UC-NRLF 'i I ^1 Sin I u i « • cT gOldBoc LIBRARY OF THE University of California, GIFT OF Class Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementsofgeometOOsmitrich da. Elements of Geometry CONTAINING BOOKS I. TO VI. AND PORTIONS OF BOOKS XL AND XII OF EUCLID BY J. HAMBLIN SMITH, M.A., 0} Gonville and Caius College, and late Lecturer at St. Peter's College, Cambridge. WITH A SELECTION OF EXAMINATION PAPERS, THOS. KIEKLAND, M.A., Science Master, Normaf School, Toronto. FOURTH CANADIAN COPYEIGHT EDITION. ^xttl^on?,cb bij t^c UUnisttr of (Sbucation. TORONTO : ADAM MILLER & CO., 1879. Entered accordiac to the Act of th« Parliament of Canada, in the year on*. thousand eight hundred and seventy-seven, by Adam Millbk & Ca, Id the OffioM of the Minister of Agriculture. PREFACE. To preserve Euclid's order, to supply omissiom, to remove defects, to give short notes of explanation and simpler method^ of proof in cases of acknow- ledged difficulty — such are the main objects of this Edition of the Elements. The work is based on the Greek text, as it is given in the Editions of August and Pe}Tard. To the suggestions of the late Professor De Morgan, published ia the Companion to the British Almanack for 1849, I have paid constant deference. A limited use of symbolic representation, wnerem the symbols stand for words and not for operations, is generally regarded as desirable, and it is certain that the symbols employed in this book are admis- sible in the Examinations at Oxford and Cambridge. I have generally followed Euclid's method of proof, but not to the exclusion of other methods recommended by their simplicity, such as the de- monstrations by which I propose to replace the difficult Theorems 5 and 7 in the First Book. I 1H4005 PREFACE. have also attempted to render many of the proofs, as, for instance, those of Propositions 2, 13, and 35 in Book L, and those of 7, 8, and 13 in Book II., less confusing to the learner. In Propositions 4, 5, 6, 7, and 8 of the Second Book I have ventured to make an important change in Euclid's mode of exposition, by omitting the diagonals from the diagrams and the gnomons from the text. In the Third Book I have deviated with even greater boldness from the precise line of Euclid's method. Thus I have given new proofs of the Pro- positions relating to the Contact of Circles : I have used Superposition to prove Propositions 26 to 29, so as to make each of those theorems independent of the others ; and I have directed the attention of the learner to the Intersection of Loci, and to the conception of an Angle as a magnitude capable of unlimited increase. In the Fourth Book I have made no change of importance. My treatment of the Fifth Book was suggested by the method first proposed, explained, and de- fended by Professor De Morgan in his Treatise, on the Connexion of Nuinher and Magnitude. The method is simple and rigorous, presenting Euclid's PREFACE. tH reasoning in a clear and concise form, by means of a system of notation, to which, 1 think, no valid objection can be taken. I have altered the order of the Propositions in this Book, so as to give promi- nence to those which are of chief importance. The only changes in the Sixth Book to which 1 desire to call the I'eader's special attention, are the applications of Superposition in the proofs of Pro- positions 4 and 19. The diagrams in Book XI. form an important feature of this Edition. For them I am indebted to the kindness of Mr. Hugh Godfray, of St, John's College, Cambridge. The Exercises have been selected with consider- able care, chiefly from the University and College Examination Papers. They are intended to be pro- gressive and easy, so that a learner may be induced from the first to work out something for himself A complete series of the Euclid Papers set in the Cambridge Mathematical Tripos from 1848 to 1872 will be found on pp. 198-210 and 342-349. I have made but little allusion to Projections, because that part of the subject is fully explained by Mr. Kichardson in his work on Conic Sections treated Geometrically, forming a part of Rivington's Mathematical Series, irffl PREFACE. During the two years in which I have been en- gaged on this work, I have received from Teachers of Geometry in all parts of the country so much encouragement to proceed, and so much assistance at each step of my progress, that I feel justified in asserting that no text-book on Elementary Geometry is likely to meet with general support in England, if it involve any wide departure from the Euclidean model. It only remains for me to offer my thanks to the friends who have improved this work by their advice, and to assure each reader of the book that any suggestion for its furl her improvement will be thankfully received by me. J. HAMBLIN 8MITH. 42 Trumpinoton Stbbw^ Oaubrhvoe, \'Si% CONTENTS Introductory Remarks on Solids, Surfaces, AND Points . . . , LiNBS. EUCLID'S ELEMENTS— BOOK L Definitions I. to XXVI. .... Postulates • • Axioms . . . . , , « Symbols and ABBRBviATioNa • « • SECTION I.— ON THE PROPERTIES OF TRIANGLES— Pp. 10 to 48, tions Euclid's Propositions I. to IV. . Note 1. On the Method of Superposition Note 2. On the Conditions of EguALrrv of two Tri- angles Propositions A, B, 0, in place of Euclid's Proposi v., VI., VIL, VIIT. Euclid's Propositions IX. to XII. Miscellaneous Exercises on Props. L to XII. Euclid's Propositions XIII. to XV. Note 3. On Euclid's Definition of an Angle Evclid's Propositions XVI. and XVII. Note 4. On the Sixth Postulate 10 14 16 16 20 24 25 28 29 CONTENTS. Euclid's Propositions XVIII. to XXIIT , Proposition D . ^ . . . Euclid's Propositions XXIV to XXV J. Miscellaneous Exercises on Props. 1. ro KX^"!. Proposition E PAGE 31 37 38 41 42 SECTION II.— ON THE THEORY OF PARALLEL LINES Pp. A* to 56. FnTRODUCTORY R F.MARKS . . . , Euclid's Propositions XXVII. and XXV[IL Note 5. On the Sixth Postulate Euclid's Propositions XXIX. to XXXIII. . Miscellaneous Exercises on Sections I. and II. 44 45 47 48 56 SECTION III.— ON THE EQUALITY OF RECTILINEAR FIGUHkH IN RESPECT OF AREA— Pp. 57 to 70. Introductory Rkmakks . ... 9 • 67 Definitions XXVIl. to XXXIII. . . . . « 56 Exercisp:s on Definitions 5f* litrcLiD's Propositions XXXIV. to XLV. , . . 6<) Miscellaneous Exercises on Props. XXXIV. to XLV. 72 Euclid's Propositions XLVI. to XLVIIL ... 73 EUCLID'S ELEMENTS— BOOK IL Introductory Remarks . . . , Proposition A i Euclid's Propositions I. to VI. . , ♦ Proposition B . . . . . • Euclid's Propositions VII. to XIV. . , Miscellaneous Exercises on Book II. . , Note 6. On the Measurement of Arkas . Note 7. On Pro.jections .... Note 8. On Loci ..... 77 77 78 84 85 93 95 102 103 CONTENTS. Note 9. On the Methods employed in the Solution OF Problems .... Note 10. On Symmetry .... Note 11. Euclid's Proposition V. of Boor I. Note. 12. Euclid's Proposition VI. of Book I. Note 1.3. Euclid's Proposition VII. of Book I. Note. 14. Euclid's Proposition VIII. of Book I. Note 1 5. Another Proof of Euclid I. 24 . Note 16. Euclid's Proof of Prop. XXVI. of Book I, MtSCELLANEOUS EXERCISES ON BoOKS I. AND II. 105 107 108 110 110 112 113 114 116 EUCLID'S ELEMENTS— BOOK IIL Ol^cI^ Postulate and Definitions I. to VL . Euclid's Propositions I. to V. Note 1. On the Contact of CiRCLEr: , Definition VII Euclid's Propositions VI. to X. . , Proposition A (Eucl. III. 25) , , Proposition B (Eucl. IV. 5) , , Euclid's Propositions XI. to XV. . Definition VIII Definitions IX. to XI. Euclid's Propositions XVI. to XX. Note 2. On Flat and Reflex Angles Proposition C Definition XII Euclid's Propositions XXI. and XXII. N(ri.e. 3. On the Method of Superposition TO Circles Definition XIII Euclid's Propositions XXVI. to XXiX. Note 4. On the Symmeirkal Fkoi-erties of the Cikcl with regard to its DlAMKTER AS AiTLiED 121 123 128 128 129 134 iSf) 136 139 142 143 149 150 150 151 15.^ 154 16f 159 xi\ CONTEr^TS. W< PAOl EaonD's Propositions XXX. to XXXVIL . . .160 Miscellaneous Exeuolses on Book IK. . . . 169 Euclid's Propositions XXIII. and XXIV. of Book III. 175 Another Proof OF III. 22 177 Another Proof of III. 31 , , , , , , 178 EUCLID'S ELEMENTS— BOOK IV. Introductory Eemakks 179 •^ Euclid's Propositions I. to IV. . . , , . 180 Euclid's Propositions VI. to XVI. , . , ,184 Miscellaneous Exercises on Book JV. , , , 196 kkr^^Uwi^ APPENDIX TO BOOKS I.— IV. t-^W-^r/jSucLiD Papers set in the Cambridge Mathematical ^ ' Tripos from 1848 TO 1872 198 ..-— EUCLID'S ELEMENTS— BOOK V. SECTION T.—ON MULTIPLES AND EQUIMULTIPLES— Pp. 211 fo 214, Defin itions I. 11. . . , 211 Postulate . . . • , • . • .211 Method of Notation , • , • • . .211 Scales of Multipi ks . , , • • . . 212 Axioms 212 Note 1 213 Proposition I. (Eucl. V. 1) 213 Proposition II. (P:ucl. V. 2). . . .. . .214 Proposition HI. (Eucl. V. 3) . . . . 214 SECTION II.- ON RATIO AND PROPORTTOfi" Pp. 215 to 229 Definition IIT, . . > . . .215 Note 2 ......... 215 NoteSi . .216 CONTENTS. PAOS DEPiNiTioirs TV. V .217 Notes 4 and 5 218 Definition VI , , • .219 Definition VIL 220 Note^ ,...,.. .220 SECTION TIT.— CONTAINING THE PROPOSITIONS MOST FRE- QUENTLY REFERRED TO IN BOOK VI.— Pp. 221 to 226. Note 7 Eitclid's Pko position TV Proposition V. (Eucl. V. 11) Proposition VI. (Eucl. V. 7) Proposition VII. (Eucl. V. 8) Proposition VIII. (Eucl. V. 9) PpnposiTiON IX. (Eucl. V. 10) Proposition X. (Eucl. V. 12) Proposition XI. (Eucl. V. 15) SECTION IV,— ON PROPORTION BY INVERSION, ALTERNATION AND SEPARATION— Pp. 227 to 280. Proposition XII. (Eucl. V. b) 227 Proposition XIII. (Eucl. V. 13) 228 Proposition XIV. (Eucl. V. 14) 228 Proposition XV. (Eucl. V. 16) . . . . .229 Proposition XVI. (Eucl. V. 18) 230 SECTION v.— CONTAINING THE PROPOSITIONS OCCASIONALLY REFERRED TO IN BOOK VI.— Pp. 231 to 234. Proposition XVII. (Eucl. V. 4) Proposition XVIII. (Eucl. V. a) Proposition XIX. (E(7cl.V. d) Proposition XX. (Eucl. V. 20) Proposition XXI. (Eucl. V. 22) Proposition XXTl. (Eucl. V. 24) 231 232 232 233 234 234 CONTENTS. SECTION VJ.— CONTAINING THE PROPOSITIONS TO WHICH REFERENCE IS MADE IN BOOK VI.— Pp. 235 to 242. Proposition XXIII. (Eucl. V. 5) . Proposition XXIV. (Eucl. V. 6) . Proposition XXV. (Eucl. V. 17) . Proposition XXVI. (Eucl. V. 19) Proposition XXVII. (Eucl. V. 21) Proposition XXVIII. (Eucl. V. 23) Proposition XXIX. (Eucl. V. 25) Proposition XXX, (Eucl. V. c) Proposition XXXI. (Eucl. V. b) NO TO VI EUCLID'S ELEMENTS— BOOK VL Introductory Bemarks Definition I. .... Euclid's Propositions I. to VI. Miscellaneous Exercises on Props. I. Euclid's Proposition VII. Proposition VIII. (Eucl. VI. 9) . Proposition IX. (Eucl. VL 10) . Proposition X. (Eucl. VI. 11) , Definitions II. III. . . , Proposition XI. (Eucl. VI. 12) Proposition XII. (Eucl. VI. 8) . Euclid's Proposition XIII. . . Definition IV Euclid's Propositions XIV. to XIX. Exercises on Proposition XIX. . Proposition XX. (Eucl. VI. 21) , Proposition XXI. (Eucl. VL 20) . Proposition XXII. (Eucl. VI. 31) . Euclid's Proposition XXIII. Proposiiion XXIV. (Eucl. VL 22) CONTENTS. XV PAGE Proposition XXV. (Ettcl. VI. 33) . 2S2 Proposition B . . . . . 284 Proposition C . • . . « . 285 Proposition D . . . . i . 286 Proposition XXVI. (Eucjl. VI. 23) , . 287 Proposition XXVII. (Eucl. VI. 24) . 288 Proposition XXVIII. (Eucl. VI. 26) . 289 Proposition XXIX. (Eucl. VI. 25) . 290 Definition V. . . . . . 291 Proposition XXX. (Eucl. VI. 30) . 291 Proposition XXXI. (Eucl. Vt. 32) . 292 Mlsckllaneous Exercises on Book VL . 293 EUCLID'S ELEMENTS— BOOK XL Introductory Remarks Definitions I. to XXX. Postulate .... Proposition I. (Eucl. XI. 2) . Proposition II. (Eucl. XI. 3) Euclid's Propositions IV, to XXI Miscellaneous Exercises on Book XL 307 307 310 311 313 314 334 EUCLID'S ELEMENTS— BOOK XIL Lemma 337 Euchjd's Propositions I. and II 338 Papers on Euclid (Books VI. and XI.) set in the Cam- bridge Mathematical Tripos 342 IVERSI OF ELEMENTS OF GEOMETRY. INTRODUCTOR\ REMARKS. When a block of stone is hewn from the rock, we call it a Solid Body. The stone-cutter shapes it, and brings it into that which we call regularity of foim ; and then it becomes a Solid Figure. Now suppose the figure to be such that the block has six flat sides, each the exact counter- \ part of the others ; so that, to one who stands facing a corner of the block, the three sides which are visible present the appearance re- x ' presented in this diagram. Each side of the figure is called a Surface; and when smoothed and polished, it is called a Plane Surface. The sharp and well-defined edges, in which each pair of sides meets, are called Lines. The place, at wliich any three of the edges meet, is caUed a Point A Magnitude is anything which is made up of parts in any way like itself. Thus, a line is a magnitude ; because we may regard it as made up of parts which are themselves lines. The properties Length, Breadth (or Width), and Thickness (or Depth or Height) of a body are called its Dimensions. We make the following distinction between Solids, Surfaces, Linea, and Points : ;' A Solid has three dimensions, Length, Breadth, Thickness, ]jl A Surface has two dimensions. Length, Breadth. / A Line has one dimension, Length. I . A point has no dimensions. ' f 8. X. BOOK I DEFINITIONS. I. A Point is that which has no parts. ' This is equivalent to saying that a Point has no magnitade, since we define it as that which cannot be divided into smaller parts. II. A Line is length without breadth. We cannot conceive a visible line without breadth ; but we can reason about lines as if they had no breadth, and this is what Euclid requires us to do. III. The Extremities of finite Lines are points. A point marks position, as for instance, the place where a line begins or ends, or meets or crosses another line. IV. A Straight Line is one which lies in the same direction from point to point throughout its length. V. A Surface is that which has length and breadth only. VI. The Extremities of a Surface are lines. VII. A Plane Surface is one in which, if any two points be taken, the straight line between them lies whoUy in that surface. Thus the ends of an uncut cedar-pencil are plane sur&ces ; but the rest of the surfetoe of the pencil is not a plane surfeu^, ^ce two points may be taken in it such that the straight line joining them will not lie on the surface of the pencil In our introductory remarks we gave examples of a Surface, a Line, and a Point, m we know them through the evidence of the senses. s Book I.] DEFINITIONS. II The Surfaces, Lines, and Points of Geometry may be regarded as mental pictures of the surfaces, lines, and points which we know from experience. It is, however, to be cbservod that Geometry requires us to conceive the possibility of the oxistenoo of a Surface apart from a Solid body, of a Line apart from a Surface. of a Point apart from a Line. VIII. When two straight lines meet one another, the inclina- tion of the lines to one another is called an Angle. When two straight lines have one point common to both, they are said to form an angle (or angles) at that point. The point is called the vertex of the angle (or angles), and the lines ire called the ourim of the angle (or angles). ■3 m Thus, if the lines OAy OB are terminated at the same point 0, they form an angle, which is called the angle at 0, or the angle AOB, or the angle BOA, — the letter which marks the vertex being put between those that mark the arms. Again, if the line CO meets the line DE at a point in the line DE, so that is a point common to both lines, CO is said to make with DE the angles COD, COE ; and these (as having one arm, CO, common to both) are called adjacent angles. Lastly, if the lines FG, HK cut each other in the point 0, the lines make with each other four angles FOH, HOG, GOK, KOF ; and of these GO II, FOK are called vertically opjiosit*: angles, as also are FOII and GOK. EUCLID'S ELEMENTS. LBook I. When iliriQ or more straight lines as OJ., OB, 00, OD have ft point common to all, the angle formed by one of them, OD, with OA may be regarded as being made up of the angles AOB BOG, COD ; that is, we may speak of the angle AOD as a whole, of which the parts are the angles AOB, BOG, and COD- Hence we may regard an angle as a Magnitude, inasmuch as any angle may be regarded as being made up of parts which are themselves angles. The size of an angle depends in no way on the length of the arms by which it is bounded. We shall explain hereafter the restriction on the magnitude of angles enforced by Euclid's definition, and the important results that follow an extension of the definition. IX. When a straight line (as AB) meeting another straignt line (as CD) makes the adjacent ^^ • angles (ABC and ABD) equal to one another, each of the angles is called a Right Anole ; and each line is said to be a Pbb- PBNDICULAR tO the OthCF. X. An Obtuse Angle is one whidi is greater than a right angle. XI. An AcuTB Angle is one which is less than a right angle. XII. A Figure is that which is enclosed by one or more boundaries. Book I.] DEFINITIONS. XIII. A Circle is a plane figure contained by one line, which is called the Circumference, and is such, that all straight lines drawn to the circumference from a certain point (called the Centre) within the figure are equal to one another. XIV. Any straight line drawn from the centre of a circle to the circumference is called a Radius. XV. A Diameter of a circle is a straight line drawn through the centre and terminated" both ways by the circumference. Thus, in the diagram, is the centre of the circle AB(W^ OA, OBy OCy OD are Eadii of the circle, and the straight line AOD is a Diameter. Hence the radius of a circle is half the diameter. XVI. A Semicirclb is the figure contained by a diametei and the part of the circumference cut oflF by the diameter. XVII. Rectilinear figures are those which are contained by straight lines. The Perimeter (or Periphery) of a rectilinear figure is the sum of its sides. XVIII. A Triangle is a plane figure contained by three straight lines. XIX. A Quadrilateral is a plane figure contained by four straight lines. XX. A Polygon is a plane figure contained by more than four straight lines. When a polygon ha« all its sides equal and all its angles equal it is called a regular polygon. EUCLID'S ELEMENTS. [Book L I XXL An Equiijlteral Triangle is one which has all its sides equal. XXII. An Isosceles Triangle is one which has two sides equal The third side is often called the hose of the / triangle. /_ The term base is applied to any one of the sides of a triangle to distinguish it from the other two, especially when they have been previously mentioned. •■ XXIII. A Right-angled Triangle is I \ one in which one of the angles is a right \ angle. /." The side subtending, that is, tuhich is opposite the right ancle, it Is called the Hypotenuse. XXIV. An Obtuse-angled Triangle i* une in which one of the angles is obtuse. It will be shewn hereafter that a triangle can have only one of its angles either equal to, or greater than, a right angle. II XXV. An Acute-angled Triangle is one in which ALL the angles are acute. XXVI. Parallel Straight Lines are such as, being in the same plane, never meet when " continually produced in both directions. Euclid proceeds to put forward Six Postulates, or Requests, that he may be allowed ta make certain assumptions on the construction of fia-nres and the properties of geometrical niag- nitiules. Book I.] POSTULATES. fi Postulates. Let it be granted — ' I. That a straight line may be drawn from any one point to any other point. II. That a terminated straight line may be produced to any length in a straight line. III. That a circle may be described from any centre at any distance from that centre. IV. That all right angles are equal to one another. V. That two straight lines cannot enclose a space. VI. That if a straight line meet two other straight lines, so as to make the two interior angles on the same side of it, taken together, less than two right angles, tliese straight lines being continually produced shall at length meet upon that side, on which are the angles, which are together less than two right angles. The word rendered "Postulates" is in the original rttTJ^fiara, " requests." . In the first three Postulates Euclid states the use, under certain restrictions, which he desires to make of certain in- struments for the coTistruction of lines and circles. In Post, I. and ii. he asks for the use of the straight ruler, wherewith to draw straight lines. The restriction is, that the ruler is not supposed to be marked with divisions so as to measure lines. In Post III. he asks for the use of a pair of compasses, wherewith to describe a circle, whose centre is at one extremity of a given line, and whose circumference passes through the other extremity of that line. The restriction is, that the compasses are not supposed to be capable of conveying distances. Post. IV. and v. refer to simple geometrical facts, which Euclid desires to take for granted. Post. VI, may, as we shall shew hereafter, be deduced from a more simple Postulate. The student must defer the consideration of this Postulate, till he has reached the 17th Proposition of Book I. Euclid next enumerates, as statements of fact, nine Axioms 8 EUCLID'S ELEMENTS, tBook L or, as he calls them . Common Notions, applicable (with the exception of the eighth) to all kinds of magnitudes, and not necessarily restricted, as are the Postulates, to geometrical magnitudes. Axioms. I. Things which are equal to the same thing are equal to one another. ^^'^ II. If equals be added to equals, the wholes are equal. ^^ III. If equals be taken from equals, the remainders are ; equal IV. If equals and unequals be added together, the wholes are unequal V. If eqiials be taken from unequals, or unequals from equals, the remainders are unequal VI. Things which are double of the same thing, or of equal , things, are equal to one another. VII. Things which are halves of the same thing, or of equal things, are equal to one another. VIII. Magnitudes which coincide with one another are equal to one another. '.^^ IX. The whole is greater than its part. With his Common Notions Euclid takes the ground of authority, saying in effect, " To my Postulates I request, to my Common Notions I claim, your assent." Euclid develops the science of Geometry in a series of Propositions, some of which are called Theorems 'and the rest Problems, though Euclid himself makes no such distinction. By the name Theorem we understand a truth, capable of demonstration or proof by deduction from truths previously admitted or proved. By the name Problem we understand a construction, capable of being effected by the employment of principles of construc- tion previously admitted or proved. A Corollary is a Theorem or Problem easily deduced from, or effected by means of, a Proposition to which it is attached. We shall divide the First Book of the Elements into three sections. The reason for this division will appear in the course of the work. Book L] SYMBOL:^ AND ABBKEVJATIONS. SYxMBOLS AND ABBREVIATIONS USED IN BOOK I •.' /or because .• therefore =» is (or are) equal to L angle A triangle equilat equilateral extr exterior intr interior pt point rectil rectilinear © for circle Oce circumference II parallel O parallelogram ± perpendicular reqd requirea rt right sq square sqq squares 8t straight It is well known that one of the chief difficulties with learners of Euclid is to distinguish between what is assumed, or given, and what has to be proved in some of the Pro- positions. To make the distinction clearer we shall put in italics the statements of what has to be done in a Problem, and what has to be proved in a Theorem. The last line in the proof of every Proposition states, that what had to be done or proved has been done or proved. The letters q. b. f. at the end of a Problem stand for Q;mA erai faciendvm. The letters q. e. d. at the end of a Theorem stand for Quod erai demonstrandum. In the marginal references : Post, stands for Postulate. Def. Definition. Ax Axiom. I. 1 Book I. Proposition L Hyp. stands for Hypothesis, supposition, and refers to ^Kiething granted, or assumed to be true. lO liUCLIiys ELEMENTS. [Book 1 SECTION I. On the Properties of TriangU$, Proposition I. Problem. To defterihe an ^^uilateral triangle on a gi/ven nircdght Un«. Let ABhe the given st. line. It is required to describe an equilat A on AB. With centre A and distance AB describe © BCD. With centre B and distance BA describe © ACE. From thept. C, in which the ©s cut one another, draw the st. lines CA, CB. Then will ABC be an equilat. A . For Post. 3. Post. 3. Post. 1. •.* ^ is the centre of © BCDy .-. AC=AB. Bef. 13. And ••• B is the centre of © ACE, .'. BC=^AB. I>ef. 13. Now •.• AC, BC nre each =ABf .: AC^BC. Ax. 1. Thus AC, AB, BC are all equal, and an equilat. a ABC has been described on AB. 0. F- F. ::^ooic I.] PROPOSITION II. il I Proposition II. Problem. From a given point to draw a straight line eqiud to a given gtradght line. Let A he the given pt., and BC the given h*. line. It is required to draw from A a st. line equal to BC. From ^ to J^ draw the st. line AB. Post. I. On AB describe the equilat. a ABD. I. 1. With centre B and distance BC describe © CGH. Post. 3. Produce DB to meet the Qcc CGH in G. With centre D and distance DG describe © GKL. Produce DA to meet the Qce GKL in L. Then will AL=-BG. For *.• B is the centre of © CGH, .'. BC=BG. And *.• D is the centre of ® GKLy .'. DL=DG. And parti of these, DA and DBj are equal. .*. remainder J.I/= remainder BO. Bnt BC-=BG; .'. AL^BC. Thus from pt. A a st. line J[i hna been drawn » Br. Q. K. F. Postu 3. Def. 13. Def. 13. Def. 91. Ax. 3. Ax. 1. 12 EUCLID'S ELEMENTS. [Book I. Proposition III. Problem. From the greater of two given straight Une» to end off a part equal to the less. Let AB be the greater of the two given st. Hnes AB, CD, It is required to cut off from AB a part = CD. From A draw the st. line AE=CI>. I. 2. With centre A and distance AE describe © jBFH, cutting AB in F. Then vfAlAF^ CD. ~'^S For *.• J^ is the centre of © FFH, .: AF=AE. But AE=CD; .-. AF=CD. Ax. 1. Thus from AB a part AF has been cut off—CD, Q. EL p. Exercises. 1. Shew that if straight lines be drawn from A and B in the diagram of Prop. i. to the other point in which the circles intersect, another equilateral triangle will be described on AB. -■' 2. By a similar construction to that in Prop. i. describe 'Y on a given straight line an isosceles triangle, whose equal sides shall be each equal to another given straight line. 3. Draw a figure for the case in Prop, ii., in which the given point coincides with B. 4. By a construction similar to that in Prop. in. produce the less of two given straight lines that it may be equal to the greater. /! '/' Book L] PROPOSITION IV. 13 Proposition IV. Theorem. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to one another, they must have their third sides equal ; and the two triangles must be equal, and the other angles must be equal, each to each^ viz. those to which tJie equal sides are opposite* 0_ In the A s ABC, DEF, let AB=DE, and AC^DF, and z BAC^^ L EDF. Then must BG^EF and A ABC = A I)EF, and the other A s, to which the equal sides are opposite, must be equ,al, that is, L ABG= L DEF and l ACB= l DFE. For, if A ABC be applied to a DEF, so that A coincides with D, and AB falls on DE, then •.* AB=DE, .'. B will coincide with E. And •.* AB coincides with UE, and z BAC= z EDF, Hyp. .-. A C will fall on DF. Then •.* AC=DF, .'. C will coincide with F. And '.* B will coincide with E, and C with F, .'. BC will coincide with EF ; for if not, let it fall otherwise as EOF : then the two st. lines BC, EF will enclose a space, which is impossible. Post. 5. .*. BC will coincide with and .'. is equal to EF, Ax. 8. and A ABC a DEF, and z ABC z DEF, and z ACB ^ DFE. Q. E. ». 14 EUCLID'S ELEMENTS. [Book L Note 1. On the Method of Superposition. Two geometrical magnitudes are said, in accordance witli Ax. VIII. to be equal, when they can be so placed that the boundaries of the one coincide with the boundaries of the other. Thus, two straight lines are equal, if they can be so placed that the points at their extremities coincide : and two angles are equal, if they car be so placed that their vertices coincide in position and their arm« in direction : and two triangles are equal, if they can be so placed that their sides coincide in direction and magnitude. In the application of the test of equality by this Method of Superposition, we assume that an angle or a triangle may be moved from one place, turned over, and put down in another place, without altering the relative positions of its boundaries. We also assume that if one part of a straight line coincide with one part of another straight line, tlie other parts of the lines also coincide in direction ; or, that straight lines, which coincide in two points, coincide when produced. The method of Superposition enables us also to compare magnitudes of the same kind that are unequal. For example, suppose ABG and DEF to be two given angles. B ■ V hi W Suppose the arm BC to be placed on the arm EV^ and the vertex B on the vertex E. Then, if the arm BA coincide in direction with the arm ED, the angle ABG is equal to BFjF. If BA fall between ET> and ZF in the direction EB, ABC is less than DEF. If BA fall in the direction EQ m that ED is between EQ and EF, ABC is greater than DEF. Book LI NOTE II. 15 Note 2. On the Conditions of Equality of two Triangles. A Triahgle is composed of six parts, three sides and three angles* When the six parts of one triangle are equal to the six parts of another triangle, each to each, the Triangles are said to be equal in all respects. There are four cases in which Euclid proves that two tri- angles are equal in ail respects ; viz., when the following parts are equal in the two triangles. 1. Two sides and the angle between them. I. 4. 2. Two angles and the side between them. I. 2t ^ 3. The three sides of each, I. 8. 4. Two angles and the side opposite one of them. I. 2g. The Propositions, in which these cases are proved, are the most important in our First Section. The first case we have proved in Prop. iv. Availing ourselves of the method of superposition, we can prove Cases 2 and 3 by a process more simple than that em- ployed by Euclid, and with the further advantage of bringing them into closer connexion with Case 1. We shall therefore give thiee Propositions, which we designate A, B, and 0, in the Place of Euclid's Props, v. vi. vii. viii. The displaced Propositions will be found on pp. 108-112. Proposition A corresponds with Euclid I. 5. B I. 26, first part. EUCLID'^ ELEMENTS. [Boot L Proposition A. Theorem. // vwo mAMi of a triangle he equalj the angles opposite tho^ »ides mvst also be equal. In the isosceles triangle ABC, let AC=AB. (Fig. 1.) Then must l ABG= i ACB. ♦ Imagine the aABG to be taken up, turned round, and set lown again in a reversed position as in Fig. 2, and designate the angular points A', B', C. Then in £,% ABC, A'CW, -.' AB=A'Cy and AG=A'B', and i BAC'^ L CTA'B',!^ i ^*^ .-. z ABG= L A'G'B'. 14 Bu* /. A'G'B'= L AGB ; .•. lABG^lAGB. Ax. 1. OoR. Hence every equilateral triangle ia also equiangular. Note. When one side of a triangle is distinguished from / 1 1 the other sides by being called the jBa.se, the angular point op- U posite to that side is called the Vertex of the triangle. Book L] PROPOSITION B. 17 Proposition B, Theorem. Ij two triangles luive two angles of the one equal to two angles of the other, each to each, and the sides adjacent to the equal angles in each also equal ; then must ilie triangles he eq ual in all respects. as/ a B In A 8 ABC, DBF, let z ABG^ L DBF, and z ACB^ i DFE, and BG=EF. /rhen must AB==DF, and AG^DF, and l BAG= i EDF. For if A DBF be applied to A ABG, so that E coincides with B, and BF falls on BG \ then */ EF=BG, .: F will coincide with G ; and •.• I DEF= l ABG, .'. ED will fall on BA ; .*. D will fall on BA or BA produced. Again, •.* z DFE= z AGB, .*. FD will fall on GA ; .*. D will fall on GA or GA produced. .*. D must coincide with J., the only pt. common to BA and GA. .'. DE will coincide with and .*. is equal to ABy and Di^ AG, and ^EDF. iBAG, and i^DEF lABG; and .'. the triangles are equal in all respects. Q. E. D. CoR. Hence, by a process like that in Prop. A, we can prov« the following theorem : If two angles of a triangle he equal, the sides which, svhUn' them arc oho equal. (End. I. fi.) y. EUCLIjyS ELEMENTS. [Book L /I Peoposition C. Theorem. If two triangles have the three sides of the one equal to the three sides of the other, each to eachy the triangles must he equal in all respectt. Let the three sides of the AS-4BC, DEF be equal, each to each, that is, AB=DE, AC=I)F, and BC=EF. Then must the triangles he eqiial in all respects. Imagine the A DEF to be turned over and applied to the A ABC, in such a way that EF coincides with BC, and the vertex D falls on the side of BG opposite to the side on which A falls ; and join AD. , Case I, When AD passes through B0» Then in aABD, v BD=BA, .\ i BAD= L BDA, I. A. And m aAGD, '.' CD=GA, ,'. l GAD= l GDA, I. A. .*. sura of L bBAD, GAD=snm of z s BDA, CD A, Ax. 2. that is, aBAG^' I BDG. Hence we see, referring to the original triangles, that z P^(7= z EDF. /., by Prop. 4, the triangles are equal in all respects. Book M PROPOSITION C, >0 Case II. When the line joining the vertices does not pass through BG, Tlien in A ABT), v BD==BA, :. L BAD= L BDA, I. A. And in aACD, v CD = GA, .'. l CAD= l CD A, I. A. Hence since the whole angles BAD, BDA are equal, and parts of these CAD, CD A are equal. .'. the rem!(inders BAG^ BDG are equal. Ax. 3. Then, as in Case I., the equality of the original triangles may be proved. Cask IIL When AG and CD are in the same straiglit line. A Then in a ABD, v BD=BA, .\ l BAD= l BDA, I. A. that is, lBAC= lBDC. Then, as in Case I., the equality of the original triangles may be proved. Q. E. D. P.VCLID\S ELEMENTS. [BooK L Proposition IX. Problem. To bisect a given angle. I Let BAChe the given angla It is required to bisect l BAO. In AB take any pt. D. In AG make AE=AD, and join DE. On /?j&, on the side remote from -4, describe an equilat. a DFE. I. 1 Join AF. Then AF will bisect z BAO. For in £,bAFD,AFE, '.' AD=AE, and ^JP is common, and FD^FEy .\ L DAF= L EAF, I. c. that is, z BAG is bisected by AF. Q.B. r. ^ >i / Ex. 1. Shew that we can prove this Proposition by means r of Prop. IV. and Prop, A., without applying Prop. 0. / Ex. 2. If the equilateral triangle, employed in the construc- ' Y tion, be described with its vertex towards the given angle ; shew that there is one case in which the construction will fall, and two in which it will hold good. C Note.— The line dividing an angle into two equal parts is '•^led the Bisector of the angle. Book t-t PROPOSITION X. i\ I Proposition X. Problem. Wo bisect a given finite strmght lint, a Let AB be" the given st. line. * It is required to bisect AB. On AB describe an equilat. aACB. I. 1. Bisect lACBhj the st. line CD meeting AB in D ; I. 9. then AB shall be bisected in D. For m £iS ACD, BCDy \' AC=^BCy and CD is common, and z ^0D= z BCD, .-. AD=BD ; I. 4. .*. AB is bisected in D. Q. E. F. Ex. 1. The straight line, drawn to bisect the vertical angle ^ of an isosceles triangle, also bisects the base. Ex. 2. The straight line, drawn from the vertex of an y. isosceles triangle to bisect the base, also bisects the vertical / Ritgle. /Ex. 3. Produce a given finite straight line to a point, such . that the part produced may be one-third of the line, which ia ^ made up of the whole and the part produced. i^ EUCLID'^ ELEATEMTS. t^ook I Proposition XI. Problem. To draw a straight line at right angles to a ifwen straight line from a given point in (he same* If Let AB be the given st. line, and a given pt. in it* It is required to draw from G a st. line J. to AB. Take any pt. 2> in ^0, and in CB make GE=CD, On DE describe an equilat. A DFE. Join FG. FG shall be ± to AB. LI. Forin AsDOF, jEJOF, \' DG=GE, and GF is common, and FD=FE, .'. jlDGF=/.EGF; I. g and .-. ^0 is X to AB. Def. 9. Q. B. P. CoR. To draw a straight line at right angles to a given ■ straight line AG from one extremity, C, take any point Din AG, produce ^C to E, making GE'=^GD, and proceed as in the proposition. / Ex. 1. Shew that in tlje diagram of Prop. ix. ^Fand EB y intersect each other at right angles, and that ED is bisected ' by^J?*. Ex. 2. If be the point in which two lines, bisecting AI I and AG, two sides of an equilateral triangle, at right angles I meet ; shew that OA, OB, OG are all equal. * / Ex. 3. Shew that Prop. xi. is a particular case of Prop, ix Book L] PROPOSITION XIL «3 Proposition XIL Problem. To d/raw a straight line perpendicular to a given Hraighi Une of an unlwnted length fro'm a given point without ii. Let AB be the given st. line of unlimited length ; G the given pt. without it. It ia required to draw from C a st line. ± to AB. Take any pt. D on the other side of AB, With centre C and distance CD describe a © cutting dB in E and F. Bisect EF in 0, and join CE, CO, CF, L 10. Then CO shaU be ± to AB, For in as COE, COF, V EO'^FOf and CO is common, and CE^CF^ ,: jlCOE=iCOF; Lc. /. 00 is ± to AB. Bet 9. Q. E. F. Ex. 1. If the straight line were not of unlimited length, how might the construction fail ? Ex. 2. If in a triangle the perpendicular from the vertex on the base bisect the base, the triangle is isosceles. Ex. 3. The lines drawn from the angular points of an equilateral triangle to the middle points of the opposite sides are equal 24 EUCLID'S ELEMENTS. [Book L Miscellaneous Exercises on Frops. I. to XII. (, / 1. Draw a figure for Prop. ii. for the case when the given J point A is (a) below the line BG and t© the right of it. / (/3) below the line BG and to the left of it. V 2. Divide a given angle into four equal parts. J 3. The angles B, G, at the base of an isosceles triangle, are i bisected by the straight lines J5I>, OD, meeting in D ; shew I that BDG is an isosceles triangle. /4. D, E, F are points taken in the sides BC, GA, AB, of an equilateral triangle, so that BD—GE=^AF. Shew that the triangle DEF is equilateral. / 5. In a given straight, line find a point equidistant from f two given points ; 1st, on the same side of it ; 2d, on opposite sides of it. 6. ABG is any triangle. In BA, or BA produced, find i a point D such that BD=GD. / 7. The equal sides AB, AG, of an isosceles triangle ABG fy are- produced to points F and G, so that AF=AG. BG and f GF are joined, and II is the point of their intersection. Prove that BE=GH, and also that the angle at A is bisected by AE. , / 8. BAG, BDG are isosceles triangles, standing on oppo- ^ site sides of the same base BG. Prove that the straight line from A to D bisects BG at right angles. 9. In how many directions may the line AE be drawn in Prop. III. ? . , 10. The two sides of a triangle being produced, if the "* j, angles on the other side of the base be equal, shew that the triangle is isosceles. / 11. ABG, ABD are two triangles on the same base AB .: and on the same side of it, the vertex of each triangle being outside the other. If AG=AD, shew that BG cannot =BB. j 12. From G any point in a straight line AB, GD is drawn J Y At at right angles to AB, meeting a circle described with centre ' A and distance A Bin D ; and from AD, AE is cut off = AG'. shew that A EB is a right angle. Book I.] FROPOSiriON XIIL 25 \ Proposition XIII. Theorem. Tiu angles which one straight line makes with another upon one side 0/ it are eiUier two right angles, or togetJter equal to two right angles. / Let AB make with CD upon one side of it the 1 8 ABC^ ABD. Then must th'ese be either two rt. L 8, or together equal to two rt l s. First, if z ABC= i ABD us in Fig. 1, each of them is a rt. z . Def . 9. Secondly, if z ABG be not= z ABD, as in Fig. 2, from B draw BE ± to CD. I. 11. I'iien sum of z s ABG, ABD =sum of z s BBC, EBA, ABD, and sum of z s EEC, EBD =&um of z s EEC, EBA, ABD ; .-. sum of z s ABC, ABD =mm of z s EEC, EBD ; Ax. 1. .*. sum of z s ABC, ABD=s\im of a rt. z and a rt. z ; /. z 8 ABC, ABD are together=two rt. z s, Q. E. D. Ex. Straight lines drawn connecting the opposite angular points of a quadrilateral figure intersect each other in 0. Shew that the angles at are together equal to four right angles. ^ Note (1.) If two angles together make up a right angle, each is called the Complement of the other. Thus, in fig. 2, z ABD is the complement of z ABE. (2.) If two angles together make up two right angles, each - called the SurriF.MRNT of the other. Thus, in both figures, : ABD is the supploinmt of z ABC. / 26 EUCLID'S ELEMENTS. [Book 1. Proposition XIV. Theorem. v J/, ai a point in a straight line, two other straight lines, ttpon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines must be in one cmd the same stradght line. At the pt. B in the st. line AB let the st. lines BGy BD, on opposite sides of AB, make z s ABG, ABD together=two rt. angles. Then BD must be in the same st. line with BG. For if not, let BE be in the same st. line with BG. Then z s ABG, ABE together = two rt. z s. I. 13. And z 8 ABG, ABD together = two rt. z s. Hyp /. Bum of z s ABGy ABE=snm of z s ABG, ABD. Take away from each of these equals the z ABG ; then z ABE=^ z ABD, Ax. 3 that is, the less = the greater ; which is impossible, ,.*. BE is not in the same st. line with BG. Similarly it may be shewn that no other line but BD is lis the same st. line with BG. ,•. BD is in the same st.' line with BO. Q. E. D Ex. Shew the necessity of the words the opposite sides Ln the enunciatioTi. Book I.] PROru- .1 1 iL '- V XV, 27 Proposition XV. Theorem, Lf two straight lines ciU orie a/iiotloarf tiie v&riicaily opposite angles must he equal. -3 Let the st. lines AB, CD cut one another in the pt. E. Then must l AEC=- l BKD and l AEI>=^ l BEC. For •.• AE meets CD, .*. sum of z s AECf AED =two rt. z s. I. 13 And •.• DE meets AB, .-. sum of z s BED, AED=two rt. z s ; I. 13. .-. sum of z s AEC, AED^sum of z s BED, AED ; .-. z AEC= L BED. Ax. 3. Similarly it may be shewn that l AED=^ l BEC. Q. E. D. Corollary I. From this it is manifest, that if two straight lines cut one another, the four angles, which they make at the point of intersection, are together equal to four right angles. Corollary II. All the angles, made by any number 0/ straight lines meeting in one point, are together equal to four right angles. y Ex. 1. Shew that the bisectors of AED and BEC are in J/ the same straight line. / Ex. 2. Prove that z AED is equal to the angle between "^two straight lines drawn at right angles from E to AE and ^ L EC, if both lie above CD. ^-y Ex. 3. If AB, CD bisect each other in E ; shew that the W •^triangles AED, BEC are equal in oil respects. 88 EUCLIiyS ELEAIENTS. [Book 1 Note 3. On J^Jv^licCs definition of an Angle. Euclid directs us to regard an angle as the inclination of two straight lines to each other, which meet, but are not in the same straight line. Thus he does not recognise the existence of a single angle equal in magnitude to two right angles. , The words printed in italics are omitted as needless, in Def. VIII., p. 3, and that definition may be extended with advantage in the following terms : — Dep. Let WQE be a fixed straight line, and QP a line which revolves about the fixed point Qj and which at first coincides with QE. Then, when QP has reached the position represented in the diagram, we say that it has described the angle JUQP. When QP has revolved so far as to coincide with QW, we say that it has described an angle equal to two Hijht angles. Hence we may obtain an easy proof of Prop. xiii. ; for what- ever the position of PQ may be, the angles which it makes with WE are together equal to two right angles. Again, in Prop. xv. it is evident that z AED^ L EEC, since each has the same supplementiiry z AEG. We shall shew hereafter, p. 149, how this definition may be extended, so as to embrace angles greater than two right angles. . BooJc I.] PROPOSITION XVI. 29 Proposition XVI. Theorem. if ane side of a triangle be produced, the exterior angle ii reater than either of the interior opposite angles. Let the Hide BC of a ABG be produced to D. Then miist L A CD be greater thin either L GAB or L ABC. Bisect ACmE, and join BE. I. 10. Produce BE to F, making EF=BE, and join FC. Tiienin A&BEA,FEC, :■ BE=FE, and EA =^EC, and z BEA = z FEG, I. 15. .-. z ECF= L EAB. I. 4. Now z ^OD is greater than z ECF ; Ax. 9. .*. lACI>\s> greater than z EAB, that la, z ^CD is greater than z (7^B. Similarly, if J[(7 be produced to G^ it may be shewn that z BCQ is greater than z J.jBC. and z 500^= z ^CD ; I. 15. .% z ^Oi> is greater than z -4^(7. Q. K D, / Ex. 1. From the same point there cannot be drawn more than two equal straight lines to meet a given straight line. Ek. 2. If, from any point, a straight line be drawn to a .' given straight line making with it an acute and an obtuse ■^J angle, and if, from the same point, a perpendicular be drawn to the given line ; the perpendicular will ffill on the side of the acute angle. 30 EUCLID'S ELEMENTS. [Book I. Proposition XVII. Theoiiem. Any two angUi 0/ a triangle cure together lets than two right angle*. a 15 Let ABG be any A , Then must any two of its z a be together lets than two rt. Ls. Produce BG to D. Then z ^ CD is greater than z ABG. I. 16. .'. z s AGD, AGB are together greater than z bABG, AGB. But I 8 AGD, AGB together = two rt. z s. I. U. .*. z sABGf AGB are together less than two rt. z s. Similarly it may be shewn that z s ABG, BAG and also that z s BAGf AGB are together less than two rt. z s. Q. E. D. Note 4. On the Sixth Postulate. We learn from Prop. xvii. that if two straight lines BM and GNf which meet in A, are met by another straight line DE in the points 0, P, fD the angles JtfOP and NPO are together less than two right angles. The Sixth Postulate asserts that if a line DE meeting two other lincH BM, GN makes MOP, NPO, the two interior Gook I.] PkOPOSlTION XVin. angles on the same side of it, together less than two right angles, BM and CW shall meet if produced on the same side of DB* on which are the angles MOT and NFO, Proposition XVIII. THBORmt If one side of a triangle be grecUer (Jian a tecond, the angle opposite the fwst must he greaier iham, that opposite th^ second. B ^ In tkABGy let side ^Obe greater than AK Then w,ust i ABC he greater than l ACB, From 4 cut off ^ D= ^ J5, and join BD. I. 3. Then \'AB=AD, .