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U.I dee symboiee suivents apparattra sur la damiAre image da cheque microfiche, seion le caa: le symbols — ^ signifie "A SUIVRE". le symboie ▼ signifie "FIN". Maps, plataa. charts, etc.. may be filmed at diffei ent reduction ratios. Thoae toe large to be entirely included in one expoeure are filmed beginning in the upper left hand comer, left to right and top to bottom, aa rnany frames aa required. The following diagrams illustrate the method: Lee cartee. planclies. tableeux. etc.. peuvent itre filmto i dea taux da riduction diff Grants. Lortqua la dlocument eat trop grand pour Atre reproduit en un saul clicli*, il set film* A partir da i'angia sup^riaur gauche, de gauche i droite, et de haut en baa. en prenant le nombre d'imegea n^cessaira. Lee diagrammas suivents illuetrent la m4thode. 1 2 3 1 2 3 4 5 6 ■ '-'H . J" ^^^''••^mmmmiimr'mm E w DEI JO THE Elements of Euclid Books I. to VI. WITH DEDUCTIONS. APPENDICES AND HISTORICAL NOTES BY JOHN STURGEON MACKAY. M.A., F.R.S.E. MATHEMATICAL MASTER IN THE BDIHBUROH ACADEMY TORONTO: THE HUNTER, ROSE CO., LIMITED ^m • k Entered accordinir to Act of the Parliament of Canada, in the year one thousnnd eight hundred and eighty-seven, hy W. Sc R. Chambbrs, at the Department of Agriculture. 7^3© I PREFACE In this text-book, compiled at the request of the publishers, a F^i-.f .T°''/?: ^'^''' ^^°^'°°'« well-known editions of Euclid 8 Llemmts has not been observed; but no change has been made on Euclid's sequence of propositions, and com- para ively little on his modes of proof. Here and ihere useful coroUaiies and converses have been inserted, and a few of Simson'8 additions have been omitted. Intimation of such insertions and ucissions has been given, when it was deem^ necess^, m the proper place. Several changes, mostly, now- ever, of arrangement, have been made on the definitions. By a slight alteration of the lettering or the construction of the figure an attempt has been made throughout, and par- cularly in the Second Book, to draw the attention ofTh^ reader to the analogy which exists between certain pairs o' propositions. By Euclid this analogy is well-nigh ignored l^i'l^^'.T"'^ °^.^°''' congruent and similar figures, care has been taken to write the letters which denote corLponding points m a corresponding order. This is a matter of minor oftLTTa'se '' '"' '°' '""" *^ '^ "^«^^^^«^> - - ^00 nnlr''^''l?'-j,''"',°' ^^'"'^'"^ ^PP^^^^d to the various pro- positions riders,' as they are son.etimes termed) have been rCdlh^at"?''""' '"',:"u*'^ '^^«* Book, numerous. It In £ %'""er8, who have little confidence in their own reasoning power, will thereby be encouraged to do more han merely learn the text of Euclid. It is hoped alsoXt eelrtlnr"" '" '"" '''"'' ''' «" ^'---^ begL 1^ seeing that the questions, deduction., and corollaries to b^ ^msBma^ m if PRITAOB. proved number coiuiderably over fifteen hundred. It shonld be ntated that when a deduction iu repeated once or oftener, in the name words, a different mode of proof is expected in each case. In the appendices, much curtailed from considerations of space, a few of the more useful and interesting theorems of elementary geometry have been given. It has not been thought expedient to introduce the signs + and — , to indicate opposite directions of measurement The important advantages which result from this use of these signs are readily apprehended by readers who advance beyond the * elements/ and it is only of the ' elements ' that the present manual treats. The historical notes, which are not specially intended for beginners, may save time and trouble to any one who wishes to investigate more fully certain of the questions which occur throughout the work. It would perhaps be well if such notes were more frequently to be found in mathematical text-books : the names of those who have extended the boundaries, or successfully cultivated any part of the domain, of science should not be unknown to those who inherit the results of their labour. Though the utmost pains have been taken by all concerned in the production of this volume to make it accurate and workmanlike, a few errors may have escaped notice. Cor- rections of these will be gratefully received. The editor desires to express his thanks to Mr J. R. Pairuan for the excellence of the diagrams, and to Mr David Traill, M.A., B,Sc., and Mr A. Y. Fraser, M.A., for valuable hints while the work was going through the pteaik Edikbxjroh Acadxuy, AprU 1884. I OONTEISTTS. BOOK I. DEFINITIONS. ^"^^f PoerruLATBS. ,. AXIOM& ' .. QUK8TI0NS ON THE DEFINITIONS, POSTULATES, AXIOMS..... ! 13 EXPLANATION OF TERMS ""' ,<. SYMBOI^ AND ABBREVIATIONa 19 PROPOSITIONS 1-48 WITH RIDERS .*.*'* 21 PROPOSITION A !.........,....!. 61 APPENDIX L PROPOSITIONS 1-5. ga i**" ^^■■■3;;;::::;:::;;;::::;::::;io2 DEDUCTIONS |Qg BOOK II. RECTANGLES AND SQUARES- DEFINmONa jjn PROPOSITIONS 1-14 WITH RIDER& 116 APPENDIX n. PROPOSITIONS 1-3 1^. _^„ 147 "^^ 149 DEDUCTIONa ; i-j BOOK III CIRCLES- DEFINITIONS. ', ^Kk PROPOSITIONS 1-37 WITH RIDERS '.!...!.!!!!!!!!!!!!.." IfiS APPENDIX m. KADIOAL AXIS. gU DSDucnoNs. «1« 51991 4' OONTBNn. BOOK IV. CIRCLES AND THEJR SCRIBED REGULAR POLYGONS REGULAR POLYGONS AND THEIR SCRIBED CIRCLES— Page DBnNrnoN& ^ 223 PROPosmoirs ^-16 with riders .I..22^ JLPFEtfDUL IV. PROPOSITIONS 1-2. 250 raDuonoNs. ............[,. 25%^ BOOK V. PROPORTION IN GENERAL— DBUNmONS. 281 AxioBfa '*265 PROfosmoKB 1-24 ........",.. .....2«? PROPOSITIONS A, B, C. ........!."!!!.!!...26^ PROPOSITION D ................21^ BOOK VI. PROPORTIONAL SECTION AND SIMILAR FI0URE8- DKPINITIONa 286 PROPOSITIONS 1-33 WITH RIDERS 1.!!!!..!. !!!!!!!.!287 PROPOSITION A 294 PROPOSITIONS B, C, D ........*...".......!! 338 APPENDIX VI. TJa-^JTStVERSALS. ^ 3^3 ;«IARMONICAL PROORBSSiCN .!!!!. .."..347 OROVOTIOIIB .........!.! ".',362 I EUCLID'S EI^EMENTS. BOOK I. DEFINITIONa 1. A point has position, but it has no magnitude. A point ia indicated by a dot with a letter attaclied. m the point P. p The dots employed to represent points are not strictiy geometrical points, for they have some size, else they could not be seen But in geometry the only thing connected with a point, or its representa- tive a dot, which we consider, is its position. 2. A line has position, and it has length, but neither breadth nor thickness. Hence the ends of a line are i)oints, and the intersection of two lines is a point A Hne is indicated by a stroke with a letter attached, as the hne G. ^ Oftener, however, a letter is phvced at each end of the line as the line AB, ^ ' " The strokes, whether of pen or pencil, employed to represent lines are not stnctly geometrical lines, for they have some breadth and some tWcktess. But in geometry the only things connected with a hne which we consider, are its position and its lengtL 3. If two lines are such that they cannot coincide in any two points without coinciding altogether, each of them ia called a straight line. Hence two straight lines cannot irclose a space, nor can they nave any part in common. Thus the two lines ABC and ABD^ ^ which have the part AB in common, cannot both be straight lines. . Euclid's definition of a straight line ~" B" ~"^ i« 'that which lies evenly to the points within itsell' buclid's elements. 4. A curved line, or a cunre, is a line of which no pan isstraiglit. B - Thua A BC is a curve. 5. A surface (or superficies) has position, and it "has lengtli and breadth, but not thickness. Hence the bound- aries of a surface, and the intersection of two surfaces, are lines. Thus AB^ ACB, and DE are lines. 6. A iilane surfi&ce (or a plane) is such that if any two points whatever be taken on it, the straight line joining them lies wholly in that surface. This definition (which is not Euclid's, but is due to Heron of Alexandna) aflfords the practical test by which we ascertain whether a given surface is a plane or not. We take a piece of wood or iron with one of Its edges straight, and apply this edge in various positions to the surface. If the straight edge fits closely to the surface m every position, we conclude that the surface is plane. 7. When two straight lines are drawn from the same point, they are said to contain a plane angle. The straight lin«s are called the arms of the angle, and the point is called the vertex. Thus the straight lines IB, ^C drawn from A are said to contain the angle BAG; AB and ^C7are the arms of th« angle, and .4 is the vertex. An angle is sometimes denoted by three letters, but these letters must be placed so that the one at the vertex shall always be between the aZ Z« ^«« the given angle is called BAG or GAB, never ABG, AGB, GBA, BGA. When only one angle is formed at a vertex it is often denoted by a single letter, that letter, namely, at Book I.] DiSli'INll'IONS. tvhea there are several angles at the Mme vertex, it ia necessary, in order to avoid ambiguity, to use three letters to express the angle intended. Thus, in the annexed figure, there are three angles at the vertex A, namely, BAC, CAD, BAD Sometimes the arms of an angle have \^ drJSll '•'**""• ''**"*^'^ to them; in which case the cnglo ma?b« denoted m various ways. ^ Fig. 1. Fig.! TTius the angle^ (fig. i) may be called AFC or BFC indifferently • ^e angle O (fig 2) may be called AOB or COB; the andei (fig. 3) may be caUed BAG, FAG, DAE, FAC, GAB, and so on It IB important to observe that aU these ways of denoting any ingle i^^r Tn'^r"^ ^r^.^y *^.""^"« ^* *^« '"'gl^ ^^^, or the on th« 1 Si, ? ■?*^*'' ''°'^' ^^'^ "^^ °^ "^ '^"gl^ does not depend IZZT "^^ "^"^ *° *^1 '^"^ ^^ °f ^^^t^^r «>«le. the angles themselves are not necessarily equal . *^ As a further illustration, the angles A, B, C with unequal arms mtclid's elements. [Book t ^ i are all equal ; of the angles D, E, F, that with the shortest arms is the largest, and that with the longest arms is the smallest. 8. If three straight lines are draw^ from the same point, three different angles are formed. Thus AB, AG, AD, drawn from A, form the three angles BAC, GAD, BAD, The angles BAG, GAD, which have a common arm AG, and lie on opposite sides of it, are called adjacent angles; and the angle BAD, which is equal to angle ^^Cand angle GAD added t /'^rher, is called the sum of the angles 5.4 C and GAD. . ace the angle BAD is obtained by adding together the two angles BAG and GAD, the angle GAD will be obtained by subtracting the angle BAG from the angle BAD; and similarly the angle 5^ C will be obtained by subtracting the angle GAD from the angle BAD. Hence the angle GAD is called the difference of the angles BAD and BAG; and the angle BAG is called the difference of the angles BAD and GAD. 9. The bisector of an angle is the straight line that divides it into two equal angles. Thus (see preceding fig.), if angle BAG is equal to angle CAD, ACm called the bisector of angle BAD. The word bisect, in Mathematics, means always, to cut into two eqvMl parts. 10. AVhen a straight line stands on * another straight line, and makes the adjacent angles equal to each other, each of the angles is called a right angle ; and the straight line which stands on the other is called a perpendicular to it. Thus, \i AB stands on CD in such a manner that the adjacent angles ABC, ABD are eqiml to one another th« B D Bd0k L] DEFINITIONS!. these angles are caUed right angles, and .!£ is said to be perpen- dicular to CD. yl\. An obtuse angle is one which is greater than a right /angle. Thus ^ is an obtuse angle. 12. An acute angle is one which is less than a right angle. Thus B is an acute angle. 13. When two straight lines intersect each other, the opposite angles are called vertically opposite angles. Thus AEG and BED are vertically opposite angles ; and so are AED and BEG. 14. Parallel straight lines are such as are in same plane, and being produced e ever so far both ways do not \ « meet. £ the A. B D Thus AB and CD are parallel straight lines. If a straight line EF intersect two parallel straight lines A By CD, the angles AQH, GHD are called alternate angles, and so are angles BQH, OHG ; angles AGE, BGE, CHF, DHF are called erterlor angles, and the Interior opposite angles correspondinir to these are GHG, DUG, AGH, BGH. I 16. A figure is that which is inclosed by one or more boundaries ; and a plane figure is one bounded by a line or lines drawn tipon a plane. The space contained within the boundary of a plane figure is called its surface ; and its surface in reference to that of another figure, with which it is compared, is called its af«a. buclid's ELEMmrra. [Book f. according to the ^ D\ 'G definition, would not be figures. The word is, however, very frequently B^" — ^0 e\— j^ BurfJl* ''''*''' '""'^ ^ '°'*° """^ combination of points, lines, or ^ 16. A circle is a plane figure contained by one (curved) line which 18 called the circiimference, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal to one another. This point IS called the centre of the circle. jaus ABCDEFG is a circle, if all the straight lines which can be drawn from to the circumference, fluch as OA, OB, OG, &c., are equal to one another; and is the centre of the circle. Strictly speaking, a circle is an , inclosed space or surface, and the cir- B cumference is the line which incloses it. Frequently, however, the word circle is employed instead of circumfer- ence. It is usual to denote a circle by three ^ - letters placed at points on its circumference. The reason for this will appear later on. 17. A radius (plural, radii) of a circle is a straight line drawn from the centre to the circumference. Thus OA, OB, 00, &c. are radii of the circle ACf. 18. A diameter of p. circle is a straight line drawn through the centre, and terminated both ways by the cir- cumference. Thus in the preceding fi^rure 5^ ii;angle is one that has a right angle. ITiuii, if ABO is a right angle, the triangle ABC is right-angled. 27. An obtuse-angled triangle is one that has an obtuse angle. A.y Thus, if ABC is an obtuse angle, the triangle ABC is obtuse-angled. B^ ^s»»0 28. An acute-angled triangle is one that has three acute angles. . ^ Thus, if angles ^. B, C are each of them acute, the triangle AnOla aoute-angled. 29. Any side of a triangle may be called the base. In an isosceles triangle, the side which is neither of the equal sides is usually called the base. In a right-angled triangle, one of the sides which contain the right angle is often called the base "id the other the perpendicular; the side opposite the right angle is called the hypotenuse. Any of the angular points of a triangle may be called a vertex. If one of the sides of a triangle ^as been called the base, the angular point opposite that side is usually called the vertex. Thus, if 5C is called the base of a triangle ABC, A is the vertex. 30. If the sides of a triangle be prolonged both ways, nine angles are formed in addition to the angles of the triangle. Book I.] DBPINITIONS. 9 Thus at the iwint A there are the angles CAHt HAFt FAB/ at B, the angles ABG, QBD, DBC: at C, the angles F BOK, KGE, EGA. \ ^ f Of these nine, six only are called exterior angles, the three which are not so called being HAFy GBD, KCE. Angles G ABC, BCA, CAB are sometimes called the Interior angles 6f the triangle. OLASSIFICATICN OF QUADRILATERALS. 31. A rhombus is a quadrilateral that has all its equal. / Thus, if AB, BG, CD, DA are all equal, the quadrilateral A BOD is la rhombtis. The rhombus ABGD is sometimes namea by two letters placed at opposite corners, as AGotBD. EucUd defines a rhombus to be 'a quadrilateral that has all its sides equal, angles.' 32. A square is a quadrilateral that has all its sides equal, and all its angles right angles. Thus, if AB, BG, GD, DA are all equal, and the angles A, B, G, D right angles, the quadrilateral ABGD is a aquare. The square ABGD is sometimes named by two letters placed at opposite corners, as AG or BD; ftnd it is sftid to be described on any one of its four sides. 10 buolid's elements. [Book X. ThuB, if AB is panJlel to CD, and AD oaraUpl fn an *». ^ nlateral J5(7i) is a paraUelog^am. ^ ^^' *^' "^"^ T^e paraUelogram ^5C/> ia some tuues named by two letters placed at opposite corners, as ^C or BD ; and any one of its four sides may be caUed the base on which it stands. B^ 34. A recta^le is a quadrilateral whose opposite sides are paraUel, and whose angles are right angles. Thus, if AB is parallel to CD, AD A, paraUel to BC, and the angles A,B,C,D right angles, the quadrUateral ABCD is a rectangle. The rectangle ABCD is sometimes named by two letters placed at opposite corners, as ^C or BD. In B' ,„ books on mensuration, BC and AB wnnl^ k« n j ^i_ , and the breadth of 4e Ltende '^ " ^ « .'*"'^ "^^ ^'°«** J^A trapezium is a quadrilateral that ha. two sides Thus, if AD is parallel to BG, the quadrilateral ABCD is a trapezium. The word trapezoid is some- times used instead of tra- pezium. 36. A diagonal of a quadri- lateral is a straight line joining any two opposite comers. Thus AC tuid BD are diagonals of th« quadrilateral ABCD, )pp08ite sides Book I.] DEFINITIONS, POSTULATES, AXIOMB. H POSTULATES. Let it be granted : 1. That a straight line may bo drawn from any one point to any other point. 2. That a terminated straight line may be produced to any length either way. 3. That a circle may be described with any centre, and at any distance from that centre. The three postulates may be considered as stating the only mstruments we are aUowed to use in elementary geometry. These are the ruler or straight-edge, for drawing straight lines, and the compasses for describing circles. The ruler is not to be divided at Its edge (or graduated), so as to enabl^us to measure orf particular lengths; and the compasses are to be employed in describing circ.es only when the centre of the circle is at one given point, and the circumference must pass through another given point Neither ruler nor compasses can be used to carry distances If two points A and B are given, and we wish to draw a straight ! ^^A ' '* ^ "^''*^ ^ ^y «^°»Ply 'J'^^ ^^•' To produce a straight line, means not to make a straight line when there is none, but when there is a straight line already, to make it longer. The third postulate is sometimes expressed, 'a circle may be described with any centre and any radius.' That, however, is not to be taken as meaning with a radius equal to any given straight hue. but only with a radius equal to any given straight line drawn irom the centre. [The restrictions imposed on the use of the ruler and the com- passes, somewhat inconsistentiy on EucUd's part, are amvmt adhered to in practice.] . f**-* "« i*vw AXIOMS. 1. Things which are equal to the same thing are equal to one another. 2. If equals be added to equals, the sums are equal 3. .f equals be taken from equals, the remainders ar. equal. 12 mjCLTD*n BLBMBMTS. [BoA L 4. If equals be added to unequala, the sums are unequal, the greater sum being obtained from the greater unequal. 5. If equals be taken from unequals, the remainders are unequal, the greater remainder being obtained from the greater unequal 6. Things which are doubles of the same thing are equal to one another. 7. Things which are halves of the same thing are equal to one another. 8. The whole is greater than its part, and equal to the sum of all its parta 9. Magnitudes which coincide with one another are equal to one another. *' 10. All right angles are equal to one another. 11. Two straight lines which interaect one another cannot be both parallel to the same straight line. An axiom is a self-evident truth, or it is a statement the truth of which is admitted at once and without demonstration. Some of Euclid's axioms are general— that is, they apply to magnitudes of all kinds, and not to geometrical magnitudes only. The first axiom, which says that things which are equal to the same thing are equal to one another, applies not only to lines, angles, surfaces, and solids, but also, for example, to numbers, which are arithmetical, and to forces, which are physical, magnitudes. It will be seen that the firat eight axioms are general, and that the last three are geometrical It ought, perhaps, to be noted that some of -the axioms are often applied, not in the general form in which they are stated, but in particular cases that come under the general form. For example, under the general form of Axiom 2 would come two paiticular cases : If equals be added to the same thing, the sums are equal ; and If the same thing be added to equals, the sums are equal Again, a particular case coming under the general form of Axiom 4 would be : If the same thing be added to unequals, the sums are unequd Book L] AXIOMS, QUESTIONS. 18 alT^ I T ^8 «^^«l from the greater unequal. Axiom. 6 and 7. on the other hand, are only particular cases of more general ones-namely. Things which are double of equals are equ^and Things which are halves of equals are equal ; and th^ aiioms agajn are only particUar cases of still more general ones: SimUar multiples of equals (or of the same thing) are equal, and Similar fractions of equals (or of the same thing) are equal ^1°"" ^i' fr °*"^ ^°^^^'' definition or test of equality and tiie method of ascertaining whether two magnitudes aw equal by seemg whether they coincide-that is. by mentally applyinTthe one to the other, is called the method of superpositdoa Two magnitudes (for example, two triangles) which coincide are said to be conflrrum^; and tWs word, if it is thought desirable, may be used instead of the phrase, 'equal in every respect' Axiom 10 is, strictly speaking, a proposition capable of proof. The proof is not fvTh« T' T *Vi;'' t^u^ """"^^ ^'^"^ °°* ^ ^""y appreciated of ff« r^ K t^' \^^^ and undeiBtood tiie definitions of tiie third book, he will probably be able to prove it for himself Axiom 11 frequently referred to as Playfair's axiom (though Playfair stetes that it is assumed by others, particularly bv Ludl«S m hi.livdtments of Mathematics), has been substituted for th^ given by Euchd, which is proved as a corollary to Proposition 29 QUESTIONS ON THE DEFINITIONS, POSTULATES, AXIOMS. 1. How do we indicate a point ? 2. What is the only thing that a point has ? What has it not ' 3. Could a number of geometrical points placed close to one another form a line ? Why ? 4. Draw two lines intersecting each other in two points. 6. Could two straight lines be drawn intersecting each other in two points ? 6. What is Euclid's definition of a ' straight ' line » 7. Could a number of geometrical lines placed close to one another form a surface ? Why? ^ •"ubuer * " tSi*",'r."Jr "," « !•)!»« -"rf"* -d the" P«xi«ced, ,. „.^ .^_ piuuucea parii lie ? u EUOLIDS ELEMENTS. [BO0k L 10. Would it be possible to draw a straight line upon a surface thai was not plane ? If so, give an example. 11. How many arms has an angle ? 12. What name is given to the point where the arms meet ? 13. When an angle is denoted by three letters, may the letters be arranged in any order ? 14. If not, in how many ways may they be arranged, and what precaution must be observed ? 15. When is it necessary to name an angi* by three letters ? 16. How else may an angle be named ? 17. OA, OB, 00 axe three straight lines which meet at O. Name the three angles which they form. 18. Name the angle contained by OA and OB ; by OB and OG ; by OG and OA, 19. OA, OB, OG, OD awj four straight lines which meet at O. Name the SIX angles which they form. 20. Name the angle contained by OA and OB; by OB and OG ; by OG and OD; by OA and OG ; by OB and OD; hjOA and OD. 21. Write down all the ways in which the angle A can be named. 22. If the arms of one angle are respectively •qual to the arms of another angle, what inference can we draw regarding the sizes of the angles ? 23. In the figure to Question 17, if the angles ' AOB and BOG are added together, what angle do they form ? 24. In the same figure, if the angle AOB is taken away from the angle A OG, what angle is left ? 2ft In the same figure, if the angle BOG is taken away from the angle AOG, what angle is left ? 26'. The following questions refer to the figure to Question 19 : (a) Add together the angles AuB and BOG; AOB and BOD^ AOG and GOD ; BOG and GOD. (6) From the angle AOD subtract successively the angles COD^ ^i;/X>, ^i.\J\J, I3\JjJ. Book L] QITBBTlOmi » 30. C (c) From the angle BOD subtract the angles COD, BOO id) To the sam of the angle. AOB and BOC ad.i the diflfereno. of the angles BOD and BOG; and from the sum of AOB and BOC subtract the diflference of BOD and COD. 27. Draw, as well as you can, two equal angles with unequal wms. ^®- " " two unequal „ equal ., 29. If two adjacent angles are equal, must they necessarily be right angles? Draw • figure to illustrate your answer. If two adjacent angles are equal, what name could be given to the arm that is common to the two angles ? 31. When an angle is greater than a right angle, what is it oftUadt ^ M less N ^ >J3. „ equal to n « 84. In the accompanying Hguro, name two right angles, two acute angles, and one obtuse angle. 35. What are angles AEC, AED called with reference to each other? angles AEC, BED? angles AEC, BEC? angles BEC, AEDf angles BEC, BED? 36. Would it be a suflBcient defini- tion of parallel straight lines to say that they never meet though produced indefinitely far either way? Illustrate your answer by reference to the edges of a book, or otherwise. 37. Draw three straight Unes, every two of which are pareUel. 38. Draw three straight lines, only two o^ which are paraUeL 39. Draw three straight lines, no two of which are parallel. 4a What is the least number of lines that will inclose a space? Illustrate your answer by an example. 41. How many radii of a circle are equal to one diameter? 42. How do we know that all radii of a circle are equal ? 43. Prove that all diameters of a circle are equal. 44 Are all Unes drawn from the centre of a circle to the dxxms^ ference equal to one another ? 45. What is the distinction between a circle and « dromik f erencd ? iA. Ta • -.'Mc rrviil CTCr USCU iOT XiikG "tJIPf f 16 suolid's blimsnts. [Book t 18 47. How many letters are generally used to denote a cirele? 48. Would it be a sufficient definition of a diameter of a circle to say that it consists of two radii ? 49. Prove that the distance of a point inside a circle from the centre is less than a radius of the circle. 50. Prove that the distance of a point outside a circle from the centre is greater than a radius of the circle. dl. What is the least number of straight lines that will inclose a space? 02. What name is given to figures that are contained by straight lines? 63. Could three straight lines be drawn so that, even if they were produced, they would not inclose a space ? 54. What is the least nimiber of sides that a rectilineal figure can have ? 65. ABC is a triangle. Name it in fire other ways. 66. If ^^ is equal to ^(7, what is triangle ABO called ? 57. If AB, BG, C^ are all equal, what triangle ABC called ? 58. U ABt BG, CA are all imequal, what is triangle ABO oaUed? 59. What name is given to the sum of AB, BG, and GA ? 60. Which side of a triangle is called the base ? 61. Which side of an isosceles triangle is called the base ? 62. When the hypotenuse of a triangle is mentioned, of what sort must the triangle be ? 63. What names are sometimes given to those sides of a right< angled triangle whicl) contain the right angle ? 64. Would it be a sufficient definition of aa acute-a^ ':^ed say that it had neither a right nor an obtuse tkti u 66. ABO is a triangle. Name by one letter the angles respectively opposite to the sides AB, BG, GA. 66. Name by three letters the angles respec tively opposite to the sides AB, BG, GA. 67. Tv'srmo the sides respectively opposite to the angles A, B,0. S8. y^imy'i'. by one letter and by three letters the angle contained by AB and AG; by AB and BG ; by AG and BG. ^riangle to ? IMk L] QUB8TION8. If 69. 70. 71. 72. 73. 74. 76. Nuae ftll the triangles in the accompanying figux«. Name the additional triangles that would be formed if AD were joined. Name by three letters all the angles opposite to BG; to BE; to CE. Name all the sides that are opposite to angle ^; to angle D. Name all the angles in the figure that are called exterior angles of the B triangle BEC; of the triangle AEB; of the triangle GED. ABGD is a quadrilateral Name it in seven other waya. K the diagonals AG, BD be B^ \ 76. 77. 7a 79. 80. 81. 82. 83. 84. SSL 86. 87. 88. 89. 90. 91. 92. 93. drawn, and E be their point of intersection, how many triangles will there be in the diagram ? Name them. Name the two angles opposite to the diagonal AO. •» •• " BD. n through which the diagonal A G passM. " •• II II BD H CJould a square, with propriety, be called a rhombus ? Could a rhombus be called a square ? Could a rectangle bo called a parallelogram ? Could a parallelogram be called a rectangle ? Would it be a sufficient definition of a parallelogram to say that it is a figure whose opposite sides are parallel ? Why ? Could a parallelogram or a rectangle be called a trapezium ? Could a trapezium be called a parallelogram or a rectangle ? What is a diagonal of a quadriUteral, and how many diagonals has a quadrilateral ? How many sides has a polygon ? Which postuUte allows us to join two points ? " H produce a straight line? w If describe a circle ? In what sense is the word 'circle ' used in the third postuUte? Wha<; are the only instruments that may be used in elementary plane geometry? Under what restrictions are they to h% used? What is an axiom ? Give an example of one. 94. 9& State Euclid's axiom about rt fim 18 buclid's elements. [Book L 96. Would it be correct to say, magnitudes which fill thj same spac^ instead of magnitudes which coincide ? lUustrate your answer by reference to straight lines, and angles. What is Euclid's axiom about right angles ? What is the axiom about parallels ? Would it be correct to say, two straight lines which pass through the same point cannot be both parallel to the same straight line ? 100. Could two straight lines which do not pass through the sam* point be both p-^rallel to a third straight line ? 97. 98, 99. EXPLANATION OF TERMS. Proposition* are divided into two classes, theorems and problems. A theoxAm is a truth that requires to be proved by means of other ■bruths already known. The truths abeady known are either cxioms or th-^jrems. A praiem is a construction which is to be made by means of ce-tain instruments. The instruments allowed to be used are (see the remarks on the postulates) the ruler and the compasses. A corollaiy is a truth which is (more or less) easUy inferred from "^proposition. xn the statement of a theorem there are two parts, the hypotheslB and the conclusion. Thus, in the theorem, 'If two sides of a tri- angle be equal, the angles opposite to them shall be equal,' the part, I if two sides of a triangle be equal,' is the hypothesis, or that which IS assumed; the other part, 'the angles opposite to them shall be equal,' is the conclusion, or that which is inferred from the hypo- thesis. The converse of a theorem is derived from the theorem by inter- changing the hypothesis and the conclusion. Thus, the converse of the theorem mentioned above is, 'If in a triangle the angles opposite two sides be equal, the sides shall be equal.' When the hypothesis of a theorem consists of several hypotheses, there may be more than one converse to the theorem. In proving propositions, recourse is sometimes had to the following method. The proposition is supposed not to be true, and the coa- Sook I.] QUESTIONS, TERMS, SYMBOLS. 19 sequences of this Buppoaition are then examined, till at lencrth a result 18 reached which is impossible or absurd. It is therefore ujferred that the proposition must be true. Such a method of proof IS called an indirect demonstration, or sometimes a reductlo ad aterardum (a reducing to the absurd). fi SYMBOLS AND ABBREVIATIONS. +, read plm, is the sign of addition, and signifies that the magni- tudes between which it is placed are to be added together. -,read minm, is the sign of subtraction, and signifies that the magnitude written after it is to be subtracted from the macni. tude written before it. <-, read difference, is sometimes used instead of minus, when it is not known which of the two magnitudes before and after it is the greater. i= is the sign of equality, and signifies that the magnitudes between which It IS placed are equal to each other. It is used here a« an abbreviation for 'is equal to,' 'are equal to,' 'be equal t«. and 'equal to.' stands for 'perpendicular to,' or 'is perpendicular W ' parallel to,' or ' is parallil to.' nm •angle.' 'triai;i::e. © QCO I ' parallelogram.' I ' circle.' I 'circumference,* .-. .. 'therefore.' This symbol turned upside down (-.-i which IS sometimes used for 'because' or 'since.' I have not introduced, partly because some writers use it for ' therefore ' and partly because it is easily confounded with the other. AB^ stands for ' the square described on AB: AB . £C stands for ' the rectangle containedby AB and BOJ A : B stands for ' the ratio of ^ to ^.' \A:BJ stands for 'the ratio compounded of th« ,.of;«- -^t a .„ n « ^' : 0' ) and ii to a' " ^""^ '" " *° ^ so EUCLID S ELEMENTS. i ■ III i . 1^ [Book t A :B = C :D stands for the proportion *A is to i? as C is to iX' The small letters a, b, c, m, n, p, &c. stand for numben. App. stands for ' appendix.' Ax. •1 • axiom.' Const. •1 ' construction.* Cor. It ' corollary.' n^/. II * definition.' Hyp. II 'hypothesis.* P08t. II * postulate.' m. II 'right' In the references given at the right-hand side of the page (Eaclid gives no references), the Roman numerals indicate the number of the book, the Arabic numerals the number of the proposition. Thus, I. 47 means the forty-seventh proposition of the first book. In the figures to ceutain of the theorems, it will be seen that some lines are thkh, and some dotted. The thick lines are those which are given, the dotted lines are those which are drawn in order to prove the theorem. [In a few figures this arrangement has been neglected to attain another object.] In the figures to certain of the problems, some lines are ihickf some th\ and some dotted. The thick lines are those which ai« given, the thia lines are those which are drawn in order to effect the construction, and the dotted lines are those which are necessary for the proof that the construction is correct. In' the figures which illiutrate definitions^ the mwm an al«aas^. aavariably thin. if Book LJ ABBRlvTlATIONS, PROPOSITION 1. 21 PROPOSITION 1. Pboblbm. To i^nU an eqmUxteral triangle m a given straight line. Let AB be the given straight Kne : it is requvred to describe an equilateral Marble < /?♦ /jj^ i.« • • j prove that aBF ia an e,„ilater:i ,^.^''' ^^ "* '"■•'«^ 3. Show how to mako a rhomhas ha™g one of it. diagonab wiuJ to a given straight line. "»goii«u equal V ' >iii 2? EUCLID 8 ELEMENTS. [Book I 5. If AB be produced both ways to meet the two circles again at D and E, prove that the straight hue DE is equal to the sum of the three sides of the triangle A BC. 6. Show how to find a straight line equal to the sum of the thnt sides of any triangle. Show how to find a straight line which shall be : 7. Twice as great as a given straight line. 8. Thrice » » %% 9. Four times ti n h 10. Five It 4a PROPOSITION 2. Probldi. From a given point to draw a straight line equal to a givmi straight line. Let A be the giren point, and BC the given straight line : it is required to draw from A a straight line = BC. Join AB, Post. 1 and on it describe the equilateral A DBA. I. 1 With centre B and radius BC, describe the GEF; Post. 3 and produce DB to meet the Q** 0£F in E. Post. 2 [Book L gain atD the sum "^ the thnt Book L] PROPObinoN ai. \ agtvmi ;ht line : Po«^. 1 ■ 1 Post. 3 ■ Post 2 U 2» With centre D, and radius Z)JS;, describe the EGH; Poet. 3 and produce DA to meet the Q* -S^^J'Z?' in G. ' Post. 2 ilG^ shall = BC. Because Z)J: = DG, being tadii of EGH, I. Def. 16 and DB = D^l, being sides of an equi- lateral triangle ; /. Def. 23 .-. remainder BE = remainder AG. /. Ax. 3 But BE = 5(7, being radii of CEF; I. Def. 16 il(? = BG. L Ax. 1 1. If the radius of the large circle be double the radius of the small circle, where will the given point be ? 2. AB ia t^ given straight linti ; show how to .baw from A any number of straight lines equal to AB. 3. AB i& a, given straight line ; show how to draw from B any number of straight lines equal to AB. 4. AB is a given straight line ; show how to draw through A any number of straight lines double of AB. 6. AB u& given straight line ; show how to draw through B any number of straight lines double of AB. 6. On a given straight line as base, describe an isosceles triangle each of whose sides shall be equal to a given straight line. May the second given straight hne be of any size ? If net, how large or how small may it be ? Give the construction and proof of the proposition— 7. When the equilateral triangle ABD is described on that side of AB opposite to the one given in the text, a When the equilateral triangle ABD is described on the same side of ^5 as in the text, but when its sides are produced through the vertex and not beyond the base. 9. When the equilateral triangle ABD is described on that side of AB opposite to the one given in the text, and when its sides are produced through the vertex. 10. When the given point A is joined to G instead of 5. /fake diagrams for all the cases that can arise by descnbmg the equilateral triangle on either side of AG, and producing ito wdes either beyond the base or through the vertex cuglid's blsmbnts. [Book I PROPOSI'^ION 3. Problbm. From the greater of two given straight lines to cut of a pan equal to the less. i Let AB and C be the two given straight lines, of which AB is the greater : it is required to cut of from AB apart = G. From A draw the straight line AD = C; L 2 with centre A and radius AD^ describe the DEF, Post. 3 cutting AB at E. AE shaU = G. For AE = AD, being radii of O DEF. I. Def. 16 B^t AD^G; Const. •'• ^E==G. LAx,\ 1. Give the construction and the proof of this proposition, using the point B instead of the point A. 2. Produce the less of two given straight lines so that it may be equal to the greater. 3. If from AB (fig. 1 and fig. 2) there be cUt oflF AD and BE, each equal to C, prove AE = BD. Fig.l. Fig. 2. D £ B B O- 4 Show how to find a straight line equal to the sum of two given straight lines. Botfk t\ PROPOSITIONS 3. 4. 3f^ 5. Show how to find a straight hue equal to the difference of two given straight hnes. 6. Show that if the difference of two straight lines be added to the sum of the two straight lines, the result will be double of the greater straight line. 7. Show that if the difference of two straight lines be taken away from the sum of the two straight Unes, the result will be double of the less straight line. PROPOSITION 4. Theorem. If two sides and the contained angle of one triangle be equa to two sides and the contained angle of another triangle, the two triangles shall he equal in every resjyect^-that, is, (1) The third sides shall be equal, (2) The remaining angles of the one triangle shall be oqual to the remaining angles of the other triangle, (3) The areas of the two triangles shall be equal A D It IS required to prove BC= EF, lB = ^ E ^ r~ J ir AABC^ABEF. ,^^~LJ', If A ABC be applied to A DEP^ 80 that A falls on D, and so that AB faUs on DE • then B wiU coincide with E, because AB = DE. ' And because AB coincides with DE, and lA~'i n /. ^C'wiUfalloni^J^. ^^-LU, And because AC = DF, -" l^^-■ TTiLii JJ , Byp, ZUCLIDS BLEA1ENT8. [Book ]fe / Def. 3 /. Ax. 9 /. Ax. 9 Now, since B coincides with E^ and C with F^ .*. BC will coincide with EF ; .'. BG = EF. Hence also z. B will coincide with z. ^; .'. L B = L E; and ^ (7 will coincide with l F; .'. lC == l F; I, Ax. 9 and A ABC will coincide with A DEF; .'. A ABC = A iD^iP. /. Ax. 9 In the two a b ABG, DEF, 1. It AB = DE, AC = DF, but z ^ greater than z A where would AG fall when J^C is applied to DEF as in the proposition ? 2. If AB = DE, AG=DF, but z ^ less than z A where would ^Cfall? a If v4i? = DE, z 4 = z Z), but AG greater than Di?; where would C fall? 4. If AB^ DE, lA = z A but AG less than DF, where would C fall? 5. Prove the proposition beginning the superposition with the point B or the point G instead of the point A. -3. If the straight line GD bisect the straight line AB perpendicu- larly, prove any point in ^D equidistant from A and B. 7. GA and GB are two equal straight lines di vn from the point C, and GD is the bisector of z AGB. Prove that any point in GD is "'equidistant from A and B. * The straight line that bisects the vertical angle of an isosceles triangle bisects the base and is perpendicular to the base. Sn. ABGD is a quadrilateral, one of whose diagonals is BD. If AB = GB, and BD bisects z ABG, prove that AD = CD, and that BD bisects also z ADG. l(X Prove that the diagonals of a aauare ar^ finiia]„ BMkLI FROPOSmONS 4, i. 27 la If two quadrilaterals have three consecutive .ide. and the two contained angles in the one respectively equal to three consecutive sides and the two contained aiglet in the oth« the quadrilaterals shall be. equal in every respect. * PROPOSITIONS. Theorem. Tfi^ ambles at the base of an isosceles triangle are e i I I 'I • ! ■■; I I f. I- 1 1 1 ■ «§ mXOUD'S BLBHBNTS. iBttfk L (2) Because the whole AF = whole AG, Const aui, the part AB = part ^IC; Hyp. ttie remainder 5^ = remainder CG. I. Ax. 3 . ( BF= CG Proved in (2) (3) In As BFG, CGBA FG = GB Proved in (1) ( L BFO = L CGB; Proved in (1) L BCF = L CBG, Bind L FBC = L GGB, /. . 4 (4) Because whole l ABG = whole l ACF, Proved in (1) and the part l GBG = part l. BGF; Proved in (3) .-. the remainder l ABG = remainder l AGB; L Ax. 3 and these are the angles at the base. But it was proved in (3) that l FBG = l GGB; and these are the angles on the other side of the base. Cor. — If a triangle have all its sides equal, it will also have all its angles equal ; or, in other words, if a triangle be equilateral, it will be equiangular. 1. If two angles of a triangle be unequal,*the sideg opposite to them will also be unequal. 2. Two isosceles triangles ABG^ DBG stand on the same base 5C, and on opposite sides of it ; prove / ABD = l AGD. S. Two isosceles triangles ABG, DBG stand on the same base BC, and on the same side of it ; prove z ABD = l AGD. 4k In^the figure to the second deduction, if ^2> be joiutd, prove that H will biieot th« angles at A and D. Book L] PBOPpSITIONS 6, 9. « . 29 point. D,E^ ukA eq^/dUwf,! l;^*.^0,two the triancIes^BP ^/^n V? ^ -^ ' P™™ th«t .ppliad to the trZ '° >« '*«1 up. turned over, „d ''i::t:tL"Lrh°Liti:"''^«» •>-"«-!-« the «,,...* PROPOSITION 6. Theorem. I B^ 10 In A ABChi L ABC = l ACB- 't IS required to prove AC = AB. iiiid from it cut off BD = AC '"iJ join Z)a . ' ^' ^ Post, i -^ ^B DBC, ACB,] BC=CB \ lDBC^ ^ ACB. Hyp 80 BuouD'fl SLmnm. M!i [B(7 is greater than l BCD. Much more then is l BDC greater than l BCD. But because BC = BD, .