has even gone farther, and has added a fird and fourth book, entirely new :+—Since Dr. Simson’s (ath, two other restorations of the Determinate Section Ive been published ;—one by Mr. William Wales, which is enexed to Lawson’s Translation of Snellius ;£ and another | Giannini,§ an [talian Geometer, who seems to have trodden ithe steps of Apollonius more closely than Mr. Wales has (ne, inasmuch as he has given two solutions of each of the jncipal problems, one by straight lines—the other by se- reircles, which, according to Pappus, was the method pur- s2d by Apollonius: with respect, however, to these Resto- rions, it may be sufficient to observe, that, independent of t» curious information in Dr. Simson’s valuable Preface, the gat superiority of his work, as a Restoration of Apollonius, mst be obvious to every Geometrician. V.—On the two Books of Apollonius, de Tactionibus, or | on Tangencies. | After the two books of Apollonius on Determinate Sec- tia, Pappus places, as the uext in order, two others by the sae author, on Tangencies, the object of which was the icowing very general Problem, branched into a great viiety of cases, and marked with various limitations :— Juppose three things to be given in position, in the same pine, consisting of puints, straight lines, and of cireles given uinagnitude, and which may be taken of any one of these kids, exclusively ;—or, of any two of the kinds, indif- fently ;—or, lastly, of all the three kinds, promiscuouly ;— its required to describe a Circle, which shall pass through eur Of the given points, (when points are given,) and also toh each of the given lines, (whether they be straight lines, orireles, or both ). - j _ Snellius’s Restoration was translated into English by Mr. Lawson, anoublished at Loridon, in 1772, in a thin quarto volume.---This and the btir Mathematical Works of Mr. Lawson may be had of the publisher. _ ‘See his Opera. Reliqua. _ ‘the two books of Apollonius Pergzus, concerning Determinate Selon, as they have been restored by Willebrordus Snellius ; by John aon, B.D. To which are added the same two books, by William Wis, being an entire new work, 4to, London, 1772. ‘See Opuscula Mathematica, auctore Petro Giannini, Parma, 1773. 9) 73 Q74 ACCOUNT OF THE LOST WRITINGS OF It is evident that the Data, in this general enunciatio may be so combined, taking them by threes, as to give ri to ten distinct propositions,—for, ’ 1. Three points may be given .eseecseccceee 2. Three straight lines ....ceseccecoesccces 3. Two points and a straight line ......ceeeee A, Two straight lines and a point .....ecceees 5. Two points anda circle seeessescececesves 6G. Two circles and a point ....cecsescccceece Two circles anda straght line sssceeeeeces . A point, a straight line, anda circle sesseees, G. Two straight lines and acircle ...seseeeees | | 10. Three circles 2. .cccceccececccccsccncces UB The solutions of the first and second of these pete: ; given in Euclid’s Elements, (B. TV. Prop. 4 and 5), following six constitute the first Book of the Treatise on r gencies, the two remaining Problems (namely, | | 0, ¢ © 0 0,) occupy the second Book. We are informed by Pappus, that there was another Tr. tise, containing a collection of Problems, nearly allied to \ Tangencies here described, and to which it was considere¢: introductory. The Propositions it contains may all be - cluded in the following general enunciation, which, it willé perceived, is more circumscribed than the former, with spect to Hypotheses, while, at the same time, it mcliiy additional condition :— Suppose two things to be given in position, in the su plane, consisting of points, straight lines, and circles, t iis required to describe a circle, given in magnitude, that sil pass through the given point, (or points, if such be giv, and also touch the given line (or lines, whether tralia Bi or cercles. ) _ Here it is evident that six distinct Mig hyve may) formed, by uniting the different Data in pairs, and a eire given in magnitude, may be described to answer all 1 cases of the ceneral problem, which are the following :—_ 1. When two points aré GIVEN e+e secre ceeereeeee | , oS * —_- me @ ee, S = —"s Ps © — See = 2. two straight lines. ....sseeceeccse so tme 3. dwo circles ....'5..s000.440:0.0's «cs 9s ae 4. a point and a straight line ........e++e0e | 2. & Puint.and a CITCle « o/ o,,010¢'s/0¢~ og: 9 5s G. a circle and a straight line........+seees \ The two Books of Apollonius on Tangencies have she the same fate as\ most of his other writings, no vestig either of them has come down to us, excepting the Lemnté and the preceding general description of them, vn WV EUCLID, APOLLONIUS, &e. yeen preserved by Pappus. Aided, however, by this descrip- jon, several of the Mathematiciaus of later times have attempt- id restitutions, either of the whole, or of particular proposi- ions ; for an account of the principal of these attempts, see the joies to Prob. 42... .48 (p. 240, &c.) of this work. A trea- ‘ise on this subject, by John William Camerer, was published t Gotha and Amsterdam, in 1795, (small 8vo,) containing the emmas of Pappus, an edition of Vieta’s Treatise before men- ioned, with notes and additions, and a curious history of the woblem, which is highly interesting, from the accounts it jontains of the labours of some foreign mathematicians on bis problem, which are little known in this country. te | VI.—On the two lost Books of Apollonius, de Inclina- tionibus, or on Inclinations. ' The Subjects of these Books was the following general ‘roblem:—** Two lines being given in position, it is required to insert a straight line between them, of a given length, | that shall also tend, or incline, to a given point.” ‘Some cases of this problem evidently belong to the ‘gher geometry, for the given lines may be either straight ‘nes, or circles, or conic sections, or of the higher orders ;— / is manifest, therefore, that certain cases will be plane, hers solid, and others, again, linear:—those cases, how- er, in which the Data are limited to straight lines, or to raight lines and circles, admit of elegant solutions by hans of the elements only. The following are among the ses which appear to have been investigated by Apol- mius :— ‘1. A circle being given in position and magnitude, it is quired to insert. a straight line of a given length, that ‘all also tend, or incline, to a given point.* 2. A circle being given in position and magnitude, and ‘straight line being drawn perpendicular to its diameter— ‘insert a straight line of a given length, between the per- ndicular and the circumference, that shall also tend, or icline, to the extremity of the diameter. 3. An angle being given, and a point in the straight line lecting it being also given, to place a straight line, so that } : ] | \ ' Ghetaldus, Apollonius redivivus, prob. 1.—D’Omerique, Anal. (om. 37. 1,—Leslie’s Geometrical Analysis, 19.11. , Ghetaldus, prob. 2.—D’Omerique, 38. I.---Simpson’s Evercises, Bb. 10.—Reuben Burrow’s Inclinations, prob. 1.---Leslie’s .Geom. ul, 23. 11. | T2 ” BN 276 ACCOUNT OF THE LOST WRITINGS OF the part intercepted by the sides of the given angle, may be of a given length, and also tend to (i, e. pass through) the given point. This problem is sometimes expressed in the following terms;—a rhombus being given, and one of its | sides being produced, to imsert in the external angle a_ straight line of a given length, that shall tend, or incline, to | the opposite or internal angle.* ve | A, Two circles, having their diameters in the same_ straight line, being given, to insert a straight line of a given. length between them, that shall tend, or incline, to a given | point in either diameter. by Pappus informs us, that, “ the first Book of the Treatise on Inclinations, contained solutions of the problem con- cerning a circle, in two cases—of the problem concerning a’ circle and a straight line perpendicular to its diameter, in | four cases—and lastly, of the problem of the rhombus, in two cases. The single problem concerning two | cireles | formed the subject of the second book, and was divided into ten cases, and these again into many dioristic sub. divisions.”’ mal The lost Treatise on Inclinations was first restored by Marinus Ghetaldus, of Ragusa, (Anno. 1607,) some of whose | defects were supplied by Alex, Anderson, of Aberdeen; ‘ other solutions of some of the problems, were given by. Des Cartes, Gregory St. Vincent, Renaldine, D’Omerique, Huygens, Thomas Simpson, and Dr. Rob. Simson. In the’ year 1770, Dr. Horsley, late Bishop of St. Asaph, printed, in Latin, a restoration of the two Books on Inclinations, ina quarto volume, and nine years afterwards the same task wa | performed, with much greater ability, and with far more simplicity and conciseness, by Mr. Reuben Burrow. , | He) pe oe << After Tangencies, we have, in three Books, Euclid’s Porisms, a most curious collection of many things which relate to the analysis of the more difficult and general pro-- blems, of which, indeed, Nature affords a great plenty. Now. these contain a subtile and natural theory, very necessary, and universal, and highly entertaining to those who are able On the Porisms of Euclid. * Ghetaldus, prob. 3. 4.---D’Omerique, 42. I.---Burrow, prob. 2.--- Leslie, 24. 25. 11.