-. z ADB=- L ABB, L a. And *.• CD, a side of A JBDO, is produced to A, .'. L ABB k greater than lAGB\ L 16. /. also L ABB is greater than l ACB. Much more is i ABC greater than z AGB. Q. B. D. 1/ Ex. Shew that if two angles of a triangle be equal, the ^^ sides which subtend them are equal also (Eucl. I. 6). 3a EUCLID'S ELEMENTS. [Book 1. Proposition XIX. Theorem. If one angU of a triangle be greater than a second^ the side opposite the f/rst must be greater than that opposite the second. In A ABC, let z ABC be greater than l ACB. Then must AG be greater than AB. For if AG he Bot greater than ABy AG must either =J.5, or be less than AB. Now AG cannot = J. J5, for then I. a. d. ABG would = z AGB, which is not the case. And J. (7 cannot be less than AB, for then I. 18. i ABC would be less than z AGB, which is not the case; .'. -40 is greater than AB. Q. E. D. ^7 Ex. 1. In an obtuse-angled triangle, the greatest side is )pposite the obtuse angle. Ex. 2. BGj the base of an isosceles triangle A (7, is pro* iuced to any point D ; shew that AD is greater than AB. yEx. 3. The perpendicular is the shortest straight line, which can be drawn from a given point to a given straight line ; and ' of others, that which is nearer to the perppnrliVnl.ir i"^ loss than the more remote. Book 1.1 PROrOSlTIOJSl XX. 33 Proposition XX. Theorem. Any two sides of a triangle a/re together greater Hfum the third side. -D Let ^BO be a A. Then any two of its sides must he together greater than the third side. Produce BA to D, making AD=ACf and join DC. Then V AI)=AO, .'. lAGD^i ADC, that is, z BDO, L a. Now z BCD is greater than z ACD ; .'. L BCD is also greater than l BDO ; /. BD is greater than BO. L 19. (* But BD=BA and AD together ; that is, BD=BA and AC together ; .*. BA and AC together are greater than BG. Similarly it may be shewn that AB and BC together are greater than AG, andBGandCA AB. Q. E. D. / Ex. 1. Prove that any three sides of a quadrilateral figure ^^ ire together greater than the fourth side. Ex. 2. Shew that any side of a triangle is greater than the difl'erence between the other +wo sides. Ex. 3. Prove that the sum of the distances of any point from the angular points of a quadrilateral is greater than half the perimeter of the quadrilateral. Ex. 4. If one side of a triangle be bisected, the sum of the vo other sides shall be more than double of the line joining he vertex and the point of bisection, s. E. 3 . 34 EtrCLWS ELEMENTS. [Book 1. Proposition XXI. Theorem. I/, frmn, the ends of the side of a tricmgle, there he d/rawn two straight lines to a point ivithin the triangle; these ivill be together less than the other sides of the triangle, but will contain a greater angle. Let ABC he a A, and from P, a pt. in the A, draw st. lines to JB and C. Then will BD, DC together he less tlmn BA, AC, but L BBC will be greater than l BAC. Produce BD to meet ^0 in ^. Then BA, AE are together greater than BE. I. 20. Add to each EC. Then BA, ^(7 are together greater than BE, EC. Again, DE, EC are together greater than DC. I. 20. Add to each BD. Then BE, EC are together greater than BD, DC. And it has been shewn that BA, AC are together greater than BE, EC ; .*. BA, AC are together greater than BD, DC, Next, •.' z BDC is greater than z DEC, I. 16. and z DEC is greater than i BAC, L 16. .*. z BDC is greater than z BAC Q. E. D. Ex. 1. Upon the base AB of a triangle ABC is described quadrilateral figure ADEB, which is entirely within the T M triangle. Shew that the sides AC, CB of the triangle are together greater than the sides AD, DE, EB of the (Quadri- lateral. a. Book l.J PRO POSITION XXII. 35 1^ .^—^^—^-^————^—~—^—- 1/ Ex, 2. Shew that ihe sum of tlie straight lines, joinhij^ the angles of a triangle with a point within the triangle, is less than the perimeter of the triangle, and greater than lialf the perimeter. Proposition XX TI. Problem. To make a triangle, of which the sid^<< shall be equal to three given straight lines, any tioo of which a/re together greater than tJiA third. Lrt A, B, f! he the three given lines, any two of which are together greater than the third. It is required to make a A having its sides ^ A, B, C Take a at, line DE of unlimited length. In DE make DF= A, FG-=B, and (?fl"= G. , I. 3. With centre F and distance FD, describe © DKL. With centre G and distance GH, describe © HKL. Join FK and GK. Then lKFG has its sides =^A, B, G respectively. YotFK=FD\ Def. 13 .-. FK=A ; and GK=GH; Def. 13. .\ GK=C; mdFG=B; ,\ a t^KFG has been described as reqd. q. e. p. /Ex. 1. Draw an isosceles triangle having each of the equal sides double of the base. 3& EUCLID'S ELEMENTS. [Book L Proposition XXIII. Problem. Ai a given point in a given straight linej to mdke an %ngU equal to a given angle. Let A be the given pt., BC the given line, DEF the given L . It is reqd. to make at pt. A an angle «= i DEF. In ED, EF take any pts. D. F ; and join DF. In AB, produced if necessary, make AG=DE. In AC, produced if necessary, make AH=EF. In HC, produced if necessary, make HK=FD. With centre JL, and distance AG, describe © GLM. With eentre jff, and distance HK, describe © LKM. Join AL and HL. Then •.' LA = AG,.\ LA=DE ; Ax. 1. and ••• HL=HK, .'. HL=FD. Ax. 1. Then in as LAH, DEF, • • LA=DE, and AB-=EF, and EL^FD ; .-. I LAH= L DEF. I. G /, an angle LAH has been made at pt. A as was reqd. Q. E. p. Book I.] PROPOSITION D. 37 Note.— We liere give the proof of a theorem, necessary to the proof of Prop. XXIV. and applicable to several proposi- tions in Book III. Proposition D. Theorem. ^^^ Every straight Kjw, drmim from the vertex of a triangle to ^^ the base, is less than the. greater of the ttoo sides, or than either ^ ^ if they be equai. ^ In the A ABCf let the side ^0 be not less than AJk Take any pt. D in BG, and join AD, Then must AD he less ihcm AC, For '.• -4(7 is not less than AB ; .'. I ABD is not less than i ACD, But I ADO is greater than z ABD ; /. I ADC is greater than z A CD ,'.ACia greater than AD. L A. and 16. L16. 1.19. Q. S. C. C OF THE "^ UNIVERSITY OF SS ' EUCLID'S ELEMENTS. I uoli I. Proposition XXIV. Theorem. Ij two triangles have two sides of the. one eqiial to two sides of the other, each to each, hut the angle contained by the two sides of one of them greater than the angle contained by the two sid.es equal to them of the other ; the base of that which has the greater angle must be greater than the base of the other. In the L% ABC, DEF, let AB=DE and AC==DF, Mid let z BAC be greater than z EDF. Then must BG be greater than EF. Of the two sides DE, DF let DE be not greater than DF.* At pt. D in St. Hne ED make z EDG= l BAC, I. 23. and make DG= AC or DF, and join EG, GF. Then •.' A B^DE, and AC=DG, and iBAG'^L EDG, .\BC=EG, 1.4. Again, vDG=DF, .'. L DFG=^ L DGF ; T. 4. /. z EFG is greater than z DGF ; much more then z EFG is greater than z EGF ; .-. EG is greater than EF. I. 19. But ^6? = ^0; .*. BQ is greater than EF. Q. E. D. *This line was added by Simson to obviate a defect in Euclid's proof. Without this condition, three distinct cases must be discussed. With the condition, we can prove that F must lie below EO, For since DFis not less than DE, and DG is drawn equal to DF, DO is not less than DE. Hence by Prop. D, any line draAvn from D to meet EG is less than DG, and therefore DF, being equal to DG, must extend beyond EG. For another method of proving the Proposition, see p. 113. Book I. J PROFO::^n.L).^ XXV. 39 Proposition XXV. Theorem. //■ tv)o triangles have two sides of the one equal to two sides of the other^ each to each, hut the hose of the one greater than the base of th.e other ; the angle also, contained^ by the sides of that ivhich has the greater base, must be greater than the angle contained by the sides equal to them of the oilier. In the A B ABC, DBF, let AB^BE and AC=DF, and let BG be greater than EF. Then must L BA C be greater than l EDF. For z BAG b greater than, equal to, or less than z EDF. Now I bag cannot = z EDF, for then, by i. 4, BG wou\d= EF ; which is not the case. . And /iBAG cannot be less than z EDF, for then, by i. 24, BG would be less than EF ; which ia not the case ; .*. z BAG must be greater than z EDF. Q. E, D. Note. — In Prop. xxvi. Euclid includes two cases, in which two triangles are equal in all respects ; viz., when the following parts are equal in the two triangles : 1. Two angles and the side between them. 2. Two angles and the side opposite one of them. Of these we have already proved the first case, in Prop, b, so that we have only the second case left, to form the subject o£ Prop. XXVI., which we shall prove by the method of superposition. For Euclid's proof of Prop, xxvi., see jw?. 114-116. 40 ^ ^" EUCLID'S ELEMENTS. [BQOk I. Proposition XXVI. Theorem. If two triangles have two angles of the one egiml to two angles of the other, each to eachy and one side equal to one side, those sides being opposite to equal angles in ea^h ; tlien miLst Die triangl\ right anglfiB, thia case of the propoeition would be identical with I. 26. 44 EUCLWS ELEMENTS. [Book 1 SECTION II. The Theory of Parallel Lines. INTRODUCTION. We have detached the Propositions, in which Euclid treats of Parallel Lines, from those which precede and follow them in the First Book, in order that the student may have a clearer notion of the difficulties attending this division of the subject, and of the way in which Euclid proposes to meet them. We must first explain some technical terms used in this Section. If a straight line l&F cut two other straight lines AB^ CD, it makes with those lines eight angles, to which ptirticular names ore given. Iw The angles numbered 1, 4, 6, 7 are called Interior angles 2,3,5,8 Exterior The angles marked 1 and 7 are nd 5, 2 and 6, 4 and 8, 3 and 7 are called corresponding angles. NoTK. From I. 13 it is clear that the angles 1, 4, 6, 7 arp tngethpv equal to four ricrlit .merles. Book I.J PROPOSITION XXVII. 45 pRopoemoH XXVII. Theorem. If a straight liiu, falling upon two other straight lines, make (^ aitemate angles equal to one another; ihejie two straight lines TMtst be paraUeL Let the st. line EF^ faUing on the st. lines AB, CD, make the alternate z s AGH, QHD equal. Then must ABbeW to CD. For if not, AB and CD will meet, if produced, either towarda B, D, or towards A, C. Let them be produced and meet towards B, DiaK. Then GHK is a a ; and .'. z AGH is greater than z GHD, 1. 16. But z AGH=' L GBDy Hyp. which is impossible. .*. ABy CD do not meet when produced towards By D. In like manner it may be shewn that they do not meet when produced towards A , C. .: AB and CD are parallel Def. 26. Q. K, D. 46 EUCLJ/y.S r.LEMENTS. [Book 1. Proposition XXVIII. Theorem. . If a straight linCf falling upon two oilier straight lineSy make the exterior angle equal to the interim' and opposite upon the same side of the line, or make the interior angles upon the sa/me side together equal to two right angles; the two straight lines are parallel to one another. y- Let the at. line EF, falling on st. lines J.B, CD, make I. L i?(TJ5 = corresponding L GHDj or II. I 8 BGH, GHD together=two rt. z s. Then, in either case., AB must he || to CD. I. •.• z EGB is given = z ORD, Hyp. and I EGB is known to be= z AGH, L 16. .'. lAGH= iGUD', and these are alternate l s ; .-. AB is irto CD, I. 27. a. •/ z s BGH, GHD together = two rt. ^ », Hyp. and z 8 BGR^ AGII together=two rt. z s, I. 13. .'. 19 BGH,AGff together = z s BGH, GHD together ; ,: iAGH= iGHDi .'. AB is B to CD. L 27. Bock L] NOTEV' ON THE SIXTH POSTULATE. 47 Note 5. 0)i the, Sixth Postulate. In the place of Euclid's Sixth Postulate many modern writers on Geometry propose, as more evident to the senses, the following Postulate : — " Ttvo straight lines which cut one another cannot both be laraUel to the mwe straight line.'" Li this be assumed, we can prove Post. 6, as a Theorem, thus : Let the line EF falling on the lines AB, CI) make the ^ s BGH, GHD together less than two rt. z s. Then must AB, CD meet when produced towards J5, D. For if not, suppose AB and CD to be parallcL Then *.• z s AGH, BGH together = two rt. z s, I. la and z 8 GHI), BGH are together less than two rfc. z », .: z AGff is greater than z GHD. Make z MGH= z GHD, and produce MG to N. Then •.* the alternate z s MGH, GHD are equal, .-. MN is II to CD. I. 27. Thus two lines M2V, AB which cut oiie another are «both parallel to CD, which is impossible. .'. AB and CD are not parallel. It is also clear that they meet towards B, D, because GB lies between GN and HD. Q. K. D. 48 EUCLID'S ELEMENTS. iBook I. Proposition XXIX. Theorem. // a straight line fall upon two parallel straight lines, it makes the two interior angles upon the sanu side together equal to tivo right angles, and ahofthe alternate, angles equal to one another^ a/nd also tlie exterior angle equal to the interior and opposite upon the same side. ^ II. Ill Let the st. line EF fall on the parallel st. lines AB, CD* Then must ^ I. z s BGE, GHD together = two rt z s. 11 z ^(?if= alternate z GSD. III. z ii'GfjB= corresponding z GHD. L z s BGH, GHD cannot be together less than two rt. z s, ior then AB and CD would meet if produced towards B and D, Post. 6. which cannot be, for they are parallel Nor can z s BGH, GHD be together greater than two rt. z 8, for then z s AGH, GHC would be together less than two rt. z s, I. 13. and AB, CD would meet if produced towards A and C Post. 6 which cannot be, for they are parallel, .*. z s BGH, GHD together = two rt. z s. z 8 BGH, GHD together = two rt. z s, and z s BGH, AGH together- two rt. z s, I. 13. z 8 BGH, AGH together^ z a BGH, GHD together, and .-. z AGH= z GHD, Ax. 3. z AGH= L GHD, and lAGH== lEGB, I. 15. L EGB^ z GHD. Ax. 1. ^ £. D. Bodk L] PROPOSITION XXIX. 49 ^ 6d Exercises. 1. If through a ppint, eqiiidistaDt from two parallel straight lines, two straight lines be drawn cutting the panillel ,/ \j/ Q straight lines ; they will intercut equal portions of those lines. 2. If a straight line be drawn, bisecting one of the angles / of a triangle, to meet the opposite side ; the straight lines / •^ / drawn from the point of section, parallel to the other sides and terminated by those sides, will be equal. 3. If any straight line joining two parallel straight lines be bisected, any other straight line, drawn through the point of bisection to meet the two lines, wiU be bisected in that point Note. One Theorem (A) is said to be the converse of another Theorem (B), when the hypothesis in (A) is the conclusion in (B), and the conclusion in (A) is the hypothesis in (B). For example, the Theorem 1. a. may be stated thus . Hypothesis. If two sides of a triangle be equal. Conclusion. The angles opposite those sides must also be equal The converse of this is the Theorem 1. b. Cor. : Hypothesis. If two angles of a triangle be equal. Conclusion. The sides opposite those angles muBt also b€> equal The following are other instances ; Postulate VI. is the converse of L 17. L 29 is the converse of L 27 and 28. && 50 RUCLIiys ELEMENTS, [Book 1. Proposition XXX. Theorem. Straight lities which are pa/rcdld to the smne straight lint cure pa/rallel to the interior angles. Book L] PROPOSITION XXXIIL 55 • 'to, eight: Prove that the interior angles of a hexagon are equal Z eight right angles. Ex. 3. Shew that the angle of an equiangular pentagon is | ^. a right angle. Ex. 4. How many sides has the rectilinear figure, the sum 3f whose interior angles is double that of its exterior angles ? ^ Ex. 5. How many sides has an equiangular polygon, four whose angles are together equal to seven right angles ? r/of Proposition XXXIIE Theorem. Tht straight lines which join the extremities of two equal and parallel straight lines, towards the samu parts, care also ihefm- selves equal and paralld. O Let the equal and il st. lines AB, CD be joined towards the same parts by the st. lines A C, BD. Then must AC and BD he equal amd I. Join BC. Then .• ^5 is || to CD, .-. z ^jBC=altemate i DCB, L 29. Tnenin h.^ ABC, BCD, •/ AB=CD, and BC is common, and l ABC= l DCB, .-. AC=BD, and iACB^^a DBC. L 4. Then '.• BC, meeting AC and BD, makes the alternate z s ACB, DBC equal, .-.AC is \\ to BD, Q. 1. D. 56 EUCLID'S ELEMENTS. fBoOfe L M.i&cdlanw*M Exercises on Sections I. and II. -^ / 1. If two exterior angles of a triangle be bisected by A> 1 straight lines which meet in ; prove that the perpendiculars from on the sides, or the sides produced, of the triangle are / equal. Tfc f 2. Trisect a right angle. 3. The bisectors of the three angles of a triangle meet in one point. 4. The perpendiculars to the three sides of a triangle drawn from the middle points of the sides meet in one point. Y^ y 5. The angle between the bisector of the angle BAG of the / ^ triangle ABC and the perpendicular from A on BG^ is equal to half the difference between the angles at B and G. 6. If the straight line AD bisect the angle at A of the \ / / triangle ABC^ and BDE be drawn perpendicular to ADy and y meeting AC,ov AG produced, in JB; shew that BB is equal UiDE. i 7. Divide a right-angled triangle into two isosceles tri- I 8. AB, GD are two given straight lines. Through a point / E between them draw a straight line GEH, such that the in- tercepted portion OH shall be bisected in E. 9. The vertical angle of a triangle OPQ is a right, acute, or obtuse angle, according as OR, the line bisecting PQ, is equal to, greater or less than the half of PQ. 10. Shew by means of Ex, 9 how to draw a perpen- j dicular to a given straight line from its extremity without pro- ducing it / Book L] EUCLIJDI'S ELEMENTS. 57 SECTION III. On the Equality of Rectilinear Figures in respeet of Area. Thb amount of space enclosed by a Figure is called the Area of that figure. Euclid calls two figures tqml when they enclose the same amount of space. They may be dissimilar in shape, but if the areas contained within the boundaries of the figures be the same, then he calls the figures egiwiZ. He regards a triangle, for example, as a figure having sides and angles and area, and he proves in this section that two triangles may have equality of area, though the sides and angles of each may be unequal Coincidence of their boundaries is a test of the equality of all geometrical magnitudes, as we explained in Note 1, page 14 In the case of lines and angles it is the only test : in the case of figures it is a test, but not the only test ; as we shall shew in this Section. The sign =, standing between the symbols denoting two figures, must be read is equal in area to. Before we proceed to prove the Propositions included in fhis Section, we must complete the list of Definitions required in Book T. , continuing the numbers prefixed to the definitions in page <$. EUCLID'S ELEMENTS, [Book I. ^ Dbpinitions. XXVII. A Parallelogram is a four-sided figure whose opposite sides are parallel. J^'or brevity we often designate a parallelogram by two letters only, which mark opposite angles. Thus we call the figure in the margin the parallelogram AG. XXVIII. A Rectangle is a par- allelogram, having one of its angles a right angle. Hence by I. 29, oZZ the angles of a rectangle are right angles. XXIX. A Rhombus is a par- allelogram, having its sides equal XXX. A Square is a paral- j lelogram, having its sides equal and one of its angles a right \ angle. 1 Hence, by I. 29, oR the angles of a square are rigni angles. XXXI. A Trapezium is a four-sided figure of which two sides only are parallel. XXXIT. A Dtaoonal of a four-sided fijTurc Is \h(' straight line joining two of the opposite anoninr points. Book I.] EXERCISES ON DEFINITIONS 27-33. 59 XXXIII. The Altitude of a Parallelogram is the perpen- dicular distance of one of its sides from the side opposite, regarded as the Base. The altitude of a triangle is the perpendicular distance ol one of its angular pointp from the side opposite, regarded as the base Thus if ABGD be a parallelogram, and AE a perpendicular let faU from A to OD, AE is the altitude f the parallelogram, and also of the triangle ACD. A S If a perpendicular be let fall from B to DC produced, meet- ing DC in F, BF is the altitude of the parallelogram. Exercises. Prove the following theorems : 1. The diagonals of a square make with each of the sides an angle equal to half a right angle. 2. If two straight lines bisect each other, the lines joining their extremities will form a parallelogram. 3. Straight lines bisecting two adjacent angles of a paral- lelogi-am intersect at right angles. 4. If the straight lines joining two opposite angular points of a parallelogram bisect the angles, the parallelogram is a rhombus. 6. If the opposite angles of a quadrilateral be equal, the quadrilateral is a parallelogram. 6. If two opposite sides of a quadrilateral figure be equal to one another, and the two remainmg sides be also equal to one another, the figure is a parallelogram. 7. If one angle of a rhombus be equal to two- thirds of two right angles, the diagonal drawn from that angular point divides the rhombus into two equilateral triangles. 6o EUCLID'S ELEMENTS, [Book L Proposition XXXIV. Theorem. The. opposUe sides and angles of a paralldogram are equal U one anotJuTf and the diagonal bisects iL a D Let ABDG be a O, and BC a diagonal of the O. Then must AB=DO and AC^DB, and I BAG^ l CDS, and z ABI}= l AGD - and i^ABC^LDGB. ¥oT '.' ABisW to GDj and BG meets them, .-. I ^5(7= alternate i BGB , - I 29 and •.• AG is Hto BD, and BG meets them, .'. z J[CB= alternate l DBG, L 29. Then in as ABG, DGB, '.' L ABG-= L BGB, and l AGB^ l BBG, and BG is common, a side adjacent to the equal z. s in each ; /. AB=DG, and AG=DB, and i BAG'= L GDB, and A ABG== A DCB I. b.* Also '.• z ^J5C= z I>CB, and z i)BC= z ^OB, .-. z 8 ABG, DBG together = z s J)GB, AGB together, that is, z ABD== z ^CZ). Q. £. D. Ex. 1. Shew that the diagonals of a parallelogram bisect each other. ~ Ex, 2. Shew that the diagonals of a rectangle are equal. Book L] PROPOSITION XXXV. 61 Proposition XXXV. Theorejt. Parallelograms on tJie sa/tae base and between the tame parallels are equal. t^y Let the Os ABCDf. EBCF he on the same base BC and between the same l|s AF, BC. Thm must ZZ7 ABCD^CJ EBCF. Case 1. If ADj EF have no point common to bo+li, Then in the as FDC, EAB, '.' extr. z FDG^'miv. i EABy I, 29. and intr. z D1^0= extr. z AEBy 1. 29. and DC=AB, I. 34. A lFDC=i^EAB. L96. Now CDABCD with a i?'DO=figure ABCF % and O JSJBOJ' with A ^^B=%ure ABCF; /. O -llfOD with A FDC=^CJ EBCF with A ^^.P • .-. O ABCD=CJ EBCF. Oau n. If the sides AD, EF overlap another, the same method of proof appatxs. 62 EUCLID'S ELEMENTS, [Book L Case III. If the sides opposite to BC be terminated in the same point i/. ^ a the same method of proof is applicable, but it is easier to reason thus : Each of the Os is double of A BBG ; I. 34 .-. O ABCD=CJ DBCF. Q. B. D. Proposition XXXVI. Theor^.m. Parallelograms on equal hasesy and between the paralleUf at« isqiud to one another. Let the Os ABCD, EFOE be on equal bases JBC, FG, and between the same \\^ AH, BG. Then must EJ ABCD^CJ EFGH. Join BE, CH. Tim •.• BC=FG, Hyp. aadER=FG; 1.34. .'.BG^'Eff; and BO is || to EH. Hyp. .-. EB is II to CE ; I. 33. .*. EBCH is a parallelogram. No w O EBCII = EJA^CD, I. 35. •/ they are on the same base BC and between the same ||s ; and OI EBCR^CDEFGH, v I. 35, •/ they are on the same base EK and between the same || s ; rj ABCD=OJEFGE Q. E. D. Book L] PROPOSiriOK XXXVII. 63 Proposition XXXVII. Theorem. Triangles upon the scume basej and between tht saim parallels, are eqtial to one another. Let A 8 ABO, DBG be on the same base BC and between the same ||s AD, BG. Then must A ABG^ i^DBG, From B draw BE j| to GA to meet DA produced in E. From C draw GF |1 to BD to meet AD produced in F. Then EBGA and FGBD are parallelograms, BudCJ EBGA^LJ FGBD, I. 35. *.* they are on the same base and between the same ||b. Now A ABG is half of O EBGA, I. 34 and A DBG is half of O FGBD ; I. 34. .-. aABG==aDBG. Ax. 7. Q. E. D. Ex, 1. If P be a point in a side AB of a parallelogram ABGD, and PG, PD be joined, the triangles FAD, FBG are together equal to the triangle PDG. Ex. 2. Two straight lines AB, GD intersect in E, and the triangle AEG is equal to the triangle BED. Shew that BC is parallel to AD. Ex. 3. If A, B be points in one, and G, D points in another of two parallel straight lines, and the lines AD, BC intersect in E, then the triangles AEG, BED are equal. i EUCLIiys ELEMENTS, [Book 1 Proposition XXXVIII. Theorem. TriaTigles upon equal baseSf cmd between the samepwraXkh e equal to one another. Let AS ABC, DEF be on equal bases, JBC, EF^ and between the same {| s BF, AD. Then must A ABC=^ A DEF. From B draw BG || to GA to meet DA produced in G. From F draw FH || to ED to meet AD produced in H. Then CG and EH are parallelograms, and they are equal, •.* they are on equal bases jBO, EF, and between the same i]s BF, GH. I. 36. Now A ABC is half of O Cfll, and LDEFh&hdMoiEJ EH\ .\ aABG^aDEF. Ax. 7. Q. E. D. Ex. 1. Shew that a straight line, drawn firom the vertex of a triangle to bisect the base, divides the triangle into two equal parts. Ex. 2. If the triangles in the Proposition are not towards the same parts, shew that the straight line, joining the vertices of the triangles, is bisected by the line containing the bases. Ex. 3. In the equal sides ABj AC of sen isosceles triangle ABC points D, E are taken such that BD-= AE. Shew that the triangles CBD, ABE are equal. Book L] PROPOSITION XXXIX. 65 Proposition XXXIX. Theorem. EqtuU trianalex upon the sa/me hase^ cmd upon the same oj it, are bettoeen the same parallels. ^ r "^ Let the equal a s ABG, DBG be on the same base BO, and on the same side of it. Join AD. Then must AD he \\ to BC. For if not, through A draw AG \\ to BGj so as to meet BD, or BD produced, in 0, and join OG. Then *.• Aa A BG, OBG are on the same baae and between the same ((s, ,\aABG^aOBC. 137 But aABG^aDBG; Hyp ,\ aOBG=aDBG, the less ss the greater, which is impossible ; •. ^0 is not II to ^0. In the same way it may be shewn that no other line passing through A but AD is || to BG ; .-. ^D is 11 to BC. Q. S. D. Ex. 1. AD is parallel to BG ; AG, BD meet in ^; £C is produced to P so that the triangle FEB ]& equal to the triangle ABG : shew that PD is parallel to AG. Ex. 2. If of the four triangles into which the diagonals divide a quadrilateral, two opposite ones are equal, the quad- rilateral has two opposite sides parallel tf. is. . 66 EUCLID'S ELEMENTS, [Book L PBOPOSITION XL. THEOREaK. Equal triangles upon equal ha.if.Sy in the samie straight Une^ and touxM-ds the somie yarts, are between the same parallels. Let the equal A s ABCj DEF be on equal bases BCj Ei in the same st. line BF and towards the same parts. Join AD. Then must AD be || to BF. i:'or if not, through A draw ^ || to BF, so as to meet ED. or ED produced, in 0, and join OF. Then a ABG= a OEF, '.' they are on equal bases and between the same ||s. I. 38. But A ABG= A DEF ; Hyp. .-. ^OEF= aDEF, the less = the j^^eater, which is impossible. .-. ^Ois notllto^i'l In the same way it may be shewn that no other line passing through A but AD is || to BF, .-. AD IB \\ to BJf. Q. ■> D- / Ex, 1. If the triangles be not towards the same parts, shew r that the straight line joining the vertices of the triangles is bisected by th^ line containing the bases, Ex. 2. The straight Ime, joining the points of bisection of two sides of a triangle, is parallel to the base. Ex. 3. The straight lines, joining the middle points of the sides of a trianjile, divide it into four eqmil triangles. Book L] PROPOSITION XLl 67 Proposition XLL Theorem. If a parallelogram and a triangle lye upon the same base, a/nd between the same ■pa/ralleUf ike pa/raUelogram is double of the triangle. Let the O ABCD and the A EBC be on the same base BC nd between the same ||s AE, BC. Then must EJ ABCD be double of A EBG. Join A C. Then A ABG= A EBC, '.' they are on the same base and between the same Ik ; L 37. and O ABCD is double of a ABC, v AG is a diagonal of ABCD ; L 34. .-. O ABCD is double of A EBC. a E. D. Ex. 1. If from a point, without a para Hologram, there be d/awn two straight lines to the extremities of the two opposite Sides, between which, when produced, the point does not lie, the diflference of the triangles thus formed is equal to half the p^raileJograiiL Ex. 2. The two triangles, formed by drawing straight lines from any point within a parallelogram to the extremiLies of It? opposite Slides, are together half of the parallelogram. 68 EUCLID'S ELEMENTS. [Book L Proposition XLII. Problem. To describe a parallelogram that shall he ^qual to a g'iven trmnglef and have one of its angles equal to a given angle. O Let ABC be the given A , and B the given / . It is required to describe a O equal to a ABC^ hamng one of its Ls^ iD. Bisect BG in E and join AE. I. 10. At E make i CEF= l D. I. 23. Draw AFG \\ to BG, and from G draw GG \\ to B^. Then FEGG is a parallelogram. Now A^^i^=A.4^0, '.' they are on equal bases and between the same ||s. I. 38, .-. A ABG is double of A AEG. But O FEGG is double of A AEG, V they are on same base and between same II h. 1. 41. .'.CJFECG^aABG; Ax. 6, and ZZZ FEGG has one of its z s, CEF=^ l D. .*. CJ FEGG has been described as was reqd. Q. E. F. Ex. 1. Describe a triangle, which shall be equal to a riven parallelogram, and have one of its angles equal to a given rectilineal angle. Ex. 2. Construct a parallelogram, e^ual to a giver tna-^gle, and such that the sum of its sides shall be equal to the sum of the sides of the triangle. Kx. 3. The perimeter of an isosceles triangle is greater than the perimeter of a rectangle, which is of the same altitude with. !ind equal to, tV' ;:iven triirigle. Book L] PROPOSITION XLIII. 69 Proposition XLIII. Theorem. Tlie complement /ihfew that the triangle MBN is equal to the triangle MDC. TO EUCLIiyS ELEMENTS, [ftook 1. Proposition XLIV. Problem. To a given straight line to apply a parallelogram^ which shall he equal to a given triangle^ and have wte of its angles eqtubl to a given angle. BL Let AB be the given st. line, G the given A, D the given z . It ia required to apply to AB a EJ — lQ and ha/ving one of its Ls— lD. Make a 0=« A C, and having one of its angles = z D, I. 42. and suppose it to be removed to such a position that one of the sides containing this angle is in the same st. line with AB, and let the O be denoted by BEFG. Produce FQ to J?, draw AH tl to BG or EF^ and join BE. Then •.' FH meets the ||s AH, EF, .*. sum of z s AHF, HFE=' two rt. z s ; I. 20. .'. sum of z s BHG, HFE is less than two rt. z s ; .*. HB, FE wiQ meet if produced towards B, E. X'osi. €, Let them meet in K. Through K draw KL \\ to EA or FH, and produce HA, GB to meet KL in the pts. Z, M. Then HJFKL is a O, and HK is its diagonal ; and AG, ME are Os about HK, /. complement jBi^= complement BF^ T. 43. .-. OJ5X=aO. Also \\\Q CD BL has one of its z s, ABM=^ L EFG, a*id /. equal to z D. Q. K. F. BooJi:?.;^ PROPOSITION XLV. 71 Proposition XLV. Problem. To describe a parallelogram, which shall he equal to a given rectilinear figure, and have one of its angles equal to a given angk. MA It Let ABCB be the given recti), figure, and E the given L . It IS required to describe a EJ = to ABCD. hmnng one of its LS= L E. Join AC. Describe a O FGHK^ A ABG, having z FKH^ l E. I. 42. To GH apply a CJ GHML-= a CD A, having l GHM= l E. L 44. Then FKML is the O reqd. For ••• z GEM and z ^iCif are each= i E ; .-. lGHM^lFKH, ,\ sum of z s GIL^r, GHK=-Bnm of z s FKH, GHK = two rt. zs ; L 29. .-. KHM is a st. line. I. 14. Again, •.' HG meets the l|s FG, KM, z FGH'^ L GUM, .'. sum of Z s FGH, LGH ^smn of z s GHM, LGH = two rt. zs; I. 29. .-. iJ'(?L is a St. line. L 14. Then •.' KF is n to ffG, and HG is || to LM .'. ^J?' is II to LM ; I 30. and /TM has been .>hewn to be || to FL, .'. FKML k a parallelogram, and •.• FH= a ABC, and GM= a OD^, .-. O ir'Jf- whole rectil. %. A BCD, and O i^M has one of its z s, FKM= z ^. lii ulit' same way a IIJ may be constructed equal to a given rectil. fig. of .my number of sifles, and having one of its angles equal to a given angle, Q. F- F. 7?i EUCLIjyS ELEMENTS. [Book L Miscellaneous Exeroijiei, 1. If one diagonal of a quadrilatei-al bise«t the other, it divides the quadrilateral into two equal triangles. 2. If from any point in the diagonal, or the diagonal pro- duced, of a parallelogram, straight lines be drawn to the opposite angles, they will cut off equal triangles. 3. In a trapesdum the straight line, joining the middle points of the parallel sides, bisects the trapezium. . . 4. The diagonals AG, BD of a parallelogram intersect in 0, and P is a point within the triangle AOB ; prove that the difference of the triangles AFB, CPD is equal to the sum of the triangles APC, BPD. 5. If either diagonal of a parallelogram be equal to a side of the figure, the other diagonal shall be greater than any side of the figure. 6. If through the angles of a parallelogram four straight lines be drawn parallel to its diagonals, another parallelogram will he formed, the area of which will be double that of the original parallelogram. 7. If two triangles have two sides respectively equal and the included angles supplemental, the triangles are equal. 8. Bisect a given triangle by a straight line drawn from a given point in one of the sides. 9. If the base of a triangle ABC be produced to a poiui D such that BD is equal to AB, and if straight lines be dmwn from A and D to E, the middle point of BC ; prove that the triangle ADE is equal to the triangle ABC. 10. Prove that a pair of the diagonals of the parallelograms, which are about the diameter of any parallelogram, are parallel to each other. BoiU I.] PROPOSITION XL VL 73 Proposition XLVI. Problem. ^o dAgcribe a souare upon a given straight Unt, C Let AB be the given st. Vme. It is required to Jssci'ihe a square 07i AB, From A draw AC 1. to AB. ' In AC make AD =AB. Through D draw DE || to AB. Through B draw BE i| to AD, Tlven AE is a parallelogram, and .-. AB==ED, aud AD=BE. Bazjitk^AD; .-. AB, BE, ED, DA are all equal ; .'. AE is equilateral. And I BAD is a right angle. .*. ^JBJ is a square, and it is described on AB. L U. Cor L31 1.31 L34. Def. XXX Q. E. p. iSx. i. Shew how to construct a rectangle whose sides art equal t4> *,wo given straight lines. Ey 2. Shew that the squares on equal straight lines ai*e ecuaU Ev. 3. Shew that equal squares must be on equal straight :ip«»'k KooE. The theorems in Ex. 2 and 3 are assumed by Euclid in the proof of Prop, xlviii. 74 EUCLID'S ELEMENTS. Book 1 Proposition XLVII. Theokkm. In any 'right-angled triangle the square which %s attended on the side subtending the right angh is ecual to the squares described on the sides which contain the right angle. Lot ABC be a right-angled A , liaving the rt. X.^C. Then mud sq. on BG= sum of sqq. on BA , AG, On BC, GA, AB descr. the sqq. BDEG, CKITA, AGFB Through A draw AL \\ to BD or CE, and join Alf, h'V. Then •/ i BAG and z BAG are both rt. l s, \ ^ .'. GAGh'd, St. line ; 1. 14, and *.• I BAG and i GAR are both rt. ^ s ; ^. .-. BAII is a St. line. I. 14. Now '.• L DBG^ L FBA, each being art. z , adding to each l ABG, we have z ABD= L FBG. Ax. 9 Then in /^s ABD, FBG, V AB=^FB, and BD^BG, and z ABD-' L FtsO, .'. lABD^lFBG. I. «. Now CD BL is double of J\ABD, on same base BD auc^ between same || s AL, BD, J. *-'. and sq. BG is double of a FBG, on same base FB and \,^ tween same l! s FB, GG ; I. C J' Book I.] PROPOSITION XLVII. 75 Similarly, by joining AE, BK it may be shewn that O ai = 8q. AK. Nov ^ iSu BC=&mn of £7 BL and O GL, B=sum of sq. jB6? and sq. AK, •—sum of sqq. on BA and AG. Q. K. D. Ex. 1. Prove that the square, described upon the diagonal of any f^ven square, is equal to twice the given square. Ex. 2. Find a line, the square on which shall be equal to the sum of the squares on three given straight lines. Ex 3. If one angle of a triangle be equal to the sum of the other two, and one of the sides containing this angle being divided into four equal parts, the other contains three of those parts ; the remaining side of the triangle contains five such parts. Ec i The triangles ABG, DBF, having the angles AGB, DFE right angles, have also the sides AB, AG equal to jDi?, DF, each to each ; show that the triangles are equal in every respect. Note. This Theorem has been already deduced as a Co- rollary from Prop. E, page 4.3. Ex. 5. Divide a given straight line into two parts, so that the square on one part shall be double of the square on the other. Ex. 6. If from one of the acute angles of a right-angled triiU^'lt' a line be drawn to the opposite side, the squares on that side and on the line so drawn are together equal to the sum of the squares on the segment adjacent to the right angle ind on the hyp(jtenuse. Ex, 7. In any triangle, if a line be drawn from the vertex at right angles to the base, the difference between the squares on ^\t. sutes IS equal to the difference between the squares on the r^nients of tlv.' b?t.!=;e. EUCLID'S ELEMENTS. [Book 1 Proposition XL VI II. Theorem. 1^ (Jve square described upon mie of the. sid.e^ of a triangle ht eq'iMi. to the squares described wpon the other two sides of it, itu ar^U contained by those sides i- a right angle. B a Cief tha sq. on JB(7, a side of lABC, he eqnal to the sum of the sqq. on AB, A G. Then must L BA Cbe a rt. angle. '^'rom pt. A draw ADi.io AG. I. 3 ! . Make AD=AB, and join DC. rhon '.'AD^AB, C&SX'^ .;. sq. on AI)=&q, on AB ; I. 46, Ex. 2. add to each sq. ow AG . *-lr^n sum of sqq. on AD, J.O=sum of sqq. on AB, AG, But •.' I DAG is a rt. angle, .'. sq. on i)6'=sum of sqq. on AD, AG \ I. 47 and, by hypothesis, sq. on J5(7=sum of sqq. on AB, AG ; .*. sq. on i)0=sq. on BG ; .-. DO==BC. I. 46, Ex. 3. Acniii LB ABG, ADG, •: AB=AD, and ^Ois common, and BG=^DG, .'. iBAG==lDAG; I c «ind I DAG is a rt. angle, by construction ; .-. I BAG is -A rt. angle. Q. >'. ^ BOOK II. INTRODUCTORY REMARKS. The geometrical figure with whicli we are chiefly concerned in this book is the Rectangle. A rectangle is said to be con- tained by any two of its adjacent sides. Thus if ABGD be a rectangle, it is said to be contained by AB, ADf or by any other pair of adjacent sides. We shall use the abbreviation rect. AB, AD to express the words " the rectangle contained by AB, AD." We shall make frequent use of a Theorem (employed, but not demonstrated, by Euclid) which may be thus stated and proved . Proposition A. Theorem. If the adjacent sides of one rectangle be equal to the adjacent sides of jLiioUwr rectangle, each to each, the rectangles are equal in area Let ABCD, EFGH be two rectangles : and let AB=EF and BC=FG, 4 D E ^ B n -F Then must rect. ABCD=rect. EFGH, For if the rect. EFGH be applied to the rect ABCD, so that EF coincides with AB, then FG will fall on BC, '.' l EFG=- l ABC, Similarly it may be shewn that H will coincide with D, .'. rect. EFGH coincides with and is therefore equal to rect ABCD. Q. E. D. 78 EUCLID'S ELEMENTS. [Book IL Proposition I. Theorem. If there be two straight lines, one of which is divided into any number of parts, the rectarigle contained by the two stradgM lines is equal to the rectangles contained' by the v/ndivided line and the several parts of the divided liiie. U 2C Let AB and CD be two given st. lines, and let CD be divided into any parts in E, F. Then must recL AB, CD = sum of red. AB, CE mid rect AB, EF and re^t. AB, FD. From G draw CG x to CD, and in CQ make CH^AB. Through U draw HM \\ to CD. I. 31. Through E, F, and D firaw EK, FL, DM \\ to CH. , Then EK and FL, being each=CJI, are each =^5. Now CM^^sum of CK and EL and FM, And OM=rect. AB, CD, :• CH=AB, CK^rect. AB, CE, v CH=AB, EL=rect. AB, EF, '.' EK=AB, i^'Jf =rect. AB, FD, ',' FL=AB ; . . rect. AB, CD = sum of rect. AB, CE and rect AB, EF wid rect. AB, FD. Q. E. D. Ex. If two straight lines be each divided into any number of parts, the rectangle contained bj the two lines is equal to the rectangles contained by all the parts of the one taken sepaiately with all the parts of the other. Book ILJ PROPOSITION II, 79 Proposition II. Theorem. If a straight line he divided into any two jjarts, the rectargh' contained by the whole and each of the parts are together equal to the squa/re on the whole Une. A. a B J> 1 7 2 5 Let the st. line AB be divided into any two parts in 0* Then miist sq. on AB—svm of red. AB, AC and rect. ABf CB. On AB describe the sq. ADEB I. 46. Through G draw CF ii to AD I. 31. Then ^.E=sum of AF and CE. Now AE is the sq. on ABy AF=Tect. AB, AC, '.' AD=^AB, C^=rect. AB, CB, V BE=^AB, '. sq. on y! B=«8um of rect. AB, AG and rect AB, CB Q. E. D. Ex. The square on a straight line is equal to four times the square on half the line 8o EUCLIjys ELEMENTS. [Book IT. Proposition 111. Theorem. If a straiyht line bedimded into any two 'parts, the rectangle contained by the ivhole and one of the parts is equal to the rect- angle contained by the two parUi together with the square on the aforesaid part. Let the st. line AB be divided into any two parts in G. Then must red ABy CB=sum of rect. AG, CB and sq. on GB. On GB describe the sq. GDEB. 1. 16 From A draw AF || to GD, meeting ED produced in F. Then ^^=sum of AD and GE. Now ^^=rect. AB, GB, v BE= GB, AD:^rect. AG, GB, / GD=GB, GE=Bq. on GB. .'. rect. AB, 05= sum of rect. AG, GB and sq. on GB. Q. E. D. Note. When a straight line is cut in a point, the distances of the point of section from the ends of the line are called the segments of the line. If a line AB be divided in C, AG and GB are called the internal segments of AB. If a line AG he produced to B, /^ and GB are called the external segments of AG. Book II.j PROrOSITION TV 8i Proposition IV. Theorem. If a straight line he divided into any two parts, the squa/re on th£ whole line is equal to the squares 07i the two pa/rts together vith twice th£ rectangle contained by the po/rts. D O IS Let the st. line AB be divided into any two parts in C. Then must q. on AB—s^im, of sqq. on AG, CB and Itoice rect. AC, CB. On AB describe the sq. A DEB. I. 46 From AD cut off AH = CB, Then HD==AC. Draw CG || to AD, and HK \\ to AB, meeting CG in F. Then •/ BK=AH, .'. BK= CB, Ax. i. •. BK, KF, FC, CB are all equal ; and KBC is a rt. z ; ,-. CK is the sq. on CB. Def. xxx Also HG^sq. on A C, '.• HF and HD each =^AC. Now ^jR:=8um of HG, CK, AF, FE, AE=sq. on AB, and HG=sq. on AC, CS:=sq. on CB, AF =rect. AC, CB, FE=Tect. AC, CB, CF=CB, FG=^AGajidFK'=CB. .'. sq. on -4jB«"8um of sqq. on AC, CB and twice rect AC, CB. Q. E. D. Etc. In a triangle, whose vertical angle is a right angle, a straight line is drawn from the vertex perpendicular to the base. Shew that the rectangle, contained by the segments of the base, is equal to the square on the perpendicular. EUCLID'S ELEMENTS, [Book n. Proposition V. Theorem.- If a straight line be divided into two equal imrts and also into two unequal parts, the rectangle contained by iiie unequal parts, together with the square on the li7ie between the points of section^ is equal to tJie square on half the line. Let the st. line AB be divided equally in C and unequally inD. Then must red. AD, DB together with sq. on. CD=sq. on CB. On GB describe the sq. CEFB. L 46. Draw DO \\ to GE, and from it cut off DH=DB. I. 31. Draw HLK \\ to AD, and AK \\ to DH. I. 31. Then rect. Djp'^rect. AL, Also LG=sq. on GD, V BF=AG, and BD=GL. V LII^ GD, and ffO=^ GD. Then rect. AD, DB together with sq. on GD »=^AR together with LG — Hiim of AL and GH and LG «sum of DF and GB. and LO ^GF ■esq. on GB. a B. D. book IL] PROPOSITION vr. 83 Proposition VI. TuEORtJi. If a straight litie he bisected and produced to any point, tJie rectangle contained by the whole line thus produced and the j:art of it produced, together with the square on half the line bisected, is equal to the square on trie straight line whicli is made up of the half and the part produced. Ji B J> ja. \ X L B '- JP u Let the st. line AB be bisected in C and produced to D. Then must rect AD, DB together with sq. on CB=sq. on CD. On CD describe the sq. CEFD. Draw BG \\ to C^, and cut off BU'^BD, Through 1? draw KLM lU> AD Through A draw AK \\ to CE, JMow •.• BG^ CD and BH=BD ; .\HG=^CB\ .'. rect. Af(?=rect. AL. Then rect. AD, DB together with sq. on CB = sum of A M and LG = sum of AL and CM and LG =8um of MG and CM and LO ■i«8q. on CD. Q. B. D. 1.46. I. :U. I. 31. Ax. 3. ir.A. ^ LUCLID'S ELEMENTS. [Cock IL Note. We here give the proof of an important theorem, which is usually placed as a corollary to Propositioii V. Proposition B. Theorkm. 'Hit difference between the squares on any two straight lines M equal to the rectangle contained by the sum and difference of those lines. ^ ^ JSl' Let AG, CD be two st. lines, of which ^0 is the greater, and let them be placed so as to form one st. line AD. Produce AD to B, making CB^AC. Then ^4/>=the sum of the lines AC, CD, and DB= the difference of the lines AC, CD. Then must difference between sqq. on AC, CD— red. AD, DB. On CB describe the sq. CEFB. I. 46. Draw DO \\ to CE, and from it cut off DH=DB. I. 31. Draw HLK \\ to AD, and AK \\ to DH. I. 31. Then rect. i>^=rect. AL, .'BF=AC, and BD=CL. Also LGf=sq. on CD, :' LH= CD, and HG= CD. 'Chen difference between sqq, on AC, CD = difference between sqq. on CB, CD -sum of CE and DF ■ssum of CH and AL m.AH ^rect. AD, DH .^lecL AD, DB. Q. K D. Ex. Shew that Propositions V. and VI. might be deduced from this Proposition. Book IL] PROPOSITION VII. 85 Proposition VII. Theorem. If a straight line he divided into any two 'parts, the squa/res on the whole line and on one of the iiarts are equal to twice the rectangle contained by the whole and thai part together vrith tJie aqua/re on the other ya/rt. O B Let AB be divided into any two parts in CL Then must sqq. on AB, BG= twice rect. AB, BG together with sq. on AC. On AB describe the sq. ADEB. I. 46. From AD cut off AH= CB. Draw CF \\ to AD and HGK \\ to AB. I. 31. Then B.F=sq, on AC, and CK=sq. on CB. Then sqq. on AB, BG =sum of AE and CK =sum of AK, HF, GE and CK -=8um of AK, HF and GK Now AK-^rect AB, BG, ••• BK=BC ; CE=rect. AB, BG, v BE^^AB ; HF=sq. on AG. .'