-. lBDC = l BCD; /. 5 that is, L BDC is greater than and equal to / BCD which is impossible. ' The third case needs no proof, because BC is not = BD Hence two triangles on the same base and on the same side of it cannot have their conterminous sides equal. 1. On the same base and on the same side of it there can be onlv one equilateral triangle. ^ 2. On the same base and on the same side of it there can be only rae isosceles triangle having its sides equal to a giv«a straigU 3. Two circles cannot cut each other at more than one point either above or below the straight line joining their centrw^ PROPOSITION 8. Theorem. Tf three sides of one triangle he respectiv«ly equal to threa sides of another triangle, the two triangles shall he equal in every respect; that is, (1) The three angles of the one triangle shall he respectively equal to the three angles of the other triangle, W The areas nf ihj>. ijju% yWy>«>^7^. ^i.^ji t- ■, 32 Euclid's elements. [Book I II III Hi In ZVb ABC, DEF, let AB = DE, AC ^ DF, BG = ^i^.- «Y w required to prove i-A = l D, l B = lE lG =- lF and A ABC = A Z>^i^. ' ' If A i4i?C be applied to A DEF, so that B falls on E, and so that ^C falls on EF; then a will coincide with F, because BG = EF, * Hyp Now since BG coincides with EF, .'. BA and -4C liiust coincide with ED and DF. For, if they do not, but fall otherwise as EG and GF ; then on the same base EF, and on the same side of it ' there wiU be two As DEE, GEE, having equal pairs of conterminous sides, which is impossible. r y .*. BA coincides with ED, and AG yrith. DF. Hence l A will coincide with l D, .- . lA= lD; LAx.^ and L B will coincide with lE, .. lB^ lE; I. Ax. 9 and z. a will coincide with lF, .: lG=^ lF; I. Ax. 9 and A ABC will coincide with A DEF • ^^BC^^DEF . ' j.^^.9 1. The straight hne which joins the vertex of an isosceles triande to the nuddle point of the base, is perpendicular to the bwe. and bisects the vertical angle. 2. The opposite angles of a rhombus are equal. 3. Either diagonal of « rhombus bisects the angles through whick it passes. 4. ABCD is a quadrilateral having AB = BC and AD = DC • prove that the diagonal BD bisects the angles through which it passes, and that i A =i /. C. -41 ■ m Book Lj PROPOSITIONS 3, 9. 88 5. Two isosceles tnan,^ stand on the same base and on opposite sides of It ; prot rthat the straight line joining their v^Z bisects both vertical angles. g 'ineir vertices 6. Two isosceles tri .gles stand on the same base and on the same bit J P7;;.*l^*t the straight hne joining their vertices, being produced, bisects both vertical angles. 7. In the figures to the fifth and sixth deductions, prove that the straight Ime joming the vertices, or that straight line produced, bisects the common base perpendicularly 8. Hence give a construction for bisecting a given straight line. 9. The diagonals of a rhombus or of a square bisect each other per- pendiciilarly, *^ 10. If any two circles cut each other, the straight line joining their pomts of intersection is bisected perpendicularly by the straight Ime joming their centres. J J' wie 11. ftove the proposition by applying the triangles so that they may fall on opposite sides of a common base. Join the two PROPOSITIO]sr 9. Problem. To bisect a given rectilineal angle. Let ACB be the given rectilineal angle it is required to bisect it. In AO take any point Z), and from nR />nf .^ap nc^ ^ /^r> —^tv -_-ii ^.,-XV — \jj^. /. 3 H mtJOLlD^S ELBMBNtS. IBotfkL \l Join DEy and on DE, on the side remote from O, describe the equilateral A DEF. /. 1 Join CF. CF shall bisect l ACB. ' (DG^EG Const. In As DGF, EGF, } GF = GF [DF^EF; LDef.^Z .-. L DGF= L EGF; /. g that is, GF bisects l AGB, 1. Prove that CJ^ bisects angle DFE. 2. If the equilateral triangle DEF were described on the same aide of DE as C7 is, what three positions might F take ? 3. Show that in one of these positions the demonstration remains the same as in the text. 4. Would an isosceles triangle DEF described on the base DB answer the purpose as well as an -equilateral one ? If so, why? 6. Prove the proi)osition and the first deduction, using I. 6 and I. 4 instead of I. 8. 6. Divide a given angle into 4 equal parts. 7. Could the number of equal parts into which an angle may be divided be «Ktended beyond 4? If so, <^numerate the numbers, t. Prove from an equilateral triangle that if a right-angled triangle have one of the acute angles double of the other, the hypot- •nuse is double of the side opposite the least angle. ieokt] W»POSiflON8 9, 10. ^» PEOPOSITION 10. .Probmm. To bisect a given straigfU tine. O /.I 7.9 /. Be/. 23 let AB be the given straight line : it is required to bisect it. On AB describe an equUateral A ABO and bisect ^ ACB by CD, which msets ^B at D, AB «hall be bisected at D. In As ACD, BOD, J OD = OD .-. AD = 5Z>,. ^' ^**«^- that is, AB is bisected at D. ^' ^ 1. Would an isosceles triangle described on AB as base answer th« purpose as well as an equilateral one ? If s^ X * *^' 3 iThet* """J' ^^'^^ '^^^^*^"« ^^' « perpenkx^ufa to AB SlTlL?'^^* '"-'^ ^ «^-P^« -^od of bisecting a given numben. '' '''' ' enumerate ttie se buclld's elements. fBookL H •f. 1 I 1 li H ! ill 11 ^^n liS ■- 1 i ■ ■ 1 i 9. Find a straight line half as long again as a given straight Una. 10. Find a straight line equal to half the sum of two given straight lines. 11. Find a straight line equal to half the diflFerence of two given straight haeb. 12. If, in the figure to the proposition, z ^ is bisected by AF^ which meets BG at F, prove BF = BD, and AF = CD. PROPOSITION 11. Problem. To draio a straight line verpendicular to a given straight line from a given point in the same. D C W I^t AB be the given straight line, and C the given point in it: it is required to draw from C a perpendicular to AB. In ^C take any point Z), and from CB cut off CE = CD. /. 3 On DE describe the equilateral A DEF^ /. 1 and join GF. CF shall be X AB. (DC^ EG • Const In As DGF, EGF, } CF = CF i DF = EF; L Def 23 .-. L DGF = L EGF; I 8 :GFia±AB. I. Def 10 1. Would an isosceles triangle described on DE as base answer the purpose as well as an equilateral one ? If an, whv ? Book I.] PROPOSITIONS 11, 12. 37 t. If th« giren point were situated at eitiier end of the given straight line, what additional constructio. would be necwsary m order to draw a perpendicular ? & At a given point in a given straight line make an angle equal to half of a nght angla ® * 4 At a given point in a given straight line make an angle equal to one-fourth of a right angle. 5. Construct an isosceles right-angled triangle 6. Construct a right-angled triangle whose base shall be equal to nail the hypotenuse. 7. Find in a given straight line a point which shall be equally distant from two given points. Is this always possible ? If not, when is it not ? 8. ABC is any triangle ; AB is bisected at L, and AG at K. From i; there is drawn LO perpendicular to AB, and from ^,A:0 perpendicular to^C, and these perpendiculars meet at a Prove that OA, OB, OCsire all equal. 9. Compare the construction and proof of I. 9 with those of I. H and show that the latter proposition is a particular case of PROPOSITION 12. Problem. To draw a straight line perpendicular to a given straight line from a given point without it. o Let AB be the given straight line, and C the given point without it : or it is required to draw from C a perpendicular to AB. ■ Take any point Z) on the other side of ^5/ with centre G and radius CD, describe the Q EDF, cutting ■4J5. or AB nrorlnpprJ of TP r^r^A jp ^ S8 KTOUd'B ELEMBNTt. [Sookt I o Biaect EF at G ; and join CO. Join CE, CF. (EG In As CGE, CGF, \ OC (CE ,\ L CGE = L GGF; ,-. GG is ± AB. CG shall be X il J7. CF; I 10 Obn«/. /. D^. 16 7.8 /. i)e/. 10 1. Is GEF an equilateral triangle ? 2. Prove that GG bisects /: JE^C^. 3. Instead of bisecting EF at G and joining CO', would it answer the purpose equally well to bisect i EGF by CG ? 4. Instead of taking D on the other side of AB, would it answer equally well to take D in AB itself ? 6. Two points are situated on opposite sides of a given straight line. Find a point in the straight line such that the straight lines joining it to the two given points may make " equal angles with the given straight line. Is this always possible ? 6. Use the tenth deduction on I. 8 to obtain another method of drawing the perpendicular. PROPOSITION 13. Thborbm. The angles which one straight line makes loith another on one side of it are together equal to tioo right angles. Let AB make with CD on one side of it the l a ABC, ABD- a is required to prove l ABC + l ABD = 2 rt, ls. I.] PROPOMTIONB 12, la. 39 /. Def. 10 /. 11 I, Ax. 9 /. Ax. I Ob do it) If L ABC = L ABD, then each of them is a right angle ; .-. L ABC + L ABD = 2 rt. lb. (2) If L ABC be not = l ABD, from B draw BE A. CD. Then l s EBC, EBD are 2 rt. l a. But z. ABC + A ^5/) = i_ EBC + z. ^5Z> • -'^ i. ABC + L ABD = '^ It. 1,8. Cor. l.^Hence, if two straight lines cut one another. Uie four angles which they make at the point where they cut are equal to four right angles. For L AEG+ l AED = 2tt. z.8, /. 13 *id d.BED+ L BEC-.2Tt.La. I. 13 .'. L AEC+ L AED + L BED + l BEG^irt. La. of^J^ ^;7^^^ ^^ successive angles made by any number of straight hnes meetmg at one point are together equal to four nght angles. ^ Let OA, OB, OC, OD, which meet at O, make the successive angles AOB, BOC, COD, DOA: it is required to prove these l9 '= i rt. Ls. Pm,h -■\j .1:3. v.r w Jit, I '" BUOLIDS ELEMENTS. ^^^^^ j Then ^ AOB + l BOG + l GOD + a Z)0^ = {lAOB^. l BOE) + ( L EOD + ^ DOA) = 2rt. ^8 . +2rt ^s. /. 13 = 4 rt. z. 8. Def^.-Two angles are called supplementary when their ««nir I ff^ '"^^''^ '^^ ^^*^^^ ^^gl^ i« «aM the supplement cf the other. Thus, in the figure to the proposition, l ABG and l ABD are supplementary; l ABG is the supplement of l ABD and I. ABD is the supplement of L ABG. * DBF.— Two angles are called complementary when their sum 18 one right angle; and either angle is called the complement of the oth^er. Thus, in the figure to the proposition, l ABD and L ABE^e complementary; z. ABD is the complement ot L ABE, and l ABE is the complement of l ABD, 1. In the figure to Con 1, name all the angles which are aupple- mentary to z AEG,to , AMD,to l BED,to l BEG, tary to i AOB, L BOE, i GOE, i EOD, l AOD. a In the figure to I. 6 name the angles which are supplementary 4. In the accompanying figure, L AOB \a right. Name the angles which are complementary to l AOG, l AOD, L BOD, L BOG. 5. In the same figure, \i i AOG = l BOD, prove I AOD = l BOG; and if L AOD = L BOG, prov« L AOG = L BOD. «. Im the figure to the proposition, if z s ^5C and ^57> be big«cted, prove, that the bisectors are perpendicular to each other. 7. If the angles at the base of a triangle be equal, the angles on the other side of the base must alan ho ^,.«\ Book I] PROPOSITIONS 13, 14. 41 8. If the base of an isosceles triangle be produced both ways, the exterior angles thus formed are equal. 9. ^^C' is a triangle, and the sides AB, AG are produced to D 10 A ^f Y- '^^^=' ^(^B, prove A ^5C isosceles. rf .u * '^^^' ^°^ *^^ ^^« ^^ i« produced both ways. If the exterior angles thus formed are equal, prove L ABC 1S0SCG16S* PROPOSITION U. Theorem. If at a point in a straight line, two other straight lines m opposite sides of it make the adjacent angles together equal to two right angles, these two straight lines shall be m one and the same straight line. B o b"^ b Atthe point B in AB, let BC and BD, on opposite sides of AB, make l ABC + l ABD = 2 rt. as: it is required to prove BD in the same straight line with BG. If BD be not in the same straight Hne with BC, produce ^^ ^ ^' Post, 2 then BE does not coincide with BD. Now since CBE is a straight line, .'. ABC + L ABE = 2 rt. z. s. /. 13 But L ABC+ L ABD=2 rt. z.s; Hyp. L ABC + L ABE = L ABC + l ABD. I. Ax. 1 Take away from these equals l ABC, which is comn)()n ; l.ABE=lABD, LAx,^ which IS impossible ; .*. BE must coincide with BD; that is, BD is in the same straight linR wi+h Tin it buolib's elbmbntb. [!Bo(flEt le ABOD, EFOH are two squares. If they be placed so that F fallg on (7, and FE along CD, show that FG will either fall along CB, or be in the same straight line with it 2. If in the straight line AB^Sk point E be taken and two straight lines EG, ED be drawn on opposite sides of AB, making I AEC = L BED, prove that EC and ED are in the same straight lin& .3. If four straight lines, AE, CE, BE, DE, meet at a point E, so that L AEG = L BED and z AED = l BEG, then AE and EB are in the same straight line, and also GE and ED. 4. f* is any point, and AOB a right angle ; PM is drawn perpen- dicular to OA and produced to Q, so that QM = MP; PN is drawn perpendicular to OB and produced to R, so that RN = NP. Prove that Q, O, R lie in the same straight line. 6. If in the enunciation of the proposition the words 'on opposite sides of it' be^omitted, is the proposition necMMurily trot? Draw a figure to illustrate your answer. PROPOSITION 15. Theorem. If two straight lines cut one another^ the vertically opposite angles shall he equal. Let AB and CD cut one another at E: it is required to prove l AEG = l BEBy and l BEG « L AED. Because GE stands on AB^ L AEG + L BEG = 2 rt. z. s. /. 13 Because BE stands upon CZ), L BEG + L BED = 2rt.z.s; /. 13 L AEG + L BEG = L BEG + l BED. I. Ax. 1 Book I.] PROPOSITIONS ,15, 16. 4^ Take away from these equals l BEC, which is common • ^ AEG= L BED. /. ^a. 3 Hence also, z. BEC = l AED. 1. Prove . AEC = . BED making z ^i^T) the common angle -. .. / BEG = / AED, „ L AEG „ „ f», 1 il u-^^'"'*'^ *'^ ^^' ""^^ ^^ '« Pr*>'J"«ed to GT, prove that EG bisects z 5jE^a *^ ^" " ^ifrS '^^^''t'^ ^^ ^^' '"'^ ' ^^^ ^'««^*«d by QE, prove /'-ft and QE m the same straight line -> ^ v '■ " line's "'e^Ib^I ^f ' ' ^""* ^ '^ **'^°' ^"^ *-« «*^-gl^t L Iwr' \.l^n^^^'' ? °PP°'^*" «^^«« °f ^^. "taking straHurne. '^^''' '"" *'^* ^^ ^^' ^^ ^^ ^ *^« -- 7. ABC is a triangle. ^A CE straight lines drawn making equal angles w.th^C. and meetin,,. the opposite sides in i) and E and each other m F; prove that if z AFE =. l AFD the triangle is isosceles. ' ^^ PROPOSITION 16. Theorem. ' If one dde of a triamjJe he produced, the exterior anrjle shall be greater than either of the intenor opposite angles. •D G^ Let ABC be a triangle, and let BC be produced to D • It IS required to prove l ACD greater than i, BAG and aiso greater than ^ ABC. ' Bisent AC at E ' ^' 7.10 44 BUOLIDS ELBIIKNTB. yl [Book X B \ join BE^ and produce it to F^ making EF = BE; and join CF. ( AE= CE In As AEB, CEF, | EB = EF ^ i L AEB = L CEF; L EAB = L EOF. But L A CD is greater than l EOF; ,\ L ACD is greater than l EAB. Hence, if 4 C be produced to G^ L BGG is greater than l ABG. But L ACD =- L BGG; ,'. L ACD is greater than t. ABG. 1. Prove / A less than AEF, BEG, ACD, BGG. L 3 Gonst. Const. I. 15 1. 4 / Ax. 8 /, 15 2. 3. 4. 6. 6. 7. 8. 9. II I F „ FCD, FCG, BEG, AEF. L ABE „ AEF, BEG, ACD, BGG. I GBE „ AGD, BGQ, AEB, GEF. L AGB „ AEB, CEF. I BEG V AGD, BGG. L BGE „ AEB, CEF. L EGF „ AEF, BEG. Draw three figures to show that an exterior angle of a triangle may be greater than, equal to, or less than the interior adjacent angle. 10. From a point outside a given straight line, there can be drawn to the straight line only one perpendicular. 11. ABG is a triangle whose vertical / ^ is bisected by a straight line which meets BG at D ; prove L ADC greater than l DAG, and i ADB greater than z BAD. Bock t] PROPOSITIONS 16, 17. 49 12. In the figure to the propoaition, if ^Z* be joined, prove : {\) AF = 5a (2) Area of a ^i9C=area of a BCF. (3) Area of A ABF = area of a ACF. 13. Hence construct on the same base a series of triangles of equal area, whose vertices are equidistant 14. To a given straight line there cannot be drawn more than two equal straight lines from a given point without it. 15. Any two exterior angles of a triangle are together greater than two right angles. PROPOSITION 17. Theorem. The sum of any two angles of a triangle is less than two right angles. Let ABC be a triangle : it is required to prove th^ mm of any two of its angles less than 2 rt. ls. Produce BG to D. Then l ABCia less than l AGD. / ig .-. L ABG + L ACB is less than l AGD + z. AGB But L AGD+ L AGB = 2 rt. ^ s ; / 13 .*. L ABG + L AGB is less than 2 rt. a e Now '. ABC and l AGB are any two angles of the triangle ; .-. the sum of any two angles of a triangle is less than 2 rt. L s. 1. Prove that in any triangle there cannot be two risht angles or two obtuse angles, or one right and one obtuse angla ° 4« euolid's blembnts. [Book L 2. Prove that in thy triangle there must be at least two acute angles. 3. From a point outside a straight line only one perpendicular can be drawn to the straight line. 4. Prove the proposition by joining the vertex to a point inside the base. 5. The angles at the base of an isosceles triangle are both acute. 6. All the angles of an equilateral triangle are acute. 7. If two angles of a triangle be unequal, the smaller of the two must be acute. 8. The three interior angles of a triangle are together less than three right angles. 9. The three exterior angles of a triangle made by producing the sides in succession, are together greater than three right anglea Prove by indirect demonstrations the following theorems : 10. The perpendicular from the right angle of a right-angled triangle on the hypotenuse falls inside the triangle. 11. The perpendicular from the obtuse angle of an obtuse-angled triangle on the opposite side falls inside the triangle. 12. The perpendicular from any of the angles of an acute-angled triangle on the opposite side falls inside the triangle. 13. The perpendicular from any of the acute angles of an obtuse- angled triaugle on the opposite side falls outside the triangle. PROPOSITION 18. Theorem. The greater side of a triangle has the greater angle oppoeite to it. A '' f '®i Let ABC be a triangle, having AC greater than AB.- it is required to prove l ABC greater than l 61 From AC cut off AD = AB, i. 3 and join BD» i! ♦ 4r /. 16 /. 5 B0<* t] PROPOSITIONS 18, 19. Because l ADB is an exterior angle of A BCD .'. L ADB Ml greater than l C. But L ADB = L ABD, since AB = AD; .'. L ABD is greater than l C. Much more, then, is l ABC greater than l a 1. If two angles of a triangle be equal, the sides opposite them must also be equal. 2. A scalene triangle has all its angles unequal. 3. If one side of a triangle be less than another side, the angle opposite to it must be acute. 4. ABCD is a quadrihiteral whose longest side is AD, and whose shortest is BG. Prove z ABC greater than i ADC and i BCD greater than i BAD. 6. Prove the proposition by producing -45 to A so that AD shaU be equal to -4 C, and joining DC. 6. Prove the proposition from the following construction : Bisect z A hy AD, which meets BC at D; from AC out off AS == AB, said joia DK m PROPOSITION 19. Theorem. The greater angle of a triangle has the greater side opposite to it. Let ABC he a triangle having l B greater than z. it is required to prove AC greater than AB. If AC he not greater than AB, then AC must be = AB, or less than AB. If AC = AB, then l B = l C. But it is not : .'. Ada not - AB. 0.' I. 5 48 EUCLID S ELEMENTS. If? [Book L If ^Cbe less than -45, then l B must be less than z. C. /. 18 But it is not ; .•. ^C is not less than AB. Hence AC must be greater than AB. Cor. — ^The perpendicular is the shortest straight line that can be drawn from a given point to a given straight line; and of others, that which is nearer to the perpendicular ia lass than the more remote. F D EG From the given point, A^ let there be drawn to the given •traight line, BG^ (1) the perpendicular AD^ (2) AE and AF equally distant from the perpendicular, that is, so that DE = DF, (3) AG more remote than AE or AF : it is required to prove AD the least of these straight lineSj and AG greater than AE or AF. ( AD = AD In As ADE, ADF, J DE = DF ^ Hyp. I L ADE = L ADF; I. Ax. 10 .*. AE = AF. ' /. 4 Because l ADE ia right, .'. l AED is acute; /. 17 .*. -4 jE7 is greater than ^Z>. /. 19 Hence also AF is greater than AD. Because L AEG is greater than l ADE, /. 16 .'. jL AEG is obtuse ; ^ /, L AGE 18 acute ; L 17 ,\ AG is greater than AE. J. 19 Hence also AG is oTeater than AF^ and than AD, Book L] PBOPOSITIONS 19, 20. 49 1. The hypotenuse of a right-angled triangle ia greater than either of the other sides. 2. A diagonal of a square or of a rectangle is greater than any on« of the sides. 3. In an obtuse-angled triangle the side opposite to the obtuse angle is greater than either of the other sides, 4. From ^,one of the angular points of a square A BCD, a straight line IS drawn to intersect 5Cand meet DO produced at E; prove that AE is greater than a diagonal of the square. 6. From a point outside not more than two equal straight lines can be drawn to a given straight line. 6. The circumference of a circle cannot cut a straight Kne in more than two points. 7. ABC is a triangle whose vertical angle A is bisected by a straight Ime which meets BC at D; prove that AB is greater than BD, and AG greater than CD. PROPOSITION 20. Theorem. The mm of any two sides of a Inangle is greater than the third side. Let ABC be a triangle : it is required to prove that the sum of any two of its sides is greater than the third side. Produce BA to D, making AD = AC, /. 3 and join CD. Then l ACD = l D, since AD = AQ, J, But L BCD is greater than /l ACD; .*. L BCD is greater than L D; .*. Bf) ia c/rAflhAT iha-n Tin mm 9e buolid's blements. ! I'll ? n> [Book I. B\itBD = BA + AG; .'. BA + AC is frroiiter than BC. Now BA and ^C are any two sides ; .-. the sum of any two sides of a triangle is greater than the third side. Cor.— :{ he difference of any two sides of a triangle is less than the third side. For BA + AC is greater than BO. /, 20 Takinc' AC from each of these unequals, there remains BA greater than BC - AC; /. Ax. 5 tliat is, the tJiird side is greater tlian the difference between the other two. 1. Prove the proposition by producing CA instead of BA. ^' " "^ di-awing a perpendicular from the vertex to the base. ^ " " bisecting the vertical angle. 4. In the first figure to I. 7, the sum of AD and BC is greater than the sum of ^C and BD. 5. A diameter of a circle is greater than any other straight line m the circle which is not a diameter. 6. Any side of a quadrilateral is less than the sum of the other three sides. 7. Any side of a polygon is less than the sum of-the other sides. 8. The sum of the distances of any point from the three angles of a triangle is greater than the semi-perimeter of the triangle. Divcuss the three cases when the point is inside the triangle, when it is outside, and when it is on a side. 9. The semi-perimeter of a triangle is greater than any one side, and less than any two sides. 10. The sum of the two diagonals of any quadrilateral is greater than the sum of any pair of opposite sides. •mn Book I.] PROPOSITIONS 20, 21. 5i 11. The perimeter of a quadrilateral is greater than the su and less than twice the sum of the two diagonals. 12. The sum of the diagonals of a quadrilateral is less than the sum of the four straight lines which can be drawn to the four angles from any other point except the intersection of the diagonals. 13. The sum of any two sides of a triangle is greater than twice the median * drawn to the third side, and the excess of this sum over the third side is less than twice the median 14. The perimeter of a triangle is greater, and the semi-perimeter is less, than the sum of the three medians. PROPOSITION 21. Theorem. Tffrom the ends of aiiy side of a triangle there he drawn two straight lines to a point within the triangle, these straight lines shall be together less than the other two sides of the triangle, but shall contain a greater angU. Let ABC be a triangle, and from B and C, the ends of BG, let BD, CD be drawn to any point D within the triangle : it is required to prove (1) that BD + CD is less than AB + AC; (2) that l BDC is greater than l A. * DBF.— A median line, or a median, is a straight line drawn fr 'ertex of a triangle to the middle point of the opposite aide. V. 52 l» buclid's elements. A [Book L Froduce BD to meet AO a,t K (1) Because BA + AE is greater than BE; J. 20 add to each of these unequals EO ; .-. BA + ^C is greater than BE + EC. I. Ax. 4 Again, CE + ^Z) is greater than CD ; /. 20 add to each of these unequals DB ; .'. CE + EB is greater than CD + DB. I. Ax. 4 Much more, then, is BA + ^ a greater than CD + DB. (2) Because CED is a triangle, .-. L BDCia greater than l DEC; I ' and because BAE is a triangle, .•. iL -D-SC is greater than /L ^ / much more, then, is l BDC greater than l A 1. Prove the first part of the proposition by produ.iuc GD instead of BD. 2. Prove the second part of the propositicn by joining AD and producing it. 3. In the second figure to I. 7, prove that the perimeter of the triangle AGB is greater than that of ADB. 4. Prove the same thing with respect to the third figure to I. 7. 5. If a point be taken inside a triangle and joined to the three vertices, the sum of the three straight hues so drawn shall be less than the perimeter of the triangle, 6. If a triangle and a quadrilateral stand on the same base, and on the same side of it, and the one fia;ure fall within the other, that which has the greater surface shall have the greater perimeter. Book I.] PROPOSITIONS 21, 22. 53 . • PROPOSITION 22. Problem. To make a triangle the ndes of which shall he equal to three given straight Ivrm, hut any two of these must he greater than the third. ^ I. 3 Let A, B, a be the three given straight lines, any two of which are greater than the third : it is required to make a triangle the sides of which shall be respectively equal to A, B, C. Take a straight line DE terminated at D, but unlimited towards^; and from it cut off DF = A, FG = B, GH = C. With centre jF^and radius FD, describe'the DKL; with centre G and radius GH, describe the © HKL, cutting the other circle at A'; join KF, KG. KFG is the triangle required. Because FK = FD, being radii of DKL, I. Def 16 FK = A. Because GK = GH, being radii of HKL, I. Def 16 GK = G. y J And FG was made = B ; .'. A KFG has its sides respectively equal to A, B, C. 1. Could any other triangle be constructed on the base FQ f ulf *\i. Jfilling II H EUOLID S ELEMENTS. [Book L 2. U A, B, O he aU equal, which preceding proposition shall we be enabled to solve ? 3. Draw a figure showing what will happen when two of the given straight lines are together equal to the third. 4. Draw a figure showing what will happen when two of the given straight lines are together less than the third. 5. Since a quadrilateral can be divided into two triangles by dt-awing a diagonal, show how to make a quadrilateral whose sides shall be equal to those of a given quadrilateral 6. Since any rectilineal figure may be decomjwsed into triangles, show how to make a rectilineal figure whose sides shall be equal to those of a given rectilineal figure. if M %,\' 1! PROPOSITION 23. Problbm. At (% given point in a given straight line, to make an angle equal to a given angle. j^ o Let AB be the given straight line, A the given point in it, and l C the given angle : it is required to make at A an angle = lO. In CD, CE, take any points D, E, and join DE. Make A AFG such that AF= CD, FG = DE, GA = EC. I. 22 A is the required angle. AF = CD Const. In As AFG, CDE,\AG = CE Const FG =DE; Const. L A= L a /. 8 BookL] PROPOSITIONS 23, 24 55 1. A* a given poxat in a given straight line, to make an angle equal to the supplement of a given angle. 2. At a ^ven i^int in a given straight line, to make an angle equal to the complement of a given angle. & If on. angle of a triangle is equal to the sum of the other twa the triangle can be divided into two isosceles triangles 4. The straight line OG bisects the angle AOB ; prove that if OD be any other straight line through O without the angle AOB th« sum of the angles DOA and DOB is double of the ancle' DOG. ® 6. The straight line OG bisects the angle AOB; prove that if OD be any other straight line through within the angle AOB the diflFerence of the angles DOA and DOB is double of the angle DOG. Construct an isosceles triangle, having given : 6. The vertical angle and one of the equal sides. 7. The base and one of the angles at the base. Construct a right-angled triangle, having given : 8. The base and the perpendicular. 9. The base and the acute angle at the baae. Construct a triangle, having given : \Q. The base and the angles at the base. U. Two Eudes and the included angle. /2. fhe Daw, an angle at tiie base, and the sum of the other two sides. 13. The base, an angle at the base, and the diflference of the other two sides. PROPOSITION 24. Theorem. If two triangles have two sides of the one respectively equal to two sides of the other, hut the contained angles umqual, ths hose of the triangle which has the greater contained angle shall be greater than the base of the other.* ♦The proof given in the text is different from Euclid's, which iii 66 Euclid's elements. D [Book L Let ABC, DBF be two triangles, liftring AB AG = DF, but L BAG greater than l EDF: it is required to prove BG greater than EF. At D make l EDG = l BAG; cut off DO = AG or DF, and join EG. Bisect L FDG by i>H, meeting EG at ^; and, if F does not lie on EG, join Z'^. 5^ = ED In As 4^(7, Z>^G?, •. BG = j;(?. In As /T)^, G^Z)/r, BAG = z. Jg;i>G«; i^2)= GD DH = DH FDH = L GDH; .'. FH= GH. Hence EH ^- FH = EH -v GH = EG, But ^fl" + FH is greater than EF ; - .'. J^G' is greater than EF ; .'. BGS& greater than EF. 1. ABG is a circle whose centre is 0. If L AOB is greater than z 50C, proye that A Bis greater than BC. 2. In the same figure, prove that AG k greater than ^5 or BG. 3. A BGD is a quadrilateral, having AB — GD, but z BGD greater than /.ABG; prove that BDia = z) j;, /. 23 /. 3 7.9 Hyp. Const. Comt. /. 4 Const^ Comt. 7.4 Z 20 tha B greater Book L] PROPOSITIONS 24, 25. w 4. ^5C' is an igoBcele. triangle, having AB = AG. AD drawn to PROPOSITION 25. Theorem. /^^/z^o /mwf7?e* have two sides of the one respedively equal to two sides of the other, hut their hoses unequal, the angle contained hy the two sides of the triangle which has the greater hose shall he greater than the angle contained hy the two sides of the other. ^ D Let ABC, DEF be two triangles, having AB = DE AC = DF, but base 5(7 greater than base EF: ' it is required to iJrove l A greater than l D. If ^^ be not greater than l. D, it must be either equal to L D, or less than l D. But A ^ is not = z. D, for then base BC would be = base EF, , . which it is not. rr' . , . . Hyp. And L A\^ not less than l D, for then base BC would be less than base EF, which it is not. .*. L A must be greater than l D. '* ^''^u'' ^£1^^ *° *^^ ^'* deduction on I. 24, if AB is greater tx.as ^o, prove that i AOB is greater than / BOG. ~ I. 24 Hyp. II 58 BU0LID8 ELEMENTS. [Book L 2. A BCD is a quadrilateral, having AB = CD, but the diagonal BD greater than the diagonal AC; prove that i DCB is greater than z ABC. 3. A BCD is a quadrilateral, having ^J5 = CD, but i BCD greater than z A BC ; prove that l DA B is greater than l A DC. 4. A BCD 16 a quadrilateral, having u47i -- CD, but i DAB greater than z ADC; prove that z 5C/> is greater than z ^i/C. 5. ABC ifl a triangle, having AB less than .40. D is the middle point of BC, and ^Z) is joined ; i)rove that z ^D/i is acute. 6. ABC is an isosceles triangle, having AB ^ AC. D\b any jioint such that BD is greater than DC; prove that AD does not bisect ^ A. 7. .^ZJO is a triangle, having ^^ less than AC, and ^Z) is the median drawn from A; prove that G, any jwint in AD, is nearer to B than to C. PKOPOSITION 26. Theorem. If two angles and a side in one triangle he respectv'ehj egiial to two angles and the corresponding side in another triangle^ the two triangles shall he equal in every respect ; that isy (1) The remaining sides of the one triangle shall be equal to the remaining sides of the other. (2) The third angles shall he equal. (3) The areas of the two triangles shall he equal. Case 1. In As ABC, DEF let l ABO = l DEF, l AOB : 1 nWE and EH = EF.' PROPOSITION prove AB iJ6. 6d DE,AC^Dt lA'^l a BuOK LI it 18 required to A ABG = A DEF. T l^f^^^.T^ ' ^^' '"' '^ *^^"^ ^^^^ ^"^ *-^ grater, i^et AB be the greater, and make BG ^ DE • 7 «? and join OC. ' ^' '^ i GB= DE In As GBC, DEF, | BC = j5;f ( L B= I. E; .'. c (?e5 = ^ DFE. But L AGB = L DFE; .-. ^ G^Cfi = z. ^C^, which is impossible. Hence AB is not unequal to DE, tljut is, AB = 7)^. NowinAs^^ai.^/'j i^:f^' ~ ^ ^ \ ^ B = ^ E; Him Case 2. C07liit. Hyp. Hyp. Hyp. C E ^i? is required to prove BG = EF, AG ^ DF / FtAP = L EDF, A ABG = A DEF. r f fi^^^^ r* ^^^' ''"' ^^ *^^"^ ^^"«* ^e the greater. l.et BC be the greater, and make Bff =>^ EF - t % and join AH. '' ''' "* 6t BUOUD'S £L£HBNTS. [Boorc ^'ii !l at « 1 WW ■1 In As ABH, DEF, AB = DE BH= EF L B = L E; Hyp. Const. Hyp. I. 4 Hyp. I. 16 .-. L AHB = L DFE. But L ACB = LflFE; .'. L AHB = L ACB, which is impossible. Hence BC is not unequal to EF, that is, BC = EF. i AB = DE Hyp. 5fow in A? ABC, DEF, ] BC = EF Proved ( L B = L E; Hyp. ,\AC=DF,lBAC=lEDF,AABC = ADEF. Li 1. Prove the first case of the proposition by superposition. 2. The straight Hne that bisects the vertical angle of an isosceles triangle bisects the base, and is perpendicular to the base. 3. The straight line drawn from the vertical angle of an isosceles triangle perpendicular to the base', bisects the base and the vertical angle. 4. Any point in the Insector of an angle i» equidistant from the arms of the angle. 5. In a given straight line, find a point such that the per- pendiculars drawn from it to two other straight lines may be equal. 6. Through a given point, draw a straight line which shall be equidistant from two other given jioints. 7 Tirough a given point, draw a straight line which shall form with two given intersecting straight lines an isosceles triangle. [loove. BOtfkLj PROPOSITIONS 26, A. 6) Hyp, Const. Hyp. I. 4 Hyp. L 16 Hyp. .Proved Hyp. 1. n isosceles ir to the ,n isosceles 36 and the t from the the per- aes may be h shall be shall form a. isosceles PROPOSITION A. Theorem. I/two sides of one triangle be respectively equal to two sides of another triangle, and if the angles opposite to one pair of equal sides be equal, the angles opposite the other pair of equal sides shall either be equal or sup- plementary. In As ABG, DEF let AB = DE, AC = DF, l B = L E: it is required to prove either lC=zlF, otlC-^-lF = 2rt. L s. L AiQ either = l D, ov not. Case L— When i. A = l D, A D Ll A == L D In As ABC, DEF, }lB=lE Hyp { AB = DE; Hyp. • '- As ABC, DEF are equal in all respects, and lC= lF ^ ^ 26 Case 2.— -When z. ^ is not = l D. A jy At n make l EDG = l BAG, I, 23 62 EUCLID a BLEMBNTS. [Book I B E F and let EFy produced if necessary, meet DG at G, lBAG = L EDG lABG ^ AB --= DG, md lG = L G. In As ABG, DEG, DEG DE; .'. AG Xow AG = DF: Const. Hyp. Hyp. I. 26 Hyp. I. 5 7.13 DF = DG; .'. L DFG = L DGF. But L DFE\% supplementary to l DFG; .'. L Z^i^'J^ is supplementary to l DGF, and consequently to z. C. NoTK. — It often happens that we wish to prove two triangles equal in all respects when we know only that two sides in the one are respectively equal to two sides in the other, and that the angles opposite one pair of equal sides are equal. In such a case, since the angles opposite the other pair of equal sides may either be equal or supplementary, we must endeavour to prove that they cannot be supplementary. To do this, it will be sufficient to know either (1) that this pair are both acute angles, or (2) that they are both obtuse angles, or (3) that one of them is a right angle, since the other must then be a right angle whether it be equal or supplementary to it. We can tell that this pair of angles must be both acute in certain cases. (o) When the pair of angles given equal are both right angles. (6) i» It tt II obtuse II («) II II equal sides opposite the given angles are greater than the other pair of equal sides. Henoe the followincf imnort-ant GoroUarv : Bo0kL] PROPOSITIONS A, 2'7. If the hypotenuse and a siJe of one right-angled triangle U respectively equal to the hypotenuse and a side of another riuht- angled tnangle, the triangles shall be equal in all respects. Const. Hyp. Ilyp. I. 26 Hyp. I. 5 7.13 PROPOSITION 27. Theorem. If a straight line cutting two other straight lines mak^ tht alternate angles equal to one another, the two straight lines shall be parallel Let EF, which cuts the two straight lines AB, QD, make L AGH = the alternate l GHD: it is required to prove AB \\ CD. UAB is not il CD, AB and CD being produced wiU meet either towards A and C, or towards B and D. Let them be produced, and meet towards B and D a.t K Then KGH is a triangle ; .-. exterior l AGH is greater than the interior opposite L GHD. , ,« Bvit L AGH =^ L GHD ; ^^^ which is impossible. ^^" .-. ^^ and CD, when produced, do not meet towards B and D. Hence also, AB and CD, when produced, do not meet towards A and C; AB is II CD, J.. W, 14 1 'i:. 64 WCLID^S ELEMENTS. [Bodkt In the figure to 1. 16 : 1. Prove ^5 II CF. 2. Join AF, and prove AF \\ BC. In the tigiire to I. 28 : 3. It I AGE = A DHF, prove AB \\ CD. 4. If L BGE = L CHF, prove AB \\ CD. 5. II L AQE -^ I CHF = 2 Ft z s, prove AB || CD. 6. If L BGE + L DHF = 2 rt z s, prove AB \\ CD. 7. The opposite sides of a square are parallel. 8. The opposite sides of a rhombus are paralleL 9. Tb« quadrilateral whose diagonals bisect each other is a !{• y li. . PROPOSITION 28. Theorem. If a straight line cutting two other straight lines make (1) an exterior angle equal to the interior opposite angle on the same side of the cutting line, or (2) tlie two interior angles on the same side of the cutting line together equal to two right angles, the two straight lines shall be parallel. Case 1. 'Let EF, which cuts the two straight h'nes AB, CD, make the exterior L EGB = the interior opposite l GHD : it is required to prove AB \\ CD. Becsause l SOB = l GHD, Hyp. kMkt Book I] PROPOSITIONS 28, 29. (1) an on the lienor (jether lall be make Hi/p. and L EGB = l AGH, being Tsiliically opposite j .-. L AGH= L GHD; and they are alternate angles ; .-. AB is II CD. 65 /. 15 /. 27 Hyp. I. 13 Case 2. Let EF, which cuts the two straight lines AB, CD, make L BGH + L GHD = 2 rt. ^ s : it is required to prove AB \\ CD. Because l BGH + l GHD = 2 rt. ^ s, and L AGH + l BGH = 2 rt. ^ s; .-. L AGH + L BGH = L BGH + l GHD. From these equals take l BGH, which is common ; .-. L AGH=lGHD; LAx.S and they are alternate angles ; .-. AB is II CD. /. 27 Cor. — Straight lines which are perpendicular to the same straight line are parallel. 1. If z BGE + L DHF = 2 rt. z s, prove AB %li L AGE + I GHF = 2 rt. z s, prove AB 3. li I AGE^L DHF, prove ^5 || CD. 4. K z BGE = L GHF, prove AB \\ GD. 6. The opposite sides of a square are parallel. 6. ABGD is a quadrilateral having z A and z J5 «iupplementary, as well as z jB and z C ; prove that it is a |p». CD. Ci>. PROPOSITION 29. Theorem. If a straight line cut two parallel straight lines, it shall make (1) the alternate angles equal to one another; (2) any exterior angle equal to the interior opposite angle on the same side of the cutting line ; (3) (lie two interior angles on the same side of the cutting line equal to two right angles. 66 EUCLID 8 ELEMENTS. [Book L Let EF cut the two parallel straight lines AB, CD: it 18 required to prove : (1) L AGH = alternate l GHD; (2) exienor l EGB = interior opposite l GHD • (3) L BGH + L GHD = 2rt, ls. (1) If L AGH be not = l GHD, make l KGH = L GHD, J 23 and produce KG to L. Because l KGH = alternate l GHD, Const .-. KL II CD. J 27 But ^5 is also II CZ)/ ff^p .'. AB and KL, which cut one another at G, are both || CD, which is impossible. /^ j^g^ 2 [ .'. L AGH is not unequal to z. GHD; .-. z. ^G!^ = z. GHD. (2) Because ^ ^(?^ = l GHD, ' Proved and L AGH = l EGB, being vertically opposite : /. 15 .-. L EGB = L GHD. (3) Because l AGH = l GHD; Proved to each of these equals add l BGH ; .'. L AGH+ L BGH= L BGH+ l GHD. I. Ax. 2 But L AGH + L BGH = 2 rt. z. s; /. 13 .". -L BGH + L GHD = 2 It. LB. Book I.] PROPOSITION 29. er Cor— If a straight line meet two others, and make with them the two interior angles on one side of it together less than two right angles, these two other straight lines will if produced, meet on that side. ' Let KL and CD meet EF and make l KGH + l CHG less than 2 rt. La: it is required to rave ^hat KG ami CH will, if produced, meet towards K and C. If not, KL and CD must either be parallel, or meet towards L and D (1) KL and CD are not parallel ; for then l. KGH + l CHG would be = 2 rt. l s. /. 29 (2) KL and CD do not meet towa-ds L and D • for tlien LaLGH, DHG would form angles of 1 triangle /. 17 I. 13 Hyp, and would .-.be togeth. r less than 2 rt. l s. If ow since the four l s KGH, CHG, LGH, DHG are together = 4 rt. z. s, and the first two are less than 2 rt. z. s • .-. the last two must be git^ater than 2 rt. l s. Hence KL and CD must meet towards K and G. [This Cor. is the converse of I. 17.] 1. In the diagram to I. 28, if AB is || CD, prove l AGE = l DHF and L BQE+ l DHF=2vt. zs. ' 2. If a straight line be perpendicular to one of two parallels it is also perpendicular to the other. ' 3. A straight line drawn parallel to the base of an isosceles triangle and meeting the sides or the sides produced, forms with them another isosceles triangle. 4. If the arms of one angle be respectively parallel to the arms of another angle, the angles are either equal or supplementary. iJistmgmsh the cases. 5. Is it always true that if two angles be equal, and an arm of the one IS parallel to an arm of the other, the other arms must be parallel ? ni 'r li 6d EUCLID 8 ELEMENTS. [Book I. 6. If any straight line joining two parallels be bisected, any other straight line drawn through the point of bisection and terminated by the parallels will be bisected at that point. 7. The two straight lines in the last deduction will intercept equal portions of the i)arallels. 8. If through the vertex of an isosceles triangle a parallel be drawn to the base, it will bisect the exterior vertical angle. 9. If the bisector of the exterior vertical angle of a triangle be parallel to the base, the triangle is isosceles. 10. The diagonals of a ||'" bisect eacli other. 11. Prove that by the following construction ^ ACS is bisected : In AC take any point D; draw DE ± AC, and meeting CB at E. Prom E draw EF ± DE and = EC; join CF. , i PROPOSITION 30. Theorem. Straight lines lohich are parallel to the mme straight line are parallel to one another. A- C- -D E- Let AB and CD be each of tliem |i EF : it is required to prove AB \\ CD. If AB and CD be not parallel, they will meet if pro- duced ; and then two straight lines wl\ich intersect eacli other will both be || the same straight line, which is im- possible. . /. Ax. 11 .