—Huygens, Opera Varia, Tom. 1. prob. 6: 7. page 397, &c. yee + Alex. Anderson, Supplementum Apollonii Redivivi.—Burrow, prob. 5.7.---Burrow’s Diary, 1779.—Gentleman’s Diary, 1785. | EUCLID, APOLLONIUS, &c. to understand and investigate them. They are, however, in “species, neither theorems nor problems, but in some sort, ofa ‘middle nature, between both ; so that their propositions may ‘be formed either as theorems or problems ; whence it has ‘arisen, that some Geometers have esteemed them in kind to be theorems, and others problems, having respect only to the form of the proposition. But it is manifest from their definitions, that the ancients better understood the difference between these three; for they said, that a theorem was that in which something was proposed to be demon- strated—a problem that in which something was proposed to be constructed—but a porism was that in which something was proposed to be investiguted. Now this definition of a ‘porism is changed by the moderns, who could not investi- gate all of them, but, using these elements, showed only what it was that was sought, but did not investigate it. And though they were confuted by the definition, and by what has been said, yet they formed their definition from an acci- dent, in this manner: ‘A porism is that which is deficient in hypothesis from a local theorem, (i. e. a porism is a local theorem, deficient, or diminished, in its hypothesis.) Now, of this kind of porism, geometrical loci are a species, of which there is great store in the books concerning analysis, and, being collected apart from the porisms, are delivered ander their proper titles, because this species is much more diffuse and copious than the rest. For of Loci, some are olane, some soltd, and others linear, and besides these, shere are Joci ad medietates, (or which arise from proportion- als’) The above is part of Pappus’s Account of the Porisms, vhat follows has suffered so much from the injuries of time, hat all we canimmediately learn from it is, that the ancients out a high value on the propositions which they called po- jisms, and regarded them as a very important part of their inalysis. Several defects in Pappus’s text, might, however, have been supplied, if the enumeration, which he next gives, ‘f Euclid’s propositions, had been entire; but on account ‘f the extreme brevity of his enunciations, and their refer- ince to a diagram which is lost, and for the constructing of vhich no directions are given; they are all, except one, per- ectly unintelligible. For these reasons, the fragment in juestion is so obscure, that even to the learning and pene- ‘ration of Halley, it seemed impossible that it could ever be : xplained; and he therefore concluded, after giving the areek text with all possible correctness, and adding the vatin translation, * Thus much for the description of Po- isms, which can neither be useful to me nor to the reader, 277 278 - detined porisms only ‘ab accidente,”’ viz. Porisma esl ACCOUNT OF THE LOST WRITINGS OF on account of the want of the scheme of which mention is, made, from whence many right lines here treated of, without, any alphabetical marks, or any other characteristic’ distine- tion, are confounded one with another, and also on account} of some things, omitted and transposed, or otherwise vitiated,, in the exposition of the general proposition, from whence] am not able to guess what Pappus meant. Add moreover, a mode of expression too concise, and which ought never to be used ona difficult subject, such as this 1s.” i It is true, however, that, before this time, Fermat had| attempted to explain the nature of Porisms,* and_ not alto- gether without success. Guiding his conjectures by the definition which Pappus censures as imperfect, because if quod deficit hypothesi a Theoremate Locali,”’ he formed t¢ himself atolerably just notion of these propositions, and illus. trated his general description by propositions that are, I effect, porisms. But the honour of completing the disco: very was reserved for Dr. Simson, whose restoration of the Porisms appeared in the Collection of his Posthumous Works, printed in 1776, at the expense of Earl Stanhope: The account Dr. Simson gives of his progress in this work and the obstacles he encountered, will always be interesting to Mathematicians. In the Preface to his Treatise de Poris: mutibus, he says, ‘ but after I had read in Pappus, that thy Porisms of Euclid were a most curious collection of many things that related to the analysis of the more difficult ant general problems, I was earnestly desirous of knowing some! thing about them, wherefore often, and by various ways I endeavoured to understand and to restore, as well Pappus’, General Proposition, lame and imperfect as it was, as als: the first Porism of the first Book, which is the only one ou of the three books that remains entire: but my labow was in vain, for I made no proficiency ; and when thes thoughts had consumed much of my time, and at Jengt had become very troublesome, I firmly resolved not to mak apy inquiry for the future, especially as that -best of Geo meters, Halley, had given up all thoughts of und ors ing them ; therefore, as often as they occurred to my min¢ I endeavoured to put them by : yet, afterwards, it happene that they seized me unawares, forgetful of my resolutior and detained me so long, until some light broke in, whic * Porismatum Euclidzorum renovata doctrina, et sub forma Isagog) exhibita, Opera varia, p. 116, | en a —_—~= | ~ EUCLID, APOLLONICUS, &c. ave me hopes of at least finding out Pappus’s General Pro- »sition, which, indeed, not without much investigation, I ‘length restored. Now this, soon after, together with the vst Porism of Book I. was printed in the Philosophical ansactions for 1723, No. 177.’’* ‘Dr. Simson’s restoration has all the appearance of being jst; it precisely corresponds with Pappus’s description of Jem. All the Lemmas which Pappus has given for the better aderstanding of Euclid’s Propositions are equally appli- ble to those of Dr. Simson, which are found to differ from ical theorems precisely as Pappus affirms those of Euclid awe done. They require a particular mode of analysis, id are of immense service in geometrical inquiries. ) Dr. Simson defines a ~porism to be ‘‘.a proposition in bich itis proposed to demonstrate that one or more things ye given, between which, and every one, of innumerable ‘her things not given, but assumed according toa given iw, a certain relation, described inthe proposition, is shown ‘take place.” This definition, it must be confessed, is not ilittle obscure, but will be plainer if expressed thus :— )A porism is a proposition affirming the possibility of finding ich conditions as will render a certain problem indetermi- ute, or capable of innumerable solutions.”’ This definition srees with Pappus’s idea of these propositions, so far, at ast, as they can be understood from the imperfect state of ue text; for the propositions here defined, like those which 3 Racribes, are, strictly speaking, neither theorems nor roblems, but of an intermediate nature between both; for vey neither simply enunciate a truth to be demonstrated, or propose a question to be resolved, buat are affirmations ’a truth in which the determination ‘of an unknown quan- ty is involved. In as far, therefore, as they assert that a irtain problem may become indeterminate, they are of the vture of theorems; and in as far as they seek to discover /e conditions by which that is brought about, they are of be nature of problems. : Another celebrated Scotch Professor, the late Mr. Play- ‘ir, whose learning embraced a wider range, has, inthe third olume of the Edinburgh Transactions, given a compreheu- ve and luminous view of this intricate subject. There is so an ingenious Paper on Porisms, ‘¢ withexamples of their se in the solution of problems,’’ in the fourth volume of € same publication, by Mr. William Wallace, now Pro- + ' Dr. Simson’s Restitution of the Porisms was partly translated by r. Lawson, and may be had of the Publisher of this Work. 279 280 ACCOUNT OF THE LOST WRITINGS OF fessor of Mathematics in the University of Edinburgh; an in the Philosophical Transactions for 1798, there is ‘a Co lection of general Theorems, chiefly Porisms, in the high Geometry,’’ by that distinguished ornament of the senaj and the bar, Henry Brougham, Esq. M. P. and F.R.S. ‘ On the two Books of Apollonius, on Plane Loci. | One of the most interesting of all the works of Apolli nius was that on Plane Loci, comprised in two books. Tl object of this work was ‘ to investigate the conditions undi which a point, varying in its position, is yet limited to th description of a straight line, or a circle, (i.e. a plane locus. given in position.”” These books were about the year 165( restored, in a sort of algebraic form, by Francis van Schor ten,* then Professor of Mathematics at Leyden ; and mo) elegantly, though only partially, by the ingenious Fermat, Counsellor of the Parliament of Thoulouse. But these ‘ tempts, however skilful, are entirely eclipsed by the finishe production De Locis Planis,§ which Dr. Robert Simsoi the great restorer of the Ancient Geometry, published ¢. Glasgow in the year 1749. If this work differ at all fror_ that which it is intended to replace, it seems to do so onl by its greater excellence ; thus much at least is certain, the the method of the ancient Geometers does not appear t greater advantage in the most entire of their writings, tha in the restoration above-mentioued ; and that Dr. Simso has often sacrificed the elegance to which his own Analys_ would have led, in order to tread more exactly in what th Lemmata of Pappus pointed out to him as the track whic Apollonius had pursued. | The following propositions are those preserved by Pappu‘ as having formed the subject of the Treatise on Plane Loc’ many of them, it will be seen, are very general, and branc into a variety of cases, each of which, as was the custom ¢ the ancient Geometers, was separately considered by Apol lonius; the investigations may be seen in the restoration ( Dr. Simson, already mentioned. a , : 1 * Schooten, Ewercit. Math. lib. iii. 4to. Lug. Bat. 1657. i + Fermat, Opera Varia Mathematica, fol. Tolos, 1679. 