. sqq. on AB, jBC= twice rect AB, £0 together with sq. on ^G Q. E. D. Ex. If straight lines be drawn from G to B and from to D, shew that BCD is a straight line. S6 EUCLIiys ELEMENTS. [Book II Proposition VIII. Theorem. If a straight line be divided into any two parts, foui times the rectangle coiitained by the whole line and one of tht pari% together ndth the square on the other part, is equal to the square on the straight line which is made up of the whok and the first part. A 1, u JL I. :f Let the st. line AB be divided into any two paits in 0. Produce AB to D, so that BD^BC. Th^n must four times red. AB, BC together with sq. on AC'=>sq. on AD. On AD describe the sq. AEFD. I. 46. From AE cut off AM and MX each = OB. Through G, B draw OH, BL \\ to AE. I. 31. Through M, X draw MGKN, XPRO \\ to AD. I. 31. Now •.• XE=AG, and XP=AG, .'. ZH«=sq. on AG. Also AG=MP=PL=BF, IL a. and GK^GE==BN=KO ; IL a. .'. sum of these eight rectangles "■four times the sum of AO, CK •-four times AK «=four times rect. AB, BC. Then four times rect. AB, BC and sq. on AG =8uui of the eight rectangles and XjS = AEFD =sq. on AD. «• b. d. Book n.] PROPOSITION IX. 87 Proposition IX. Theorem. If a straight line be divided into two equal, and also into two unequal partSj the squares on the two unequal parts am together double of the square mi half th€ line and of thr square on the line between the points of section. Let AB be divided equally in Oand unequally in IX Then must turn of sqq. on AD, I) B— twice sum of sqq. on ACy CD. Draw CE^- AC at rt. i s to AB, and join JEA, EB. Draw DF at rt. L%io AB, meeting E^ in F. Draw FG at rt. z s to EC, and join AF. Tlien ••• L ACE is a rt i, .'. sum of z s AEC, EAC'^a, rt. i ; and V £.AEC= lEAC, .'. z^^C- half art. z. So also z BEC and z EBC are each = half a rt. z . Hence z AEF is a rt. z . Also, *.• z GEF is half a rt. z , and z EGF is a rt, z .-. z EFG is lialf a rt. z ; .-. z Ei^(?= z 6f^i^, and /. EG^GF. So also z BED is half a rt. z , and BD=DF. 1.32. La. I. B. Cor. Now sum of sqq. on AD, DB = sq. on AD together with sq. on DF =sq. on AF I. 47. ==sq. on AE tocrether with sq. on EF I. 47. «=sqq. on AC, EC locrpther with sqq. on EG, GF I. 47. = twice ?q. en AC toojether with twice sq. on GF =twice sq. on AC together with twice sq. on CD. Q. E. D. EUCLID'S ELEMENTS. pBook n. Proposition X. Theorem. If a stratfjht line he bisected and prodiiced to any pointy the square on the ivhole line thus produced and the square on the part of it produced are together double of the square wn half the line bisected and of the square on the line made v/p of the half and the paH produced. Let the St. line AB be bisected in C and produced to I), Then must sum. of sqq. on AD, BD'=' twice sum of sqq. on AC, CD. Draw CB± to AB, and make CE=AC. Juin BA, EB and draw EF || to AD and DF l| to CE. Then '.• z s FEB, EFD are together less than two rt. i s, .*. EB and FD will meet if produced towards B, D in iome pt. 0. Join AG. Then '.• z ACE is a rt z , .•. / B EAC, AEG together = a rt. L , and •.• lEAC= a A EG, .-. z^^0= half art. z. So also z 8 BEG, EBC each = half a rt z . /. z AEB is a rt. z . Also z DBG, which = z EBC, is half a rt. / , and .*. z BGD is half a rt. z ; .-. BD==DG. I Again, •.* z jF'G'£'=half a rt. z , and z JS?^(? is art z .-. z i?'Ji^6?=half a rt. z , and EF^FG. I. b. Cor. Then sum of sqq. on AD, DB -^yum of sqq. on AD, DG =sq. on AG ' L47. =sq. on AE together with sq. on EG r%7. ==sqq. on AC, EC t<2gethei with sqq. on EF, FG I. 47. = twice sq. on AG together with twice sq. on EF = twice sq. on ^0 together with twice sq. on CD. Q. f. o La. B. Cor. , L34. Book n.] PROPOSITION XI. 89 Proposition XI. Problem. ' To divide, a given straight line into two j^arts, so that the rect- angle contained hy the whole and one of the parts shall be equai to the aqwure (W th^ ether ya/rU F . (t Let JJ5 be tb*^ given pt. Ifrte. On AB descr the sq. ADCB, Bisect AD in E and joir EB. Produce DA to F, n^iking EF=EB, On AF descr. the sq. AFGB. Then AB is divided in H so that red. AB, BH=s^. on AH. Produce GH to K. Then •.• DA is bisected in E and produced to F^ .'. rect. DF, FA together with sq. on JE n=sq. on EF =sq. on EB, \- EB^EF, =8um of sqq. on AB, AE. Take from each the square on AE. Then rect. DF, FA=sq. on AB. Now FK^iect. DF, FA , v FG=Fa .'. FK=AC. Take from each the common part AK. Then FH^HC', that is, sq. on J^if=rect. AB, BH^ Thus AB is divided in H as was reqd. Q. E. p. Ex. Shew that the squares on Liiu whole line and one of th*> parts are equal to three times the square on the other part. 1.46. I 10. 1.46. n. 6 1.47. JBO-JR 90 EUCLIIXS ELEMENTS. [Book IL Proposition XII. Theorem. In ohtufe-a7}fjled triangles, if a perpendicular be draicn from either of the, acute angles to the opposite side 'produced, the, sqv/ire on the side subtending the obtuse angle is greater than the squares on Uu sides containing the obtuse angle, hy tivice the rectangle contained by the side, upon which, when produced, the perpendir eular falls, and the straight line intercepted without the triangh between th^ perpendicular and the obtuse angle. Let ABC be an obtuse-angled A , having z ACB obtu.^. From A draw AD ± to jBO produced. Then rnuM sq. on AB he greater than sum of sqq. on BC, CA by tunce rect. BG, CD. For since BD is divided into two parts in G, sq. on J5I>=sum of sqq. on BG, CD. and twice rect. BG, GD. II. 4. Add to each aq. on DA : then sum of sqq. on BD, DA — ^\\m of sqq. on BG, GD, DA and twice rect. BG, CD. Now sqq. on BD, /)^=sq. on AB, I. 47. and sqq. on GD, Z)^=sq. on GA ; I. 47. .*. sq. on ^i>=sum of sqq. on BC, GA and twice rect. BG, GD. .'. sq. on AB is greater than sum of sqq. on BG, GA by twice rect. BC, CD. Q. E. I). Ex. The squares on the diagonals of a trapezium are together equal to the squares on its two sides, which are not parallel, and twice the rectangle contained by the sides, which arc parallel. Book II.] PROPOSITION XIIi. 91 Proposition XIII. Theorem. In every trianyle, the square on the side subtending any of the acAite angles 'is less than the squares on the sides containing that angle, by tivice the rectangle contained by eitJier of these sides and the straight line intercepted between the perjiendicular, let faU wpon it from the opposite angle, and the acute angle. 7? C Let ABC be any A , having the z ABC acute. From A draw AD JL to BC or BC produced. Then must sq. on AC be less tha,n the smn of s^q. on AB BC, by twice rect. BC, BD. For in Fig. 1 BC i? divided into two parts in D, and in Fig. 2 BD is divided into two parts in C; .'. ill both cases sum of sqq. on BC, BD= sum of twice rect. BC, BD and sq. on CD. 11. 7. Add to each the sq. on DA, then sum of sqq. on BC, BD, DA^enm of twice rect. BC, BD and sqq. on CD, DA ; .-. sum of sqq. ou BC, AB^aum of twice rect BC, BD and sq. on AC] 1-47 .*. sq. on ^Ois less than sum of sqq. on AB, BChj twice rect. BC, BD. The case, in which the perpendicular AD coincides with AC needs no proof. Q. R D Ex. Prove that the sum of the squares on any two sid-.* A a triangle is equal to twice the sum of tlie squares on half tue base and on the line joining the vertical angle with the middle point of the base. 92 EUCLID'S ELEMEhrrS. [Sook n. Proposition XIV. Problem. To de^yri^e a square that shall be equal to a given rectilinear figure. Let A be the criven rectil. figure. It is reqd. to describe a square that shall=A. Describe the rectangular O BCDE=A. I, 45. Then if BE=ED the O BCDE is a square, and what was reqd. is done. But if BE be not = ED, produce BE to F, so that EF=ED, Bisect BF in G ; and with centre G and distance GBy describe the semicircle BHF. Produce DE to H and join GH. Then, *.• BF is divided equally in G and unequally in E, ,'. rect. BE, EF together with sq. on GE =sq. on'6?i^ II. 5. =sq. on GH •=sum of sqq. on EH, GE. I. 47. Take from each the square on GE. Then rect. BE, EF==^q. on Efl. But rect. BE, EF=BD, v EF^ ED ; .'. sq. on EH =BD ; .*. sq. on EH=rectii.. figure A. Ex. Shew how to describe a rectangle equal to a given sqnnre, and having one of its sides equal to a given straight line. Book n.] MISCELLANEOUS EXERCISES. 93 Miscellaneous Exercises on Book II, 1. In a triangle, whose vertical angle is a right angle, a straight line is drawn from the vertex perpendicular to the base ; shew that the square on either of the sides adjacent to the right angle is equal to the rectangle contained by the base and the segment of it adjacent to that side. 2. The squares on the diagonals of a parallelogram are to- gether equal to the squares on the four sides. 3. If ABCD be any rectangle, and any point either within or without the rectangle, shew that the sum of the squares on OA, OG is equal to the sum of the squares on 0J5, OD. 4. If either diagonal of a parallelogram be equal to one of the sides about the opposite angle of the figure, the square on it shall be less than the square on the other diameter, by twice the square on the other side about that opposite angle. 5. Produce a given straight line AB to C, so that the rect- angle, contained by the sum and difference of AB and AC^ may oe equal to a given square. 6. Shew that the sum of the squares on the diagonals of any quadrilateral is less than the sum of the squares on the four sides, by four times the square on the line joining the middle points of the diagonals. 7. If the square on one perpendicular from the vertex of a triangle is equal to the rectangle, contained by the segments of the base,"the vertical angle is a right angle. 8. Produce a given straight line so that the rectangle con- tained by the whole line thus produced and another given straight line may be equal to the square on the produced part. 9. ABC is a triangle right-angled at A ; in the hypote- nuse two points Z), E are tnken such that BD = BA and CE=CA ; shew that the square on DE is equal to twice the rectangle contained by BE, CI). 94 EUCLID'S ELEMENTS. [Book II. 10. In any quadrilateral the squares on the diagonals arc together equal to twice the sum ot tne squares on the straight lines joining the middle points of opposite sides. 11. If straight lines be drawn from each angle of a triangle to bisect the opposite sides, four times the sum of the squares on those lines is equal to three times the sum of the squares on the sides of the triangle. 12. CD is drawn perpendicular to AB^ a side of the triangle ABC^ in which AG=AB. Shew that the square on CD is equal to the square on BD together with twice the rectangle AD, DB. 13. If in any triangle BAG 2k line AD be drawn bisecting BC in D, shew that the sum of the squares on AB, AC x"^ equal to twice the sum of the squares on AD, BD, 14. If ABC be an equilateral triangle, and AD, BE bh perpendiculars to the opposite sides intersecting in F ; shew that the square on AB is equal to three times the square on AF. 15. Divide a given straight line into two parts, so that the rectangle contained by them shall be equal to the square described upon a straight line, which is less than half the iin© divided. Books i & 11.] ON THE MEASUREMENT UP AREAS 95 Note 6. — On the Measurememi of Areas. To measure a Magnitude, we fix upon some magnitude of the same kind to serve as a standard or unit ; and then any magnitude of that kind is measured by the luiniber of times it contains this unit, and this number is called the Measurk of the quantity. Suppose, for instance, we wish to measure a straight line AB. We take another straight line EF for our standard, A B C D E F and then we say ii AB contain EF three times, the measure ot AB is 3, if four 4, if X X. Next suppose we wish to n)easu)e two straight lines AB, CD by the same standard EF. If AB contain EF m times and CD n times, where w and n stand for number?, whole or fractional, we say that AB and CD are com/mensurable. But it may happen that we may be able to find a standard line EF, such that it is contiiined an exact number of tunes in AB ; and yet there is no number, whole or fractional, which will express the number of times EF is contained in CD. In such a case, where no unit-line can be found, such that it is contained an exact number of times in each of two lines AB, CD, these two lines are called incommensurable. In the processes of Geometry we constantly meet with incommensurable magnitudes. Thus the side and diagonal of a square are incommensn rabies ; and so are the diameter and circimiference of a circle. 96 EUCLID'S ELEMENTS, [Boolts I. & II Next, suppose two lines AB, J.0 to be at right angle? to each other and to be commensurable, so that AB contains four times a certain unit of linear measurement, which Js contained by -40 three times. Divide ABj AG into four and three equal parts respectively, and draw iine^ through the points of division parallel to AC, AB respectively ; then the rectangle ACDB is divided into a number of equal squares, each constructed on a line equal to the unit of linear measurement. If one of these squares be taken as the unit of area, the meas^ire of the area of the rectangle ACDB will be the number of these squares. Now this number will evidently be the same as that obtained by multiplying the measure of AB by the measure of AC ; that is, the measure of AB being 4 and the measure of AC 2, the measure of ACDB is 4 X 3 or 12. (Algebra, Art. 38.) And generally, if the measures of two adjacent sides of a rectangle, supposed to be commensurable, be a and b, then the measure of the rectangle will be ab. (Algebra, Art. 39.) If all lines were commensurable, then, whatever might be the length of two adjacent sides of a rectangle, we might select the unit of lengtli, so that the measures of the two sides should be whole numbers ; and then we might apply the processes of Algebra to establish many Propositions in Geometry by simpler methods than those adopted by Euclid. Take, for example, the theorem in Book ii. Prop. TV. If all lines were commensurable we might proceed thus : — Let the measure of AG he aj, . ofCB ... y. Then the measure of A B is x-^y. Now (sc + y)'^ = x'^ + if + 2xy, which proves the theorem. 8ookh L & II.] .ON THE MEA:SU-REMENT OF AREAS. 97 But, inasmuch as all lines are not commensurable, we fiavf in Geometry to treat of magnitudes and not of m^as7ires : that is, when we use the symbol A to represent a line (as in I. 22), ^1 stands for the line itself and not, as in Algebra, for the number of units of length contained by the line. The method, adopted by Euclid in Book II. to explain the relations between the rectangles contained by certain lines, is more exact than any method founded upon Algebraical prin- ciples can be ; because his method applies not merely to the case in which the sides of a rectangle are commensurable, but also to the case in which they are incommensurable. The student is now in a position to understand the practical application of the theory of Equivalence of Areas, of which the foundation is the 35th Proposition of Book I. We shall give a few examples of the use made of this theory in Men- suration. Area of a Parallelogram. The area of a parallelogram ABGD is equal to the area of the rectangle ABEF on the same base AB and between the same paiallels A By FG. jp J) B Cf Now BE is the altitude of the paralleloj^ram ABCD if 4jB be taken as the base. Hence area of O ABCD==Tect. AB, BE. If then the measure of the base be denoted by 6, and altitude h, the measure of the area of the ZZ7 will be denoted by ok That is, when the base and altitude are commensurable, measure of area = measure of base itito mejusure of altitude. 8. K. 98 EUCLID'S ELEMENTS. [Books L & a Arta of a Triangle. If from one of the angular points A of a triangle ABC, a [)erpendiculiir AD be drawn to jBC, Fig. 1, or to EG produced, K.g. 2, ji jy c MO and if, in both caseis, a parallelogram ABCJil be completed of which AB, BG are adjacent sides, area of A vl 50= half of area of O ABGE.' Now if the measure of BG be b, and AJJ... h, measure of area oiCJ ABCE is hh ; .*. measure of area of A ABG is — ■• Arm of a Rhombus. Let ABGD be the given rhombus. Draw the diaffonals AG and BD, cutting one another in n It is easy to prove that A G and BD bisect each other «t fight angles. Then if the measure of AG he x, and BD...V, measure of ar a of rhombus = twice measure of A AGD. =twice ^ 4 Books L & ilj AR£A OF A 'rKAJ'JsZJUM. 99 Area of a Trapezium. > Let A BCD be the given trapezium, having the sides AB, CD parallel. Draw AE at right angles to CD. A ^ 3 1 ^"^^ \ Ann / P V .^ h W.^ c ^j Produce DC to F, making CF=AB. Join AF, cutting JBC in 0. Then in as AOB, COF, V aBAO= I CFO, and z ^0^= z FOC, an/^ AB='CF; .: hCOF=^LAOB. 1.26. Hence trapezium ABCD= lADF. Now suppose the measures of AB^ CD, J E to be tn, n, |3 respectively ; .-. measure oiDF^m + n^ v CF=AB. Then measure of area of trapezium =i (measure of DF X measure of AE) -»J(m + ?i)X j?. That is, the measure of the area of a trapezium is found by multiplying half the measure of the sum of the, parallel sides by the measure of the perpendicular distance between the parallel sides. EUCLID'S ELEMENTS. [Books I. & II. Area of an Irregular Polygon. There are three methods of finding the area of an irregular polygon, which we shall here briefly notice. I. The 'polygon may he divided into triangles, and the area of each of these triangles be found separately. E D Thuq the area of the irregular polygon ABODE is equal W) the sum ci the areas of the triangles ABE, EBD, DBG. II. The polygon may he converted into a single triangle of equal area. If ABODE be a pentagon, we can convert it into an equivalent quadrilateral by the following process : Join BD and draw OF parallel to BD, meeting ED pro- duced in F, and join BF. Then will quadrilateral ^£i^^= pentagon ABODE. For A BDF= A BOD, on same base BD and between same parallels. Tf, then, from the pentagon we remove A BOD, and add lBDF to thp remainder, we obtain a quadrilateral ABFE eatiivalent to the pentaj^on A BODE. 2oo!rs t. & IL] AREA OF AN IRREGULAR POLYGON, loi The quadrilateral may theii,l)y a siiuilar process, be con- verted into an equivalent triangle, and thus a polygon of any number of sides may be gradually converted into an equiva- lent triangle. The area of this triangle may then be found. III. The third method is chiefly employed in practice by Surveyors. Let ABCDEFG be an irregular polygon. Draw AE, the longest diagonal, and drop perpendiculars on AE from the other angular points of the polygon. The polygon is thus divided into figures which are either right-angled triangles, rectangles, or trppeziuma ; and the areas of each of these figures may be readiJy calculated. I02 EUCLID'S ELEMENTS. [Books L & IL Note 7. On Proje/^iant. The projection of a point B, on e explained by the following examples. Our first example of the Analytical process shall be the 31s< Proposition of Euclid's First Book. io6 EUCLID'S ELEMENTS. [Books I. & II Ex. 1. To draw a straight line through a given point parallel to a given straight hne. Let A be the given point, and BG be the given straight line. Suppose the problem to be effected, and EF to be the straight line required. K 4 JT Now we know that any straight line AD drawn jfrom A to meet BG makes equal angles with EF and BG. (i. 29.) This is a fact from which we can work backward, and arrive at the steps necessary for the solution of the problem ; thus : Take any point D in BG, join AD, make z EAD= L ADC, and produce EA to F : then EF must be parallel to BG. Ex. 2. To inscribe in a triangle a rhombus, having one of its angles coincident with an angle of the triangle. Let ABC be the given triangle. Suppose the problem to be effected, and DBFE to be the rhombus. Then if -BB be joined, z DBE=^ l FEE. This is a fact from which we can work backward, and deduce the necessary construction ; thus : Bisect L ABG hy the straight line BE, meeting AGin E. Draw ED and EF parallel to BC and AB respectively. ' Then DBFE is ihe rhombus required. (See Ex. 4, p. 69.) Books I. & n.] SOLUTION OF PROBLEMS. 107 Ex. 3. To determine the jpoint in a given straight line, at which straight lines, drawn from two given points^ on the same mle of the given line, make equal angles with it. Let CD be the given line, and A and B the given points. Suppose the problem to be effected, and P to be the point required. "21 We then reason thus : If BP were produced to some point J.', I CPA', being= z BFD, wiU be= z APO. Again, if PA' be made equal to P-4, AA' will be bisected by CP at right angles. This is a fact from which we can work backward, and find the steps necessary for the solution of the problem ; thui : From A draw ^0 ± to CD. Produce -40 to A', making OA'^OAt Jom BA', cutting CD in P. Then P is the point required. Note 10. On Symmetry. The problem, which we have just been considering, Buggestt the following remarks : If two points, A and A', be so situated with respect to a straight line CD, that CD bisects at right angles the straight line joining A and A\ then A and A' are said to be symmetrical with regard to CD. The importance of Bymmetrioal relations, as suggestive of methods for the solution of problems, cannot be fully shewn io8 EUCLID'S ELEMENTS. [Books I. A; IL to a learner, who is unacquainted with the properties of the circle. The following example, however, will illustrate this part of the subject sufficiently for our purpose at present. Find a point in a given straight line, such that the sum of its distances from two fixed points on the same side of the line is a minimwnhj that is, less than the sum of the distances of a/ny other point in the line from the fixed points. Taking the diagram of the last example, suppose CD to be the given line, and A, B the given points. Now if A and A' be symmetrical with respect to CD, we know that every point in CD is equally distant from A and A'. (See Note 8, p. 103.) Hence the sum of the distances of any point in CD from A and B is equal to the sum of the distances of that point irom A' and B. But the sum of the distances of a point in CD from A' and B is the least possible when it lies in the straight line joining A' and B. Hence the point P, determined as in the last exa/rnple, is the point required. Note. Propositions ix., x,, xi., xii. of Book I. give good examples of symmetrical constructions. Note 11. Euclid! s Froof of L ft. The angles at the base of in isosceles triangle cure equal to one another ; and if tJie equal sides he produced, the angles upon the other side of the base shall be equal. Let ABC be an isosceles A, having AB" 40 Produce AB, AG to D and E. Then must L ABC= l ACB, and L DBC= l ECB Books I. & II.] EUCLWS PROOI^ OF I. 5. 109 In BB take any pt. ^. From AE cut off AG^AB. JoinFCsiiidGR Than in ah AFC, AGS, '.' FA = GA, and AC=AB, and lFAC=l GAB, .'. FC^GB, and z AFC= l AGB, and lACF^ L ABG. I. 4 Again, •/ AF=AGy of which the parts A B, AG axe equal, .*. remainder ^J*™ remainder OG. Ax. 3 Then in as BFG, CGB, '.' BF=CG, and FG=GB, and l BFC= l CGB, .-. L FBC'= L GCB, and z JB(7J?'= l CBG, L 4. jSow it haa been proved that L ACF=^ i ABG, of which the parts i BCF and z CBG are erjual ; .*. remainmg z ^C5= remaining L ABC At^ H. /Uso it has been proved that z FBC= z G'C^, that is, z r>BC=- z ^CJ5. Q. K. n. I lo ttU CUD'S ELEMENTS. [Books I. & H Note 12. Eudids Proof of I. 6. If two a/rigles of a triangle he equal to one another^ the sides also, which iubtend the equal anglti^ shaU b« equai to onecmoUm; 5 In A ABG let l ACB= l ABiX Then must AB=AG. Wot if not, AB is either greater or less than ACL Suppose AB to be greater than AC. From AB cut off BD=AC, and join DO. Then in £^a DEC, A CB, '/ DB'^AGy and BG is common, and z DBG= I AGB, .: aDBG=£.AGB; 14. that is, the less = the greater ; which is absurd. .'. AB is not greater than AG. Similarly it may be shewn that AB is not less than AC; .'. AB=^AC. Q. E. D. Note 13. Eudid's Proof of I. 7. Upon the same base and on the same side of tt, there cannot be two triangles that have their sides which are ter- minated in one extremity of the base equal to one another^ and their sides which are terminated in the other extremity of tivt ba^e equal also. If it be possible, on the same base AB, and on the same side of it, let there be two a s ACB, ADB, such that AC=AD, and also BC=BD. Join CD. Books I. & II.] EUCUUS PROOF OF I. 7. I.ti First, when the vertex of each of the as is outside the other A (Fig. 1.) ; Via 1. Fro. 2. C_ P V AD=AG, /. z ACD= L ADO, I. 5. Bat I ACD is greater than z BCD ; ,-. z ^J)(7 is greater than z jBCD ; much more is z ^DC greater than z .BCD. Again, •.• BG=BD, .-. z jBi)C= z BCD, chat is, z BD(7 is both equal to and greater than z BCD ; which is absurd. Secondly, when the vertex D of one of the A s falls tvithin the other A (Fig. 2) ; Produce AC and AD to .E and J* Then '.- AC=AD. .'. L ECD= z FDG. h 6. But I BCD is greater than z jBCD ; .-. z FDC is greater than z BCD ; much more is z 5D0 greater than z BCD, Again, -.♦ BC=BD, ,', L BDC== L BCD ; that is, z BDCT is both equal to and greater than z BCD . which is absurd. Lastly, when the vertex D of one of the As falls on a side BC of the other, it is plain that BC and BD CJinnot be equal Q. e. d. 112 EUCLID'S ELEMENTS. [Books I. & II Note 14. EucMs Proof of I. 8. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their ba^es equal, the angle which is contained by the two sides of ifie one must be equal to the angle contained by the two aides of the other. Let the sides of tho A s ABG, DEF be equal, each to each^ that is, AB=DE, AG=DF and BC=EF. Then must lBAG= ^ EDF. Apply the t^ ABO to the i^DEF, so that pt. J5 is on pt. E, and BC on EF. Then •/ BG=EF, ,\ G will coincide with F, and BO will coincide with EF. Then AB and AG'mnBt coincide with DE and DF, For if AB and AG hare a different position, as GE, GF, then upon the same base and upon the same side of it there can be two A s, which have their sides which are terminated in one extremity of the base equal, and their sides which are ter- minated in the other extremity of the base also equal : which is impossible. I. 7„ .*. since base BO coincides with base EF, AB must coincide with DE, and AG with DF ; /. z BA coincides with and 's equal to z EDF, Q. E. D. Books !. & II.] ANOTHER PROOF OF I. 24. nj Note 15. Another Proof of I. 24. In the AS ABC, DEF, let AB=DE and AG^^DFyS^id let ABAC he greater than z FDF. Tfien mvst BC he greater than EF, Apply the A DEF to the A ABC so that JD-E coincides with AB. Then •.• z jKjDi'' is less than z J5^0, D^ will fell between BA and ^0, and F will fall o/i, or above, or 6eZow, jB(7. I. If F fall on JBO, BF is less than £(7 ; .*. EF is less than BC. n. If^falla6(M?eJ5(7, JB^*, FA together are less than BC, CA, find FA==CA', .. ^jPisless than jBC; .-. EF is less than BO, m. If FfsMhdowBC, let AF cut BGin 0. Then BO, OF together are greater than BF, I. 20 and OC, AO AC; I. 20. .\BC,AF. £i^, ^C together, and AF=AC, .'. BC is greater than BF ; and .*. EF is less than BC. Q. b. d. [14 EUCLIiyS ELEMENTS. [Books L & H Note 16. Euclid's Proof of I. 26. If two triangles have two angles of the one equal to two angles of the othery each to ea^h, and one side equal to one side, viz.y either the sides adjacent to the equal angles^ or the sides opposite to equal angles in each ; th^n shall the other sides be equal^ each to each ; and also the third angle of the one to the third angle of the other. 3 a M^ In i\^ ABC, DEF, Let L ABC = L DEF, and z ACB = z DFE; BJidfvrstf Let the sides adjacent to the equal z s in each be equal, that is, let BG=EF. Then must AB=DE, and AC=DF, and z BAC = z EDF. For if AB be not=D^, one of them must be the greater. Let AB be the greater, and make GB=DE, and join GG. Then in AS GBC, DEF, V GB^DE, and BC=EF, and z GBC = z DEFy .-. z GCB= . DFE. 1. 4. But lACB= I DFE by hypothesis ; /. lGCB= lACB; that is, the less = the greater, which is impossible. .*. AB is not greater than BE. In the same way it may be shewn that AB is not less than DE', .'. AB=DE. Then in As ABC, DEF, '.' AB=DE, and BC=EF, and z ABC=' l DEF, .'. AG=DF, and z BAC^-- l EDF. T. 4. Books L & II.] EUCLID'S PROOF OF I. 26. 115 ]^e,xi, let the sides which are opposite to equal angles in each triangle be equal, viz., AB=I)E. Tlien must AG=DF, and BC=EF, and L BAG = l EDF. For if BG be not=^J', let BG \t^ *)»e greater, and make BH= EF, and join AH. Then in A s AB3, DEF, V A B^DE, and BH=EF, and z ABff^ l BEF, /. L AHB= L DFE. L 4. But I ACB='/ DFE, by hypothesis, .-. z ABB = z AGB ; that is, the exterior z of A ^JEC is equal tr '*;1>*» A»tArior and opposite z ^0J5, which is impossible. /. BG is not greater than EF. In the same way it may be shewn that BG is not le«8 thAP EF) .'. BG=EF. Then m AB ABGy DEF, -.' AB=DE, and BG=EF, and z ABG=^ L D£F, ,\ AG^DF, and z J5^(7= z EDF. L 4 Q. B. D. i i6 EUCLID'S ELEMENTS. [Books L & IL Miscellaneous Exercises on Books I. and II. 1. AB and GD are equal straight lines, bisecting one another at right angles. Shew that ACBB is a square. 2. From a point in the side of a parallelogram draw a line dividing the parallelogram into two equal parts. 3. Draw through a point, lying between two lines that intersect, a line terminated by the given lines, and bisected in the given point 4. The square on the hypotenuse of an isosceles right-angled triangle is equal to four times the square on the perpendicular from the right angle on the hypotenuse. 5. Describe a rhombus, which shall be equal to a given triangle, and have each of its sides equal to one side of the triangle. 6. Shew how to describe a square, when the difference between the lengths of a diagonal and a side is given. 7. Two rings slide on two straight lines, which intersect at right angles in a point 0, and are connected by an inextensible string passing round a peg fixed at that point. Shew that the rings will be nearest to each other when they are equidistant from 0. 8. ABCD is a parallelogram, whose diagonals AC, BD in-. tersect in ; shew that if the parallelograms AOBP, DOCQ be completed, the straight, line joining P and Q passes through 0. 9. ABCDy EBCF are two parallelograms on the same base BCy and so situated that CF passes through A. Join DF^ and produce it to meet BE produced in K ; join FBy and prove that the triangle FAB equals the triangle FEK. 10. The alternate sides of a polygon are produced to meet ; shew that all the angles at their points of intersection together with, four right angles are equal to all the interior angles of the polygon. 11. Shew that the perimeter of a rectangle is always greater than that of the square equal to the rectvigle. 300K8 L & IL] MISCELLANEOUS EXERCISES. 117 12. Shew that the opposite sides of an equiangular hexagon are parallel, though they be not equal ; and that any two sides that are adjacent are together equal to the two which are purallel. 13. If two equal straight lines intersect each other anywhere at right angles, shew that the area of the quadrilateral formed by joining their extremities is invariable, and equal to one-half the square on either line. 14. Two triangles ACB^ ABB are constructed on the same side of the same base AB. Shew that if AG=BD and AD=BCy then CD is parallel to AB ; but if AG=BC am] AD=BD, then CD is perpendicular to AB. 16. AB is the hypotenuse of a right-angled triangle ABC : find a point D in AB, such that DB may be equal to the per- pendicular from J> on AC. 16. Find the locus of the vertices of triangles of equal area on the same base. 17. Shew that the perimeter of an isosceles triangle is less than that of any triangle of equal area on the same base. 18. If each of the equjd angles of an isosceles triangle be equal to one-fourth the vertical angle, and from one of them a perpendicular be drawn to the base, meeting the opposite side produced, then will the part produced, the perpendicular, and the remaining side, form an equilateral triangle. 19. If a straight line terminated by the sides of a triangle be bisected, shew that no other line terminated by the same two sides can be bisected in the same point. 20. From a given point draw to two parallel straight lines two equal straight lines at right angles to each other. 21. Given the lengths of the two diagonals of a rhombos^ con- struct it. 22. ABCD is a quadrilateral figure : construct a triangle whose base shall be in the line AB, such that its altitude shall be equal to a given line, and its area equal to that of the quadrilateral 23. If ABC be a triangle in which (7 is a right angle, shew how to draw a straight line parallel to a given straight line, so as to be terminated by CA and CB and bisected by AB. tx8 EUCLIiyS ELEMENTS. [Books I. & a 24. If ABC be a triangle, in which is a right angle, and DE be drawn from a point D in J. (7 at right angles to ABy prove that the rectangles ABy AE and ACy AD are equal 25. A line is drawn bisecting parallelogram ABCD, and meeting AD, BG'm E and F : shew that the triangles EBF, GED are equal. 26. Upon the hypotenuse BC and the sides CA, AB of a rightrangled triaugle ABGy squares BDEG, AF and AG are described ; shew that the squares on DG and EF are together equal to five times the square on BG. 27. If from the vertical angle of a triangle three straight lines be drawn, one bisecting the angle, the second bisecting the base, and the third perpendicular to the base, shew that the first lies, both in position and magi;iitude, between the other two. 28. If ABG be a triangle, whose angle ^ is a right angle, and BEy GF be drawn bisecting the opposite sides respectively, shew that four times the sum of the squares on BE and GF is equal to five times the square on BG. 29. Let AGBy ADB be two right-angled triangles having a common hypotenuse AB. Join GD and on GD produced both ways draw perpendiculars AEy BF. Shew that the sum of the squares on GE and GF is equal to the sum of the squares on-DE&n^DF. 30. In the base AG oi a triangle take any point D: bisect ADy DGy ABy BG at the points Ey F, G, H respectively. Shew that EG is equal and parallel to FE[. 31. If AD be drawn from the vertex of an isosceles triangle A BG to a point D in the base, shew that the rectangle BD, DC is equal to the difference between the squares on AB and A D. 32. If in the sides of a square four points be taken at equal distances from the four angular points taken in order, the figure contained by the straight lines, which join them, shall also be a square. 33. If perpendiculars AF, BQy GR be drawn from the anj^u/Air points of a triangle ABG upon the sides, shew that tlv»y ^'*P im&»4. the angles of the triangle PQR. Books L&IL] MISCELLANEOUS EXERCISES. 119 34. If of the four triangles, into which the diagonals divide a quadrilateral, any two opposite ones are equal, the quadrila- teral is a trapezium. 35. ABGDj AECF are two parallelograms, EAy AD being in a straight line. Let FG, drawn parallel to AC^ meet BA produced in G. Then the triangle ABE equals the triangle ADG. 36. From AG, the diagonal of a square ABGD, cut off AB equal to one-fourth of AG, and join BE, DE. Shew that the figure BADE is equal to twice the square on AE. 37. If ABG be a triangle, with the angles at B and G each double of the angle at A, prove that the square on AB is equal to the square on BG together with the rectangle AB, BG. 38. If two sides of a quadrilateral be parallel, the tnangle contained by either of the other sides and the two straight lines drawn from its extremities to the middle point of the opposite side is half the quadrilateral. 39. If two opposite angles of a quadrilateral be right angles, the angles subtended by either side at the two opposite angular points will be equal. 40. If the sides of a triangle taken in order be produced to twice their original lengths, and the outer extremities be joined, the triangle so formed will be seven times the original triangle. 41. If one of the acute angles of a right-angled isosceles triangle be bisected, the opposite side will be divided by the bisecting line into two parts, such that the square on one will be double of the square on the other. 42. ABG is a triangle, right-angled at B, and BD is drawn perpendicular to the base, and is produced to E until EGB is a right angle ; prove that the square on BG is equal to the sum of the rectangles AD, DG and BD, DE. 43. Shew that the sum of the squares on two lines is greater than twice the rectangle contained by the lines. 44. In any triangle the sum of the squares on the straight lines, drawn from the angles to the middle points of the I20 EUCLID'S ELEMENTS. [Books I. & II. opposite sides, is equal to three-fourths of the sum of the squares on the sides of the triangle. 45. If any number of parallelograms be constructed having their sides of given length, shew that the sum of the squares on the diagonals of each will be the same. 46. ABCD is a right-angled parallelogram, and AB is double of BC ; on AB an equilateral triangle is constructed : shew that its area will be less than that of the parallelogram. 47. A point is taken within a triangle ABC, such that the angles BOG, COA, AOB are equal ; prove that the squares on BCj CA, AB are together equal to the rectangles contained by OB, 00 ; 00, OA ; OA, OB ; and twice the sum o/ the squares on OA, OB, 00. 48. In any quadrilateral the squares on the diagonals to- gether equal four times the sura of the squares on the lines joining the middle points of opposite sides. 49. If a straight line be divided into three parts, the square on the whole line is equal to the sum of the squares on the parts together with twice the rectangle contained by each two of the parts. 50. Given one side of a rectangle which is equal in area to a given square, find the other side. 61. AB, AC are the two equal sides of an isosceles triangle ; from B, BD is drawn perpendicular to AC, meeting it in I) ; shew that the square on BD is greater than the square on CD by twice the rectangle AD, CD. 62. Assuming that in any triangle ABC, the lines drawn from the angular points, to the middle points of the opposite sides meet in a point G, shew that three times the sum of the squares on AG, BG, CO is equal to the sum of the squares on BC, CA, AB. // BOOK IIL Postulate. h POINT is within, or without, a circle, according as it8 distance from the centre is less, or greater than, the radius of the circle. Def. I. A straight line, as PQ, drawn so as to cut a circle ABCDf is called a Secant. That such a line can only meet the circumference in two points may be shewn thus : Some point within the circle is the centre ; let this be 0. Join OA. Then (Ex. 1, i. 16) we can draw one, and only one, straight line from 0, to meet the straight line PQ, such that it shall be equal to OA. Let this lino be 00. Then A and C are the only points in PQ, which are on the circumference of the circle. fl B. If. 122 EUCLID'S ELEMENTS. , [Book III. Def. II. The portion ^OVf the secant p^^ intercepted by the circle, is called a Chord. Def. III. The two portions, into which a. chord divides the circumference, /as ABC and ADCj&re called Arcs. Def. IV. The two figures into which a chord divides the chrcle, as ABC and ADC, that is, the figures, of which the boun- daries are respectively the arc ABC and the chord AC, and the arc ADC and the chord AC, are called Segments of the circle. Def. V. The figure AOCD, whose boundaries are two radii and the arc intercepted by them, is called a Sector. Def. VI. A circle is said to be described about a rectilinear figure, when the circumference passes through each of the angular points of the figure*. And the figure is said to be inscribed in the circ?s. Book III.J FROPOSiriON I. 23 f Proposition I. Theorem. The lint, which bisects a chord of a cvrcle aJb right angles, nrnat contam the centre. Let ABC be the given ©. Let the st. line CE bisect the chord AB at rt. angles in JX. Then the centre of the © must lie in CE. For if not, let 0, a pt. out of CE, be the centre"; » and join 0^, OB, OB. Then, in as ODA, ODB, %* AD = BD, and DO is common, and OA = OB ; .-. z ODA = z ODB ; L a and .-. I ODB is a right z . L Def. 9 But z CDB is a right z , by construction ; ,•. z ODB — L CDB, which is impossible ; .*. is not the centre. Thus it may be shewn that no point, out of CE, can be the centre, and .*. the centre must lie in CE. Cor. // the chord CE be bisected in F, then F is the centre of thf n'rde. 124 EUCLID'S ELEMENTS. [Book III. Proposition II. TnEOREM. / j If any two 'points he talcen in the circuwfcrfMce of a circle, ji the straight line^ which joins them, must fall within the J J circle. Let .4 and B be any two pts. in the Oce of the © ABG. TJien must the st line AB fall within tM ® . Take any pt. D in the line AB. Find the centre of the © . HI. 1, Cor. Join OA, OD, OB. Then'.' lOAB = lOBA, La. and z ODB is greater than z OAB, I. 16. .*. I ODB is greater than z OB A ; and .'. OB is greater than OD. I. 10. /. the distance of D from is less than the radius of the © , and .'. JD lies within the ©. Post, And the same may be shewn of any other pt. in AB. .*, AB lies entirely within the ®. a B> »' Book m.] PROPOSITION III. 125 Proposition III. Theorem. If a straight line, drawn through the centre of a cirde, bisect jl a choi'd of the circle, which does not pass through the centre, it/ j micst cut it at right angles : and conversely f if it cut it at righv / angleSf it must bisect it. In the ® ABCf let the chord AB, which does not paas through the centre 0, be bisected in E by the diameter CD. Then must CD be A. to AB. Join OA, OB. Then in a s AEO, BEG, V AE=BE, and EG is common, and GA = GB, .: L OEA= L GEB. I.e. Hence GE is j. to AB, L Def. £f; that is, CD is ± to AB. Next let CD be jl to AB. Then must CD bisect AB. For *.* OA = GB, and GE is common, in the right-angled As AEG, BEG, .: AE^BE, I. E. Cor. p. 43. that is, CD bisects AB. q. e. d. Ex. 1. Shew that, if CD does not cut AB at right angles, it cannot bisect it. Ex. 2. A line, which bisects two parallel chords in a circle, is also perpendicular to them. Ex. 3. Through a given point within a circle, which is not the centre, draw a chord which sdinll be bisected in that point. 126 EUCLID'S ELEMENTS. [Book III Proposition IV. Theorem. If in a circle two chords^ which do not both pass through ihe emtn, cut one another , they do not bisect each other. Let the chords AB, CD, which do not both pass through the centre, cut one another, in the pt. E, in the © AGBD. Then AB, CD do not bisect each other. If one of them pass through the centre, it is plainly not bisected by the other, which does not pass through the centre. But if neither pass through the centre, let, if it be possible, AE=EB and CE=ED; find the centre 0, and join OE, Then *.• OE, passing through the centre, bisects AB, .-. z OEA is art. I. III. 3. And *.• OE, passing through the centre, bisects CD, .'. L OEC is a rt. z ; III. 3. /. I OEA = z OEO, which is impossible ; •% AB, CD do not bisect each other. q. e. d. Ex. 1. Shew that the locus of the points of bisection of all parallel chords of a circle is a straight line. Ex. 2. Shew that no parallelogram, except those which are rectangular, can be inscribed in a circle. Book ni.] PROPOSITIOM V. \'2.'j Proposition V. Theorem. y two circles cut one another, they cannot ha/ue the same centre. A. If it be possible, let be the common centre of the ®s ABC, ADC, which cut one another in the pts. A and C. Join OA, and draw OEF meeting the © s in j& and F. Then *.• is the centre of © ABC, .'.OE=OA; I.Def. 13. and *.* is the centre of © ADG, .:OF=OA; LDef. 13. /. 0E= OF, which is impossible ; •'. is not the common centre. Q. E. D. Ex. Two circles, whose centres are A and B, intersect in ; through C two chords DGE, FOG are drawn equally in- clined to AB and terminated by the circles : prove that DF and FG are equal. Note. Circles which have the same centre are called Con- centric 128 EUCLID'S ELEMENTS, IBook III. Note 1. On the Contact of Circles. Def. VII. Circles are said to touch each other, which meet b')t do not cut each other. One circle is said to touch another internally, wnen one point of the circumference of the former lies mi, and no point without, the circumference of the other. HetH'e for internal contact one circle must be smaller than the other. Two circles are said to touch externally, when one point of the circumference of the one lies on, and no point ivithin the circumference of the other. N.B. No restriction is placed by these definitions on the number of points of contact, and it is nor. till we reach Prop, iiii. that we prove thai there can be hut one point of coiVw^t. Book m.] PROPOSITION VI 129 Proposition VI. Theorem. If one ci/rcle touch another internally ^ they cannot ham the U samecenke* " lie* ADE touch © ABC internally, and let ^ be a point of contact Then some point E in the Oce ADE lies within © ABO, Def. 7. If it be possible, let be the common centre of the two s. Join OA, and draw GEO, nieetincr the^ Oces in E and C. Then •.• is the the centre of ABG, .:OA = OG; I. Def. 13. and •.* is the centre of © ADE, .'.OA = OE. I. Def. 13. Hence GE= GG, which is impossible ; •'• '0 is aot the common centre of the two ©s. I30 EUCLID'S ELEMENTS. [Book UI. W Proposition VII. Theorem. If from any point within a circle, which is not the centre, straight lines he drawn to the circumference, the greatest of these lines is that which passes through the centre. Let ABC be a © , of which is the centre. From P, any pt. within the © , draw the st. line PA, pass- " ing through and meeting the Oce in -4. Th^n must PA be greater than any other st. Une, drawn from P to the Oce. For let PB be any other st. line, drawn from P to meet the Oce in B, and join BO. Then '.• AO=BO, .'. ^P=sum of BO and OP. But the sum of BO and OP is greater than BP, 1. 20. and .'. AP is greater than BP. q. e. d. Ex. 1. If AP be produced to meet tlje circumference in D, shew that PD is less than any other straight line that can be drawn from P to the circumference. Ex. 2. Shew that PB continually decreases, as B passes from A to D. Ex. 3. Shew that two straight lines, but not three, that shall be equal, can be drawn from P to the circumference. Book III.] PROPOSITION VIII. t%\ Proposition VIII. Theorem. If from, any point without a circle straight lines he drawn to the d/rcumference, the least of these lines is that which, when pro- duced, passes through the centre^ and the greatest is thai vjhich passes through the centre. II Let ABC be a © , of which is the centre. From P any pt. outside the © , draw the st. line PA OG, meeting the Oce in J. and C. Then must PA be less, and PC greater, than any other st. tine drawn from P to the Oce. For let PB be any other st. line drawn from P to meet the Oce in ^, and join BO. Then •/ sum of PB and BO is greater than OP, I. 20. .'. sum of PB and J50 is greater than sum of AP and AG. But BO=AO; .'. PB is greater than AP. Again •.