-. AB is II CD. 1. Two 11™ are situated either on the same side or on different sides of a common base. Prove that the sides of the \\^^ which are opposite the common base are |1 each other. 2, Prove the proposition in Euclid's manner by drawing a straight line GHK to cut AB, CD, and EF. and applying I. 29, 27. Book Ll PROPOSITIONS 30, 31, 32. 69 PROPOSITION 31. Problem. Through a given point to draw a straight line parallel to a given straight liiie. E- B- D ■0 Let A be the given point, and BC the given straight line : it is required to draw through A a straight line \\ BC. In BC take any point D, and join AD; at A make l DAE = l ADC ; j 23 and produce j;^ to /: ^i^ shall be || J5a * Because the alternate l s EAD, ADC are equal .-. ^i^isll^C. 7.27 1. Give another construction for the proposition by means of I. 12. 11, and a proof by means of I. 28. 2. Through a given point draw a straight Hne making with a given straight hne an angle equal to a given angle. 3. Through a given point draw a straight line which shall form with two given mtersecting straight lines an isosceles tri- angle. 4. Through a given point draw a straight line such that the part of It mtercepted between two parallels may be equal to a given straight hne. May there be more than one solution to this problem? Is the problem ever impossible ? PROPOSITION 32. Theorem. If a side of a triangle he produced, the exterior angle is equal to the sum of the two interior opposite angles, and the sum of the three interior angles is equal to two right angles. 11 7a if li BUOLId's BLBHBNIB. A [^kL /. 31 7.29 /. 29 Proved Let ABC be a triangle having BC produced to D: it is required to prove {\) l A CD = l A + L B ; {'i) L A-\- lB-^ L ACB ~ 2 rt. Z.S. Through C draw CE \\ AB. (1) Because AC meets the parallels AB, CE, .'. L A = alternate l ACE. Because BD meets th^ parallels AB, CE, .*. interior l B >= exterior z. ECD ; ,\ L A + L B = L ACE + L ECD, = L ACD. (2) Because l A + l B = l ACD; adding l ACB to each of these equals, .\ L A + L B + L ACB = L ACD + l ACB, = 2 rt. ^ s. /. 13 Cor. 1. — If two triangles have two angles of the one respectively equal to two angles of the other, they are mutually equiangular. For the third angles differ from 2 rt. ^ s by equal amounts ; .*. the third angles are equal. Cor. 2. — The interior angles of a quadri- lateral are equal to four right angles. FcT the quadrilateral ABCD may be divided into two triangles by joining ^C; and the six angles of the two As ABC, ^^ ACD = 4 rt. z. s. /, 32 ti ti Book I.] PROPOSITION 32. 71 But the six angles of the two triangles = the interior angles of the quadrilateral ; .•. the interior angles of the quadrilateral = 4 rt ^s. CoR. 3. — A five-sided figure may be divided into three (that is, 5-2) triangles by drawing straight lines from one of its angular points. Similarly, a six-sided figure may be divided into four (that is, 6-2) triangles ; and generally a figure of n sides may be divided into (w - 2) triangles. Hence, by a proof like that for the quadrilateral, the interior z. s of a five-sided figure = 6 rt. ^ s ; n H six-sided » = 8 rt. ^ s ; and • rr figure with n sides = (2n - 4) rt. i. s. 1. If an isosceles triangle be right-angled, each of the base angles is half a right angle. 2. If two isosceles triangles have their vertical angles equal, they are mutually equiangular. 3. If one angle of a triangle be equal to the sum of the other two, it must be right 4. If one angle of a triangle be greater than the sum of the other two, it must be obtuse. 5. If one angle of a triangle be less than the sum of the other two it must be acute. 6. Divide a right-angled triangle into two isosceles triangles. 7. Hence show thit the middle point of the hypotenuse of a right- angled triangle is equidistant from the three vertices. 8. Hence also, devise a method of drawing a perpendicular to - given straight line from the end of it without producing the straight line. 9. Each angle of an equilateral triangle is two-thirds of a right angle. 10. Henee show how to trisect * a right angle. * It is sometimes stated that the problem to trisect anp angle is beyond the power of Geometry. This is not the case. The problem is beyond the power of Elementary Geometry, which allows the use of only the tulcr ui'id the compasses. 72 EUCLID 8 ELEMENTS. [Book L 11. Prove the second part of the proposition by drawing through A a straight line DAE || BC. (The Pythagorean proof.) 12. If any of the angles of an isosceles triangle be two thirds of a right angle, the triangle must be equilateral. 13. Each of the baae angles of an isosceles triangle equals half the exterior vertical angle. 14. If the exterior vertical angle of an isosceles triangle be bisected, the bisector in \\ the base. 15. Show that the space round a point can be filled up with six equilateral triangles, or four squares, or three regular hexagons. 16. Can a right angle be divided into any other number of equal parts than two or three ? 17. In a right-angled triangle, if a perpendicular be drawn from the right angle to the hypotenuse, the triangles on each side of it are equiangular to the whole triangle and to one another. 18. Prove the seventh deduction indirectly ; and also directly by producing the median to the hypotenuse its own length. 19. If the arms of one angle be respectively perpendicular to the arms of another, the angles are either equal or supplementary. 20. Prove Cor. 3 by taking a point inside the figure and joining it to the angular points. ! ill PEOPOSITION 33. Theorem. The straight lines which join the ends of two equal and parallel straiyld lines towards the same parts, are themselves equal and parallel. iB C^ -jj Let AB and CD be equal and parallel : a is required to prove AC and BD equal and parallel. Join BC. Because BC meets the parallels AB, CD, ' L. ABC — oHomafo ' nan t oq Book L] rROPOSITIONS 33, 34. fS ( AB^ DC Hyp. In As ABC, ^'h, ^ BC = CB - ABC = L DCB; Proved .'. AC = DB, - A{:o = l DBG. /. 4 Because CB meek Au and Z//;, and makes the alter- nate L8ACB, BBC equal; Proved .-.AOieWBD. i 27 1. State a converse of this proposition. 2. If a quadrilateral have one pair of opposite sides equal and parallel, it is a H"", 3. What statements may be made about the straight lines which join the ends of two equal and parallel straight lines towards opposite parts ? PROPOSITION 34. Theorem. A parallelogram has its opposite sides aiid angles equal, and is bisected hy either diagonal. A_ .B O Let ACDB be a H" of which BC is a diagonal : it is required to prove that the opposite sides and angles oj ACDB are equal, and that A ABC = A DCB. Because BG meets the parallels AB, CD, .-. L ABC = alternate l DCB ; /. 29 and because BC meets the parallels AC, BD, .'. L ACB = alternate l DBG. L 29 I. ABC = ^ DCB Proved In As ABC, DCB, ] l. ACB - l DBO Proved BC « CBi »'^^! H I III bcclid's elements. A. _B [Book I. O' ^D .-. AB = DC, AG = DB, l BAG = l GDB, ■ A ABG = A DGB. J 26 Again because l ABG was proved = l DGB, I, 29 and L DBG WRs proved = l ACB; /. 29 .-. the whole z. ABD = the whole l. DGA. Cor. — If the arms of one angle be respectively parallel to the arms of another, the angles are either (1) equal or (2) supplementary. For (1) L EAGhos, been proved = z. GDB ; and (2) if BA be produced to E, L EAG, which is supplementary to l BAG, /. 13 must be supplementary to l GDB. 1. If two sides of a ||» which are not opposite to each other be equal, all the sides are equal. 2. If two angles of a r which are not opposite to each other be equal, all the angles are right. 3. If one angle of a \\^ be right, all the angles are right. 4. If two irs have one angle of the one = one angle of the other, the ir» are mutually equiangular. 6. If a quadrilateral have its opposite sides equal, it is a !|m 6. If a quadrilateral have its opposite angles equal, it is a ||™. 7. If the diagonals of a H™ be equal to each other, the ||n> is a rectangle. 8. If the diagonals of a H" bisect the angles through which they pass, the ll^ is a rhombus. 9. If the diagonals of a H"" cut fach other perpendicularly, the ||m is a rhombus. 10. If the diagonals of a H" be equal and cut each other perpen- dicularly, the ||n» is a square. 11. Show how to bisect a etk-ight line by means of a pair of parallel rulera. Book L] PROPOSITIONS 34, 36. 76 18 12. Every straight line drawn through the intersection of the diagonals of a |r, and terminated by a pair of opposite sides, is bisected, and bisects the ||™ 13. Bisect a given |p by a straight line drawn through a given point either within or without the ||™ 14. The straight line joining the middle points of any two sides of a triangle is || the third side and = half of it 16. If the middle points of the three sides of a triangle be joined n ^*^ ^^^^ ^*^^^' *^® ^**"^ triangles thence resulting are equal. 16. Construct a triangle, having given the middle points of its three sides. PROPOSITION 35. Theorem. Parallelograms on the same base and between the same parallels are equal in area. ^ ^ ^ faedf T . Bo Be Let ABCD, EBCF be |r on the same base BC, and between the same parallels AF, BG : it is required to prove ||'" ABCD = W^ EBCF. Because AF meets the parallels AB, DC ,'. interior l A = exterior z. FDC ; and because AF meets the parallels FB, FGy .*. exterior l AEB = interior l F. ( L EAB = L FDC In As ABE, DCF, ] l AEB = l DFC ( AB = DC; .'. A ABE = A DCF Hence quadrilateral ABCF - A ABE = quadrilateral ABCF - A DCF - ir ^BCF =. ir ABCD. F /. 29 /. 29 Proved Proved /. 34 /. 26 76 kuclid's elements. [Book t 1 Note. — This proposition affords a means of measuring the area of a II™ ; thence (by I. 34 or 41) the area of a triangle ; and thence (by I. 37, Cor.) the area of any rectilineal figure. For the area of any ll™ = the area of a rectangle on the same base and between the same parallels ; and it is, or ought to be, explained in books on Mensura- tion, that the area of a rectangle is found by taking the product of its length and breadth. This phrase 'taking the product of its length and bread ih/ means that the numbers, whether integral or not, which express the iongth and breadth in terms of the same linear unit, are to bfc litultiplied together. Hence the method of fin«ling the area •■>£ s, jj'" is to take the product of its base and altitiide, the altitude being defined to be the perpendicular drawn to its base from any point in the side opposite. \. Prove the proposition for the case when the points D and JS coincide. 2. Equal 11™* en the same base and on the same side of it are between the samo parallels. 3. If Ihruugh the vertices of a triangle straight lines be drawn || the opposite sides, and produced till they meet, the resulting figure will contain three equal ||™'. 1. On the srvme base and between the same parallels as a given jl", construct a rhombus = the ||°». 6r Prove the equality of As ABE and DGF in the proposition by I. 4 (as Euchd does), or by I. 8, instead of by I. 26. PROPOSITION 36. Theorem. Parallelograms on equal basee and beticeen the same parallels arc equal in area,. ^ A D E IT E CI G Let ABCD, EFGH be r* or equal bases BG, FG, and batween the same parallels, AH^ : 'G : it is required to prove ^ ABCD ■- |r EFGH. Hyp., I 34 /. 33 /. Def. 33 Book L] PROPOSITIONS 36, 37. f 7 Join BE, CH. Because BG = FG, and FG = EH, .-. BG = EH. And because BG is || EH, .'. EB is (I HG, .-. EBGH is a If^. Now ir ^5CZ) = II- EBGH, being on the same base BG, and between the same parallels BG, AH; I. 35 and r ^J^GH = T ^^C^, being on the same base EH, and between the same parallels EH, BG; I. 35 .-. 11"" ABGD = ir ^i^6^zr. 1. Prove the proposition by joining AF, DO instead of BE, GH. 2. Divide a given H"" into two equal ||«", 3. In how many ways may this be done ? 4. Of two r^ which are between the same parallels, that is the greater which stands on the greater base. 5. State and prove a converse of the last deduction. 6. £qual H*^ situated between the same parallels have equal bases. PROPOSITION 37. Thborbm. Triangles on the same base and between the same parallels are equal in area. Let ABG, DBG h?. triangles on the same basb BO, and between the same narahels AD, BG: it is required to prov9 A ABG = A DBC n buolid's elements. [Bookt, ■»! Through B draw BE || AG, and through C draw CF II BD; I' 31 and let them meet AD produced at E and F. Then EBGA, DBCF are |r ; /. Def. 33 and ir EBGA = |r />5(7i^, being on the same base BG, and between the same parallels BG, EF. I. 35 But A ABG = half of |r EBGA, I. 34 and A DBG = half of H™ I^i^Ci?'; /. 34 .-. A ABG = A DBG Cor. — Hence any rectilineal figure may be converted into an equivalent triangle. \r '■i V 1 t Let ABGDE be any rectilineal figure : it is required to convert it into an equivalent triangle. Join AG, AD; through B draw BF || AG, through E draw EG \\ AD, I. 31 and let them meet OD produced at F and G. Jom AFj AG, AFG is the required triangle. M PROPOSITIONS 37, 38. 79 look I.] For A AFC ^ A ABC, and A AGD - A AED ; I. 37 .-. A AFC + A ACD + A AQD = A ABC + A AGD + A AED. .-. A ^F(^ = figure ABCDE. 1. ^5C' is any triangle ; DE is drawn || the base BG, and meets AB, AG at D and J2J; 5-fi? and CD are joined. Prove A DBG = A EBG, A BDE = a C^Aand a ABE = a ^CD, 2. A BCD is a quadrilateral having ^^ || GD ; its diagonala -4C, BD meet at 0. Prove l ADD = t^ BOG 3. In what case would no construction be necessary for the proof of this proposition ? 4. Convert a quadrilateral into an equivalent triangle. 6. ABC is any triangle, D a point iuAB; find a point E in BG produced such that a DBE = A ABG. PROPOSITION 38. Theorem. Triangles on equal bases and between the same parallels are , equal in area. A D „ B E F Let ABG, DEF be triangles on equal bases BC, EF, and between the same parallels AD, BF: it is required io ^ywe A ABG — A DEF. Through B draw BG || AC, and through F draw FH\\DE; /. 31 and let them meet AD produced at G and H. Then QBGA, DEFH are 1^" ; and ir GBGA = T DEFH, being on equal bases BC, EF, 80 Euclid's elements. [BookL I't. < i I and between the same parallels BF, GH. But A ABC = half of |r GBCA, and A DEF = half of l^ DEFH ; .'. A ABC = A DEF, 7.36 /. 34 /. 34 Cor. — The straight line joining any vertex of a triangle to the middle point of the opposite side bisects the triangle. Hence the theorem : If two triangles have two sides of the one respectively equal to two sides of the other and the contained angles supplementary, the triangles are equal in area. 1. Of two triangles which are between the same parallels, that is the greater which stands on the greater base 2. State and prove a converse of the last deduction. 3. Two triangles are between the same parallels, and the base of the first is double the base of the second; prove the first triangle double the second. 4. The four triangles into which the diagonals divide a H™ are equal. 5. If one diagonal of a quadrilateral bisects the other diagonal, it also bisects the quadrilateral. •. 6. ABCD is a II™ ; E is any point in j4D or AD produced, and F any point in BG or EG produced ; AF, DF, BE, GE are joined. Prove A AFD = A BEG. 7. ABG is any triangle ; L and K are the middle points oi AB and AG; BK and GL are drawn intersecting at O, and AO is joined. Prove A BGG ~ A AGG = L AQB. 8. ABGD is a 11=' ; P is any point in the diagonal BD or BD pro- duced, and PA, PG are joined. Prove A PAB = A PGBy and A /^^D = A PGD. 9. Bisect a triangle by a straight line drawn from a given point ia •ne of the sides. [Book L Book L] PROPOSITIONS 38, 39. 81 7.36 /. 34 /. 34 triangle triangle. 70 sides her and ,re equal Is, that ia e base of the first sire equal. Eigonal, it ed, and F 1, GE are ts of AB ', and^G^ • BD pro- : A PCB, Q point ia PROPOSITION 39. Theorem. EqvM triangles on the same side of the same base are between the same parallels. A ug 3rD 7.31 7.37 Hyp. Let As ABO, DBC on the same side of the S8' te ba«c EG be equal, and let AD be joined : it is required to prove AD \\ BC. If AD is not II BO, through A draw AE \\ BO, meeting BD, or BD produced, at E, and join EC. Then A ABO = A EBG. But A ABO = A DBO; A EBO = A DBO; which is impossible, since the one is a part of the other. .-. AD is II BO. 1. The straight hne joining the middle points of two sides of a triangle is || the third side, and = half of it. Hence prove that the straight line joining the middle point of the hypotenuse of a right-angled triangle to the opposite vertex = half the hypotenuse. The middle points of the sides of any quadrilateral are the vertices of a i|™, whose perimeter =■ the sum of the diagonals of the quadrilateral. When will this !P be a rectangle, a rhombus, a square ? If two equal triangles be on the same base, but on opposite sides of it, the straight line which joins their vertices will be bisected by the base. 5. Use the first deduction to solve I. 31. 6. In the figure to I. 16, prove AF \\ BC. 7. If a quadrilateral be bisected by each of its diagonals, it ia a H"". 8. Divide a given triangle into four triangles which shall be equal in every respect. 2. 3. 4. d2 BtOLID*S BLIMBKTS. [BookL PROPOSITION 40. Thborbm. Equal triangles on the same side of equal bases which are in the same straight line are between the same parallels. A D ^.._ /. 31 I. 38 Hyp. B O E F Let As ABC, DEF, on the same side of the equal bases BG, EF, which are in the same straight line BF, be equal, and let AD be joined : it is required to prove AD || BF. If AD is not II BF, through A draw AG \\ BF, meeting DE, or DE produced, at G, and join GF. Then A ABC = A GEF. But A ABC = A DEF; A GEF = A DEF; which is impossible, since the one is a part of the other. .•,ADis\\BF. 1. Prove the proposition by joining AE and AF. 2. Prove the proposition by joining DB and DC 3. Any number of equal triangles stand on the same side of equal b. 'es. If their bases be in one straight line, their vertices ■will also be in one straight hne. 4. Equal triangles situated between the same parallels have equal bases. 6. Trapeziums on the same base and between the same parallels are equal if the sides opposite the common base are equal. 6. The median from the vertex to the base of a triangle biaeotf every parallel to the base. 7. Hence devise a method of bisecting a given straight line. Bock X.] PROPOSITIONS 40, 41. 83 PROPOSITION 41. Theorem. If a parallelogram and a triangle be upon the same base and between the same parallels, the parallelogram shall be double of the triangle. A r B B Let ihe H" ABCD and the A EBC be on the same base BC, and between the same parallels AE, BC : it is required to prave ||"' ABCD = twice A EBC. Join AC. Then A ABC = AEBC. J. 37 But II'" ABCD = twice A ABC; I. 34 II"' ABCD = twice A EBC. 1. Prove the proposition by drawing through G a parallel to BE. 2. If a tl"' and a triangle be on equal bases and between the same parallels, the ||™ shall be double of the triangle. 3. A II™ and a triangle are equal if they are between the same parallels, and the base of the triangle is double that of the ||">. 4. State and prove a converse of the last deduction. 5. If from any point within a H™ straight lines be drawn to the ends of two opposite sides, the sum of the triangles on these sides shall be equal to half the ||'". Is the theorem true when the point is taken outside ? Examine all the cases. 6. ABCD is any quadrilateral, AC and BD its diagonals. A H" EFOH is formed by drawing through A, B, C, D parallels to AC and BD. Prove ABCD = half of EFOH. 7. Hence, show that the area of a quadrilateral = the area of a triangle which has two of its sides equal to the diagonals of the quadrilateral, and the included angle equal to either of si H tot' 'i ¥i 84 buolid'b elements. [Book t the Migles at which the diagonals intersect; and that tw» quadrilaterals are equal if their diagonals are equal, and ftlso the angles at the intersection of the diagonals. PROPOSITION 42. Problem. To describe a parallelogram that shall be eqtml to a given triangle, and have one of its angles equal to a given angli. A F O B" W C Let ADC be the given triangle, and D the given angle : it is rer/uired to describe a jp «^i"«^ <^ ^ ^^^' ^^^ having one of its angles equal to L D. Bisect BC at E; I- 1^ and at E make L CEF = L D. I- 23 Through A draw AG \\ BG ; through Odraw CO \\ EF. I. 31 FECG is the |1"' required. Join AE. The figure FECG is a ll"" ; and IP FECG = twice A AEC. But -nee A ABE = A AEC, A ABO = twice A AEO; ||'» FECG = A ABC, and L. CEF was made = L D, 1. Describe a rectangle equal to a given riangle. 2. Describe a triangle that shaP be eqtal to a given ||™ and have one of its angles equal to a given angle. 3. On the same base as a |p consoruct a. right-angled triangle = ther. 4. Construct a rhombus = a given triangle. /. Def. 33 /. 41 /. 38 .1 Book X.] PROPOSITIONS 42, 43. 86 PROPOSITION 43. Theorem. The complements of the parallelograms which are about a diagonal of any parallelogram are equal. AH D I. 34 /. 34 /. 34 Let ABCD be a H", and AC one of its diagonals ; let EH, GF ])e |p" about AC, that is, through which A(j^ l)asse's, and BK, KD the other ||'"* which till up the figure ABCD, and are therefore called the complements : it is required to prove complement BK = complement KD. Becauie EH is a H" and AK its diagonal, A AEK = A AHK. Similarly A KGG = A KFC; .'. A AEK + A KGC = A AHK + A KFC. But the whole A ABC = whole A ADC ; .'. the remainder, complement BK = the remainder, com^- plement KD. 1. Name the eight H""' into which ABCD is divided by EF and GH, and prove that they are all equiangular to ||"» ABCD. 2. Prove \r AG = \r ED, and l|"» BF ■-= ir DG. 3. If a point K be taken inside a ||»' ABCD, and through it parallels be drawn to AB and BC, and if ||"» BK = |r A' A the diagonal AC passes through K. (Converse of I. 43.) 4. Each of the ||"'» about a diagonal of a rhombus is itself a rhombus. 5. Each of the ||'"* about a diagonal of a square is itself a square. 6. Each of the H'"" about a square's diagonal produced is itself a square. 7. When are the complements of the H""* about a diagonal of any H"" equal ia every respect ? I IMAGE EVALUATION TEST TARGET (MT-S) / O <^ 1.0 I.I J- KiS III 2.2 ^ 1^ 112.0 u 1.8 pS 1.4 1.6 ■• f ► % ^ ^>7 c5 > :>> # ^^ 7 Hiotographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4 J03 U ^ \ 4- 86 buolid's ^ilbmbnts. Dioelt!. PROPOSITION 44. Problbm. On a given straight line to describe a parallelogram which shall be equal to a given tnangle, and have one ofiU angles equal to a given angle. F E K D /. 42 /. 31 Let AB be the given * straight line, (7 the given triangle and D the given angle : * it is required to descnbe m AB a |J- = A C, awl having an angle = l D. . Describe the r BEFO = A C, and having l EBG = ^ D; and let it be so placed that BE may be in the same straight line with -^5. Through A draw AH \\ BG or EFy and let it meet FG produced at H; join HB. Because HF meets the parallels AH, EF, .-. L AHF + L HFE = 2 rt. ^ s j .-. L BHF + L HFE is less than 2 rt. z. s ; .-. HB, FE, if produced, will meet towaids B, E. I. 29, Cor. Let them be produced and meet at ^; through K draw KL || EA or FH, ' 73] and produce HA, GB to L and M. ABML is the jj-" required. For FHLK is a r, of which HK i , a diagonal, and AG, ME are U"- about HK ; 7.29 Book L] PROPOSITIONS 44, 45. 87 \ complement BL = complement BF^ 7.43 = A C. ' And L ABM = l EBG, /. 15 = L D. 1. On a given 8*^raight line decicribe a rectangle equal to a given triangle. 2. On a given straight line describe a triangle equal to a given ||m, and having one of its angles equal to a given angle, a On a given straight line describe an isosceles triangle equal to a given II™. 4. Cut oS from a triangle, by a straight line drawn from one of tke vertices, a given area. PROPOSITION 45. Problbm. To describe a parallelogram equal to any given rectilineal figurey and having an angle equal to a given angle. ^ F G L ^ K H M Let ABCD be the given rectilineal figure, E the given angle : U is required to describe a \^ = ABCD, and having an angle =^ L E, Join BD, and describe the H"" FH = A ABD, and having l K= l E; / 42 on GH describe the [j"" GM - A BCD, and having L QHM ^ L E. / 44 FKML is the (I" required. 8b EUCLID'S ELEMENTS. [Book I. F G L A^ /. 29 /. 14 /. 30 K H M Because L K = l GHM, since each = l E; to each of these equals add L GHK; L K+ L GHK = L GUM + L GHK. But L K+ L GHK = 2 rt. Z.8; .-. L GHM+ L GHK = 2 rt. ^s; .*. KH and HM are in the same straight line. Again, because FG and GL drawn from G are both || KM,- .*. FG and GL must be in the same straight line. /. Ax. 11 I^ow because KF and ML are both || HG, .-. KFis\\ML; ami KM is \\FL; .-. FKML is a ||"». But ir i^iSTifL = ir JFw + r (^iif, = A ABD + A ^CA Cbw«#. = figure ABGD; andL K=lK Const. 1. Could two |in« have a common side and together not form one II™ ? Illustrate by a figure. 2. Describe a rectangle equal to a given rectilineal figure. S. On a given straight line describe a rectangle equal to a given rectilineal figure. 4. Given one side and the area of a reotangle ; find the other side. 5. Describe a ||'" equal to a given rectilineal figure, and having an angle equal to a given angle, using I. 37, Cor. 6. Describe a ||>" equal to the sum of two given rectilineal figures. 7. Describe a |j« equal to the difierence of two given rectUineal figures. took LI PBOP08ITI0NR 45. 4ft, Sf PROPOSITION 46. Probibm. On a given straight line to describe a square. D /. 11,3 7.31 /. 31 Let AB be the given straight line ; it is required to describe a square on AB. From A draw AC ± AB an^ ^ AB ; through G draw CD \\ AB, and through B draw BD \\ AC. ABDC is the iquare required For ABDC is a (^ j ^ ^^^ 33 .-. AB = CZ) and AJ = J5Z). But AB = AC; .'. the four sides AB, BD, DC, CA are all equal Because 4 C meets the parallels AB, CD ^'^ L A + L C=2Tt L8. But L A ia right ; .'. ii Cis also right. Now L A^ L Dand L C~ L B; .*. the four L8A,B, D, Cane right; .*. ABDC IB a square. j^ 2)e/; 32 1. What is redundant in EucUd's definition of a square ? 2. If two squares be equal, the sides on which they are described are equal. a ABDC is oonstnicted thus: At A and B draw AG and BD A . J-^^^ «d = ^5, and join C2>. ^5i>C7 is a square. 4. ABDG IS constructed thus : At A draw ^C±^5and = ^5- with ^ and C as centres, and a radius = AB or AG, describe /. 34 Const. /. 29 7.34 M mCUD'H ULBMBNT8. two circles intersecting at D ; and join BD. DG. ABDO ia a square ft. Describe a square navmg give?i a diagonal. PROPOSITION 47. Theorem. The square descnhed on the hypoti'ime of a ri, /'/JC, ] BD = BG ( L ABD = L FBC; .*. A ABD = A i^'^C. But II™ BL = twice A ABD, being on the same base BD, and between the same ||» BD, AL ; /. 41 and square Bii = twice A FBC, being on the same base BF, and between the same y j?^, CG; J, 41 .-. Il"' BL = square 5(?. Similarly, if AE, BK be joined, it may be proved that II'" CL = square CH ; .-. \r BL + r CL = square ^G^ + square CH, that is, square on BC = square on BA + square on ^C. [It is usual to write this result BG'^ = 5^^? + AC'^; but see p. 113.] Cor. — The difference between the square on the hypoten- use of a right-angled triangle and the square on either of the sides is equal to the square on the other side. For since BC'^ = BA^ + AC^ .-. BC^ - BA^ = AC^y and BC - AC^ = BAK NoTK.— This proposition is an exceedingly important one, and numerous demonstrations of it have been given by mathematicians, 3omo of them such as easily to afford ocular proof of the equality asserted in the enunciation. With respect to Euclid's method of i)i()of (which is not* that of the discoverer), it may be remarked tliat he has chosen that position of the squares when -^Viey are all exterior to the triangle. The pupil is advised to makt the seven other modifications of the figure which result from placing the squares in different positions with respect to the sides of the iriangle, and to adapt Euclid's proof thereto. It will be found that AO and -4C, as well &;& AH and AB, will always be in the um» * S«« Friedlein's Proclm, p. 420. O .$ M Euclid's elements. [Book t aTJ^^WZ' T 5i"''*'n^ °^ ^"'"^ ^^*^" '" «PP-«it« directions fn>m of equal a^d that fl 7. ''"^f"""' ^ supplementary instead oi equal, and that then the equality of ab JBD und t Fin «,;ii foHow, not from I. 4, but from I. 38 Cor ^ ""'" All the different varieties of figure are obtained thus : other sidef B^rr ^'^ '''^'^"'"' "^ ^"'^ ^ *^^ ^^^^ ^ *^« (1) X outwardly, F outwardly, Z outwardly. (2) (3) (4) (6) (6) (7) (8) inwardly, II inwardly II II II outwardly, M II II inwardly. inwardly. outwardly. inwardlj'. outwardly. inwardly. outwardly. inwardly. The following methods of exhibiting how two squares may be dissected and put together so as to form a third square, arrnrobablv pt»n ! "' '"*"* ^^^^^ ^'^^^^^ y^^ S^-" "^ tC cerbrS FIRST METHOD. K Tif wfii . ^' '^'^'^ ^"'^^^*^' '^"^^ ^ '"oves «iung line ime z^ii, it will come to occupy the position I^FP Ti, two squares ABOH and Rrpp^^iu 4.u r'*'J'^^^^^^ ^/'ii. Ihe sqn^r^ DEKH BCEF y^xM then be transformed into the [This method is substantially that given bv Srhnn+-« • v Exercitatwnes MathematiccB imi) d 111 Th.f? 7 .'" ^" ly be bably rated that join of a If. the tion ans* 3KC0ND METHOD. , ! 4'> The square ABFO is then d.vMed to ISl^i 1^^' ^ f^' every resnecfc. Thrnnrrh +1, • 1 1, . ^uaoruaterals equal in parts 1, 2, 3, 4, 5 will be fnnnTl f '""^ *■«""• ''''«■' «■« 4-, 5'. '''•'"""'« «»"M to comoide exactly with 1', 2-, 3', [This method i» due to Henry ftrig.1, F.E.A.a, «,d w«, di.. 94 BUOLID's BLEMBNT8. [Book I. See The Messenger of Mathematics, new series, covered about 1830. voL ii. pp. 103-106.] 1. Show how to tind a square = the sum of two given dquareii. * •• 11= „ three „ •• «• 11= the difference of two u ■» •• •» double of a given square. 6. M M half II 6* •• M triple II 7. The square described on a diagonal of a given square is twice the given square. 8. Hence prove that the square on a straight line is four times the square on half the line. 9. The squares described on the two diagonals of a rectangle are together equal to the squares described ou the four sides. 10. The squares described on the two diagonals of a rhombus are together equal to the squares described on the four sides. 11. If the hypotenuse and a side of one right-angled triangle be equal to the hypotenuse and a side of another right-angled triangle, the two triangles are equal in every respect. 12. K from the vertex of any triangle a perpendicular be drawn to the base, the difference of the squares on the two sides of the triangle is equal to the diflference of the squares on the segments of ine bas& 13. The square on the side opposite an acute angle of a triangle is less than the squares on the other two sides. 14. The square on the side opposite an obtuse angle of a triangle is greater than the squares on the other two sides. 15. Five times the square on the hypotenuse of a right-angled triangle is equal to four times the simi of the squares on the medians drawn to the other two sides. 16. Three times the square on a side of an equilateral triangle is equal to four times the square on the perpendicular drawn from i,ny vertex to the opposite side. 17. Divide? a given straight line into two parts such that the sum of their squares may be equal to a given square. Is this always possible ? 1& Divide a given straight line into two parts such that the square on one of them may be double the square on the other. 19. If a straight line be divided into any two parts, the square on the whole line is gi eater than the sum of the squares on the two parts. t Book t] PROPOSITION 47. 95 26. 26. a(X The sum of the squares of the distances of any point from two opposite corners of a rectangle is equal to the sum of the squares of its distances from the other two corners. The following deductions refer to the figure of the proposition in the text. They are all, or nearly all, given in an article in Leyboum's Mathematical Repository, new series, vol. iii. (1814), Part II. pp. 71-80, by John Bransby, Ipswich. 21. What is the use of proving that AO and ^C are in th« same straight line, and also AB and AH? 22. A F and A K are in the same straight line. 2a BO is II CII. 24. Prove As ABD, FBC equal by rotating the former rounds through a right angle. Similarly, prove AbACE, KCB equal. Hence prove AD A. FG, and AE i. KB. L 8 ABQ and DBF are supplementary, as also are l%ACB and ECK. * > 27. Hence prove lb FBD, KCE = A ABO. 28. FG, KH, LA all meet at one point T. 29. abAOH, THO, a a T, HTA are each = A ABC. 30. If from D and E, perpendiculars />(/, ^Fbe drawn to FB and KC produced, as UBD and VEG are each = a ABC. Prove by rotating. 31. DF^ + EK'' = 5 BC^. 32. The squares on She sides of the polygon DFOHKE = 8 BC^. 33. If from F and K perpendiculars FM, KN be drawn to BG produced, and /be the point where AL meets BG, A BFM = A ABT, and a GKN = A AGI. 34. FM + KN= BG, and BN=GM=AL. 38. If DB and EG produced meet FO and KH at P and Q, prove by rotating A ABG that it = each of the a s FBP, KGQ. 36. If PQ be joined, BCQP is a square. 37. ABPT is a ir, and = rectangle BL ; AGQT ia & r, and = rectangle GL. ADBT is a ir, and = rectangle BL ; AEGT is a ||«», and = iGctangle GL. DFPU and EKQV sltq |r», and each = 4 a ABG. ADUH and AEVQ are !r», and each = 2 A ABG. BKia ± GT, and CF ± BT. 42. Hence prove that AL, BK, GF meet at one point O. (See App. I. 3.) 38. 39. 40. 41. 96 41 Bt70LID*S ELEMENTS. [Book I 44. 4fi. 46. If RK meet A C in X, and CF meet A Bin W, £,» BHX CO W are each = a ABC. ' ^ >K = AX. A ^CJK= A ^CX.and A ABX=. a 5Clf. Quadrilateral A WOX = a i?OC. 47. If from (? and // iHjrpendiculars OR, HS be drawn to bC or 56* produced, and. if these perpendiculars meet ABt^nd AC in Y A^ _*"d^'J>»-«^e by rotating a ABCth^tit = a OA For a -^^^ «iFI^""'u \T^\ *^''*'"«^ ^' ^'^' P«>d'»««8, aiid AH^ + AC^ = BO^/ .-. cy/^ = nc^; ,'. CA> = Ba (BA = Z)i4 In As BAG, DAC\ < AC ^ AO \BC = DG; \ L BAG = L DAG, = a right angle. 1. In the construction it is aud, draw AD L AC. Would if not be simpler, and answer tlie same purpose, to say, produce AB to/). Why? 2. Prove the proposition indirectly by drawing AD ± AC, &nd on the same side of ^C^ as ^ if, and using I. 7 (Proclus). a If the square on one side of a triangle be less than the su.n of the squares on the other two sides, the angle opposite that side is acute. 4. If the square on one side of a triangle be greater than the sum of the squares ou the other two sides, the angle opi^site that side is obtuse. 5. Prove that the triangle whose sides are 3, 4, 5 is right-angled.* C. Hence derive a method of drawmg a perpendicular to a given straight line from a point in it. 7. Show that the following two rules.f due respectively to Pytha- goras and Plato, give numbers representing the sides of right-angled triangles, and show also that the two rules are fundamentally the same. (o) Take an odd number for the less tide about the right angle. Subtract unity from the square of it, and halve the remain- der ; this will give the greater side about the right angle. Add unity to the greater side for the hyiM)tennse. (6) Take an even number for one of the sides about the right angle. From the square of half of this number subtract unity for the other side about the right angle, and to the square of half this number add unity for the hypotenuse. * This is said by Plutarch to have been known to the early Egyptians. + See Friedlein'8 Froclus, p. 428, and Hultsch's Eeronis . . . relimioi pp.56, 57. ^ 96 Euclid's elements. [Book I APPENDIX L Proposition 1. The atraiqht line joining tfie middle points of any two artea of a irtangh is parallel to the third aide and equal to Uie hxjlj ofii Let ABO be a triangle, and let L, K be the middle points of AB, AG: ^ it is required to prove LK || BG and = half ofBG. Join BK, GL. Because AL = BL, .• . a BLG = half of a ABG; /. 38 and because AK = GK, .-. a BKG = half of a ABG: I. 38 .-. £,BLG= ^BKG. .'. LK is II BG. J 39 Hv^nce, if ^ be the middle of BG, and HK be joined, HK is || AB • .'.BHKLv,^\r; ^ I.DefZ^ .'.LK = Bff= half of BG. /. -4 Cor. 1.— Conversely, The straight line drawn through the middle point of one side of a fnangle parallel to a second side bisects the third side.* Cor. 2.—AB is a given straight line, G and D are two points, either on the same siue of ^5 or on opposite sides of AB, and such that ^C and BD are parallel. If through E the middle point of AB, & straight line be drawn \\ AC or BD to meet CD at F, then * The corollaries and oonverses given in the Appendices shovld be proved to be true. Many of *h«m are not obvious. Book t] APPENDIX I. 99 F is the middle point of CD, and JtlF is equal either to half the sum oi AC and BJJ, or to half their difference. f Proposition 2. The straight lines dravm perpendicular to the sides of a triangle from the middle points of the sides are concurrent {thai is, pass through the same point). See the figure and demonstration of IV. 5. If 8 be joined to H, the middle of BC, then SH is x BC. I. 8 NoTB.— The point S is called the circumsaibed centre of a ABC. Proposition 3. The straight lines drawn from tlie vertices of a triangle perpendicular to tiie opposite sides are cortcurrent.* Let AX, BY, GZ be the three perpendiculars from A, B, C on the opposite sides of the A ABC; It is required to jtrove AX, B Y, GZ concurrent. Through A, B, G draw KL, LH, HK \\ BC, CA, AB. I. 31 Then the figures ABCK, ACBL are H™* ; T. Def. 33 .•.AK = BC=AL, /. 34 that is, A is the middle point of KL. ♦Pappus, VII. 62. The proof here given seems to be due to F. J. Servois : see his Solutions peu connuea de diffiremt prnhlAm^$ de G^omitrie- pratique (1804), p. 15. It is attributed to Gauss by Dr R. Baltzer. i 100 Euclid's element's. [Book u Hence also, H and Care the middle points of L/f and //A'. «ut since AX, BY, CZ are respectively j. BC, GA, AB, ' Const. they must be respectively i. KL, LII, HK, /. 29 and.', concurrent. ^pp j 2 Note.— The point is called the orthocentre of the a ^J9C(an expression due to W. H. Besant), and a XYZ, formed by joming the feet of the perpendiculars, is called sometimes WMpedal, sometimes the orthocentric, triangle. Proposition 4. The medians of a triangle are concurrent. ^ tl Q Let the medians BK, CL of the is ABO meet at G: it M required to prove that, if H be the middle point of BC, the median AH mil pass through O. Join AG. Because BL = AL .-. a BLC == ^ ALC, »n^ ^BLG = lALG; 7.38 .'. LBGC = t.AGG, . ^.^^.3 = twice A CKG ; / 33 .'. BG=Wv^GK,(irBK = i\mQQGK, that is, the median CL cuts BK at its point of trisection remote from B. Hence also, the median AH cuts BK at its point of trisection remote from B, that is, AH passes through G. Cor.— If the points H, K, L be joined, the medians of thp A HKL are concurrent at G. NoTJL— The point G is called the cmt^-oid of the a ABC (an APPENDIX I. 101 expresftiou due to T. S. Davies), and a HKL may be caUed the median triangle. The centroid of a triangle is the same point as that which in Statics is called the centre of gravity of the triangle, and may be found by drawing one median, and trisecting it. Proposition 6. Th£ ortJiocentre, the centroid, and the circumscribed centre of a triangle are coilinear {tfiat is, lie on the same straight line), and the distance between the first two is double of the distance between the tost tiuo* Let ABChe& triangle, O its orthocentre determined by drawin AX &nd BY ± 5(7 and CA ; 8 its circumscribed centre determine i by drawing through ^ and ^ the middle points of BG aiid CA, h8 and K8 ± BC and GA ; and AH the median from A : U is required to prove that if SO be joined, U wiU cut AH at the ceMroid. Let 80 and AH intersect at O ; join P and Q, the middle points of OA, GO; " ^ " ^^ » II OA, OB; and join HK. Because H and K are the middle pomts of CB, GA • .'. HK is II AB and = half AB. Because U and V are the middle points of OA OB • .-. £;■ F is II AB and = half AB, .'. HKiaW C7^rand= VV. App. 1. 1 App. L 1 * First given by Euler in 1765. See N; ' ,. .^ join DE and GB. ^' '" Because ^^ joins the middle points of two sides oi a ACB But (Tff being the radius of a fixed circle, is a fixed length i^^' .• . DE, Its half, IS also a fixed length. Again, since A and O are fixed points, ,'. ACiaa fixed straight line ; .-. A the middle point ofAC,ia& fixed point • ^mlLl.'i^rA'^"" '"^''' '" """*'' >*'«'«• ^"^ But AB was any straight line drawn from ^ to the o«* • A the middle points of all other straight lines drawn from A to the O- must be situated at the same fixed distance from the fixed pomt .'. the locus of the middle points is the 0« of a circle, who^e centre is A and whose radius is half the radius of the fixed drcle IT- . .^"""^ '* """^ ^^ '^^° *^^* ^* ^« immaterial whether AB ti.^ J I i f .*^' ""'^^^^ P^^"* °^ ^^' **^«« ^^ = i ^C, tbttt 18 = half the radius of the fixed circle ; .•. the locua of JT is the same o«> as befow. Book I.] APPENDIX I. 105 fThe reader is requested to make figures for the cases when the given point A is inside the given circle, and when it is on the o** of the given circle.] INTERSECTION OF LOOL Since two conditions determine a point, if we can construct the locus satisfying each condition, the point or points of intersection of the two loci will be the iK)int or points required. A familiar example of this method of determining a point, is the finding of the position of a town on a map by means of parallels of latitude and meridians of longitude. The reader is recommended to apply this method to the solution of I. 1 and 22, and to several of the problems on the construction of triuagles. DEDUCTIONS. 1. The straight line joining the middle points of the non-parallel sides of a trapezium is || the parallel sides and = half their sum. 2. The strai;,'ht line joining the middle points of the diagonals of a trapezium is || the parallel sides and -- half their difference. 9* The straight line joining the middle points of the non-iiarallel sides of a trapezium bisects the two diagonals, 4.' The middle points of any two opposite sides of a quadrilateral and the middle points of the two diagonals are the vertices of a ir. 6. The straight lines which join the middle points of the opposite sides of a quadrilateral, and the straight line which joins the middle points e>f the diagonals, are concurrent. 6. If from the three vertices and the centroid of a triangle perpen- diculars be drawn to a straight line outside the triangle, the perpendicular from the centroid = one-third of the sum of the other perpendiculars. Examine the cases when the straight line cuts the triangle, and when it passes through the centroid. 7. Find a point in a given straight line such that the sum of its distances from two given points may be the least possible. Examine the two cases, when the two given jMjints are on the same side of the given line, and when they are on different sides. 106 euoud's elements. [Book t 12. 13. & Find a point in a given straight line such that the diflference of Its distances from two given points may be the greatest possible. Examine the two cases, 9. Of all triangles having only two sides given, that is the greawst in which these sides are per{)endicular. 