7 _§ Apollonii Pergeei Locorum Planorum, lib. ii. restituti a Robert Simson, M.D. Matheseos in Academia Glasguensi Professore, 4to. 4 German translation of this excellent work, by J. W. Camerer, was pul lished at Leipsic, in 1796, in one vol. Svo. "It has also been translate — into French, by Simon L’Huillier, of Geneva. » a EUCLID, APOLLONIUS, &c. Book 1. : 1, If two straight lines be drawn, either from the same gen point—or, from two given points—or, which are either ¢ directum—or, parallel—or, comprehend a given angle,— ed, if the lines so drawn have—either—a given ratio to one = BA, * For (Euclid Cor. Prop, 8. B. 6,) BC’ is a mean propor- tional between B A and BO, whence SECT. III. EQUILIBRIUM OF ARCHES. 59 Therefore BO? : Agq::BA’: Cp, that is, AqxBA’ Cp=—B or PROPOSITION VI. To determine the height of the Extrados above every point of a Catenarian Arch. a Lee 4 = The same analogy will obtain as in the other Propositions, the radius of curvature, &c. being found as directed in Sect. 2, Prop. 5. Or let the tension of the curve at the vertex be called X. Then (by the nature of the catenary) X will be constantly to A O, As X—Aq to qP. That is X:AO:: X—Aq: Pq. q B’ BC? BO —AB; and squaring both sides, we have 60. EQUILIBRIUM OF ARCHES. SECT. III. AOxX—AOxAq mee But Cp=AO+Aq—Pq. Whence Pq= Therefore —A A Soa W Oe ip en eatneaesecemncia 88 = Cea Aq x(X+A0) ERG Se Te » which being reduced becomes C p = OBSERVATIONS. The subjoined table will shew at one view the height pC of the superincumbent wall, over any point C of the intrados; and by referring to the figures in plates 1 and 2, we shall see the general form of the extrados to each of the curves that have been treated of. In Emerson’s Fluxions the reader will find, besides the curves already mentioned, the mode of determining the extrados to the logarithmic curve and cissoid ; but as these curves would be seldom, if ever used in practice, it has been considered unne- cessary to give them a place in this Treatise, which is designed to instruct the practical artist. SECT. III. EQUILIBRIUM OF ARCHES. 61 Curve. | Value of C p at any point C. The height at the crown sti by the cube of pe the radius, and the product O R® divided by the cube of the vertical height of the point C (from the horizontal diameter. The vertical height of the eerie every where equal. The height at the crown multiplied by the cube o Parabola. |PC=A q or the semiconjugate axis, and the product divided by the cube of the vertical height Aq xAB?* Ellipsis. me reyteh es of the point from the trans- (verse axis. ¢ The height at the crown multiplied by the cube of the semitransverse axis, and the product divided by the cube Aq XA B* Hyperbola.| 5a + OA)orBO! TERA (of the semitransverse axis added to the abscissa). The height at the crown multiplied by the square o AqxAB (AB—AOP AOFe ing circle, and the product Cycloid. divided by the square (0 the diameter aforesaid less the abscissa). — | L It the diameter of the generat- L ¢ The height at the crown “ multiplied into the sum o Catenary. sales ee ie Dye the tension at the vertex and the abscissa, and the product bdigiderebysthejtension- 62 EQUILIBRIUM OF ARCHES SECT. III. Thus it appears that all arches which spring perpendicularly, or whose secants at the springing are infinite, as the semi-circle, semi-ellipsis, and eycloid, will require to be loaded with an infinite weight over that point. The extrados of a parabola will be another parabola, since pC is every where equal. And in the hyperbola the extrados continually ap- proaches the intrados; but the scantiness of the haunches of these two curves, independent of their extreme meagre aspect, renders them un- fit for the purposes of a bridge. Where great weights are required to be discharged (under par- ticular circumstances) from the weakest parts of an edifice, as is often necessary in warehouses, and sometimes under apertures inverted, in order to bring an equal and uniform pressure on the foundations, the parabolic arch may with great propriety be introduced. The catenarian curve will seldom be admissible, more especially in bridges; for where an horizon- tal roadway is required, the height at the crown must be very great. An arch of this kind of 31,3, feet span, and 15% feet rise, with an horizontal extrados, must be ten feet high at the crown (see fig. A, plate 2). In general, however, by using segments of the ellipsis, cycloid, and circle, we may obtain SECT. III. EQUILIBRIUM OF ARCHES. 63 convenient roadways or extradosses, little differ- ing from the conditions of the theory. In a seg- ment of a circle, for instance, of 90 degrees, if the height at the crown be to its radius as 1 to 64, that is, about one-ninth of the span, the roadway will be nearly horizontal (see fig. B, plate 2). The numerous instances in which arches spring perpendicularly, without the infinite load which this theory requires over the springing point being given to them, seem to induce doubts of its truth; let us therefore shortly consider the apparent variance. Take for instance two semi-arches A C, C H, of a bridge (segments of circles); EI K LD 1s _the extrados likely to be applied in practice, 64. EQUILIBRIUM OF ARCHES. SECT. III. in which the part K L D is the actual extrados over AOC; but the extrados which the theory requires is ML PD. Now DPL it will be seen by inspection nearly coincides with it as far as L; but the part of the extrados over A O appears deficient in the quantum of weight with which it ought to be loaded to obtain an equi- librium. But IH AOL, .connected as it is by the bond of the stones of which it is com- posed, and the cement which unites them, may be considered as a solid mass of stone acting by its absolute gravity in a vertical direction upon AOQ, the part of the arch which requires the weight K L MN, and therefore may be con- sidered as exerting a pressure equal to that of the infinite loading which theory requires. Thus in most cases the mode of constructing arches is not very far out of the conditions of- the theory. SECT. IV. EQUILIBRIUM OF ARCHES. 65 SECTION IV. Of the Method of finding an Intrados to any gwen Kutrados: If the extrados or roadway over an arch be considered polygonal, (as we considered the in- trados in the first Section,) and the direction of the first line of the intrados be given, we shall have an easy and tolerably correct method of finding the rest of the lines which compose the intrados, so as to equilibrate with the extrados ; for the sides of both may be increased in num- ber, so as to coalesce with the curves of which these lines are the chords (see pages 22 and 23, Sect. I.) ol |, | mm nn 0,0.p, p qs (hey J, plate 3) be the sides of an extrados, and let C D be the direction of the first side of the intrados, cutting a vertical let fall from p’ in C, D q’ being the height at the vertex D. Make the anglesq Dt, D qt, equal to the angle q D C, make D q equal to gq’ D, and F 66 EQUILIBRIUM OF ARCHES. SECT. IV. draw the lines D t, q t, intersecting each other in t. Produce (if necessary) C D indefinitely to- wards s, and in it take Cs equal to qt, and in the vertical p’ C produced, make C p equal p’ C, and join p s. Through C parallel to ps draw C B, cutting a vertical from o’ in B, and produce it towards r, then making B r equal to ps, and joming or, proceed as above for the heights under the points of the extrados n,m,1; or is the direc- tion of the side under n o’. Through all the points thus found, as also” those of the extrados, bend a flexible ruler and draw the curve lines, which will be similar to — q Gand D EK. 3 The operations for finding an intrados to any ~ other extrados, will be precisely the same; as_ may be seen fig. 2, plate 3, where the extrados — is horizontal. ; This construction is evidently the converse of that in Prop. 6, Sect. 1; and the demonstration i of both depends upon the lemma to Prop. 5. Sect. 1; for in both it is supposed that the arch consists of an assemblage of beams kept in equilibrio by weights placed at the angular’ points. SECT. I. EQUILIBRIUM OF ARCHES. 