* PB is less than the sum of PO, OB, I. 20. .*. PB is less than the sum of PO, OC ; .•. PB is less than PC. q. e. d. Ex. 1. Shew that PB continually increases as B passes from J to O. Ex. 2. Shew that from P two straight lines, but not three, that shall be equal, can be drawn to the circumference. Note. From Props, vii. and viii. we deduce the following Corollary, which we shall use in the proof of Props, xi. and xiii. Cor. Tfa 'p oint be taken, within or without a circle^ o f all straight hnes drawn from it to t he circumference^ ^ hA prpjiipjjt i^ thai wJiAchmecfy thr ,-, ,c>i iiif> rcncr after passivq through the centra i 132 EUCLID'S ELEMENTS. [Book IIL Proposition IX. Theorem. If a point he taken within a circle, from which there fall more than two equal straight lines to the circumference, that pwni m tht centre of ike circle. Let be a pt. in the ABC from which more than two st lines OA, OB, OC, drawn to the Oce, are equal. Then must be the centre of the ©. Join AB, BC, and draw OD, OE ± to AB, BG Then •.* OA = OB, and OD is common, in the right-angled as AOD^ BOD, .-. AD=DB ; L s. Got. p. 4a .', the centre of the is in DO, III. 1. Similarly it may be shown that the centre of the © is EO ; «% is the centre of ti^ ^ D. Book III.] PROPOSITION X, 133 Proposition X. Theorem. nnot ha/ve Tuore bifthf wiihout coinciding erUvrehf. Two circles cannot hawe more than two points eonvnum to If It it be possible, let ABC and ADE be two ©s which have more than two pts. in common, as A, B, G» Join AB, BG. Then I* AB is a chord of each circle, .'. the centre of each circle lies in the straight line, which bisects AB at right angles ; III. 1. and •.* BC is a chord of each circle, .'. the centre of each" circle lies in the straight line, which bisects BO at right angles. III. 1. .*. the centre of each circle is the point, in which the two straight lines, which bisect AB and BC at right angles, meet. .'. the ©8 ABC, ABE have a common centre, which is impossible : III. 6 and 6. .'. two ©8 cannot have more than two pts. common to both. Q. E. D. Note. We here insert two Propositions, Eucl. iii. 25 and IV. 5, which are closely connected with Theorems 1. and x. of this hook. The learner should compare with this portion of the subject the note on Loci, p. 103. i34 Mt/CLWS ELEMENTS. [Book 111, / Proposition A. Problem. (Eucl. iii. 26.) An cure of a circle being given, to complete the circle of wMch %tisa part. Let ABC be the given arc. It it required to complete the © of which ABC is a part Take B, any pt. in arc ABGj and join AB, BC. From D and Ey the middle pts. of AB and BC, draw DO, EO, ± s to AB, BO, meeting in 0. Then •.* AB is to be a chord of the © , .'. centre of the ® lies in DO ; III. 1. and *.* BC is to be a chord of the ® , .*. centre of the © lies in EO. III. 1 Hence is the centre of the © of which ABC is an arc, and if a © be described, with centre and radius OA, this will be the & required. Q. B. F. Book m.] PROPOSITION B, 135 Proposition B. Problem. (Eucl. iv. 6.) To describe a circle about a given trianglx Let ABC be the given A . It is required to describe a © ahout the A. From D and E, the middle pts. of AB and AG, draw DO, BOf xa to ABf AC, and let them meet in 0. Then •.* AB is to be a chord of the ©, .'. centre of the © lies in DO. Ill, 1. And *.* AC is to be a chord of the ®, .'. centre of the © lies in EG, III. 1. Hence is the centre of the © which can be described a,bout the A , and if a © be described with centre and radius OA, this will be the © required. Q. E. F. Ex. 1. If BAC be a right angle, shew that will coincide with the middle point of BC. Ex. 2. If BAC be an obtuse angle, shew that will fall on the side of BC remote from A. 136 EUCLID'S ELEMENTS. [Bools III. Proposition XI. Theorem. If one circle touch another internally at any point, the emtre of the interior circle must lie in that radius of the other circle which passes through that point of contact. Let the © ADE touch the © ABC internally, and let A \m a pt. of contact. Find the centre of © ABC, and join OA. Then must the centre of © ADE lie in the radius OA. For if not, let P be the centre of © ADE. Join OP, and produce it to meet the Oces in D and B. Then •.* P is the centre of © ADE, and from are drawn to the Oce of ADE the st. lines OA, OD, of which OD passes through P, .-. OD is greater than OA. III. 8, Cor. But OA^OB; .'. OD is greater than OP, which is impossible. ,*, the centre of © ADE is not out of the radius OA. .'. it lies in OA. Q. E. D. \ Book m.] PROPOSITION XIL 137 Proposition XII. Theorem. Lj two circles touch one another externally at any point, the It straight line joining the centre of one with that point of contact 11 must when produced pass through the centre of the other. Let ® ABC touch © ADE externally at the pt J. Let be the centre of © ABC. Join OA, and produce it to E. Then must the centre of © ADE lie in AE. For if not, let P be the centre of ADE. Join OP meeting the Qsin B, D ; and join AP. Then v OB =0 A, and PD=AP, .'. OB and PD together=0^ and AP together ; .*. OP is not less than OA and AP together. But OP is less than OA and AP together, I. 20. which is impossible ; .'. the centre of ®.ADE cannot lie out of AE. Q. B. D. Ex. Three circles touch one another externally, whose centres are A, B, C. Shew that the difference between AB and AC is half as great as the difference between the diameters of the circles, whose centres are B and C, 138 EUCLID'S ELEMENTS. [Boofc m. Proposition XIII. Theorem. Ofvt cvrcle cannot touch another at more points than is nearer to the centre than ^JV Q. E. T) Ex. 1. Draw a chord of given length in a given circle, which shall be bisected by a given chord. Ex. 2. If two isosceles triangles be of equal altitude, and the sides of one be equal to the sides of the other, shew that their bases must be equal. Ex. 3. Any two chords of a circle, which cut a diameter in the same point and at equal angles, are equal to one another. Dep. IX. A straight line is said to he a Tangent to, or to touch, a circle, when it meets and, being produced, does not cut the circle. From this definition it follows that the tangent meets the circle in one point only, for if it met the circle in two points it would cut the circle, since the line joining two points in the circumference is, being produced, a secant. (III. 2.) Def. X. If from any point in a circle a line be drawn at right angles to the tangent at that point, the line is called a Normal to the circle at that point. Def. XI. A rectilinear figure is said to be described about a drde, when each side of the figure touches the circle. And the circle is said to be inscribed in the figure. Book m.] PROPOSITION XVL i43 Proposition XVI. Theorem. The, straight line drawn at right angles to the dia/meter vf a eircUf from the extremity of it, is a tangent to the circle. \ Let ABC be a © , of which the centre is 0, and the diameter A OB Through B draw DE at right angles to AOB. I. 1 1 Then must DE he a tangent to the ®. Take any point P in DE, and join OP^ Then, •. • z OBP is a right angle, /. I OPB is less than a right angle, I. 17 and .-. OP is greater than OB. I. 19 Hence P is a point without the © ABC. Post. In the same way it may be shewn that every point in DE, or DE produced in either direction, except the point B, lies without the © ; Def. 9. Q. S. D. *. DE is a tangent to the . / 144 J£VCUD'S ELEMENTS. [Book ItL Proposition XVII. Problem. To draw a straight line from a given point, either without or ON iA^ circumference, which shall touch a given circle. Let A be the given pt., toithout the © BCD, I Take the centre of © BGD, and join OA. Bisect OA in E, and with centre E and radius EO describe » ® ABOD, cutting the given © in ^ and D. Join AB, AD. These are tangents to the ® BCD Join BO, BE. Then •.• OE=BE, .-. z OBE= l BOE ; I. a. .-. z ^^5 = twice z OBE ; I. 32. and ••• AE=BE, .'. z ABE= z J5^J? ; I. a. .-. z OEB = twice z ^£^ ; I. 32. .*. sum ofzs AEB, OEB = twice sum ofzs OBE, ABE, that is, two right angles = twice z OB A ; .'. z 05J. is a right angle, and .-. AB is a tangent to the © BCD. 111. 16. Similarly it may be shewn that AD is a tangent to © BCD. Next, let the given pt. be on the Qce of the ®, as B. Then, if BA be drawn ± to the radius OB, BA is a tangent to the ® at B. III. 16, Q. E. D. Ex. 1. Shew that the two tangents, drawn from a point with- out the circumference to a circle, are equal. Ex. 2. If a quadrilateral ABCD be described about a circle, shew that the sum of AB and CD is equal to the sum of AC and BD Book lll.j PROPOSITION XVIIl HS Proposition XVI II. Theorem. If Oj straight line touch a circle^ the straight line dn-avm from t» the centre to the point of contact must he jp&rpendiculm to thjj Une touching the circle. f / Let tie st. line DE touch the © ABC in the pt (X Find the centre, and join OC. Then must 00 he ± to DE. For if it be not, draw OBFa. to DE, meeting the Oce in B. Then •/ z OFO is a rt. angle, *. L OCF is less than a rt. angle, I. 17. and .-. OG is greater than OF. I. 19. Bnt 00= OB, /. OB is greater than OF, which is impossible ; .*. OF is not ± to DE, and in the same way it may be shewn that no other line drawn from 0, but OC, is i. to DE ; .'. OG is ± to DE. Q. E. D. Ex. If two straight lines intersect, the centres of all circles touched by both lines lie in two lines at right angles to each other. Note. Prop, xviii. might be stated thus : — All radii of a circle are normals to the circle at the points where they meet the cwcv/mference. 146 EUCLID'S ELEMENTS. [Book in Proposition XIX. Theorem. If a straight liv/i touch a circle, and from the point of con tact a straight line be dravm at right angles to tiie touching lincy Ihi centre of the circle imist be in that line. Let the st. line DE touch the © ABC at the pt. (7, and from (7 let CA be drawn ± to DE. Then must the centre of the © be in CA. For if not, let F be the centre, and join FC. Then •.• DCE touches the ©, and FCis drawn from centre to pt of contact, .-. z FCE is a rt. angle. III. 18. But z ACE is a rt. angle. .*. I FCE = I ACE, which is impossiWa In the same way it may be shewn that no pt. out of CA can be the centre of the © ; .'. the centre of the © lies in CA. Q. B. D. Ex, Two concentric circles being described, if a chord of the greater touch the less, the parts of the chord, intercepted between the two circles, are equal. Note. Prop. xix. might be stated thus -.— Every normal to a circle passes through the cevtre. Book III.] PROPOSITION XX. ^An Proposition XX. Theorem. y Tht angle at the centre of a circle is double of the angle at the ji circvmference. subtended by the same mrc. Let ABC be a © , the centre, B(7any arc, A any pt. in the Oce. Thm must l BOG = twice l BAG. First, suppose to be in one of the lines containing the lBAG, Then •.• OA = OC, :.lOGA = lOAG\ ► «mi of z s OGA, GAG = twice i GAG. Batz 500= sum of z s GCA, GAG, .'. L BOG = twice z GAG. that is, z BOG ^ twice z BAG. L A. 1.32. I4S EUCLID'S ELEMENTS. [Book III. Next, suppose to be within (fig 1), or without (fig. 2) the lBAG. Join AOf and produce it to meet the Oce in D. Then, as in the first case, L COD = twice z GAD, and lBOD= twice z BAD ; .-., fig. 1, sum of z s GOD, BOD = twice sum of z s CAD, BAD, that is, z BOG = twice z BAG. And, fig. 2, difference of z s GOD, BOD = twice diflference of z s GAD, BAD, that is, z BOG = twice z ^^0. Q. E. D. Er- 1. The centre of the circle GBED is on the circum- ference of ABD. If from any point A the lines ABC and A ED be drawn to cut the circles, the chord BB is parallel to CD. Ex. 2. From any point in a straight line, touching a ciTfde, a straight line is drawn through the centre, and is terminated by the circumference ; the angle between these two straight lines is bisected by a straight line, which intersects the straight line joining their extremities. Shew that the angle between the last two lines is half a right angle. Book III.] NOTE II, 149 Note 2. On Flat and Bejlex Angles. We have already explained (Note 3, Book I., p. 28) how Euclid's definition of an angle may be extended with advan- tage, so as to include the conception of an angle equal to two right angles : and we now proceed to shew how the Definition given in that Note may be extended, so as to embrace angles greater than two right angles. xe Let WQ be a straight line, and QE its continuation. Then, by the Definition, the angle made by WQ and QE, which we propose to call a Flat Angle, is equal to two right angles. Now suppose QP to be a straight line, which revolves about the fixed point Q, and which at first coincides with QE. When QP, revolving from right to left, coincides with QW, it has described an angle equal to two right angles. When QP has continued its revolution, so as to come into the position indicated in the diagram, it has described an angle EQP, indicated by the dotted line, greater than two right angles, and this we call a Reflex Angle. To assist the learner, we shall mark these angles with dotted lines in the diagrams. Admitting the existence of angles, equal to and greater than two right angles, the Proposition last proved may be extended, as we now proceed to shew. »5o EUCLID'S ELEMENTS. [Book m. Proposition C. Theorem. j The. angle, not less than two right angles, at the centre of a j circle is double of the angle at the circumference, subtended by tiie same arc. Fig.1. Fig.S. In the © ACBD, let the angles ^ OP (not less than two right angles) at the centre, and ADB at the circumference, be subtended by the same arc ACB. Then must z AOB=tvdce l ADB. Join DO, and produce it to meet the arc ACB in G. Then •.• z ^00= twice z ADO, III. 20. and z BOC= twice z BDO, III. 20. /. 8um of z s AOC, 50(7= twice sum of z s ADO, BDO, that is, z ^05= twice z ADB. Q. E. D. NoTB. In fig. 1, z AOB is drawn a flat angle, idJuJSg. 2, 1 AOB is drajpi^fc-Teflex-faigle. i j/ Def. XII. The angle in a segment is the angle contained by Iftwo straight lines drawn from any point in the arc to the ex- Irtn^mities of the chord. Book m.] PROPOSITION XXL 151 Proposition XXI. Theorem. Th^ angles in the same segment of a circle a/re equal to ont tmoihet. I1»L Let BA Cj BDC be angles in the same segment BADC, Then must i BAC= /L BDC. First, when segment BADC is greater than a semicircle, From 0, the centre, draw OB, OG, (Fig. 1.) Then, •.• z £00=twice z BACy III. 20. and z J50C= twice z BDC, III. 20. .:^BAC= I BDC. Kext, when segment BADC is less than a semicircle, Let E be the pt. of intersection of AC, DB. (Fig. 2.) Then '.'_lABE= l DCE, by the first case, and z BE A = z 0^1), I. ] 5. .-. z EAB= L EDC, 1. 32. that is, /.BAC= ^ BDC. q. e. d. Ex. 1. Shew that, by assuming the possibility of an angle being greater than two right angles, both the cases of this proposition may be included in one. Ex. 2. AB, AC are chords of a circle, D, E the middle points of their arcs. If DE be joined, shew that it will out off equal parts from AB, AC. Ex. 3. If two straight lines, whose extremities are in the circumference of a circle, cut one another, the triangles formed by joining their extremities are equiangular to each other 15* EUCLID'S ELEMENTS. [Book IIL Proposition XXII. Theorem. TIm opposite angles of any quadrilateral figure, inscribed in • drdsj cure together equal to two right angles. Let ABCD be a quadrilateral fig. inscribed in a ®. Then must each pair of its opposite is be together equ^ to two rt. L s. Draw the diagonals AO, BD. Then *.• z ADB= i ACS, in the same segment, III. 21. and z BDC= i BAG, in the same segment ; III. 21. .-. sum of z s ADB, BDG=Bxim of z s ACB, BAC ; that is, z ADC =8um of z s ACB, BAG, Add to each z ABG. Then z 8 ADC, ABG together =sum ofzs ACB, BAG, ABG; and .'.13 ADC, ABG together = two right z s. I. 32. Similarly, it may be shewn, that z s BAD, BCD together = two right z s. Q. E. D. Note. — Another method of proving this proposition is given on page 177. Book m.] PROPOSITION XXII. 153 Ex 1. If one side of a quadrilateral figure inscribed in a circle oe produced, the exterior angle is equal to the opposite •ingle of *-.he quadrilateral. Ex. 2. Jf the sides AB, DG of a quadrilateral iasor^bed in a circle be produced to meet in E, then the triangles EBCf BAD will be equiangular. Ex. 3. Shew that a circle cannot be described about a rhombus. Ex. 4. The lines, bisecting any angle of a quadrilateral figure inscribed in a circle and the opposite exterior angle, meet in the circumference of the circle. Ex. 5. AB, a chord of a circle, is the base of an isosceles triangle, whose vertex G is without the circle, and whose equal sides meet the circle in D, E : shew that GD is equal to GE. Ex. 6. If in any quadrilateral the opposite angles be to- gether equal to two right angles, a circle may be described about that quadrilateral. Propositions xxiii. and xxiv., not being required in the method adopted for proving the subsequent Propositions in this book, are removed to the Appendix. Proposition xxv. has been already proved. Note 3. O/* the Method of Superposition^ as applied to Gircles. In Props, xxvr. xxvii. xxviii. xxrx. we prove certain relations existing between chords, arcs, and angles in equal circles. As we shall employ the ISIethod of Superposition, we must state the principles which render this methv^d appli- cable, MS a test of equality, in the case of figures with circalar boundaries. 154 EUCLID'S ELEMENTS. [Book IIV Def. XIII. Equal circles are those, of which the radii a/ri equal. a o' For suppose ABG^ A'B'C to be circles, of which the radii are equal. Then if © A'B'd be applied to © JBC, so that (7, the centre of A'B'C, coincides with 0, the centre of ABG^ it is evident that any particular point A' in the Qce of the former must coincide with some point A in Qce of the latter, because of the equality of the radii O'J.' and OA. Hence Qce A'B C must coincide with Qce ABC, thatis, © J['jB'0'=©^5C. Further, when we have applied the circle A'B'C to the circle ABC, so that the centres coincide, we may imagine ABC to remain fixed, while A'B'C revolves round the common centre. Hence we may suppose any particular point B' in the circumference of A'B'C to be made to coincide with any par- ticular point B in the circumference of ABC. Again, any radius CA' of the circle A'B'C may be made to coincide with any radius OA of the circle ABC. Also, if A'B' and AB be equal arcs, they may be made to coincide. Again, every diameter of a circle divides the circle into equal segments. For let AOB be a diameter of the circle ACBD, of which is the centre. Suppose the segment ACB to be ap- plied to the segment ADB, so as to keep AB a common boundary : then the arc ACB must coincide with the arc ADB, because every point in each is equally distant from 0. Book III.] PROPOSITION^ XXVL 155 / ' , ^ Proposition XXVI. Theorem. In equal circles, the arcs, which subtend equal a/ngl^, whether they be ai the centres or at the circumferences, must be equal. Let ABC, DEF be equal circles, and let z s BGC, EHF at their centres, and z s BAG, EDF at their Qces, be equal. Thert must arc BKG=arc ELF. •For, if © ABG be applied to © BEF, so that G coincides with H, and GB falls on ETB, then, *.* GB=nE, .'. B will coincide with E. And ••• z BGO= z EHF, .\ GG will fall on HF ; and •/ GG=HF, .: G will coincide with F. Then •.* B coincides with E and with F, .*. arc BKC will coincide with and be equal to arc ELF. Q. £. D. Cor. Sector BGCK is equal to sector EHFL. Note. This and the three following Propositions are, and will hereafter be assumed to be, true for the same e\.rcle as well as for equal circles. £56' EUCLID'S ELEMENTS. [Book III. ^dii^^"^ Proposition XXVII. Theorem. In equal circles, the angles, which are subtended by equal arcs, whether they are at the centres or at the circumferences, must he equal. I Let ABC, DEF be equal circles, and let z s BGG, EHF at their centres, and z s BAG, EDF at their Qces, be subtended by equal arcs BKG, ELF. Then must z BGG= z EHF, and lBAG==i EDF. For, if © ABG be appUed to © DEF, so that G coincides with H, and GB falls on HE, then •.• GB=HE, .'. B will coincide with E ; and *.• arc BKG=a,rc ELF, .*. G will coincide with F. Hence, GG will coincide with HF. Then •.• BG coincides with EH, and GG with HF, .*. z BGG will coincide with and be equal to i EHF. Again, '.• z jB^O=half of z BGG, III. 20. and z ^Di^=half of z EHF, III. 20. .-. iBAG= lEDF. I. Ax. 7. Q. E. D. Ex. 1. If, in a circle, AB, GD be two arcs of given magni- tude, and AG, BD be joined to mee\i in E, shew that the angle AEB is invariable. . Ex. 2. The straight lines joining the extremities of the chords of two equal arcs of the same circle, towards the same parts, are parallel to each other. Ex. 3. If two equal chords, in a given circle, cut one an- other, the segments of the one shall be equal to the segments of the other, each to each. Book III.] PROPOSITION XXVIII. 157 r « Proposition XXVIII. Theorem. In equal circles, the arcs, which are subtended hy equal ^chords, must he equals the greater to the greater, and the less to the less. Let ABCf DEF be equal circles, and BC, EF equal chords, subtending the major arcs BAG, EDF, and the minor arcs BGG, EHF. Then must arc BAG = arc EDF, and arc BGG = arc EHF. Take the centres K, L, and join KB, KG, LE, LF. Then V KB=LE, and KG=LF, and BG=-EF, .'. L BKG = z ELF. I. G Hence, if © ABG be applied to © DEF, so that K coincides with L, and KB falls on LE, then ••• z BKG = z ELF, .«. KG will fall on LF ; and •.* KG = LF, .: G will coincide with F. Then *.• B coincides with E, and G with F, .'. arc BAG will coincide with and be equal to arc EDF, &ndsivcBGG EHF. Q. B. D. F.x. 1. If, in a circle ABGD, the arc AB be equal to the arc DG, AD must be parallel to BG. Ex. 2. If a straight line, drawn from A the middle point of an arc BG, touch the circle, shew that it is parallel to the chord BG. Ex. 3. If two chords of a circle intersect at right anglea, the portions of the circumference taken alternately are together equal to half the circumference. «S8 EUCLID'S ELEMENTS. [Book III. Proposition XXIX. Theorem. I tn equal circles, the chords, which subtend equal a/rcs, must he equal. Let ABC, DEF be equal circles, and let BC, EF be chords subtending the equal arcs BGG, EHF. Then must chord BC — chord EF. Take the centres K, L. Then, if © ABC be applied to © DEF, so that K coincides with L, and B with E, and arc BOC falls on arc EHF, •/ arc BGG=a.TC EHF, .'. C will coincide with F. Then *.* B coincides with E and G with F, ,\ chord BG must coincide with and be equal to chord EF. Q. B. D. Ex. 1. The two straight lines in a circle, which join the extremities of two parallel chords, are equal to one another. Ex, 2. If three equal chords of a circle, cut one another in the same point, within the circle, that point is the centre. BorK lU.] ArOT£ 4. «5^ ( Note ^ On the Syrmrutriccd properties of the Circle with regard to its diameter. The brief remarks on Symmetry in np. 107, 108 may now be extended in the following way : A figure is said to be symmetrical with regard to a line, when every perpendicular to the line meets the figure at points which are equidistant from the line. Hence a Circle is Symmetrical with regard to its Diameter, because the diameter bisects every chord, to which it is per- pendiculaE. Further, suppose AB to be a diameter of the circle ACBD, of which is the centre, and CD to be a chord perpendicular to AB. Then, if lines be drawn as in the diagram, we know that AB bisects (1.) The chord CD, III. 1. (2.) The arcs CAD and CBDy III. 26. (3.) The angles CAD, COD, CBD, and the reflex angle DOG, I. 4 Also, chord CB = chord DB, I. 4. and chord ^C= chord AD, I. 4. These Symmetrical relations should be carefully observed, because they are often suggestive of methods for the solution of problems. i6o EUCLID'S ELEMENTS. [Book lil Proposition XXX. Problem. To bisect a given cure. D a Let ABC be the given arc It is required to bisect the a/rc ABC, Join AGf and bisect the chord AC in D, 1. 10. From D draw DB± to AC. L 11. Then will the arc ABC he bisected in B, Join BA, BC. Then, in a ADB, CDS, V AD=CDj and DB is common, and z ADB = z CD5, .-. BA=BC. I. 4. But, in the same circle, the arcs, which are subtended by equal chords, are equal, the greater to the greater and the less to the less ; III. 28. and *.* BD, if produced, is a diameter, /, each of the arcs BAy BC, is less than a semicircle, and .*. arc jB^=arc BC. Thus the arc ABC is bisected in B. Q. E. F. Ex. If, from any point in the diameter of a semicircle, there be drawn two straight lines to the circumference, one to the bisection of the circumference, and the other at right angles to the diameter, the squares on these two linea are together double of the square on the radius Book m.] PROPOSITION XXXI. i6i , Proposition XXXI. Theorem. mj In a circle, the angle in a semicircle is a right angle ; cmd ft (he angle in a segment greater than a semicircle is less than a / right angle ; and the angle in a segment less than a semicircle f is greater than a right angle. Let ABC be a © , its centre, and BG a. diameter. Draw AC, dividing the © into the segments ABC, ADC. Join BA, AD, DC, AO. Then must the l in the semicircle BAC be a rt. i , and l in segment ABC, greater than a semicircle, less than art. i , and i in segment ADC, less than a semicircle, greater than art. L . First, •.• BO=AO, .: i BAO= l ABO ; I. a. . •. z CO A = twice a BAO ; I. 32. and •.• CO=AO, .'. l CA0= l ACO ; I. a. .-. z BOA = twice i CAO ; I. 32. .-. sum of z s COA, BOA=twice sum of z s BAO, CAO, that is, two right angles = twice z BAC. "^ .: L BAC is a right angle. Next, v7 BAC is a rt. z , .*. z ABC is less than a rt. z . I. 17. Lastly, *.• sum of z s ABC, ADC=two rt. z s, IIL 22. and z ABC is less than a rt. z , .'. z ADC is greater than a rt. z . Q. e. d. Note. — For a simpler proof see page 178. 12 i62 EUCLID'S ELEMENTS. [Book HI. Ex. 1. If a circle be described on the radius of another circle as diameter, any straight line, drawn from the point, where they meet, to the outer circumference, is bisected by the in- terior one. Ex. 2. If a straight line be drawn to touch a circle, and be parallel to a chord, the point of contact will be the middFe point of the arc cut off by the chord. Ex. 3. If, from any point without a circle, lines be drawn touching it, the angles contained by the tangents is double of the angle contained by the line joining the points of contact, and the diameter drawn through one of them. Ex. 4. The vertical angle of any oblique-angled triangle inscribed in a circle is greater or less than a right angle, by the angle contained by the base and the diameter drawn from the extremity of the base. Ex. 5. If, from the extremities of any diameter of a given circle, perpendiculars be drawn to any chord of the circle that is not parallel to the diameter, the less perpendicular shall be equal to that segment of the greater, which is contained between the circumference and the chord. Ex. 6. If two circles cut one another, and from either point of intersection diameters be drawn, the extremities of these diameters and the other point of intersection lie in the same straight line. Ex. 7. Draw a straight line cutting two concentric circles, so that the part of it which is intercepted by the circumference of the greater may be twice the part intercepted by the circum- ference of the less. Ex. 8. Describe a square equal to the difference of two given squares. Ex. 9. If from the point in which a number of circles touch each other, a straight line be drawn cutting all the circles, shew that the lines, which join the points of intersection in each circle with its centre, will all be parallel Book III.] PROPOSITION XXXII, 163 Proposition XXXII. Theorem. If a straight line touch a circle, and from the point of contact a straight line he drawn cvMing the circle, the angles made by this line with the line touching the circle must he equal to the angles, which a/re in the alternate segments of the ci/rde. Let the st. line AB touch the © CDEF in F. Draw the chord FD, dividing the ©into segments FCD, FEL. Tlien must l DFB= l in segment FCD, and L DFA = /. in segment FED.* From F draw the chord FCjl to AB. Then FC is a diameter of the . 111. 1£. Take any pt. E in the arc FED, and join FE, ED, DC. Then •.* FDO is a semicircle, .'. z FDC is a rt. z ; III. 31. .-. sum of z s FCD, CFD=a rt. z . I. 3P,. Also, sum of z s DFB, CFD=3. rt. z . .-. sum of z s DFB, CFD =sum of z s FCD, CFD, and .-. z DFB= z FCD, that is, z DFB= z in segment FCD. Again, *.* CDEF is a quadrilateral fig. inscribed in a , .-. sum of z s FED, FCD = two rt. z s. J II. 22. Also, sum of z s DFA, DFB=two rt. z s. I. 13. /. sum of z s DFA, DFB=^snm of z s FED, FCD ; and z DFB has been proved = z FCD ; .'. I DFA= I FED, that is, z DFA = z in segment FED. Q. E. D. Ex. The chord joining the points of contact of parallel tan- gents i? a dinnioUT. if>4 EUCLID'S ELEMENTS. [3oo: Proposition XXXIII. Troblem. On a given straight line to describe a segment of a Cf/rcle capable of containing an angle equal to a given angle. Let AB he the given st. line, and C the given z . It is required to describe on AB a segment of a ® which shall contain an i = a G. At pt. A in St. line AB make z BAD= l C. I. 23. Draw AE± to AD, and bisect AB in F. From F draw FGi. to AB, meeting AE in G. ThenmABAGF,BGF', V AF=BF, and FG is common, and z AFG= z BFG ; .-. GA = GB. I. 4. With Gf as centre and GA as radius describe a © ABH. Then will AHB be the segment reqd. For *.* AD is±to AE, a line passing through the centre, .-. AD is a tangent to the © ABH. III. 16. And *.* the chord AB is drawn from the pt. of contact A, .'. L BAD= L in segment AHB, III. 32. that is, the segment AHB contains an z = z C, and it is described on AB, as was reqd. Q. E. F. Ex. 1. Two circles intersect in A, and through A is drawn a straight line meeting the circles again in P, Q. Prove that the angle betw^n the tangents at P and Q is equal to the angle between the tangents at A. Ex. 2. From two given points on the same side of a straight line, given in position, draw two straight lines which shall con- tain a given angle, and be tortninated in the given liue. Bo-.klll.J PROiJ^IlIONXXXiy. l6c \ Proposition XXXIV. Problem. To cut off a segment from a given circkf cajpahU of eojir ining an angle equal to a given angle. Let ABC be the given © , and D the given z . It is required to cut off from © ABC a segment capable Oj containing an i = l D. Draw the st. line EBF to touch the circle at B. At B make z FBC = z D. Then *.• the chord BC is drawn from the pt. of contact By .'. I FBC = z in segment BAC, III. 32. that is, the segment BAC contains an z = iD\ and .*. a segment has been cut off from the © , as was reqd. Q. B. F. Ex. 1. If two circles.touch internally at a point, any straight line passing through the point will divide the circles into seg- ments, capable of containing equal angles. Ex. 2. Given a side of a triangle, its vertical angle, and the radius of the circumscribing circle : construct the triangle. Ex. 3. Given the base, vertical angle, and the perpendicular from the extremity of the base on the opposite side : construct the triangle. 16:- EUCLID'S ELEMENTS. fBook 111. Proposition XXXV Theorem. 11 If two chords in a circle cut one another, the rectangle con- I mUned by the segments of one of them, is equal to the rectangle j f contained by the segments of the other.^ i Let the chords AC, BD m the © ABOD intersect in the pt. P. Then must rect. AP, PC=rect. BP, PD. From 0, the centre, draw OM, ON ±& to AC, BD, and join OA, OB, OP. Then '.* ^0 is divided equally in M"and unequally in P, .*. rect. AP, PC with sq. on M'P=sq. on AM. II. 5. Adding to each the sq. on MO, rect. AP, PC with sqq. on MP, ilfO=sqq. on AM, MO ; .-. rect. AP, PC with sq. on 0P= sq. on OA. I. 47. In the same way it may be shewn that rect. BP, PD with sq. on OP =sq. on OB. Then *.• sq. on OA=sq. on OB, /. rect. AP, PC with sq. on OP = rect. BP, PD with sq. on OP; .'. rect. AP. PO=rect. BP, PD. q. e. d. Ex. 1. A and B are fixed points, and two circles are described passing through them ; PCQ, P CQ' are chords of these circles intersecting in 0, a point in AB ; shew that the rectangle GP, CQ is equal to the rectangle OP', CQ'. Ex. 2. If through any point in the common chord of two circles, which intersect one another, there be drawn any two other chords, one in each circle, their four extremities shall all lie in the circumference of a circle. Book ni.] PROPOSITION XXXVI. i67 Proposition XXXVI. Theorem. //, frmn, any point vnthout a circle, two straight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle cant dived hi the whole line which cuts the circkf and the part of it without the drcUj must be et^aal to the square on the liiie which touches it. Let D be any pt. without the © ABGy and let the st. lines DBA, DC be drawn to cut and touch the ® . Then must rect. AD, DB=sq. on DC. From 0, the centre, draw DM bisecting AB in M, and join OB, DC, OD. Then •.* AB is bisected in M and produced to D, .*. rect. AD, DB with sq. on MB=sq. on MD. II. 6. Adding to each the sq. on MO, rect. AD, DB with sqq. on MB, MO =sqq. on MD, MO. Now the angles at M and C are rt. z s ; III. 3 and 18. .'. rect. AD, DB with sq. on OB=sq. on OD ; ,\ rect. AD, DB with sq. on 05= sqq. on 00, DC. I. 47. And sq. on 0J5=sq. on OG ; .'. rect. AD, DB=sq. on DC. Q. e. d. Ex. 1. Two circles intersect in A and B ; shew that AB produced bisects their common tangent. Ex, 2. If the circle, inscribed in a triangle ABC, touch J5Cin D, the circles described about A BD. A CD will touch each other i68 EUCLID'S ELEMENTS. [Book III. Proposition XXXVII. Theorem. Jf, from a point without a circle, there be drawn two straight lines, one of which cuts the circle, and the other meets it ; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square on the line which msets itj the line u'hichjmeBta 'iiLuat touclh tlis circle. Let ^ be a pt. without the © BCD, of which is the centre. From A let two st. lines AGD, AB be drawn, of which A CD cuts the © and AB meets it. Then if red. DA, AC=sq. on AB, AB must touch the ©. Draw AE touching the © in E, and join OB, OA, OE. Then *.' ACD cuts the ©, and AE touches it, .-. rect. DA, AC=&(\. on AE. III. 36. But rect. DA, ^C=sq. on AB ; Hyp. .'. sq. on AB=&(\. on AE \ .-. AB=AE. Then in the A s OAB, OAE, V OB=OE, and OA is common, and AB=AE, .-. z ABO = z AEO. But z AEO is a rt. z ; .'. z ABO is a rt. z . Now BO, if produced, is a diameter of the © ; .*. AB touches the ®. I.e. III. 18. III. 16. Q. E. D. Ex. If two circles cut each other, and from any point in the straight line produced, which joins their intersections, two tangents be drawn, one to each circle, they shall be equal. Book in.\ MISCELLANEOUS EXERCISES. 169 Miscellaneous Exercises on Booh III. 1. The segments, into which a circle is cut by any straight Hue, contain angles, whose dift'erence is equal to the inclination to each other of the straight lines touching the circle at tlie ex- tremities of the straight line which divides the circle. 2. If from the point in which a number of circles touch each other, a straight line be drawn cutting all the circles, shew that the lines which join the points of intersection in each circle "with its centre will be all parallel. 3. From a point ^ in a circle, QN is drawn perpendicular to a chord PP\ and QM perpendicular to the tangent at P : shew that the triangles NQP', QPM are equiangular. 4. If a circle be described round the triangle ABG, and a straight line be drawn bisecting the angle BAG and cutting the circle in D, shew that the angle DGB will equal half the angle BAG. 5. One angle of a quadrilateral figure inscribed in a circle is a right angle, and from the centre of the circle perpendiculars are drawn to the sides, shew that the sum of their squares is equal to twice the square of the radius. ~ 6. AB is the diameter of a semicircle, D and E any two points on its circumference. Shew that if the chords joining A and B with D and E, either way, intersect in F and G, the tangents at D and E meet in the middle point of the line FG, and that FG produced is at right angles to AB. 7. If a straight line in a circle not passing through the centre be bisected by another and this by a third and so on, prove that the points of bisection continually approach the centre of the circle. 8. If a circle be described passing through the opposite angles of a parallelogram, and cutting the four sides, and the points of intei-section be joined so as to form a hexagon, the straight lines thus drawn shall be parallel to each other. 9. If two circles touch each other externally and any third circle touch both, prove that the difference of the distances of 170 EUCLID'S ELEMENTS. [Book IIL the centre of the third circle from the centres of the other two is invariable. 10. Draw two concentric circles, such that those chords oi the outer circle, which touch the inner, may equal its diameter. 11. If the sides of a quadrilateral inscribed in a circle be bisected and the middle points of adjacent sides joined, the circles described about the triangles thus formed are all equal and all touch the original circle. 12. Draw a tangent to a circle which shall be parallel to a given finite straight line. 13. Describe a circle, which shall have a given radius, and its centre in a given straight line, and shall also touch anothei straight line, inclined at a given angle to the former. 14. Find a point in the diameter produced of a given circle, from which, if a tangent be drawn to the circle, it shall be equal to a given straight line. 15. Two equal circles intersect in the points A^ J5, and through B a straight line GBM is drawn cutting them again in 0, M. Shew that if with centre G and radius BM a circle be described, it will cut the circle ABG in a point L such that arc ^L=arc AB. Shew also that LB is the tangent at B. 16. AB is any chord and AC Vk tangent to a circle at A ; ODE a line cutting the circle in £> and E and parallel to AB. Shew that the triangle AGD is equiangular to the triangle EAB. 17. Two equal circles cut one another in the points J., B ; BG is a chord equal to AB ; shew that ^(7 is a tangent to the other circle. 18. In any two circles, which cut one^ another, the straight line joining the extremities of any two parallel radii cuts the line joining the centres in the same point. 19. J., B are two points ; with centre B describe a circle, svicb that its tangent from A shall be equal to a given line. Book III.] MISCELLANEOUS EXERCISES. 171 20. If perpendiculars be dropped from the angular points of a triangle on the opposite sides, shew that the sum of the squares on the sides of the triangle is equal to twice the sum of the rectangles, contained by the perpendiculars and that part of each intercepted between the angles of the triangles and the point of intersection of the perpendiculars. 21. When two circles intersect, their common chord bisects their common tangent. 22. Two circles intersect in A and B. Two points G and D are taken on one of the circles ; CA, CB meet the other circle in E, F, and DA, DB meet it m G, H : shew that FG is parallel to EH, and FH to FG. 23. A and B are fixed points, and two circles are described passing through them ; CP, CP' are drawn from a point C on AB produced; to touch the circles in P, P' ; shew that GP=CP\ 24. From each angular point of a triangle a perpendicular is let fall upon the opposite side ; prove that the rectangles con- tained by the segments, into which each perpendicular is divided by the point of mtersection of the three, are equal to each other. 26. If from a point without a circle two equal straight lines be drawn to the circumference and produced, shew that they will be at the same distance from the centre. 26. Let 0, 0' be the centres of two circles which cut each other in A, A'. Let B, B' be two points, taken one on each circumference. Let 0, C be the centres of the circles BAB', BA'B'. Then prove that the angle CBO is equal to the angle OA'a. 27. The common chord of two circles is produced to any point P , PA touches one of the circles in A ; PBG is any chord of the other : shew that the circle which passes through A, B, G touches the circle to which PA is a tangent. 28. Given the base of a triangle, the vertical angle, and the length of the line drawn from the vertex to the middle point of the base : construct the triangle. 172 EUCLID'ii ELEMENTS. [Book IIL 29. II a circle be described about the trian' J. and cutting the circle in D, shew that the angle TJGB will be equal to half the angle BAG. 30. If the line AD bisect the angle A in the triangle ABC, and BD be drawn without the triangle making an angle with 5(7 equal to half the angle BAQ^ shew that a circle may be described about ABCD. 31. Two equal circles intersect in A, B : PQT perpendicular to AB meets it in Tand the circles in P, Q. AP, BQ meet in R ; AQ, BP in S ; prove that the angle UTS is bisected by TP. 32. If the angle, contained by any side of a quadrilateral and the adjacent side produced, be equal to the opposite angle of the quadrilateral, prove that any side of the quadrilateral will subtend equal angles at the opposite angles of the quadrilateral. 33. If DE be drawn parallel to the base .BC of a triangle ABGj prove tliat the circles described about the triangles ABC and ADE have a common tangent at A. 34. Describe a square equal to the difference of two given squares. 35. If tangents be drawn to a circle from any point without it, and a third line be drawn between the point and the centre of the circle, touching the circle, the perimeter of the triangle formed by the three tangents will be the same for all positions of the third point of contact. 36. If on the sides of any triangle as chords, circles be de- scribed, of which the segments external to the triangle contain angles respectively equal to the angles of a given triangle, those circles will intersect in a point. 37. Prove that if ABC be a triangle inscribed in a circle, such that BA = BC, and A A' be drawn parallel to BC, meeting the circle again in A', and A'B be joined cutting JIO in E, BA touches the circle described about the triangle AEA'. 38. Describe a circle, cutting the sides of a given square, so that its circumference may be divided at the points of inter- section into eight equal arcs. Book III.] MISCELLANEOUS EXERCISES. 173 39. A is the extremity of the diameter of a circle, any point in the diameter. The chord which is bisected at sub- tends a greater or less angle at A than any other chord through 0, according as and A are on the same or opposite sides of the centre. 40. Shew that the square on the tangent drawn from any point in the outer of two concentric circles to the inner equals the difference of the squares on the tangents, drawn from any point, without both circles, to the circles. 41. If from a point without a circle, two tangents PT, PT', at right angles to one another, be drawn to touch the circle, and if from T any chord TQ be drawn, and from T a perpen- dicular rMbe dropped on T^, then TM^^QM. 42. Find the loci : (1.) Of the centres of circles passing through two given points. (2.) Of the middle points of a system of parallel chords in a circle. (3.) Of points such that the difference of the distances of each from two given straight lines is equal to a given straight line. (4.) Of the centres of circles touching a given line in a given point. (5.) Of the middle points of chords in a circle that pass through a given point. (6.) Of the centres of circles of given radius which touch a given circle. (7.) Of the middle points of chords of equal length in a circle. (8.) Of the middle points of the straight lines drawn from a given point to meet the circumference of a given circle. 43. If the base and vertical angle of a triangle be given, find the locus of the vertex. 44. A straight line remains parallel to itself while one of its extremities describes a circle. What is the locug of the other extremity ? 174 EUCUjys ELEMENTS. [Book IIL 45. A ladder slips down between a vertical wall and a horizontal plane : what is the locus of its middle point ? 46. AB is the diameter of a circle ; ACD is a chord pro- duced to Z), so that A C= CD. Find the locus of the point in which BG and the line joining D to the centre intersect. 47. ABC is a line drawn from a point A, without a circle, to meet the circumference in J5 and G. Tangents are drawn' to the circle at B and G which meet in D. What is the locus ofD? 48. Two circles intersect in the ppints A, B ; any straight line GDEF is drawn cutting the circles in 0, D, E, F ; prove that AG intersects BD and AE intersects BF in points, which lie on a circle passing through A and B. 49. The angular points A, Oof a parallelogram ABGD move on two fixed straight lines OA, OG, whose inclination is equal to the angle BGD ; shew that the points B^ D will move on two fixed straight lines passing through 0. 60. On the line AB is described the segment of a circle, in the circumference of which any point G is taken. If ACy BC be joined, and a point P taken in ^C so that GP is equal to GB, find the locus of P. 51. Find the locus of the centre of the circles circumscribing two trapeziums, into which a parallelogram is divided by any line equal to one of its shorter sides. 52. If a parallelogram be described having the diameter of a given circle for one of its sides, and the intersection of its diagonals on the circumference, shew that the extremity of each of the diagonals moves on the circumference of another circle of double the diameter of the first. 53. One diagonal of a quadrilateral inscribed in a circle is fixed, and the other of constant length. Shew that the sides will meet, if produced, on the circumference of a fixed circle. E£.ok in.] EUCLtns PROOF OF TIL 23. 175 We here insert Euclid's proofs of Props. 23, 24 of Book III. first observing that he gives the following definition of similar segments : — Def. Similar segments of circles are those in which the angles , /. z ABD and z J.i)B are each = twice z 5^D ; and thus an isosceles A ABD has been described as was reauired. o. k. f. Book IV.] PROPOSITION XI. 189 Proposition XI. Problem. To mscriJbe a regular pentagon in a given drcU, Let ABODE be the given 0. It is required to inscribe a regular pentagon in the © . Make an isosceles A FGH, having each of the z s at G', fl double of z at F. In © ABODE inscribe slaACD equiangular to A FGH, iv. 2. having z a at A, 0, D=thezs at F, G, H, respectively. Then z ^[1)0= twice z DAO, and z ^GD= twice z DAO. Bisect the z s ADO, ACD by the chords DB, CE. Join AB, BO, DE, EA. Then will ABODE be a regular pentagon. For •.