10. The perimeter of an isosceles triangle is less than that of any other triangle of equal area standing on the same 11. Of all triangles having the same vertical angle, and the bwes of which pass through the same given point, the least is tnat which has its base bisected by the given point Of all triangles formed with a given angle which is contained by two sides whose sum is constant, the isosceles triangle hma the least perimeter. * The sum of the perpendiculars drawn from any point in the' base of an isosceles triangle to the other two sides is cbn- Btant. Examine the case when the point is in the baae produced. 14. The sum of the peT>endicular8 drawn from any point inside an equilateral triangle to the three sides is constant. Examine the case when the point is outside the triangle 15. The sum of the perpendiculars from the veri;ices of a triangle on the opposite sides is greater than the semi-perimeter and less than the perimeter of the triangle. 16. If a perpendicular be drawn fromlhe vertical angle of a triangle to the base, it will divide the v^ical angle and the b^e into parts such that the greater is next the greater side of the triangle. 17. The bisector of the veridical angle of a triangle divides the base into segments such that the greater is next the greater side of the triangle. 18. The median from the vertical angle of a triangle divides the vertical angle into parts such that the greater is next the less side ot the triangle. 19. If from the vertex of a triangle there be drawn a perpendicular to the opposite side, a bisector of the vertical angle and a median, the second of these lies in position and magnitude between the other two. 20. The sum of the three angular bisectors of a triangle is greater ^ the semiperimeter, and less tkan th« perimeter of the Book I.] APPENDIX I. 107 21. If one side of a triangle be greater than another, the perpen- dicular on it from the opposite angle is less than the »;oiiedpondJng perpendicular on the other side. 22. If one side of a triangle be greater than another, the median drawn to it is less than the median drawn to the other. 23. If one side of a triangle be greater than another, the bisector of the angle opposite to it is less than the bisector of the angle opposite to the other. 24. The hypotenuse of a right-angled triangle, together with the perpendicular on it from the right angl^ is greater than the sum of the other two sides. 25. The sum of the three medians is greater than three-fourths of the perimeter of the triangle. 26. Construct an equilateral triangle, having given the perpendicular from any vertex on the opposite side. Construct an isosceles triangle, having given : 27. The vertical angle and the i)erpendicular from it to the base 28. The perimeter and the perpendicular from the vertex to the base. Construct a right-angled triangle, having given : \ 29. The hypotenuse and an acute angle. 30. The hypotenuse and a side. 51. The hypotenuse and the sum of the other sides. 32. The hypotenuse and the difference of the other sides. 33. The perpendicular from the right angle on the hypotenuse and a side. 34 The median, and the perpendicular from the right angle, to the hypotenuse. 35. An acute angle and the sum of the sides about the right angle. 36. An acute angle and the difference of the sides about the right angle. Construct a triangle, having given : 37. Two sides and an angle opposite to one of them. Examine the cases when the angle is acute, right, and obtuse. One side, an a»gle adjacent to it, and the sum of the other two udes. One side, an angle adjacent to it, and the difference of the other two sides. 40. One side, the angle opposite to it, and the sum of the o«h9r twa sides. H 38 39, 108 *""^J'S BLBMENT8. [Book L r- 41. One aide, the angle opposite to it, .^nd the difference of the other two sides. 42. An an;^'le, ita bisector, and the perpendicular from the angle on the opposite side. 43. The angles and the sum of two sides. 44. The angles and the difference of two sides. 45. The perimeter and the angles at tlie base. 4l). Two sides and one mediau. 47. One side and two medians. 48. The three medians. 49. 60. Construct a square, having giren : The sum of a side and ;i the area and perimeter of which shall = the area and peritneter of a given triangle. 59. The diagonals of all the ||>"» inscribed* in a given |h intersect one another at th*^ same point. 60. In a given rhombus inscribe a square. 01. In a given right-angled isosceles triangle inscribe a square. 62. In a given square inscribe an equilateral triangle having one of its vertices coinciding with a vertex of the square. AA', BB\ GC are straight lines drawn from the angular points of a triangle through any point O within the triangle, and cutting the oi)posite sides at A', B', C. AP, BQ, CM are cut off from AA', BB', CC, and = OA , OR, OG'. Prove A A'BG' = A PQR. * One figure is inscribed in another when the vertices of the first figure are on the sides of the second. 63 Book L] APPENDIX I 109 04. On AB, AC, sides of a ABC, the |r- ABDE, ACFO are described ; DE and FO are produced to meet at H, aod AH is joined ; through B and C, BL and CM are drawn || AH and meeting DE and /"(? at L and if. If LM be joined', -BCJfZ, is a ir, and = r BE + |r C(?. (Pappus, IV. 1.) 65. Deduce I. 47 from the preceding deduction. 66. If three concurrent straight lines be respectively perpendicular to the three sides of a triangle, they divide the sides into segments such that the sums of the squares of the alternate segments taken cyclically (that is, going round the triangle) are equal ; and conversely, 67. Prove App. I. 2, 3 by the preceding deduction. 68. If from the middle point of the base of a triangle, perpendiculars be drawn to the bisectors of the interior and exterior vertical angles, these perpendiculars will intercept on the sides segments equal to haH the sum or half the diflFerence of the sides. 69. In the figu™ to the preceding deduction, find all the angles which are equal to half the sum or half the diflFerence of the base angles of the triangle. If the straight lines bisecting the angles at the base of a triangle, and terminated by the opposite sides, be equal, the triangle is isosceles. Examine the case when the angles belpw the base are bisected. [See NouvelUs Annates de Mathematiques (1842), pp. 138 and 311 ; Lady's and Gentleman's Diary ior 1857, p. 68 ; for 1859, p. 87 ; for 1860, p. 84 ; Lond of a given circle consists of the o<^' of two circles con- centric with the given circle. Examine whether the locus will always consist of two 0<*". [The distance of a point from the circumference of a circle is measured on the straight line joining the point to the centra ©5 the circls.] no ■C0LID8 ELEMENTS. [ r* a V 3. The locus of the points equidistant from two given straight lines which intersect, consists of the two bisectors of the angles made by the given straight lines. 4. What is the locus when the two given straight lines are parallel ? fl. The locus of the vertices of all the triangles which have the same base, and one of their sides e(iual to a given length, consists of the O"" of two circles. Determine their centres and the length of their radii. 6. The locus of the vertices of all the triangles which have the same base, and one of the angles at the base equal to a given angle, consists of the sides or the sides produced of a certain rhombus. 7. Find the locus of the centre of a circle which shall pass through a given point, and have its radius equal to a given straight line. I 8. Find the locus of the centres of the circles which pass through two given jwints. 9. Find the locus of the vertices of all the isosceles triangles which stand on a given base. 10. Find the locus of the vertices of all the triangles which have the same base, and the median to that base equal to a given length. 11. Find the locus of the vertices of all the triangles which have the same base and equal altitudes. 12. Find the locus of the vertices of all the triangles which have the same base, and their areas equal. 13. Find the locus of the middle points of all the straight lines drawn from a given point to meet a given straight line. 14. A series of triangles stand on the same ^ase and between the same parallels. Find the locus of the middle po.ut- vt their sides. 15. A series of H""" stand on the same base and Tbetweeii tlie same parallels. Find the locus of the intersection of their diagonals. 16. From any point in the base of a triangle siraight lines are drawn parallel to the sides. Find the locus of the intersection of tbe diagonals of every H"" thus formed, iV. StrEigho 'ines are drawn ^larallel to the base of a triangle, to meet the sides or the sides produced. Find the locus of th«ir middle poiuts. t] APPENDIX I. HI 18. Kmd tho locuB of the angular iwint opposite to the hypotenuse of all the nght-angled triangles that have the same hypoteu- use. 19. 20. 21 A ladder stands upright against a perpendicular wall. The foot of it 18 gradually drawn outwards till the ladder lies on the ground. Prove that the middle point of the ladder haa described part of the 0~ of a circle. Find the locus of the points at which'two equal segments of » straight line subtend equal angles. A straight line of constant length remains always parallel to Itself, whUe one of its extremities describes the O" of a circle Find the locus of the other extremity. 22. Find the locus of the vertices of all the triangles which have the . same base 3C, and the median from B equal to a given 23. The base and the difference of the two sides of a triangle are given ; findthe locus of the feet of the perpendiculars drawn from the ends of the base to the bisector of the interior vertical angle. The base and the sum of the two sides of a triangle are given • hnd th-^ locus of the feet of the perpendiculars drawn from' the ends of the base to the bisector of the exterior vertical angle. Three sides and a diagonal of a quadrilateral are given : find the locus (1) of the undetermined vert;ex, (2) of the middle point of the second diagonal, (3) of the middle point of the straight Ime which joins the middle points of the two diagonals. (Solutions raimnnSes des ProhUmes inoncSs dans let EUmenU «• G6o7rUtrie de M. A. Amiot, 76me ed. p. 120 24 25 112 BOOK II. DEFINITIONS. 1. A rectangle (or rectangular parallelogram) is aaid to be contained by any two of its conterminous sides. Thus the rectangle A BCD is said to M — iB be contained hy AB and BG; or by BC end CD : or by CD and DA ; or by j^ DA and AB. " ^ The reason of this is, that if the lengths of any two conterminous sides of a rectangle are given, the rectangle can be constructed ; or, what comes to the same thing, that if two conterminous sides of one rectangle are respectively equal to two conterminous sides of another rectangle, the two rectangles are equal in all respects. The +.ruth of the latter statement may be proved by applying the one rectangle to the other. 2. It is oftener tbe case than not, that the rectangle con- tained by t"^o straight lines is spoken of when the two straight lines do not actually contain any rectangle. When this is so, the rectangle contained by the two straight lines will signify the rectangle contained by either of them, and a straight line equal to the other, or the 'rectangle contained by two other straight lines respectively equal to them. F«. 1. A 0- B E Fig. 2. Ci iD F £ Fig. 8. E H A G B A -B C— Book U] DBFINITIONS. US Thu3 ABEF (fig. 1) may be considered the rectangle contwned by AB and CD, if BE = CD ; CDEF (fig. 2) may be coiujid..ed the rectangle contained hy AB and CD, if DE = AB : and EFOn (fig. 3) may be considered the rectangle contained by AB and CD if EF = AB and ^G? = Ci)„ ^ . 3. As the rectangle and the square are the figures which the Second Book of Euclid treats of, phrases such as * the rectangle contained by AB and AC,' and 'the square described on AB; will be of constant occurrence. It is usual, therefore, to employ abbreviations for these phrases. The abbreviation vrhich will be made use of in the present text-book* for 'the rectangle contained by ^^ and BC* is AB-BC, and for * the square described on AB,' AB\ 4. When a point is taken in a straight line, it is often ' called a point of section, and the disiances of this point from the ends of the line are called segments of the line. D -B Thus the point of section D divides AB into two segments AD and BD. In this case AB ib said to be divided internally at D, and AD and BD are called internal segments. The given straight line is equal to the sum of itg internal sec- menta ; for AB — AD + BD. 5. When a point is taken in a straight line produced, it ib also celled a point of section, and its distances from the eiids of the line are called segments of the line. B -I- D D A -+- Thus D is called a point of section of AB, and the segments into which it is said to divide A B axe AD and BD. * In certain written examinations in England, the only abbreviation aUowed for 'the rectangle contained by ^^ and BC is rect AB BO and for ' the square described on AB,' sq. on AB ; the pupil, therefore if preparing for these examinations, should practise himself in the use'of «uoh abbreviations. MMi^lM r.T 111 ik *H Euclid's elements. [Book n. In this case, AB ia said to be divided externaUy at D, and A I), HI) are called exlirnal segments. The uiven straight line is equal to the difference of its external segments ; iot AB = AD - BD, or BD - AD. 6. When a straiglit line is divided into two segments, such that the rectangle (Contained by the whole line and one of the segments is equal to the square on the other segment, the straight line is said to be divided in medial section.* H — 4— B Thus, if ^5 be divided at IT into two segments AH and BH such that AB.BH = AH\ AB is said to be divided in medial section at H. It will be seen that AB h internally divided at H • and in general, when a straight line is said to be divided in medial section It IS understood to be internally divided. But the definition need not be restricted to internal divisioa H' A B Thus, if J « be divided at H' into two segments AH' and BH' such that AB . BH' = AH'\ AB in this case also may be said to bJ divided in medial section, 7. The projection t of a point on a straight line is the foot of the perpendicular drawn from the point to the straight line. B- D Thus D is the projection of A on the straight line BC. 8. The projection of one straight line on another straight * The phrase, 'medial section,' seems to be due to Leslie. See his Elements of Geometni (1809), p. 66. t Sometimes the a.liect.ve ' orthogonal' is prefixed to the word pro-" 'Ottion, to distinguish this kind from othen. Book n.] DEFINITIONS. 116 line is that portion of tne second intercepted between per- pendiculars drawn to it from the ends of the first. F«. 1 While the straight hne to be projected must be hmited in length is^r fttr ^ '^ ^'''' '' '' '^ '^ ^-^-^^' -t be considtd' 9. If from a parallelogram there b^ taken away either of the parallelograms about one of its diagonals, the remaining figure is caUed a gnomon. ^ C B A __C! B r- about the diagonal BD, the figure which remains Z^uHfI CK 18 talteu away from ADEB is caUed a momon In th. ,■ . ca,e,when /^^ is taken away, the gnomon'^SG^" rt' I'^a /^Oii., in the second case, when CK is taken aw^Tr fi,^ ^i^^iiTGrC would similarly be called AFK otCHe''' ^°'°*°'' The word 'gnomon' in Greek means, among other thin«« carpenter's square/ which, when the ^ ADEB i^:i^Tor a *i:r!''?i''^^"°^'? ^^^^.^ ^««' fr«"^ its shape, called bv th« »n.-.«* g.«x....«, .u«BiioemaKeraiu»ife.^ See Pappus, IV. section 14. 116 EUOLIDS ELEMENTS. [Book n. rectangle, the figure AKF resembles. The only gnomons mentioned by Euclid in the second book are parts of squares. The more general definition given by Heron of Alexandria, that a- gnomon is any figure which, when added to another figure, produces a figure similar to the original one, will be partly understood after the fourth proposition has been read. PROPOSITION 1. Theorem. // there he two straight lines, one of winch is divided inter- nally into any number of segments, the rectangle con- tained by the two straight lines is equal to the rectangles contained by the undivided line and the several segments of the divided lii^e. C E D G •B Let AB and CD be the two straight lines, and let CD be divided internally into any number of seg- ments CU, EF, FD: it is required to prove AB-CD = AB-CE + AB-EF + AB.FD. From Cdraw CG ± CD and = AB ; ' I. 11, 3 through G draw GH \\ CD, and through E, F, D draw EK, FL, DH \\ CG. I. 31 Then CH= CK + EL-{- FH; % I. Ax. 8 that is, GC' CD = GC-CE -{- KE- EF + LF - FD. But GC, KE, LF are each = AB ; Const, I. 34 .-. AB- CD = AB- CE + AB EF A T) .a.j-j Book n] PROPOSITIONS 1, 2. 117 ALOBBRAICAL ILLUSTRATION. Let AB « o, C7i) = 6, CE = c,EF^ d, FD - «; then 6 = c + d + e. Now^J5.CZ)= aft, and ^5 . 0J& + ^5 . i?i^ + il5 . /'D = oc + ad + a* But since h = c + d + e, .: ab = ac + ad + ae ; .\ AB-GD = AB.GE+AB.EF+AB.FD. 1. The rectangle contained by two straight lines is equal to twice the rectangle contained by one of them and half of the other. 2. The rectangle contained by two straight lines is equal to thrice the rectangle contained by one of them and one-third of the other. a The rectangle contained by two equal straight lines is equal to the square on either of them. 4. If two straight lines be each of them divided internally into any number of segments, the rectangle contained by the two straight lines is equal to the several rectangles contained by all the segments of the one taken separately with all the segments of the other. PROPOSITION 2. Thborbm. If a straight line be divided internally into any two segments, tiie square on the straight line is equal to the sum of the rectangles contained by the straight line and the two segments. .JL. P E Let AB be divided internally into any two segmenta AC, OB: it is required to prove AB* = ABAC + AB- CB, «;l 118 Euclid's elements. [Book n. A r- I > D B I ! — J E On AB describe the square ADEB, J. 46 and through C draw CF || AD, meeting DE at K I. 31 Then AE = AF + CE; /. Ax. 8 that is, AB^ = DA • AC + EB ■ CB. But DA and EB are each = AB ; .•.AB^ = AB-AC+AB-CB. \ ALGEBRAICAL ILLUSTRATIOIT. JjetAC=a,CB = b; then AB = a + b. Now, AIi^ = (a + 6)2 = a2 + 2a6 + 62, and ^5 . ^C + ^5 . C5 = (a + 6) a + (a + 6) 6 = a2 + 2a6 + 6»; .•.AB^= ABAC + ABCB. 1. Prove this proposition by taking another straight line = AB, and using the preceding proposition. 2. If a straight line be divided internallj' into any three segments, the square on the straight line is equal to the 'sum of the rectangles contained by the straight line and the three segments. 3. If a straight line be divided internally into any number of segments, the square on the straight line is equal to the sum of the rectangles contained by the straight line and the several segments. Show that the proposition is equivalent to either of the following : 4 The square on the sum of two straight lines is equal to the two rectangles contained by the sum and each of the straight lines. 5. The square on the greater of two straight lines is equal to the rectangle contained by the two straight lines together with the rectangle contained by the greater and the difference between the two. Book n.] PROPOSITIONS 2, 6. 119 PROPOSITION 3. Theorem. If a straight line he divided externally into any two segments the siiuare on the straight line is equal to the difference of the rectangles contained by the straight line and the two segments. B i i D E"""F Let ^5 be divided extemaUy into any two segments it is required to prove AB^ = AB ■ AC - AB ■ CE On AB describe the square ADEB, j ^g Mid through G draw OF \\ AD, meeting DE produced at Then AE=AF- CE ; j^ j^^g that is, AB^ = DA- AC - EB - CB. But DA and EB are each = AB ; .-. AB^ = AB.AC- AB. CB. ' Note.— The enunciation of this proposition usually given is • If a straight line be divided into any two parts, the reciangle contained by the whole and one of the parts is equal to the rectangle sTd ^^ ^^^ *^^^*^^' "^^^ *^^ ''l"*'^ ^"^ *^« ^«^e. That is, in reference to the figure, AG-AB = AE^ + AB-BC, an expression which can be easUy derived from that in the text. ALGEBRAICAL ILLUSTRATION. Let^C=a, GB = h; then AB = a - h. , -.., _ y^ - ^j^ — a- - -zao + o', 120 Euclid's elements. I r f\ m n'r [Book a And AB ■ AG - AB • CB = {a - b) a - (a - b) b .•.AB* = ABAG-AB'CB. 1. Prove this proposition by taking another straight line = AB, and using the first proposition. Show that the proposition is equivalent to either of the following : 2. The rectangle contained by the sum of two straight lines and one of them is equal to the square on that one together with the rectangle contained by the two straight lines. 3. The rectaijgle contained by two straight lines is equal to the square on the less together with the rectangle contained by the less and the difference of the two straight lines. PROPOSITION 4. Theorem. 1/ a straight line be divided internally into any two ^ffments, the sq^mre on the straight line is equal to the squares on the ttoff segments increased by twice the rectangle conr tained by the segments. D F B Let AB be divided inteirally into any two segments ACy GB: it is required to prove AB^ ^^C^ + CB^ + 2 AC - CB. On AB describe the square ADEB, and join BD. I. 4G Through C draw CF || AD, meeting DB at G; and through O draw HK || AB, meeting DA and EB at H and K. /, 31 Because GG \\ AD, .-. i. GGB = l ADB ; I. 29 wad because AD = AB, .-. l ADB = :. ABD ; I. 5 Book nj PROPOaiTION A-. 121 .-. L CGB = L ABD, = L CB(Jt i CB = CG. J g Hence the \r CK, having two adjacent sides equal, has all its sides equal. ^ „ . But the r CK has one of its angles, KBC, riglit since L KBC is the same as l ABE; ' .-. it has all its angles right; " j^a .'. the II" CK is a square, and = CB^. /. xjJ 39 Similarly, the f HF is a square, and = HG^ = AC^ ' ' Again, the |r AG = AC- CG = AC- CB; GE = AC- CB; ' / 40 -Now AB^ = ^/)^^, = HF+CK + AG+GE, L Ax. S = ^C^ + C^2 + 2 ^6'. C5. Cor. 1.— The square on the sum of two straight lines is equal to the sum of the squares on the two straight lines increased hy twice the rectangle contained by the two straight lines. "^ For if ^C and CB he the two straight lines, then their sum = AC + CB = AB. Now since AB^ = AC' + CB' + 2 AC- CB IT ± .-. {AC + CBf = AC-^ + CB^ + 2 AC CB. Cor. 2.-The \r about a diagonal of a square are them- selves squares. [It is recommended that II. 7 be read immediately after II. 4.] OTHERWISE : •B /. 5 AB^= AB.AG + j^B BG = (AC -AG + BG-AG) + {AG . BG + BG . BG\ -^AG^ + BG'+2AG.Ba ^* II. 2 IL3 128 ■trCfL1>r<« IMitfHKNTS. [Book U. u ALOEBRAia^ i::JtaTRATroN. Let AC = a, CB = b; then AH = a + b. Now A B'^ = (a + 6)2 = a^ + 2ah 4- b^. and AG^ + CB^ + 2AG-CB = a-' + b^ + 2ab; .- . AB» = AC^ + CB^ + 2 AC • CB. 1. Name the two figures which form the sum of the squares on AO and CB. 2. Name the figiire which is the square on the sum of AC and CB. 3. Name the figure which is the ditference of the squares on ^^ and AC. 4. Name the figure which is the difference of the squares on ^J5 and BC 5. Name the figure which is the square on the difference oi AB and AC. 6. Name the figure which is the square on the difference oi AB and BC. 7. By how much does the square on the sum of ^C and GB exceed the sum of the squares on ^ C cand CB ? 8. Show that the proposition may be enunciated : The square on the sum of two straight lines is greater than the sum of the squares on the two straight lines by twice the rectangle contained by the two straight lines. 9. The square on any straight line is equal to four times the square ou half of the line. VO. If a straight line be divided internally into any three segments, the square on the whole line is equal to the squares on the three segments, together with twice the rectangles contained by every two of the seyments. 1 1. Illustrate the preceding deduction algebraically. liiiii PEOPOSITION 5. Theorem. Tf a drai(jM line he divided info livo equal, and also internally into two unequal ser/meiits, the rectangle contained by the unequal segments is equal to the difference between the sduare on half the line and the square on the dine between the joints of section. Book a] PROPOSITION 5, m — j -sr— " M i ../ E -J J Q F Let AB be divided into two equal segments AO, CB, and also internally into two unequal segments Ad\ DB: it is required to prove AD-DB = CB^ - CL^. On GB describe the square CEFB, and join BE I. 46 Through D draw DHQ \\ CE, meeting EB and EF at // and O; through /r draw MHLK \\ AB, meeting /!fi and EC lit i/andZ; and through A draw ^^ || CL. Then AD-DB = ADDH, = ^^, = -4Z. + Ciy, = CJf + iTiP; = gnomon GMG. But 052 -. CL^ = CB'^ - LH\ = gnomon CMG. .-. AD.DB = CB^ ~ CD\ 7.31 //. 4. Cot, 2 /. il«. 8 /. 36, 43 /. Ax. 8 /. 34 /. Ax, 8 CoR.— The difference of the squares on two straight lines is equal to the rectangle contained by the sum and the differ- ence of the two straight lines. Let ^C and CD be the two straight lines ; it is required to prove AC' - CT2 ^ (^^ _^, ^jy^ . ^^^ ^ ^^^ 134 buoud's elements. [look n. A0-¥ CD = AD, and AG - CD ^ CB - CD .-. (AC + CD) . {AC - CD) = DB; = AD . DB, = CB^ - CD^, ^ AC^ - CD'. n. > ALGEBRAICAL ILLUSTBATIOK. LetAG=CB = a,GD=b; then AD = a + 6, and DB = a- b. Now AD DB = (a + b) (a - b) = a* - 1^, and C7fi» - GD^ = a" - 62 . .\ADDB = GB»- GD'. . , ( 1. 1 1. By how much does the rectangle AG • GB exceed the rectangle AD • DB P The rectangle contained by the two interna^ segments of a straight line is the greatest possible when th» segments are equal. (Pappus, VII. 13.) 2. The rectangle contained by the two internal segments oi Zj straight line grows less according as the point of section is removed farther from the middle })oint of the straight line. (Pappus, VII. 14.) . ^ 3. Prove that AG = half the sum and CD = half the difference of AD and DB. 4. Name two figures in the diagram, each of which = the rectangle contained by half the sum, and half the difference of ^i> and DB. 5. Name that figure in the diagram which is the square on half the sum ci AD and DB. , 6b Name that figure in the diagram which is the square on half the difference of ^i> and DB. Book XL] PBOPOaiTIONS 6, 6. 1319 7. Hence Bhow that th. propoaition may be enuneutted : The r^ •ngle contamod by any two straight line, ia equal toihi *" "^Iqra^l^^. *'' '"^*^«^' ^i> • i>^ = the perimeter of the 9. Hence show that if a square and a rectangle Bare equal peri- meters, the square has the greater area. ^ ^ ^^!21* '^''°^'' '^"** *" *^' ^^^-^^^^ o^ *^o given 11. By means of the first deduction above, and II. 4, .how that the ^Zlu :r""" "" *^' '^"^ ^««'"«'^'- ^^ » -t^ight line is least when the segments are equal 12. The square on either of the sides about the right angle of a rjght-angle.1 triangle, is equal to the rectangle coZned bv . «^-um and the difference of the hypotenuse' and ttth'^ PROPOSITION 6. Theorem. 1/ a straight line be divided into two equal, and also exter- nalh/ into two unequal segments, the rectangle cm- tained by the unequal segments is equal to the difference between the square on the line between the points of section and the square on half the line. K M A L H -1 # -iD ..J G i^: i. B F V, Let AB be divided into two equal segments AG, CB, and also externally mto two unequal segments AD, BD: a is required to prove AD • DB =-- CD^ - CB^ On CB describe the square CEFB, and join B£. I. 46 yi ' It'* 't f 1 > 'II 11 !! ! N f^iil 126 BUOLID 8 BLBMBNTS. K L M H [Bookn E F O Through D draw HDO \\ CE^ meeting EB and EF produced at fTand G; through H 6.mv HMLK \\ AB, meeting FB and EC pro- duced at M and L ; and through A draw AK \\ CL. ThoL AD'DB = AD'DHy = AL+ CH, = CM+ HF, = gnomon CMG. But 07)2 - CB^ = Lfl^* - Ci^^ = LEGH - CEFBy = gnomon CMG. .\AD'DB= CD^ - GB^. 7.31 //. 7, Cor. 2 /. ila;. 8 /. 36, 43 /. Ax. 8 /.34 /. ila;. 8 Cor. — The diflference of the squares on two straight lines is equal to the rectangle contained by the sum and the difference of the two straight lines. Let AC and CD be th« two straight lines : it is required to prove CB« - AC^ = {CD + AC) . {CD - AG). 0D + AC = AD, and CD - AC = CD - CB = DB ; .-. {CD + AC) ' {CD - AC) = ADDB, = GD» - CB', = CD» - AC'-. 11^ Book a] PROPOSITION 6. ivr OTHIRWISX:* R- — - c B -♦- Let ^^ be divided into two equal segments AC, CB,uid also externally into two unequal segments AD, DB: it is required to 2Jrove AD • DB = CD' - CB^. Produce BA to E, making AE = BD r o Then EC = CD, and E'J^AD. Now, because ED is divided into two ^ual segments EC, CD, and also mternally mto two unequal segments EB BD .-. EB.BD = CD^ - CB^; ' ' „ - .♦. AD . BD = CD^ - CB\ ALGEBRAICAL ILLUSTRATIOir. Let ^C = CJ? = a, CZ) = 6; then AD = h-\-a, and DB='b- a. Now AD.DB=(b + a){b-a) = b'^ a* and CD^ - CB^ ^ 1,'i _ ^i . ** ' .\AD'DB= CD'^ - cm. 1. Does the rectangle ^2) . DB exceed the rectengle AG^OBf Mamme the various cases. ^ ^«? 2. The rectangle contained by the two external segments of a straight line grows greater according as the point of section » ^i^moved farther from the middle point of the st'^M 4. Name two figur^ la the diagram each of which = the rectanrfe ^utamed.by half the sum and half the diffex^xce of T^S ^ ""zVotiriiiVB ""''''"^ ^'^^' " *^^ «>"- - ^ t^« ' ""^eref rji\*^^ ^T"- -^^^ ^ *^« «^- o» ^ the Due to Mauriffliia R •no/itna f^C /-t ri » in Paris (nTnhMJ^~^'Z"^'' T :-'^^"""«/. a professor of Mathematics xn i-aris (probably about the end of the sixteenth century). 'i» . \m 111 it 128 buolid's elements. r^ook n. 7. Hence, ahow that the proposition may be enunciated : The rectangle contained by any two straight lines is equal to the square on half their sum diminished by the square on half their difference. 8. The perimeter of the rectangle AD • DB = the perimeter of the square on CD. PROPOSITION 7. Theorem. 1/ a straight line be divided externally into any two seg- ments^ the square on the straight line is equal to the squares on the two segments diminished by twice the rectangle contained by the segments. K H r- B z. o 40 L._. E .4 P Let AB be divided externally into any two segments AC, CB: it is required to prove AB"^ = AC^ ■[■ CB^ - 2AC' CB. On AB describe the square A DEB, and join BD. I. 46 Through C draw CF \\ AD, meeting DB produced at G ; and through G draw HK \\ AB, meeting DA and UB pro- duced at H and K. . /. 31 Because CG || AD, .-. l CGB = l ADB ; and because AD = AB, .'. l ADB = /. ABD; .-. L CGB = L ABD, = L CBG; CB = CG. I. 29 /. 5 /. 15 /. 6 Hence the H*" CK, having two adjacent sides equal, has all its sides equal. i. 34 ^ook ilJ PftOPOSlTION 7. 129 But the r CK has one of its angles, KBO, right since L KBC = l ABE; *^ ' ' » » ^^^ .• . it has all its angles right ; / 34 .-. the r C7^ is a square, and = CM / i,J 30 Similarly, the p BF is a square, and = HG^ ^ AG' Again, the |r AG « AC- CG = AC- CB • O^ = ^C CB; ' J Ao AG + GE = 2 AC. CB. Now AB^ = ADEB, -HF+CK-AG-GE, L Ax. d> Cor. l.—The square on the difference of two straight ines IS equal to the sum of the squares on the two straight mes diminished by twice the rectangle contained by the two straight lines. For if ^C and CB be the two straight lines, then their difference = AC - CB = AB Now since AB' = AC' + Cm - 2 AC- CB //. 7 .-. {AC - CBf = AC' + CB^-2 AC- CB. Cor. 2.— The |r about a square's diagonal produced are themselves squares. OTEUCBWISE: ^52= AB-AC - AB.BG = UG-AG - BG.AG) - {AG.BG - BC-BO) ^AG^ + BG*-2AG-BG. ^' ALGEBRAICAL ILLUSTBATION Let ^(7= a, GB = b; then AB = a - b. NowAB>={a- hy = a^-2ab + bK ^j88 = 4C7» + Gfi2_2^C //. 3 //. 2, 3 GB. ^f 130 tetoLtD^S BLtilMiKTd. [Book li. 1. N»m« the two JSgures which form the sum of the sqiutres on AG and GB. 2. Name the figure which is the square on the difference of AC and CB. 3. Name the figure which is the difference of the squares ou AB &nd AG. 4. Name the figure which is the square on the difference of AB and AG. 5. By how much is the square on the difference of ^C and GB exceeded by the sum of the squares on ^C and GB ? 6. Show that the proposition may be enunciated : The square on the difference of two straight lines is less than the sum of the squares on the two straight lines by twice the rectangle contained by the two straight lines. 7. The sum of the squares on two straight lines is never less than twice the rectangle contained by the two straight lines. 8. If a straight line be divided internally into two segments, and if twice the rectangle contained by the segments be equal to the sum of the squares on the segments, the straight line ia bisected. PROPOSITION 8. Theorem. 2%e aqmre on the sum of two straight lines diminished by the square on their difference, is equal to four times the rectangle contained by the two straight lines. £ Hf- N I I • ) I I I • E I K D Let AB and BC be two straight lines : it is required to prove {AB + BG)^ - (AB - BCY % ^X> ' JjLf. feook n] PROPOSITION a 131 Place AB and BC in the same straight line, and on AC describe the square ACDE r ar From CD DE EA cut off CF, DG, EH each = AB; I I through i? and G^ draw BL, ON \\ AE, and through F md H draw FM, HK \\ AC. /. 31 Then all the |[- in the figure are rectangles. / 34 Cor Now because CD, DE, EA are each = AC / Def^. ^ CF, DG, EH are each = AB; CoZf DF, EG, AH are each = BC • ^AB^Jr '''*'^^^'' ^^' ^^' ^^' ^^ ^'^ «»^^ Because AC = AB + BC, ••• ^CDE = ^(72 = (^^ + BC)^ Because ^Z, i^Jf, GN, HK are each = ^5 / 34 and ^^, /x, (?ilf, HN are each = ^(7/ f 34 ^Z,, ZJ/, il/iv; NK are each = ^i^'_ BC • .-. the rectangle ^ZJfi^is a square, and = (AB - BCV^ Hence ^AB + BC)^ - (^^ . le/= ACDE- k!mn, = ^^K" + cz + DM + je;a; = iAB.BC. OTHERWISB : (AB + BO)* = AiP + BO^ + 2AB.B0 ^^^-B0)^-AB^^BC^-2AB BC. *v . Jo* *^^ ^^''''''^ equaUty from the first • then (^£ + £(7)^ - (AB - BC)^ = 4 /i^^^a //. 4, Obr. 1 //. 7, Cor. 1 ALGEBRAICAL ILLUSTRATION. Let ^5 = a, 5C = d . then ^5 + 5C = a + J, and ^5 - 5C7=a - & I <'^^ + ^C')'^-U5-^)8=4^£ BC. 1^2 atrOLID^S BLEM8NTS. [Booktt 1. Name the figure which is the square on the sum of ^5 and BO, 2. Name the figure which is the square on the difierence of AB and BC. 3. Name the figures by which the square on the sum oi AB and BC exceeds the square on the r'jffbvence oi AB and BG. 4. By how much does the square on u e m m of ^^ and BC exceed the sum of the squares on AB ana JC? 5. By how much does the sum of the squares on AB and BC exceed the square on the difference of ^^ and BOf PROPOSITION 9. Thbobem. If a straight line he divided into two equal, and also inter- nally into two v(nequal segments, the mm of the squares on the two unequal segments is double the sum of the squares on half the line and on the line between the , points of section. E ! \ A D B Let AB be divided into two equal segments AG, 05, and also internally into two unequal segments AD, DB : it is required to prove AD^ + DB^ =- 2AG^ + 2 01^. From C draw C£J ± AB, and = AC or OB, I. 11, 3 and join AE, EB. Through D draw DF \\ CE, meeting EB at F; through F draw FG \\ AB, meeting EC at G; said join AF. (1) To prove l AEB right Because l ACE is right, /. 31 /. 31 L CAE + L CEA = a right angh /.32 1*1, 133 1.6 ■•* n.j PROPOBITION d. But L CAE ^ L CEA; .-. each of them is half a right angle. Similarly, l CBE and l CEB are each half a right angle • .-. L AEB IB vighi. ^ *^ ' (2) To prove EG = GF. L EGF is right, because it => z. EOB; and L GEF was proved to be half a right angle; .". L GFE is half a right angle ; .-. L GEF = L GFE; EG = GF. (3) To prove DF = DB. L FDB is right, because it = z. ECB; and L. DBFia half a right angle, being the same qm .'. ^ Z>iF!fi is half a right angle ; .-. L DBF = L DFB; DF = DB. AF^, * , AE^ + FFi^ «= i4C2 + CE^ + EG^ + GF^, 2AC^ + 2GF\ 2AC^ + 2aD« 1.29 1.32 I. 6 /. 29 CBE; /. 32 /. 6 (3) /. 47 /. 47, (1) /. 47 Const., (2) /. 34 OTHERWISS : Consider A C and CD as two straight lines • then AD=.AG+ CD, atd DB=CB - GD = AC- CD. ^^f^AD»=.(AC+GDY = AG^ + GD^+2AC.CD IT 4 Hnr i Add the second equaUty to the first ; then AD» + DB» = 2AG^ + 2 Gd( ALOKBRAICAL ILLUSTRATION. LetAG=GB=a,CD=b; then AD = a + h, and BB = a =^ Now AD» + D£f»=(a + b)' + (a - 6)2 = 2a2 + 2M 184 mrOLID'S ELEMENTS. [Book XL and 2AC^ + 2 CD^ = 2a» + 26» ; .-. AD^ + Dm = 2AG^ + 2GD». 1. Show that the proposition may be enunciated : The square on the 8JUU together with the square on the diflference of two straight lines = twice the sum of the squares on the two straight lines. Or, The sum of the squares on two straight lines = twice the square on half their sum together with twice the square on half their diflference. 2. By how much does AD'^ + DB^ exceed ^C* + C5*? 3. The sum of the squares on two internal segments of a straight line is the least possible when the straight line is bisected. 4. The sum of the squares on two internal segments of a straight line becomes greater and greater the nearer the point of section approaches either end of the line. (Euclid, x. Lemiua before Prop. 43.) 6. Prove that AD'^ + D& = 4 CD"^ +2 AD- DB. 6. In the hypotenuse of an isosceles right-angled triangle any jwint is taken and joined to the opposite vertex ; prove that twice the square on this straight line is equal to the aum of the squares on the segments of the hypotenuse. PROPOSITION 10. Theorem. If a straight line he divided into two equal, and also exter- nally into two unequal segments, the sum of the squares on the two unequal segments is double the sum of the squares on half the line and an the line between the points of section. E A^ Jo .B -iD G' :-v-=:,'':Jf Let AB be divided into two equal segments AC, CB, and also externally into two unequal aecTnents J ri DR . /. 31 7.31 /32 /. 5 •®«* D-] PROPOSITION 10. 135 it is required to prove AD^ + DE^ = iAC^ + 2 CD\ From Cdraw CJ:± ^5, and = 4CorCfi / 11 ^ and join ^J7, ^5. ' ., ' Through D draw DF\\ CE, meeting ^5 produced at F; through i^draw ^G^ || AB, meeting reproduced at O; and join ili^. (1) To prove l AEB. right. Because l AGEiq right, e*. ^ CAE + z. CJ;^ = a right angle. But L CAE = L CEA; .'. each of them is half a right angle. SimUarly, l CBE and l CEB are each half a right angle • .-. A AEB is right. K gie , (2) To prove EG = (?^. z. J7(?r is right, because it = ^ ECB; and L GEFw&8 proved to be half a right angle ; .*. L GFE is half a right angle ; , .-. L GEF = L GFE; EG = GF. (3) To prove DF = DB. L FDB is right, because it = .'. ECB; and L DBF is half a right angle, being = l CBE; .*. L DFB is half a right angle ; .V L DBF ^ L DFB; DF = DB. yow AD^ + DB^^ AD^ .■ DF\ AF\ « AE^ + EF\ = AC^ + GE^ + EG^ + GF\ 2AC^ + 2GF^, I. 29 /. 32 7.6 /. 29 /. 16 7.32 2An2 4. 9/7n2 7 6 (3) 7 47 /. 47, (1) I. 47 Const, (2) I. 34 130 Euclid's elements. [BMkn OTHERWISE: Consider A O and CD as two straight liftes ; then AD = CD + AG, and DB = GD-CB=GD- AG. ^^riQ6AD'^=(GD + AGY^GD» + AG^ + 2GD-AG; IIA.Cor.l ?A ^^^=^^^-^(^^'=CID' + AG^-2CD-AC. IL1,Cor.l Add the second equality to^the first ; then AD» + DB» = IGI^ + '2,AC\ or:* E- O -4- -4- Let ^£ be divided into two equal segments AG^ GBy and also externally into two unequal segments AD, DB: it is required to prove AD^ + DB^ =2AC^ + 2 GD^, Produce BA to E, making AE =» BD, / 3 Then EG = GD, and EB = AD. Now because ED is divided into two equal segments EG, CD, and also internally into two unequal segments EB, BD ; .; EB» + BDi = 2EG^ +2GB'; //. 9 .'. AD» + Bm = 2 GD» + 2AGK y ALGEBRAICAL ILLUSTRATIOIT. Let AC =GB = a, GD = b; then AD = b + a, and DB =h - a. Now AD* + DIP = (6 + a)a + (6 _ «)» = 262 + aA and 2 ^(78 + 2 CZ)2 = 2 a2 + 262 ; .-. AD^ + i)52 = 2 ^(72 + 2 cm • Ctow Commwtorta in Euclidia Elmenta Geometrica (1612). p. 98^ Book H] PROPOSITIONS 10, 11. ISf 1. Show that the proposition may be enunciated : The square on the sum together with the square on the diflference of two straight lines = twice the sum of the squares on the two Btraight hues. Or, The sum of the squares on two straight lines = twice the square on half their sum together with twice the square on half their difference. 2. By how much does AD^ + D& exceed AC^ + C&l 3. The sum of the squares on two external segments of a straight line becomes less and less the nearer the point of section approaches either end of the line. 4. Prove that AD'^ + DIP = 4 GD» -2 AD- DB. 6. In the hypotenuse produced of an isosceles right-angled triangle, any point is taken and joined to the opposite vertex ; prove that twice tlie square on this straight line is equal to the aam of the squares on the segments of txie hypotenuse. PROPOSITION 11. Problem. To divide a given straight line internally and externally* in medial section. ff G' F G A H r^ E _^i^ ^ 1 [« B Let AB be the given straight line : * u required to divide it in medial section. * The second part of this proposition is not given by Eufll',r. that is, CF' .F'G'= Am; that is, CG' = ^lA To each of these equals add Al! ; F'H' = H'D; that is, AH"^ = DB.BH\ = AB . BH'. Cor. 1.— If a straight line be divided internally in medial section, and from tlie greater segment a part be cut off equal to the less segment, the greater segment wUl be divided in medial section. For in the proof of the proposition it has been shown that CF • FA = ^ija, that is = AG-i; .'. GFia divided internally in medial section at A. Now, from AB, which = AG, the greater segment of CF, a part AH has been cut off = ^i^, the less segment of CF; and ^i? has been shown to be divided in medial section at H. Let yl^ be divided internally in medial section at G, so that AG is the greater segment. Q F ' i I E D ■+- •B ^m ^C- cat off AD = BC; then AC is divided in medial section at JJ, and AB la the greater segment. /r"':^?^''.**'? ^^= ^^'' *h«" -4^ i« divided in medial .ectio. at £,, and ^^ ig the greater segment. From^^out off ^Z*::: DE: thfin A S! i« a^a^^a : j-.i _ ,. at /<, and AF is the greater segment. I 149 EUCLIDS ELEMENTS. [Book n. From A Font off AG = EF ; then AFis divided in medial section Mi O, and ^4 O is the greater segment. This process may evidently be continued as long as ve please, and it will be seen on comparison that it is equivalent to the arithmetical method of linding the greatest common measure. That method, if applied to two integers, always, however, comes to an end ; unity, in default of any other number, being always a common measure of any two integers. In like manner any two fractions, whether vulgar or decimal, have always some common ujeasure, for instance, unity divided by their least common denominator. From these considera- tions, therefore, it will appear that the segments of a straight line divided in medial section cannot both be expressed exactly either in integers or fractions ; in other words, these segments are incom- mensurable. lU. Cor. 2.