17 Now, in the triangle C pr, from the proportion of the sides to the sines of the opposite angles, we have C p= Ie ee DE CTO Therefore Se PL egy the weight represented by C p will be propor- tional to this expression ; and because C r varies as —_____, if we substitute this last term in the fae cr Op above value of € p, we shall have the weight or Selah hey ‘ BAe Cap x So eer Gs buts. 2 Crp=s. 2 XCs, ors.zrCs. There- Si Cis s.2pCsxs.2rCp We shall therefore have the following propor- tions for finding the weights or lines representing them; viz. s. 2 tDt’ aR) HE s. 2 7rCs ) -——$___—_———_____: Dg:; _~=~* §2tDqxs.zq Dt’ Sine px s: 2 pCs force C p proportional to fore C p is proportional to Cp. In numbers thus: Let 2 CD E=130° - 00’ ab Cl) 165, 4100) | Z ABC=145°: 00’ And let the weight or force represented by _ Dq=1-00; then, because in the figure we have | supposed 2 CD E to be bisected by D q, we shall | 3 C 18 EQUILIBRIUM OF ARCHES. SECT. I. have Z qDt’=65°; and because 2 pCs is the supplement of 2 q Dt, or 2 q Dt’, we have Z pCs=115. 00 Z rCp=(165°—115°) = 50°. 00’ We may now determine the value of C p, Bo, &c. in which the operation would be facilitated by using a table of logarithmic sines: but, to be more intelligible, we shall take the values of the trigo- nometrical terms from a table of natural sines, the labour of calculating such sines being too trouble- some for practice. Thus, | The natural sine of 2 CDE ( = 130°) or of 2zrCp (=50°), these angles being supplements of each other The nat. sin. 2 qDt’ or 2 tDq = 9063078 = 7660444 ( =65°) or of its supplement 4 pCs(=115) The nat. sin. Z BCD (=165°) =2588190 W hence 7660444 2588190 9063078 x 9063078 7660444 x 9063078 1:00: °39972 nearly the value of C p. In like manner B 0 is found to be 3.1019 nearly. SECT. I. EQUILIBRIUM OF ARCHES. 19 PROPOSITION VI. PROBLEM. The weights or lines representing them, D q, C p, Bo, being given, as also the direction of the first line DC, or the angle C dq, which it makes with the vertical D q; it ts required to find the direction of the other sides, so that an equilibrium between. the parts of the assemblage may take place. : \ A G Complete the parallelogram upon the diagonal _Dq, make Cs equal to Dt, join s p, and make BC parallel thereto; then we shall have the | C2 20 EQUILIBRIUM OF ARCHES. SECT. where X is put to repre- sent the tension of the curve at the Vertex e- .)0lnw Lec) and it will be the tangent required, and will = O T? + OE Gee At C draw DC, ® at right angles to CT, and in it take * It will be necessary to notice, that in the catenary there SECT. II. EQUILIBRIUM OF ARCHES. Al C R, a third proportional to X : X +AO, that is, As X :X+A0::% + AO:CR (the ra- dius of curvature to C)=175°37. isa peculiar property denominated its ten- Shire B B’ sion, that is, the degree of stretching it i x exerts at any point; and the ordinate to any of its abscissz cannot be found with- A out involving this tension or stretching in the calculation. The tension at the vertex A may be found by the summa- tion of the nee 2 series, calling it X. X=} AB x (ae ep 8AB?i, 691 AB! _ S8851ABS) AB 45B'B-' 8780B'B! 453600BB", &c., which is a series that will serve for every case, taking care to change the real values of A B and B’B, according to the height and semi-span given. Thus, Let AB = 40 BiBe— 50 Then it will stand thus : 40, le pee 8 eh TOS 922090 2a 16 1125 § 2362500 7087500000 Put these fractions into decimals, to sum them up more easily, and then X = 20 X (1:5625 + °3333 —:1137 + :0749—-0138, &c.= 20 X 1°8432 = 36°864; but if the series had been car- ried a term further (which is not necessary for practical pur- poses) it would have come out 36°88, or 36:9 nearly, as Dr. Hutton makes it in his Principles of Bridges (page 37, second edition, London, 1801). Any ordinate O C to a given abscissa, may be found by a 42 EQUILIBRIUM OF ARCHES. SECT. II. Hence 2 x 175°37=350°74 (the diameter of curvature DC). In the above the height AB=50, B B=30, and AO = 25. summation of the following series. Let it be to an abscissa A O—40-00. The constant tension (= X) being as above 36-9 nearly. Then 1 40 , SA0? 15A03 eee) OC=v7 2xXxAO x ( — 2x ' 160X? 2688x3 " 55296X!/)’ Putting these into numbers as before, we shall find O C=49°835 or nearly 50°00. SECT. III. EQUILIBRIUM OF ARCHES. 43 SECTION III. On the comparative strength of arches, and the method of finding the extrados of an arch from a gwen intrados. From what has been observed at page 25, it may be inferred, that the strength of one part of an arch to that of another, will be propor- tional to the greatest weights those parts are capable of bearing; that is, as the cube of the secant of the curve’s inclination to the horizon at those points, divided by the radii of curva- ture; and when two arches formed of similar materials have the same span and height, their comparative strengths at any two corresponding points will be also in the same proportion; but, since if one part of an arch fails, the whole will fall to ruin, and as the crown is the weakest part in all arches, it will only be necessary to compare them with each other at that point. Now as all curves at their vertices have no in- clination to the horizon, the angles thereat will always be 0°: 00’, and the secants become the 4.4 EQUILIBRIUM OF ARCHES. SECT. III. radii of those angles, and are the same or equal in every curve; whence it follows, that the comparative strength at the crowns, of two arches having the same span and height, is reciprocally as the radii of curvature at those points. Taking for example an arch 100 feet span and 40 rise; the radii of curvature at the crowns in the different curves will be as follow: FEET. Seoment of a circle Bleed Parabola : - 30°125 Ellipsis : Ben hap Hyperbola_. 37.417 Catenary HES EO Hence, ceteris paribus, at the crowns of a parabola and an ellipse, the strength of the former to the latter will be as 62°5 to 30°125, &e., &e. This will be more easily perceived if we con- sider the arches as polygonal, the sides being infinitely small. When the two first sides form a very large angle with each other, so that their mutual inclination approaches nearly to a straight line, the curvature 1s very small, and SECT. III. EQUILIBRIUM OF ARCHES. A5 they will evidently have less power to resist a pressure on the angle at the vertex, than when inclined to each other in a smaller angle, wherein the curvature is increased, and the radius of cur- vature consequently diminished. There are few, if any, instances of elliptical arches of large span in proportion to their height, in which the crowns have not sunk considerably. 46 EQUILIBRIUM OF ARCHES. SECT. III. PROPOSITION I. To find the height of the superincumbent wall, or extrados, above every point of a circular arch, so that by its pressure all the parts of the arch may be kept in equalibrio. R Let B’ A B’ be the segment of a circle, whose span is B’ B’, and height A B; R the centre, and A Rthe radius. Also let A q be the height of the wall at the vertex, and A O an abscissa to the ordinate O C at the point C, above which the height of the wall is to be found. SECT. IV. EQUILIBRIUM OF ARCHES. 67 If it be required to find, arithmetically, the heights of the several lines p’ C, o’ B, &c. the following is an easy process. Let q D = 1-00 op = 490 jee tet kes Ze Galen S57. 005 EmOn Das 80° 00 zCDq = 80°00 Having let fall the verticals q’ D, p’ C, o’ B, draw p’ c, C e, and o’ b, at right angles thereto. Then by plane trigonometry, eyed) 8.2 Ped. es gee pe= WET fe ak F 2 68 EQUILIBRIUM OF ARCHES. SECT. IV- Ce(=p'c) xs. 2 DCe Di ETE Vt eee e eA DT 0°7465 cq PAS Sea Oe ye Vise ce=p C=pC=Dq'+De—q c=10+0°7465 — 03703 = 13762. Now D Ponto oreo OE AB oY erates Wt a th eke oo C s. of (180° —2 2 CDq) cea ree Also Cp+Cs:Cs—C p:: tangent 3 sum of the angles C ps,Cs p: tangent 4 difference of those angles. So that ZC ps=56° 31’ ZC 8. p=23° 29’ dry Si) a DP -dbsie : S. LTRS As the angle B C a is equal to the angle C p s, draw B a parallel to o b, then Ba C is a right angle; consequently 2 a BC=90°— 56° Blea 33° 29’, SECT. IV. EQUILIBRIUM OF ARCHES. 69 And ‘on yee Dial (OD axe sri a BiG Sata 7 st 706 Cre, e mept— ba) — Cp +Ca — bp — 1:°376 + 3192 —°7941 =3°774. The whole extent is Ba + Ce (= p’c) = 4°323 + 4°233 = 9°056, and height = Ca + e D = 3:192 + 0°7465 = 3:9385. Many intradosses may be found by the above method to a single given extrados, the span of each depending upon the degree of inclination given to the first line D C; that is, if the angle q’ DC were only 92° 00’, the intrados will be of less height or rise in proportion to any given span, than if the said angle were 95° 00’. Thus the architect has the means of choosing the most con- venient and pleasing form for his intrados as cir- cumstances may require *. * If, having an extra- dos of a given curve, it ' q be required to adapt an S| intrados thereto, whose span and height must be of certain given dimen- sions, the following ope- ration will be necessary ; Let F p’ q' G be an extrados, as a segment of a circle, el- lipse, &c. &c. whose ordinate p’c, and abscissa q’ c, are known at every point thereof. 70 EQUILIBRIUM OF ARCHES. SECT. IV. If, instead of treating the arch as an assem- blage of beams with weights at the angles, we Let q’ D (height at crown) sseccossseeeees =~ 7:00 q’ c the abscissa ...seeserereesseseeeeres mer LL Ce, and p’ c its common ordinates (supposing, in the present example, the extrados to be the segment of a circle whose radius is 230) will = 34°00 B B’ the half span .....sesseceseveveees == 150-00 D B the height or rise ....... he ti ae To find D e. PutC e(=p' ey BD meni DIB st ot Die selves ee *¢ 2 ofl i & s I. p Coe c= C x22 =5%* fluxion of 5 Sg It is evidente c= q D+De—qe= S x fluxion of 5+ 5 II. Assuming 7 = we have z = Gand x fluxion of J * Cz z ay x ? That is 7 + *—8 =~ %” Czz , . : Andx+42-—.