* z s ADO, AOD are each = twice z DAG, and z s ADO, AOD are bisected by DB, OE, ,\ L s ADB, EDO, DAO, EOD, AOE, are all equal ; and .-. arcs AB, BO, OD, DE, EA are all equal ; III. 26. and .-. chords AB, BO, OD, DE, EA are aU equal. III. 29. Hence, the pentagon ABODE is equilateral. Again, '.* arc CD = arc AB, adding to each arc AED, we have arc AEDO= a,rc BAED, and .-. z ABO= z BOD. III. 27. Similarly, z s ODE, DEA, EAB each= z ^BC. Hence, the pentagon ABODE is equiangular. Thus a regular pentagon has been inscribed in the © . Q. B. F. Ex. Shew that OE is parallel to BA. 190 EUCLID'S ELEMENTS. [Book IV. Proposition XII. Problem. To describe a tegular pentagon about a gwen cirdtk Let ABODE be the given ®. It is required to describe a regular pentagon ahdut the © . Let the angular pts. of a regular pentagon inscribed in the © be at A, B, C, D, E, so that the arcs AB, BC, CD, DE, EA are all equal. Through A, B, (7, D, E draw GH, HK, KL, LM, MG tangents to the © ; take the centre 0, and join OB, OK, OG, OL, OD. Then in AS OBKj OCK, V OB=OG, and OK is common, and KB=KC, I. B. Cor. /. z BKO= L CKO, and z BOK= z COK, that is, z J5E:0= twice z CKO, and z jB0(7= twice z COK. So also, z DiO- twice z OLO, and z 1)0(7= twice z OOi. Book IV.] PROPOSITION XII. 191 Now •.' arc jBC-^arc CD, .-. L BOC= L DOG, and /. z COK= l COL. RencemAB OCK,OCL, V I GOK" L COL, and rt. i OCK=xt. l OGL, and OG is common, .-. I GK0= L GLO, and OK=GL, I. b. and .-. I HKL= l MLK, and KL=twice KG. Similarly it may be shewn that /. s KHG, HGM, GML each = z EKL, .'. the pentagon GHKLM is equiangular. And since it has been shewn that ^i= twice KG, and it can be shewn that HK ^twice KB, and •.• KB=KG, I. E. Cor. ,-. HK=KL. In like manner it may be shewn that HG, GM, ML, each .', the pentagon GHKLM is equilateral. Thus a regular pentp.gon has been described about the . 192 EUCLID'S ELEMENTS. [Book IV. Proposition XIII. Problem. To mscrihe a circle in a given regular pentagon. Let ABCDE be the given regular pentagon. It is required to inscribe a ® in the loentagon. Bisect I s BCD, CDE by the st. lines CO, DO, meeting in 0. Join OB, OA, OE. Then, in LsBCO,DCO, V BC=DC, and CO is common, and l BC0== l DCO, .-. z OBC= L ODC. I. 4. Then, •.• z ABC= z CDE, Hyp. and z CDJ^=twice z ODG, .'. A ABC=twice I OBC. Hence OB bisects z ABC. In the same way we can shew that OA, OE bisect the z s BAE, AED. Draw OF, OG, OH, OK, 0L± to AB, BC, CD, DE, EA, Then, in AS GOC, HOC, V L GCO= L HCO, and z OGC= l OHO, and OC is common, .'.OG=OH. 1.26. So also it may be shewn that OF, OL, OK are each = OG^or OB"; .-. OF, OG, OH, OK, OL are all equal, a © described with centre and radius OF will pass through G, H, K, L, and will touch the sides of the pentagon, •.' the z s at ^, G, H, K, L are rt. z s. III. 16. Thus a © will be inscribed in the pentagon, q. e. f. Book l^J PROPOSITION XIV. 193 Proposition XIV. Problem. To describe a cirde about a given regulcvr pentagont .4. Let ABCDE be the given regular pentagon. It is required to describe a © about the pentagon. Bisect the z s BCD, CDE by the st. lines 00, DO, meeting inO. Join OB, OA, OE. Then it may be shewn, as in the preceding Proposition, that OB, OA, OE bisect the z s CBA, BAE, AED. And •/ z BCD= L CDE, and z 00i)=half z BCD, and z ODO=half z Oi)^, .-. z 00i)= z ODO, and.-. 0I>=- 00. In the same way we may shew that OB, OA, OB each= OD or OC ; .-. OA, OB, OC, OD, OE are all equal, and a © described with centre and radius OA will pass through B, 0. D, E, and will be described about the pentagon. Q. B. F. 194 EUCLID'S ELEMENTS. [Book TV- Proposition XV. Problem. To imcribe a regular hexagon in a given circle* A L32. 1.13. Let ABCDEF be the given © , of which is the centre^ It is required to inscribe a regular hexagon in the ©. Draw the diameter AOD, and with centre D and radius DO describe a © EOCG Join EO, CO, and produce them to B and F. Join AB, BG, CD, DE, EF, FA. Then *.• is the centre of ACE, .: OE=OD; and •.• D is the centre of GCE, .'. OD=DE ; .'. OED is an equilateral A, and .*. z EOD = the third part of two rt. z s. So also z D0(7=the third part of two rt. z s, and .•. z BOG = the third part of two rt. z s. Thus z s EOD, DOG, BOG are all equal ; and to these the vertically opposite z s BOA, A OF, FOE are equal ; I. 15. .-. z s AOB, BOG, GOD, DOE, EOF, FOA, are all equal, and .-. arcs AB, BG, CD, DE, EF, FA are all equal. III. 26. and .-. chords AB, BG, CD, DE, EF, FA are all equal. Ill 29. Thus the hexagon ABGDEF is equilateral. Also *.' each of its z s= two-thirds of two rt. Z s, .*. the hexagon ABGDEF is equiangular. Tnus a regular hexagon has been inscribed in the © . Q. B. P. Book IV.] PROPOSinOA XVI. 195 Proposition XVI, Problem. To mscrihe a regula/r quindecagon in a given cireU. Let ABC be the given ©. It w required to inscribe in the ® a regular quindecagon. Let AB be the side of an equilateral A inscribed in the © , IV. 2. and AD the side of a regular pentagon inscribed in the . IV. 11. Then of such equal parts as the whole Oce ABC contains fifteen, arc ADB must contain five, and arc AD must contain three, and .*. arc DB, their difference, must contain two. Bisect arc DB in E. III. 30. Then arcs DE, EB are each the fifteenth part of the whole Oce. If then chords DE, EB be drawn, and chords equal to them be placed all round the Oce, IV. 1. a regular quindecagon will be inscribed in the © . Q. E. p. firricy 196 EUCLID'S ELEMENTS. [Book IV. Miscellaneous Exercises on Booh TV, 1. The perpendiculars let fall on the sides of an equilateral triangle from the centre of the circle, described about the triangle, are equal. 2. Inscribe a circle in a given regular octagon. 3. Shew that in the diagram of Prop. X. there is a second triangle, which has each of two of its angles double of the third. 4. Describe a circle about a given rectangle. 5. Shew that the diameter of the circle which is described about an isosceles triangle, which has its vertical angle double of either of the angles at the base, is equal to the base of the triangle. 6. The side of the equilateral triangle, described about a circle, is double of the side of the equilateral triangle, inscribed in the circle. 7. A quadrilateral figure may have a circle described about it, if the rectangles contained by the segments of the diagonals be equal. 8. The square on the side of an equilateral triangle, inscribed in a circle, is triple of the square on the side of the regular hexagon, inscribed in the same circle. 9. Inscribe a circle in a given rhombus. 10. ABC is an equilateral triangle inscribed in a circle ; tangents to the circle at A and B meet in M. Shew that a diameter drawn from M bisects the angle AMB, and is itself trisected by the circumference. 11. Compare the areas of two regular hexagons, one in- scribed in, the other described about, a given circle. 12. Inscribe a square in a given semicircle. 13. A circle being given, describe six other circles, each of them equal to it, and in contact with each other and with the given circle. Book IV.] MISCELLANEOUS EXERc/SES. Mr. 14. Given the angles of a triangle, and the perpendicp'^js from any point on the three sides, construct the triangle. 15. Having given the radius of a circle, determine its centre, when the circle touches two given lines, which are not parallel. 16. If the distance between the centres of two circles, which cut one another at right angles, is equal to twice one of the radii, the common chord is the side of the regular hexagon, inscribed in one of the circles, and the side of the equilateral triangle, inscribed in the other. 17. Construct a square, having given the sum, or the differ- ence, of the diagonal and the side. 18. If from 0, the centre of the circle inscribed in a triangle ABC, OD, OE, OF be drawn perpendicular to the sides BC, CAy AB, respectively, and from any point P in OP, drawn parallel to AB, perpendiculars PQ, PR be drawn upon OD and OE respectively, or these produced, shew that the triangle QBO is equiangular to the triangle ABC. 198 EUCLID'S ELEMENTS. [Books I. to IV. Euclid Payers set in the Mathematical Tripos at Cambridge from 1848 to 1872. Questions arising out of the Propositions, to which they are attached, have been proposed in the Enclid Papers to Candidates for Mathematical Honours since the year 1848. A complete set of these questions, so far as they refer to Books i.-iv,, is here given. The figures preceding each question denote the particular Proposition to which the question was attached. It is expected that the solution of each question is to be obtained mainly by using the Proposition which precedes it, and that no Proposition which comes later in Euclid's order should be assumed. Of some of the questions here given we have already made use in the preceding pages. As examples, however, of what has been hitherto expected of Candidates for Honours, and in order to keep the series of Papers complete, we have not hesitated to repeat them. 1848. I. c. How does it appear that the two triangles are equiangular and equal to each other ? I. 34. If the two diagonals be drawn, shew that a paralVlogram will be divided into four equal parts. In what case will the diagonal bisect the angle of parallelogram ? m. 16. Shew that all equal straight lines in a circle will be touched by another circle. m. 20. If two straight lines AEB, CEB in a circle intersect in E, the angles subtended hj AC and BD at the centre are together double of the angle AEC. Books I. to IV.] SENATE-HOUSE RIDERS. 199 1849. I. 1. By a method similar to that used in this pro- blem, describe on a given finite straight line an isosceles triangle, the sides of which shall be each equal to twice the base. IL 11. Shew that in Euclid's figure four other lines beside the given line, are divided in the re- quired manner. IV. 4. Describe a circle touching one side of a triangle and the produced parts of the other two. 1350. I. 34. If the opposite sides, or the opposite angles, of any quadrilateral figure be equal, or if its diagonals bisect each other, the quadrilateral is a parallelogram. II, 14. Given a square, and one side of a rectangle which is equal to the square, find the other side. UL 31. The greatest rectangle that can be inscribed in a circle is a square. III. 34. Divide a circle into two segments such that the angle in one of them shall be five times the angle in the other. IV. 10. Shew that the base of the triangle is equal to the side of a regular pentagon inscribed in the smaller circle of the figure. 1851. 1. 38. Let ABG^ ABD be two equal triangles, upon the same base AB and on opposite sides of it : join CD, meeting AB in B ; shew that CE is equal to BB. I. 47. If ABG be a triangle, whose angle J. is a right angle, and BE, CF be drawn bisecting the opposite sides respectively, shew that four times the sum of the squares on BE and CF is equal to five times the square on BC. m. S2. If a polygon of an even number of sides be in- scribed in a circle, the sum of the alternate angles together with two right angles is equal to as many right angles as the figure has sides. 200 EUCLID'S ELEMENTS. [Books I. to IV, 1851. IV. 16. In a given circle inscribe a triangle, whose angles are as the numbers 2, 5 and 8. 1852. I. 42. Divide a triangle by two straight lines into three parts, which, when properly arranged, shall form a parallelogram whose angles are of given magnitude. n. 12. Triangles are described on the same base and having the difference of the squares on the other sides constant : shew that the vertex of any triangle is in one or other of two fixed straight lines. IV. 3. Two equilateral triangles are described about the same circle : shew that their intersections will form a hexagon equilateral, but not gene- rally equiangular. 1853. i,B.Cor. If lines be drawn through the extremities of the base of an isosceles triangle, making angles with it, on the side remote from the vertex, each equal to one third of one of the equal ingles, and meeting the sides produced, prove that three of the triangles thus formed are isosceles. L 29. Through two given points draw two lines, form- ing with a line, given in position, an equi- lateral triangle. n. 11. In the figure, if H be the point of division of the given line J.JB, and DA be the side of the square which is bisected in B and produced to i^, and if DR be produced to meet B¥ in i, prove that BL is perpendicular to BF^ and is divided by BE similarly to the given line, UL 32. Through a given point without a circle draw a chord such that the difference of the angles in the two segments, into which it divides the circle, may be equal to a given angle. IIL 36. From a given point as centre describe a circle cut- ting a given line in two points, sp that the rect- angle contained by their distances from a fixed point in the line maybe equal to a given square Books I. to IV.] SENATE-HOUSE RIDERS. 201 1854. L 43. If K be the common angular point of the paral- lelograms about the diameter, and JBD the other diameter, the difference of the paral- lelograms is equal to twice the triangle BKD. II. 11. Produce a given straight line to a point such that the rectangle contained by the whole line thus produced and the part produced shall be equal to the square on the given straight line. ITL 22. If the opposite sides of the quadrilateral be pro- duced to meet in P, Q, and about the tri- angles so formed without the quadrilateral circles bo described meeting again in B, shew that P, li, Q will be in one straight line. IV. 10. Upon a given straight line, as base, describe an isosceles triangle having the third angle treble of each of the angles at the base. 1855. r. 20. Prove that the sum of the distances of any point from the three angles of a triangle is greater than half the perimeter of the triangle. L 47. If a line be drawn parallel to the hypotenuse of a right-angled triangle, and each of the acute angles be joined with the points where this line intersects the sides respectively oppo- site to them, the squares on the joining lines are together equal to the squares on the hypo- tenuse and on the line drawn parallel to it. U. 9. Divide a given straight line into two parts, such that the square on one of them may be double of the square on the other, without employing the Sixth Book. in. 27. If any number of triangles, upon the same base BC, and on the same side of it, have their vertical angles equal, and perpendiculars meeting in D be drawn from jB, C upon the opposite sides, tind the locus of D, and shew that all the Imes which bisect the angle BBC pass through the same point. 202 EUCLID'S ELEMENTS. | Books I. to IV 1855. IV, 4. If the circle inscribed in a triangle ^ BO touch the sides AB^ AC'va the points D, E, and a straight line be drawn from A to the centre of the circle, meeting the circumfeiencein G^ shew that G is the centre of the circle in- scribed in the triangle ABE. 1856 I. 34. Ol all parallelograms, which can be formed with diameters of gi'^en length, the rhombus is the greateot. IL 12. If AB, one of the eqaal sides of an isosceles triangle ABQ^ be prod iced beyond the base to D, so that BI)=AB) shew that the square on CD is equal to the square on AB together with twice the square on BG. rv. 15. Shew how to derive the hexagon fi jm an equi- lateral triangle inscribed in the circle, and from this construction shew that tb*} side of the hexagon equals the radius of the circle, and that the hexagon is double of the tri- angle. 1867. I. 35. ABC is an isosceles triangle, of which A is the vertex: AB, AC are bisected in D and E respectively ; BE, CD intersect in F : shew that the triangle ADE is equal to three times the triangle DEF. IL 13. The base of a triangle is given, and is bisected by the centre of a given circle, the circum- ference of which is the locus of the vertex : prove that the sum of the squares on the two sides of the triangle is invariable. m. 22. Prove that the sum of the angles in the four segments of the circle, exterior to the quadri- lateral, is equal to six right angles. IV. 4. Circles are inscribed in the two triangles formed by drawing a perpendicular from an angle of a triangle upon the opposite side, and analo- gous circles are described in relation to the two other like perpendiculars prove that the Books L to IV. J SENA TE-IJOUSE RIDERS. 203 sum of the diameters of the six circles toge- ther with the sum of the sides of the original triangle is equal to twice the sum of the three perpendiculars. 1858. L 28. A-Ssuming as an axiom that two straight lines cannot both be parallel ta the same straight line, deduce Euclid's sixth postulate as a corollary of the proposition referred to. n. 7. Produce a given straight line, so that the sum of the squares on the given line and the part produced may be equal to twice the rectangle contained by the whole line thus produced and the produced part. in. 19. Describe a circle, which shall touch a given straight line at a given point and bisect the circumference of a given circle. 1859. I. 41. Trisect a parallelogram by straight lines drawn from one of its angular points. IL 13. Prove that, in any quadrilateral, the squares on the diagonals are together equal to four times the sum of the squares on the straight lines joining the middle points of opposite sides. in. 31. Two equal circles touch each other externally, and through the point of contact chords are drawn, one to each circle, at right angles to each other : prove that the straight line, joining the other extremities of these chords, is equal and parallel to the straight line joining the centres of the circles. ly. 4. Triangles are constructed on the same base with equal vertical angles : prove that the locus of the centres of the escribed circles, each of which touches one of the sides externally and the other side and base produced, is an arc of a circle, the centre of which is on the circumference of the circle circumscribing the triangles. 204 EUCLID'S ELEMENTS. [Books I. to IV I860. L 36. If a straight line DME be drawn through the middle point M of the base BG of a triangle ABC, so as to cut oif equal parts AD, AE from the sides AB, AC, produced if neces- sary, respectively, then shall BD be equal to CE. IL 14. Shew how to construct a rectangle which shall be equal to a given square ; (1) when the sum, and (2) when the difference of two ad- jacent sides is given. IIL 36. If two chords AB, AC ho, drawn from any point A oi a. circde, and be produced to D and E, so that the rectangle AC, AE is equal to the rectangle AB, AD, then, if be the centre of the circle, AG is perpendicular to DE. IV. 10. If A be the vertex, and BD the base of the constructed triangle, D being one of the points of intersection of the two circles employed in the construction, and E the other, and AE be drawn meeting BD produced m F, prove that FAB is another isosceles triangle of the same kind. 1861. L 32. If ABC be a triangle,- in which 6 is a right angle, shew how, by means of Book I., to draw a straight line parallel to a given straight line so as to be terminated by CA and GB and bisected by AB. TL 13. If ABC be a triangle, in which C is a right angle, and DE be drawn from a point D in AG at right angles to AB, prove, without using Book III., that the rectangles AB, A E and AG, AD will be equal. IIL 32. Two circles intersect in A and B, and CBD is drawn perpendicular to AB to meet the circles in G and D ; if EAF bisect either the interior or exterior angle between GA and DA, prove that the tangents to the circles at E and F intersect in a point on AB produced. Books I. to IV.J SENA TE-f/OUSE RIDERS. 205 lbt>i. iV. 4. Describe a circle touching the side BO of the triangle ABC, and the other two sides pro- duced, and prove that the distance between the points of contact of the side BC with the inscribed circle, and the latter circle, is equal to the difference between the sides AB and AC. 1863. I. 4. Upon the sides AB, BC, and CD of a parallelo- gram ABCD, three equUateral triangles are described, that on BC towards the same parts as the parallelogram, and those on AB, CD towards the opposite parts. Prove that the distances of the vertices of the triangles on AB, CD, from that on BC, are respectively equal to the two diagonals of the parallelo- gram. ISL 10. Divide a given straight line into two parts, so that the squares on the whole line and on one of the parts may be together double of the square on the other part. QL 28. A triangle is turned about its vertex, until one of the sides intersecting in that vertex is in the same straight line as the other previously was : prove that the line, joining the vertex with the point of intersection of the two positions of the base, produced if necessary, bisects the angle between these two positions. If. 10. Prove that the smaller of the two circles, em- ployed in Euclid's construction, is equal to the circle described about the required tri- angle. 1863. L 47. Two triangles ABC, A'B'C have their sides respectively parallel. BBi, CCi are drawn perpendicular to B'C; CC^, AA^ to CA'; and AAz, BB3 to A'B'. Prove that the sum of the squares on ABi, BC^, CAz together, is equal to the sum of those on ACi, BA^, CB^ together. n. 11. Divide a given straight Ime into two parts, such 2o6 EUCLID'S ELEMENTS. [Books I. to IV. that the rectangle contained by the whole and one part may be equal to that contained by the other part and a given straight line. 1863. III. 28. Two equal circles intersect in A^ B \ PQT perpendicular to AB meets it in T, and the circles in P, Q. AP, BQ meet in i^ ; AQf BP in 8 : prove than the angle BT8 is bi- sected by TP. J864. I. 38. If a quadrilateral figure have two sides parallel, and the parallel sides be bisected, the lines joining the points of bisection shall pass through the point in which the diagonals cut one another. n. 14. Divide a given straight line (when possible) into three parts such that the rectangle con- tained by two of them shall be equal to a given rectilineal figure, and that the squares on these two parts shall together be equal to the square on the third. III. 36. If from a given point A without a given circle any two straight lines APQ, ABS, be drawn, making equal angles with the diameter which passes through A, and cutting the circle in P, Q, and B, S, respectively, then PS, QB, shall cut one another in a given point. IT. 11. If a figure of any odd number of sides have all its angular points on the same circle, and all its angles equal, then shall its sides be equal. 1865. I. 20. Give a geometrical construction for finding a point in a given straight line, the difierence of the distances of which from two given points on the same side of the line shall be the greatest possible. n. 12, The base EG of an isosceles triangle ABC is produced to a point D ; AD is joined, and in AD a point E is taken, such that the rect- angle AD, AE, is equal to the square on either of the equal sides AB, AC, of the triangle : Bcoks I. to IV.] SENATE-HOUSE RIDERS, ao7 prove that the rectangle BD, CD is equal to the rectangle AD, ED. 1865. m. 18. A given straight line is drawn at right angles to the straight line joining the centres of two given circles : prove that the difference be- tween the squares on two tangents drawn, one to each circle, from any point on the given straight line, is constant. IV. 5. Having given one side of a triangle, and the centre of the circuuiscribed circle, determine the locus of the centre of the inscribed circle. 1866. I. 33. Prove that a quadrilateral, which has two op- posite sides and two opposite obtuse angles equal, is a parallelogram. Shew that the figure is not necessarily a paral- lelogram, if the equal angles are acute. n. 9. Prove this also by superposition of the squares or their halves. IIL 22. If four circles be drawn, each passing through three out of four given points, the angle be- tween the tangents at the intersection of two of the circles is equal to the angle between the tangents at the intersection of the other two circles. IV. 2. In a given circle inscribe a triangle such that two of the sides of the triangle shall pass through given points and the third side be at a given distance from the centre of the given circle. 1867. I. 16. Any two exterior angles of a triangle are together greater than two right angles. L 43. What is the greatest value which these comple- ments, for a given parallelogram, can have ? . n. 11. Divide a given straight line into two parts such that the squares on the whole line and on one of the parts shall be together double of the square on the o*:her part 2o8 EUCLID'S ELEMENTS. [Books I. to IV. L867. III. 22. If the chords, which bisect two angles of a triangle inscribed in a circle, be equal, prove that either the angles are equal, or the third angle is equal to the angle of an equilateral triangle. 1868. I. 41. OKBM and OLDN are parallelograms about the diameter of a parallelogram ABCD. In MNj which is parallel to BA, take any point F and prove that, if PC, produced if neces- sary, meet KL in Q, BP will be parallel toD^. n. 12. In a triangle ABC, D, E, F are the middle points of the sides BC, CA, AB respectively, and K, L, M are the feet of the perpendi- culars on the same sides from the opposite angles. Prove that the greatest of the rect- angles contained by BC and BK, CA and EL, AB and FM, is equal to the sum of the other two. m. 36. Through a point within a circle, draw a chord, such that the rectangle contained by the whole chord and one part may be equal to a given square. Determine the necessary limits to the magni- tude of this square. IV. 4. If two triangles ABC, A'B'O be inscribed in the same circle, so that AA' BB' CC meet in one point 0, prove that, if be the centre of the inscribed circle of one of the triangles, it will be the centre of the perpendiculars of the other. I. 40. ABC is a triangle, E and F are two points ; if the sum of the triangles ABE and BCE be equal to the sum of the triangles ABF and BOF, then under certain conditions EF will be parallel to AC. Find these conditions, and determine when the difference instead of the sum of the triangles must be taken. Books 1. to IV.] SENATE HOUSE RIDERS. 209 i«b"9. II. 1 1. Shew that the point of section lies between the extremities of the line. III. 33. An acute-angled triangle is inscribed in a circle, and the paper is folded along each of the sides of the triangle : Shew that the circumferences of the three segments will pass through the same point. State the equivalent proposition for an obtuse-angled triangle. IV. 11. Shew that the circles, each of which touches two sides of a regular pentagon at the ex- tremities of a third, meet in a point. 1870. I. 26. ABCD is a square and E a point in BG ; a straight line EF is drawn at right angles to AE, andjiieets the straight line, which bisects the angle between CD and BG produced in a point F : prove that AE is equal to EF. IL 9. The diagonals of a quadrilateral meet in E, and F is the middle point of the straight line joining the middle points of the diagonals : prove that the sum of the squares on the straight lines joining E to the angular points of the quadrilateral is greater than the sum of the squares on the straight lines joining F to the same points by four times the square on^i^. UL 3?.. AB, GD are parallel diameters of two circles, and AG cuts the circles in F, Q: prove that the tangents to the circles at P, Q are parallel. IV. 10. Hence shew how to describe an equilateral and equiangular pentagon about a circle with- out first inscribing one. 1871. I. 38. Through the angular points -4, B, 0, of a triangle are drawn three parallel straight lines meeting the opposite sides in A', B'j (7 re- spectively : prove that the triangles AB'C, BOA', GA'B' are all equal n. 10. Produce a given straight line so that the square on the whole line thus produced may be double the square on the part produced. 2IO EUCLID'S ELEMENTS. [Books' I. to IV. 1871. iiL 32. The opposite sides of a quadrilateral inscribed in a circle are produced to meet in P, Q, and about the four triangles thus formed circles are described : prove that the tangents to these circles at P and Q form a quadrilateral equal in all respects to the original, and that the line joining the centres of the circles, about the two quadrilaterals, bisects PQ. IF. 5. A triangle is inscribed in a given circle so as to have its centre of perpendiculars at a given point : prove that the middle points of its sides lie on a fixed circle. 1872. L -47. If CE, BB be the squares described upon the side AC, and the hypotenuse AB, and if EB, CD intersect in F, prove that AF bi- sects the angle EFD. n. 14. If the given rectilineal figure be that of Euclid i. 47, shew how to determine the required square graphically. m. 22. Two circles intersect m A, B: PAP'y QAQ' are drawn equally inclined to AB to meet the circles in P, P\ Q, Q' : prove that PP is equal to QQ'. If, 4. Having given an angular point of a triangle, the circumscribed circle, and the centre of the in- flcribed drcie, construct the tnangie. BOOK V. SECTION I. On Multiples and Equimultiples Dbf. I. A GREATER magnitude is a Multiple of a less magni- tude, when the greater contains the less an exact number of times. Def. II. A LESS magnitude is a Sub-multiple of a greater magnitude, when the less is contained an exact number of times in the greater. These definitions are applicable not merely to Geometrical magnitudes, such as Lines, Angles, and Triangles ; but also to such as are included in the ordinary sense of the word Magni- tude, that is, anything which is made up of parts like itself, such as a Distance, a Weight, or a Sum of Money, Postulate. Any one magnitude being given, let it be granted that any number of other magnitudes may be found, each of which is equal to the first Method of Notation. Let J. -represent a magnitude, not as one of the letters used in Algebra to represent the ineasure of a magnitude, but let A stand for the magnitude itself. Thus, if we regard A as repre- senting a weight, we mean, not the number of pounds con- tained in the weight, but the weight itseit EUCLIjyS ELEMENTS. LBook V. Let the words A^ B together represent the magnitude obtained by putting the magnitude B to the magnitude A, ^ Let J., A together be abbreviated into 2/1, Ay A, A together 3J, and so on. Let A, A repeated m times be denoted by mA, m standing for a whole number. Let mAj mA repeated n times be denoted by nmJ., where nm stands for the arithmetical product of the whole numbers n and m. Let (m+n) A stand for the magnitude obtained by putting nA to mA, m and n standing for whole numbers. These, and these only, are the symbols by which we propose to shorten and simplify the proofs of this Book : capital letters standing, in all cases, for tnagnitudes ; and small letters standing for whole numbers. Scales of MuLTiPLEa By taking a number of magnitudes each equal to A, and putting two, three, four of them together, we obtain a set of magnitudes, depending upon A, and all known when A is known ; namely, A, 2Ay 3Af 4Af 5 A and so on ; each being obtained by putting A to the preceding one. This we call the Scale op Multiples of A. If m be a whole number, mA and mB are called Equi- multiples of A and B, or, the same multiples of A and B respectively. Axioms. 1. Equimultiples of the same, or of equal magnitudes, are equal to one another. 2. Those magnitudes, of which the same, or equal, magni- tudes are equimultiples, are equal to one- another. Book v.] ON MULTIPLES AND EQUIMULTIPLES. 213 3. A multiple of a greater magnitude is greater than the same multiple of a less. 4. That magnitude, of which a multiple is greater than the same multiple of another, is greater than that other magni- tude. Note 1. If A and B be two commensurable magnitudes, it is easy to show that there is some, multiple of A^ which is equal to sovtie, multiple of JB. For let ilf be a common measure of A and B ; then the scale of multiples of iW is M, 2M, 3M, Now one of the multiples in this scale, suppose pM, is equal to A, and one suppose qM, B. Hence the multiple qpM is equal to gA, V. Ax. 1. and the same multi))le is equal to pB ; and therefore qA = pB. L Ax. 1. Proposition I. (Eucl. v. 1.) If any number of magnitudes be equimultiples of as many, each of each ; whatever multiple any one of them, is of its sub- multiple, the same multiple must all the first magnitudes^ taken together, be of all the other, taken together. Let A be the same multiple of C that B is of D. Then must A, B together be the same multiple of G, D together that A is of C. Let A = C, C, C repeated m times. Then B = D,D,D repeated m times. .'. A,B together = C,D ; C,D; G,D; repeated m times. .'. A, B together is the same multiple of C, D together tb** OF THE ^ UNIVERSITY OF 214 EUCLID'S ELEMENTS. [Book V Proposition II. (Eucl. v. 2.) If thef,rst be the same multiple of the second that the third is of the fourth, and the fifth the same multiple of the second that the sixth is of the fourth ; the first together with the, fifth must he the same multiple of the second, that the third together with the sixth is of the fourth. Let A, B, G, D, E, F be six magnitudes, such that A is the same multiple of B, that C is of Z>, and E is the same multiple of B, that F is of D. Then must A, E together be the same multiple of B, tfiat G, F together is of D. Let A == B, B, B, repeated m times ; then G = D,D,D, repeated m tinies. Also, let E = B, B, B, repeated n times ; then F = D,D,D, repeated n times. .'. A, E together = B, B, B, repeated m+n tinies, and 0, F together = D,D,D, repeated m+n times. .'. A, E together is the same multiple of B, that 0, F together is of D. Q. B. D. Proposition III. (EucL v. 3.) If the first he the same multiple of the second that the third «s of the fourth ; and if of the first and third there he talcum equimultiples, these must be equimultiples, the c ne of the second, and the other of the fourth. Let A be the same multiple of B that is of 1> ; and let E and F be taken equimultiples of A and G. Then must E and F be equimultiples of B and B. For let A — B, B, repeated m times=mjB ; then G = D, D, repeated m times=wiJ>. Again, let ^ = A, A, repeated n times ; then F == G, G, repeated n times. .*. E = mB, mB, repeated n times=rwn.JB ; and F — mD, mD, repeated n times = nml). /. E is the same multiple of B that F is of D. Q. E. D. Book v.] ON RATIO AND PROPORTION. 215 SECTION IL CS j^ciw and Proportimu 1535F. III. if A vi^X B W magiiitjides of the liame kind, the relative greatn<^ss of A vfiih i wpect to B is called the ratio of -A to B. Note 2. When A and B are coinmems'urable, we can estimate their relative greatness by considering what multiples they are of some common standard. But as this method is not appli- cable when A and B are incomraeuburdble, we have to adopt a more general method, applicat)lte-6oih to commensurable and incommensurable magnitudes. If A and B be magnitudes of tne saate kind, commensurable or incommensurable, the scale of multiples of A is A,2A.. .mA, (m+ l)A...2mA, (Sra 4 l)A...SmA...miiA.. . and the Ratio of B to A is estimated by considering the posi- tion which B, or some multiple of By occupies among the multiples of A. If A and B be commensurable, a multiple of jB can be found, such that it would occupy the same place among the multiples of A, which is occupied by some one of the multiples of A ; that is, this particular multiple of B represents the same magnitude as that, which is represented by some one of the multiples of A. See Note 1, p. 213. If, for example, the 7th multiple in the scale of B represents the same magnitude as that which is represented by the 5th multiple in the scale of Aj or in other words, if 7B = 6A, we rt^e enabled to form an exact notion of the greatness of B ' ^latively to A. 2i6 EUCLID'S ELEMENTS. [Book V. When A and B are incommensurable, the relation mA^nU can have no existence ; that is, no pair of multiples, one in each of the scales of multiples of A and B, represent the same magnitude. But we can always determine whether a par- ticular multiple of B be greater or less than some one of the multiples of A ; that is, we can always find between what two successive multiples of A any given multiple of B lies. Hence, whether A and B be commensurable or incommen- surable, we can always form a third scale, in which the multiples of B are distributed among the multiples of A. Suppose, for example, we discover the following relations between particular multiples of A and B : B greater than A and less than 2 J., 2B greater than 3 A and less than 4 A, SB greater than 6A and less than 7 A, and so on ; the third scale will commence thus A, B, 2A, 2A, 2B, AA, 5^, QA, 3B, 7 A, and so on ; the scale not being formed by any law, but con- structed by special calculations for each term. Such a scale we call the Scale of Eelation of A and B, and we give the following Definition : — The Scale of Relation of two magnitudes of the same kind is a list of the multiples of both ad infinitum, all arranged in order of magnitude, so that any multiple of either magnitude being assigned, the scale of relation points out between which multiples of the other it lies. Note 3. It may here be remarked that, if A and B be two finite magnitudes of the same kind, however small B may be, we may, by continuing the scale of multiples of B suffi- ciently far, at length obtain a multiple of B greater than A. Also, if B be less than A, one multiple at least of the scale of B will lie between each two consecutive multiples of the scale of A. From these considerations we shall be justified in assuming Book v.] ON RATIO AND PROPORTION. iij (1.) That we can always take mB greater than A or than pA. (2.) That we can always take nB such that it is greater than pA but not greater than qA, provided that B is less than A, and p than q. We can now make an important addition to Definition iii., so that it will run thus : — If A and B be magnitudes of the same kind, the relative greatness of A with respect to B is called the Ratio of A to B, and this Ratio is determined by, that is, depends solely upon, the order in which the multiples of A and B occur in the Scale of Relation of A and B. Def. IV. Magnitudes are said to have a Ratio to each other, which can, being multiplied, exceed each the other. This definition is inserted to point out that a ratio cannot exist between two magnitudes unless two conditions be ful- filled :— first, the magnitudes must be of the same kind ; secondly, neither of them may be infinitely large or infinitely small. See Note 3. Def. V. When there are four magnitudes, and when any equimultiples of the first and third being taken, and any equi- multiples of the second and fourth, if, when the multiple of the first is greater than that of the second, the multiple of the third is greater than that of the fourth, and when the multiple of the first is equal to that of the second, the multiple of the third is equal to that of the fourth, and when the multiple of the first is less than that of the second, the multiple of the third is less than that of the fourth, then the first of the original four magnitudes is said to have to the second the same ratic* which the third has to the fourth. 2l8 EUCLID'S ELEMENTS. [Book V. Note 4. — To make Def. v. clearer we give the folloM-ing illustration. Suppose A, B, G, D to be four magnitudes ; the scales of their multiples will then be — A, 2A, SA.. mA , By 2B, 3B nB , C, 20, 3C mC , D, 2D, 3D nD ; where mA, mC stand for any equimultiples of A and G, and ihB, nD stand for any equimultiples of jB and D: then the Definition may be stated more briefly thus : A is said to have the same ratio to B which G has to D, when mA is found in the same position among the multiples of B, in which mG is found among the multiples of D ; or, which is the same thing, wheM the order of the multiples of A and B in the Scale of Relation of A and B, is precisely the same as the order of the multiples of G and D in the Scale of Relation of G and D ; or, when every multiple of A is found in the same position among the multiples of B, in which the same multiple of G is found among the multiples of D. Note 5. The use of Def. v. will be better understood by the following application of it. To show that rectangles of equal altitude an-e to one anA)theT as tlieir bases. a u f fj h. Let AG, ac be two rectangles of equal altitude. Let B, B' and R, R' stand for the bases and the areas of these rectangles respectively. Take AD, DE, EF, m in number, and all equal, And ad, de, ef,fg, gh, n in number, and all equal. Book v.] ON RATIO AND PROPORTION. 219 Complete the rectangles, as in the diagram. Then base A¥ = mi, base ah = nB^. rectangle AP — mR^ rectangle ap = nR\ Now we can prove, by superposition, that if ^F be greater than ah, AP will be greater than op, and if equal, equal ; and if less, less. That is, if mB be greater than nB\ mR is greater than nR'; and if equal, equal ; and if less, less. Hence, by Def. v., B is to B' as R is to R\ Hence we deduce two Corollaries, which are the foundation of the proofs in Book vi. Cor. I. Parallelograms of equal altitude are to one another as their hoses. For the parallelograms are equal to rectangles, on the same b&ses and between the same parallels. Cor. II. Triangles of equal altitude are to on^ another as their bases. For the triangles are equal to the halves of the rectangles, on the same bases and between the same parallels. N.B. — These Corollaries are proved as a direct Proposition in Eucl. VI. 1. Cor. 11. could not, consistently with Euclid's method, be introduced in this place, for it assumes Proposi- tion XI. of Book V. Def. VI. Magnitudes which have the same ratio are called Proportionals. If Af Bf C, D be proportionals, it is usually expressed by saying, J. is to 5 as is to D. The magnitudes A and G are called the Antecedents of the ratios. — B andD Consequents The antecedents are said to be homologous to one another, that IS, occupying the same position m tne ratios (6fi6\oyoi), and the consequents are said to be homologous to one another. 220 EUCLID'S ELEMENTS. [Book V. Def. VII. When of the equimultiples of four magnitudes, taken as in Def. v., the multiple of the first is greater than [or is equal to] the multiple of the second, but the multiple of the third is not greater than [or is less than] the multiple of the fourth, then the first is said to have to the second a greater ratio, than the third has to the fourth. Note 6. The meaning of Def. vii. may be expressed, after taking the scales of multiples as in the explanation of Def. v., thus : — A is said to have to JB a greater ratio than has to D, when two whole numbers m and n can be found, such that vfiA is greater than rtJ5, but mO not greater than nD ; or, such that tikA is equal to nB^ but vtiG less than nD, Booli: v.] PROrOSiriONS CITED IN BOOK VI. 221 SECTION IIL Containing the Propositions most frequently referred to in Book VI. Note 7. The Fifth Book of Euclid may be re^rarded in two aspects : first, as a Treatise on the Theory of Ratio and Propor- tion, complete in itself, and depending in no way on the pre- ceding Books of the Elements ; and secondly, as a necessary introduction to the Sixth Book. If we make the number of references in Book vi. a test of the importance of particular Propositions in Book v., they will be arranged in the following order : — Froposition v. is referred to 23 timei^ »» ^- » 14 „ „ VIII. ^ 7 „ y> XXI. ^ ** » » „ XVIII. „ „ XII. „ 2 „ Propositions x., xi, xv., xvi., xix., xxii., are referred to and as E is to ^... Then must A be to B as A, G, E. ..together is to B, D, F. . .together. Take oi A, G, JE,...smj equimultiples mA, mG, mE... and of J5, D, J^...any equimultiples nB, nD, nF... Then •.* ^ is to 5 as is to D and as E is to F... .'. if mA be greater than nB, mG is greater than nD, and mE is greater than nF. . . ; and if equal, equal ; if less, less. V. 4. .*. if mA be greater than nB, mA, mG, mE .. .together are greater than nB, nD, «- J?',.. together ; and if equal, equal; if less, less. Now mA and mA, mG, mE. . .together are equimultiples of A and A, G, E .. .together. V 1. And nB and nB, nD, nF... together are equimultiples of B and B, D, /^...together. .'. -4 is to ^ as A, G, ^...together is to B, D, J?*... together. V. Def. 5. Q. E. D. Proposition XI. (Eucl. v. 15.) Magnitudes have the same ratio to one another which iheii equimultiples have. Let A be the same multiple of G that jB is of D. Then must G be to D as A to B. Divide A into magnitudes E, F, G^,...each equal to G, and B into magnitudes H, K, i,...each equal to D, the number of the magnitudes being the same in both cases, because A and B are equimultiples of G and D. Then •.' E, F, G are all equal, and if, K, L are all equal. .'. ^is toH, asi^toE:, as G^ toi... V. 6 .: E \a to H as E, F, G^... together is to H, K, L . together, V. 10 that is, jE? is to J3" as J. to ^ ; and '.• ^ = 0, and ^ = D, /. is U) D as JL to J5. Book v.] PROPOSITIONS CITED IN BOOK VI. 22^ SECTION IV. On Proportion by Inversion , Alternation, and Separation Proposition XII. (Eucl. v. B.) If four mcbgnitudes be proportionals, they imist also be pro portionals when taken inversely. Let ^ be to J5 as (7 is to D. Then inversely B must he to A as D is to G, Take of A and C any equimultiples mA and m(7, pnd of B and D any equimultiples nB and nD. Then •/ ^ is to J5 as is to D, .'. if mA be greater than nB, mC is greater than nD ; and if equal, equal ; if less, less. ?. 4. Hence, if nB be greater than '(nAj nD is greater than mC ; and if equal, equal ; if less, less. ,\£iAtoAaADiaioO, V. Def. 6. Q.S. D. 228 EUCLID'S ELEMENTS. [Book V Proposition XITI. (Eucl. v. 13.) If the first has to the second the same ratio which the third ha^ to the fourth, but the third to the fourth a greater ratio than the fifth has to the sixth ; the first must also have to the second a greater ratio than the fifth has to the sixth. Let A have to B the same ratio that C has to D, but to D a greater ratio than E has to F. Th.en must A have to B a greater ratio than E has to F. For *.