— If a straight line be divided internally in medial section , ar 1 to the given straight line a part be added equal to the greater segment, the whole straight line will be divided in medial section. For this process is just the reversal of that described in Cor. I, as will be evident from the following. (See fig. to Cor. 1.) Let AF he divided in medial section at G, so that AG ia the greater segment. To AF add FE, which = AG ; then AE is divided in medial section at F, and AF is the greater segment. To AE add ED, which = AFj then AD is divided in medial section at E, and AE is the greater segment. To AD add DC, which = AE ; then ^C is divided in medial section at D, and ^D is the greater segment. To .4C add CB, which = AD; then A'B is divided in medial section at C, and .4(7 is the greater segment. ALOEBRAICAL APPLICATION. Let AB^ a; to find the length of ^.fl^ or Aff. Denote AH by x ; then BH = a - x. Now, since AB . BH = Am .'. a (a - x) = x\ a quadratic equation, which being solved gives a(\/5-l) -a(\/W+\) ic = or Book n.] PROPOSITION 11. Ul The first value of », which is leas than o, since ^- ' ia i^ th»» unity, correeponds to AH; aud the second ralue^of x, which is numerically greater than a, since ^» is g^ter tiian unity. ZZrZ^' ^^^' J^' "«°i««ance.of the - in the second value Zf i^ "'"^T'S!^ ^''' ' ^* ^"^ ^^ ^"^'^g^ *° «*y *!>** it indicates ^ * if"*. ^^^ measured in opposite directions from A. .tririf 1 T^/^/'°''''°''*^°" *° *^« ^'^^"^ °^ *he segments of a straigh Ime divided mtemaUy in medial section, is given in Leslie's Elements^ Gentry (4th edition, p. 312). and attributed to Girard, a Flemish mathematician (17th cent.). Take the series 1, 1, 2, 3, 6, 8, 13. 21. 34, 55, 89, 144, &c.. where each term u, got by taking the sum of the pr;ceding two. iTany term be considered as denoting tiie length of the straight line, the two preceding terms wUl approximately denote the lengths of its «o^l\ "^ V^ .'" ^'''''^^'^ internally in medial section Thus if 89 be the length of the line, its segments will be nearly 34 and Si because 89 x 34 = 3026, and 65^ = 3025. If 144 be the ten ^ of 2. 3. 4. 5. A TT '^ J^^^ construction that a side of the square Tf lT"^i? "'^ ^^^ '"^ ''"'^^^'^'^^ ^i*^ ^^- Prove this. It nS i T '^^'' P''^''® *^** ^^^ i« greater than Bff. itLH be produced, it will cut BF at right anglea The pomt of intersection of BE and CH in the projection of A on O/i. It is assumed in the proof of the second part that a side of the Tb'vZ^^J''''^^'^''^ ^ ^ *^''*'"' '*'*^«^* ""«^*h s il . produced, it will cut BF' at right angles. ^^'a/ "'*'"''*^"° ^^ ^^ ^^'l ^^ « the projection of U2 buoud's elements. [Book n. >a 111 PROPOSITION 12. Theorem. In obtuse-angled triangles, the square on the side opposite the obtuse angle is equal to the sum of the squares on the other two sides increased by twice the rectangle contained by either of those sides and the projection on it of the other side. Lot ABC be an obtuse-angled triangle, having the obtuse angle ACB; and let CD be the projection of CA on EG: it is required to prove AB^ = BG^ + GA^ + 2BC- CD. Because BD is divided internally into any two segments BG, CD, .'. BD^ = BC^ + CD^ + 2 BG' GD. II. i Adding DA^ to both sides, BD^ + DA'^ = BG^ + GD"^ + Z>il2 + 2 BG- GD; .-. AB^ = BG^ + GA^ + 2BG' GD. I. 47 ALGEBRAICAL APPLICATION. Let the sides opposite the ^a A, B, Che denoted by a, b, e, ao that AB = c,BG = a, GA=b; then, since ^52 = 5C2 + C^a + 2 5C.CA - //• 12 d^ = d^ JtV^ + 2a'CD; c^ - a3 - 63 CD = 2a BD= BC + CD-a + 2a 2a Hence, if the three sides of an obtuse-angled triangle are known, we can calculate the lengths of the segments into which either side .about the obtuse angla is divided by a perpendicular from one of the acute angles. Book tL] PBOPOBITIONS 12, 13. 148 1. If from B there be drawn BE ± AO produced, thee BO - CD = AG'GE. 2, A BCD is a r having i ABC equal to an angle of an equilateral triangle ; prove BD^ = BG^ + CD* + BC ■ CD. 3.U AB» = AC* + 3 CD* (figure to propomtion), how will the perpendicular AD divide BCf ^ U I ACB become more and more obtuse, till at length A falls on BC produced, what does the proposition become ? PROPOSITION 13. Thborbm. In every triangle the square on the aide opposite an acute angle is equal to the sum of the squares on the other two sides diminished by twice the rectangle cmitained hy either of those sides and the projection on it of the other side. Let ABO he any triangle, having the acute angle ACB ; and let CD be the projection of CA on BO: it is required to prove AB^ = BO^ + OA^ - 2B0' OD. Because BD is divided externally into any two segments BL/f UDf BD^ = BO^ + OD^ -2BC.CD. // 7 Adding DA^ to both sides, BD^ + DA^ = BG^ + CZ>» + DA^ - ^BC- CD; .*. Am =.BC^+ CA^ -2BC.CD, 7.47 ALOXBRAIOAL APPLIOATIOir. As before, let AB = c, BC = a, CA =b; then, sinoa AB* =" A/^a j. nja o d^ nn. i! I: :i I in i.i III r 144 EUOUDS ELEMENTS, c" = a" + 62 - 2a . CD ; [Book XL CD = 2^^ ' .'. (fig. 1) and (fig. 2) BD=:BG-CD = a- o2 + 62_cS oa_52 + ca 2a 2a 2a t" - c» - g' 2a ; Hence, from the results of this proposition and the preceding, if the three sides of any triangle are known« we can calculate the lengths of the segments into which any side is divided by a perpendicular from the opposite angl& Hence, again, if the three sides of any triangle are known, we cap calculate the length of the perpendicular drawn from any angle of »• triangle to the opposite side. For example (fig. 1), to find the length of AD. "^"( — 2^ — ; _ 4oa6a_(a2^fta _ ca)a 4a2 * _ (2a6 + pa + 68 - c2) (2a6 - o2 - y + c») _ {(gg + 2a6 + 6^) - c«} {c^ - (a" - 2o6 + b^) 4g2 _ {(a + b)i-c^\y-{a-b)'\ "■ 4a2 _ (g + 6 + c) (a + 6 - c) (c + g - 6) (g - g + 6) .*. AD = ^ v'<« + 6 + c) (g + 6 - c) (g - 6 + c) (6 4- c - g). This expression for the length of AD may be put into a ihortei and more convenient form, thus : Denote the semi-perimeter of the a ABG by s; then g + 6 + c = the perimeter = 2«; .-. a + 6-c = a + 6 + c-2c=2«-2c = 2(«-c), a-6 + c = g + 6 + c-26 = 2«-26 = 2(»-6), and 6 + c-g=a + 6 + c-2g = 2«-2g = 2(«-a). iyt'i fiook a] , 1»»0P081T10N 13. Henoe AD m ^ V2«.2(#- c).2(« - i^2(« - •). 14ft -- V«(« - o) (« - 6) (« - c). Similarly, the perpendicular from BonGA= -|\/« («-a)(« -i)(«-"c) **><* " H Con il5 = -Va (« - a) (« - i) (« - c). c Hence, lastly, if the three sides of a triangle are known, we can calculate the area of the triangle. For the area of a ABC = ^ £G ' AD, /. 41, 35 = V« (» - a) (« - b){8 - c); which expression may be put into the form of a rule, thus : From half the sum of the three ?ides, subtract each side separ- ately ; multiply the haJ.f sum and the three remainders together, and the square root of the product will be the area.* 1. If from B there be drawn BE i. AC or AC produced, then BG-GD = AG'CE. 2. ABGD is a ir having z ^5(7 double of an angle of an equilateral triangle ; prove BD^ = BG* + GD^ - BG • GD. a If AB> = AG^ + 3CD^ (fig. 1 to proposition), how will the perpendicular AD divide BG? ^li I AGB become more and more acute till at length A falls on GB or GB produced, what does the proposition become ? 6, If the square on one side of a triangle be greater than the sum of the squares on the other two sides, the angle contained by these two sides is obtuse. (Converse of II. 12.) 6. If the square on one side of a triangle be less than the sum of the squares on the other two sides, the angle contained by these two sides is acute. (Converse of II. 13.) 7. The square on the base of an isosceles triangle is equal to twice the rectangle contained by either of the equal sides and the projection on it of the base. * The discovery of this expression for the area of a triangle is due to Heron of Alexandria. See Hultsoh's Heronit AUxandrini . . . religuio! ue Ktolid's elements. [Book a I I' i' PROPOSITION 14. Problem. To describe a square that shall he equal to a given rectilineal figure. 7.45 Let A be the given rectilineal figure : it is required to describe a square = A. Describe the rectangle BODE = A. Then, if BE = ED, the rectangle is a square, and what was required is done. But if not, produce BE to F, making EF = ED. I. 3 Bisect ^i^in G; /. 10 with centre G and radius GF describe the semicircle BHF ; and produce DE to H. EH^ = A. Join GH. Because BF is divided into two equal segments BG, GFf and also internally into two unequal segments BE, EF; BE'EF= GF^ - G^^, = Gm - GE\ = EW^. BD = Em ; A = Em. 1. Prom any point in the arc of a semicircle, a perpendicular is drawn to the diameter. Prove that the square on this per- pendicular = the rectangle contained by the segments into which it divideb the diameter. 17.5 7. 47, Co^ Book tt] PROPOSITION 14, APPENDIX II. ' , I47 ^ ^lilf * «^^*\«*«^;«^* ^i°« i"t«rnaUy into two Begmente, such that the rectengle contained by them may be equal to the square on another given straight line. What limit- are there to the length of the second straight line ? ^* ^^^\ » «i^«" "trwght line extemaUy into two segments, such that the rectangle contained by them may be equal to the TrrH'^Yl ^'^''^ '*'"«*^* ^^ ^« ther7any limits to the length of the second straight line » J ^ 4. Describe a rectangle equal to a given square, and having oae of Its sides equal to a given straight line. APPENDIX II. Y Proposition 1. y^e «,m 0/ the squares on two sides of a triangle is dovMe (he qfthe squares on half the base and on the median to the base.^ B Deo Let ABG be a triangle, AD the median to the base BO' U M reqmred to prove AB^ + AG^ = 2 BD^ + 2AD\ Draw AE ± BG. Then AB^ = BD» + AD^+2BD. DE and AC^=. GD^ + Alfl - 2 GD . DE. .*. Am + AG^ = 2 BD^ + 2 AD^. CoR._The theorem is true, however near the vertex Am&vh^ta toe base ira When A falls on BG, the theoi^rWomef XL 9 when A falte on BC produced, the theorem becomes JXlO # ^..^ tTT-r fAA I. 12 //. 12 //. 13 148 lli ll# IB' Euclid's blbments. [Book a KoT«.— It may be well to remark that the converse of the theorem, 'If ^5C be a triangle, and from the vertex A a straight line ^D be drawn to the base BC, so that AB> + AG* ^ 2BD^ + 2AD\ then D is the middle point of BG,' is not always true. A App. IL 1 B DO' • B^ D^ C _ For, let ABG, ABG' he two triangles having AG= AG'. Find D, the middle point of BG. D must fall either between B and G', between G and C", or on C". In the first case, join AD. Then AB^ + AG^ =^2BD"' + ^AD"^; A£^ + AG'^ = 2BD^ + 2AB^; and we know that D is not the middle point of BG\ In the second case, find D' the middle point of BG\ and join Aiy ihen Am + AG'^^2 BD'^ + 2AD^i App. II. { AB^ + AG^ =2BU^ + 2AD^i and we know that U is not the middle point of BO, The third case needs no discussion. Proposition 2. Th* difference of the squares on ttvo sides of a triangle is double the rectangle contained by the base and the distance of its middle point from the perpendtcuiar on it from the vertex,* B DEO B D OB Let ^5C be a triangle, Z> the middle point of the base BO, and AE the perpendicular from A on BG.- it is required to prove AB^ - AG^ = 2 BG • DS. ' * Pappus, vn. lao. It look n.J APPENDIX II. 149 /.47 For A» - AG^ - {Bm + AW^ - {BC* + AE^ = JS^a - E0\ « (fiJ^ + EC) (BE - EG), 11. 6, 6, C7orr ^BG.2DE in fig. 1; or ^2 DEBG in fig. 2, ^2BC'DE. PaoposmoN a ^%'7JfAn'''nn ^f^^"^ '"'^^ '^ ««y twop^taG and Jit«^AQ.BD + ADBO=»AB.OD* ' -S — g p ^'AC.BD + BD.BC + AB .'bG, II. 1 «^i).UC + ^C) +AB.BC,II.l '^BD.AB ■{■AB.BO, LOCI. PEOPoamoN 4. i%id a = iJlfa _ Bm Now i if 2 is a constant magnitude, and so is BI>», being the mum. on half the given baae ; -n"~w .'. 4 if" - i?Z>2 must be constant ; .'.AD' must be constant And since AD^ is constant, AD must be equal to a fixed length • that IS, the vertex of any triangle fulfilling the given condition^ i. always at a constant distance from a fixed point D, the middle of tne given base. Hence, the locus required is the 0«* of a circle whose centre is the middle point of the basa ,J''^ *^!Sf ""'f !x*^* ^?^"' completely, it would be necessary to find the^ length of the radius of the circle. This may be left to the ' Pkoposition 6. Find the locus of the vertices of all the triangles which have the same base, and the difference of the squares of their sides equal to a gtven square. M / 1 1 \ \ \ B D E C Let BG be the given base, if a the given square. Suppose ^ to be a point situated on the required Ioctul Join AB, AG; bisect 50 in A and draw AE ± BG or BG produced. /. 10, 12 Then, since .^ is a point on the locus AB^ - AG^ = M^i rr-,^ But Am -AC* = 2BC. DE; App // 2 .\2BGDE = MK ^PP- n. i Now M' is a constant magnitude, and so is 2 BG; .*. DE must be constant ; .-. a perpendicular drawn to BG from the vertex of any triangle fulfiUing the given conditions wiU cut ^ at a fixed point n.] APPENWX a. 151 nn ^^^ ~^^^ ^'' *^® P«'Pendicul*r from .4 on ^C' will « »«„ .1 ^cjy»^2qs^ + BC'\, i^;;"^--"-'* — * --- ^ ^^^ 162 ■UCLID's BLEMINTS. [Book H 11. Thrica the lum of the aquares on the aidee of any pentagon » the lum of the squares on the diagonal! together with four time* the turn of the aquares on the five straight lines joining, in order, the middle points of those diagonals. 12, UA,Bhe fixed points, and O any other jwint, the sum of the squares on OA and 0£ is least when O is the middle point of AB. la Prove 11. 9, 10 by the following construction : On AD describe A rectangle AEFB whose sides AE, J)F are each = AO or CB. According aa D ia in AB, or in AB produced, from DF, or D^ produced, cut off FO < T)B ; and join EC, CO, OE. Show how these figures may be derived from those in the text. 14. If from the vertex of the right angle of a right-angled triangle a ^perpendicular be drawn to the hypotenuse, then (1) the square •n this perpendicular is equal to the rectangle contained by the segments of the hypotenuse ; (2) the square on either side is equal to the rectangle contained by the hypotenuse and the segment of it adjacent to that side. 15. The sum of the squares on two unequal straight lines is greater than twice the rectangle contained by the straight lines. 16. The sum of the squares on three unequal straight lines is greater than the sum of the rectangles contained by every two of the straight lines. 17. The square on the sum of three unequal straight lines ia greater than three times the sum of the rectangles contained by every two of the straight lines. 18. The sum of the squares on the sides of a triangle is less than twice the sum of the rectangles contained by every two of the sides. '■ 19. If one side of a triangle be greater than another, the median drawn to it is less than the median drawn- to the other. 20. If a straight line AB be bisected in C, and divided internally at D 9ji6. E, D being nearer the middle than E, then AD.DB = AEEB + CD.DE+CE' ED. 21. ABC is an isosceles triangle having each of the angles B and G = 2A. BD is drawn j. AC; prove AD^ + DC^ = 2 BD^ 22. Divide a given straight line internally so that the squares on the whole and on one of the segments may be double of the square on the other segment. Book H] ABt^rijx II. 153 23. GiT»n that AB is divided internally at //, and externally at If, in medial aection, prove the following : (1) AH .BH = {AH + BH) • (AH - BH); AH' . BH' = {BH' + AH) . (BH - AH'). (2) AH . {AH - BH) = BH'^; AH • {AH + BH) = BH^. (2,)AB» + BH^ >3AH'i AB' + BH'^ = 3AH*. (4) {AB + BH)' =6AH'i {AB + BH')^ = 6 AH'\ (5) {AH - BH)' = 3 Bm - AH' ; {BH - AH)'= 3 AH' - BH' (6) {AH-\-BH)' = 3AH' -BH'; {AH' + BH')' = 3BH' - AH' (7) {AB + AH)' = BAH'-3BH'; {AH -AB)'=8AH'-3BH'' (8) ABi + AH' = 4 AH' - BH'; AB' + AH'=4AH'' - BHK 24. In any triangle ABC, if BP, CQ be drawn ±CA,BA, produced if necessary, then shall BC = AB • BQ + AC - CP. 25. If from the hypoteniwe of a right-angled triangle segments be cut off equal to the adjacent sides, the square of the middle segment thus formed = twice the rectangle contained by the extreme segment*. Show how this theorem may be used to find nuDibers expressing the sides of a right-angled triangle. (Leslie's Elements of Geometry, 1820, p. 315.) Loci. 1 Given & a ABC; find the locus of tlie points the sum of the squares of whose distances from B and C, the ends of the base, is equal to the sum of the squares of the sides AB, AC. 2. Given a A ABC; find the locus of the i>nints the difference of ths squares of whose distances from B and C, the ends of the base, is equal to the difiFerence of the squares of the sides ^^,^ a 5. C* the A ABC, the base BC is given, and the sum of the sides AB,AC; find the locus of the point where the perpendicular from C to AC meets the bisector of the exterior vertical angle at A. Ot the A ABC, the base BC is given, and the difference of the sides AB, AC; find the locus of the point where the per- pendicular from O to .4 C meets the bisector of the interior vertical angle at A. 6. A variable chord of a given circle subtends a right angle at a fixed point ; find the locus of the middle point of the chord. Examine the cases when the fixed point is inside the circk. ouu 4. ou the cj^* lo4 BOOK III DEFINITIONS. i. A circle is a plane figure contained by one line which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal. This point is called the centre of the circle, and the straight lines drawn from the centre to the circumferen'^ are called radii. Cor. 1. — If a point be situated inside a circle, its distance from the centre is less than a radius ; and if it be situated outside, its distance from the centre is greater than a radius Thus, in fig. 1, Fig. 1. Fig. 3. OP, the distanc« of the point P from the centre O, is less than the radius OA; in fig. 2, OP is greater than the radius OA. Cor. 2. — Conversely, if the distance of a point from the centre of a circle be less than a radius, "the point must be situated inside the circle ; if its distance from the centre be greater than a radius, it must be situated outside the circle. Cor. 3. — If the radii of two circles be equal, the circum- ferences are equal, and so are the circles themselves. This may be rendered evident by applying the one circle to ihfo other, so that their centres shall coincide. Since the radii of the oae airol9 ivre squai to those of the other, every point in the ciroiua* Botik m.] 3 circum- DBPINITIONS. will coincide 155 with a point in the E ference of the one circle circumference of the other; therefore, the two circumferences coincide and are equal. •Jonsequently also the two circles coincide and are equal. Cor. 4.— Conversely, if two circles be equal, their radii are equal, and also their circumferences. This may be proved indirectly, by supposing the radii unequal. Cop. 5.— a circle is given in magn..ude when the length of Its radius is given, and a circle is given in position and magnitude when the position of its centre and the length of its radius are given. (Euclid's Data, Definitions 5 and 6.) Cor. 6.— The two parts into which a diameter divides a circle are equal. This may be proved, like Cor. 3, by superposition. The two parts are therefore called semicircles. Cor. 7.— The two parts into which a straight line not a iiameter divides a circle are unequal. Thus if -45 is not a diameter of the circle ABC, the two parts AGB and ADB into which AB divides the circle are unequal For if a diameter AE hQ drawn', the part ACB\?i less than the semicircle ABE, and the part ADB is greater than the semicircle ADE. 2. Concentric circles are those which have a common centre. b a o 3. A straight line is said to touch a circle, or to be a tangent to it, when it meets the circle, but being produced does not cut it. Thus BC is a tangent to the circle ADE. ■\ 166 Euclid's elements. [Book m. 4. A straight line drawn from a point outside a circle^ and cutting the circumference, is called a secant. Thus EGA and EBD are secants of the circle ABC. If the secant EGA were, like one of the hands of a wa^i^h, to revolve round ^ as a pivot, ^e points A and G would approac' one another, and at D^ length coinci^' . When the points A and G coincided, the secant would have become a tangent. Hence a tangent to a circle may be defined to be a secant in its limiting position, or a secant which meets the circle in two coincident points. This way of regarding a tangent straight line may be applied also to a tangent circle. 5. Circles which meet but do not cut one another, are said to touch one another. Fig. 2. Thus the circles ABG, ADE, which meet but do not intersect, are said to touch each other. In fig. 1, ^the circles are feaid to touch one another internally, although in strictness only one of them touches the other internally; in fig. 2, they are said to touch one another externally. 6. The points at which circles touch each other, or at which straight lines touch circles, are called points of contact. Thus in th« figni:^ to definitioug 3 and 5, the points A are points oi' «oniaot. Book m.] DEFINITIONS. 157 7. A chord of a circle is the straight line joining any two points on the circumference. Thus ABiaa chord of the circle ABG. 8. An arc of a circle is any part of the circumference. Thus ACB is an arc of the circle ABO • so is ADB. ' 9. A chord of a circle which does not pass through the centre divides the circumference into two unequal arcs. Ihese arcs are caUed the major and the minor arcs, and they are said to be conjugate to each other. Thus the chord AB divides the circumference of the circle ABG into the conjugate arcs ADB, ACB, of which ADB is a major arc. and AuB a minor arc. 10. Chords of a circle are said to be equidistant from the centre when the perpendiculars drawn to them from the centre are equal ; and one chord is farther from the centre than another, when the perpendicular on it from the centre IS greater than the perpendicular on the other. Thus in the circle ABC, whose centre is O, if the perpendiculars OO, OH on the chords AB, CD are equal, AB and CD are said to be equidistant from 0; if the perpendicular OL on the chord EF is greater than CG or OH, the chord EF is said to be farther from the centre than AB or CD. 11. A segment of a circle is the figure contained by a chord, and eitlicr of the arcs into wliich the chord divides the circumference. The se-ments are called major or minor segments, according as their arcs are major or minor arcs. Thus (see figure to definition 7) the figure contained by the minor g,rc. ACB a.rn] iht- rh-^-A *n- • ««"icu uy sue «i- ^Lij ann ttit ciioid ^-^^ m a minor segment; the figure i 168 EUCLID S ELEMENTS. If .!| i m i\ [Book m. contained by the major arc ADB and the chord AB is a major segment. It is worthy of observation that a segment, like a circle, is generally named by three letters; but the letters may not be arranged anyhow. The letters at the ends of the chord must be placed either first or last. 12. An angle in a segment of a circle is the angle contained by two straight lines drawn from any point in the arc of the segment to the ends of the chord. Thus AGB and ADB are angles in the segment AGB. 13. Similar segments of circles are those which contain equal angles. Thus 11 the angles ~^^ C and F are equal, the segment ACB is ■aid to be similar to ^ the segment DFE. 14. A sector of a circle is the figure contained by an arc and the two radii drawn to the ends of the arc. Thus if be the centre of the circle ABD, the figure OAGB is a sector ; so is OADB. It is obvious that, when the radii are in the same straight line, the sector becomes a semi- circle. 15. The angle of a sector is the ^gle contained by the two radii. ^ Thus the angle of the sector OAGB is the angle AOB. 16. Two radii of a circle not in the same straight line divide the circle into two sectors, one of which is greater and the other less than a semicircle j the former may be called a major, and the latter a minor sector. Thus OADB is a major sector, and OAGB is a minor sector. 159 Book m.] DEnNITIONS. 17. Sectors have received particular names according to the size of the angle contained by the radii. When the contained angle is a right angle, the sector is called a quadrant ; when the contained angle is equal to one of the angles of an equilateral triangle, the sector is called a sextant. B ^ Thus if ^05 is a right angle, or one-fourth of four right angles, the sector OAB ia a quadrant; if AOG is two-thirds of one right angle (see p. 71, deduction 9), or one-sixth of four right angles, the sector OAG is a sextant. 18. An angle is said to be at the centre, or at the circumference of a circle, when its vertex is at the centre or on the circumference of tlie circle. Thus BI^ is an anyle at the centre, and BAG &n angle at the circumference of the circle ABG. 19. An angle either at the centre or at the circumference of a circle is said to stand on the arc intercepted between the arms of the angle. Thus the angle BEG at the centre and the angle BAG at the circumference both stand on the same arc BDG. In respect to the angle BEG at the centre of the circle ABG, it may readily occur to the reader to inquire whether the minor arc BDG is the only arc intercepted by EB and EG, the arms of the angle. Obviously enough EB and EG intercept also the major arc BAG. What, then, is the angle which stands on the major arc BAG? This inquiry leads us naturally to reconsider our definition of an angle. 20. An angle may be regarded as generated (or described) by a straight line which revolves round one of its end points, the size of the angle depending on the amount of rovolution. I m> « 1 i ;| 1 V 160 buolid's slembnts. f&odkin Thus if the straight line OB occupy at first the position OA, and then revolve round O in a manner opposite to that of the hands of a watch, till it conies into the position OB, it will have generated or described the angle A OB. If OB continue its revolution round O till it occupies the position OD, it will have generated the angle AOD; if OB ■till continue its revolution round O till it occupies successively the jjositions OF, OH, it will have generated the angles AOF, AOH. The angles AOB, AOD, AOF, AOH, being successively generated by the revolution of OB, are therefore arranged in oMer of magnitude. AOD being greater than AOB, AOF greater than AOD, and AOH gresAer th&n AOF. It is plain enough that OB, after reaching the position OH, may continue its revolution till it occupies the position it started from, when It will coincide again with OA. OB will then have described a complete revolution. If the revolution be supposed to continue, the angle generated by OB will grow greater and greater (since its size depends on the amount of revolution), but OB itself will return to the positions it occupied before; and therefore in its second revolution OB wiU not indicate any new direction relatively to OA which It did not indicate in its first Hence there is no need at present to consider angles greater than those generated by a straight line m one complete revolution. 2L In the course of the revolution of OB from the position of OA round to OA again, OB wUl at some time or other occupy the position OE, which is in a straight line with OA ; the angle AOE thus generated is called a straight (or sometimes a flat) angle. When OB occupies the position OC midway between that of OA and OE (that is, when the angles AOO and COE are equal), the angle AOC thus generated is called a right ^f!^ ^^^^^ ^ straight angle is equal to two right angles When OB occupies the position OG which is in a straight line with 00, the ar.gle AOG thus generated is an angle of three right angles ; when OB again coincides ^iih OA, it has took mi DEJINITIONS. 161 generated an angle of four i.ght angles. Hence angle AOb IS less than a right angle; angle AOD is greater than one right angle, and /ess than two; angle vlOT^ is greater than two and less than throe right angles ; angle AOH is ,areater than three and less than four right angles. 22. It has been explained how OB, starting from the position OA, and k volving in a man- ner opposite to that of the hands of a watch, generates Uie angle AOB, less than a right a^ -le when it reaches the position OB. But we may suppose that OB, starting from OA, reaches the position OB by revolving round O in the same maimer as the hands of a watch; it will then have generated another angle AOB greater than three right angles. Thus it appears that two straight lines drawn liom a point contain two angles having common arms and a common vertex. Such angles are said to be conjugate, the greater being called the major conjugate, and the less the minor conjugate angle. When, however] the angle contained by two straight lines is spoken of, the minor conjugate angle is understood to be meant. 23. It will be apparent from the preceding that the sum of two conjugate angles is equal to four right angles; and that when two conjugate angles are unequal, the minor conjugate must be less than two right angles, and the major conjugate greater than two right angles. When two con- jugate angles are equal, each of them must be a straight angle. Major conjugate angles are often called reflex angles, and to prevent obtuse angles from being confounded with reflex angles, obtuse angles may now be defined to be angles greater than one right angle, and less than two right angles. m bucud's elements. [Book tOe Ik t uii;i PROPOSITION 1. Problem. To find t/ie centre of a given circle. c Let ABC be the given circle : it is required to find its centre. Draw any chord AB, and bisect it at D/ / 10 from D draw DC ± AB, j n and let DC, produced if necessary, meet the O** at C and E Bisect CE at F. j^q F is the centre of ABC. For if F be not the centre, let O be the centre • and join GA, GD, GB. ' i AD - BD Const In As ADG, BDG, )dG=DG (GA = GB; III, Def. 1 .-. L ADG = L BDG; ' j g •*• ^ ^DG is right. / jy^j^ \q But z.^Z>(7 is right; " Qmist. .'. L ADG = L ADC, which is impossible j .*. G i^ not the centre, Now G is any point out of CE; .'. the centre is in CE. But, since the centre is in CE, it must be at F, the middle point of CE. PROPOSITION 1. 168 Book m.] Cor. 1.— The straight line which bisects any chord of a circle perpendicularly, passes through the centre of the circle. Cor. 2.— Hence a circle may be described which shall pass through the three vertices of a triangle. For if a circle could be described to pass through ABO the vertices of the triangle ABC, AB and AC woild' be chords of this circle ; .-. DF, which bisects AB perpendicularly, would past through the centre. jjj | q^^ , Similarly EF, which bisects AC perpendicularly,' would pass through the centre. /// 2 Qfj^. j Hence F wiU be the centre, and FA, FB, ot'FC the radius. 1. Show how, by twice applying Cor. 1, to find the centre of a given circle. 2. Similarly, show how to find the centre of a cm^le, an arc only of which is givea 3. Describe a circle to pass through three given points. When is this impossible ? 4. Describe a circle to pass through two given points, and have Its centre in a given straight line. When is this impossible ? 6. Describe a circle to pass through two given points, and have Its radius equal to a given straight line. When is this impossible ? 6. A quadrilateral has its vertices situated on the 0<» of a circle Prove that the straight lines which bisect the sides perpen- dicularly are concurrent. 7. From a point outside a circle two equal straight lines are drawn to the 0<*. Prove that the hispnf^r nf t^h^ a««i^ *i-__ contain passes through the centre of the oircle. i if. i [■iff n ** I 164 buolid's elements. [Book m a Show also that chn auu. thing is true when the point is taken either witliiu th*' .«;,•. 'e or on the 0«». 9. Hence give auother method of finding the centre of a «iven circle. PROPOSITIO^^ 2. TuKOREM. If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle.* Let ABC be a circle, A and B any two points in the Q*^ 5 it is required to prove that AB shall fall within the circle. Find D the centre of the ABC; take any point E in AB, and join DA, DE, DB. Because DA = DB, .-. l A = l B. But L DEB is greater than l A; .'. L DEB is greater than l B; JD^ is greater than Z)^. Now since DE drawn from the centre of the © ABC is less than a radius, E must be within the circle. III. Def 1 , Cor. 1 But E is any poinjb in AB, except the end points A and B; .'. AB itself is within the circle. 1. Prove that a straight line cannot cut the o«« of a circle in more than two points. ♦Eudid'B proof is indirect. The one in the text is' found in CUivii Commentaria in JEunlidia Klj>ntj>ntr, ft R-to\ „ -inn /. 5 /. 16 /. 19 anok m. I pftOPoaiTiONs 2, 3. M 2 Describe a circle whose O- shall pasa through a given ,K,int t^ ir Tfr^!''*" "^ •" "°« «'^«" «^-«^' «- -d Zae raduw shall be equal to another given straight line. May more than one circle be so drawn? If so, how many ? When Will there be only one, and when none at all ? PROPOSITION 3. Theorems. If a etrmght line drawn through the centre of a cirde bisect a chord which does not pass through the centre, it shall cut it at right angles. Cmverselg: If it cut it at right angles, it shall bisect it. (1) Let ABC be a circle, F its centre ; and let CE, which passes through F, bisect the chord AB which does not pass through F.- ^ it is required to prove CE J. AB. Join FA, FB. (AD = BD In As ADF, BDF, \dF=DF { FA = FB; .*. i- ADF = L BDF; .'. CE is X AB. (2) In ^^Clet C^be ± AB.- it la required to prove AD = BD. Byp. III. Def. 1 /. 8 /. Def, 10 166 t!^ i BUOLID'u i£LSMBNTS. |Book OL Join FA, FB. . ^ ( ^ ADF = ^ 5Z>i^ ff^^, In As ADF, BDF, } l FAD = l FED J, 5 ( DF = DF; .'.AD=^BD. ^26 1. In the figure to the proposition, C and ^ are on the o<* Need they be so ? 2. The 0<^ of a circle passes through the vertices of a triangle Prove that the straight lines drawn from the centre of the circle perpendicular to the sides will bisect those sides. 3. Two concentric circles intercept between their O«««two equal portions of a straight line cutting them both. 4. Through a given point within a circle draw a chord which shall be bisected at that point. 5. If two chords in a circle be paraUel, their middle points will lie on the same diameter. 6. Hence give a method of finding the centre of a given circle. 7. If the vertex of an isosceles triangle be taken as centre, and a circle be described cutting the base or the base produced, the segments of the base intercepted between the o** and the J ends of the base will be equal, s/ a If two circles cut each other, any two parallel straight lines drawn through the points of intersection to the o"" will be equal. ,y 9. If two circles cut each other, any two straight lines drawn tnrcngh one of the points of intersection to the o««' and Making equal angles with the line of centres will be equal BOOK ULJ PBOP08ITION8 3, 4. w PPO ^TTION i. Theorem. If two chords c , . c. cut one another and do not both pass through tt, 3 cm •■, they do not bisect me another. IS^ Let ABC be a circle, AC, BD two chorda which cut one another at E, but do not both pass through the centre : It u required to prove that A C, BD do not Insect one another. ,v ^*\u ' °''! °! ^^'"''^ P^'' *^'°"8h the centre, it may bisect the other which does not pass through the centre; but it cannot bo itself bisected by that other. ^^JJ^ "f'*^^^' ""^ ^^^"^ P^^ *^'«"g^ *^e centre, let AE = EC. and BE = ED. Find F the centre of ABC, ttt i and join FE. ^ Because FE passes through the centre, and bisects AC .'. L FEA is right. jjj 3 Because FE passes through the centre, and bisects BD .'. L FEB is right; jjj^ .-. L FEA = L FEB, which is impossible. *'. AC, BD do not bisect one another. ^' " tTemtef ' °^ ^ ''^''^^ ^'^^'^ '^'^ °*^''' ''^*'' °^"«* ^°**» °' 2. No r whose diagonals are unequal can have its vertices on the O** of a circle. ^ ^^'crcir^^* * J-ectangle can have its vertices on the o«« of « ':^ ^ > SI ; I! (I ifil •.! f i M 168 ^iuolid's ahmsssfn. fBook HI PROPOSITION 5. Theorem. r/ two circles cut one another, they cannot have the same centre. Let the ©s ABC, ADE cut one another at A : it is required to prove that they cannot have the same centre. If they can, let F be the common centre. Join FA, and draw any other straight line FCE to meet the two O''*' Then FA = FG, being radii of ABO, HI. Def. 1 and FA - FE, being radii of ADE; III. Def. 1 FC = FE, which is impossible. .•.0s ABC, ADE cannot have the same centre. 1. If two circles do not cut one another, can they have the same centre ? 2. If two circles cut one another, can their common chord be a diameter of either of them ? Can it be a diameter of both ? 3. li- the common chord of two intersecting circles is the diameter of one of them, prove that it is ± the straight lino joining the centres. 1 If two cmjles cut one another, the distance between their centres is less than the sum, and greater than the difference of their radii. ff. Prove the converse of wiie preceding deduction. Book m.j PBOPOSITIONS 5, 6. 169 PROPOSITION 6. Theorem. If two circles touch one another internally ^ they cannot have the same centre. Let the 0s ABC, ADE touch one another internally at A .• it is required to prove that they cannot have the same centre. If they can, let F bo the ct)iunion centre. Join FA, and draw any other straight line FEG to meet the two O"^"- Then FA = FG, being radii of ABG, IIL Def, 1 and FA = FF, being radii of ADE j IH, Def. 1 FG = FE, which is impossible. .*. 0s ABG, ADE cannot have the same centre. 1. If two circles touch one another externally, can they hare the same centre ? 2. Enunciate III. 5, 6, and the preceding deduction in one state- ment. 3. If one circle be inside another, and do not touch it, the distance between their centres is less than the difference of their radii, 4 If one circle be is less than Pa (4) At O make z. POL = l POC and join PL. ' T A ( PO=PO In As POL, POC, j 0L= OC ..PL. PC ^^^^^=^^00; ft tt o" =Va '''" ''"'''' '^^ "^ ^^ ^^^- f-- For if PJ/ were also = PC, then PM = PZ, which is impossible. . ^''?;T.^^ ^'^"^ IP°^^^ ^^^^^^ ^ circle more than two eaual For another proof of this Coiv, see III. 9. 1. Prove PCgre^^er than PD, using I. 20 instead of I. 20 Cor TBG\t '" V.^ *^'^"' P^'^^^^^^ ^* ^« inside the drcle fl.ff ' I f "■ *^' ^'■"^*^^* ^"d *^« i«««fc straight li !e8 3 FiJ^ '"? ' ^ •-^"'^ ''"^ ^* *° *^^« O"^ i« constant. 3. Fxnd arv^ber point whose greatest and least distances from the O ^ ar ^ respectively := those of P from the Qce VfZ auchp..t3a.e there? Where do th^ IS' * "^ '"'"^ 'tw'pr'^'^' "^" ^'^^ "°^ ^^^ ^« ^"fi'^itel" thin triandea. t^^t^A I. ^:.ater than PB, and P(7 greater th J P^'^bJ 172 Euclid's elements. w [Book ni PROPOSITION 8. Theorem. If from any point without a circle straight lines be drawn to the circumference, of those which fall upon the concave part of the circumference the greatest is that which passes through the centre, and of the others that tvhich is nearer to the greatest is greater than the more remote: but of those which fall on the convex part of the circumference the least is that which, when produced, passes through the centre, and of the others that which is nearer to the least is less than the more remote; and from the given point straight Hues which are equal to one another can be drawn to the circum- ference only in pairs, one on each side of the diameter. Let ABC be a circle, and P any point without it ; from P let there be drawn to the Q** PDA, PEB PFC of which PDA passes through the centre O : it is required to prove (1) that PA if. greater than PB ; (2) that PB is.greater than PC; (3) that PD is less than PE ; (4) that PE is less than PF ; (5) that only one straight line can be drawn from P to the Q"' = PF Join OB, OC, OE, OF. (1) Because OB = OA, being radii of the same circle ■ PO + OB --= PO + OA, ov PA. But PO + OB is greater than PB ; /. 20 PA is greater than PB. • • Book in.] PROPOSITION 8. .1 173 (2) In As POB, POO, }0B = 0C m. Oef. 1 .-. PB is greater than PC ^ '^^^'■^S'^^'^^'han ^ POO, (3) Because OP - OE is less than PE T ->.n'n^ OP 'nn' T'"^ '^'" °* "^^ -- »^-le; . . C/ - ODis less than P^/ .'. PZ> is less than PE ^^- (PO = PO (4) In As FOE, POF, lOE = OF jji ^,y j . DC • 1 ,, ' ^ -^^^^ is less than l POF • .'. PE IS less than PF '- j^ ^r , ' I 24 (5) At make i POO = ^ P0#, r „ and join PO. ^- ^* T A ( PO = PO In As POG, POF, 00 = 0^ „^. ^,^ j ... PC. = PE. ^^ '''"'-- POE; ■pfZ^^'^S^;;. °*" ^"''*' ""^ can be drawn fL' For if PH were also = PF, then PZf = P(^, which is impossible. 1. Prove PE greater than P^ using I. 20 instead of I 20 Cor 3 WW "T " '"^ *'"^ ^^' "^^°^ ^- 21 instead n: 24 ^I?rtf %T* "^ '^ */^"^' P^^^^^*^^ ^* "^ -^-*^« the circle wf;w rr' "^ ^^^ S^^"*^«* ^^^ the least straight hnes that can be drawn from it to the Qce is constant 4. Compare the enunciations of the last deduction and of the analo- gous one from II. 7, and state and prove the corresponding theoren. viien the point P is on the o- of the © ABC 6 IfT ^'-i"" r ^"^*" *^"" ^^' ^"^ ^^greate?than^i. ^6. If the strajgl. Ime PFC he sxipposed to resolve round P as a "^ pivot txl the points F and C coincide, what 4ould fca! straight line PPCr become? ^<*i' wouia the ^ 7. The tangent to a circle from any external point is less than anv secant U> the circle from that point, ind great" Zl the external segment of the secant. 174 Euclid's elements. Rl rl< [Book m. 8. Could a line be drawn to separate the concave from the convex part of the o <» of the Q ABC viewed from the point Pf How ? PROPOSITION 9. Theorem. If from a point within a circle more than two equal straight line» can he drawn to the circumference, that point is the centre.* Let ABC be a circle, and let three equal straight lines DA, DB, DC be drawn from the point D to the 0°* : it is required to prove that D is the centre of the circle. Join AB, BC, and bisect them at JE, F; /, lo and join DE, DF. CAE = BE Const. In As AFD, BED, I ED = ED (dA = DB; ' Hyp. .'. L AED = L BED; j g .-. DE is X AB; .-. DE, since it bisects AB perpendiqularly, must pass through the centre of the circle. J//, i^ Cor. 1 Henc« also DF must pass through the centre ; .-. D, the only point common to DE and DF, is the centre. Prove the proposition by using the eighth deduction from III. 1. * In the MSS. of Euclid, two proofs of this proposition occur, only the second of which Simson inserted in his edition. The one given in tho t«xt is the first. Book in.] PROPOSITIONS 9, 10. 17ft Hyp. I. 8 PROPOSITION 10. Theorem. One circle cann>ot cut another at more than twopoinU^* A If it be possible, let the ABC cut the EBC at more than two points — namely, at B, C, D. Join BC, CD, and bisect them at i^'and (?; /. 10 through i?' and G draw FO, 00 J. BC, CD, /. 11 and let FO, 00 intersect at 0. Because BC is a chord in both circles, and FO bisects it perpendicularly, .*. the centres of both circles lie in FO. III. 1, Cor. 1 Hence also the centres of both circles lie in GO ; .*. is the centre of both circles, which is impossible, since they cut one another. ///. 5 .•. one circle cannot cut another at more than two points. 1. Two circles cannot meet each other in more than two points. 2. If two circles have three points in common, how must they be situated ? a Show, by supposing the radius of one of the circles to increase indefinitely in length, that the first deduction from III. 2 is a particular case of this proposition. * In the MSS. of Euclid, two proofs of this proposition occur, only the second of which Simson inserted in his edition. The one given in th« lext is the first. 176 Euclid's elements. [Book in. I !'K m. 9BS PROPOSITION 11. Theorem. If two circles touch one another infernally at any point, th€ straight line which joins their centres, being produced, shall pass through that point. A Let the two Os kBC, ADE, whose centres are F and G, touch one another internally at the point A .