—»orv@ez 4+ 4e= Oz z. att Cc? 3 2 The fluent of this expression is = +42= : SECT. IV. EQUILIBRIUM OF ARCHES. PL were to consider it as covered by a loading dif- fused over it, it is evident, from the observations or 2.09 + 162 2-627, on x? +. 8 2 =-C 2’: from which z = EES. III. Now as 7 = , = eS yi a iy Seal ey C C Pp ee + ee x aa The fluents of which are y = VW C X* hyperbolic logarithm of (a +44 7 2 + 8 x) but at the vertex where z = 0 we have y = Vv C x hyperbolic logarithm of 4; So that the corrected fluent becomes y = VW C xX hyp. log. NDE ui peer aR 4 of IV. To obtain the constant quantity C. When e arrives at B, we have x = D B which call A, y = BB which call s, and the abscissa of the extrados = 5°5. Then repeating the above operation, we obtain RY 5 ee heey iS'h 175 h+1°5 h? +3 Consequently ” C = s + hyp. log. of ee eee sa WC x hyp. log. of V. Hence we obtain, * The hyperbolic logarithm of any number may be found by multiplying its common or Briggs’s logarithm by 2°302585, &c. the product being the hyperbolic logarithm of the number required, 79 EQUILIBRIUM OF ARCHES. SECT. IV. at the end of Section I. that a nearer approxima- tion to the true form of the intrados would be s X hyp. log. shia Pek Mind ast La se! 4, y— a hyp. log. hA+154+vy7h+3h 15 hyp. log. sfoaves 1! Pa o> = ere eee tat = s Hy pulopee mine ane Pe Sh Hence ¥ x h. 1. Ga Se SN a Ss ‘ “5 Sa we Suppose the last expression 7 x Sy ee Ae s when put into numbers to be a hyperbolic logarithm, and find the natural number corresponding to it, which call N ; YAS pea feabcat sca a Fi ; or4Nmr+44+ YY 2? + 82, and 4N—«—4—= of 2 4 82x. Squaring both sides, we have 16N?—8Nex—82N42°4+824+16=27 +4 8a, or 16 N?—82N+4+ 16=8Nza; 2 whence x = EeBING sce Ne sced N : It is only in the second and third steps that a change is ne- cessary ; namely, where the value of the abscissa of the ex- trados varies according to the circumstances of the data. Any SRGTS IV: EQUILIBRIUM OF ARCHES. 73 obtained by multiplying the heights D q, C p’, B o’, &c., found as above, by radius and by the secants of the angles D Ce, C Ba, &c., re- spectively. person the least acquainted with common algebra may carry this operation through. In numbers as follows : B84 —.¢8, and as =40, the expression h+ls + thot Veoh s 50 1-5 40 + 1°5 + Y 1600 + 120 ote ite b RANTES bolic logarithm of which is 4°01297, this being multiplied by 68 gives 2°72884, and looking into a table of hyp. logs. we have the natural number corresponding to this = 13°13, which we call N ; 2N?—4N +2_344:8- 52°52 + 2_ 9.4 N 13°13 for the length of the abscissa D e, whence the value of ce, that is of p’ C, is known. becomes = 55°31, the hyper- then + = For an horizontal extrados (see Hutton’s Principles of Bridges, sect. 3.) the equation to the curve of the intrados is found to be ata+t+Y2ar+ 2’ a go s , ee where a ex- ath+/2ahk + h® a hyp. log. of hyp. log. of presses q’ D, the height at the crown. 74 * EQUILIBRIUM OF ARCHES. SECT. IV. CONCLUDING OBSERVATIONS ON THE EQUILI- BRIUM OF ARCHES. It has long been a subject of complaint, that the theories of the statical equilibrium of bodies afford little assistance in the execution of any works where it is of importance to obtain a re- quisite degree of stability with the least possible expense of materials. Yet it can hardly with justice be considered as a reproach to mathe- maticians that they have not been able to dis- cover theories which should comprehend all the relations of the subjects involved, and which should be applicable to the actual circumstances of every case, since this would require a more intimate knowledge of the laws of mechanical action than we at this time possess. Under the disadvantages arising from our im- perfect acquaintance with the properties of bo- dies, the only thing which can be done is to assume the absolute perfection of all the quali- ties of the materials employed, to investigate the conditions of equilibrium according to that supposition, and then, to leave to the persons who are to apply the conditions practically, SECT. IV. EQUILIBRIUM OF ARCHES. 75 the care of making such modifications in them as are indicated by the results of the best ex- periments that have been tried. The objections that are made to the theory of arches just delivered are, that the voussoirs are supposed indefinitely thin, and that the loading above is supposed to act only in the vertical direction. Now, with respect to the first, the smallest excess of pressure on any part of the arch, above what the theory assigns, would force the voussoirs from their places; and with respect to the second, when the load- ing consists of loose material, as rubble or gravel, it exerts pressures laterally as well as vertically. The latter point may be dispatched in a few words; the principal part of that lateral pres- sure is counteracted by the resistance of the piers, and the remainder is too inconsiderable to deserve notice. It may also be observed that, of late, the practice of fillmg up the haunches with loose material seems to be abandoned, and that of solid walls, parallel to the length of the bridge, to be adopted. This work not only presses vertically upon the arch; but, by its co- hesion, renders an excess of weight at the crown less likely to force up the arch at the haunches. 76 EQUILIBRIUM OF ARCHES. SECT. IV. The weakness arising from the other circum- stance would be of serious consequence, if it were not that the artists have invariably given to the voussoirs a certain degree of thickness, in order that the divergency of the joints may be sufficient to allow them to keep their places. This is certainly an indispensable requisite ; and as the tendencies of these stones downward upon their joints produce pressures, which are different from those produced by the vertical loading in the catenarian theory, it may be ex- pected that attention should be paid to some- thing more than the equilibrium produced by such vertical loading. The simplest method of adapting the catena- rian theory to practice, seems to be that of mak- ing the courses of voussoirs balance themselves, as in Mr. Atwood’s theory, viz. by equalising the tendency which any given course has to rise, by a resultant of the lateral pressure of the superior courses, with the tendency downward in an opposite direction by a resultant of the weight of the given course. The weights of the voussoirs, and consequently their lengths being determined by such means, we may consider the extrados of these voussoirs as a new intra- dos, and proceed to determine a new extrados SECT. IV. EQUILIBRIUM OF ARCHES. me for the roadway by the heights of the columns of vertical loading, according to the catenarian theory: and as the exterior curve of the vous- soirs will always be more flat than the interior curve, and will always rise at an acute angle with the horizon, it is manifest that the final extrados will approximate nearer to a right line, and will consequently be more convenient for a roadway than that which would he obtained either by the catenarian theory in its simple state, or by the method of equilibrial voussoirs. Let it also be recollected, that the voussoirs or stones of which the arch is composed, have flat surfaces in contact with each other, and that the cement and friction add to their stability. It is true that no theory will shew what addi- tional weight (in practice) an arch can sustain on its weakest part; but an arch of equilibration is certainly better able to sustain a greater weight accidentally placed on any part than another arch would be. Strict mathematical precision cannot always be obtained; but by attention to what has been shewn, the architect will be better able to combine the requisite strength with that beauty which it will ever remain in his province alone to impart to the design. 6 78 EQUILIBRIUM OF ARCHES. SECT. IV. Of the thickness given to the voussoirs at the crown, there have been many differing examples; in general, we may put it down from J, to 77 of the span of the arch, varying according to parti- cular circumstances. Palladio gives the thickness of them in the ancient* bridge at Rimini, at 35 of the span to the middle arch, and 4 to the side arches. Again in the bridge at Vicenza, he gives the voussoirs to the middle arch ;4 of the span, and the side arches 1. But in a design by himself, he makes the thickness of the voussoirs in the middle arch ;4, and that of those of the side arches =1, of the span. L. B. Alberti recommends | them to be of the largest and hardest stones; and directs no stone to be used that is not at least +. of the chord of the arch; nor (says Alberti) should the chord itself be longer than six times the thickness of the pier, nor shorter than four times. In Westminster bridge the voussoirs are about 7; of the span, and at Blackfriars bridge nearly the same. ‘ il lor modeno e per la decima parte della luce de’ maggiori e per |’ ottava parte della luce de’ minori. Cap. 11. lib. 3. | } Lib. 3. cap. 14. + Lib. 4. cap. 6. SECT. IV. EQUILIBRIUM OF ARCHES. 