* has to D a greater ratio than E has to F, we can find such equimultiples of and E, suppose mOand mEy and such equimultiples of D and F, suppose nD and nF, that mC is greater than nl), but mE not greater than nF. V. Def. 7. Then '.' A is to B SiS C is to 2>, Hyp« and mC is greater than nD, ,'. mA is greater than nB. V. 4. And mE is not greater than nF. .*. A has to ^ a greater ratio than E has to F. V. Def. 7. Q. E. D. Proposition XIV. (Eucl. v. 14.) If the first has to the second the same ratio which the third has to the fourth; then, if thefi/rst he greater than the third the second must be greater thun the fov/rth ; and if equal, equal ; and if less, less. Let A have the same ratio to B that has to D. Then if A be greater than 0, B must be greater than D, For '.' A'lB greater than C, and B is any other magnitude, .*. A has a greater ratio to B than C has to B. Y. 7. Book v.] PROPOSITIONS CITED IN BOOK VI. 229 But A is to 5 as is to D. .-. has a greater ratio to D, than has to B. V. 13. .*. JB is greater than D. V. 9. Similarly it may be shown that if A be less than 0, B must be less than D ; and that if J. be equal to C^B must be equal to D. Q. B. D. Proposition XV. (Eucl. v. 16.) If four magnitudes of the same kind be proportionals, thxy miist also be proportionals when taken alternately. Let A, B, C, D he four magnitudes of the same kind, and let ^ be to E as is to D. Then alternately A must be to C as B is to D, Take of A and B any equimultiples 'iiiA and mJ?, and of C and D any equimultiples nC and nD. Then *.• mA is to mB as A is to B, V. 11. and (7 is to D as A is to B, Hyp. .*. mA is to mJB as is to D. V. 6. But nC is to nD as C is to JD ; V. 11. and .'. mA is to mB as nC is to nD. V. 6. If .*. mA be greater than nO, m^ is greater than nD ; and if equal, equal ; if less, lesai V. 14. .% iiBtoOas^BJatolA V. De£ 6^ 23© EUCLID'S ELEMENTS. [Book V. Proposition XVi. (Eucl. v. 18.) If magnitudes taken separately he proportionals, they miist be proportionals also when taken Let A have the same ratio to B that G has to D. The7i must A, B together have the same ratio to J5, that G, D together has to D. First, when all the magnitudes are of the same kind, ••• ^ is to jB as G is to D, .'. J: is to as 5 is to D. V. 15. .*. A, B together is to 0, D together as B is to D, V. 10. and .'. A,B together is to B as G, D together is to D. V. 15. Next, when all the magnitudes are not of the same kind, we may employ a method of proof which includes the former case : thus — Take of A, B, G, D any equimultiples 7nAy mB, mG, mD, and of jB and D take any equimultiples nB, nD. Then '.• J. is to jB as is to J>, .*. if mA be greater than nB, mG is greater than nD ; and if equal, equal ; if less, less. V. 4. If then mA, mB together be greater than mB, nB together, mG, mD together is greater than mG, nD together ; and if equal, equal ; if less, less. I. Ax. 2, 4. Now mA, mB together is the same multiple of A, B together that mG, mD together is of G, D together ; V. 1. and mB, nB together is the same multiple of B that mD, nD together is of D. V. 2. .*. Af B together is to .B as 0, D together is to 2>. Y. Def. 5. Q. £. D. Book v.] PROPOSITIONS CITED IN BOOK VI, 231 SECTION V. Containing the Propositions occasionally referred to in Book VI. Proposition XVII. (Eucl. v. 4.) If thejvrst of four magnitudes has to the second the same ratio which the third has to the fourth, and any equimultijdes of the first and third be taken, and also any equimultijjles of the second and fourth, then must the multiple of the first have the same ratio to the multiple of the second which the multiple of the third has to that of the fourth. If ^ be to 5 as is to D, and mA, mC be taken equimultiples of A and G, and nB, nD of B and D, Uien must mA be to nB as mC is to nD. Take of mA, mC any equimultiples jJ'mA, pmC, and of nB, nD qnB, qnD, Then pmA, pmC are equimultiples of A and 0, V. 3. and qnB, qnD of £ and i>. V. 3. And ••• A is to JB as is to J), .'. if jmiA be greater than qnBy pmC is greater than qnD ; V. 4. and if equal, equal ; if less, less. Then •.• pmA, pmC are equimultiples of mA, mC, andqnB, qnD of nB, nD, .: mA is to ri^ as mC is to nD- V. Def. 5. Q. B. D. 232 EUCLID'S ELEMENTS. [Book V. Proposition XVIII. (Eucl. v. A.) If the jvrst of four magnitudes have the same ratio to the second that the third has to the fourth, then, if the fvrst he greater than the second, the third must be grealor than the fourth ; and if equal, equal ; and if less, less. Let ^ be to J5 as is to D. Then if A be greater than B, G must be greater than D ; and if equal, equal ; and if less, less. Take any equimultiples of each, mA, mB, mC, mD. Then :• A is to B as is to D, .'. if mA be greater than mB, mC is greater than 7nD ; and if equal, equal ; and if less, less. V. 4. First, suppose A greater than B, then mA is greater than mB, V. Ax. 3. and .-. mG is greater than mD, and .-. is greater than D. V. Ax. 4. Similaily the other cases may be proved. a s. o. Proposition XIX. (Eucl. v. D.) If the first he to the second as the third is to the fourth, and if the first he a multiple, or a submultiple, of the second, the third must be the samie multiple, or the same submultipUf of the fourth. Let ^ be to J5 as is to D, and, first, let J[ be a multiple of B. Then must G he the same multiple of D. Let A =mJB,and take mD the same multiple of D that -4 is of jB. Then •.• J. is to jB as is to D, .'. Jl is to mB as is to mD. V 17. But A = mB, and .-. = mD. . V. 18. Fook v.] PROPOSITIONS CITED IN BOOK VI. 233 Next, let J. be a submuUiple of B. TJien must G be the same suhmultijjle of D» For •. • J. is to -B as is to I>, .-. j5 is to ^ as D is to C, V. 12. Now 5 is a multiple of A, and .". i> is the same multiple of G, by the first case. Hence G is the same submultiple of D, that ^ is of J5, Q. B. D. Proposition XX. (Eucl. v. SO.) If there he three Tnagnitvdes, arid other three, which have tht same ratio, talcen two and two, then, if the first he greater than the third, the fov/rth must be greater than the sixth ; and if equal, equal ; if less, less. Let A, B, G be three magnitudes, and D, E, F other three, and let ^ be to 5 as D is to E, and J5 be to as ^ is to F. Then if A be greater than G, D must be greater than F ; and if equal, equal ; if less, less. First, if A be greater than G, A has to J5 a greater ratio than G has to B. V. 7. But has to jB the same ratio that F has to E, Hyp. & V. 12. . •• A has to ^ a greater ratio than F has to E, .'. D has to ^ a greater ratio than F has to E, V. 13. .'. r) is greater than F. V. 9. Similarly the other cases may be proved. 234 EUCLirys ELEMENTS. [Book V. Proposition XXI. (Eucl. v. 22.) If there be any number of magnitudes, and as many others^ which have the same ratio taken two and two in order, thefvrst must have to the last of the first magnitudes the same ratio which the first of the others has to the last of these. First, let there be three magnitudes A, B, 0, and other three 2), E, F. And let ^ be to jB as D is to E, and J5 be to as E is to F. Then must A be to C as D is to F. Take of A and D any equimultiples mA, mD^ of 5 and E nB, nE, of O&ndF pC,pF. , Then •.* ^ is to 5 as D is to E, .'. mA is to fiB as m,D is to nE. V. 17. So also, nB is to pC as nE is to pF. .: if mA be greater than pG, mD is greater than pF, ind if equal, equal ; if less, less. V. 20. .: A \% to C-A&D is to F. V. Def. 5. The proposition may be easily extended to any number of magnitudes. q. e. d. Proposition XXII. (Eucl. v. 24.) If the first have to the second the same ratio which the third has to the fourth, and the fifth have to the second the same ratio which the sixth lias to the fourth, then the first and fifth together must have to the second the same ratio which the third and sixth together have to the fourth. Let J. be to J5 as G is to i), and jE; be to J5 as ^ is to D. Then must A, E together be to B as G, F together is to D. For •.• ^ is to 5 as i^ is to D, .'. Bisto E as D is to F. V. 12. And '.'A is to JB as is to D, and 5 is to j& as D is to i?*, ' - ,.'. A is to E as G is to F. V. 21. .-. A, E together is to jS as 0, jF together is to F, V. 16. and J5? is to jB as F is to D ; '. A, E together is to B as C, F together is to D. V. 2L SECTION VI. Containing the Propositions to which no reference is made in Book VI. Proposition XXIII. (Eucl. v. 5.) If one rrmgnitvde be the srnne multiple of another, which a magnitude taken from the first is of a magnitude taken from the other, the remainder must be the same multiple of the remainder, that the whole is of the whole. Let 5 and D be the magnitudes which are taken away, and A and the magnitudes which remain, then A, B together, and G, D together will be the wholes. And let A, B together be the same multiple of C, D together, that B is of D. Then must A be the same multiple of C that A, B together is of Gj D, together. Take E the same multiple of G that B is of D, Then E, B together is the same multiple of 0, D together that ^ is of I). V. 1. But 4, B together is the same multiple of C, D together that B is of D. .'. Ey B together = A, B together, V. Ax. 1. a- \.'.E = A. I. Ax. 3. .*. ^ is the same multiple of G that ^ is of D. Q. B. D. 236 EUCLID S ELEMENTS, [Book V. Proposition XXIV. (Eucl. v. 6.) If two magnitudes he equimultijAes of two otherSj and if equimultiples of these be taken from the first two, the remainders are either equal to these others, oi' equimultiples of them. Let B and D be the magnitudes which are taken away, and A and G the magnitudes which remain ; then A, B together and C, D together will be 4;he wholes. Let A, B together be the same multiple of P, that G, D together is of Q, and let B be the same multiple of P, that D is of Q. Then must A and G he equal respectively to P and Qy or A and G be equimultiples of P and Q. For let Aj B together = P, P repeated m + n times, then 0, D together = Q, Q repeated m+n limes. Also, let B= P, P repeated n times, then D 'B Q^ Q repeated n times. Hence A = P, P repeated m times, and G^Q, Q repeated m times. If then A = P,m = l, and .: G = Q ; and if ^ be a multiple of jP, ib the same multiple of Q, Book v.] PROPOSITIONS NOT CITED IN BOOK VI. 237 Proposition XXV. (Eucl. v. 17.) If magnitudeSy taken joi7ttly, he proportionals, they shall also be pi'oportionals when taken separately ; that is, if two magni- tudes together have to one of them the same ratio which two others have to one of these, the remaining one of the first two must have to the other the same ratio which the remaining one of the last two has to the other of these. ^ Let Af B together have the same ratio to B that C, D together have to D. Then must A be to B as G to D. Take of A, B, C, D any equimultiples mA, mB, mG, mD, and again of B, D take any equimultiples uB, nD. Then ••• mA is the same multiple of A that mB is of B, .: mA, mB together is the same multiple of A, B together that mA is of ^. V. 1. And '.• mC is the same multiple of C that mD is of D, .*. mC, mD together is the same multiple of C, D together that mC is of (7. V. 1. But mA is the same multiple of A that mC is of G, .'. mA, mB together is the same multiple of ^, i>' together that mC, mD together is of 0, 1> together. Again, mB, nB together is the same multiple of B that mD, nD together is of D. Now, since A, B together is to B as G,D together is to D, .'. if mA,mB together be greater than mBy 71B together, mC, mD together is greater than mD, nD together ; and if equal, equal ; if less, less. V. 4. That is, if mA be greater than w-i>*, mC is greater than nD ; and if equal, equal ; if less, less. I. Ax. 3, 5. .% .4 is to £ as is to D. V. Def. 5. Q. E. D. 238 EUCLID'S ELEMEMTS. [Book V. Proposition XXVI. (Eucl. v. 19.) If a whole Tnagnitude be to a whole as a magnitude taken from the first is to a magnitude taken from the other, the re- mainder must he to the reinainder as the whole is to the whole. Let A, B together have the same ratio to C, D together that B has to D. llien must A he to C as A, B together is to (7, D together. For •.• A, B together is to 0, D together as B is to D, .'. A, B together is to jB as C, D together is to D, V. 15 and .-.AistoB as C is to D, V. 25. Hence ^ is to as -B is to D. V. 15 But Aj B together is to (7, D together as B is to D. Hyp. .*. J. is to Oas ^, J5 together is to 0, D together. V. 5. Q. «. D. Proposition XXVII. (Eucl. v. 21.) If there he three magnitudes, and other three, which have ths same ratio, taken two and two, hut in a crosa order, tlmi if the first he greater than the third, the- fourth must he greater than the sixth ; and if equal, equal ; and if less, less. Let A, B, C be three magnitudes, and i>, E, F other three, and let ^ be to -B as ^ is to F, and i> be to as D is to E. Then if A be greater than C, D must be greater than F; and if ^.ijnal, equal ; and if less, less. First, if A be greater than C, A has to JB a greater ratio than C has to 5, V. 7. and .*. B has to .^ a greater ratio than G has to B. V. 13. Now •.• 5 is to as i) is to E, Hyp. .-. Ciato Bas E is to D. V. 12. Hence E has to J?' a greater ratio than E has to D, .'.lJ\s greater than F. V. 9, Similarly the other cases may be proved. Q. B. D. Fook v.] PROPOSITIONS NOT CITED IN BOOK VL 239 Proposition XXVIII. (Euc!. v. 23.) If there be any number of magnitudes, cmd as many others, which have the same ratio, taken two and two in a cross order, the first must have to the last of the first magnitudes the same ratio which the first of the others has to the last of these. Let A, B, C be three magnitudes, and D, E, F other three, and let J. be to B as _E^ is to i\ and B be to G as D is to E. Then must A be to as D is to F. Of A, B, D take any equimultiples mA, mB, mD, and of C, E, F take any equimultiples nC, nE, nF. Now •.• J[ is to B as E is to F, .-. mA is to mB as nE is to nF ; V. 11, and V. 5. and •. • B is to as D is to Ey .'. mB is to nC as mD is to nE. V. 17. Hence, if mA be greater than nC, mD is greater than nF ; and if equal, equal ; and if less, less. V. 27. . •. ^ is to as D is to F. V. Def. 6. The proposition may be easily extended to any number of macnitadea. ^ U «^ D. 240 EUCLID'S ELEMENTS. [Book V. Proposition XXIX. (Eucl. v. 25.) If fowr Tnagnitudes of the same kind he 'proportionals, ike greatest and least of them together must he greater thorn, the other two together. Let J. be to J? jis C is to D, and let A be the greatest of the four magnitudes, and conse- quently D the least. V. 18, and V. 14. The7i must A, D together he greater than B, C together. Let A = B, P together^ and C = D, Q together. Then \' B, P together is to B sls D, Q together is to D, .-. P is to jB as Q is to D, V. 25. and B is greater than D. . •. P is greater than Q. V. 14. Hence P, P, D together are greater than Q, P, D togetLta. I. Ax. 4. /. Ap D together are greater than B, C together. 4.X.D. BOOK V.j PROPOSITIONS NOT CITED IN BOOK VI. 241 Proposition XXX. (Eucl. v. 0.) If the first he the sa/rne multiple of the second, or the same svhmulti'ple of it, that the third is of the fourth, the first must be to the second as the third is to the fowrtli. First, let A be the same multiple of B, that (7 is of D. Then must A be to B a^ C is to D. Let A =pB and .\ G = pD. » Take of A and G any equimultiples mA, mCy and of B and Dany equimultiples nB, nD. Then mA = mpB and mG = mpD. V. 3. Now if mpB be greater than nB, mpD is greater than nD ; and if equal, equal ; if less, less. That is, if mA be greater than nB, mG is greater than nD ; and if equal, equal ; and if less, less. r. A is to B a;s G is to D. V. Def. 6> Next, let A be the same submultiple of B, that is of D. Then must A be to B a^ G is to D. For •• • ^ is the same submultiple of B, that is of 2>, ,-. J5 is the same multiple of ^, that X) is of 0, .-. 5 is to -4 as D is to (7, by the first case, and.-. ^istoBafl Oisto D, V. 12. Q. & o. 242 EUCLID'S ELEMENTS, [Book V Proposttton XXXT. (Eucl. v. R) If four magnitudes he proportionals, they must also he pro- portionals hy conversion ; that is, the first must he to its excess above the secoP<^ cts the third is to its excess above the fourth. Let A, B <>ogether fee to B as G, D together is to D. Then must A, B together he to A as G, D together is to G. For -.* 4.,B together is to £ as C, X) together is to D, .'. 4 is to J5 as (7 is to D, V. 25. and .• B is to ^ as i) I? to (7, V. 12 ■ an4 J., B together is to ^ as (/, i> together is to 0. V. 16, (i i:. D. 2- Vd BOOK VI. INTRODUCTORY- REMARKS. The chief subject of this Book is the Similarity ot Kecti- linear Figures. Def. I. Two rectilinear figures are called similar, when they satisfy two conditions : — I. For every angle in one of the figures there'Tnust be a corresponding equal angle in the other. II. The sides containing any one of the angles in one of the figures must be in the same ratio as the sides containing the cor- responding angle in the otherfigure: the antecedents of the ratios being sides which are adjacent to equal angles in each figure. Thus ABC and DEF are similar triangles, if the angles at Aj B, C be equal to the angles at D, E, F, respectively, and if BA be to AG as ED is to DF, and AC be to CB as DF is to FE, and CB be to BAa&FE'm to ED. A c T, r The sides adjacent to equal angles in the triangles are thus homologous, that is, BA^ AC, CB are respectively homologous to ED, DF, FE. It will be shown in Prop. iv. that in the case of triangles the second of the above jjonditions follows from the first. In the case of quadrilaterals and polygons hoth condi- tions are necessary : thus any two rectangles have each angle of the one equal to each angle of the other, but they are not necessarily similar figures. N.B. — The very important Prop. xrv.(Eucl. vi. 33) is indepen- dent of all the other Propositions in this Book, and might be placed with advantage at the very commencement of the Book. 244 EUCLID'S ELEMENTS. [Book VI. Proposition I. Theorem. Triangles of the same altitude are to one another as thei/r bases, A Let the A s ABC, ADC have the same altitude, that is, the perpendicular drawn from A to BD. Then must A ABC be to A ADC as base BC is to base DC. In DB produced take any number of straight lines BG, GH each=BC. 1.3. In BD produced take any number of straight lines DE, KL, LM each=Da L 3. Join AG, AH', AK, AL, AM. Then •/ OB, BG, GB are all equal, .-. AS ABC, AGB, AHG are all equal I. 38. .-. A AHC is the same multiple of A ABC that EC is of BC. So also, A AMC is the same multiple of A ADC that MC is of DC. And A AHC is equal to, greater than, or less than A AMC, according as base HC is equal to, greater than, or less than base MC. I. 38. Now A AHC Sind base HC sue equimultiples of A ABC and base BC, and A AMC and base MC are equimultiples of A ADC and base DC. .-. A ABCis to A ^DO as base BC is to base DC. V. Def. 5 Q. E. D. Book VI.] PROPOSITION I. 245 Cor. I. ParalldograTas of the same altitude cure to one another as their bases. Let ACBE, ACDF be parallelograms having the same alti- tude, that is, the perpendicular drawn from A to BD. Then must 0-40^^ be to CJ ACDF as BG is to DO. Jl A P For n7ACBE=twice A ABG, L 41. and O^0I>jP'= twice A ADG. I.4\. .'.m ACBE is to O ACDF-^s A ^J50is to A ADCy V. 11. md.'.CJAGBEiBtonJACDFa^ BG h to DG. V.5. Q. B. D. Cor. II. Triangles and Parallelogramis, that h>a/oe equal altitudes, are to one another as their hoses. Let the figures be placed, so as to have their bases in the same straight line ; and having drawn perpendiculars from the vertices of the triangles to the bases, the straight line, which joins the vertices, is parallel to that, in which their bases are, because the perpendiculars are both equal and parallel to one another. I. 33. Then, if the same construction be made as in the Proposition, the demonstration will be the same. Ex. 1. ABG, DEF a.re two parallel straight lines ; show that the triangle ADE is to the triangle FBG as DE is to BG. Ex. 2. If, from any point in a diagonal of a pardllelogram, straight lines be drawn to the extremities of the other diagonal, the four triangles, into which the parallelogram is then divided, must be equal, two and two. 246 EUCLID'S ELEMENTS. [Book VI Proposition II. Theorem. If a straight line be drawn pa/rallel to one of the sides of a triangle, it must cut the other sides, or those sides producedy pro- portionaily. Let DE be drawn || to BG, a side of the A ABG. Then must BD he to DA as GE to EA. Join BE, GD. Then •/ a BDE= a GDE, on the same base DE and between the same ||s, DE, BG. I. 37. .-. A BDE is to A ADE as a GDE is to A ADE V. 6. Bui t^ BDE \& to £^ ADE as BD is to DA, VI. 1. and A GDE is to A ADE as GE is to EA ; VI. 1. BD is to DA as GE is to EA. V. 5. Ex. 1. If any two straight lines be cut by three parallel lines, they are cut proportionally. {N.B. — This is of great use.) Ex. 2. If two sides of a quadrilateral be parallel to each other, a straight line, drawn parallel to either of them, shall cut the other sides, or these produced, proportionally. Ex. 3. If two triangles be on equal bases, and between the same parallels, shew that tlie sides of the triangles intercept equal lengths of any straight line, which is parallel to their Book VI.] PROPOSITION IL 247 And Conversely, If the sides, or the sides jyroduced, be cut jyroportiofially, the straight line which joins the points of section must be parallel to the remaining side of the triangle. Let the sides AB, AC oi the A ABC, or these produced, be cut proportionally in D and E, so that BD is to DA as CE is to EA, and join DE. Then must DE be parallel to BG, The same construction being made, ••• BD is to DA as CE is to EA, and BD is to DA as A BDE is to A ADE, VT. 1. and CE is to EA as A CDE is to A ADE, VI. 1. .-. A BDE is to A ADE as A CDE is to A ADE, V. 5. and /. A BDE= a CDE ; V. 8. and they are on the same base DE ; .-. DJK is II to BC. I. 39. Q. E. D. Ex. 4. If there be four parallel straight lines, two of these lines intercepfcnpon two given lines, of unlimited length, OA, OB, parts proportional to the parts' intercepted upon OA, OB, by the remaining two parallel straight lines. Ex. 5. If the four sides of a quadrilateral figure be bisected, the lines joining the points of bisection will form a parallelo- gram. Ex. 6. A quadrilateral figure has two parallel sides : shew that the straight line, joining the point of intersection of its other two sides produced and the point of bisection of its dia^nals, bisects the two parallel sides. 248 EUCLWS ELEMENTS. [Book VI. Proposition III. Theorem. If the vertical angle of a triangle be bisected by a straight line, which also cuts the ba^e, the segments of the base must have the same ratio, which the other sides of the triangle have to one another* o o Let L BAC of A ABO be bisected by the st. line AI), which meets the base in D. ThAn must BD be to DO as BA is to AC. Througii C draw CJE || to DA, I. 31. and let BA produced meet CE in E. Then z ^J^Z>= interior z AEG, I. 29. and z CAD = alternate z ACE, I. 29. But z BAD= L GAD, by hypothesis, and .-. z AEO^ z ACE, Ax. I. and .-. AC = AE. I. b. Oor. Then '.• AD is || to EG, a side of A BEG, .'. BD is to DC as BA is to AE, VI. 2. and .-. BD is to DC as BA is to AG. V. 6 Ex. 1. Shew that in a parallelogram the diagonals do not bisect the angles, unless the sides are equal. Ex. 2. Shew how to trisect a straight line of finite length. Ex. 3. Shew that the bisectors of the angles of a triangle meet in the same point. Ex. 4. The bisectors of the angles A and B, of a triangle ABC, meet the opposite sides in the points D and F : BA ?ind BC are produced to F' and D', so that AF', AC and CD ■ - all equal : prove that F'D' is parallel to FD. Book VI.] PROPOSITION III. i% And Conversely, If the segments of the bane h(we the same ratio, which the other sides of the triangles have to one another, the straight line, dranm from the vertex to the point of section, must bisect the vertical angle. Let BD be to DC as BA is to AG, and join AD. Then must z BAD= l CAD. The same construction being made, •.• BD is to DG as BA is to AC, Hyp. and BD is to DC as BA is to AE, VI. 2. .-. BA is to AC as 5^ is to AE, V. 5. and .-. AC=AE, • V. 8. and .-. z AEC = z ACE. I. a. But z ^jE;0= exterior z BA D, I. 29. and z ^0E= alternate z 0^2), I. 29. .-. z 5vli)= z CAD. Ax. 1. Q. E. D. Ex. 5. Two straight linfes are drawn, bisecting the angles at the base of an isosceles triangle. Shew that the straight line, joining the points, in which they cut the sides, is parallel to the base. Ex. 6. If AD And AE bisect the interior and exterior angles at A, and meet the base BC in D and E, and be the middle point of BC, shew that OD is to 0^ as 0^ is bo OE. 25© EUCLID'S ELEMENTS. [Book VI. Proposition A. Theorem. If the exterior angle of a triangle, be bisected by a straight line, which also cuts the base produced, the segments, between the dividing straight line and the extremities of the base, must have the samie ratio, which the other sides of the triangle have to one another, S Letz EAC, an ext'zof the A ABC, be bisected by the St. line AD which meets the base produced in D. Then must BD be to DC as BA is to AC. Through C draw CF || to DA, meeting AB in F. I. 31. Then z FAD = interior z AFC, I. 29. and z CAD =alternate z ACF, I. 29. But z EAD= L CAD, by hypothesis. .\lAFC=lACF, Ax. 1. and .-. AC=-AF. I. B, Cor. Then •/ AD is |1 to FC, a side of A FBC^ .-. BD is to DC as BA is to AF, VI. 2. and .-. BD is to DC as BA is to AC, V. 6. Ex. 1. If the angles at the base of the triangle be equal, how is the proposition modified ? Ex. 2. If B be any point in a straight line AC, intersected by another, CD, give a geometrical construction fur detenuin- ing a point D in CD, such that AD is to DB as ^0 is to CB, Book VI.] PROPOSITION A, 251 And Conversely, If the segments of the base produced have the same ratio ^ which the other sides of the triangle have to one another, the straight line drawn from the vertex to the point of section must bisect ths exterior angle of the triangle. Let BD be to DC as BA is to AO, and join AD» Then must L CAD= l BAD, For, the same construction being made, •/ BD is to DC as BA is to AG, Hyp. and BD is to DC as BA is to AF, VI. 2. .-. BA is to AG as BA'm to AF, V. 5. and .'. AG=AF, V. 8. and .-. z AFC= l AGF. I. a. But z ^J?'(7= exterior z FAD, I. 29. and z ^Oi^= alternate z GAD^ I. 29. and .-. z 0-dX>= z ^AD. Ax.l. Q1B.D. Ex. 3. If the base be divided into two segments, having the sjime ratio with the segments specified in the Proposition, the straight lines, drawn from the two points of section to the vertex of the triangle, are at right angles to each other. Ex. 4. If the angle, between the external bisector and a side, be equal to the angle, between the external bisector and the base, the perpendicular to the greater side, through the vertex, will bisect the segment of the base, cut off between the bisectin2ktD,E,F respectively. Then must the sides about the equal l s be proportionals, those being homologous sides, which are opposite the equal l s. For suppose a DEF to be applied to a ABC, so that D coincides with A and DE falls on AB ; then •.• /. BAC = z EDF, .: DF will fall on AC. Let G and H be the points in AB and AC, or these pro- duced, on which E and F fall. Join GH. GH will be |1 to BC, v l AGH= l ABC. I. 28. Then BA is to GA as CA is to BA, VI. 2. and .'. BA is to ED as CA is to FD, V. 6. whence BA is to ^0 as ED is to DF. V. 15. Similarly, by applying the A DEF, so that the j^aat F, E may coincide with those at C, B successively, we might show that AChto CBas DF is to FE, and that CB is to BA as FE is to ED. Q. E. D. Ex. Divide a given nngle into two parts, such that the perpendiculars from any point of the dividing line upon the two arms of the angle may be in a given ratio. Book VI.] PROPOSITION V, 253 Proposition V. Theorem. If the sides of two triangles, about each of their angles, be pro ortionahy the triangles must be equiangular to one another, and must have those angles equal, which a/re apposite to the homo- logmis sides. Let the A s ABCy DEF have their sides proportional, so that BA h to AC as ED is to DF, and ^(7 is to CB as DF is to FE, and CB is to BA as FE is to ED. Then must A ABC be equiangular to A EDF, those z s being equal, which are opposite to the homologous sides, that is, L BAC= L EDF,SiTid L ABC= l DEF, and z ACB= l DFE. In AB, produced if necessary, make AG=DE, and draw GH \\ to BC, raeetina AC in H. I. 31. Then A AGH is equiangular to A ABC, I. 29. and .-. BA is to ^0 as GA is to All. VI. 4. But ED is to DF as BA\^ to AC', Hyp. and .-. ED is to D^as GA is to All. V. 5. But ED=GA, and .-. DF=AH. V. 14. So also it may be shown that GH=EF. Then in as AGH, DEF V GA=^ED, and AH=DF, and HG^FE, .: L GAH= L EDF ; z AGH= z i>J^i?' ; z AHG^ l DFE. I. c. ButzG«^7/= zi?^(7; iAGH= lABC; i AHG= i ACB, .'. L BAC== L EDF ; l ABC^ l DEF, and z ^CB= z /).F7?. (>. K, IJ. 254 EUCLID'S ELEMENTS. [Book VI. Proposition VI. Theorem. If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles ]f)roportionals, the triangles must be equiangular to one another, and must have those angles equal, which are opposite to the homologous sides. In the AS ABC, DEF, let i BAC= a EDF, and let BA he to AG as ED to DF. Then must a ABC be equiangular to A DEF, and L ABG= l DEF, and l ACB== l DFE. In AB, produced if necessary, make AG=DEf and draw GH II to BG. I 31. Then a AGE is equiangular to A ABC, I. 29. and .-. GA is to AH as BA is to AG, VI. 4. and .-. GA is to AH as ED is to DF. V. 5. But GA =ED, by construction, anA.:AH=DF. V.Y'i Then •.• GA =ED, and AH=DF aiiA l GAH= l EDF; .-. z AGH= L DEF, and z AHG= l DFE, I. 4. and .-. z ABG= z DEF, and z AGB^ z DFE. Q. E. D, l:^x. 1. If from B, G, the extremities of the base of a triangle ABC, be drawn BD, GE, perpendicular to the opposite sides, shew that the triangles ADE, ABG are equiangular. Ex. 2. A variable chord OP is drawn through a fixed point on the circumference of a circle, and Q is taken in it, so that the rectangle OP, OQ is constant, find the locus of Q. BookVL] EXERCISES ON PROPOSITIONS L-VL 255 Miscellaneous Exercises on Props. I. to VI. 1. If two triangles stand on the same base, and their vertices be joined by a straight line, the triangles are as the parts of this line intercepted between the vertices and the base. 2. If a circle be described on the radius of another circle as its diameter, and any straight line be drawn through the point of contact, cutting the two circles, the part, intercepted between the greater and lesser circles, shall be equal to the part within the lesser circle. 3. The side BG, of a triangle ABC^ is bisected in D, and any straight line is drawn through D, meeting AB, AC, pro- duced if necessary, in E, F, respectively, and the straight line thi-ough A, parallel to BC, in G. Prove that DE is to DF a&GE'mto GF. 4. If the angle A, of the triangle ABC, be bisected by AD, which cuts BC in D, and be the middle point of BC, then OD bears the same ratio to OB that the difference of the sides bears to their sum. 5. The diameters of two circles and the distances between their centres are as the numbers 5, 4, 3 ; find the proportionate distances between the points of intersection of their common tangents. 6. If P, E be points in the sides AB, AG respectively of the triangle ABC, such that the triangles DAG, EAB are equal, shew that the sides AB, AG are divided proportionally in D and E. 7. If two of the exterior angles, of a triangle ABC, be bisected by the lines GOE, BOD, intersecting in 0, and meet- ing the opposite sides in E and D, prove that OD is to OB as AD is to DB, and that OC is to OE as ^0 is to AE. 8. B, G, the angles at the base of an isosceles triangle, are joined to the middle points, E, F, of AB, AG, by lines inter- secting in G. Shew that the area BGG is equal to the area AEFG. 9. If, through any point in the diagonal of a parallelogram, a straight line be drawn, meeting two opposite sides of the figure, the segments of this line will have the same ratio as those of the diagonal. 10. The sides AB, AG, of a given triangle ABC, are pro- duced to any points D and E, and the straight line DE is divided in F, so that DF is to FE as BD ia to CE ; shew that the locus of F is a straight line. 2S6 EUCLID'S ELEMENTS. [Book VI. Proposition VII. Theorem. If two triangles have one angle of the one equal to one angle of the other, and the sides about a second angle in each propor- tionals ; then, if the third angles in each be both acute, both obtuse, or if one of them be a right angle, the triangles must be equiangular to one another, and must have those angles equal, ai)out which the sides are proportionals. In the AS ABC, DEF, let l BAG= i EDF, and let AB be to BG as DE is to EF, and let z s ACB, DFE be both acute, both obtuse, or let one of them be a right angle. Then must A s ABC, DEF be equiangular to ons another, having i ABG= l DEF, and l AGB= l DFE. For if L ABG be not= z DEF, let one of them, as z ABC, be greater than the other, and make z ABG= z DEF, I. 23. and let BG meet -40 in G. Then '.• z BAG= z EDF, and z ABG= z DEF, .'. A ABG is equiangular to A DEF, I. 32. and .-. AB is to BG as DE is to EF. VIA. But AB is to BG as DE is to EF, Hyp. .-. AB is to BGasABis to BG, V. 5. ' and .-. BG=BG, V. 8. and .-. z BGG= z BGG. L a. Book VI.] PROPOSITION VII. 257 First, let l AGB and l DFE be both acute, then z AGB is acute, and .•. i BGC is obtuse ; I. 13. .". L BGG is obtuse, which is contrary to the hypothesis. Next, let L AGB and z DFE be both. obtuse, then z AGB is obtuse, and .". z BGG is acute ; I. 13. .*. i BGG is acute, which is contrary to the hypothesis. Lastly, let one of the third Ls> AGB, DFE be a right z . If z JOjBbeart. z, then z BGG is also a rt. z ; I. A. .*. z s BGG, BGG together = two rt. z s, which is impossible. L 17. Again,, if z DFE be a rt. z , then z AGB is a rt. z , and /. z BGG is a rt ^ . I. 13. Hence z BGG is also a rt. z , La. and .'. z s BGG, BGG together = two rt. z s, which is impossible. I. 17. Hence z ABG is not greater than z DJ^jP. So also we might shew that z DEF is not greater than I ABG, :.L ABG=A DEF, and .'. z ^0.5 = z DJ?'J^. ' I. 32. Q. E. D. N.B. — This Proposition is an extension of Proposition e of Book I. p. 42. Note. — We have made a Blight change in Euclid's arrange- ment of the four Propositions that follow, because Eucl. vj. 8 is closely connected with xiiQ proof of £ucL vl 13. 258 EUCLID'S ELEMENTS. [Book VI. Proposition VIII. Problem. (Eucl. vi. 9.) Frmn, a given straight line to cut off any submultiple. Let ABhe the given st. line. It is required to shew how to cut off any submultiple from Ah. From A draw AC making any angle with AB. In ^0 take any pt. D, and make AG the same multiple oi A D that AB is of the submultiple to be cut off from it. Join BG, and draw DE \\ to BG, I. 31. Then '.• ED is || to BG, .*. GD is to DA as BE is to EA, VI. 2. and .-. GA is to DA as BA is to EA. V. 16. /. EA is the same submultiple of BA that DA is of GA. V. 19. Hence from AB the submultiple required is cut off. Q. B. F. Ex. 1. Cut off one-seventh of a given straight line. Ex. 2. Cut off two-fifths of a given straight line. Note. — This Proposition is a particular case of Proposi- ticn IX. Book VI.] PROPOSITION IX. 259 Proposition IX. Problem. (Eucl. vi. 10.) To divide a given straight liv£ similarly to a given stra/ight line, A Let ^B be the st. line given to be divided, and AO the divided st. line. It is required to divide AB similarly to AC. Let ^C be divided in the pts. D, E. Place AB, ^(7 so as to contain any yngle. Join BC, and through D, E draw DF, EG \\ to BQ. L 31. Through D draw DEK II to AB. L 31. Then '.• FlI and GK are Os, .-. FG=I)H, and GB=HK. I. 34. And •.• HE is ll to KC, .'. KB is to HD as CE is to ED, VL 2. that is, BG is to (?i^ as CE is to J&'D. Again, '.• FD is II to GE, .'. GF is to FA as ED is to DA. VL 2. Hence J.B is divided similarly to AC. Q. E. F. Ex. 1, Produce a given straight line, so that the whole pro- duced line shall be to the produced part in a given ratio. Ex. 2. On a given base describe a triangle, with a given vertical angle and its sides in a given ratio. 26o EUCLID'S ELEMENTS. HBook VI. Proposition X. Problem. (Eucl. vi. 11.) To find a third proportional to two given straight lines, A. Let AB and J.Cbe the given st. lines. It is required to find a third proportional to AB, AG. Place AB, AC so as to contain any angle. Produce AB, AC to D and E, making BD=AC. I. 3. Join BC, and through D draw DE || to BC. I. 31. Then '.• i^O is || to Dj^, .-. AB is to BDasAG is to OE, VI. 2. and .-. AB is to AC as JO is to CE. V. 6. 'J'hus CE is a third proportional to AB and J (7. Note. This Proposition is a particular case of Proposition xi. Def. II. When three magnitudes- are proportionals, the first is said to have to the third the duplicate ratio of that, which it has to the second. Thus here AB has to CE the duplicate ratio of AB to AG. Def. III. When three magnitudes are proportionals, the first is said to have to the third the ratio compounded of the rai^f), which the first has to the second, and of the ratio, which t^ ^^ second ha* to the third. Thus here AB has to CE the ratio compounded of the ratios of AB to A C and A C to CE. Book VI.} PROPOSTTION XT. 26t Proposition XI. Theorem. (Eucl. vi. 12.) To find a fourth proportioned to three given straight Unes. Let Af B, Che the three given st. lines. It is required to find a fourth proportional to A, B, C. Take DE, DF, two st. lines making an z EDFj and in these make DG=A, GE=B, and DH= 0, I. 3. and through E draw EF II to GH. I. 31. Then, •.• GH is || to EF, .-. DG is to GE as DH is to HF, VI. 2. and .-. A is to jB as is to HF. V. 6. Thus fl!F is a fourth proportional to A, B, G. Q, E. F. Ex. ABC is a triangle inscribed in a circle, and jBT) is drawn to meet the tangent to the circle at A in D, at an angle ABD equal to the angle ABC. Show that -4C is a fourth proportional to the lines BD, DA, AB, 262 EVCLW'S ELEMENTS. [Book VI. Proposition XII. Theorem. (Eucl. vi. 8.) In a right-angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle and to one another. B n ^ Let ABChe a right-angled A, having A BAG a rt. z , and from A let AD be drawn ± to BC. Then must As DBA, DAG be similar to A ABG, and to each other. For ••• rt. z jBD-4=rt. z BAG, and z ABD= z GBA, .'. lDAB= lAGB. 1.32. /. A DBA is equiangular, and .*. similar to A ABG. VI. 4 In the same way it may be shown that A DAG is equiangular, and .*. similar to A ABG. Hence A DBA is similar to A DAG. Q. E. D. Cor. I. DA is a mean proportional between BD and DG, For BD is to DA as DA is to DG. VI. 4. Cor. II. BA is a mean proportional between BG and BD, For BG is to BA as BA is to BD. VI. 4. Cor. III. CA is a mean proportional between BG and CD, For BG is to 0.4 as CM is to GD. VI. 4. Ex. j5 is a fixed point in the circumference of a circle, whose centre is G ; PA is a tangent at any point P, meeting GB pro- duced in A, and PD is drawn perpendicularly to GB. Prove that the line bisecting the angle jIPD always passes through B. Book. VI.] PROPOSITION XIII. 263 Proposition XIII. Problem. To find a mean proportional between two given straight •n Let AB and i?<7 be the two given st. lines. It is required to find a mean proportional hetwe&n. AB and EC. Place AB and BC so as to make one st. line AG^ and on ^0 describe the semicircle ADO. From B draw BD± to AG, and join AD, GD. I. 11. Then '.• z ADG is a rt. z , III. 31. and DJ5is±to AG, .'. DB is a mean proportional between AB and BG. VI. 12, Cor. 1. Q. E. F. Ex. 1. Produce a given straight line, so that the given line may be a mean proportional between the whole line and the part produced. Ex 2 Shew that either of the sides of an isosceles triangle is a mean proportional between the base and the half of the segment of the base, produced if necessary, which is cut off by a straight line, drawn from the vertex, at right angles to the equal side. Ex. 3. Shew that the i/ 10 to B£ AnQBiB to BJf, Book VI.] PROPOSITION XIV. £65 Proposition XIV. Theorem. Uqual parallelograms, which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional. ji^ F \ . Let AB, BC be equal Os, having l iBD=^ l EBG. Then must DB he to BE as GB is to BF. Place the Os so that DB and BE are in the same st, line ; then must GB and BF also be in one st. line. I. 14. Complete the ZZ7 FE. Then '.• O AB = LJ BG, and FE is another O, .-. O AB is to CJ FEdsH/ BOis to O FE. V. 6. But as ZZ7 AB is to O FE so is DB to BE, VI. 1, Cor. 1. and as O BC is to £7 FE so is GB to BF. VI. 1, Cor. I. .-. DB is to BE as GB is to BF. V. 5. And Conversely, Farallelograms, which have one angle of the one equal to one angle of the other, and their sides about the equal angles recipro cally proportional, are equal to one another. Let the sides about the equal z s be reciprocally proper tional, that is, let DB be to BE as GB is to BF. Then must CJ AB^CJBC. For, the same construction bcin^ made, •.• DB is to BE as GB is to BF, and that DB is to BE as O AB is to HJ FE, VI. 1, Cor. I. and that GB is to BF as O BG is to O FE, VI. 1, Cor. I. .-. O AB is to O i^^ as CJ BG ia to LJ FE. V. 5. and .-. £7 AB=CJ BC. V. 8 Q. K. p. 266 EUCLID'S ELEMENTS, [Book VI. Proposition XV. Theorem. Equal triangles, which have one angle of the one equal to one angle of the other, have tluir sides ahov,t the equal angles recip- rocally proportional. Let A BC, ABE be equal as, having z J5J.C= z DAE, Then must CA he to AD as EA is to AB, Place the a s so that CA and AD are in the same st. line ; then must EA and AB also be in one st. line. I. 14. Join BD. Then V A ABC= A ADE, and ABD is another A , /. A A BC is to A ^ BD i\s a ADE is to A ABD. V. 6. But as A A BC is to a ABD so is CA to AD, VI. 1. and an A ADE is to A ABD so is EA to AB. VI. 1. .-. CA is to AD as EA is to AB. V. 5. Ex. 1. Shew that, provided the sides of one of the triangles be made the extremes, it is indifferent, so far as the truth of the Proposition is concerned, in what order the sides of the other triangle are taken as the means of the four pro- portionals. Ex. 2. ABb, AcG are two given straight lines, cut by two others BC, be, so that the two triangles ABC, Abe may be equal ; shew that the lines BC, be divide each other propor- tionally. Book VI.] PROPOSITION XV. 267 And Conversely, Triangles, which have one angle of the one equal to one a%gle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another. Let the sides about the equal z s be reciprocally proportional, that is, let CA be to AD as EA is to AB. Then must A ABG= A ADE, For, the same construction being made. *.• GA is to AD as EA is to AB, and that CA is to AD as a ABC is to A ABDj VI. 1. and that EA is to ^JB as A ADE is to a ABD, VI. 1. .-. A ABC is to A ABD as a ADE is to A ABD. V. 6. and . •. A ABC== A ADE. V. 8. Q. E. D. Ex, 3. Through the extremities of the base BC, of a triangle ABC, draw two parallel lines, BE and CD, meeting AC and AB produced in E and D respectively, so that BCD may be equal in area to ABE. Ex. 4. P is any point on the side AC, of the triangle ABC; CQ, drawn parallel to BP, meets AB produced in Q ', AN, AM are mean proportionals between AB, AQ, anH AC, AP, respectively. Shew that the triangle AJsM is equal fco the triangle ABC. 268 EUCLID'S ELEMENTS. [Book VI, Proposition XVI. Theorem. If four straight lines he proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means. ._7f -JI M Let the four st. lines AB, CD, EF, GH be proportionals, so that AB is to CD as EF is to GH. Then must reel. AB, GH^^rect. CD, EF. Draw ^ M" ± to AB, and GN± to CD ; I. 1 1 . and make AM==GH, and CN^EF; and complete the ZZ7s BM, DN. I. 31 . Then •/ AB is to CD us EF is, to GH, and that EF=CN, and GH=AM, .-. AB is to CD as CN is to ^if. V. 6. Thus the sides about the equal z s of the equiangular Os BM, DN are reciprocally proportional, and .-. O BM=CJ DN ; VI. 14. that is, rect. AB, ^M=rect. CD, GN. .: rect. J.5, GH =rect. OX), J^i^. Ex. 1. If E be the middle point of a semicircular arc AEB, and EDC be any chord, cutting the diameter m D, and the circle in C, prove that the sqiiare on CE is equal to twice the quadrilateral AEBC. Book VI.1 PROPOSITION XVI. 269 And Conversely, If the rectangle contained hy the extremes he equal to the rect- angle contained by the means^ the Jour straight lines are pro- portionals, liBt rect. AB, GH=rect. CD, EF. Then must AB he to CD as EF is to QH, For, the same construction being made, •.• rect. AB, GH =rect. CD, EF, .'. rect. AB, AM^rect. CD, CN, that is, CJ BM^CJ DN. and these Os are equiangular to one another, and .'. the sides about the equal z s are reciprocally proportional, VI. 14. and .*. AB is to CD as CN is to AM, and .-. AB is to CD as EF is to GH. V. 6. Q. E. D. Ex. 2. If, from an angle of a triangle, two straight lines be drawn, one to the side subtending that angle, and the other cutting from the circumscribing circle a segment, capable of containing an angle, equal to the angle, contained by the first drawn line and the side, which it meets ; the rectangle, con- tained by the sides of the triangle, shall be equal to the rect- luigie, containec^ by the iiucs thui> dittwa. 270 EUCLID'S ELEMENTS. [Book VI. Proposition XVII. Theorem. If three straight lines he proportionals, the rectangle contained hy the extremes is equal to the square on the mean. Let the three st. lines A, B, be proportionals, and let J. be to JB as ^ is to G. Then must red. A, C=sq. on B. Take D=B. Then •.* ^ is to jB as JB is to C, .: A is to J5 as D is to (7, * V. 6. and .-. rect. A, (7=rcct. B, D, VI. 16. that is, rect. A, 0=sq. on B. And Conversely, If the rectangle contained hy the extremes be equal to the square on the mean, the three straight lines are proportionals. Let A, B, Che three straight lines such that rect. A, 0=sq. on J5. Then must A he to B as B 'is Co 0, For, the same construction being made, *.* rect. A, 0=sq on B, and B=D, .'. rect. A, C'=rect. B, D ; and .-. ^ is to B as D is to 0, VL 16. that is, ^ is to j5 as B is to O. V. G. Q. K. T>. Book VI.] PROPOSITION XVIII 271 Proposition XVIII. Problem. Upon a given straight line lo describe a rectilinear figure similar and similarly situated to a given rectilinear figure. Let AB be the given st. line, and CDEF the given rectil, fig. of four sides. It is required to describe on AB a fig. similar and simila/rly situated to CDEF. Join DF, and at A and B, make z BAG = z DGF, and z ABG= L CDF ; then A BAG is equiangular to a DCF. At G and B, make z 5Gfir= z DJ'^, and z Gj5^= z JPl^^: ; then A 0//5 is equiangular to A FED. Then •.• z ^(?5= z 0#A and z 5(?ir= z X>^^, .