- it is required to prove that FG iwoduced passes through A. If not, let it pass otherwise, as FGHL. Join FA, GA. Because FA = FL, being radii of ABC, III. Def. 1 and GA = GH, being radii of ADE; III. Def. 1 .-. FA- GA = FL- GH, = FG + HL; .'. FA - GA\Q greater than FG by HL. But FA - GA is less than FG ; /. 20, Cor. .'. FA - GAis both greater and less than FG, which is impossible ; .'. FG produced must pass through A. 1. If two circles touch internally, the distance between their centres 18 equal to the difference of their radii. 2. Two circles touch internally at a point, and through that point a straight line is drawn to cut the o=«« of the two circles. If the points of intersection be joined with the respective centres, the two straight lines will be parallel. 3. This proposition is a particular case of the tenth deduction from I. a Book m.] PROPOSITIONS 11, 12. in PROPOSITION 12. Theorem. Jf two circles toach one another extenmllij at any point, the straight line which joim their centres shall pass through that point. Let the two 0s ABC, ADE, whose centres are F and O, touch one another externally at the point A : it is required to j., rove that FG passes through A, If not, let it pass otherwise, as FLHO. Join FA, GA. Because FA = FL, being radii of ABC, III. Def. 1 and GA = GH, being radii of ADE; III. Def. 1 .-. FA ^ GA ^ FL + GH, = FG - 'HL; .'. FA + GA is less than FG by HL. Buti^^ + GA is greater than FG; /. 20 .-. FA + 6^^ is both less and greater than FG, which is impossible ; .'. FG must pass through A. 1. If two circles touch externally, the distance between their centres is equal to the sum ol their radii. 2. Two circles touch externally at a point, and through that point a straight line is drawn to cut th.j o<^ of the two circles. If the points of intersection be joined with the respective centres, the two straight lines will be parallel. 3. This proposition is a particular case of the tenth deduction from I. a 'i'i 1 'i 178 Euclid's elements. [Book m. PEOPOSITION 13. Theorem. Two circle, carmot touch each other at more point, than m» whether internallij or exlenuiUy. * A For, if it be posfeible, let the two ©s ABC. BDC touch each other at the points B and G. Join BC, and draw AD bisecting BC perpen- dicularly. J j^^ j^ Because B and Care points in the Q"- of both circles .-. ^C is a chord of both circles. ' * And because AD bisects BO perpendicularly, Const .-. AD passes through the centres of both circles ; e . AD passes also through the points of contact ^^""^^^ ///. 11 12 which is impossible. * ' Hence the two 0s ABO, BDO cannot touch each otlier at more points than one, whether intemaHy or externally. 1. If the distance between the centres of two circles be equal to the sum of their radii, the two circles touch each other externally. 2. If the distance between the centres of two circles be equal to the difference of their radii, the two circles touch each other mternally. Book m.] PR0P08ITI0NS 13, 14. 179 PROPOSITION 14. Theorems. Bq^al chords in a circle are equididani from the centre tmvers.Jy: Chonh in a circle which are equidistant from the centre are equal. FF^J;'^\J^'\^.^ ^^T^ '^°'^' ^^ *^^ O ABC, and -fiij, ioCr their distances from the centre E : it is required to prove EF = EG. Join EA, EC. Because EF drawn through the centre ^ is ± ^^ .-. EF bisects AB, that is, AB is double of AF '/// 3 Hence also CD is double of C6^ Now since AB . CD, ,. AF = CG, and AF^ = CG^ -tJut because EA = ^a, .-. EA^ = ^C^ • Take away AF^ and CG^2 ^^ich are equal • , .-. FE^ = GE^, and i^^j; == GE. EF\r'\yf\T ^\"^''^' ^^ *^^ ® ^^^' ^d let ^i;, EG, their distances from the centre E, be equal • It 18 required to prove AB = CD. Join JSr^, ^C. It may be proved as before that AB = 2AF CD - ^ na and that ^^^2 + ^^2 ^ CG^ + GEK ~ * IMAGE EVALUATION TEST TARGET (MT-3) 1.0 11.25 "* 14.0 u 2.0 I 1.4 1.6 ^ w V) ^? ■> •> .<." ^4 lis '■^' '/ Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER. N.Y. 14580 (716) S73-4503 m ^ (V ookia PROPOSITION 18. Thhobbm. The radius of a circle drawn to the point oj contact of a tangent ia perpendicular to the tangent, A Q E Let ABC be a circle whose centre is F. and DE a tanceut to It at the point C\- **^^ M ia required to prove tlmt the radiua FCia ± DE. If not, from F draw FG ± DE, and meeting tiie O* ** /. 12 Because l FGG is a right angles Const. .'. FG IS less than FG. r la n^ ButFG=FB; j^/l^'"-^ .-. FG is less than FB, ^' which is impossible ; .-. i^Cmust be ± DE. ' I. o Tangents at the ends of a diameter of a 9irole are paraUeL If a series of chords in a circle be tangents to another concentric circle, the chords are all equal. If two circles be concentric, and a chord o'f the greater be a 8 A ri^^T^ **? *^^ ^^^' '* " ^^^^ ** **»« point of contact [4. Through a given point within a circle draw a chord whick shall be equa^ to a given length. May the given point be outside n nJ T^^- What are the Umits to the given length? ' "^tSlt fira^:^^-- ^ ^' '^ -^^^^^ ^« tangent i>i. J m.j PB0P08IT10N8 18, 1ft 187 of a PROPOSITION 19. Theorbm. The straight line drawn from the point of contact of a tangent to a circle perpmdieular to the tangmi pastes through the centre of the circle. J Let DE be a tangent to the O ABC at tke point C, and let CA be X DE: it is required to prove that CA passes through the centre. if not, let F be the centre, and join FC. ihen l FCEiBiighi, jjj jg But z. ^CJ7 is right; ^^^ .'. L FCE = L ACE J which is impossible; /. CA must pass through the centre of the cirde. 1. In the figure, -4 is on the o««. Need it be so ? 2. This proposition is a particular case of III. 1, Cor. 1. ;i. A series of circles touch a given straight line at a siyMi po Where will their centres all lie? 4. Describe a circle to touch two given straight lines at two given points. When is this problem possible ? 5. If two tangents be drawn to a circle from any point, the angle J contjuned by the tangents is double the angle contained by the chord of contact and the diameter drawn through eithar poiat of contact . * ■IM 168 Euclid's ELSlkENTs. PROPOSITION 20. Thborhm. [Book nL An angle at the centre of a circle is double of an angle at ifhe circumference which stands on the same are. F ^F^ B In the ABC let l EEC at the centre and l BAG at the O** stand on the same arc BC: U is required to prove l BEG = twice l BAG. Join AE and produce it to F. Because EA = EG, .-. l. EAG = l EGA; t, o .*. L EAG + L EGA = twice l EAG. But L FEG = L EAG + l EGA; /, 32 L FEG = twice l EAG. Similarly l FEB = twice l EAB. Hence, in figs. 1 and 2, L FEG + L FEB = twice z. EAG + twice l EAB, that is, L BEG = twice l BAG,' and in fig. 5, z. FEG - L FEB = twice l EAG - twice z. EAB, that is, ^ jB^a = twice l BAG. 1. In the figures to the proposition, F is on the O •*. Need it be so ? 2. The angle in a semicircle is a right angle. %. B and C are two fixed points in the O** of the circle ABG. Prove that wherever A be taken on the arc BAG, the magni- tude of the angle BAG is constant m.] PROPOSITIONS 20, 21. m 'd PROPOSITION 21. Thborbmb. Angles in the same segment of a circle are equal. Conversely: If two equal angles stand on the same arc, and the vertex of one of them he an the conjugate arc, the vertex of the other will also be m it.* ^ JO A. .n A (1) Let ABD be a circle, and z.s il and C in the same segment BCD : it is required to prove L A^ l C. Find F the centre of the ABDy JII. i and join BF, DF. Then l BFD = twice l A, ' HI. 20 and L BFD = twice l G; ///. 20 L A^ lG. (2) Let L& A and C, which are equal, stand on the same arc BD, and let the vertex A be on the conjugate arc BAD : it is required to prove that the vertex C will also he on it. If not, let the arc BAD cut BC or BG produced at G • join DO. Then l A = l BGD. jji 21 Bu^ LA = ^ C; Hyp. .'. L BGD = L C, which is impossible. / 16 Hence Cmust be on the circle which passes through^. A, D. * The Becond part of this propogition is not given by Euclid, *^ !Ji(p.|ipiJ .*i.'" ido BUOUD's ILBIIBNT8. [Book m. 1. In the aguw to III i if AB, (3D be joined, a il^5 is eoui. angular to b. DBO. ^ 2. If from a point B outiide a circle, two eecanti EGA EBD bo t'jRSo"'* '*^' ^^ *^ ^''^*^ ^ ^^^ " equiangular to 8. Oiren three point* on the o- of a circle; fbd any number d other pointi on the o- without knowing the centre 4. Two tangents AB, AO are drawn to a circle from an external pomt A; Z) ie any point on the o~ outside the ▲ ABO Show that the nun of ^ B ^ BD, AGDia consUnt. 6. Is the last theorem true when D lies elsewhere on the o* ? 6. Segments cf two circles stand upon a common chord AB Thioujih O, any point in one segment, are drawn the straight lines ACB, BOD meeting the other segment in B, - Prove that the length of the arc i)^ is invariable wherever the point C/be taken. "^ I PROPOSITION 22. Thbobbm& Ths opposite angles of a quadrilateral inscribed in a dreU are supplementary. Conversely: If the opposite angles of a quadrilateral be supplementary, a circle may be circumscnbed about the quadrilateral.* A . . (1) Let the quadrilateral ABCD be inscribed in the ABC: it is required to prove that l A + l C = 2 rt. ^s. Find Fthe centre of the ABD, jjj i and join BF, DF. • The second part of this proposition is not given by BuoUd. and he fKorvt the first part by joining AC, BD. ^ , -uu ao PROPOSITION SJL 111 ///. 20 ///. 20 •L O. III. Def. 23 Book m.] Then L BFD = twice l A, and the reflex l BFD =-. twice l C; .'. the sum of the two coiyuga;^ l a BFD ^ . ^, = twice L A + twice uut the sum of the two conjugate l a BFD - 4 rt. La; z. i4 + z. c = 2 rt L a. nSl'.^^ f' ^.^^ ^' ^^^^^ ^^ Bupplementary, be opposite angles of the quadrilateral ^A(7i? and the vertex ^ be on an arc BAD 4ich ;^ »i8o through BrndD: ~ it is required to prove (hat the vertex C mil be on the cm- jugate arc If not, let the arc conjugate ^o BAD cut BO or BC produced at G; .jj\ 7. ^^ join DQ. ^'^' ^' ^^^- ^ Then l A is supplementary to l BOD. ///; 22 But L A'\a supplementary to l C; j^^ .-. L BOD ^lC, which is impossible. /. fe Hence Cmust be on the circle which passes through B, A, D. Cor.— If one side of a quadrilateral inscribed in a circle be produced the exterior angle is equal to the remote interior angle of the quadrilateral. For each is supplementary to the interior a^acent angle. • /. 13, ///. 22 1. If a r be inscribed in a circle, it must be a rectanrfe 2. If from a po.nt^ outside a circle, two secants EGA, EBD be 3. If fPolygon of an even number of sides (a hexagon, for example) be mscribed m a circle, the sum of its al Wte angles iTK the sum of all its angles. ^ ^ " SLTw ^ '^'''''^f -"^^ *"y ^° P*^ *^« -«"> *>f tl»e ^I'glea ia toe two segments is constant * id2 suolid's blemekw. [Book m. t^ a. DiTide a circie mto two segments, such that the angle in the one segment shall be (a) tw-ce, (6) thri.., (c) five times, {d) .even times the angle m the ether segment. ^ 6. ACB is a right-angled triangle, right-angled at C, and is the point of mtersection of the diagonals of the square described ?/(S^^^ *° **"* triangle; prove that CO bisects 7. What modification must be made on the last theorem when the square is described on AB in'vardly to the triangle » tr ^'Jv"^' "''* °f °"' P^^ ^'^ "^"^•^ «««^«^*« from two circles, the other pair of segments they cut oflF are also similar Gxven three points on the o- of a circle : find any number ef other points on the O"* without knowing the centre. ). ABC IS a triangle ; AX, BY, CZ are the three perpendiculars from the vertices on the opposite sides, intersecting at O. Prove the following sets of four points concyclic (that is. «tuated on thfe o- ^ a cin^le) : A,Z. O, Y; B,X,0,Z; <^»^*0,X;A,B,X,Y;B,C,Y,Z;0,A,Z,X. 9. Z PROPOSITION 23. Theorem. On the same chord and on the same side of it there cannot be two similar segments of dries not coinciding toith me another. It It be possible, on the same chord 'AB, and on the same side of it, let there be two similar segments of Os ACB ABB not coinciding with one another. " ' Dra;7 any straight line ABC cutting the arcs of the segments at B and C; and join BC, BB. Because segment ABB is similar to segment A CB Hvt? .^. L ABB = L ACB, IILBafU which IS iTnnnaaihlp - - ^ Book m.] PROPOSITIONS 23, 24. 198 Hence two similar segments on the same chord and on the same side of it must coincide. 1. Of all the segments of circles on the same side of the same chord, that which is the greatest contains the least angi& Z Prove by this proposition the second part of IIL 2L PROPOSITION 24. Theorem. Similar segments of circles on equal chords are eguak E Y Let AEB, CFD be similar segments on equal chords AB, CD: ^ ' it is required to prove segment AEB = segment CFD. If segment AEB be applied to segment CFD, 80 that A falls on (7, and so that AB falls on CD; then B will coincide with D, because AB = CD. Hyp. Honce the segment AEB being similar to the segment CFD, must coincide with it ; /// 23 .'. segment AEB = segment CFD. 1. Similar segments of circles on equal choids are parts of equal circles. 2. ABC, ABC are two as such that AG = AG\ Prove that the circle which passes through A, B, C is equal to the circle which passes through A, B, C. S. UABGD is a \\% and 5^ makes with AB, l ABE = i BAD, **»d ^meets DC produced in E, the circles described about Ln BCD, BED will be equal. iH Euclid's BLsij^NTS. [Book in. PROPOSITION 25. Problem. An arc of a circle being given, to complete the circle. I. 10 /. 11 Let ABC be the given arc of a circle : It w required to complete the circle. Take any point B in the arc, and join AB, BO. Bisect ^5 and Z?C"at i) and ^ • draw DF and £F respectively ± AB and BG and let them meet at F. ' ^ Because Z>^ bisects the chord AB perpendicularly, . . DF passes through the centre. /// \ n^ , Hence also, EF passes through the centre • ' .". F\% the centre. ' Hence, with /-as centre, and FA, FB, or ^£7 as radius th. eirde may be completed. ^ 1. ftove that DF^M EFmnat meet. BL^yTf?^"^** =°°"'''' «»°rtmction, which i,. Book m.J l^ROPOSlTlONfl 25, 26. 195 PROPOSITION 26. Theorem. In equal circles, or in the name circle, if two angles, whether at the centre or at the circumference, be equal, the arcs on which they stand are equal. A D Let ABC, DBF be equal circles, and lei lb G and H at the centres be equal, as also z. s ^ and i> at the Q- : it is required to prove that arc BKC = arc ELF. Join BC, EF. Because Os ABC, DEF nre equal, .*. their radii are equal. ( BG = EH In As BGC, EHF, ] QC = HF (l G = lH; .'. BC = EF. But because l A = l D, .'. segment BAC is similar to segment EDF ; III. Def. 13 and they are on equal chords BC, EF, .-. segment BAC = segment EDF. Jil 24 Now O ABC = O DEF; ffyp .'. remaining segment BKC = remaining segment ELF- .-. arc BKC = arc ELF. CoR.--In equal circles, or in the same circle, those sectors are equal which have equal axigies. Hyp. Ill Def. 1, Cor. 4 Hyp. L 4 196 BU0LID*8 ELEMENTS. TBookin. 1.UAB and CD be two parallel chords in a cirole AGDB prove arcJf^ arc BD, and arc ^ /> =. arc BG. 2. In eqiuJ circles, or in the same circle, if two angles, whether at the centre or at the O"" be unequal, that which is the greater stands on the greater arc. 3. If two opposite angles of a quadrilateral inscribed in a circle be equal, the diagonal which does not join their vertioea is a diameter of the circle. 4. Any segment of a circle containing a right angle is a semicircle. 5. Any segment of a circle containing an acute angle is greater than a semicircle, and one containing an obtuse angle is less than a semicircle. 6. If two angles at the o«« of a circle are supplementary, the sum of the arcs on which they stand = the whole o*. 7. Prove the proposition by superposition. y/8. li two chords intersect within a circle, the angle they contain is equal to an angle at the centre standing on half the sum of the intercepted arcs. ^ 9. If two chords produced intersect without a circle, the angle they contain is equal to an angle at the centre standing on half the difference of the intercepted arcs. 10. Skow how to divide the o<« of a circle into 3, 4, 6, 8 equal parts. PROPOSITION 27. Theorem. /n equal circles, or in the same circle, if two arcs he equal, the angles, tohether at the centre or at the circu7nference, which stand on them are equal. PROtOSItlON 27. m L EHF, III. 26 Hyp. III. 20 ///. 20 Book rXL] Let ABCy DEFhe equal circles, and let arc BO = arc EF: it is required to prove that l BOC = l EHF, and l A =^ L D. If L BGC be not = l EHF, one of them must be the greater. Let L BOC be the greater, and make l BGK = l EHF. I 23 Because the circles are equal, and z.- BOK .'. arc BK = arc EF. But arc BC = arc EF; .'. arc BK = arc BC, which is impossible. Hence i. BGC must be = z. EHF. Now, since l A = half of z. BGC, and ^ Z> = half of ^ EHF, .-. z. >4 = z. i>. C«R. — In equal circles, or in the same circle, those sectors are equal which have equal arcs. 1. UACasidBDhe two equal arcs in a circle ACDB, prove chord AB II chord CD. 2. In equal circles, or in the same circle, if two arcs be unequal, that angle, whether at the centre or at the 0«», is the greater which stands on the greater arc. 3. The angle in a semicircle is a right angle. 4. The angle in a segment greater than a semicircle is less than a right angle, and the angle in a segment less than a semi- circle is greater tban a right angle. 5. If the sum of two arcs of a circle be equal to the whole 0*», the angles at the O"* which stand on them are supplementary. 6. Prove the proposition by superposition. 7. Two circles touch each other internally, and a chord of the greater circle is a tangent to the less. Prove that the chord is divided at its point of contact into segments which subtend equal angles at the point of contact of the circles. 198 IUOLID's BLBMBNm [Bookm. PROPOSITION 28. Theorem. In equal circles or in the same circle, if two chords he equal the arcs they cut off are equal, the major arc equalt the major arc, and the minor equal to the minor ^ D J^DF, and minor arc Bac = minor arc EHF, Find K and L the centres of the circles. /// i and join BK, KC, EL, LF. Because Os ABC, DEF are equal, .*. their radii are equal. CBK^EL In As BKC, ELF, ]kC == LF i BC = EF/ .'. L K = L L; .'.arc BGO = arc EHF. But O'"' ABO = O"* BEF; Hyp. IIL Def. 1, Cor. 4 Hyp. L 8 ///. 26 ///. Def. I, Cor. 4 . . remammg arc BAC = remaining arc EDF. a Hence devise a method of drawing through a given pomt a straight Ime paraUel to a given straight hne. *° I^* » Book m.] PROPOSITIONS 28, 29. 199 3. If two equal circles cut one another, any rtraight Une drawn through one of the points of intersection will Let the circ^ again in two points which are equidistant from the other point of intersection. PROPOSITION 29. Theorem. In equal circles, or in the same circle, if two ares be equal the chords which cut them of are equal. « H EHF:"^^^' i^^i^be equal circles, and let arc BGC = arc it is required to prove that chord BC = chord EF. Find K and L the centres of the circles ttt i and join BK, KG, EL, LF, Because the circles are equal, n- .-. their radii are equal. ' /// ^^^ j J^^ And because the circles are equal, and arc BGG ^'arc EHF In Ab BKG, ELF,} KG ^ LF [lK^ lL; .-. BG = EF, 7.4 1. K^Cand5Z?betwoequalarcsinaciitj|o^CD»«^^ ». ^ ^ 2) = chord BC. ^i^VB, prove chord % Xt?o?^ the proposition by superposition. 200 buolid's blbmbnts. [Book HL PROPOSITION 30. Problem. To bisect a given arc, D B Let ADB be the given arc : it is required to bisect it. Draw the chord AB, and bisect it at C; from G draw CD ± \AB, and meeting the* arc at D. Bia the point of bisection. Jom AD, BD. In As ACD, BCD, ,\ AD = BD. AC = BC CD = CD L ACD = L BCD; I. 10 /. 11 Const. 1.4 But in the same circle equal chords cut off equal arcs, the major arc being = the major arc, and the minor =' the minor ; and AD and BD are both minor arcs, since DC if produced would be a diameter ; ^ jjj j q^ | .-. arc AD = arc BD. ' 'jjj gg 1. If two circles cut one another, the straight" line joining their centres, being produced, bisects all the four arcs. 2. A diameter of a circle bisects the arcs cut off by all the chords to which it is perpendicular. 3. Bisect the arc ADB without joining AB. 4. Prove £, DAB greater than any other triangle on the suae base 4i9( and having its vertex on the arc ADB. ,Book m.j PROPOSITIONS 30, 31. 301 PROPOSITION 31. Theorem. An angle in a semicircle w a right angle ; an angle in a segment greater than a semicircle is less than a right angle ; and an angle in a segment less than a teinicirde is greater than a right angle. A. Let ABC be a circle, of which E is the centre and BG a diameter; and let any chord AC he drawn dividing the circle into the segment ABC which is greater than a semi- circle, and the segment ADC which is less than a semi- circle : it is required to prove (1) L in semicircle BAG = art. z. ; (2) L in segment ABC less than art. l; (3) L in segment ADC greater than art. l. Join AB ; take any point D in arc ADC, and join AD, CD. (1) Because an angle at the O"" of a circle is half of the angle at the centre which stands on the snme arc ; ///. 20 .-. L BAG = lialf of the straight z. BEG, = half of two rt. l s, ///. Def. 21 = a rt, L. (2) Because l BAG + l B is less than two rt. l s. /. 17 and L BAG = a rt. l ; ID Iaqci 4-l-»rt»* £\ «^ Air Z3S £ vsTK? ytiaii a X V* ^» 4 202 BUOLID'b ELBVEirr& [Bo«k za (3) Because ABCD is a quadrilateral inscribed in the circle, • '- L B + L D = ivio vi. LB, jjj 22 But L Bib less than a rt. z. ; .'. z. Z> is greater than a rt. z. . 1. Ciwies described on the equal sides of an isosceles trianrfe m diameters intersect at the middle point of the base 2. Circles described o« any two sides of a triangle as diameters mter8«5t on the third side or the third side produced. 3. Use the first part of the proposition to solve I. 11, and I. 12. 4. Solve III. 1 by means of a set square. 5. Solve III17, Case 3, by the following construction : Join AB aT fiT„r^ T*"' J*T^^' * "'•'^" ''''^^^ *^« «i^«° circle ' at B and G. B and C are the points of contact of the tangents from A. 6. If one circle pass through the centre of another, the angle in the exterior segment of the latter circle is acute 7. If one circle be described on the radius of another circle, any chord in the latter drawn from the point in which the circles . meet IS bisected by the former. "i« "rcies 7 8. If two circles cut one another, and from one of the points of intersection two diameters be drawn, their extremities and o TT !u 7 T'""^ ""^ mtersection will be in one straight line I !t' r^ P*"* 1 *^' proposition to find a square equal Vto tne difference of two f^ven squares. 10. The middle point of the hypotenuse of a right-angled triangle is equidistant from the three vertices. ""angw is 11. State and prove a converse of the preceding deduction. 12. Two circles touch externally s.t A ; B and are points of ^"?f* .* "^"^"^ ^^''fi*^* ^ **^« *^<» circles. Prove l BAG Tight. m.] PROPOSITIONS 31, 8i. 203 PROPOSITION 32. Theorem. // a straight line be a tangent to a circle, and from tJie pmnt of contact a chord he drawn, the angles which the chcrrd makes with the tangent shall be equal to the angles in the altematc segments of the circle. A B p Let ABC he a circle, EFa tangent to it at the point A and from B let the chord BD be drawn : it is required to prove l DBF = the l in the segment BAD, and L DBE = the l in the segment BCD. From B draw BA J. EF; j ,, take any point C in the ^ro BD, and join BC, CD, DA. Because BA is drawn j. the tangent EF horn the point of contact, ^ .*. BA passes through the centre of the circle ; ///. 19 .'. L ADB, being in a semicircle, = a rt. l •' ITT •?i .-. L BAD ^ L ABD =. 2. Ti. L, ' j;32 = L ABF. From these equals take away the common l ABD • .-. L BAD = L DBF. /r?.in, because ABCD is a quadrilateral in a circle, ^ A-\- L G = 2Ti, Ls. ur >9 But lDBF+l DBE = 2 rt. lb; i^ jj lA-\-lC^l. DBF+ L DBE, N 9U Now I. A .*. L G ■troUD'S ELBMBNTS. |Book IK L DBF; L DBE. 1. The chord which joins the pointe of contMt of paraUel tantrenta to a circle is a diameter. yy 2. li two circles touch each other extemaUy or internally, any straight line passing through the point of contact cuts off pairs of similar segments. ^^Z. U two circles touch each other externally or internally, and two straight lines be drawn through the point of contact, the chords joining their extremities are parallel. i^i. U two tangents be drawn to a circle from any point, the angle •^ contained by the tangents is double the angle contained by the chord of contact, and the diameter drawn through either ' point of contact 5. Enunciate and prove the converse of the proposition. S. A And B axe two points on the o«* of a given circle. With B •8 centre and BA as radius describe a circle cutting the given circle at G and AB produced at D. Make arc DE =: n au^ ?^' *°^ ^"""^ ^^' ^^ ^8 a tangent to the given circle. 7. Show that this proposition is a particular case either of IIL 21 or of IIL 22; Cor. * PROPOSITION 33. Problem. Oh a given straight line to describe a segment of a circle which shall contain an angle equal to a given angle, H H f PROPOSITIONS 32, 33. 306 /. 23 . 11 /. 10 /. 11 Const. JL> \ Book m.] Let ^/^ be the given straight line, l C the given angle : it w re(iuired to describe on AB a segment of a circle which shall contain an angle = l C. At A make l BAD = l G. From A draw AE ± AD; bisect AB at F, and draw FG ± AB, JoinBG. ( AF= BF In As AFG, BFG, | FG = FG { L AFG = L BFG: .-. AG = BG; .-. a circle described with centre G and radius ilG^ will pass through B. Let this circle be described, and let it be AHB. The segment AHB is the required segment. Because AD is ± AE, a diameter of the O AHB, .*. -4Z> is a tangent to the circle. ///. jg Because AB is a chord of the circle drawn from the point of co.itact A, /. the angle in the segment AHB = l BAD, III. 32 = L a 1. Show that the point O could be found equally well by making at 5 an angle = i BAE, instead of bisecting AB perpendicu- larly. Construct a triangle, having given : 2. The base, the vertical angle, and one side. 3. The base, the vertical angle, and the altitude. 4. The base, the vertical angle, and the perpendicular from one end of the base on the opposite side. 5. The base, the vertical angle, and the sum of the sides. & Tbo bi«p, the vertical angle, and the difference of the sides. [Severn other methods of solving this proposition wUl be found in T. S. Davies's edition (12th) of Button's Course of Mathemattct roL L pp. 389. 390.1 ' I 306 Euclid's blkments. [Book II PROPOSITION 34. Problem. From a given circle to cut off a segment which shall contain ' an angle equal to a given angle. E B p I.et ABC be the given circle, and l D the given angle • tt 18 required to cut off from Q ABC a segment which shall contain an angle = lD. Take any point B on the 0% and at B draw the tangent EF. jjj ,y At B make l FBC = l D j 23 The segment ^^Cis the required segment. Because EF is a tangent to the circle, and the chord BC IS drawn from the point of contact B, .*. the angle in the segment BAC = l FBC, III, 32 = L D. ' Through a given pomt either within or without a given circle draw a straight line cutting oflF a segment contaiiiing a given angle. Is the problem always possible ? PROPOSITION 35. Theorems. If two chords of a circle cut one another, the rectangle con- tained hy the segments of the one slrnll he equal to t/ie rectangle contained by ike segments of the other. Sook lU.] PROPOSITIONS 34, 36. aoy Conversely : If two straight lines cut one another so that the rectangle contained by the segmeiits of the one is equal to the rectangle contained by the segments of the other, the four extremities of the two straight lines are coney die* III. 1 /. 12 ///. 3 B o (1) Let AC, ED two chords of the circle ABC cut one another at E : it is required to prcyve AE • EC = BE • ED. ' Find F the centre of the O ABC, and from it draw FG ± AC, and FH ± BD. Join FB, FC, FE. Because FG drawn from the centre ia ± AC, .'. ACia bisected at G. Because AC is divided into two equal segments AG, GC, and also internally into two unequal segments AE, EC, .-. AE.EC= GC' - GE\ H. 5 = (FC^ - FC/^) - (FE^ - FG^), 1. 47, Cor. FC - FE\ Similarly, BE.ED = FB^ - FE^ But FC^ = Fm ; .-. FC^ - FE^ = FB' - FE^ ; AE'EC=BEED. (2) Let the two straight lines AC, BD cut one another at E, so that AEEG = BE • ED it is reauired in ryrnup. fh.p. ■frniv nrtinia A n #» #i ^yv>.....^.,^ n n rk ^. * Th« Moond part of this proposition is not given by Eudid. ,7,'^ { s« When the point is situated" on the o«»of f},« • i -. with respect to the circle is zero "''''^' '^ P<»*«o°y four exteraities a™ croyc^a ' ' '" ^'' "'"'«• """'' ^tfrt!l: :nL" 4^//'..^^ the Pe^„die,„.„ ,„. i^ ^O . o;f > fiO oTI CO oi '°""'°""« »' O. Prove the b«e. auch that ri2tl'*7J^'''r*^";''^'°" -AGAB. ■•"^i'-iAC& Prove AS. AD 6. Through a point i» within a circle a chord ^ pb ., that ^P. PZ( = a ^„- ^ ™* ""hf" '^'^^ ■» drawn such 6. Preve VI. B, and VI C ^'termuie the a^uM* PROPOSITION 36. Thkobem. »fm«< «»,7T ' , "'^'"'V^o cmtamed by the 211 ■OO* ™-] PROPOSITION 36. Let ^^C7 be a circle, and from the point E without it Jet there be drawn a secant EGA and a tangent EB : it is required to prove AE- EC = EB"\ Find F the centre of the O ABC^ jji j and from it draw EG ± AC. /■ *io Join FB, EC, FE. ' Because FB is drawn from the centre of the circle to B the point of contact of the tangent EB * .'. L FBEk Tight ' jjj^Q Because EG, drawn from the centre, is ± AC, .*. -4(7 is bisected at G^. * III 3 Because ^C is divided into tjvo equal segments AG GC and axso externally into two unequal segments AE EC * .-. AE.EC= GE^ - oc^ ' 'jj g = {FE^ - FG^) - (FC^ _ >G^2), /. 47 ^j^^ FE^ - jrc^^ FE^ - ^B% ^^' ' I.i7,Cor. I. Prove the proposition when the secant passes through the centre of the circle. (Euclid gives this particular case ) -. If two circles intersect, their common chord produced bisects their common tangent& 3. If two circles intersect, the tangents drawn to them from any point in their common chord produced are equal 4. ^^C is a triangle, AX, BY, CZ the perpendiculars from its vertices on the opposite sides. Prove AC-AY -AB.AZ BG.BX = BA.BZ,GA.CY=CB.CX. ^ " ^^ ^^' U From a given point as centre describe a circle to cut a given straight Ime m two points, so that the rectangle contLed by their distances from a fixed point in the straight line may be equal to a given square. ^ a Show by revolving the secant EBD (fig. to III. 35, Cor.) round Ml, tix&t this proposition is a particular case of IIL 35, Cor. 212 bcoud's elements. [Book IXL PROPOSITION 37. Thborbm. ' If from a point without a circle two straight lines he draim, one of which cuts the circle, and the other meets it, and if the rectangle contained by the secant and its external segment be equal to the square on the line ■ which meets the circle^ that line shall be a tangent. G Let ABC be a circle, and from the point E without it let there be drawn a secant EC A and a straight line EB to meet tlie circle ; also, let AE • EC = EB^ : it is required to prove that EB is a tangent to the G ABC. Draw EG touching the circle at O, and join the centre F to B, G, and E. Then l FGE = art /.. Now, since EGiaa tangent,,, and ECA a secant, EG^ = AE.EC, ' = EB^; EG = EB. CEB^EG In As EBF, EGF, ] BF = GF [EF = EF; . . L EBF = L EGF, = a rt. L \ . EB is a tangent to the ABQ. III. 17 ///. 18 IIL 36 Hyp. /. 8 TTT X 16 Book m.] PROPOSITION 37. 918 1. Prove tke proposition indirectly by supposing EB to moot the circle again at D. rr a «•• 2. Prove the proposition indirectly by drawing the tangent EG oo the other side of EF, and using I. 7. 3. Describe a circle to pass through two'given points, and touch a given straight line. 4. Dewribe a circle to pass through one given point, and touch two given straight lines. Show that to this and the previous problem there are in general two solutions. 6. Describe a circle to touch two given straight lines and a given circle. Show that to this problem there are in general four solutions. tt. Describe a circle to pass through two given points, and touch » given circle. Show that to this problem there are in ceneral two solutions. 7. ^5 is a straight line, Cand Z>two points on the same side of It ; find the point in AB at which the distance C£> subtends the greatest angia [The third, fourth, fifth, and sixth deductions, along with IV 4 5 ^ cases of the general problem of the Tangencies, a subject on which Apollonius of Perga (about 222 b.o.) composed a treatise, now lost Ihis problem consists in describing a circle to pass through or touch any three of the foUowing nine data: three points, three strai«h. lines^ three circles. It comprises ten cases, which, denoting a pSnt S" ' *onn^^^i'°' ^y ^' ^^ * ^^'•'^^^ ^y ^' ^y ^ symbolised OCC. An excellent historical account of the solutions given to ^/''t w,,^ '^ '"'^^^ °^"' ^^ ^ fo"°d in an article n Tt'^' .^'i'''?^^"' *^^ Tactionibus,' in the Trama^tims of the Jluttonc Society of Lancashire and Cheshire (1872). To the authonties there mentioned should be added Das Problem des ^P^nius, by C. HeUwig (1866) ; Das Problem des Pappus vc^ den Beriihmngen, by W. Berkk i (1867); *The Tangencies of Circles and of Spheres,' by Benjamin Alvord, published in 1865 in ^he wn voJ. of the Smithsonian Contributions, and 'The lnterseotto«^ of Circles and the Intersection of Spheres,' by the same author m the ^menoan Journal of Mathematics, voL v., pp, 25-44.] 914 buclid's elements. L>ook m. APPENDIX IIL Badical Axis, hnS!'!n^~"TI''' ^T^ °^ * P°^* ''*'°"* potencies (both extern^ « both mternal) with respect to two oiwle. are equal, is caUed the raduMl aoeui* of the two cirdeg. PROPOSmON 1. The radical axis of two circles is a straight line perpendicular to «JU hue 0/ centres cfthe two circle*. f^^-u^t to wm Let A and B be the centres of the given circles, whose radii n. a and h, and suppose C to be any point on the required locus. Join CA, GB, and from G draw GD ± AB the line of centres. Since the potency of G with respect to circle A ~ AC^ - a* Def. and since the potency of G with respect to circle B = BG^ - Ui' nlf .\AG^~a^ =BG^-l^' «^.-t^«/i .-. AG^ - BG^ = a2 - 6»/ foMlTt ^ *^^ "^^^^^^ ^ *°'*^ *** ^'"^^' **'®^ ^*^ ^« *°<^ *) «• .-. the squares on the radii (a^ and W) are constant ; .-. the difference of the squares on the radii (a^ - 6») is constant- .*. AG^ - BG^ is constant. ' Hence the locus of G is a straight line x AB. App. IT, 5 * This DMne as weU as that of 'radical centi«/ was introduced by L. Gaultier de Toura. See Journal de VEcoU polytechnique, 16. oahis^ ^ ~/y X**** -*w| i-xva m.] APPENDIX m ?)» Cor. 1. — Tangents drawn to the two circles from any point in their radical axis are equal. CoK.' 2. — The radical axis of two circles bisects tl^eir common tangents. Hence may be derived a method of drawing the radical axis of two circles. Cor. 3. — If the two circles are exterior to each othet- and have n« common point, the radical axis is situated outside both circles. C!oR. 4. — K the two" circles touch each other either externally or internally, their radical axis consists of the common tangent at the point of contact. C!oR. 5. — If the two circles intersect each other, their radical aTi> oooflists of their common chord produced. Cor. 6. — ^If one circle is inside the other and does not touch 1% their radical axis is situated outside both circles. Cor. 7. — The radical axis of two unequal circles is nearer to the centre of the small circle than to the centre of the large one, but nearer to the O** of the large circle than to the 0<* of the small Propositiok 2. I%e radical axes of three circles taken in pairs are ooneurreni,* Let A, B, (7 be three circles, whose radii are O) ft, c; t^ is required to prove that the radical axis of A and B, that of B and C, and that of G and A aU meet at one point. teti. This theorem, in one of its cases, is attributed to Monge (1746-1818), «vnoci$t's "fopfictSs PtoJccUoCs dc9 Figwssy § 71. I AB BUCLlDg ELEMENTS. [Book ni '^Z^f *^f ^,^*1*^ °f ^ »"d C and Di^, the radical axis of C and A, will meet at some point D ; for they are respectively, ± BG and CJ, and BC and (74 .re not in the same straight line. Since D is a point on the radical axis of 5 and (7 • .-. BL»~l^ = GIfi- c3. ' Since D is a point on the radical axis of Cand 4 • .-. GD^-.Tbe radical centre of three circles which have no common 3. nEDirCTIONS. 1. Find a point inside a triangle at which the three side. .haU mibtend equal angles. Is this always possible ? 2. G'ven two intersecting circles, to draw, through one of the points such tlT T;.* '•*""«^* "°« terminated.by i^e cin^leHnd may = a given length. Of all the straight lines which can be drawn from two given pomts to m^t on the convex o- of a circle, the sum o" those two wiU be the least, which make equ^ anglel wiUi the tiuigent at the point of concourse ^ *■ "^iV'^fotf *'^' ''■ *^', '""^*^^ •^^ '^ ««---^« - centres, any ..wo other semicircles are drawn to-vhing each other extertaUy and a straight line is draw, to touch them both s'rcl^fe** *"" '''-''''' '*•• -'' ^^- *-»^ *^« origi^i & Find a point in the diameter pmiuced of a given circle such tha^a tangent drawn from it to the circle fhall be 1^'^t 6. ^^^^ »*;»°gje having z BAO acute; prove ^CMess tha. If + ^?' by twice the square on the tangent drawn from A to the circle of which £G is a diameter. ^^C IS a triangle, AX, BY, GZ, the perpendiculars from ita vertices on the opposite sides. Prove thatVhese perpenSlarl bisect the angles of a XYZ, and that ^sAYZ, XBZ, XYG ^^C are mutually equianguhir. . '^^^, 8. If the perpendiculars of a triangle be produced to meet the circle cu^umscnbed about the triangle, the segments of Z e perpendicular between the orthocentre and the o- a^ bisected by the sides of the triangle. 9. If O be the orthocentre of a ABG, the circles circumscribed irt Yf f"""^ t\^^^' ^^^' ^^^' ^^^ a^« «q»^. nif'f^n !u"**'^ respectively on 5(7, C^. ^5, the sides of A ^^C, the ooee of the circles circumscribed about the i^ AS AEF, BFD, CDE^m pass through the nZ . 7. iook m.] /PPBNDIX UI. fit 11; If on the three sides of any triangle equilateral triangles be dear^ribed outwardly, the straight lines joining the oircum* scribed centres of these triangles will form an equilateral triangle. Construct a triangle, having given the base, the vertical angle, and 12. The perpendicular from the vertex to the base. 13. The median to the base. 14. The projection of the vertex on the base. 15. The imnt where the bisector of the vertical angle meets the base. IG. The sum or difference of the other sides. 17. Construct a triangle, having given its orthocentrio triangle. 18. Draw all the common tangents to two circles. Examine the various cases. (One pair are called direct^ the other pair transverse, common tangents.) 10. Of tlie chords drawn from any point on the o** of a circle to the vertices of an equilateral triangle inscribed in the circle, the greatest = the sum of the other two. 20. If two chords in a circle intersect each other perpendicularly, the sum of the squares on their four segments = the squart on the diameter. (This is the 11th of the Lemmas ascribed to Archimedes, 287-212 B.a) 21. A quadrilateral is inscribed in a circle, and its sides form chorda of four other circles. Prove that the second points of inter- section of these fom* circles are concyclic. 22. If fpur circles be described, either all inside or all outside of any quadrilateral, each of them touching three of the sides or the sides produced, their centres will be concyclic 123. The opposite sides of a quadrilateral inscribed in a circle are produced to meet. Prove that the bisectors of the two angles thus formed are ± each other. ■ki. If the opposite sides of a quadrilateral inscribed in a circle be produced to meet, the square on the straight line joining the points of concourse = the sum of the squares on the two tangents from these points. (A converse of this is given in Matthew Stewart's Propositionea Oeometricce^ 1763, Book i., Prop. 39.) !^5. If a circle be circumscribed about a triangle, and from the ends of the diameter ± the base, perpendiculars be drawn to the other two sides, these perpendiculars will intercept on the sides segments = half the sum or half the differenceof the sides, 9 220 buclid's elements. [B Q taken on ^P such that AP • AQi& constant 22. The hypotenuse of a right-angled triangle is ^iven ; find the loci of the comers of the squares described outwardly on the sides of the triangle. 23. A variable chord of a given circle passes through a fixed point, and tangents to the circle are drawn at its extremities ; prove that the locus of the intersection of the tangents is a straight line. (This straight line is called the ^laroi the given fixed point, and the given fixed point is called the poky with refer- ence to the given circle. See the reference to Desargues on p. 221.) 24. Examine the case when the fixed point is outside the circle. *i33 BOOK IV. DEFINITIONS. 1. Any closed rectilineal figure may be called a polygon. Thus triangles and quadrilaterals are polygons of three and four sides. Polygons of five sides are called pentagons ; of six sides, hex- agona ; of seven, heptagons ; of eight, octagons ; of nine, nonagons or enneagons ; of ten, decagons ; of eleven, undecagons or hen- decagons ; of twelve, dodecagons ; of fifteen, qumdecagons or pente- decagons ; of twenty, icosagons. Sometimes a polygon having n sides is called an n-gon. 2. A polygon is said to be regular when all its sides are equal, and all its angles equal. It is important to observe that the triangle is unique among polygons. For if a triangle have all its aides equal, it must have all its angles equal (T. 5, Cor.) ; if it have all its angles equal, it must have all its sides eqn il (I. 6, Cor.) Polygons with mote than three sides may have all their sides equal without having their angles eciual ; or they may have all their angles equal without having their sides equal. A rhombus and a rectangle are illustrations of the preceding remark. Hence in order to prove a polygon (other than a triangle) reguhw, it must be proved to be both equilateral and equiangular. 3. When each of the angular points of a polygon lies on the circumference of a circle, the polygon is inscribed in the circle, or the circle is circumscribed about the polygon. 4. When each of the sides of a polygon touches the cir cumference of a circle, the polygon is circumscribed about the circle, or the circle is inscribed in the polygon. 5. The diagonals of a polygon are the straight lines which join those vertices of the polygon which are not consecutive. 234 l!U0LID^4 JiLBMENfS. [look nr. PROPOSITION 1. Problem. In a given circle to pla^ a chord equal to a given straight line which is not greater than the diameter of tlie circle. in, 1 Hyp. /.3 Let D be the given straight line which is not greater than the diameter of the given O ABQ : it is required to place in the © ABC a chord = D. Draw BO any diameter of the ABO. Then if BO = Z>, what was required is done. But if not, BO is greater than D. Make OE = D; with centre O and radius OE, describe the AEF • jom OA. Then OA = OE, being radii of the AEF, ^ ^' ' Oonst. 