79 It is to be observed that in most of, if not in all, their bridges, the ancients did not increase the dimensions of their voussoirs from the crown to- wards the springing or coussinet, but made them of an equal thickness throughout ; in this they were followed by Palladio, and all the Italian architects. They sometimes, however, made the alternate voussoirs larger than the others, as Mr. Labelye has done in Westminster bridge. While the voussoirs are considered as inde- finitely short, and are held in a state of tottering equilibrium by the vertical pressure of the supermceumbent loading alone, as the theory, in its simple state requires, the arch would not be calculated to support any extraneous weight. In practice, the voussoirs are of considerable length, and their adjacent surfaces are in contact; and when an additional weight is brought to act upon any part, suppose at the crown, it will cause the joints at such part to open on the concave side; the haunches will consequently be forced up, and their joints will open on the convex sides: but while the imaginary lines, expressing the directions of pressure passing through the voussoirs, are not so much dis. torted as to be thrown out of the limits of the 80 EQUILIBRIUM OF ARCHES. SECT. IV: surfaces of the blocks, the arch will stand, though loaded at the crown with a certain de- gree of weight beyond what the strictness of the theory allows; when these imaginary lines are removed out of the limits of the adjacent blocks or voussoirs, the arch will be completely de- stroyed. It therefore follows that the voussoirs should be as large as may conveniently be got; the larger they are the more may the distortion be increased, without endangering the structure, since the directions of their pressures will be less likely to exceed the limits of their magni- tude. In general, while a line can be drawn from the crown to the haunches, passing entirely within the surfaces of the voussoirs, the arch will stand; but, when any part of the line falls out of their surfaces, the stability of the arch is instantly destroyed. Enough, it is hoped, has been said to convince the artist of the necessity of balancing his arch with caution, and as exactly as circumstances will allow; and where an equilibrium according to this theory cannot be obtained, the fertile mind of an ingenious artist will naturally furnish expe- dients to obviate the inconvenience likely to arise from a want of it. SECT. V. EQUILIBRIUM OF ARCHES. 81 SECTION V. On the horizontal drift or shoot of an arch, and the thickness of the piers. PRELIMINARY OBSERVATIONS. If a bridge should consist of but one arch between the abutments, and each abutment does not immediately rest against an immove- able object, such as the bank of a river, it will be evident, if the materials composing the arch have a tendency to yield to any pressure in the direction of the length of the bridge, that the effect of this lateral thrust will be either to push the piers off horizontally, or to overturn them. The former effect may take place if the curve of the arch should begin near the foundations ; the latter, if it rest upon piers considerably ele- vated. In either case the thrust must be coun- teracted, by giving a proper thickness to the abutments. If a bridge consist of several arches, and we choose to consider the lateral thrust of each arch alone, without regard to the contrary G 82 EQUILIBRIUM OF ARCHES. SECT. V. thrusts of those with which it is connected, the counteraction must be effected by the same means. No theory, purely mathematical, has yet been discovered for determining the equilibration of | an arch in this respect, nor have all the circum- stances attending this species of pressure been satisfactorily ascertained. The principle which has been generally assumed as the basis of the investigation is, that the materials of which each semi-arch as A C DB (see the figures to Plate 1) is composed, have a freedom of motion and a tendency to roll or slide over the intrados_ BD, by a resultant of the action of gravity, and that they are retained by the side B A of the wall or pier. The equivalent of all these forces being estimated in some particular direc- tion, is considered as the force which tends to push off or overturn the pier or abutment. To determine this force, it becomes necessary to find the centre of gravity of any longitudinal section of a semi-arch; for in that point, by the — laws of mechanics, the mass of materials may be considered as collected. Or, if the force is - to be estimated in a horizontal direction, it will be merely necessary to determine the situation of a vertical line passing through the centre of SECT. V. EQUILIBRIUM OF ARCHES. 83 gravity, which is more easy than to determine that of the centre of gravity itself. If the arch be perfectly equilibrated by its loading, according to the rules delivered in this treatise, and if only such a portion of the arch is employed as will permit its extrados to serve accurately for the intended roadway, then the two following propositions will shew how, in a - very easy manner, to determine the situation of the vertical passing through the centre of gra- vity, and how from thence to determine the lateral thrust exerted by the arch. ~ ‘Op ¢ yd 84 EQUILIBRIUM OF ARCHES. SECT. V. PROPOSITION I. PROBLEM. To find the horizontal distance B’' f from the abutment or springing B’, of the centre of gravity of the materials with which the equilibrial arch B’ D ts loaded. At B’ and D draw the tangents B’ t, D t, intersecting each otherint. The cen- tre of gravity will be in a vertical line passing through t. From the point t of intersection let fall the vertical t f, cutting B’B in f. Then B’ f is the horizontal distance of the centre of gravity of the semi-arch B’ F q D, from the abutment B’, and f B the horizontal distance from q B. For the arch line B’ D with its loading may be considered as resting on the points B’ and D, SECT. V. EQUILIBRIUM OF ARCHES. 85 and exerting a pressure there in the directions of the tangents t B’, t D; the re-actions of the materials which support the half-arch at those points are consequently to be considered as forces in the opposite directions B’ t, Dt. But in me- chanics, where a body is sustained by two forces, those forces produced will meet, either in the centre of gravity of the body supported, or the centre of gravity will be in a vertical line passing through that point of intersection. ' EXAMPLE. lie 2iB t= 846 AG; and ZB’ ft = 90° 00’ HEE meyimliyg 8O!s Then t f=B D being = 40° 00’ and B B = 50° 00’ By plane trigonometry, We shall have Wenders. tor tte a nae ss et Saree fips ‘andf B = 40—3°699=36°301. = 3°699 When the curve thus loaded to equilibrium is one of those mentioned in note to Prop. 1, 86 EQUILIBRIUM OF ARCHES. SECT. V. Sect. 3, the angle f B’t may be found by the means there shewn; but for any other curve than those, other means to find that angle must be used. PROPOSITION II. PROBLEM. To find the horizontal shoot or drift of any semi-arch B’ D loaded to equilibrium (that ts, the force it exerts in an horizontal direction at B’). Let B’ F q D be the semi-arch. Find (by the last prop.) the horizontal distance B’ f of the centre of gravity of the arch loaded to equilibrium and draw f t perpendicular to B’ B. SECT. V. EQUILIBRIUM OF ARCHES. 87 Find the area of B’ F q D which may repre- sent the weight of the semi-arch, since the weight is proportional to the area of the section. Then by mechanics as t f: B’ f:: weight of the semi-arch : its shoot or horizontal drift in the direc- tion f B’. OBSERVATIONS. The method of estimating the lateral thrust, contained in the two preceding propositions, must not be employed when the arch rises per- pendicularly, or nearly so to the horizon. In fact, the intersection t of the tangents D t, Bat would then take place upon, or very near to the vertical line B’ F, and the semi-arch would ap- pear to have no lateral thrust, a circumstance which, though true in theory, must not be ad- mitted in practice. For as the extrados cannot, on account of its great height over the spring- ing courses, serve for a road, it is evident that - by cutting off the part above the line of the in- tended road, the position of the centre of gravity is altered, and to determine the vertical line t f, the centre of gravity of the semi-arch must be found by one of the usual methods. The easiest 88 EQUILIBRIUM OF ARCHES. SECT. V. would undoubtedly be to cut the form of the half arch in pasteboard, and make it support itself upon a point, the point thus found will be the place of the centre of gravity, which sup- pose to be at G: then the horizontal shoot or drift may be found as in Prop. 2, of this Sec- tion. Now to determine the thickness of a pier ne- cessary to counteract the horizontal thrust of the arch, we must consider that this thrust is to be resisted by the friction which the stones com- posing the pier experience in sliding upon each other. From sundry experiments it has been found, that in some kind of stone the friction of one block moving horizontally upon another, is equal to one third of the weight of the moving block. If we adopt this determination, the weight of the pier ought to be equal to three times the horizontal drift of the arch to produce an equilibrium. If A =the area of the semi-arch, which area may represent its weight: then, by Prop. 2, we have a =the horizontal thrust; hence 3BixA_., area of the vertical section F R tf of the pier, which area may represent the weight SECT. V. EQUILIBRIUM OF ARCHES. 