•.iAGH= iCFE. Ax. 2. So also z ABE= z OD^. And we know that i BAG = i DCF, and that z (JIT^ = z FED, .'. rectil. fig. ABHG is equiangular to fig. CDEF. Also, •.* A BAG is equiangular to A DCF, .: BA is to ^6? as DO is to CF ; VI. 4. and •.* aBGH is equiangular to a DFE, .'. GB is to air as FD is to ^^. VI. 4. Also, ^(? is to Gf^ as CF is to 2?'D. .: AG is to GH Bs CF is to FE. V. 21. Similarly, it may shown that GH is to B^B as FE is to ^D, and that HB is to BA as ^D is to DC. .-. the rectil. figs. ABHG and CDiZ-F are similar. \ 272 EUCLID'S ELEMENTS. TBook Vi. Next. Let it he required to describe on AB a fig., similar and similarly situated to the rectil. fig. GDKEF. Join DE, and on AB describe the fig. ABHG^ similar and similarly situated to the quadrilateral CDEF. At B and H make z HBL= i EDK, and z BHL= L DEK ; then A HLB is equiangular to A EKD. Then •/ the figs. ABHG, CDEF are similar, .-. z GHB= L FED ; and we have made z BHL= z DEK ; .-. whole z GHL=\v\\o\e z FEE. Ax. 2. For the same reason, z ABL= z CDZ^. Thus the fig. AG HLB is equiangular to fig. CFEKD. Again, '.• the figs. J(?^J5, OFi/D are similar, .-. GB is to HE as J^J5; is to ED : also we know that HB is to HL as Ei) is to EK, VI. 4. .-. (5^JT is to HL as jPjE: is to EK. V. 21. For the same reason, AB is to BL as OD is to DK. ■ And BL is to iif as DK is to iTj^ ; VI. 4. .-. the five-sided figs. AGHLB, CFEKD are similar. In the same way a fig. of six or more sides may be described, on a given line, similar to a given fig. Q. E. F. Bock VI.1 PROPOSITION XJX. Proposition XIX. Theorem. Similar triangles are to one another in ths duplicate ratio o^ th^ir homologous sides. Let ABC, DEF be similar as, having z s at A, B, C= i s at D, E, F respectively, so that BG and EF are homolog-ous sides. Then must a ABC have to a DEF the duplicate ratio o; (hat which BC has to EF. Suppose A DEF to be applied to a ABC, so that E lies on B, ED on BA, and .-. EF on BG. Let P and Q be the pts. in BA, BG on which D and F fall. Join AQ. Then a ABC is to a ABQ as BC is to BQ, VL 1. and A ABQ is to a PBQ as AB is to BP. VI. 1. Bat JJ5 is to J5P as BC is to BQ, VI. 4. .-. A ABQ is to A PBQ as £0 is to BQ. V. 5. Hence a ABC is to a ^£^ as a ABQ is to a PBQ. V. 5. .'.A ABC has to A P5Q the duplicate ratio of A ABC to A ABQ ; VL Def. 2. .-. A ^Ul.' had 10 A PBQ the duplicate ratio of BG to BQ. V. 5. that is, A ABC has to A DEF the duplicate ratio of BG to jSJ". Q. E. D. Cor. If MN be a third proportional to 5(7 and EF, BC has to MN the duplicate ratio of PC to EF, VL Def. 2. and .'. BC is to MN as a ^PO is to a DEF. 274 EUCLID'S ELEMENTS, [Book VI. Exercises on Proposition XIX. Ex. 1, Prove this Proposition without drawing any lin« inside either of the triangles. Ex. 2. In the figure, if 50 be equal to FD, shew that the triangles will be in the ratio of AB to JEF. Ex. 3. Cut ojff the third part of a triangle by a straight lins parallel to cjpe of its sides. Ex. 4. AB, AG are bisected in D and E. Prove that the quadrilateral DBGE is equal to three times the triangle ADE. Ex. 5. ABC is a line passing through the centre of the circle BCD, and AD a tangent to the circle. If CE be drawn parallel to BD, shew that the triangles ACD, AGE are to one another -ds AC to AB. Ex. 6. A straight line drawn parallel to the diagonal BD of a parallelogram A BCD meets AB, BC, CD, DA, in E, F, G, H. Prove that the triangles AFG, CEH are equal. Ex. 7. If two triangles have an angle equal, and be to each other in the duplicate ratio of adjacent sides, they are similar. Ex. 8. The circle B'C (centre 0) touches the circle ABC in- ternally, and AB'B touches B'C in B'. Shew that if BD be perpendicular to the common diameter, AB, B' divides AB into segments, which are in the duplicate ratio of OG to OD. Ex. 9. From the extremities A, B, of the diameter of a circle, perpendiculars AY, BZ, are let fall on the tangent at any point G. Prove that the areas of the triangles AGY, BCZ are together equal to that of the triangles ACB. Ex. 10. If to the circle, circumscribing the triangle ABC, a tangent at G be drawn, cutting AB produced in D, shew that AD is to DB in the duplicate ratio of ^ C to GB. Ex. 11. Construct a triangle which shall be to a givei) triangle in a given ratio. Book VI. j rjvUJ uoll/UA' X.'.. 275 Proposition XX. Theorem. (Eucl. vi. 21.) Rectilinear figures^ which are similar to the same rectilineaf figure^ are also similar to each other. Let each of the rectilinear figures A and B be similar to the rectilinear figure C. Then must the figure A be si7)iilar to the figure B. For '.• A is similar to C, .: A is equiangular to G, and A and C have their sides about the equal z s pro- portionals. VI. Def. 1. Again, *.* B is similar to G, .'. JB is equiangular to (7, and B and G have their sides about the equal z s pro- portionals. VI. Def. 1. Hence A and B are each equiangular to G, and have the sides about the equal z s of each of them and of G pro« portionals. .*. A is equiangular to i?, Ax 1. and A and B have their sides about the equal z s pro- portionals. V. 5. .'. the figure A is similar to the figure B. VI. Def. 1. Q. £. D. 276 EUCLID'S ELEMENTS. [Book VI. Proposition XXI. Theorem. (Eucl. vi. 20.) Similur polygons may he divided into the same number of simMar triangles, having the same ratio to one another, which the i^olygons have ; and the polygons are to one another in the duplicate ratio of tlieir hom,ologous sides. A F n c ^< -^ Let ABODE, FGHKL be similar polygons, and let AB be the side homologous to FG. I. The polygons mny he divided into the same numher of similar As. II. Thcf^e A s have each to each the same ratio ivhich the poly- gons have. III. The polygon ABODE has to the polygon FGHKL the duplicate ratio of that u-Jiich the side AB has to the side FG. Join BE, EG, GL, LH : then I. '.* the polygon ABODE is similar to the polygon FGHETL, .: L BAE = / GFL, and BA \s.\.q AE as GF is to FL, :. A ABE is similar to A FGL. VI. 6 and 4. and .-. L ABE = z FGL. VI. Def. 1. Again, '.• the polygons are similar, .-. L ABO = z FGH, VI. Def. 1. and . •. z EBO = z LGH ; Ax. 3. and '.• the AS ABE, FGL are similar, .-. EB is to -AB as LG is to FG ; VI. Def. 1. also, '.' the polygons are similar, .-. AB is to BO as FG is to GH ; VI. Def. 1. and .'. EB is to BO as LG is to GHy V. 21. and .-. since z EBO = z LGH, the A EBO is similar to a LGH. VI. 6 and 4. For the same reason the A EOD is similar to A LHK. Thus the polygons are divided into the same number of similar As. 300k VI.] PROPOSITION XXI. 277 II. •.' A ABE is similar to A FGL^ .'. A ABE has to a FGL the duplicate ratio of BE to GL. VI. 19. So also, A EBG has to A LGH the duplicate ratio of BE to GL. VI. 19. .-. A ABE is to A FGL as A EBG is to A LGH. V. 5. Again, *.* A EBG is similar to a LGHj .'. A S^O has to A LGH the duplicate ratio of EC to LH. VL19. So also, A J^JOD has to a LHK the duplicate ratio of EC to LH. VI. 19. .-. A EBG is to A LGH as A ^OD is to A LHK. V. 5. But A EBG is to A LGH as a ^5^ is to a FGL. .'. as A ABE is to a FGL so is a EBG to a iGfJT, smd A ECD to £, LHK. Now as one of the antececjents is to one of the consequents so are all the antecedents together to all the consequents together, V. 10. and .*. A ABE is to A FGL as polygon ABCDE is to polygon FGHKL. III. Since t^ ABE has to A FGL the duplicate ratio of AB to FG, VI. 19. .*• polygon ABCDE has to polygon FGHKL the duplicate ratio of AB to FG. V. 5. Q. E. D. Cor. I. In like manner it may be proved, that similar figures of four or any number of sides, are to one another in the duplicate ratio of their homologous sides : and it has been already proved for triangles, vi. 1!). Therefore, universally, similar rectilinear figures are to one another in the duplicate ratio of their homologous sides. 278 E UCL ID S EL EMENl S. [Book VI. Cor. II. If M'N be a third propurtioual to AB and iG^ AB has to MN the duplicate ratio of AB to FG, VI. Def. 2. and .', AB is to MN as the figure on AB to the similar and similarly described figure on FG ; that being true in the case of quadrilaterals and polygons, which has been already proved for triangles. VI. 19 Cor. Proposition XXII. Theorem. (Eucl. vi. 31.) In right-angled triangles, the rectilinear figure, described ujMn the side opposite to the right angle, is equal to the similar and similarly described fi^gv/res upon the sides containing tlie righi angle. Let ABG be a right-angled a , having the right z BAC. Then must the rectilinear figure, described on BG, be equal to the similar and similarly described figures on BA, AG. Draw AD ± to BG. Then A ABG is similar to A DBA, VI. 12. and .-. BG is to BA as BA is to BD, VI. 4. and .*. as BG is to BD so is the figure described on BG to the similar and similarly described figure on BA, VI. 21, Cor. 2. and .'. as BD is to BG so is figure on BA to figure on BG. V. 12. For the same reason as DG is to BG so is figure on ^C to figure on BG. Hence as BD, DG together are to J50 so are figures on BA^ AG together to figure on BG. V. 22. But BD, DG together are equal to BG, and .*. figures on BA, AG together = figure on BG. V. 18. Q. E. D. Book Vl.j PROPOSITION XXIII 279 Note. — The Proposition which follows is not given by Euclid, but is necessary to the proof of Prop, xxiv. Proposition XXIII. Theorem. If two rectilinear figures he equal and also TvmilaTf Ihevr homologous sides must be equal, each to each. Let the rectil. fig>^. A BGDE, FGRKL be equal and similar, and let DC and KU be homologous sides of the figures. Then must DC=KH:. For, if not, let DC be greater than KH. Then '.• DC is to DE as KH is to KL, .*. DE is greater than KL. V. 14. Hence MlKLH be applied to L DEC, so that KR falls on DC and KL on DE (tor z HKL = z CDE\ EL will fall entirely within a D^O, .'. A RLE is less than a DEC. But V A DEC is to A KLE as figure ABCDE is to figure FGHKL, VI. 21. and figure ABCDE = figure FQEKL ,\adec=aklh, V. la or the greater = the less, which is impossible. .". DC is not greater than KH. Similarly it may be shown that DC is not less tljan KH, .'. DC=KH. Q. £. D. 28o EUCLID'S ELEMENTS. [Book VI Proposition XXIV. (Kucl. vi. 22.) If jour straight lines he proportionals, the similar recti linear figures similarly described upon them must also he pro- portionals. A 3f iV Let the four straight lines AB, CD, EF, GH be propor- tionals, that is, AB to CD as JEF is to GE ; and upon AB, CD let the similar rectilinear figures KAB, LCD be similarly described ; and upon FF, GH the similar rectilinear figures MF, NH in like manner. Then 7nust KA B he to LCD as MF is to NH. To ABf CD take a third proportional X and to FF, GH take a third proportional 0. VL 10. Then •.• AB is to CD as FF is to GH, .-. CD is to X as GH is to 0, V. 5. and .-. AB is to X as EF is to 0. V. 21. But as AB is to X so is KAB to LCD, VI. 21, Cor. 2. and as EF is to so is ifi^ to NH. VL 21, Cor. 2. /. £"^45 is to LCD as Mi'' is to NH. V. 5. BookVI.1 PROPOSITION XXIV. 28 1 And Conversely, If the similar figures, similarly described on four straight lines, he proportionals, those straight lines must he proportionals. The same construction being made, let KAB be to LCD as MF is to NH, then must AB he to CD as EF is to GH. Make as AB to CD so FF to PR, VI. 11. and on PR describe the rectilinear figure 8R, similar and simi- larly situated to either of the figures MF, NH. VI. 18. Then, by the first part of the proposition, KAB is to LCD as MF is to SR. But KAB is to LCD as ikTF is to NH. Hyp. .'.SR=]SIH, V.8. Also, SR and NH are similar and similarly situated, s^d.:PR=GH. VI. 23. Now AB is to CD as FF is to PB, and .\ AB \b to CD B3 EF iB to ^H, V. 6. 282 EUCLID'S ELEMEXrs. [Book VI. Proposition XXV. Theorem. (Eiicl. vi. 33.) In equal circles, angles, whether at the centres or the circum- ferences, have to one another the same ratio as the arcs which subtend them ; and so also Jiave the sectors. In the equal ©s ABG, DBF let the z s BGC, ERF at the centres, and the z s BA C, EDF at the circumferences, be sub- tend v3d by the arcs BC, EF. Then I. z BGC must he to z EIIF as arc BC is to arc EF. Take any number of arcs CK, KL, each — BO. and any number of arcs FM, MN, NB ea.ch=E(P. Then •/ arcs BC, CK, KL are all ecjuai, .-. z s BGC, CGK, KGL are all equal. III. 27. .*, z BGL is the same multiple of z BGC that arc BL is of arc BC. So also, z EHB is the same multiple of z EIIF that arc ER is of arc EF. And z BGL is equal to, greater than, or less than . EHB, according as arc BL is tqual to, greater than, or less than arc ER. III. 27. Now z BGL and arc BL are equimultiples of z BGC andarc BC, and z EHR and arc ER&re equimultiples of z EJIFandarcEF. .-. z BGO is to z EHF as arc BC is to arc ^2^. V. Def. 5. Book VI.] PROPOSITION XXV. 28 j II. L BAC must be to I EDF as arc BC is to arc EF. For •.• I BGC= twice l BAC, and i EI1F= twice z EDF, III. 20. /. z i?^ is to z EDF asiBGC is to z EHF, V. 11. and .-. z jB.40 is to z Ei)i^ as arc BC is to arc EB\ V. 5. III. (Sector BGG must be to sector EHF a^ arc BC is to arc EF. For sectors BGC, CGK, KGL are all equal, III. 26, Cor. and sectors EHF, FHM, MHN, NHB, are all equal, III. 26, Cor. /. sector BGL is the same multiple of sector BGG that arc BL is of arc BC, and sector EHR is the same multiple of sector EHF that arc EB is of arc EF ; also, sector BGL is equal to, greater than or less than sector EHR, according as arc BL is equal to, greater than, or less than arc ER, III. 26. and .*. sector BGC is to sector EHF as arc BC is to arc EF. Q. E. D. Cob, In the same circle, angles, whether at the centres 01 the circumferences, have the same ratio as the arcs which sub- tend them ; and so also have the sectors. 284 EUCLID'S ELEMENTS. [Book VI. Proposition B. Theorem. If an angle of a triangle he bisected by a straight line, which likewise cuts tJie base ; the rectangle, contained by the sides of the triangle, is equal to the rectangle, contained by the segments of the base, together with the square on the line bisecting the angle, A Let z B AC of the a ABC be bisected by the st. line AD. Then reel BA, AG = reel BD, DC together with sq. on AD. Describe the © ABC about the A , III. b. p. 135. produce AD to meet the Oce in E, and join EC. Then •.• z BAD = z CAE, Hyp. and /. ABD = l AEC, in the same segment, III. 21. .*. A ABD is equiangular to A AEG. I. 32. ••. BA is to AD as EA is to AG. VI. 4. A feet BA, AC=rect. EA, AD, VI. 16. =recfc. ED, DA together with sq. on AD. II. 3. wcrect. BD, DO together with sq. on AD. III. 35 q. E. D. Bock VI.] PROPOSITION C. 28$ Proposition C. Theorem. If from any angle of a triangle, a straight line he drawn 'per- pendicular to the base, the rectangle, contained by the sides of the triangle, is egual to the rectangle, contained by the per- pendicular and the diameter of the circle described about the triangle. Let ABC be a A , and AD the ± from A to BC. Describe the © ABC about the A ABC, HI. B. draw the diameter AE, and join EC. Then must rect. BA, AC = rect. EA, AD. For •.* rt. z BDA = l EC A, in a semicircle, III. 31. and L ABD = l AEC, in the same segment, III. 21. /. A ABD is equiangular to the A AEC. I. 32. .-. BA is to AD as EA is to AC, VI. 4. and .-. root. BA, ^a=rect. EA, AD. VI. 16. Q. E. D. Ex. 1. Shew that the rectangle contained by the two sidea can never be less than twice the triangle. Ex. 2. ABC'viZ. triangle, and AM the perpendicular upon BC, and P any point in BC ; if 0, 0' be the centres of the circles described about ABP, ACP, the rectangle AP, BC is double of the rectangle of AM, OC/. Ex. 3, A bisector of an angle of a triangle is produced to meet the circumscribed circle. Prove that the rectangle, con- tained by this whole line mid the part of it within the triangle, is equal to the rectangle cont-^iiied by the two sides. 2S6 EUCLID'S ELEMENTS. [Book VI. PROPOSITION D. Theorem. T)ie. rectangle, contained by the diagonals of a quadrilateral inscribed in a circle, is equal to the sum of the rectangleSy con- tained by its opposite sides. Let ABCD be any quadrilateral inscribed in a ®, Join AC, BD. Then rect. AG, BD=rect. AB, CD together with red. AD, BG. Make iABE== I DBG ; I. 23. and add to each the i EBD. Then z ABD = z CBE ; and z BDA = i BCE u\ the same segment ; .*. A ABD is equiangular to A BCE, .-. AD is to BD as GE is to BG, and .-. rect. AD, BC^rect. BD, CE. Again, •.• z ABE = z DBG, by construction, and z BAE = z BDG, in the same segment, .'. A ABE is equiangular to A BCD. .'. AB is to AE as BD is to CD, and .-. rect. AB, CD = rect. 5i>, AE. Hence rect. ^^, (ID together with rect. AD, BG =rect. BD, AE together with rect. BD, GE. «rect. AG, BD. IL 1. III. 21. I. 32. VI. 4. VI. 16. III. 21. I. 32. V.1 [.4. VI. 16. D. Ex. Tf the diagonals cut one another at an angle equal to one third of a right angle, the rectangles contained by the opposite sides are together equal to four times the qi'adrilateral figure. Book VI.] PROPOSITION XXVI 287 Proposition XXVI. Theorem. (Eiicl. vi. 23.) tlquiangular parallelograms have to one another the ratio, which t» compounded of the ratios of their sides. Let ^C and OF be equiangular Os, having z BCD = z ECG. Then must ZZ7 AC have to CO CF the ratio compounded of the ratios of their sides. Let BG and CG be placed in a straight line. Then DC and CE are also in a straight line. I. 14. Complete the O DG, and taking any st. line K, make as BC is to CG so K to L VI. 11. and make as DC is to CE so L to M. VI. 11. Then •.• K has to M the ratio compounded of the ratios of K to L and L to M, .'. X has to Mthe ratio compounded of the ratios of the sides. VI. Def. 3, p. 260. Now BC is to CG as CJ AC is to CJ CH, VI. 1. and DCis to CE as O CHis to O CF, VI. i. .-. E: is to i as O AC is to O CH, V. 5. and L is to M as CJ CH is to O CF, V. 5. Hence ^ is to M as O ^C is to O Ci^ ; V. 21. and .*. O AC has to O CjP the ratio compounded of the ratios of their sides, <^ K D. EUCLID'S ELEMENTS. [Boak VI.. Proposition XXVII. Theorem. (End. vi. 24). Parallelograms about the diameter of any 'parallelogram: are timilar to the whole parallelogram and to one another. Let ABCD be a O, of which the diameter is AG', and AEFG, FHGK the Os about the diameter. Then must these EJs he similar to ABCD and to each other. For •.' GF is II to DC, .'. l AGF = l ADC, I. 29. and •.• FF is ll to BC, .'. z AFF = i ABC ; I. 29. and each of the z s FFG, i? CD = opposite z BAD, I. 34. and .-. I FFG= z BCD. Ax. 1. Thus the ZZ7s AEFG, ABCD are equiangular to one another. Again, •.* EF is II to BC, .-. ^5 is to BC as J[^ is to EF ; VI. 4. and since the opposite sides of the Os are equal, .-. AB is to AD as AE is to AG, V. 6. and DC is to CB as 6?^ is to FE, V. 6. and OZ) is to DA as J^G' is to GA. V. 6. Thus the sides of the Os AEFG, ABCD about their equal angles are proportional. .-. OJ AEFG is similar to CJ ABCD. Similarly, O FHCK is similar to O ABCD ; and .-. O ^j&i^'G^ is similar to O J^HCiT. VI. 20. Q. E. D. Ex. Show that each of the complements of the parallelogram is a mean proportional between the parallelogrnms nbont the diameter. Book VI.] PROPOSITION XXVIII. 289 PaoPOSiTioN XXVIII. Theorem. (Eucl. vi. 26.) Ij two similar parallelograms have a common angle, and be similarly sitimted, they are about the same diameter. A a D 't ^^ 1 /|, E, of the triangle ABE, to the angles D, C\ of the triangle ADC, then these pairs of triangles shall be respectively equal to each other ; and if BE, CDy intersect in F, the triangles BFD, GFE, shall also be similar. 21. If, from the extremities of the diameter of a semicircle, perpendiculars be let fall on any line cutting the semicircle, the parts intercepted between those perpendiculars and the circumference are Qi\m\\. 22. In a given circle place a chord, parallel to a given chord , and having a given ratio to it. 23. ABC is an equilateral triangle. Thro'.igh G a line is drawn at right angles to AC, meeting AB produced in D, and u line through A parallel to BC in E. Through K, the middle point of AB, lines are drawn respectively parallel to AE, AC, and meeting DE in F and G. Prove that the sum of the squares on KG and FG is equal to three times the square on FE. 24. Find a point in the base of a right-angled triangle pro- duced such that the line drawn from it to the angular point opposite to the base, shall be to the base produced as the fv^ijiondicular to the buso \tMt 296 EUCLmS ELEMENTS. [Book VI 25. AB is a given straight line, and D a given point in it ; it is required to find a point P, in AB produced, such that AF is to PB as AB is to I)B. 26. If two circles touch each other externally, and parallel diameters be drawn, the straight line, joining the extremities of those diameters, will pass through the point of contact. 27. If two circles touch each other, and also touch a straight line ; the part of the line, between the points of contact, is a mean proportional between the diameters of the circles. 28. Two circles touch each other internally, the radius of one being treble that of the other. Shew that a point of tri- section of any chord of the larger circle, drawn from the point of contact, is its intersection with the ciicumference of the smaller circle. 21). If ABC be a right-angled triangle, and D any point in its hypotenuse AB^ determine by a geometrical construc- tion the point P, to which AB must be produced, so that FA IS to BB as AI> is to BB, 30. If a line touching two circles cut another line joining leir centres, the segments of as the diameters of the circles. .aheir centres, the segments of the latter will be to each other 31. If through the vertex of an equilateral triangle a per- pendicular be drawn to the side, meeting a perpendicular to the base, drawn from its extremity, the line, intercepted between the vertex and the latter perpendicular, is equal to the radius of the circumscribing circle. 32. If on the diagonals of a quadrilateral as bases, parallelo- grams Ije'described, equal to the quadrilateral, and each con- taining an angle equal to a given angle, find the ratio of their altitudes. 33. The opposite sides BA^ CD of a quadrilateral ABGDy which can be inscribed in a circle, meet, when produced, at E ; F is the point of intersection of the diagonals, and EF meets AD in G ; prove that the rectangle EA^ AB is to the rectangle ^Df DC as AG is to <7Z>. Book VI.] MISCELLANEOUS EXERCLHES. 297 ■__ * ?A. If from the extremities of the diameter of a circle ranoeiits be drawn, any other tangent of the circle, terminated by them, is so divided at its point of contact, that the radius of the circle is a mean proportional between the segments of the tangent. 35. If the sides of a triangle, inscribed in the segment of a ^..^ circle, be produced to meet lines drawn from the extremities of tlie base, forming with it angles equal to tlie angle in the segment, tlie rectangle contained by these lines will be equal to the square on the base. 36. Describe a parallelogram, whidh shall be of a given altitude, and equal and equiangular to a given parallelogram. 37. Two circles touch eacb other internally at the point J., and froju two points in the line joining their centres pei-pen- dicnlars are drawn, intersecting the outer circle in the points '•'», C, and the inner circle in the points D, E. Shew that AB !o to AG as AD is to AE. 38. Given of any triangle the base, and the point, where the line, bisecting tlie exterior vertical angle, cuts the base pro- duced, find the locus of the vertex of the triangle. 39. Draw a line from one of the angles at the base of a triangle, so that the part of it cut off by a line drawn from the vertex parallel to the base, may have a given ratio to the pan cut off by the opposite side. f^T ' 40. Find the point in the base produced of a right-angled triangle, from which the line dVawn to the angle opposite to the base shall have the same ratio \o the base produced, which the perpendicular has to the base itself. — C~^ 41. If the centres A, B, of two circles be joined, and B be the point in the line AB, from which equal tangents can be drawn to the circles ; the tangent drawn from any point in i. line, which passes 'through P at right angles to AB are all equal. ^ ^ 42. Construct a triangle, similar to a given triangle, and having its angular points upon +hroe given straight lines, which meet in a point. 598 EUCLID'S ELEMENTS. [Book VI, 43. Let ABCD be any parallelogram, BD its diagonal. Then the perpendiculars, from A on BD, and from B and B upon AD and AB, shall all pass through a point. 44. If a quadrilateral be inscribed in a circle, its diagonals shall be to one another as the sums of the rectangles contained by the sides adjacent to their extremities. 45. A square is described on the base of an isosceles triangle, remote from the vertex. Prove that, if the vertex be joined to the corners of the square, the middle segment of the base will be to the outer one in double the ratio of the perpendicular on the base to the base. 46. The base AB of an isosceles triangle ABC is produced both ways to D and E, so that the rectangle AD, BE is equal to the square on AG. Shew that the triangles DAC, EBC, are similar. 47. If each of the angles at the base of an isosceles triangle be double of the angle at the vertex, shew that either side is a mean proportional between the perimeter of the triangle, and the distance of the centre of the inscribed circle from either end of the base. 48. Pro^. e that, if the rectangle contained by the diagonals of a quadrilateral be equal to the sum of the rectangles con- tained by its opposite sides, the quadrilateral may be inscribed in a circle. 49. Draw a line parallel to one of the sides of a triangle, so that it may be a mean proportional between the segments into which it divides one of the other sides. 50. If an equilateral triangle be inscribed in a circle, and the adjacent arcs cut off by two of its sides be bisected, shew that the line joining the points of bisection will be trisected by the sides. 51. ABC is an equilateral triangle, BC is produced to D, and CD is made equal to BG : CE is drawn at right angles to DGB, and at A the angle GAE is made equal to the angle DCA ; DE, DA are drawn. Shew that the rectangle Dd, Book VI.J MTSCELIANEOUS EXERCISES. 299 OE is equal to the rectangle BE, AC together with the square on GB. 52. Two straight lines AB, CD, intersect in E. If when AC, BD are joined, the sides of the triangle ACE, taken in order, are proportional to those of the triangle DBE, taken in order, shew that A, C, B, D, lie on the circumference of the same circle. 53. If any triangle be inscribed in a circle, and from the vertex a line be drawn parallel to a tangent at either extremity A the base, this line will be a fourth proportional to the base and two sides. 54. If a triangle be inscribed in a semicircle, and a per- pendicular be drawn from any point in the diameter, meeting one side, the circumference, and the other side produced ; the segments cut off will be in continued proportion. 55. If ABCD be any quadrilateral figure inscribed in a circle, and BK, DL be perpendiculars on the diagonal AC, shew that BK is to DL as the rectangle AB, BC is to the rectangle AD, DC. 56. If a rectangular parallelogram be inscribed in a right- angled triangle, and they have the right-angle common, the rectangle, contained by the segments of the hypotenuse, is equal to the sum of the rectangles, contained by the segments of the sides about the right angle. 57. If from the vertex of an isosceles triangle a circle be described, with a radius less than one of the equal sides, but greater than the perpendicular from the vertex to the base, the parts of the base cut off by it will be equal 58. Through a fixed point ^ on a circle, a chord AB is drawn, and produced to a point Af, so that the rectangle con- tained by AB and AM is constant. Find the locus of M. 69. Having given a circle and a point, another point may be determined, such that the segments of any chord of the circle, drawn through eithor point, shall subtend, at the other point, anjjles which are either equal or supplementary. 300 EUCLID'S ELEMENTS. [Book VL 69. From one angle of a triangle, perpendiculars are dropped on the external bisectors of the other two angles ; prove that the distance between the feet of these perpendiculars is equal to half the sum of the sides "of the triangle. 61. A^ B, P, Q, a, are five points in the circumference oi a circle ; p, q, r, are the intersections of perpendiculars of tha triangles ABP, ABQ, ABB respectively ; prove that the triangles PQB, pqr are similar, equal, and similarly placed. 62. AD, BE, OF a,re perpendiculars from the angular points of a triangle on the opposite sides, intersecting in P. Prove that the rectangle AP, BG is equal to the sum of the rectangles PE,ACsindPF,AB. 63. ABC is a triangle, and AD. AE, are drawn to points /), E, in the base, so as to make equal angles with AB, AC, respectively. Shew that the square on AB is to the square on AC as the rectangle BD, BE is to the rectangle CD, CE. 64. Find a straight line, such that the perpendiculars, let fall upon it from three given points, shall be in a given ratio to each other. 65. Find a fourth proportional to three given similar triangles. 66. If the sides of a triangle be bisected, and the points joined with the opposite angles, the joining lines shall divide each other proportionally, and the triangle, formed by the joining lines, and the remaining side, shall be equal to a third of the original triangle. 67. Find the locus of a point, such that the distance between the feet of the perpendiculars from it upon two straight lines, given in position, may be constant. 68. ABCD is a parallelogram, AC, BD diagonals. If parallel lines be drawn through A, C, and also through B, D, the diagonals of all parallelograms so formed will pass through the same point. 69. OPQ is any triangle. OR bisects PQ in B ; PST bisects on in S, and cuts OQ in T. Shew that OQ^'WT Book VL] MISCELLANEOUS EXERCISES. 301 70. If the side 50, of a triangle ABG, be bisected by a line, which meets AB and J.0, produced if necessary, in D and E respectively, shew that AE is to EG as AJ) is to BB. 71. Two circles are drawn in the same plane, having a com- mon centre 0. If the tangent, at any point P of the inner circle, meet the outer in ^, and be produced both ways to points J., JB, such that QA, QB, are each of them equal to QC, the area of the triangle GAB will be constant. 72. From P, a point without a circle, whose centre is C, two tangents PA, PB, are drawn, and also a line, meeting the circle in D, and AB in E. If GF be perpendicular to PB, then FD is a mean proportional between FP and FE. 73. Three circles touch the sides of a triangle ABG in the points where the inscribed circle touches then), and touch each other, in the points G, H, K. Prove that A G, BH and CK meet in a point. 74. If ABG he a right-angled triangle, and EF, parallel to BG, the hypotenuse, meet AB, AG in E, F, then EH, FL, AK being drawn perpendicular to BG, shew that the difference of the rectangles GK, GH and BL, BK is equal to the differ- ence of the squares on AB, AG. 75. From a point A in the circumference of a circle two chords AB, AG are drawn, cutting off arcs greater than a quad- rant and less than a semicircle ; and from the extremity B of the greater chord, a line BD is drawn in a direction perpendi- cular to that of the diameter through A, and meets AG pro- duced in D. Shew that AD is to AB as AB is to AG. 76. Two circles intersect, and through a point of intersection two lines are drawn, terminated by the circumferences of both circles ; one of these lines remains fixed, while the other may have any position. Shew that the locus of the intersection of the lines joining their extremities is a circle. 77. If the side BG of an equilateral triangle ABG be pro- duced to any point D, and AD be joined, and if a straight line GE be drawn parallel to A B, cutting AD in E, prove that the square on AE '\^ to the rect. DA, DE as the rect. G7^, GB is to the square on DG 302 EUCLID'S ELEMENTS. [Book VI. 78. In a triangle, right-angled at A^ if the side ^Obe double of AB^ the angle B is more than double the angle 0. 79. From the obtuse angle of a triangle draw a line to the base, which shall be a mean proportional between the segments, into which it divides the base. 80. AB^ AG are two straight lines, B and C given points in the same ; BD is drawn perpendicular to AC, and DE per- pendicular to AB ; in like manner CF is drawn perpendicular to JJ5, and FG to AC. Shew that FG is parallel to BC. 81. AB is the diameter of a circle, and CD a chord at right angles to it, E any point in CD. If AE and BE be drawn and produced to cut the circle in F and G, the quadrilateral FCGD has any two of its adjacent sides in the same ratio as the remaining two. 82. ADEB is a semicircle ; AB the diameter ; DF, EG perpendiculars on the diameter ; C the centre of a circle, which touches the semicircle and these perpendiculars ; and CJH is drawn perpendicular to the diameter. Shew that CH is a mean proportional to J i'' and BG. 83. Divide a straight line in a given ratio, and produce it so that the whole line thus produced shall be to the part pro- duced in the same ratio ; shew that the circle described on the line between the two points of section, as diameter, is such, that if any point of its circumference be joined with the ex- tremities of the given line, the straight lines so drawn shall also be in the given ratio. 84. AB, CF, DE, are chords in a circle, intersecting in 0. CE, DjP joined cut AB in G and II respectively. Show that the rectangle AG, OB is to the rectangle GO, OH as the difference between AO and OB is to the difference between GO and OH. 85. Triangles on the same base, and with equal vertical angles, are to one another as the products of their sides. 86. A line ACBD is divided, so that AC \9. to CB as AD is to DB. Shew that a semicircle, described on CD, is the lociis of F, such that APh to FB as 4 C is to CB, took VI.] MISCELLANEOUS EXERCISES. 303 87. If the two diagonals of a quadrilateral, inscribed in a circle, be given, shew that the quadrilateral is greatest, when they are at right angles. 88. ABC is a triangle, 1>, E, the middle points of AB, AC, and BE, CD, meet in # : a triangle is drawn, having its sides parallel to AF, BF, CF. Shew that the lines, joining its angular points to the middle points of its opposite sides, will be parallel to the sides of the triangle ABC. 89. A circle rolls within another, of twice its radius ; if P V-e the point of contact, and A a given point of the rolling circle, PA will be constant in direction. 90. Two circles intersect ; the line AHKB joining then- centres A, B, meets them in H, K. On AB is described an equilateral triangle ABC, whose sides BC, AC intersect the circles in F, E. FE produced meets BA produced in G. Shew that as GA is to GK, so is CF to CE, and so also is GB to GB. 91. If, in Euclid's construction for forming a triangle ABC, . with each of the angles B and C double of the angle A, D is ^HXXi the point where AB is divided, and AE is taken in .1 7> equal to BD, shew that the area AEC is equal to the aic ilDC. 92. An isosceles triangle has one of its equal sides a mean proportional between two sides of another triangle. If these two sides include the same angle as the vertical angle of the isosceles triangle, shew that the triangles are equal. 93. Two triangles ABC, BCD, have the side BC common, the angles at B equal, and the angles ACB, BDC right angles. Shew that the triangle ABC is to the triangle BCD as ^£ is to^BD. 94. Given the straight line which is drawn from the vertex of an equilateral triangle to a point of trisection of the base, find the side of the triangle. 95. Straight lines being drawn from the angular points A, By G, of a triangle through any the same point, so as to cut the opposite sides respectively in a, h, c, shew that the rectangle Ab, Be is to the rectangle Ac, Ba as Cb is to Ca. 304 EUCLID'S ELEMENTS. [Book VL 96. ABCD is a quadrilateral inscribed in a circle, and its diagonals intersect in F. Shew that the rectangle AF, FD is to the rectangle BF, FC as the square on HD is to the square on BC. 97. ABCD is a quadrilateral figure whose opposite angles are not supplemental ; the circle described about ABD cuts DC in E, and the circle described about BCE cuts AE in F. Shew that the triangle ^jBi^'is equiangular to the triangle BCD, and the triangle BCF to the triangle j^BD. 98. ACB is a triangle whereof the side AC is produced to D until CD is equal to AC ; and BD is joined, shew tliat if any line drawn parallel to AB cuts the sides ^C and CB, and from the points of section lines be drawn parallel to DB, these will meet AB in points equidistant from its extremities. 99. A and B are fixed points, and AC, BD are perpendi- culars on CD, a given straight line : the straight lines AD, BC, intersect in E, and EF is drawn perpendicular to CD. Shew that EF bisects the angle AFB. 100. If be the centre of a circle circumscribed about the triangle ABC, obtuse-angled at C, and if in DC a circle be described meeting AB in D and E, then either CD or CE shall be a mean proportional between the segments into which they respectively divide AB. 101. The exterior angle CBD of the triangle ABCh bisected by the line BE, which cuts the base produced in E. Shew that the square on BE, together with the rectangle AB, BC, is equal to the rectangle AE, EC. 102. ABCD is a quadrilateral figure inscribed in a circle ; BA, CD, are produced to meet in P, and AD, BC, are pro- duced to meet in Q. Prove that PC is to PB as QA is to QB. Also, shew that half the sum of the angles at P and Q is equal to the complement of the opposite angle ABC of the quadrilateral figure. 103. Having given the vertical angle, and the ratio of the sides containing it, and also the diameter of the circumscribing circle, construct the triangle. Book VL] MISCELLANEOUS EXERCISES. 305 104. From the centre of a given circle draw a straight line to meet a given tangent to the circle, so that the segment of the line between the circle and the tangent shall be any required part of the tangent. 105. Given in any triangle the base, the ratio of the sides, and the distance between the points, in which the internal and external bisectors cut the base, construct the triangle. 106. AB is the diai^ieter of a circle, I) any point in the circumference, and C t1ie middle point of the arc AD. If AG, AD, BC, be joined, and AD cut BC in E, the circle described about the triangle AEB will touch AC, and its diameter wiU be a third proportional to BC and AB. 107. From a given point A a variable straight line is drawn, meeting a fixed straight line on P, and a point Q is taken on it so that the rectangle AP, AQ is constant. Find the locus ofg. 108. On a given base describe a rectangle, which shall be equal to the difference of the squares on two given straight lines, any two of the three given linee being together greater than the third. 109. If the exterior angles of a triangle be bisected by Rtraight lines, forming another triangle, shew that the two triangles cannot be similar, unless they be each equilateral. 110. If ABC, A'B'ir be similar triangles, and AB=-A'C, II to BC. I. 31. Then •.' AD, BC are two || st. lines, of which BC is± to the given plane, .•. AB is J. to the plane, XI. 8. and a line has been erected fioux A±io the plane. 9»& F- 324 EUCLID'S ELEMENTS. [Book XI. Proposition XIII. Theorem. From the same 'pohit in a given plane, there cannot he two straight lines at right angles to the plane, upon the same side of it ; and thsre can be but one perpendictdar to a plane from a point vdthout the i^ane. If it be possible, let two st. lines AB, AG, be at rt. z s to a given plane, from the same pt. A in the plane, and upon the same side of it. Let a plane pass through AB, AC: the common section of this with the given plane, is a st. line, passing through A. XL 2. Let DAE be the common section of the planes. Then the st. lines AB, AC, DAE are in one plane. And •.• CA is ac rt. z s to the given plane, .'. CA is at rt. z s to every st. line that meets it in that plane, XI. Def. 2. and DAE, wnich is in that plane, meets it ; \ L CAE is a rt. z . / So also, I BA £J is a rt. z . .-. z CAE — L BAE, in the same plane ; which is im- Also, from a pt., without a plane, there can be but one perpendicular to that plane ; for if there could be two, they would be parallel to one another ; which is impossible. XI. 6. Q. K. D Book XI.J PROPOSITION xrv. 325 Proposition XIV. Theorem. Planes, to vMch the same straight line is perpendicula/Tf are parallel to one another. Let the st. line AB be± to each of the planea CD, BF. Then must CD he parallel to EF. If not, let thein meet, and let the st. line GH be their com- mon section. In GH take any pt. K, and join AK, BK. Then -.* AB is± to the plane EF, .'. AB is± to BK, a st. line in that plane, XL Def. 2. and .-. z ABK is a rt. z . So also, z BAK is a rt. z . Hence two z s of the a ABK are together w two rt. z s ; which is irapo«is\bl'? I. 1 7. .-. the planes CD, EF do not lueot when produced, and .-. CD is II to EF. XI. De£ 7. 326 EUCLID'S ELEMENTS. [Book XI. Proposition XV. Theorem. £f two straight lines, meeting one another, he 'parallel to two other straight lines, which meet one another, hut are not in the same plane with the first two ; the plane, which passes through these, must he parallel to the plane passing through the others. Let AB, BC, two st. lines meeting one another, be i| to DEy EF, which meet one another, but are not in the same plane with AB, BC. Then must the plane AG be \\ to the plane DF. From B draw BG ± to the plane DF, meeting it in G. XI. 1 1. Through G draw Gil \\ to ED, and GK || to EF. I. 31. Then •.' BG is ± to the plane DF, .'. BG is ± to GH and GK, lines in that plane, XI. Def. 2. and .-. each of the z s BGH, BGK is a rt. z . Again •/ BA and GH are both || to ED, .'. BA is II to GH, and .-. z s GBA, BGH together = two rt. i «. But z BGH is a rt.. i . .'. L GBA is a rt. z . Hence GB is ± to BA ; and GB is ± to BG, for the same reason ; .-. GB is ± to the plane AG. XI. 4. Also, GB is ± to the plane DF ; Constr. .-. the plane ^0 is || to the plane DF. XI. 14. Q. E. D. XT. 9. 129. Book XI.] PROPOSITION XVL 327 Proposition XVI. Theorem. If two parallel planes be cut by another plane^ sections wiik it are poflralld. coTnmon A V Let the parallel planes AB^ CD be cut by the plane EFHG, and let their common sections with it be EF^ GJS. Then must EFbe\\ to GH. If they be not ||, let them meet in K, Then '.• EF is in the plane AB, .'. K is a, point in the plane AB. XI. Def. 1. So also, jK^ is a point in the plane GD. XI. Def. 1. .*. the planes AB, GD meet, if produced. But they do not meet, for they are parallel. .*. EF and GH do not meet, when produced. And EFy GH are in the same plane EFGH. \ EF is II to GH. I. Def. 26. Q. E. D. 32^ EUCLms ELEMENTS. [Book XL Proposition XVII. Theorem. if two straight lines be cut by parallet planes, they mufi be cut in the same ratio. Let the st. lines AB, CD be cut by the II planes GH, KL, MN in the pts. A, E, B , C, F, D. Then must AE he to EB as CF is to FD. Join AC\ BD, AD. Let AD meet the plane KL in the pt. X; and join EX, XF. Then '.• the j| planes KL, MN, are cut by the plane EBDX, .% EX is II to BD. XL 16. And the H planes GH, KL, are cut by the plane AXFO, ,\Xi^isl|to^a XL 16. Now '.^ EX is II to BD, a side of A ABD, .'. AE is to EB as AX is to XD; VL 2. and •/ XF is |1 to ^0, a side of a ADG, .\ J X is to XD as CF is to FD. VL 2. Hence AE is to EBa&CF is to FD. V. 5. ^E. T>. Boo'^ ni.] PROPOSITION XVIII. 3'^9 Ifa which Proposition XVIII. Theorem. line be at right angles to a ylane, every plane, through it, must he at right angles to that plans. Lot the St. line AB be ± to the plane CK. Then must every plane passing through AB he JLt^ the 2Jlane CK. Let any plane DE pass through AB, and let CE be the common section of the planes DE, CK. Take any pt. F in CE. In the plane DE draw FG ± to CE. L 11. Then •.• AB is±to the plane CK. .'. AB is± to CE, a st. line in that plane ; XI. Def. 2. and .-. z ABF is a rt. z . Now z GFB is a rt. z , by construction ; .*. FG is II to AB. L 28. And AB is ± to the plane CK, .-. FG is± to the phine CK XL 8. Then *.- FG, a st. line in the phme DE, drawn x to 0^, the common section of DE and OX", is ± to CK", .-. the plane DE is J. to the plane CK XI. Def. 3. So it may be proved that all planes, which pass through AB, are ± to the plane CK. Q. E. D. 330 EUCLim ELi:..>ii:..\ 1 ^. [Book XI. Proposition XIX. Theorem. Jf two planes, ivhich cut one another^ he each of them perpen- dicular to a third plane, their common secHon mvst beperpen- dimlar to the same plane. 7 Let the two planes AB, BC be each ± to a third plane, and let Bd be the common section of AB and BG. Then must BD be ±. to the third plane. If it be not, draw, in the plane AB, the st. line DE X to AD, the common section of AB with the third plane ; I, 11. and draw, in the plane BC, the st. line DF i. to DC, the common section oi BC with the third plane. T. 11. Then •.* the plane AB is± to the third plane, and DE is drawn in the plane ABl. to the common section, .-. DE is± to the third plane. XI. Pef. 3. So also, DF is ± to the thiid plane. Hence, from the pt. D, two st. lines are drawn ± to the third plane, and on the same side of it; which is impo.ssible. XI. 13. .*. no other line but BD can bo-Lto the third plane at D : .•. BD is J. to the third plane. Q. E. D. Book XL] PROPOiilTION XX. • 331 Proposition XX. Theorem. // a solid angle be co7itained by three plaiu angles, any two of them must be together greater than thi third. n B :/s c liet the solid z at ^ be contained by the three plane z a BAC, CAD, DAB. Any two of these must be together greater than the third. If the z s BAC, CAD, DAB, be all equal, any two of them are together greater than the third. If they are not equal, let BAC be that z , which is not less than either of the other two, and is greater than one of them, DAB. At A, in the plane passing through AB, AC, make lBAE= iDAB, " 1.23. and make AE—AD, and through ^draw the st. line EEC, cutting AB, AC, in tlie pts. B, C ; and join DB, DC. Then in AS ABD, ABE, :• AD = AE, and AB is common, and z BAD = z BAE, .'.DB=BE. 1.4. Then-.- DB, DC together are greater than BC, I. 20. and DB=BE, a part^-of BC, .-. DC is greater than EC. Then in i^s ADC, AEC, '.' AD=AE, and AC is common, and DC greater than J5J(7, .