1. How ma^y chords can be placed in the circle equal to the civei. straight line D? * 2. Place a chord in the © ABO equal to the given straight line D, and so that one of its extremities shaU be at a given point A u ° °'*' ™*°^ ^^^^^ ^^^ ^^ so placed ? 3. About a given chord to circumscribe a circle. How many circles can be so circumscribed? Where wiU their centres all be? What hmits are there to the lengths of the diameters of all !!tch cir'>'««' ' fiook TV,] I>R0P081tI0NS 1, 2. 225 4. About a given chord to circumscribe a circle having a given radius. How many circles can be so circumscribed ? Place a chord in the A£G equal to the given straight line D and so that it shall ' 6. Pass through a given point within the circle. "• " " II without II 7. Be parallel to another given straight line. 8. Be perpendicular „ „ „ PROPOSITION 2. Problem. In a given circle to inscribe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle : it is required to inscribe in ABC a triangle equiangular to Za jDEF, Take any point A on the Q** of ABC, and at A draw the tangent OAH. Make l HAG = lE, and l GAB = l F; join BC. ABC is the required triangle. Because the chord AC \^ drawn from Ay the contact of the tangent QAH, L B = L HAC, = L E. Similarly, l C =^ l GAB, -^ lF; III. 17 /. 23 point of III. 32 Const. III. 32 Const, 226 BUOLID's ELBMBNT8. remaining l BAC = remaining l D; A ABG is equiangular to A DEF. [Book Vf. /. 32, Cor. 1 1. Show that there may be innumerable triangles inscribed in the ABO equiangular to the given a DEF. 2. If the problem wer.% Jr. a given circle to inscribe a trianc^lo equiangular to a fe;'/<;j a DEF, and having one ol its verticva at a given points ju the 0<*, show that six different iiositious of the inscribed triangle would be possible. 3. Given a. q ABC ; inscribe in it an equilateral triangle. 4. Two ^aABG, LMN are inscribed in the © ABG, each of them equiangular to the a DEF; prove As ABC, LMN equal in Ml respecta PROPOSITION III. Problem. About a given circle to cirtumscrlhe a triangle equiangular to a given triangle. ?kJ\ Let ABOhQ the given circle, and DEF i\iQ given triangle : it is required to circumscribe about ABO a triangle equU angular to A DEF. Produce EF both ways to G and H. Find the centre of the O ABC, Jji \ and draw any radius OB. Make l BOA = l DEG, and z. BOO = l DFH ; I. 23 and at A, B, O, draw tangents to the circle intersecting each N. LMN is the required triauglo. ofhoti of T T14' VVAX^'X. tw XJ. iCX Book IV.] 1»R0P08IT10M fl, 3. 227 But and /. 32, (hr. 2 111. 18 /. 13 Const. /. 32, Cor. 1 Be(!au8e 0AM H is a quadrilateral, . • . tlu) sum of its four z. s - 4 rt. ^l s. But L 0AM + L OHM = 2 rt. z. s ; . •• L M m supplementary to l BO 4. L DKF is supplementary to l DEO, BOA = L DEG; .'. L M = L DEF. Similarly, l N = l DFE; .-. remaining l L = remaining l D ; .'. A LMN is equiangular to A LfEF. 1. It is assumed in the proposition that the tangents &t A, B, C will meet and form a triangle. Prove this. 2. Show that there may be innumerable triangles circumscribed about the © ABC equiangular to the given a DEF. 3. Given a © ABC; circumscribe about it an equilateral triangle. 4. If the points of contact of the sides of the circumscribed equi- lateral triangle be joined, an inscribed equilateral triangle will be obtained. & A side of the circumscribed equilateral triangle is double of a side of the inscribed equilateral triangle, and the area of the circumscribed equilateral triangle is four times the area of the inscribed equilateral triangle. Supply the demonstration of the proposition from the following constructions, which do not require EF to be produced : 6. In the given circle, whose centre is O, draw any diameter BOO Make z OOA = L E, l Q0G= l i^, and at/., J?. O draw tan- gents mtersecting at L, M, N. LMN is the required triangle. 7. At any point B on the o** of the given circle draw a tangent PBQ, and on the tangent tak^ any points P, Q, on opposite sides of B. At P make L QPR ^ l E, and at Q make L PQR = L F. Assuming that Pi?, QR do not touch the given circle, from O the centre draw perpendicidars to PR, QRy and let these perpendiculars, produced if necessary, meet the circle at A and C. At A and G draw tangents LM, LN to the circle. LMN is the required triangle. 8. In the given circle inscribe a i^ ABC equiangular to a DEF. -'*■", xj?y, i^ji^ ana at suu puiuis oi Disecttion draw tangents. m IBUOLID^S BLBMBNTS. tDook 17, 9. Any rectilineal %ire ABODE is inscribed in a circle. Bisect the arcs Ali, liC, CD, DK, EA, and at the points of bisection draw tangents. The resulting figure is equiangular to 10. Two trianirles are circumscribed about the (v ABC, each of them equiangular to a DEF ; prove that they are equal in all respeots. Describe a triangle equiangular to a given triangle, and such that a given circle shall be touched by one of its sides, and by the other two produced. Show that there are tliree solutions of this deduction. 11 PROPOSITION 4. Problem. To imcr/he a circle in a given triangle. Let ABC be the given triangle : it is required to imcnbe a circle in A ABG. Bisect L s ABC, ACB by BI, CI, which intersect at /; /. 9 from /draw ID, IE, IF X BC, CA, AB'. I 12 ( L IDB = L IFB In As IDB, IFB, \ l IBD = l IBF Const. { IB = IB; • • ^^ = ^^. /. 26 Similarly, ID = IE ; m Tn rrr rn ii i* » . Jj^, xiu, ijj axe ail equal. iseot ition • to hem I aU luoh and lupee look IV.] PROPOSItlOK 4. m r.9 12 I8t. 26 With centre / and radius ID describe a circle, which will pass through the points D, K, F. Of this circle, ID, IE, IF will be radii ; and since BG, CA, AB are ± ID, IF, IF, Comi. .-. BC, GA, AB will be tangents to the O DEF ; III 16 .-. the DEF IB inscribed in the A ABG. NoTB. — This proposition is included in the more general one, to describe a circle which shall touch three given straight lines. See Appendix IV. 1, p. 250. 1. It is assumed in the proposition that the bisectors BI, CI will meet at some point /. IVove this. 2. If M be joined, it will bisect /. BAG. 3. The centre of the circle inscribed in an equilateral triangle is equidistant from the three vertices. 4. The centre of the circle inscribed in an isosceles triangle is equidistant from the ends of the base. 5. Prove AF + BD + GE = FB -t DC + EA = semi-perimeter of A ABO. 6. Prove AF + BO = BD + GA = CE + AB = semi-perimeter of A ABG. 7. With A,B,G, the vertices of a ABC aa centres, describe three circles, each of which shall touch the other two. 8. Find the centre of a circle which shall cut oflF equal chords from the three sides of a triangle. 9. If through / a straight line be drawn || BC, and terminated by AB,AC, this parallel will be equal to the sum of the segments of A B, AC between it and BC. £xamine the cases for L, L, Is, in Appendix IV. I. 10. If D, E, F, the points of contact of the inscribed cir6le, be joined, a DEF is acute-angled. 11. The angles of a DEF are respectively complementary to half the opposite angles of a ABC. 12. ABC is a triangle. D and E are points in AB and AC, or m AB and AC produced. Prove that the vertex A, and the centres of the circles inscribed ia ^a ABC, A DE, are collinear. la Draw a straight line which would bisect the angle between two straight lines which are not parallel, but which cannot be produced to meet. 290 BtJOUD's SLBMXNTS. [Book ZV. PROPOSITION 5. Problem. To circumscribe a circle about a given triangle, A Let ABC be tl^e given triangle : it is required to circumscribe a circle about A ABQ. Bisect AB at L &.n^ AG at K; from L and K draw LS J. AB and KS ^_ AC, and let LS, KS intersect at S. Join SA ; and if ;S be not in BC, join SB, SG. AL = BL In As ALS, BLSy { I. 10 /. 11 Const. /. 4 LS = Z5f lALS= l BLS; .-. /Sf^ = SB. Similarly, SA = SG; .'. SAy SB, SCare all equaL With centre S and radius SA, describe a circle ; this circle will pass through the points A, B, C, and will be circumscribed about the A ABC. • Cor. — From the three figures it appears that S, the centre of the circumscribed circle, may occupy three positions ; (1) It may be inside the triangle. (2) It may be on one of the sides. (S) It mav be ontsidfi fho frinncrlA ^ /• >/ - ' 0~"* Book IV.] PROPOSITION S. 231 1,4 In the first case, when S is inside the triangle, the l s ABC, BCA, CAB, being in segments greater than a semi- circle, are each less than a right angle ; ///. 31 .*. the triangle is acute-angled,. In the second case, when S is on one of the sides as BG, L BAC, being in a semicircle, is right ; ///. 31 .*. the triangle is right-angled. In the third case, when /Sis outside the triangle, l BAC, being in a segment less than a semicircle, is greater than a right angle; ///. 31 .*. the triangle is obtuse-angled. And conversely, if the given triangle be acute-angled, the centre of the circumscribed circle will fall within the triangle ; if the triangle be right-angled, the centre will fall on the hypotenuse; if the triangle be obtuse-angled, the centre will fall without the triangle beyond the side opposite the obtuse angle. 1. It is assumed in the proposition that the perpendiculars at L and K will intersect Prove this. 2. With which proposition in the Third Book may this proiwsition be regarded as identical ? , 3. Give an easy construction for circumscribing a circle about a right-angled triangle. 4. An isosceles triangle has its vertic il angle double of each of the base angles. Prove that the iiameter of its circumscribed circle is equal to the base of th 3 triangle. 5. A quadrilateral has one pair of opposite angles supplementary. Show how to circumscribe a circle about it. 6. If a perpendicular SH he drawn from Sto BC, it will bisect BG. 7. If the perpendicular in the preceding deduction meet the circle below the base at Z>, and above the base at E, prove (a) I BSD = L CSD = l BAG; (6) L BSE = L GSE = L ABG -i- l AGB; (c) L ASE=^ I ABG - L AGB; (d) that AD and AE bisect the interior and exterior vertical angles at A» 932 Euclid's blbmfnts. [Book IV. 9. 10. a The angle between the circumBoribed radius drawn to the vertex of a triangle, and the perpendicular from the vertex on the opposite aide, is equal to the diflference of the angles at the base of the triangle. The centre of the circle circumscribed about an equilateral triangle is equidistant from the three sides. The centre of the circle circumscribed about an isosceles triancle 18 equidistant from the equal sides. 11. When the inscribed and circumscribed centres of a triangle coincide, the triangle is equilateral. 12. When the straight line joining the inscribed and circumscribed centres of a triangle passes through one of the vertices, the tnangle is isosceles. la If Hhe the middl6 point of BC, what will the point ^ be in reference to A HKL ? 14. 8A SB SO are respectively ± the sides of the orthocentrio tnangle of a ABC. 15. The straight line joining the inscribed centre of a triangle to any vertex bisects the angle between the circumscribed radius to that vertex, and the perpendicular from that vertex on the opposite side. PROPOSITION 6. Problem. To inscribe a square in a given circle. A Let ABC be the given circle : it is required to inscribe a square in ABC. Find the centre of the O ABa IIL 1 Hook IV.] PROPOSITION 6. 393 and through draw two dianioters AC, BD J. each other; j\\ join AD, BC, CD, DA. A BCD is the required square. (1) lb prove ABCD equilateral. C AO = AO In As AOB, AODA OB = OD ( L AOB = L AOD; .*. AB = AD. /. 4 Hence also AB = BC, BC = CD/ .*. i4/yCZ> is equilateral. (2) To prove ABCD rectangular. Because ^s ABC, BCD, CDA, DAB are right, being angles in semicircles ; ///, 31 .*. ^/^CZJ ifi rectangular j .*. ABCD is a square. /. Def. 32 Cor.— If the arcs AB, BC, CD, DA be bisected, the points of bisection along with A, B, C, D will form the vertices of a regular octagon inscribed in the circle. If the arcs cut off by the sides of the octagon be bisected, the vertices of a regular figure of 16 sides inscribed in the circle will be obtained. Repeated bisections will give regular figures of 32, 64, 128, 256, &c. sides inscribed in the circle. All these numbers 4, 8, 16, 32, 64, &c. are com- prised in the formula 2", where n is any positive integer greater than 1. 1. Prove that ABCD is equilateral by using III. 26, 29. 2. The square inscribed in a circle is double of the square on the radius, and half of the square on the di,inieter. 3. All the squares inscribed in a circle are equal. 4. If the ends of any two diameters of a circle be joined consecu- tively, the figure thus inscribed is a rectangle. 5. What is the magnitude of the angle at the centre of a circle subtended by a side of the inscribed square ? 6. If r denote the radius of the given circle, then the side of the I 334 .*r buolid's blbmbnts. [Book I?. PKOPOSITION 7. Problem. To circumacnbe a square about a given circle, A Let ABO be the given circle : it is required to circun^cribe a square about ABO. Find the centre of the ABC, Jij \ and through draw two diameters AC, ^D ± each other. J J, At A, B, C D, draw EF, FO, QH, HE, tangents to the circle. jjj jy EFGH is the required square. (1) To prove EFGH equilateral. Because EF dindi GH are both ± AC, and BD is also ± AC; .-. -fi^i^; BD, and (^^iTare all parallel. Hence also FG, AC, and HE are all parailol, .'.all the quadrilaterals in the figure are ||™». Hence EF and GH are each = i?i>, and FG and /f^ are each = AC. But ^C = BD ; .-. ^i^, i^(?, GH, HE are all equal (2) To prove EFGH rectangular. Because OE is a Jl"", ,-. lE= l AOD ; •*• L E\b righi ///. 18 Const. I 28, Cor. I. Def. 33 /. 34 /. 34 LU Book ZV.] PROP08IT10N8 7, 8. 235 Hence also ls F, G, H ure right. .*. ^i'^C?// is a square. / jjgf 32 1. It is assumed in the proposition that the four tangents at A, B O, D will form a closed figure. Prove this. ' * 2. The square circumscribed about a circle is double of the square inscribed in the circle. 3. All the squares circumscribed about a circle are equal. 4. If a rectangle be circumscribed about a circle, it must be a square. 6. If tangents be drawn at the ends of any two diameters of a circle and produced to meet, the figure thus circumscribed is a rhombus. 6. What is the magnitude of the angle at the centre of a circle, subtended by a side of the circumscribed scjuare ? 7. If r denote the radius of the given circle, then the side of the oiroumacribed square will be denoted by 2r. PROPOSITION 8. Problkm. To imcnhe a circle in a given squnrg. Let ABCD be the given Svjuare : it is required to inscribe a Q in ABCD. 4 Join A O, BD intersecting at O ; and from O draw OE, OF, OG, OH X the sides of the gquaro. 236 Euclid's blbmbnts. [Book IF. F o In Aa BAG, DACA AG = AG iBG = DG; LDef.n .-. L BAG = L DAG, and l BGA = l DGA; I. 8 .*. the diagonal AG bisects z. s BAD, BGD. Hence also, the diagonal BD bisects ^ s ^45(7, ^i>a ' ( L OEB = ^ 0i!5 In As OEB, OFB, ] l QBE = l OBF ( OB = OB; .'. OE = OF. J 2g Hence also OF = 0(?, OG = Oi7/ .'. 0^, OF, OG, OH&ve aU equal. With centre and radius OE, describe a circle which will pass through the points E, F, G, H. Of this circle, OE, OF, OG, OH will be radii; and since AB, BG, CD, DA are ± OE, OF, OG, OH, Gomt. .-. AB, BG, GD, DA wiU be tangents to QEFGH; ^ jjj^Q .-. O EFGH ia inscribed in the square ABGD. 1. Could O, the centre of the inscribed circle, be found in any other way than by joining AG, BD ? 2. Show that a circle cannot be inscribed in a rectangle unless it be a square. 3. Insbribe a circle in a given rhombus. 4. Enumerate the ||«« in which circles can be inscribed. fi. If a denote a side of the given square, then the radius of the inscribed circle will be denoted by \a. a Book IV.] PROPOSITIONS 8, 9. 357 PEOPOSmON 9. Problem. To drcumscnbe a circle about a givers square. L Def. 32 /. 8 Let ABOD be the given square : it is required to circumscribe a circle about ABCD, Join AC, BD intersecting at O. In As BAG, DAG, \ag = AG {bG=DG; I Def m •• L BAG= L DAG, and l BGA =. l DGA, .*. the diagonal AG bisects l s BAD, BGD Hence also, the diagonal BD bisects I s ABG ADG. ^ToltoB^ = ^ OBA, each being half a rt. z. ,' Hence also OB = OG, and OG = OD; .-. OA, OB, OG, OD are all equal. With centre and radius OA, describe a circle which will pass through the points A, B, G, D, and .-. will be circumscribed about the square ABGD. a Circumscribe a circle about a given rectangle. 3. Enumerate the ||m. about which circles can be circumscribed 4 If a denote a side of the given sa„ar.. thon fZ !1 '"^'^;., circumscribed circle wiU be denoted by ^avt" '""'"" " /. 6 ■wjwn ^^Esar:* 338 EUOLIO'S ELEMENTS. tBook IV. \ i t i PROPOSITION 10. Problem. To deacribe an isosceles triangle having each oj the angles ai the base double of the third angle. Take any straight line AB, and divide it internally at (7 so that AB - BC =^ A C^. //. 1 1 With centre A and radius AB, describe the BDE, in which place the chord BD -AC. IV. \ Join AD. ABD is the required isosceles triangle. Join CD, and about A A CD circumscribe the ACD. IV. 5 Because AB - BC = AC\ Const. ^BD"^; .'. BD is a tangent to the ACD. III. 37 Because the chord DC is drawn from D, the point of contact of the tangent BD ; lBDC=lA. IILZI Add to each the l CD A ; L BDA ^ L A + L CDA; _ DBA = ^ A ^ L CDA. L 5 But L DCB =^ L A + L CDA ; I. 32 L DBA or ^ DBC - L DCB; Book IV.] PROPOSITION 10. 239 7.6 DC^DB, =' AC. r. lA=^l CDA, ot lA + l CDA^ twice l A. /. 6 ^"* L BDA = L A + L CDA ; .', L BDA, and consequently l DBA = twice l A, 1. The A DBC is equiangular to a ABD. 2. Angle A = one-fifth of two right angles. 3. Divide a right angle into five equal pari». 4. The A GAVhM one of its angles thrice each of the other two 6. On a given base, construct an isosceles triangle havine each of Its base angles double of the vertical angle. 6. On a ^ven base, construct an isosceles triangle having each of Its base angles one-third of the vertical angle 7. The smaU cir:-le in the figure to the proposition must cut the large one. (Campanus.) a If the smaU circle out the large one at F, and DF be joined. UjP = JiJj. (Campanus.) lA ^n ^ ^ '!??.°^ * '®^'^*'' **^^*«°" inscribed in the large circle. 10. AC and CD are sides of a regular pentagon inscribed in the small circle. !o S^®«Tf^^ ''"^^^ "^ *^® ^^'^^^ circumscribed about a ABD 12. If ^^ be joined, BF is a side of a regular pentagon inscribed m the large circle. ' 13. If AF and ^C be joined, a s ADF, FAG possess the property reqmred m the proposition. 14. If DG be produced to meet the large circle at O, and BG be joined, ^C is a side of a regular pentagon inscribed in the large curcle. 16. If FO be joined, FG bisects ^C perpendicularly. 16. Divide a right angle into fifteen equal parts. ' 17. The square on a side of a regular pentagon inscribed in a circle IS greater than the square on a side of the regular decagon (EucUi xil *^* ^"""^ ''"^^^ ^^ *^® '^""® **" *^® '*^"* 1& Show, by referring to 1. 22, that the lar^ circle could be omitted trom the figure of the proposition. 19. Si>ow that the proposition could be proved without describing ^e smsii ciiiae, by dtawlag a perpendicular from i> to 340 Euclid's elements. [Boo^ nr. 20. Show that the centre of the circle circumscribed about a BGiy is the middle point of the arc CD. 21. What is the magnitude of the angle at the centre of a circle subtended by a side of the inscribed regular decagon ? 22. If r denote the radius of the circle, then the side of the inscribed regular decagon will be denoted by ^^(Vs - 1). M !f PROPOSITION 11. Problem. To inscribe a regular pentagon in a given circle. Let ABC be the given circle : it is required to inscribe a regular pentagon in ABC. Describe an isosceles A FGH^ having each of its lqG,H double of u F; IV. 10 in the O ABC inscribe a A ACD equiangular to A FOH, so that L s ACD, ADCm&y each be double of l CAD. IV. 2 Bisect La ACD, ADC by CE, D^ ; /. 9 and join AB, BC, DE, EA. ABCDE is the required regular pentagon. (1) To prove the pentagon equilateral. Because l a ACD, ADC are each double of l CAD, Const. and they are bisected by CE, DB ; .-. the five L 8 ADB, BDC, CAD, DCE, EC A are aU equal ; .• . the five arcs AB, BC, CD, DE, EA are all equal ; ///. 26 .-. the five chords AB, BC, CD, DE, EA are all equal ///. 29 Book 17.] PROPOSITION 11. 24t (2) To prove the pentagon equiangular. Since the five arcs AB, BO, CD, DE, EA are all equal .'. each IS one-fifth of the whole O"* • .-. any three of them = three-fifths of the O"" Now the five l s ABC, BCD, ODE, DEA, EAB stand each on an arc = three-fifths of the O*"; .-. these five angles are all equal. jjj 27 1. How many diagonals can be drawn in a regular pentagon ? ?' ^Z A "^^ '^T^*^ ^ " * ^'^^ ^* *^« '^g'^'^'- pentagon. 3. All the diagonals of a regular pentagon are equal sector °^ * "^^"^^ ^''*^^°'' '''* ^"^^ ^^^^'^ '° "^^»1 & The intersections of the diagonals of a regular pentagon are the vertices of another regular pentagon. 6. The intersections of the alternate sides of 'a regular pentegon are the vertices of another regular pentagon. 7. If BE be joined, show that there wiU be in the figure five pentagons each of which is equilateral but not equiangular. Z' ABODE '""'^"^' ^"* ^''''' *^ '"'^■^'^^ ^Z !^%*7?^^ deduction from IV. 10, to obtain another 11 WW 1 mscnbmg a regular pentagon in a given circle. 11. What ,8 the magmtude of an angle of a regular pentagon ? 12. Knowing the magnitude of an angle of a regular pentagon how 13 CnZr ""l '"""'^T* ^ ''^"^^^ P'"*"^"^ ^" ^ given straight line ? 14. If the alternate sides of a regular pentagon be produced te meei !oualTf '^"J""' r^^'' "* *^^ ^^'^ o^ intersection is IK „,J^"*^ *o two nght angles. (Campanus.) What IS the magnitude of the angle at the centre of a circle 16. Tf !7**°^f^^^ ^-J'^^ of *1^« inscribed regular pentagon ? la If r denote the radius of the circle, then the side of the inscribed regular pentagon will be denoted by JrVio - 2V& 8. 10. I r ttOMM'lMM* Mi><>«MMr^ 242 kuolid's elements. [Book ly. PROPOSITION 12. Problem. To circumscribe a reijiUar pentagon, about a given drcle. H c Let ABC be the given circle : it is required to circumscribe a regular pentagon about ABO. Find A, B, O, D, E the vertices of a regular pentagon inscribed in the circle ; IV. xl at these points draw FG, GH, HK, KL, LF tangents to the circle. jjj jy FGHKL is the required regular pentagon. ///. 1 /. 32, Cor. 2 ///. 18 Find the centre of the circle, and jom OB, OH, OG, OK, OD. (1) To prove tlie pentagon equiangular. Because OBHG is a quadrilateral ; .". the sum of its four z. s = 4 rt. lb. But L OBH + L OGH . = 2 rt. ^s-; .-. ^ ^^C is supplementary to l BOG. Hence also, l GKD is supplementary to l GOD. P)ut since B, G, D are consecutive vertices of an inscribed regular pentagon ; arc BG = arc GD ; ///; 28 L BOG = L GOD. iii^ 27 Hence l BUG — l GKD. Book IV.] ///. 1 PROPOSITION 12L MS III 17 Cor. I. 8 onLpeSnf ^ ^^^-^ any two consecutive angle. •'• ^^^ *^^^ angles of the pentagon are equal. (2) To prove the pentagon equilateral. (BO = 00 ^^BB0H,C0H,]0H=0ff (bH=CH; .-. lBOH=l COH; .'. L BOG is double of l HOC. Hence also, l DOC is double of 'l KOC But because l BOC ^ l DOC, .-. lH0C==l KOC T A ( ^ ^<^^ = -^ KOC In As^oa ^oc;h. oa^= z. OCR in, ig ( OC = OC: .-. HC^KC; .-. Zr/i is double of i/C. ^^ Similarly, (?zr is double of HB But since ZTJ? = HC, .-. 6^iy = jjK Xo.^ '"' '''"'" ""' '"° """^""""^^ ^'^- of *« •. aU the sides of the pentagon are equal. 1. I* ^ -^-^d in the proposition that the five tangents at - P l^ri : ^"^ ^"""^ * ^^°sed figure. Prove this H .ny regular polygon be i"«X in .oi^T^^^Z^T I vertices will form another regular Mlvlr^r;!.*^ *' '** of aide, oireumseribed about toe 0^1^ *^ '"" """'^ 2. 4. I ■|'Hl i|ii lViir iiili » l ni i S44 li 1 i /I t 4 I BtJOUO's BJtBMENTS. [Book IV. PROPOSITION 13. Pkoblbm. To inmrihe a circle in a given regular pentagon. G D Let ABODE fce tlie given regular pentagon : it is required to inscribe a circle in ABODE. Bisect L s BOD, ODE by 00, DO intersecting at 0; 1.9 join OB, and draw OF, OG ± BO, OD. /. 12 I BO = DO Hyp. In As BOO, DOO, j 00 = 00 ( L BOO = L DOO; Const .'. L OBO = L CDO. J 4 But L OpO is half of the angle of a regular pentagon ; Omst. .'. L OBO is half of the angle of a regular pentagon ; .-. 05 bisects l OB A. Hence also, OA would bisect l BAE, and OE, l AED. ( L OFO = L OGO In As OFO, OGO,]l OOF = l OOG Omst, { 00 = 00; ••• ^^= 00. J 26 Now since O is the point where the bisectors of all the angles of the pentagon meet, and OF, OG are perpendiculars on any two consecutive sides ; .-. the perpendiculars from on all the sides are equal, nenve inc Cni;ic ucacnuuu wim u as centre and OF as PBOPOSITIONS 13, 14. 7.9 7. 12 Hyp. 246 Book IV.] wduis^ wUl pass through the feet of aU the perpendiculars and wiU touch AB, BC, CD, DE, EA. ni. 16 1. Find the centre and radius of the circle inscribed in a regular pentegonbyn^eansofasqua... (A set square or T^ITi^e * '^^'.^H K^ > ''^"^*' P'"***^^ " ^^"^^ ^ *^« r^^t'^ngle con- tamed by Its semi-penmeter and the radius of the i^cribed PROPOSITION 14. Problem. To cireumscribe a circle about a given regular pentagon. ^ Let ABODE be the given regular pentagon • *tt 18 required to circumscribe a circle about ABODE. f^^ot LsBOD,ODEhyOO,DOmteTBectingatO; I 9 and ^om OB, OA, OE. & ^, i.v OB, OA, OE bisect the l s OBA, BAE, AED IV 1 S Because ^ OOD = ^ ODO, each being half of the angle of a regular pentagon ; ^ .-. 0(7=0Z>. Hence also, 0Z> = OE, 0E'=. OA, OA = OB • .-.the circle described with O as centre and OA as radius, will pass through A, B, O, D, E, and wiU be circumscribed about the pentagon ABODE. 1. Find the centre and radius of th« oiroU ^r ru., _, . regular pentagon by means of a square. i 246 buolid's elements. [iMk XT. 2. The square on the diameter of the cirole oircumicribed «bout * regular pentagon = the square on one of the sides of the pentagon together with the square on the diameter of the inscribed circle. 3. If a denote a side of the given rtgular pentagon, then the radius of the c ircumscribed cirole will be denoted by AaVfiO + 10 V 6. PROPOSITION 15. Problem. To inscribe a regular hexagon in a given circle. Let ABQ be the given circle : it is required to inscribe a regular hexagon in ABC. Find the centre of the circle, ///. 1' and draw a diameter AOD. With centre D and radius DO, describe the O EOC; join EO, CO, and produce them to B and F. Join AB, BC, CD, DE, EF, FA. ' ABCDEF is the required regular hexagon. (1) To prove the hexagon equilateral. As DOE, DOC are equilateral ; 7. 1 .-. z_ s DOE, DOC are each one-third of two rt. L s. J. 32 But L DOE + L DOC + L COB = two rt. z.s; /. 13 .'. L (70B = one-third of two rt. z.s. Hence l 3 BOA, AOF, FOE are each = one-thiid of two it. ^ 31 ^.15 /. 1 /. 32 /. 13 ar 1 K ■OrtS IV.] PROPOSITION 1«. f^f .-. the six ^8 AOB, BOG, COD, DOE, EOF, FOA are all equal ; .-. the six arcs AB, BO, CD, DE, EF, FA are aU equal; jjj gg . •. the six chords AB, BC, CD, DE, EF, FA are aU Tmrp- . ^^/. 29 (2) To prove the hexagon equiangular. Since the six arcs AB, BC, CD, DE, EF, FA are all equal, .% each is one-sixth of the whole O** ; ,'. any four of them = four-sixths of the whole O** Now the six L a FAB, ABC, BCD, CDE, DEF, EFA stand each on an arc = four-sixths of the O** ; .'. these six angles are all equal. /// 27 CoR—The side of a regular hexagon inscribed in a circle is equal to the radius. 1. If the points A,C,Ehe joined, £^ AGE is equilateral. 2. The area of an inscribed equilateral triangle is half that of a regular hexagon inscribed in the same circle. 3. Construct a regular hexagon on a given straight line. 4. The area of an equilateral triangle described on a given straight line 18 one-sixth of the area of a regular hexagon described on the same straight line. 6. The opposite sides of a regular hexagon are parallel. 6. The straight lines which join the opposite vertices of a regular hexagon are concurrent, and are each || one of the sides. 7. How many diagonals can bo drawn in a regular hexagon ? 8. Prove that six of them are parallel in pairs. 9. The area of a regular hexagon inscribed in a circle is half of the area of an equilateral triangle circumscribed about the circle. 10. The square on a side of an inscribed regular hexagon is one-third of the square on a side of the equilateral triangle inscribed in the same circle. IL What is the magnitude of the angle at the centre of a circle subtended by a side of an inscribed regular hexagon ? 12. Give the constructiond for inscribing a circle in a regiUar hexagon; and for circumscribing a regular hexagon about a firae, and a circle about a regular hexagon. I 'imm t ' i 'M UUOUDS BLBMENTS. tiMokrr. < I PROPOSITION 16. Problem. To insenbe a regular quindecagon in a given ctrci&t Let ABC be the given circle : it is required to inscribe a regular quindecagon in ABO. Find AC Q. side of an equilateral triangle inscribed in the circle; [v. 2 and find AB^ BE two consecutive sides of a regular penta- gon inscribed in the circle. IV. \\ Then arc ABE = | of the O**, and arc AC = ^ of the O"* ; arc CiEJ = (I - i), or yV, of the O**. Hence, if CE be joined, CE will' be a side of a regular quindecagon inscribed hi the O ABC. Place consecutively in the O* chords equal to CE ; IV. 1 then a regular quindecagon will be inscribed in the circle. 1. How could the regular quindecagon be obtained, if, besides AG^ a side of an equilateral triangle, only one side ^jB of the regular pentagon be drawn ? 2. How could the regidar quindecagon be obtained by making use /\f 4:1so ai/loo /\f ^\%a ««A/Vfl1nt« imo^mKa/^ l«MiwnM<%M ^^A J^^. « Book 17.] PROPOSITION 16. 34» a In a given cirele inscribe a triangle whose angles are as th« nu^^rs 2, 6, 8 ; and another wLe angles areTthTnumlSs 4 Give the constructions for inscribing a circle in a regular quin. decagon ; and for circumscribing a regular quindecCn abou 5 Hot' ^f ^ 'r^' "^"* * '^S"^'^ quindecagon.^ 6 fZJTl ^'^'°t '"^ ^' ^'^^ ^^ * regular quindecagon ^ dTago'tls" * ^'^^^^ '^^^ ^ "^^^ ^* -^ ^- 4^-3) 7. Show that the centres of the circles inscribed in, and circiun- scnbed about, any regular figure coincide, and are obS by bisecting any two consecutive angles of the figure. Jl^I^u'f^^Vf^^'' P^^y^°' ^^ 3' ^ 5, and 15 sides, and such Gauss, in 1796, found Lt a regu^ lytZ^T?, S°™"^- juscriptible, and in hi, IH^tnuUioJ AriS^^^^ii^tlm number of .ts sides was a prime number, and expresmiHy 2' + 1 Note 2.--The polygons of which Euclid treats are all of „„. such as AWD m the accompanying iigures. The reader wiU find it instmctiT. to inquire how far the properties of convex do1v«»>. (forexample, quadrilaterals) are true fofthe othe„ A^fe »~.»^n. polygons there is a class called stellale or .tar, "whfch I m 250 EUCLID 8 ELEMENTS. V ill [Book IV. are obtained thus: Suppose A, B, G, D, E (see fig. to IV. 11) to be five points in order on the 0<=* of a circle. Join AG, GE, EB, BD, DA : then AGEBD is a star pentagon. If the arcs AB, BG, &c. are all equal, the star pentagon AGEBD is regular. Similarly, if 1, 2, 3, 4, 6, 6, 7, 8, 9, 10 denote the vertices of a regular decagon inscribed in a circle, the regular star decagon (there can be only one) is got by joining consecutively 1, 4, 7, 10, 3, 6, 9, 2, 5, 8, 1. It will be found that if a regular polygon have n sides, the number of regular star polygons that may be derived from it is equal to the number of integers prime to n contained in the series 2, 3, 4, ... \{n - 1). (For more information on the subject of star polygons, see Chasles, Aperqu Hiatoriqm mr VOrigine et le Divel- oppement des M^thodes en G6omUrie, sec. 6d. pp. 476-487, and GeorgM Dostor, ThSorie 04n4rak des Polygones EtoiUa^ 1880.) III APPENDIX IV. Proposition 1. To describe a circle which shall tomh three given straight lifies. (1) If the three straight lines be so situated that every two are parallel, the solution is impossible. (2) If they be so situated that only two are parallel, there can be two solutions, as will appear from the following figure : Let AB, GD, EF be the three straight lines of which ^ JS ia !l CU. I" ! Book IV.} APPBNDiy. IV. 251 1.9 7.12 7.26 7.26 Bisect z 8 AEF, GFE by EI and FI, which meet at 7; from I draw IH, IK, IL respectively i. AB, EF, CD. Then A a lEH, lEK are equal in all respects ; .-. IH = IK. Similarly, IK = IL; .\ IH = IK = IL. Now since laat H, K, L are right, ;'" w n'l^z,'^^'^'''^^^ "^^^ ^ as' centre and IH as radius will touch AB, EF, CD. jjj [7 A similar construction on the other side of EF will give another circle touching the three given straight lines. (3) If they oe so situated that no two avfi parallel, then they will t'Ttner all pass through the same point, in which case the solution •s mpossiDie ; or they will form a triangle with its sides produced, ii\ which case four solutions are possible. GC. i";et AB, BO, CA produced be the thi^ given straight lines terming by their intersection the a ABC. If the interior z s 5 and C be bisected, the bisectors will meet ac some point /, which is the centre of the circle inscribed in the triangle, as may be proved by drawing perpendiculars ID. IE IF to the sides BC, GA, AB of the triangle. /f^ 4 n the exterior jngles at -fi and C be bisecied by 57„ Gh wttih I 262 Euclid's elements. TBooklf; ■■ lit meet at /j, and perpendiculars /iA> h^n IxF\ be drawn to the sides BC, and AG, AB produced, it may be proved that IiD^, I^E^, IiF^ are all equal, and .*. that /i is the centre of a circle touching BG, and AG, AB pro- duced. Hence also, /g, the point of intersection of the bisectors of the exterior angles at G and A, will be the centre of a circle touching GA, and BA, BG produced; /g, the point of intersection of the bisectors of the exterior angles at A and B, will be the centre of a circle touching AB, and GB, GA produced. Cor. — The following sets of points are coUinear : A, I, Ii ; B, I, h ; G, I, h ; /s, A, /s ; li, B, h ; h G, I^. In other words, the six bisectors of the interior kv/^ exterior angles at ^, jB, C meet three and three in four points, /, 7i, /g, /g, which are the centres of the four circles touching the three given straight lines. Or, the six straight lines joining two and two the centres of the four circles which touch AB, BG, GA, pass each through a vertex of the A ABG. The circles whose centres are I^, /j, /s are called escribed or exscribed circles of the A ABG, an expression which, in its French form (ex-inacrit), is said to be due to Simon Lhuilier. See his ilimena d Analyse GiomStrique et di" Analyse Alg4brique (1809), p. 198, S a 01 I'l 'i!i Book iv.J APPENDIX iV. 253 J. . zoo radius of the circumacribed circle by A * " ''" """^ ^^* THE MEDIOSCRIBED CIRCLK. Proposition 2. ■^.rpen -th ^f; Similarly, ff^, p^^„ „ ^^ HVKUtCO- A '"'r^l or'^ r « dia'™:t ""■ ""• "■ ' "' '"' *^' -* -^-ribed with /«' /// Ql I'\ H! iiiii Nil ( >ir ^1! ' Ilii 354 EUCLID S ELBMBNT8. [Book I?. Kbcce the six points H, K, L, U, T, W lie on the same circle, and BU,KV,LW are diameters of it. But since the angles at X, Y, Z are right ; .•. iC lies on the circle whose diameter is HU^ Y u , I. KV. Z » " LW; III. 31 .'. the nine specified points are concyclic. Cor. 1.— Since HU, KV, LW are diameters of the same circle, their common point of intersecti.^a M is the centre. Cor. 2.—M is midway between the orthocentre and the circum- scribed centre. Let S be the circumscribed centre, and SH be joined. Then SH is ± BC {III. 3) ; and .'. || OU. But SH = OU (App. I. 6, Cor.) ; .-. SHOU is a ir ; .-. the diagonal SO bisects HU, that is, passes through M, and is itself bisected at M. ^ CoR. 3.— The medioscribed diameter = the circumscribed radius. For SHUA is a ir ; and .-. HU = SA. Cor. 4.— a 8 ABC, A OB, BOG, CO A We the same medioscribed circle. Since the medioscribed circle of A ABC passes through U, L, T, the middle points of the sides of A AOB, and since a circle is determined by three points ; .'. the medioscribed circle of A ABC must also be the medioscribed circle of a AOB. Sin^ilarly for the other triangles. Cos, 5.— By reference to Cor. 3, it will be seen that the circles circuKascribed about as ABC A*tB, BOC, CO A must be equal (GMnot, Qiomitrie de Position, ib03, § 130.) 11 1 Book ZV.J APPENDIX IV 20ft Cor. 6.-The me. jscribed circle of a ABC i« also the medio- acnbed circle of an intiuite series of triangles. For H, K, L, the middle points of the sides of a ABC, may be taken as the feet of the jjerpendiculard of another a ABC ■ the middle points of the sides of a A'BV may be taken as the feet of the perpendiculars of a a A"B"C" ; and so on. Or, instead of the median a HKL, as XKL, YLH ZHK may '- taken as median triangles, and the triangles formed of which they are the median triangles ; and so on. (The circle HKL is generally called the nine-point circle oi t,ABC a name given by Terque.n, 'le cercle des neuf points.' Following' huwever, the suggestion of an Italian geometer, Marsano, who calls It il circolo medioscritto,' I have adopted the name Tmdioacribed. The property that one circle does pas3 through these nine points '''^,/^.,E!;? ^^^^ '"^ Gergonne's Annalea de MatUmatiques, vol. xi p. 215 (1821), in an article by Biianchon and Poncelet. See this reference, or Poncelet's Applications d\inalyse et de Ofomdtrie : m . ^^' ^* '' probable that K. W. Feuerbach of Erlangen! and T. b. Davies of Woolwich, also discovered the property inde- pendently though they were later in publication. See Feuerbach's f^genschaften etniger merkwilrdiuen Punkte des geradlinu,en Dreiecka (1822) and a paper by Davies on 'Symmetrical Properties of Plane Triuogles, m the Pkilosophical Magazine for July 1827 For other proofs, see Rev. Joseph Wolstenholme in Quarterly Jourmilof Pure and Applied Matheamtics. vol. ii. pp. 138, 139 (1858) S\ f *"* "' *^'^ i/^.s«e«i/er of Mathematics (old series), vol i i' frL kf V?'?-' "Tir ^°'"^^' '^ *^« Mathematics^ Repr^; from the Mucational Times, vol. vii. p. 86 (1867) • Desboves' Qiceattom de QeomStne Elimentaire, 26me ^d. p 146 (1875) • and Casey's^/emen<»o/^Mc^irf, p. 153(1882). ) , ^^ The proof in the text was given by T. T. Wilkinson of Burnley m the Lady s and Gentleman's Diary for 1855, p 67 It may be mentioned that it was discovered "by 'Feuerbach (see his Exgenschaften, &c. § 57) that the medioscribed circle touches the inscribed and escribed circles of a ABC. The proofs that have been given of this theorem by elementary geometry are rather comphcated : see Lady's and Gentleman's Diary for 1854 d 66 • Qmrte^ly Journal of Pure and Applied Mathematics, vol. iv (1861)' p. 246. and vol. v. i\RR>l\ t» 07n. b„u r>. ' „, ^^^y' Mathematik, vol. ii. pp. 92, 93. It is also proved by J. J. Robinaon 306 SUOLID'8 ILEMKNTS. [Book IV. ,1 : ■ ii'l'i J .M' m ■A4 I in th« Ladj^t and Oentleman's Diary for 1857 (and it seems to have been flnt noted by T. T. Wilkinson), that the medioacribed circle touches an infinite series of circles.] DEDUCTIONS. 1. Every equilateral figure inscribed in a circle is equiangular. 2. In a given circle inscribe (a) three, (6) four, (c) five, {d) six equal circles touching each other and the given circle. 3. The perpendicular from the vertex to the base of an equilateral triangle = the side of an equilateral triangle inscribed in a circle whose diameter is the base. 4. The area of an inscribed regular hexagon = three-fourths of the area of the regular hexagon circumscribed about the same circle. 5. Inscribe a- regular hexagon in a given equilateral triangle, and compare its ^rea with that of the triangle. 6. Inscribe a regular dodecagon in a given circle, and prove that its area = that of a square described on the side of an equilateral triangle inscribed in the same circle. 7. Construct a regular octagon on a given straight line. 8. A regular octagon inscribed in a circle = the rectangle contained by the sides of the inscribed and circumscribed squares. 9. The following construction is given by Ptolemy (about 130 a.d.) in the first book of his Almagest, for inscribing a regular pentagon and decagon in a circle : ^raw any diameter AB, and from Cthe centre draw CD ± AB, meeting the O"* at D : bisect v4C at jE7, and join ED. From EB cut off EF = ED, and join DF. CF will be a side of the inscribed regular decagou, and DF a side of the inscribed regular pentagon. Prove this. ^ 10. A ribbon or strip of paper whose edges are parallel, is folded up into a flat knot of five edges. Prove that the sides of the knot form a regular pentagon. U. Construct a regular decagon on a given straight line. 12. In a given square inscribe an equilateral triangle one of whose vertices may be (a) on the middle of a side, (6) on one of the angular points, of the square. Construct a triangle having given 13. The inscribed circle, and an escribed circle. 14. Two escribed circles. 16. Any three of che centres of the four contact circles. i I: Book IV.] APPENDIX IV. 267 16. Tte base, the vertical angle, and the inscribed radius. 7. The penmeter, the vertical angle, and the inscribed radius ' "tsc^LJXs.^^ '''---'' the other two sides, L &e '"■ ^Z'X'l^r'"^ properties with inspect to a ABO (see fig. s =. AEj = AF, = BD, = BF, = OD, = CE^ ^^-1= jE = AF = BD, = BF,== Gd] = CE, ^-b=AE = AF,=BD =BF =Cd\=CE s-c=AE,= AF, =BD,^BF, = CD =CE a = EE, = E^, = FF, = F,F,. b=DD,=. D,D., = FF^ = F,F^. ma^b + c=AE ^AE. + AE.+AE,^''^^^'- = AF + AF, = BD + BD, + BD, + 5i)3 = BF + BFi = CD + CD, + CA + (7Z>3 = C^ + CE = AF'^+AF,^ +AFo^ + AF» = 5/>-^ + BDi^ + BD,^ + BD,^ ' = BF^ + BF,^ + BF«^ ^ liW3 = GD^ ^ CDi^ ^ CD.^ ^ GdI^ "^^^^ ^^^' + AF^ + + BFs + + CEi + AFn BF^ CEs, (9) I = ^f ' + CEi'' + CEi' + C^3» = 3 (a2 + yi + c'O + = 5 {a' + 62 + cS).* I An + ^v + Ai;^ + J/32 + ^/- + BI{^ + ^42 + Bli^ AD^ + ADi' + Ji),2 + ^2)32 + BE^ + ^Ai2 + BE,^ + JS^32 + C7i?'^ + QF,^ + c^,3 + cp^^ uaL^L^ P* ^T'i !!*' ''^ expressions may be written more shortlv bv 258 lUOLin's ELEMENTS. [Book #. Pi |:S| I (10) TriiiuBles mutually equiangular in acta of four are : AIK,Ar,E\,AI,E,,AhE,', BIF,B/,F,,BI,F,,BI,F,', CID, Gf^Di, CI^D^, CI^Di. (11) Mention other twelve triangles which are mutually eciui- angular in seta of four. (12) Trianglf'J^ Tnutuailv equiangular in sets of three are : ArB,ACr,,LCB:BIC,BAhJ,AC;CIA,CBf,,hBA, (13) Triangles mutually equiangular in sets of four are : lyBI,, hCl., IBh, ICh; hCh, hAI,, ICh, lAh; r^AIu hBL, /A I,, IBIy 114) Express in terms of l^ A, B,C, (a) The o-ngles of as hhh, DEF ; hBC, hCA, hAB ; AEF, BFD, ODE. (6) „ subtended by AB, BG, GA at /, A, /«, 1%. (c) i„ M DE,EF,FD;hhJihMh\ IiD, I^E, hF &t I. . ) D and i>i w« equidistant from the middle point of BO; 80 are D^ and A- Similar relations hold for the E points and the ^points. 20. Of tne four points 7, /i, h, li, any one is the orthocentre of the triangle formed by joining the other three, and in each case ABG is the orthocentric triangle. 21; The orthocentre and vertices of a triangle are the inscribed and escribed centres of its orthocentric triangle. Verify in the four cases. 