89 8 BtxA of the pier, and +P eX Y — the thickness RQ required. The specific gravity of the materials compos- ing the arch and pier, is here supposed to be the same, which is generally the case, and on this account, it does not enter into the for- mula. The height X Y is given, and we sup- pose the whole pier to stand dry, or out of the water. If we could determine the conditions of equi- librium between the stability of the pier and the effort of the arch to overturn it about R, we might estimate the horizontal thrust as before ; thus let G be the centre of gravity of the half arch; draw G C parallel to B’ B, and suppose the horizontal thrust to be applied at C, at the extremity of the arm C Q of the bent lever CQR. In this case Bf x A x CQ aoe CQ press the effort of the arch to overturn the pier. But F QxR Q is equal to the area of the sec- tion of the pier, and represents its weight; and this weight is supposed to act in the vertical line X Y, passing through the centre of gravity of the pier, it consequently acts at the extremity of the lever R Y, which is equal to 4 R Q. will ex- 90 EQUILIBRIUM OF ARCHES. SECT. V. Therefore we have F Q x 4RQ’ to express the stability of the pier. Then equating these two forces and reducing, we have eh) ge 0 ESE oo NA or aie thickness required, supposing as before, the whole pier to stand out of water. But inasmuch as part of the pier is in most cases immersed in water, it thereby loses so much of its weight as is equal to the weight of a quan- tity of water, whose bulk is equal to that of the immersed part of the pier. To allow for this, let us suppose the material of the bridge to be granite, the specific gravity of which, to that of water is as 3g to 1, and let the pier stand in water as high as the springing B’. Then the stability of the pier will be expressed by Wi SoR-OexR OQ? 10-543 Osx Re@? ean the effort of the arch to overturn it will be Spe ps tex CO tx TA tf WBS f3se-@: QSnwk we have RQ = /tf35 FOB 0) (35 FQ—B Q,) The thickness of the piers determined on this principle will be found rather greater than is al- lowed by modern architects, and to reduce the results of theory nearer to the general practice, equating these terms, SECT. V. EQUILIBRIUM OF ARCHES. 91 mathematicians have brought the point of appli- cation of the thrust of the arch nearer to the foot — than the point C is; but as this is rather arbi- trary, it has been thought better to leave it in a horizontal line passing through G. It is for the architect to exercise his judgment in making such modifications of the theory. It is advisable (if possible) to construct the pier of heavier materials than the arch, as by that means it will not require so great a thickness, and a greater waterway * will consequently be obtained. If the pier be -constructed as we have sup- posed above, that is, independently of the resist- ance arising from the abutments, or from the neighbouring arches, a considerable saving of expense in the centering of a bridge, where se- veral arches are required, will accrue; for al- lowing the piers to be individually capable of * In the Philosophical Transactions 1758, is a paper by Robertson, on the fall of water under bridges, wherein the following formula is given. ((—— pee ah 1 )).— = fall of the river occasioned by the piers. 81 ¢ 3 Where b = breadth of the river, c = the waterway, v = the velocity of the river in one second, a = the fall of a heavy body in one second= 16-0899. 92 EQUILIBRIUM OF ARCHES. SECT. V. resisting the drift, the centre of one arch may be struck before another is begun, and it will serve for its opposite and corresponding one, that is, supposing them to be of equal sizes, which is, unless under particular circumstances, always the case. Otherwise the arches depend, for their stability, upon the counteracting ef- forts of each other, and additional centering, one of the most expensive pieces of machinery used in their construction, must be necessarily employed. The Italian as well as the ancient architects seem to have regulated the thickness given to the piers of bridges, by proportioning them to the span of the arches, without regard to any other circumstances. In the little bridge over the Ilyssus, near Athens, the middle arch of which is only 19 feet 10 inches span, the extraordinary thickness of nearly 8 feet 6 inches is given to the piers, being nearly five- twelfths of the span. In the bridge at Rimini, admired by Palladio for its beauty* as well as strength, nearly the same proportion is observed f. * “ Mi pare il piu bello e il piu degno di considerazione, si per la fortezza, come per il suo compartimento.” + ‘I pilastri sono grossi poco meno della metta della luce degli archi maggiori.” Lib. 3, cap. 11. SECT. V. EQUILIBRIUM OF ARCHES. 93 In the bridge over the Bacchilione, at Vicenza, the piers are one-sixth of the span of the middle arch; and in a design by Palladio himself, he makes the pier to the middle arch one-fifth of the span. L. B. Alberti says, that the piers ought not to be less than one-sixth, nor more than one-fourth of the span. _ At Westminster bridge the piers to the middle arch are a little more than one-fourth of the span, and at Blackfriars bridge a little more than one- fifth. It is evident, however, after what has been shewn, that in order to estimate the proper thick- ness, the circumstances of the span, height, forms of the extrados and intrados, must enter into the consideration. The extremities of the piers of bridges are usually terminated in points towards the current of a river, in order to diminish the pressure of the water against the piers, and lessen the eddy which the water, flowing off laterally after striking the piers, forms at the shoulders. Now, to compare the pressure sustained by a pier terminating in a head, as A B, at right angles with the stream, with that sustained by one whose extremity 94 EQUILIBRIUM OF ARCHES. SECT. V. forms an isosceles triangle as A DB on the plan: let X Y parallel to the length of the pier represent the force with which the water would strike any point of the square head A B: this force may be re- solved into X Z, which is parallel and Z Y which is perpendicular to the face AD. The portion of the force of water represented by X Z being parallel to A D produces no effect upon it, and the remaining force ZY may be resolved into the two Z W perpendicular to, and W Y coin- cident with the length of the pier: the portion of the force of the water represented by ZW is counteracted by an equal force in an opposite direction, and there only remains the force represented by Y W tending to push the pier from its place, consequently the pressure of the current against a pier terminated perpendicu- larly to the stream, as A B is to the pressure against one terminated by the triangle A D B, ase Y to VW xe aa SECT. V. EQUILIBRIUM OF ARCHES. 95 If the angle at D were a right angle, the force against A DB would'be only half the force against A B, and it is evident that the force must become less in proportion as the angle becomes more acute; and in fact it may be shewn that the force is inversely proportional to the square of the side A D. | It is evident too, that curvilinear piers oppose less resistance to the stream the nearer their faces approach to right lines; but care should be taken not to make the angle at D too acute, on ac- count of the injury which vessels may sustain by being driven against it, and the eddy which is produced when the current sets obliquely to the pier. Though some objections have been made to this theory, nothing better has yet been submitted to calculation. OF DOME VAULTING. A work of this kind would be incomplete if it did not contain some account of the principles of equilibrium in domes of masonry; it is therefore purposed to conclude this section by a few words on that species of building. 