-. z DAC is greater than z EAC. I. 25. Also, by construction, z DAB = z BAE, .-. z s DAC, DAB together are greater than z 8 BAE, i'i/.lO together ; that is, z 8 DAC, DAB together are greater than z BAC Again, z BA C is not less than either of the z s DAC, DA B, and .*. z BAC with either of them is greater than the other. Q. E, I). 332 EUCLWS ELEMENTS. (Book XI Proposition XXL Theorem. Ev&ry solid angle is contained by plane angles, which (vrt, together less than four right angles. First, let the solid i at ^ be contained by three plane l s BAG, GAD, DAB. These shall be together less than four right angles. Take, in each of the st. lines AB, AG, AD, any points B, G, D, and join BG, GD, DB. Then •.* the solid /. at B is contained by the three plane isGBA,ABD,DBG, .'. L s GBA, ABDare together greater than z DBG. XI. 20. So also, /. s BGA, AGD are together greater than z BGD, and z s CDA, ADB are together greater than z GDB. .'. the six z s GBA, ABD, BGA, AGD, GDA, ADB are together greater than the three z s DBG, BGD, GDB, and are .*. together greater than two rt. z s. Again, '.• the three z s of each of the as ABG, AGD, ADB are together equal to two rt. z s, I. 32. .-. the nine z s GBA, BAG, AGB, AGD, GDA, DAG, ADB, DBA, BAD are together equal to six rt. z s ; and of these the six zs GBA, AGB, AGD, GDA, ADB, DBA, have been proved to be together greater than two rt. z s, and .-. the three z s BAG, GAD, DAB, which contain the solid z at vl, arc together less than four rt. z s. Book XLj PROPOSITION XXI. 333 Next, let the solid l at J. be contained by any numDer of plane l s BAC, CAD, DAE, JEAF, FAB. These must be together less than four rL 1$, Let the planes, in which the z s are, be cut by a plane, and let the common sections of it with those planes be BC, CD, DB, EF, FB. Then *.• the solid /. at B is contained by the three plane z s CBA, ABF, FBC, of which any two are together greater than the third, XI. 20. .'. z s CBA, ABF'dve together greater than z FBC. So also, the two plane z s at each of the pts. C, D, E, F, which are at the bases of the A s having the common vertex A, are together greater than the third z at the same pt., which is one of the z s of the pcjlygon BCDEF. .'. all the z s at the bases of the As are together greater than all the z s of the polygon. Now all the z s of the as together = twice as many rt. z s as there are As, that is, as there are sides in the polygon BCDEF: 1.32. and all the z s of the polygon, together with four rt. z s, together = twice as many rt. z s as there are sides in the polygon. I. 32. Cor. 1 .'. all the z s of the A s together = all the z s of the polygon together with four rt. z s. But all the z s at the bases of the A s have been proved to be together greater than all the z s of the polygon ; .*. all the z s s^ the vertex A arc together less than four rt. z s. Q. E. D 334 EUCLIUS ELIlMENTS. [Book XI. Miscellaneous Exercises on Book XL 1. If two straight lines in one plane, be equally inclined to another plane, they will be equally inclined to the common section of the two planes. 2. Two planes intersect at right angles in the line AB ; from a point G in this line are drawn GB and GF in one of the planes, so that the angle AGE is equal to AGF. Shew that GE and GF will make equal angles with any line through G in the other plane. 3. ABG is a triangle ; the perpendiculars from A, B, on the opposite sides, meet in D, and through D is drawn a straight line, perpendicular to the plane of the triangle ; if J5J be any point in this line, shew that^^, BG ; EB, GA ; and EG, AB ; are respectively perpendicular to each other. 4. A number of planes have a common line of intersection : what is the locus of the feet of perpendiculars on them from a given point ? 5. Two perpendiculars are let fall from any point on two given planes : shew that the angle between the perpendiculars will be equal to the angle of inclination of the planes to one another. 6. If perpendiculars AF, A'F', be drawn to a plane from two points A, A', above it, and a plane be drawn through A perpendicular to A A', its line of intersection with the given plane is perpendicular to FF'. 7. Prove that equal straight lines drawn from a given point to a plane are equally inclined to the plane. 8. Prove that the inclination of a plane to a plane is equal to the angle between the perpendiculars to the two planes. 9. From a point above a plane two straight lines are drawn, the one at right angles to the plane, the other at right angles Book XL] MISCELLANEOUS EXERCISES. 335 to a given line in that plane : shew that the straight line join- ing the feet of the perpendiculars is at right angles to the given line. 10. In how many ways may a solid angle be formed with equilateral triangles and squares ? 11. Two planes are inclined to each other at a given angle. Cut them by a third plane, so that its intersections with the iven planes shall be perpendicular to each other. 12. AB, AC^ AD, are three given straight lines, at right angles to one another. AE is drawn perpendicular to CD, and BE is joined. Shew that BE is perpendicular to CD. 13. Two walls meet at any angle. Shew how to draw on their surfaces the shortest line joining a point on one to a point on the other. 14. Straight lines are drawn from two points to meet each other in a given plane. Find when their sum is the least possible. 15. If two parallel planes be cut by a third plane in the straight lines AB, ah, and by a fourth plane in the straight lines AC,(JM respectively, the angle BAC will be equal to the angle hoc. 16. If four points be so situated, that the distance between each pair is equal to the distance between the other pair, prove that the angles subtended at any one point by each pair of the others are together equal to two right angles. 17. Give a geometrical construction for drawing a straight line, which shall be equally inclined to three straight lines meeting at a point. 18. A triangular pyramid stands on an equilateral base. The angles at its vertex are right angles. The square of the per- pendicular from the vertex on the base is one-third of the «quare on either of the edges. 336 EUCLID'^ hLEiMENTS. iJJook XI. 19. If one of the plane angles, forming a solid angle, be a right angle, and the sum of the other two be equ;il to two right angles, and a plane be drawu, cutting off equal lengths from the two edges, containing the right angle, the sum of the squares on the three straight lines, subtending the plane angles, will be double of the squares on the three edges, con- taining them. 20. If P be a point in a plane, which meet% the containing edges of a solid angle in J., J5, C, and be the angular point, shew that the angles FOA^ FOB, POC are together greater than half the angles AOB, BOC, CO A. togethai; BOOK XIL LEMMA. If from the greater of two unequal magnitxides oj the sa/iM kind there he taken more than its half, and from the remainder more than its half, and so on, there must at length remmn a mafjnitude less tlian the smaller of the proposed magnitudes. Let A and B be two unequal magnitudes of the same kind, of which A is the greater. Thsn if from A there be taken more than its half, and from the remainder more than its half, and so on ; there must at length remain a magnitude less than B. Take a multiple of B, as mB, greater than A ; and divide A, by the process indicated, taking from it a magnitude greater than its half, and from the remainder a magnitude greater than its half, and carry this process on till there are m divisions, and call the parts successively taken away C, D, E, F Z Now mB=B, B, B repeated m times, and A is greater than the sum oi G,D, E,...Z m in number. Then Z, the last remainder, must be less than jB. For if not, since each of the preceding remainders is greater than Z, each of them would be greater than B, and the sum of C,D Z would therefore be greater than mB ; that is, A would be greater than mB, which is contrary to the Iiypothesia. .', <^ is less than B. Q. B. O. 33S EUCLID'S ELEMENTS. FBook XII. Proposition 1. I'heorem. Similar ■polygons inscribed in circles are to one another as tiie on the diameters of tJie circles. Let A99S^, FGHKL be similar polygons inscribed in two © 8, and let BM and GN be diameters of the © s. TJien must polygon ABODE he to polygon FGHKL as sq. on BM is to sq. on GN. Join AM, BE ; FN, GL. Then a BAE is equiangular to A GFL. .-. z AEB = L FLG. But z AMB~ L AEB, in the same segment, and z FNG = l FLG, in the same segment, .-. I AMB= L FNG. also, z BAM= z GFN, each being a, rt. z , .*. A ABM is equiangular to A FGN, .'. AB is to BM as FG is to GN, and .-. AB is to FG as BM is to GN. .'. the duplicate ratio of AB to FG =the duplicate ratio of BM to GN V. 21. But polygon ABODE has to polygon FGHKL the dupli- cate ratio of AB to FG. VI. 21. And sq. on BM has to sq. on GN the duplicate ratio of BM to (?iVr. VI. 21. .-. polygon ABODE is to polygon FGHKL assq.on BM is to sq. on GN. V. 5. VI. 21. III. 21. m. 31. VI. 4. V. 15. Q. E. D. Book Xn.] PROPOSITION J I. 339 Proposition II. Theorem. Circles are to one another as the squares on their (lia/meien. S Let ABCD, EFGH be two ®8, and BD, FJI theii diameters : Then must ® ABCD be to ® EFGR as sq. on BD is to sq. on FH. For, if not, sq. on BD must be to sq. on FH as ® ABCD is to some space either less than © EFGHy or greater than it. First, if possible, let it be as © ABCD is to a space S less than © EFGH. In © EFGH describe the square EFGH, IV. 6. This square is greater than half of the © EFGH. For the sq. EFGH is half of the square, which can bo formed by drawing straight lines to touch the circle at the points E, F,G,H; and the square thus formed is greater than the ©; .-. sq. tJFGH is greater than half of the ®. 340 EUCLID'S ELEMENTS. [Book XIl. Bisect the arcs E¥, FG, GH, HE at the pts. K, L, M, N, and join EK, KF, FL, LG, GM, ME, HN, NE. Then each of the as EKF, FLG, GMH, HNE, is greater than half of the segment of the circle in which it stands. For A ^Z"i^ = half of the CJ, formed by drawing a st. line to touch the © at K, and parallel st. lines through E and F ; and the /U thus formed is greater than the segment FEK ; .*; A EKF is greater than half "of the segment FEK, and similarly for the other A s. .-. sum of all these triangles is greater than half of the sum of the segments of the 0, in which they stand. Next, bisect EK, KF, etc., and form a s as before. Then the sum of these a s is greater than half of the sum of the segments of the ©, in which they stand. If this process be continued, and the A s be supposed to be taken away, there will at length remain segments of ©s, whicn are together less than the excess of the © EFGH above the space S, by the Lemma. Let segments EK, KF, FL, LG, GM, MK, EN, NE be those which remain, and which are together less than the excess of the © of the above (S. Then the rest of the © , i.e. the polygon EKFLGMHN, is greater than S. In © ABCD inscribe the polygon AXBOCPDR similar to the polygon EKFLGMHN. The polygon AXBOCPDR is to polygon EKFLGMHN aa sq. on BD is to sq. on FH, XII. 1. that is, as © -4 BCD is to the space S. Hyp. and V. 6. But the polygon AXBOCPDR is less than © ABCD, .'. the polygon EKFLGMHN is less than the space S ; V. 14. but it is also greater, which is impossible ; .'. sq. on BD is not to sq. on FH fvs® ABCD is to any space less than © EFGH. In the same way it may be shown that Book XII] PROPOSITION II. 34« sq. on FH is not to sq. on BD as © EFGH is to any space less than w ABCD. Nor is sq. on BD to sq. on FH as © ABCD is to any space greater than © EFGH. For, if possible, let it be as © ABCD is to a space T, greater than © EFGH. Then, inversely, sq. on FH is to sq. on BD as space T is to © ABCD. But as space T is to © ABCD so is © EFGH to some space, which must be less than © ABCD, because space T is greater than © EFGH. V. 14. .-. sq. on FH is to sq. on BD as © EFGH is to some space less than © ABCD ; which has been shewn to be impossible. .'. sq. on BD is not to sq. on FH as © ABCD is to any Bp'.ice greater than © EFGH. And it has been shown that sq. on BD is not to sq. on FH ns © ABCD is to any *^' «e less than © EFGH. , sq. on BD is to sq. on FH a^ © ABCD is to © EFGH. y. E. D. 342 EVCLWS ELEMENTS. [Books VI. :a. and XII Fapers on Euclid (Boohs VI., XL, and XII.) set in the Cambridge Mathematical Tripos. 1849. VI. 4. Apply this proposition to prove that the rect- angle, contained by the segments of any chord, passing through a given point within a circle, is constant. XI. 11. Prove that equal right lines, drawn from a given point to a given plane, are equally inclined to the plane. 1850. VI. 10. AB is a diameter, and P any point in the cir- cumference of a circle ; AP and BP are joined and produced, if necessary ; if from any point of ^i^ a perpendicular be drawn to AB, meeting AP and BP in points D and E respectively, and the circumference of the circle in a point F, shew that CD is a third proportional to CE and CF. 1861. VI. 3. Jf A, B, C be three points in a straight line, and D a point, at which AB and BC subtend equal angles, sliow that the locus of the point D is a circle. XI. 8. From a point E draw EC, ED perpendicular to two planes CAB, DAB, which intersect in ^i5, and from D draw DF perpendicular to the plane CAB, meeting it in i'': shew that the line, joining the points C and Fy produced if necessary, is perpendicular to AB. / 1862. VI. 2. If two triangles be on equal bases, and between the same parallels, any line, parallel to their bases, will cut off equal areas from the two triangles. Books VI. XI and XII.] SENA TE-HOUSE RIDERS. 343 1852. XI. 11. A BCD is a regular tetrahedron, and, from the vertex A, a perpendicular is drawn to the base BCD, meeting it in 0: shew that tliree times the square on J.0 is equal to twice the square on AB. 1853. VI. 6. If the vertical angle C, of a triangle ABC, be bisected by a line, which meets the base in D, and is produced to a point E, such that tlie rectangle, contained by CD and CE, is equal to the rectangle, contained by AC and CB : shew that if the base and vertical angle be given, the position of .fi? is invariable. XI. 21. li BCD be the common base of two pyramids, whose vertices A and A' lie in a plane pass- ing through BC, and if the two lines AB, AC, be respectively perpendicular to the faces BA'D, CA'D, prove that one of the angles at 4. J together with the angles at A\ make up four right angles. 1854. Vi, 1() EA, EA' aie diameters of two circles, touching each other externally at jK ; a chord AB of the forn)er circle, when produced, touches the latter at (7, while a chord A'B of the latter touches the former at C : prove that the rect- angle, contained by AB and A'B', is four times as great as that contained by BC and B'C. au 2u. Within the area of a given triangle is described a triangle, the sides of which are parallel to those of the given one : prove that the sum of the angles, subtended by the sides of the interior triangle, at any point, not in the plane of the triangles, is less than the sum of the angles, subtended at the same point by the sides of the extericr triangle. 186fi. VI. 2. A tangent to a circle, at the point A, intersects two parallel tangents in B, T, the points of 544 EUCLID'S ELEMENTS. [Books VI. XI. and Xn. contact of which with the circle are D, ^, respectively : shew that if BE^ CD, intersect in F, AF is parallel to the tangents BD, CE. 1855. XI. 16. From the extremities of the two parallel straight lines AB, CD, parallel lines Aa, Bb, Cc, JDd, are drawn, meeting a plane in a, b,c,d: prove that AB is to CD as ab is to cd, taking the case, in which JL, jB, C, D are on the same side of the plane. 1866. VI. Def. 1. Enunciate the propositions, which prove that in the case of triangles the conditions of similarity are not independent. XI. 11. Shew that the perpendicular, dropped from the vertex of a regular tetrahedron upon the opposite base, is treble of that dropped from its own foot upon any of the other bases. 1867. VI. 19. Any two straight lines, BB\ CC, drawn parallel to the base DD', of a triangle ADD', cut AD in B, C, and AD' in B', C ; BC', B'G, are joined : prove that the area ABC or AB'C varies as the rectangle, contained by BB',CCr. XL 16. A triangular pyramid stands on an equilateral base, and the angles at the vertex are rights V angles : shew that the sum of the perpendi- culars on the faces, from any point of the base, is constant. 1868. VI, 15. Find a point in the side of a triangle, from which two lines, drawn one to the opposite angle, and the other parallel to the b;ise, shall cut off, towards the vertex and towards the base, equal triangles. XL 11. Two planes intersect : shew that the loci of the points, from which perpendiculars on the planes are equal to a given straight line, are straight lines ; and that four planes may be Books VI. XL and XII. J SENATE-IIOUSh RIDERS. 345 dnivvn, each passing through two of these lines, such that the perpendiculars, from any point in the line of intersection of the given planes, upon any one of the four planes, shall be equal to the given line. 1859. VI. 31. Shew that, on a given straight line, there may be described as many polygons of different magnitudes, similar to a given polygon, as there are sides of different lengths in the polygon. 2L 20. Three straight lines, not in the same plane, intersect in a point, and through their point of intersection another straight line is drawn within the solid angle formed by them : prove that the angles, which this .straight line makes with tiie ftrst three, arc together less than the sum, but greater tlian half the sum of the angles which the first three make with each other. 1860. ▼!. A- If the two sides, containing the angle, through which the bisecting line is drawn, be equal, interpret the result of the proposition. Prove from this proposition and the preceding, that the straight lines, bisecting one angle of a triangle internally, and the other two ex- ternally, pass through the same point. XL 17. If three straight lines, which do not all lie in one plane, be cut in the same ratio by three planes, two of which are parallel, shew that the third will be parallel to the other two, if its intersections with the three straight lines are not all in one straight line. 1861. VI. 6. From the angular points of a parallelogram A BCD, perpendiculars are drawn on the diiigonals, meeting them in E, F, G, H re- 346 EVCLID'S ELEMENTS. [Books VL XI. ana XII. spectively ; prove that EFCr H \s a, parallelo- gram similar to A BCD. 1861. XI. 12. Shew that the shortest distance between two opposite edujes of a regular tetrahedron is equal to half the diagonal of the square, de- scribed on an edge. 1862. VI. 1. Lines are drawn from two of the angular points of a triangle, to divide the opposite sides in a given ratio ; prove that the line, joining the third angular point with the point of in- tersection of these two lines, either bisects the opposite side, or divides it in a ratio which is the duplicate of the given ratio. XI. 21 If four points be so situated that the distance between each pair is equal to the distance between the other pair, prove that the angles subtended at any one of these points by each pair of the others, are together equal to two right angles. 1863. VI. 4. The internal angles at the base of a triangle, and the external angle at the vertex, are bisected by straight lines ; prove that the three points, in which these straight lines meet the oppo- site sides respectively, lie on one straight line. v. 17. If each edge of a tetrahedron be equal to the opposite edge, the straight line, joining the middle points of any two opposite edges, shall be at right angles to each of those edges. 1864. VT. 23. If one parallelogram have to another parallelo- gram the ratio, which is compounded of the ratios of their sides, one parallelogram shall be equiangular. ICT. 12. On a given equilateral triangle describe a regular tetrahedron. Books VI. XI. and XII.] SENA TE-HOUSE RIDERS. 347 1865. VI. 19. The opposite sides, BA, CD of a quadilateral ABCD, which can be inscribed in a circle, meet, when produced, in ^ ; jP is the point of intersection of the diagonals, and EF meets AD in G : prove that the rectangle EA, AB is the rectangle ED, DC as AG is to GD. XI. 16. In the triangular pyramid ABCD, AB is at right angles to CD, and AC to BD : prove that AD is at right angles to BC. 1866. VI. 4. ABC is an isosceles triangle ; AE is the perpen- dicular from A on the base BC \ D is any point in AE ; and CD produced meets the side AB at F : shew that the ratio of AD to DE is double of the ratio of AF to FB. XII. 1. Give an outline of Euclid's demonstration that circles are to one another as the squares on their diameters. 1867. VI. A. Each acute angle of a right-angled triangle and its corresponding exterior angle are bisected by straight lines meeting the opposite sides ; , prove that the rectangle, contained by the portions of those sides intercepted between the bisecting lines is four times the square on the hypotenuse. M, 21. Two pyramids are described, the one standing on a square as a base, the other on a regular octagon, the vertex of each being equally distant from the angular points of its base ; if this distance be the same for each pyramid, and the perimeters of the bases be equal, prove that the plane angles, containing one solid angle at the vertex of the former, are together greater than the plane angles, con- taining the solid angle at the vertex of the latter. 1868. VI. 2. Without assuming any subsequent picposition, inove that the equiangular triangles in either 34?< EUCLID'S ELEMENTS. [Books VI. XI. and XIL of the figures of this proposition, are to each other in the duplicate ratio of the sides oppo site to tile equal angles. 1868. XI. 11. Of the least angles, which a given line in one plane makes with any line in another plane, the greatest for different positions of the given line is that which measures the inclina- tion of the two planes. 1869. XI. 20. ]f be a point, within a tetrahedron ABCD, prove that the three angles of the solid angle, subtended by BCD at 0, are together greater than the three angles of tlie soiiil angle at A. 1870. VI. 15. Two straight lines are given in position, and a third straight line is drawn so as to cut ofl' a triangle equal to a given triangle ; through the middle point of tliis third side is drawn a straight line in a given direction, termin- ated by the two given straight lines : provt- that the rectangle under the segments of tht intercepted part is constant. XL 7. In a tetrahedron each edge is perpendicular to the direction of the opposite edge ; prove that the straight line joining the centre of the sphere, circumscribing the tetrahedron, to the middle point of any edge, is equal and parallel to the straight line joining the centre of perpendiculars to the middle point of the opposite edge. 1871. VI. 2. ABC is a triangle, and lines AO, BO, CO cut the opposite sides in I), E, F ; if EF cut BC in G, prove that BD is to DC as BG is to GC. XL 11. The perpendiculars from the angular points of a tetrahedron on the opposite faces meet in a point : prove that tlie necessary and sufficient condition for this is that the sums of the squares of pairs of opposite edges be equal. Books VT. XI. aii<1 XII.] SENA TE-HOUSE RIDERS. 349 187]^ VI. 2. Draw through a point a straight line, so that the part of it intercepted between a given straight line and a given circle may be divided at the given point in a given ratio. Between what limits must the ratio lie in order that a solution may be possible ? XL 20. If the opposite edges of a tetrahedron be equal two and two, prove that the faces are acute- angled triangles. Prove also that a tetra- hedron can be formed of any four equal and similar acute-angled triangles APPENDIX. EXAMINATION PAPERS IN EUCLID SET TO CANDIDATES FOR First and Second Glass Provincial Certificates, AND TO STUDENTS MATRICULATING IN THE UNIVERSITY OF TORONTO. SECOND CLASS PROVINCIAL CERTIFICATES, 1871. TIME — TWO HOURS AND A HALF. 1. If two *riangle8 have two sides of the oue equal to two sides of the other, each to each, and have likewise their bases equal, the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sidos, equal to them, of the other. 2. Triangles upon the same base, and between the same par- allels, are equal to one another. 3. If the square described upon one of the sides of a triangle be equal to the squares described upon the other two sides of it, the angle contained by these two sides is a right angle. 4. If a straight line be divided into two equal, and also into two unequal, parts, the squares on the two unequal parts are together double of the square on half the line, and of the square on the line between the points of section. 5. If a straight line be divided into any two parts, the rec- tangles contained by the whole and each of the parts are together equal to the square on the whole line. 6. Bisect a parallelogram by a straight line drawn from a point in one of its sides. 7. Let A B C be a triangle, and let B D be a straight line drawn to I), a point in A C between A and C, then, if A B be greater than A C, the excess of A B above A C is less than that of B D above D C. 8. In a triangle A B C, A D being drawn perpendicular to the straight line B D which bisects the angle B, show that a line drawn from D parallel to B C will bisect A C. Note. — The percentage of marks requisite, in order that a candidate may be ranked of a particular giade, will be taken on the value of the above paper, omitting question 8. ii. APPENDIX. aECONB CLASS PBOVINCIAL CERTIFICATES, 1872. TIME— 2| HOURS. 1. Define a straight line, a plane rectilineal angle, a right angle, a Gnomon. Enunciate Euclid's Postulates. 2. If from the ends of the side of a triangle there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, - but shall contain a greater angle. 3. If two triangles have two angles of the one equal to angle? of the other, each to each, and one side equal to one side, namely, either the sides adjacent to the equal angles, or sides which are opposite to equal angles in each ; then shall the other sides be equal, each to each ; and also the third angle of the one equal to the third angle of the other. (Take the case in which the assumed equal sides are those opposite to equal arigles.) 4. In every triangle, the square on the side subtending an acute angle is less than the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall on it from the opposite angle, and acute angle. {Take the case where the per- pendicular falls within the triangle.) 5. If a straight line be divided into any two parts, the squares on the whole line, and one of the parts, are equal tc twice the rectangle contained by the whole and that part, together with the square on the other part. 6. Prove that, if a straight line AD be drawn from A, one of the angles of a triangle ABC, to D, the middle point of the opposite side BC, BA X AC is greater than 2 AD. 7. Let the equilateral triangle ABC, and triangle ADB, in which the angle ABD is a right angle, be on the same base AB, and between the same parallels AB and CD. Prove that 4 AD2 = 7 AB2 8 From D, a point in AB, a side of the triangle ABC, it is required to draw a straight line DE, cutting BC in E, and AC produced in F, so that DE may be equal to EF. SECOND CLASS PROVINCIAL CERTIFICATES, 1873. TIME — TWO HOURS AND A HALF. Note. — Candidates who take only Book I, will confine them- selves to the first eight questions ; those who take Books I and U, will omit the first two questions. At»I'ENmX. 111. 1. If tr.7 > angles of a triangle be equal to one another, tlie sides also which subtend, or are opposite to, the equal angles, shall be equal to one another. 2. If one side of a triangle be produced, the exterior angle shall be greater than either of the interior opposite angles. 3. The opposite side, and angles of a parallelogram, are equal to one another. 4. The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another. 5. To describe a square on a given straight line. 6. Let A B C D be a quadrilateral figure whose opposite angles ABC and ADC are right angles. Prove that, if A B be equal to A D, C B and C D shall also bo equal to one another. 7. If A B C D be a quadrilateral figure, having the side A B parallel to the side C D, the straight line which joins the middle points of A B and D C shall divide the quadrilateral into two equal parts. 8. The straight line, which joins the middle points of two sides of a triangle, is parallel to the base. 9. If a straight line be divided into any two parts, the square on the whole line is equal to the squares on the two parts, together with twice the rectangle contained by the parts. 10. In an obtuse angled triangle, is the sum of the sides con- taining the obtuse angle greater or lees than the square of the side opposite to the obtuse angle? And, by how much ? Prove the proposition. SECOND CLASS PEOVINCIAL CEKTIFICATES, 1874. TIMK— TWO HOURS AND THREB-QUABTEBS. Note. — Candidates who take only Book I. will confine them- selves to the first 7 questions. Those who take Books I. and 11. will omit questions 1, 2, and 3. 1. When is one straight line said to be perpendicular to an- other. To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it. 2. If one side of a triangle be produced, the exterior angle shall be greater than either of the interior opposite angles. 8. If two triangles have two angles of the one equal to two angles of the other, each to each ; and one sidft equal to one side, namely, sides which are opposite to |y. APPENDIX. equal angles in each ; then shall the other sides be equal, each t^ each. L What are parallel straight lines ? If a straight line, falling on two other straight lines, make the alternate angles equal to one another, the two straight lines shall be parallel to one another. 5. What is a parallelogram ? Parallelograms on equal bases, and between the same parallels, are equal to one another. 6. If two isosceles triangles be on the same base, and on the same side of it, the straight line which joins their vertices, will, if produced, cut the base at right angles. 7. Let ABC be a triangle, in which the angle ABC is a right angle. From AC cut off AD equal to AB, and join BD. Prove that the angle BAC is equal to twice the angle CBD. 8. If a bLiaiy;hL Imu be divided into two equal parts and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to, &c. (5, II.) 9. In every triangle, the square on the side subtending an acute angle is less than the squares on the sides con- taining that angle, by &c. (13,11). (It will be suf- ficent to take the case in which the perpendicular falls within the triangle.) 10. To describe a square that shall be equal to a given recti- lineal figure. 11. The square on any straight line drawn from the vertex of an isosceles triangle to the base is less than the square on a side of a triangle by a rectangle contained by the segments of the base. SECOND CLASS PROVINCIAL CERTIFICATES, 1875. TIME — TWO HOURS AND THREE-QUARTERS. Note. — Those students who take only Book I. will confine themselves to the first seven questions. Those who take Books I. and II. will omit the questions marked with an asterisk (*), namely, (1) and (2). *1. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. *2. If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, namely, the sides opposite to equal angles, then shall the other sides be equal, each to each. 3. If a straight line falling on two other straight lines make the alternate angles equal to each other, these two straight lines shall be parallel. APPENDIX. V. 4. If a straight line fall upon two parallel straight lines, it makes the two interior angles upon the same side together equal to two right angles. 6. Assuming Proposition XXXII, deduce the corollary : " all the exterior angles of any rectilineal figure, made by producing the sides successively in the same direction, are together equal to four right angles." 6. If a straight line, drawn parallel to the base of a triangle, bisect one of the sides, it shall bisect the other also, 7. Let ABC and ADC be two triangles on the same base AC and between the same parallels AC and BD. Prove, that, if the sides AB and BO be equal to one another, their sum is less than the sum of the sides AD and DC. 8. If a straight line be divided into any two parts, the rect- angles contained by the whole and each of the parts are together equal to the square on the whole line. 9. If a straight line be bisected and produced to any point, the rectangles contained by the whole line thus pro- duced, and the part of it produced, together with, etc., (6, n). 10. Divide a straight line into two parts, such that the sum of their squares may be the least possible. FIKST CLASS PROVINCIAL CERTIFICATES, 1871. TIME. — THREE HOURS. 1. To describe a square that shall be equal to a given recti- lineal figure 2. A segment of a circle being given, to describe the circle of which it is the segment. 8. If the vertical angle of a triangle be divided into two equal angles by a straight line which also cuts the , base, the segments of the base shall have the same ratio which the other sides of the triangle have to one another. 4. In a right-angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle and to one another. 6. If four straight lines be proportionals, the similar rectilineal figures similarly described upon them shaU also be proportionals. 6. Draw a straight line so as to touch two given circles. 7. Let ABC be a triangle, and from B and C, the extremi- ties of the haso B C. let line B F and C E be drawn to F and E, the middle points of A C and A B respect- VI. APPENDIX. ively, then, if B F = C E, A B and A C shall be equal to one another. 8. Describe an equilateral triangle equal to a given triangle. FIRST CLASS PROVINCIAL CEETIFICATES, 1872. TIME — TWO AND A HALP HOURS. 1. If a straight line touch a circle, and from the point of con- tact a straight line be drawn cutting the circle, the angles which this line makes with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle. 2. To inscribe a circle in a given triangle. 3. Equal triangles which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional. 4. Similar triangles are to one another in the duplicate ratio of their homologous sides. 5. In any right angled triangle, any rectilineal figure described on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle. 6. Two circles cut each other, and through the points of sec- tion are drawn two parallel lines, terminated by the circumferences. Prove that these lines are equal. 7. Let A G and B D, the diagonals of a quadrilateral figure A B C D, intersect in E. Then, if A B be parallel to C D, the circles described about the triangles ABB and C D E shall touch one another. 8. Divide a triangle into two equal parts by a straight line at right angles to one of the sides. FIBST CLASS PKOVINCIAL CERTIFICATES, 1873. TIME — THREE HOURS. 1. The angle in a semicircle is a right angle. 2. A segment of a circle being given, describe the circle of which it is a segment. 3. Give Euclid's definition of proportion ; and prove, by taking equi-multiples according to the definition, that 2, 3, 9, 13, are not proportionals. 4. Similar triangles are to one another in the duplicate ratio of their homologous sides. 5. To find a mean proportional between two given straight lines. 6. Through U, the vertex of a triangle A C B, which has the sides A C and C B equal to one another, a line D APPENDIX. VII. is drawn parallel to A B ; and straight lines, A D, D B, are drawn from A and B to any point D in CD. Prove that the angle A G D is greater than the angle ADB. 7. A B C D is a quadrilateral figure inscribed in a circle. From A and B, perpendiculars A E, B F are let fall on C D (produced if necessary) ; and from C and D, perpendiculars C G, D H, are let fall on B A (produced if necessary). Prove that the rectangles A E, B F and C G, D H, are equal to one another. 8. A B C D is a quadrilateral figure inscribed in a circle. The straight line D E drawn through D parallel to A B, cuts the side B C in E ; and the straight line A E pro- duced meets T> G produced in F. Prove, that if the rectangle B A, A D be equal to the rectangle E (', CF, the triangle A D F shall be equal to the quadrilateral ABCD. FIRST CLA.SS PROVINCIAL CERTIFICATES, 1874. TIME — THREE HOURS. 1. In equal circles, equal straight lines cut off equal circum- ferences, the greater, equal to the greater, and the less to the less. 2. To describe a circle about a given equilateral and equiangu- lar pentagon. 3. To find a mean proportional between two given straight lines. 4. What is meant by duplicate ratio ? Write down two whole numbers, which are in the duplicate ratio of J to J. What are similar rectilineal figures ? Similar triangles are to one another in the duplicate ratio of their homologous sides. 6. In any right angled triangle, any rectilineal figure described on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle. 6. To describe a triangle, of which the base, the vertical angle, and the sum of the two sides are given. 7. From A the vertex of a triangle ABC, in which each of the angles ABC and ACB is less than right angle, AD is let fall perpendicular on the base BC. Produce BC to E, making CE equal to AD ; and let F be a point in AC, such that the triangle BFE is equal to the tri- angle ABO. Prove that F is one of the angular points of a square inscribed in the triangle ABC, with one of its sides on BO. Vlll. APPENDIX. 8. Let E be the point of intersection of the diagonals of a quadrilateral figure ABCD, of which any two opposite angles are together equal to two right angles. Pro- duce HO to Gr, making CG equal to EA ; and produce AD to F, making DF equal to BE. Prove that if EG and EF be joined, the triangles EDF and ECG ar<» equal to one another. FIRST CLASS PROVINCIAL CERTIFICATES, 1875. TIME— THREE HOURS. 1. If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, namely, the sides adjacent to the equal angles in each, then shall the other sides be equal each to each. 2. From a given circle to cut off a segment, which shall con- tain an angle equal to a given rectilineal angle. 3. If the angle of a triangle be divided into two equal angles by a straight line which also cuts the base, the segments of the base shall have the same ratio which the other sides of the triangles have to one another. 4. The sides about the equal angles eqai-angular triangles are proportionals ; and those which are opposite to the equal angles are homologous sides. 5. If the similar rectilineal figures similarly described upon four straight lines be proportionals, those straight lines shall be proportionals. 6. Any rectangle is half the rectangle contained by the diameters of the squares on its adjacent sides. 7. Through a given point within a given circle, to draw a straight line such that one of the pares of it intercept- ed between that point and the circumference shall be double of the other. 8. If, from any point in a circular arc, perpendiculars be let fall on its bounding radii, the distance of their feet ia invariable. MATRICULATION, 1871. 1. State the points of agreement and disagreement of the circle, square and rhombus, with one another as appearing from their definitions. 2. Any two sides of a triangle are together greater than the third side. Show that the sum of the excesses of each pair of sides above the third side is equal to the sum of the three sides of the triangle. APPENDIX. IX. If the square described upon one of the sides of a triangle be equal to the square described on the other two sides of it, the angle contained by these two sides is a right angle. In an isosceles triangle if the square on the base be equal to three times the square on either side the vertical angle is two-thirds of two right angles. If a straiglit line be divided into any two parts the square on the whole line is equal to the square on the two parts, together with twice the rectangle contained by the parts. Is there any difference between the principle of this propo- sition and the statement (a -)- ^)^ = a^ _|_ 2ab -j- b^ . Of all the squares that can be inscribed within an- other the least is that formed by joining the bisec- tions of the side. If a straight line be divided into two equal and also into two unequal parts, the squares on the two unequal parts are together double of the square on half the line and of the square on the line between the points of section. Docs tlie statement respecting the equality of the square hold for any other division of the line. Equal straight lines in a circle are equally distant from the centre ; and conversely, those which are equally distant from the centre are equal to one another. The lines joining the extremities of two equal straight linos in a circle towards the same parts are parallel to each other. ' What is meant by the Angle in a segment of a circle ? Define similar segments of circles. Upon the same straight line and upon the same side of it, there cannot be two similar segments of circles not coinciding with one another. In equal circles the angles which stand upon equal arcs, are equal to one another whether they be at the cen- tres or circumferences. If two equal circles so intersect each other that the tan- gents at one of their points of intersection are inclined to each other at an angle of 60" shew that Radius of circle : line joining their centres : : 1 : 1/3] From a given circle to cut off a segment that shall contain an angle t qual to a given rectilineal angle. In a given circle inscribe a triangle which shall have a given vertical angle, and whose area shall be equal to a given triangle ; and shew with what limitation thia can be done. X. APPENDIX. 10. When is a circle said to be inscribed in a rectilinea figure. To inscribe a circle in a given triangle. LI. Inscribe an equilateral and equiangular pentagon in a given circle. Show how to divide a right angle into fifteen equal parts. MATRICULATION, 1872. 1. From a given point to draw a straight line equal to a given straight line. Explain what different constructions there are in this proposition. 2. If a side of a triangle be produced, the exterior angle is equal to the two interior and opposite angles ; and the three interior angles of every tdangle are together equal to two right angles. Find the number of degrees in one of the exterior angles of a regular heptagon. 3. Triangles upon the same or equal bases and between the same parallels are equal to une another. By means of these propositions prove that a line drawn parallel to the base of a triangle and cutting off one- fourth from one of its sides, will also cut off a fourth part from the other side. 4. If a straight line be divided into two equal and also into two unequal parts, the squares on the two unequal parts are together double of the square on half the line, and of the square on the line between the points of section. If a chord be drawn parallel to the diameter of a circle and from any point in the diameter lines be drawn to its extremities, the sum of their squares will be equal to the sum of the squares of the segments of the diam- eter. 5. To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square on the other part. Solve the problem algebraically. Interpret and construct geometrically the second root so obtained. Divide a given line so that one segment maybe a geomet- ric mean between the whole and the other. 6. In every triangle, the square on the side subtending either of the acute angles, is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight APPENDIX. "• XI. line intercepted between the acute angle and the per- pendicular let fall upon it from the opposite angle. In a triangle ABC, if AD be drawn to the bisection of BC, the difference between the square on BC and twice the square on AG is double of the difference between i^-.- square on AB, and twice the square on IL 7. If a straight line touch a circle, the straij^Qf- Iik *2. If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by two sides of one of them greater than the angle contained by the two sides equal to them of the other, the base of that which has the greater angle shall be greater than the base of the other. Why the restriction " Of the two sides DE, DF, let DE be the side which is not greater than the other"? * 16 *3. If two triangles have two angles of the one equal to two angles of the other, each to each, and have also the sides adjacent to the equal angles in each, equal to one another, then shall the other side bo equal, each to each ; and also the third angle of the one to the third angle of the other. (Prove by superposition.) 3 What propositions in Book I are thus proved ? 16 4. If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to the inte- rior and opposite angle on the same side ; and also the two interior angles on the same side together equal to two right angles. 8 What objection may be taken to the twelfth axiom ? 2 What is its converse ? 16 5. In any right-angled triangle, the square which is described on the side subtending the right angle is equal to the squares described on the sides which contain the right angle. 12 Prove also by dissection and superposition. 18 6. Draw through a given point between two straight lines not parallel a straight line which shall be I bisected in that point. T8 I 7. The perpendiculars from the angles of a triangle on the opposite aides meet in a point. XX. APPENDIX. 20 8. Given the lengths of the lines drawn from the angles of a triangle to the points of bisection of the opposite sides, construct the triangle. 20 ^ If a straight line be divided into two parts, the square on the whole line is equal to the squares on the parts, together with twice the rectangle contained by the parts, 20 10. In every triangle, the square on the side subtending an acute angle is less than the squares on the sides containing that angle by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall on it from the opposite angle, and the acute angle. OF THE ^ fWfVERSITY OF 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. ^^V^ REC'P LD JAN 1 3 ' S 5 9 AW ^ NOV 2 1967 5 ( g TSttt DEC 2 19 6 7 p e cs '67 -3 M l,s?AN Pii^'i; LD 21A-60m-4,'64 (E4555sl0)476B General Library- University of California Berkeley 1 84005