96 EQUILIBRIUM OF ARCHES. SECT. V. Let Figure 1, Plate 4, represent part of a ver- tical section through the centre of the dome; let A.B represent one half of the keystone, and the line A B the direction in which it presses against the adjacent stone BC, that is perpen- dicular to the joint B: find g the centre of gra- vity of the half keystone; draw the vertical line gv, and make it of any length at pleasure to represent the weight of the half-stone A B, and through v draw vf at right angles to gv, meeting A B produced in f. If g v represent the weight of the stone A B, then will gf repre- sent its pressure against the adjacent stone BC, in a direction perpendicular to the joint B. Now make Bb equal to and in the same direc- tion as of, and on an indefinite vertical line drawn through B take Be to represent the weight of the stone BC, complete the parallelo- gram cb, and draw the diagonal Bd; then will Bd represent the quantity and direction of the pressure of the two stones AB, BC upon the joint C. Again, make Ce equal to and in the same direction as Bd, and on an indefinite ver- tical line through C take Ch to represent the weight of the stone CD; complete the parallelo- eram he, and draw the diagonal C k, then will Ck represent the quantity and direction of the SECT. Vv. EQUILIBRIUM OF ARCHES. O7 pressure of the three stones A B, BC, C D upon the jot D. Make Dm = Ck, and proceed as before. It is evident from this construction that the pressure of the materials in the vertical sections of the dome changes its direction continually : viz. from A B to B C, toC D, and go on; and, if the breadths A B, B C, &c. of the stones were indefinitely small, the polygon A B, BC, CD, &e. would become a curve concave towards the axis of the dome. Now, in forming a cylindrical vault which should support itself without any loading, the above construction might be employed, and in that case, the several verticals B c, Ch, D 0, &e. which represent the weights of the arch stones, or the vertical pressures which they exert upon the next lower joints C, D, E, &c. must be equal to each other when the lengths of the arch Stones are equal, which is here supposed, and each must be equal to twice g v (g v represent- ing the weight of a half voussoir) ; consequently the polygon formed would be an approximation to the common catenary. But in forming a dome, the keystone at A rests on a circular base formed by the next inferior horizontal course, and the lower circumference of each H 98 EQUILIBRIUM OF ARCHES. SECT. V. horizontal course is greater than the upper. Therefore each horizontal course rests upon a oreater number of points than the course next superior to it, and the number of points which the courses rest upon is proportional to the cir- cumferences, or to the radii of their bases. Hence, if we have assumed any line, as g V, to represent the weight of the half arch-stone ACB; or rather to represent the vertical pressure which that half stone exerts on the oblique joint B; then it is plain that to get the weight of the arch-stone BC, or rather the vertical pressure it exerts against the oblique joint C, we must make Be equal to 2 g v diminished in the pro- portion of the radius of the horizontal course B C taken at C, to the radius of the same course, taken at B; that is zC:wB::2gv: Be. Afterwards Ch instead of being made equal to Bc, must be obtained by the following propor- tion x D:zC::Be:Ch; and in the same manner the other verticals may be found. It must be observed, however, that the radii z C, x D, &e. cannot be accurately known till the positions of the courses BC, C D, &e. are determined, but an approximation may be made to these radii by first making B e=2 g v de- termining the direction of the diagonal B d, and SECT. V. EQUILIBRIUM OF ARCHES. 99 the position of C as described above. The radius zC may then be employed to get a more correct value of Bc, from which the position of C may be determined with sufficient accuracy. In the same manner the approximate value of x D may be found, and from thence the more correct place of the point D, and so on. Now, if the dome were so constructed that the several lines of direction A B, BC, &c. were always within the thickness of the voussoirs, it would stand and be in perfect equilibration ; but if the lines of direction of the forces fall on the exterior of the curve of the dome, it is evident that the pressures of the upper courses would give the lower ones a greater tendency to slide outward at the joints than would be counter- acted by their weights, and consequently the dome would fall. Again, if the directions of the forces fell on the interior of the curve, the stones would tend to fall inward, but as this would. take place equally in every stone of each horizontal course respectively, the tendency would not be obeyed, © the breadth of the stones on the exterior being greater than on the interior. It is even consi- | dered that in proportion to the flatness of the dome, the action of gravity binds the stones in Hine 100 EQUILIBRIUM OF ARCHES. SECT. V. the horizontal courses together more firmly. In this case, however, its action in the vertical planes increases the horizontal thrust of the dome towards the exterior, in the direction of the radii, and the flatness may be such as to cause the evil arising from the tendency in the latter direction, to exceed the advantage gained by the former. If, however, the foot of the dome is sufficiently secured against giving away by the horizontal thrust outward, then the dome may be considered more firm in proportion as the curve falls within that of equilibration ; within this limit any curve, whether convex or concave, may be chosen for the vertical section of a dome. Fig. 2, Plate 4, exhibits the curve of equilibra- tion for one half of a dome, and is constructed from a table calculated by Dr. Robison. It is evident that the voussoirs of domes being bound together by forces acting both in vertical and horizontal planes, form a building, having greater stability than a cylindrical vault, in which the voussoirs are only pressed together in vertical planes: perforation may be made in any part of a dome without sensible injury, the upper part may be either left quite open as in the Pantheon, or it may be crowned by a cupola, as in the cathedrals of St: Peter and St. Paul. SECT. V. EQUILIBRIUM OF ARCHES. 101 Dr. Robison shews that when a dome is spherical, it is not safe to employ a segment of more than about 103 degrees for the vertical section: beyond this limit the lower part would fall within the lines of direction of the forces. TO DETERMINE THE HORIZONTAL THRUST OF A DOME AT ITS BASE. If the courses of which a dome is composed be indefinitely thin, or, at least, if their thick- ness bears but a small proportion to the mag- nitude of the dome, and if the directions of the pressures of the courses in a vertical plane be perpendicular to the joints, as in the dome of equilibration, then it is evident that the pres- sure upon the lowest joint all round the base of the dome will be the same as would be pro- duced by a cone of equal weight, whose slant side is coincident with the direction of a tangent to the curve of the dome in a vertical plane at the springing course. Let the curve V A, Fig. 3, Plate 4, be that by whose revolution about V B the given dome is formed, and let A C be a tangent to V A at the 102 EQUILIBRIUM OF ARCHES. SECT. V. springing. Then, if the cone formed by the revolution of AC be of equal weight with the . dome, the pressure of the cone will be equal to that of the dome. Now, let the whole weight of the dome or cone be accumulated in one point of the side of the cone; suppose at C; then the weight will be to the horizontal thrust at A, produced by that weight, as C B to A B. If, for example, the whole weight of the dome should be 1000 tons, and that A B should be equal to the half of C B, the horizontal thrust at A would be equal to 500 tons; but this thrust is distributed all round the base of the dome, and we are to find what it is equivalent to on every point of that base. - Since A B may be taken to represent the thrust of the dome, we may consider a line equal in length to A B, as expressive of a force which would resist that thrust when applied at one point; and, we may consider the whole thrust as equally diffused over a line equal to A B. Consequently the pressure which each point in such a line would have to support, will diminish as the length of the line increases; that is, the pressure on any point would be inversely proportional to the length of the line. Therefore, SECT. V. EQUILIBRIUM OF ARCHES. 103 if we suppose the thrust of the dome to be resisted by a force acting-all round the base, and tending every where towards the centre, the thrust upon each point of the line equal to A B, will be to that on each point of the base inversely, as that line is to the circumference of the base: that is inversely as the semi-diameter of a circle is to its circumference. But the radius being 1, the circumference is 6°283. Consequently, to find the horizontal thrust on every point of the base in the above example, we may say 6°283 : 1::500 tons : 794 tons, nearly the thrust required. If the dome be supported on a circular wall of masonry, whose height is A G, the above thrust must be multiplied by the arm A G of the lever, at the end of which it acts, and the product will express the force with which the dome endeavours to overset the wall. This, of course, must be resisted, as in the case of com- mon vaulting, by a force expressed by the area of the section A D of the wall, multiplied by the half breadth DH, (supposing the wall and dome constructed of materials having the same specific gravity, and the pier rectangular,) and as all the terms are given, except D G, this may also be found. 104 EQUILIBRIUM OF ARCHES. SECT. V. If the dome is intended to be hooped with iron at the base as usual, and we would give such dimensions to the ring that it shall resist the strain, we have only to find from the tables that are published, how much a bar of iron of given dimensions, (whose section is one square inch for example,) will support, without breaking, when a load is diffused uniformly over it; then the hori- zontal thrust divided by this tabular number, will give the number of square inches in a section of the intended ring. THE END. G. Woodfall, Printer, Angel Court, Skinner Street, London. rn 3 0112 017293