22. Six straight rines join the itiscribed and escribed centres ; the circles described on these as diameters pass each through two vertices of the triangle, and the centtes of these six circles lie on the 0<=» ot the circle circumscribed about the triangle. 23. Prove the second part of the last deduction without assuming the property of the medioscribed circle. 24. Prove the following properties (see fig. on p. 251) : (1) The radii /lA, hE^^ ^i^a arws concurrent at SiJ ID, I3E3, I,F^ at A, ; I,D„ IE, hF^ at B^ ; I A I^Ei, IF at G^. (2) The figures AJ^S^I^, B^I^^u^, GJ^SJi are rhombi, and AJiBiIiGJ^ is an equilatwal hexagon whose opposite sides are parallel. . (8) AS AiB^Cu hhh arc eoagrue-t, and their corracponding sides are parallel. I ^ iMk nr.] APPINDIX IT. 301 (4) The points Su A,, B,, C„ are the circumscribed centres of A8y,/j,/3, //j,/„ //.,/„ //,/,. (5) The figures AJB,J„ Ii,/(\f„ 'cjA.I, are rhombi, and /is the orcumscnbed centre of a AiBiO^. (6) The circumscribed circle of a ABC \s the. medioscribed 7c7^^"A^'''' 'H'' "'''' '^''-' ^»^>^" '^^'^>^" diOi^j, tiiAiB^ ; and its centre is the middle point of ISx. [See Davies' Symmetrical Properties, &c. quoted on p. 255.] 26. The area of ^ ABO = rs = r^ {s - a) = r, («-&) = r^ (« - c) 26. The bisector of the vertical angle of a triangle cuts the o<* of the circumscribed circle at a point which is equidistant from the ends of the base and from the centre of the inscribed circle. 27. The diameter of the circle inscribed in a right-angled triangle no rp, ^''fi®**^®^ with the hypotenuse = the sum of the other two sides. 28. Ihe rectangle under the two segments of the hypotenuse of a right-angled triangle made by the point of contact of the mscnbed circle = the area of the triangle. Twice the circumscribed diameter = the sum of the three escribed radii diminished by the inscribed radius. The sum of the distances of the circumscribed centre from the sides of a triangle = the sum of the inscribed and circum- scribed radii ; and the sum of the distances of the orthocentre from the vertices = the sum of the inscribed and circumscribed diameters. (Camot's QiomUrie de Position, § 137 ) 31. Examine the case when the circumscribed centre and orthocentre are outside the triangle. 32. If Jj^iCi be the triangle formed by joining the escribed centres A ABC; A^B^C^ the triangle formed by joining the escribed ce-tres of a A,B,C,; A,B,C, the triangle formed by joining the escribed centres of a A,B,C, ; and this process of construction be continued, the successive triangles will approximate to an equilateral triangle. (Booth's New Geo metrical Methods., ro\. ii. p. 315.) 3». if an equilateral polygon be circumscribed about a oirole. it v/ill be equiangular if the number of sides be odd. iL'xamine the case when the number of sides is even. 34. AB, CD, two alternate sides of a regular polygon, are produced TO meet at E, and is the centre of the polygon. Prove A By G, concyclic, and also Z), E, By O. ' 29. 30. It »: iij m EUOLID's ELBMBINT8. [Book If. !i '-1 111 lis 1 36. The Bum of the i)eri>endicular8 on the sideB of a regular w-gou from any point iuaide = n times the radius of the inscribod circle. Examine the case when the point is outside. Loci, The base and the vertical angle of a triangle are given ; find th« locus of 1. The orthocentre of the triangle. 2. The centre of the inscribed circle. 3. The centres of the three escribed circles. 4. The oentroid of the triangle. b, ABC IB a. triangle, and E is any point in AG. Through E a straight line DEF is drawn cutting AB &t F and BO pro- duced at D; circles are circumscribed about i\»AEF, CDE. Find the Idcus of the other point of intersection of the circles. Q. AB and AG are two straight lines containing a fixed angle; and between AB sxidi AG there is moved a straight line DE of given length. The perpendiculars from D and E to AB and AG meet at P, and the perpendiculars from D and E to AGoMdiAB meet at O ; find the loci of O and P. 7. Given the vertical angle of a triangle, and the stun of the sides containing it ; find the locus of the centre of the 'circle cir- cumscribed about the triangle. 8. A circle is given, and in it are inscribed triangles, two of whose sides are respectively parallel to two fixed straight lines. Fmd the locus of the centres of the circles inscribed in these triangles. 9. A circle is given, and from any point P on another given con- centric circle of greater radius,' tangents are drawn touching the first circle at Q and i?; find the loci of the centres of the inscribed and circumscribed circles -of the triangle PQR. 10. A point is taken outside a square such that of the straight lines drawn from it to the vertices of the square, the two inner ones trisect the angle between the two outer ones ; show that the locus of the point is the O* of the circle circumscribe^^ fttMut the square. 2tl BOOK y. DEFINITIONS. 1. A less magnitude is said to be a submultiple of a greater magnitude, when the less measures the greater; that is, when the less is contained a certain number of times exactly in the greater. 2. A greater magnitude is said to be a multiple of a less, when the greater is measured by the less; that is, when the greater contains the less a certain number of times exactly. 3. Equimultiples of magnitudes are multiples that con- tain these magnitudes, respectively, the same number of times. 4. Ratio is a relation of two magnitudes of the same kind to one another, in respect of quantuplicity (a word which refers to the number of times or parts of a time that the one is contained in the other). The two magnitudes of a ratio are called its terms. The first term is called the antecedent ; the latter, the consequent. The ratio of ^ to ^ is usually expressed A : B. Of the two terms A and B, A is the antecedent, B the consequent. 5. If there be four magnitudes, such that if any equi- multiples whatsoever be taken of the firat and third, and any equimultiples whatsoever of the second and fourth,' and if, according as the multiple of the first is greater than the multiple of the second, equal to it, or less, so is the multiple of the imm greater than the multiple of the fourth, equal Mi PffpMlllilillMfai H i i! fll r 263 lUOLID'e TSL3MENT8. [Book ▼. to it, or less ;■ then the lirst of the magnitudes has to the second the same ratio that the third has to tlie fourth,, Cor. — Conversely, if the first of four magnitudes have tc the Becond the same ratio that the third has to the fourth, and if any equimultiples whatsoever be taken of the first and third, and any whatsoever of the second and fourth ; then according aa the multiple of the first is greater than the multiple of the second, equal to it, or less, the multiple of the third shall be greater than the multiple of the fourth, equal to it, or less. 6. Magnitudes are said to be proportionals when the first has the same ratio to the second that the third lias to the fourth; and the third to the fourth the same ratio which the fifth has to the sixth ; and so on, whatever be their number, i. When four magnitudes, A, B,C, D, are proportionals, it is usual to say that ^ is to ^ as C to D, and to write them thus — A:B::G:D,OT thus, A : B = : D. 7. In proportionals, the antecedent terms of the ratios are called homologous to one another ; so also are the con- sequents. 8. When four magnitudes are proportional, they consti- tute a proportion. The first and last terms of the proportion are called the extremes ; the second and third, the means. 9. When of the equimultiples^of four magnitudes, taken as in the fifth definition, the multiple of the first is greater than that of the second, but the nviiltiple of the third is not greater than the multiple of the fourth ; then the first has to the second a greater ratio than the third magnitude has to the fourth ; and the third has to the fourth a less ratio than the first has to the second. Cob. — Conversely, if the first of four magnitudes have to the second a £,veater ratio than the third has to the fourth, two numbers m and n may be found, such that, while m times the first magnitude Ul Book v.] DEFINITIONS. 263 isgreator than n times the second, m times the third sbUl not be jcreater than n times the fourth. ^^ 10. When there is any number of magnitudes greater than two of which the first has to the second the same ratio that the second has to the third, and the second to the third the same ratio which the third has to the fourth and so on, the magnitudes are said to be continual proportionals or in continued proportion. ' 11. When three magnitudes are in continued proportion the second is said to be a mean proportional between the other two. Three magnitudes in continued proportion are sometimes said to be m geometncal progression, and the mean proportional is iJxen CftUed a geometric mean between the other two. 12. When there is any number of magnitudes of the same kmd, the first is said to have to the last of them the ratio compounded of the ratio which the first has to the second, and of the ratio which the secowi has to the third and of the ratio which the third has to the fourth, and so on to the last magnitude. Thus : A ^!/A ^' %^. \ ^''''' magnitudes of the same kind, the ratio of ^ to i) 18 said to be compounded of the ratios of A to B, R to G i A • B ' * and G to D. This is expressed A iD=\ B :G I G.D 13 A ratio which is compounded of two equal ratios is said to be duplicate of either of tliese ratios. CoR.-If the three magnitudes A, B, and C are continual nrn portionals. the ratio of ^ to G is duphcate of that of i to J o' oi i A A Vll "^ *^' ^""^ ^^^°^*^°°' *^« '•^^io of A to G is com pounded of the ratios of ^ to B, and of B to G; but the ratio rf t^- T 1'**;' "^ ^ *' ^' ^"'^^"^^ ^' ^' ^ ^re continual pro port onals; therefore the ratio of A to G, by thi. definition 1 duplicate of the ratio of ^ to ^, or of ^ to (7 aennitiion, is 264 Euclid's elements. Tboa v. :< i\ 14 A ratio which is compounded of three equal ratios is said to be triplicate of any one of these ratios. tl,? ^r^i^Z "^^^""'^^^^^ ^' ^> O, D be continual proportionals, IJ f .f • /'o *^' '**'^ of ^ to i> is compounded of the thre^ 'o on?l .h 'k^ ^ ''• ^*° ^' ^"^ *^^«^ *^^- -ti- -« equal .0 one another because A, B, G, D are continual proportionals; there ore the ratio of ^ to i> is triplicate of the ratio of ^4 to B, «; 01 Bto 6, or of Gio D. The following technical words may be used to signify certain ways of changing either the order or the magnitude of the terms of a proportion, so that they continue stiU to be proportionals : 16. By alternation, when the first is to the third as the second is to the fourth. (V. 16.) 16. By inversion, when the second is to the first, as the fourth IS to the third. (V. A.) 17. By addition, when the sum of the first and the second IS to the second, as the sum of the third and the fourth is to the fourth. (V. 18.) 18. By subtraction, when the difference of the first and the second is to the second, as the difference of the third and the fourth is to the fourth. (V. 17.) 19. By equality, when there is any number of magni- tudes more than two, and as many others, so that they are proportionals when taken two and two of each rank, and It is mferred^that the first is to the last of the first rank of magnitudes, as the first is to the last of the others Of this there are the two following kinds, which arise from the diff-erent order in which the magnitudes are taken two and two : 20. By direct equality, when the first magnitude is to the second of the first rank, as the first to the second of ill f I' . W :'. the the Book v.] DEFINITIONS, AXIOMS, PROPOSITION 1. 260 the other rank ; and as the second is to the third of the first rank, so is the second to the third of the other: and so on in a direct order. (V. 22.) ' 21. By transverse equality, when the first magnitude is to the second of the first rank, as the last but one is to the last of the second rank; and as the second is to the third of the first rank, so is the last but two to the last but one of the second rank ; and as the third is to the fourth of the first rank, so is the last but three to the last but two of the second rank ; and so on in a transverse order. (V. 23 ) AXIOMS. 1. Equimultiples of the same, or of equal magnittidea are equal to one another. 2. Those magnitudes of which the same, or equal magni. tudes, are equimultiples, are equal to one another. 3. A multiple of a greater magnitude is greater than the same multiple of a le-s. 4. That magnitude of which a multiple is greater than the same multiple of another, is greater than that other magnitude. PJ^OPOSITION 1. Theoreik. // any number of magnitudes he equhiultiples of as r^^nij other,; each of each, what mulHph soever any on of the first is of its submultiple, the sane malH . :s the jmn of all the first of the sum of all the rest. Let any number of magnitudes A, B, and C be equi- riiultiples of as many others D, E. a>i.! F, each of each • ttv required to prove that A + B -^ Cis the same multiple f/i> -^E-^FthatA is ofD. 366 Euclid's elements. [8ook y. Let A contain D, B contain E, and G contain F, each any number of times, as, for instance, three times; then A = D + D + D. Similarly, B = E + U + E, and C = F + F + F; r.A + B+0=D + E+ ^ taken three times. /. Ax. 2 Hence also, if A, B, and C were each any other equimultiple of A ^, and i^, A + B+C would be the same multiple oiD + E + F. ^ Cor.— Hence, if m be any number, mD + mE + mF = m{D + E + F). ^:?i: PROPOSITION 2. Theorem. If to a multiple of a magnitude by any number, a multiple of the same magnitude by any number be added, the sum will be the same multiple of thnjt magnitude that the sum, of the two numbers is of unity. Let A = mC, and B = nC: it is required to prove A + B = (m + n) O. Since 4 = mC, A==C+C-\-C+ repeated m times. Similarly, 5=0+ C+ C+ repeated n times^ A4-B=C+C+C+ repeated w + w times, that is, A + B = (m + n)C; ' r. A + B contains C as often as there are units in w + n. Cor. 1.— If there be any number of multiples whatsoever as A ^^mE, B = nE, C=pE, then A+B+ (7= (m + n+p)E. Cor. 2.— Since A + B+G=(m+n+ p)E, and since A = mE, B ---- nE, and C ^ pE, ,\ mE + nE +pE ^ (m + n + p)E. !«* v.] PROFOSITIONS 1, 2, 3, 4. m PROPOSITIONS. Theorem. ^f^^^M't 0/ three magnitudes contain the second as often 607^^am Me //r.,r? as often as there are units in a certain number the first will contain the third as often It/Z are units in the product of these two number,. Let ^ = mB, and B = nC- it is required to prove A = mnO. Since B = nC, But nr"'f'.yf "-"^^"(^^ repeated m timea by \^:^f " '"V^^'-tin.e. = ^multiplied oy n + n + n+ repeated m times. V. 2 Cor •> "^rlr"' repeated ,„ time, = ^:. '^ But A = mB ; PROPOSITION 4. Theorem. /r anx, e^ W/^,>/,, be taken of the antecedents of a propor^ Let ^ :5 = C:D, and let m and n be any two numbers • It .. r.'^wzn^r/ /« prove mA:nB = mC: nD. """^'^^^^ • Of mA ana mC take equimultiples by any number t, • Then he equimultiples of mA and mC7 by « are eoui multiples also of A and C, for they contain' 1 a^I^ 7^ R M III!) \l\ EUOUD'S ELEHBNTl. [BookT. V.3 often as there are units in pm ; and they are equal to pmA and pmC. Similarly the multiples of nB and nD by q are qnB, qnD. Now since A : B = G : D, sm^ oi A and C there are taken any equimultiples pmA and pmO, and of £ and i> there are taken any equimultiples qnB, qnD ; a pmA he equal to, greater, or less than qnB, then pmC is equal to, greater, or less than qnD. V. Def. 5, Cor. 'RvXpmA, pm,G are also equimultiples of mA and mC by p; and qnB, qnD are also equimultiples of nB and nD by q ; .-. m^ : wi^ = mC : nD. V, Def. 5 H ,1 'M '1 •! 'ii I I I III ' iir iitf PROPOSITION 5. Theorem. Ifim magnitude be the same multiple of another, which a magnitude taken from the first is of a magnitude taken from the other, the remainder is the same multiple of the remainder that the whole is of the whole. Let A and B be two magnitudes of which A is greater than B, and let mA and mB be any equimultiples of them : it is required to prove that mA - mB is the same multiple ofA-B that mAisofA ; that is, that mA-mB = m(A - B), Let D be the excess of A above B; then A- B = D. Adding B to both, A = D + B; ir i .'. mA = mD + mB. ^ * Taking mB from both, mA - mB = mD. Now D^ A - B; .*. wA - mB = wM - B)> H ^ '■" '1 ll i- ^5 ▼.1 raoposmoNs 4, 6, 6, A. 269 PROPOSITION 6. Theorem. If from a multiple of a magnitude by any number a multiple of the same magnitude by a less number be taken away the remainder mil be the same multiple of that mag^l tude that the difference of the numbers is of unity ^ Let mA and ;^^ be multiples of the magnitude A by the numbers m and n, and let m be greater than ^ / ^ ^^ ^^^ tt ts required to prove mA-nA =^ (m- n) A, "i^tm- n = q; thenm = n+q; .'. mA = nA + qA. ' Taking nA from both, mA ~ nA = qA ; .'.mA - nA contains A as often as there are units in a that IS, as often as there are units inm^n; .'. mA^nA =^ (m - n) A. V 1 PROPOSITION A. Theorem. Th6 terms ofapropoHim are propoHional by inverHan. IjbiA.B = C.D: it is required to prove B : A = D : C. Let m^ and mC be any equimultiples of A and C nB and nD any equimultiples of B and D. Then, because A : B = C : D a mA be less than nB, mC will be less than nD; V. Def 6, Cbr .-. if 7i5 be greater than mA, nD will be greater than mC, For the same reason, if nB = mA, nD = mC, and if nB be less than mA, nD will be less than mC. But nB, nD are any equimultiples of B and D, and mA, mC are any equimultiples of A and 6 - i» m 270 suolid's elements. [BookT. PROPOSITION B. Theorem. If the first be the same multiple or suhmultiple of the second that the third is of the fourth^ the first is to the second as the third to the fourth. Let mA, mB be equimultiples of the magnitudes A and B : it is required to prove mA : A = mB : 5, and A : mA = B.mB: Of mA and mB take equimultiples by any number /», and of A and B take equimultiples by any number 2>; these will be nmA, pA, nmB, pB. V. 3 Now if nmA be greater than pA, nm is greater than p; and if nm is greater than p, nmB is greater than pB ; .'. when nm,A is greater than pA, nmB is greater than j? A Similarly, if nmA = pA, nmB = pB, and if n7nA is less than pA, nmB is less than pB. But nmA, nmB are any equimultiples of mA and mB, and pA, pB are any equimultiples of A and B; .-. mA:A = mB: B. V. Def 5 Again, since mA : A = mB : 5, .'. il : mA = B : mB, by inversion. V. A PROPOSITION C. Theorem. Ff the first term of a proportion be a multiple or a suh- multiple of the second, the third ia the same multiple or subimdtiple of the fourth. "Let A : B = C : D, and first let A = mB: it is required to prove C = mD. Of A and G take equimultiples by any number as 3, PROPOSITIONS B, 0, 7. 271 Book T.] and of B and D take equimultiples by the number 2m • these will be 2 A, 2C, 2mB, 2mD. V. 3 Now since A = niB, 2 A = 2mB ; and since A : B = C : D, .. 2C = 2mD; V. Def. 5 .-. C= mD. "^ Next let i4 be a submultiple of B : it 18 required to prove that C is the same mbmultiple of D Since^:i?=C:/>, ^^^ B : A = D: C,hj inversion. y^ ^ But A being a submultiple of B, B is a multiple of il ; .'. Z> is the same multiple of C; .*. C is the same submultiple of D that A is of B. PEOPOSITION 7. Theorem. Equal magnitudes have the same ratio to the same magni- tude; and the same has the same ratio to equal magni- tudes. Let A and B be equal magnitudes, and C any other : it is required to prove A : C == B : C, &nd C : A = C : B. Let mA, mB be any equimultiples of A and B, and nG any multiple of O. Because A = B, mA = mB ; v. Ax. 1 .-. if mA be greater than nC, mB is greater than nC ; and if mA = nC, mB = nC; and if mA be less than nC, mB is less than nC. But 7nA and mB are any equimultiples of A and B, and nC is any multiple of C; .'. A : O — B : 0. y j^^f g Hence also C . A -^ C \ B,hy inversion. v. A i.'r 27!8 KUOLID'a ELKMBNT8. [Book ▼. PROPOSITION 8. Theorem. 0/ unequal magnitudes, the greater has a greater ratio to any other magnitude than the less has; and the same magnitude has a greater ratio to the less of two magni- tudes than it has to the greater. Let ^ + ^ be a magnitude greater than A, and C a third magnitude : it is required to prove A + B : C greater than A : C, ^nd C : A greater than C : A + B. Let m be such a number that mA and mB are each of them greater than C, and let nC be the least multiple of that exceeds mA 4- mh ; then nC - C will be less than mA + mBj that is, (w - 1)(7 will be less than m{A + B)', .'. m{A + B) is greater than {n-\)C. But because nC is greater than mA -!- mB, and C is less than mB; nC- C is greater than mA, that is, mA is less than nC - C, oi {n - \)0. Hence the multiple oi A + Bhy m exceeds the multiple of C by n - 1, but the multiple of ^4 by m does not exceed the multiple of C by w - 1 , .*. A + B \ C\B greater than A : C, V. Def. 9 Again, because the multiple of C hy n-l exceeds the multiple of A by m, but does not exceed the multiple of A + Bhy m; .'. C : A is greater than C : A + B. V. DeJ 9 Book v.] PROPOSITIONS 8, 9, 10. 973 PROPOSITION 9. Theorems. Magnitudes which have the same ratio to the same magnitude are equal to one another ; and those to which the same magnitude has the sawAi ratio are equal to one another. First l^i A:G = B.C.- it is required to prove A = B. For if A be greater than B, then A \C\B greater than B:G. y. 8 And if ij be greater than A, then B : Cia greater than A : G. Hence A = B. Next let G'.A = C i B : it is required to prove A =B. For A:C ^ B'.Cfhy inversion ; .-. A = B, V. 8 V.A PROPOSITION 10. Theorems. Tfmt magnitude which has a greater ratio than anoth&r has to the same magnitude is the greater of the two ; and that magnitude to which the same has a greater ratio than it has to another magnitude is the less of the two. Let il : C be greater than" B : C: it is required to prove A greater than B. Because A-.Cis greater than B : (7, two numbers m and n may be found such that mA is greater than nC, and mB not greater than nC; V. Def 9, Gor. .'. mA is greater than mB, .'. A is greater than B. v. Ax. 4 Next \&iG'.B be greater than CiA: U is required to prove B less than A. ^m IMAGE EVALUATION TEST TARGET (MT-3) ^/ 1.0 i.l IM |25 2.2 1^ 1^ ■" 116 i ui Hi L25 III 1.4 Mm III 1.6 $% Ta ^M / M Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. MS80 (716) 872-4503 ^ 1 •>^ f\ :\ \ %^ ^\^\ '^ <* 4 ^ S74 Euclid's elements. [Book V, For two numbers m and n may be found such that nO is greater tlian mB, and nC not greater than mA ; .'. mB is Jess than mA/ .*. J9 is less than A. y^ ^^P^ 4 PROPOSITION 11. Theorem. y^a^/os that are equal to the same ratio are equal to one another. LetA:B= C: D and : D = E : F: ii in reqiiired to prove A \ B = E : F. Take mA, mC, mE any equimultiples of A, C, and E, and nB, nD, i nF any equimultiples of B, D, and F. Because A.B = O : D, if mA be greater than nB, mC must be greater than nD. v. Def. 5 Cor. But because C : D == E : F, \i mChQ greater than nD, mE must be greater than nF ; v. Def. 5 Cor .'. if m^ be greater than uB, mE is greater than nF.' Similarly, if mA = nB, mE = nF, and if mA be less than nB, mE is less than nF • .-. ^:^ = j;:i^. ' V.Def.b PROPOSITION 12. .Theorem. If any number of magnitudes he propm-tionals, as one of the antecedents is to its consequent, so iB the sum of all the antecedents to the sum of all the consequents. l^t A : B = G : D md C : D = E : F: it is required to prove A:B = A + C+E:B + D + F. Take mA, mC, mE any equimultiples of A, C, and E, «ld nB, nD, nF any equimultiples of B, D, and F. to one Book V.J PROPOSITIONS 10, 11, 12, 13. 275 Because^ : i? = C : A if mA be greater than nB, mC must be greater than nD ; y rJ k n and because C : D = E . F, when mC is greate; thfn 'i> "^ wJ5 IS greater than w P. ^ F. ^ f 5 A _ •• if mA be greater than nB, mA + mC + m£ 'is 4atTr than nB + nD •+ uF. t,reaier Similarly, if m^ = nB, mA + mC + mE = nB + nD + uF- TB^nlf::^:-' ^^^^ ^^^ -^^ --^--^ '« Ifss^hfn ^owmA + mC+mE=m{A + C+E)- v i n -a PROPOSITION 13. Theorem. If the first have to the second the same ratio which the third has to the fourth, lut th. third to the fourth agZ^ Let A : S = C : A but (7 : Z) greater than E:F- it IS required to prme A : B greater than E:F. ' Because O : D is greater than E : F, there are two numbera m and » such that mC is greater *!,.„ nD, but mE is not greater than nF ^n)l But because A :B = C: D, it mCis greater than «Z) "^^ »iil IS greater than bB,- . F Z) A ,^ .;. ».^ is greater than nB, and m£ is not greaterTh^; uf'- ••■ A : B IS gteatei Oxsa E : F. ir n7« 276 BUCLIO'S ELEMENTS. [Book V. 1! ,& PROPOSITION 14. Theorem. If the first term of a proportion be greater than the third, the second bhall he greater than the fourth ; and if equal, equal ; and if less, less. Let^:J5= C:D: it is required to prove that if A he greater than C, B is greai&r tlian D; if A = C, B ^ D; if A he less tfian C, B is less than D. First, let A be greater than C; then A'.B is greater than C : B, But A\B = q : D; .'. C'.Dia greater than C :B/ .'. B 18 greater than D. Similarly, it may be proved that ii A = C, B and if il be less than C, B is less than D. = D; V.8 Hyp. V. 13 V. 10 Mi i|i m PROPOSITION 15. Theorem. Magnitudes have the same ratio to one another which their equimultiples have. Let A and B be two magnitudes, and m any number : it is required to prove A:B = mA : mB. Because A: B = A-.B ; • V. 7 .'. A:B = A + A:B + B, ' F. 12 ^ 2A:2B. Again, since A:B = 2 A:2 B; .: A:B = A + 2A:B + 2B, V. 12 3A:3B; and so on for all the equimultiples of A and B. M ' [Book V. Book v.] PR0P0BITI0N8 14, 16, 16, 17. 277 the thirdy 1/ and if is greater \ B is less V. 8 Hyp. V. 13 V. 10 PROPOSITION 16. Theorem. The terms 0/ a proportion, if they be all of the mme kind are proportional by alternatim. * Let A :B = 0:D.' it is required to prove A : C == B : D. Take mA, mB any eqiiimultiples of A and B, nC, nD any equimultiples of C and D. A:B = mA: mB. A:B=, C'.D; C:D=mA:mB. Again, C:Z>= nC.nD; . . mA :mB = nC . ,D. Now if mA be greater than nC, mB is greater than nD- limA = »a, mB = nD; and if mA be less than nC ' mB IS less than nD; V M .'.AiC^B.D, V.Def.b and Then But F.16 Hyp. V. 11 V. 15 V. 11 \ich their mber : V. 7 V. 12 V. 12 PROPOSITION 17. Theorem. ne terms of a proportion are propoHional by subtractim. LetA + B:B=C+D:D: it is required to prove A : B = C : D. Take mA and nB any multiples of ^ and B by the numbers m and n ; and first let mA be greater than nB. lo each of these unequals add w5; then mA + mB is greater than mB + nB. /. 4^ 4 But mA+mB^ m{A + 5), , p. '^ ^^ and m5 + n^ = (m + n)B ; y 2 .'. m(ii + B) is greater thaa {m + »)5. TT 278 ■Kclid's elements. i [Book T. Now because A + B : B = G + D : D^ if m{A + B) be greater than (m + n)B, m{C + D) is greater than (m + n) D ; V. Def. 5, Cor. or mC + mD is greater than mD + nD ; that is, taking mD from both, mC is greater than nD. Hence when mA is greater than nB^ mC is greater than nD, Similarly it may be proved that if mA = nBj mC = nD; and if mA be less than nB, mC is less than nD; .'. A :B = C: D. V. Def. 5 Cor. — The proposition is equivalent to the following : IiA:B=C:D, then A - B : B = C - D . D. Hence also, on the same hypothesis, it may be proved that A - B:A = C- D:C; that A : A - B = C: C- D; and that B : A - B = D : C - D [If it be thought desirable, any one of these changes on the proportion A :B=C:D may be denoted by the word n^itracHan.] PROPOSITION 18. Theorem. The terms of a proportion are proportional by addition, JjetA-.B^C-.D: it is required to prove A + B : B — C + D : D. Take m{A + B) and nB any multiples oi A -^ B and B, First, let m be greater than n. •■ T^ecause ^4 + J5 is greater than B; m{A + B) is greater than nB. ' Similarly m{C + D) is greater than nD ; .'. when m is greater than n, m{A + B) is greater than nB^ and m{C + D) is greater than nD. Second, let m = w. In the same manner it may be proved that in this case m{A + -B) is greater than nJ?, and m((7 -f- D) greater than nD, 1 43, [Book ▼. >) is greater >e/ 5, Cor. nD. r than nD, ^ = nD; V, Def. 5 owing : D. oved J:C-D; iges on tho ubtracHon.} 279 iddition. BmdB, r than nBf this case T than nD, ■••* ▼•) PROPOSITIONS 17, 18. Third, let m be less than n. Then m{A + B) may be greater than nB, or may be equal to it, or may be less than it. First, let m{A + B) be greater than nB ; then niA + mB is greater than nB. Take mB, which is less than nB, from both ; .*. Tn^l is greater than nB - mB, or mA is greater than {n - m)B. But because ^ : 5 = C:D ; .'. if mil is greater than (n - m)^, mC is greater than (w - m)D, that is, mC is greater than nD - mD. Add mZ) to each of these unequals ; then mC + mD is greater than nD, that is, m{G + D) is greater than nD. If therefore m(A + B) is greater than w^, m(C -^ D) L greater than nD. In the same manner it may be proved that, if m{A + 5) = nB, m{C + D) = nD; if m{A + B) be less than nB, m(G + D) is less than nD Hence A + B.B = C+ D:D. y. Def. 5 Cor.-— Hence also, on the same hypothesis, it may be proved t^&t A + B:A = G+ D:C; that A: A + B=C:C+ D- and that B:A + B = D:C+D. [If it be thought desirable, any one of these changes on the pro- portion A:B=G'.Dui&yhe denoted by the word addition. The words addUton and subtraction, as being more significant of the operations performed on the terms of the proportion, have been substituted for composition (componendo) and division (dividendo) which are the translations of the words (*i;»,, and let C7 be less than ^1 .• it is required to pi'ove A - G : B - D = A : B. Because A:B = C: Df- A-.C = B. D, by alternation ; A - 0:0 = B - D.Dyhy subtraction ; A - C:B- D = C:D, hy alternation; A - 0:B^ D=^A:B. ffpp, V. 16 V. 17 V. 16 F. 11 PROPOSITION 20. Thborbm. > If there he three magnitudes, and other three, whichy taken two and two in direct order, have the same ratio; if the first he greater than the third, the fourth shall he greater than the sixth ; and if equal, equal ; and if less, less. Let A, B, O be three magnitudes, and D, E, F other three, such that A:B = DiE, and B:G = E\F: it is required to prove that if A he greater than C, D will he greater than F; if A = C,Dmll = F; if A he less than O, D will he less thin F, First, let A be greater than ; then -4 : -B is greater than : B, V. 8 But AiB = D'.E; Hyp, /. Z> : ^ is greater than 0:B, F. 13 [lookT ■Mk v.] ide taken the other, s whole to Hyp. V. 16 V. 17 V. 16 F. 11 PROPOSITIONS 19, 20, 21. Now 381 V.A V. 13 V. 10 Now B'.C = E.F; C '.B = F'.Eyhy inversion ; •*. D :E is greater than F-.E; .'. Z) is greater than ^. Second, let A - C ; then A:B = C.B. But A:B = D : E; C:B = D:E. C:B = F:E; D:E=F:E; D =. F. Third, let A be less than C; tiien C7 is greater than A ; and, as was shown in case first, C:B = F:E, and B:A = E:D. • by case first if C7 be greater than A, F i. greater thaii />. tliat IS, if A be less than C, D is less than F, * V.7 Hyp. v.n v.u K 9 chy taken "atio; if shall be id if less, F other D will be s than Cf V.8 Hyp. V. 13 PEOPOSITION 21. Theorem. // there be three magnitudes, and other three, which, taken two and two in transverse order, have the same ratio • If the first he greater than the third, the fourth shall be greater than the sixth; and if equal, equal; and if less, less. ' a ^ j Let A, B, C be three magnitudes, and D, E, F other three, such t^B.t A:B = E : F, and B :C = D - E • it is required to prove that if A he greater than C, D will be greater timn F; if A = C, D will = F; if A be less thrni O, D will be less than F. First, let A be greater than C; then A: Bis greater than C :B, f^^ mmmm 282 EUCUDS ELEMENTS. But A\B ^ E'.F; •• . EiFis greater than C : B. Now 5:0= D:E; .'. G : B = E : D, hy inversion ; . •• E'.FiB greater than E:D : .'. D is greater than F. Second, let A — 0/ then A'.B =^ C'.B. But A:B = E.F; .-. G:B = E:F, Now 0:5 = E:D; .'. E'.F=E:D; D = F. Third, let A be less than C ; then A: Bi& less than O : 5. But A:B = E:F; .'. E: Fisleas than C:B. Now C:B = E:D; .\ E :F 18 les8 them E:D , Z) is less than F, I Bcok V. Hyp. V. 13 Hyp. r. A K 13 v. 10 K 7 ^///>- Kll r. il r. 9 r. 8 Hyp. V. 13 F 13 v. 10 \ PROPOSITION 22. Theorem. If there he any number of magnitudes, and as many others, which taken two and two in direct order, have the same ratio ; the first shall have to the last of the first magni- tudes the same ratio which the first of the others has to the last. First, let there be three magnitudes A, B, O, and other three Z>, E, F, such that A : B = D :E,and B : C = E :F.' it is required to prove A: C ~ D : F. Of A and D take any equimultiples whatever mAy mD ^ I Book V. Hyp. F. 13 Hyp. V.X V. 13 v. 10 V. 7 Hyp. Kll r. il r. 9 r. 8 Hyp. V. 13 F. 13 'v. 10 983 Book v.] • PROPOSITIONS 21, 22, 23. of B and J: any whatever nB, nE; and of (7 and -P any ■;i^hatever gC, gf^. Because A: B = D. E; ffyp mAxnB = mD : nE. y 4 Similarly nBiqG = nE : qFj y 4 .'. according as m,A is greater than qC, equal to it> or less, mD is greater than qF, equal to it, or less, V 20 '' ^''C=D:F. yj^f^^ Second, let there be four magnitudes A, B (7, D ani other four E, F, O, H, such that A'.£ ^E:k B'.C=F'.C^, G'.D^G.H: it is required to prove a : D = E : H. Since A, B, C are three magnitudes, and E, F, G, other i;^>T'ee, which, taken two and two in direct order, have the same ratio, A:G = E'.G,hy the first case. But because C : D = G:H; ^''^ = ^'Hyhythe firat case, f^'jnilarly the demonstration may be extended to any number of magnitudes. Hyp. \y others, the same tt niagni- hers has nd other = E'.F: Ay mD PROPOSITION 23. Thborbm. Tf there be any number of magnitudes, and as many o^ets, lohich taken two and two in tramwerse order, have tht same ratio; the first shall have to the last of the first magnitudes the same ratio which the first of the othen, has to the last. First, let there be three magnitudes A, B, 0, and other three D, E, F, such that A: B =: E : F, m^ B .0 ^ D .IS it is i-aquired to prove A:C=^D:F, a 284 BUOLID's BLBMENT8. [Booll V. Of A, By and D take any oquimultiples mA, mB, mD ; Vid of 0, E, and F take any equimultiples nC, nE, nF. ^ . Because A\B ^ mA: mB^ V^ 1 ttnd E:F = nE: nF, V. 15 and because A : B = EiF; Hyp, mA -.mB = vE \nF. V. \) Again, because B.C = D:E; Hyp, mB.nC = niD'.vE; V. 4 .-. according as mA is greater than nC, equal to it, or lesj^ mD is greater than nF^ equal to it, or less ; F. 21 .'. A:C=^D:F V. Def. J Second, let there be four magnitudes Ay By C. D, and other four E, F, G, H, such that A : B = Lr : H, B'.G=^F\Gy C:D = E'.F: it is required to prove A : D - E:H. Since Ay B, C are three magnitudes, and Fy (?, H otner three, which, taken two and two in transverse order, have the same ratio, A:C = F'.Hyhy the first case. But because C.D = E:F; Hyp. A:D = E : Hhy the first case. Similarly the demonstration may be extended to any number of magnitudes. ' Cor. From this proposition and the preceding it may b« inferred that ratios which are compounded o* tqual ratios are equal to one another. tuad it has beet tiiown that A:C = I):F. /" Book V. , mD ; iiF. , F. 1"' V. 15 Hyp. V. \) Hyp. V. 4 , or lesa, F. 21 ; Del u H otner ler, have Byp. [ to any 5 it may aal ratios SMk v.] PROFOSlTlOMfi 23, 24, D. 285 PROPOSITION 24. Theorem If the first has to the second the same ratio tchieh the third has to the fourth, and the fifth to the second the same ratio tohich the sixth has to the fourth ; the first and fifth together shall have to the second the same ratio wmch the third and sixth together have to the fourth, let A: B = C:D, and E:B = F:D: H is required to pro"e A + E : B = C + F:D, Because E : B = F : D ; B :E = D : F, by inversion. But A'.B = C:D; A•.E = C '. Fy by direct equality ; .-. -4 + J5; : i^ = C + /\- i^, by addition. But again, E : B = F : D ; .-. A-{- E.B ^ C ■{■ F.D^hy direct equality. V.k Hyp. V. 22 V. 18 Hyp. V. 22 PROPOSITION D. Theorem. The terms of a proportion are proportional by addition and 8id>traction. Lct^ .B= CD: it is required to prove A + B :A - B = C + D : C - D. Because A - B . B = C - D : D, hy subtraction ; F. 17 B.A~B = D.C-D,hy inversion ; V. A But ^ + i?:^= a+ i>: A by addition; F. 18 .-. A + B:A -- B = C + D:C - D,hy direct equality. y 22 [Proposition 25 has been omitted, as being of little use.] 9SU WSP*^KM 286 BOOK YT. DEFINITIONS. i. BlmilM lectilinepl figures are those whicli ha*"* theii several angles equal, each to each, and the sides about the equal angles proportional. Of the two requisites for sunilarity amonr- figures, nameiy, equi- anKularity and proportionality of sides, it wiU be seen from VL 4, 5, that if two triangles possess the one, they also possess the other. In this respect triangles are imique. Hence, in order to prove t^o rectihneal figures (other than triangles) simihir, it must be shown that they possess both requisites. 2 When any proportion is stated among the sides of two similar figures, those pairs of sides which form ante- cedents or consequents of the ratios are called homologous sides. 3 Similar figures are said to be similarly described upon given straight lines when the given straight Imes ar-j homologous sides of the figures. 4 When two simUar figures have their homologous sides parallel and drawn in the samo direction, they are said to be similarly situated; when they have them parallel and drawn in opposite directions, they are said to be oppositely situated. 5 Triangles and parallelograms which have their sideo about two of their angles proportional in such a manner that a side of the first figure is to a side of the second, as the other side of the second is to the other side of the first, aw said to have these sidiw reciprocaiiy proijorticsa*. 't f! Book VI.J DEFINITIONS, PROPOSITION 1. m i"o theii ibout the ueiy, equi- m VL 4, 5, the other, prove two be shown sides of arm ante- imologous bed upon lines ax^ gous sides ire said to rallsl and oppositely iheir sideo a manner second, as ){ the first, 6. The altitude of a triangle is the perpendicular >n the case of medial section, a straight line might be cut in extreme and mean ratio both internally and externally ; but internal division only is generally implied by the phrase. PROPOSITION 1. Theorem. Triaviihn antf parallel oy rams of the same altitude are to one another as their bases. 'I H G B D L M Let As ABG, ACD, and ||™ EC, OF have the same altitude, namely, the perpendicular drawn from A to BD, or BD produced : it is required to prove BC : CD = A ABO : A ACDy and BC : CD = |p £C : ||'" CF. Produce BD both ways, and take any number of straight lines BG, GH, HK each = BC, /. 3 and DL, LM, any number of them, each = CD; I, 3 and join A with the points K, H^ G, JD, M. S88 BUCLID*8 EI^MENTS. rseok VL K H (f B D L M Because KH, HG, GB, BC are all equal, Const. '. As AKH, AHG, AGB, ABC are all equal. 7. 38 .*. whatever multiple the base A'Cis of the base BC, the same multiple is A AKC of A ABC. Similarly, whatever multiple the base CM is of the base CD, the same multiple is A ^Cilf of A ACD. And if the base KC be equal to, greater, or less than the base CM, A AKC will be equal to, greater, or less than A ACM. I. 38 Now since there are four magnitudes BC^ CD, A ABC, A ACD; and of BC and A ABC (the first and third) any equi- multiples whatever have been taken, namely, KC and A AKC, and of CD and A ACD (the second and fourth) any equi- multiples whatever have been taken, namely, CM and A ACM; and since it has beei? shown that if KC be equal to, greater, or less than CM, A AKC is equal to, greater, or less than A ACM; .'. BC:CD = A ABC: A ACD. V. Def. 5 Again, because BC:CD= A ABC : A ACD ; BC:CD=2AABC:2AACD 7.15,11 = ir £C : IP CF. I. 41 Cor. 1. — Triangles and parallelograms that have equal altitudes are to one another as their bases. Book VL] PROPOSITION 1. 289 Cor. 2.— Triangles and parallelograms that have equal bases are to one another as their altitudes. For each triangle or H"* may be converted into an equi- valent right-angled triangle or rectangle with base and altitude = its base and altitude ; and in-these latter figures the bases and altitudes may be interchanged. 1. If two triangles or ||«n» have the same ratio as their bases, they must have equal altitudes ; if they have the same ratio as their altitudes, they must have equal bases. 2. The rectangle contained by two straight lines is a mean pro- jtortional between their squares. 3. A, B, and C are three straight lines ; prove that A has to B the same ratio as the rectangle contained by A and C7 has to the rectangle contained by B and C. 4. A quadrilateral is such that the perpendiculars on a diagonal from the opposite vertices are equal. Show that the quadri- lateral can be divided into four equal triangles by straight hnes drawn from the middle point o^the diagonal. 6. .45 is II CD, and AD, BG are joined, intersecting at E : prove AE'.ED^BE: EC. ^ 6. Triangles ABC, DEF have L A = L D,&n6L AB = DE • prove t\ ABC . i^ DEF = AC : DF. 7. AD, BE, CF drawn from the vertices of a ^^Cto the opposite sides are concurrent at 0; prove BD \ DC =r l, AOB • [\ AOG GE '.EA=t. BOG: A BOA, AF : FB = ^ COA \ a COB. 8. ^is the middle point ot AD,a. median of c^ ABC ; BE ia joined and produced to meet AC at F. Prove CF='2 AF. 9. ABC is any triangle ; from BG and CA are cut off BD = one- fourth of BG, and GE = one-fourth of CA. If AD, BE intersect at O, prove that GO produced will divide AB into two segments in the ratio of 9 to 1. 10. Perpendiculars are drawn from any point within an equilateral triangle to the three sides. Prove that their sum is constant 11. Triangles and |i«w are to one another in the ratio compounded of the ratios of their bases and altitudes. m BUOLID^S ELEMENTS. [B«dk n PROPOSITION 2. Theorems. If a straight line be drawn parallel to one side of a triangle, it shall cut tlie other sides, or those sides produced '* propoHionally. Conversely : If the sides or the sides produced he cut pro- portionally, the straight line joining the points of section shall be parallel to the remaining side of the triangle. A A (1) Let DE be drawn || BC, one of the sides of A ABC: it is required to prove that BD : DA = CE : EA. Join BE, CD. Then A BDE = A CDE, being on the same base DE, and between the same parallels DE, BC ; I. 37 .-. A BDE : A ADE = A CDE : A ADE. V. 7 But A BDE : A ADE = BD : DA, VL 1 and A CDE : A ADE = CE : EA ; VL 1 BD : DA =^ CE : EA. V. 11 (2) Let BD \ DA = CE : EA, and DE be joined : it is required to prove DE \\ BC. Join BE, CD. = CE : EA, = A BDE : A ADE, = A CDE : A ADE; Because BD : DA and BD : DA and CE : EA Hyp. VI 1 VL 1 * This useful extension was introduced by Robert Simson. feooktt] WlOPOSlTION 2. m ase DE, I. 37 V.7 VI. 1 VI. 1 VI 1 F/. 1 .•. ^BDE'.AADE = ACDEiAADE; K 11 A ^Z)^ = A CDE. V. 9 Now these tri.'-ugles are on the same base DE and on the same side of il ; .-. DEi&\\BG. J 39 1. The straight Ime which joins the xniddle points of two sides of a triangle is || the third side. 2. The straight hne drawn through the middle point of one of the sides of a triangle and || another side will bisect the third side. 3. Any two straight lines cut by three parallel straight Unes are cut proportionally. (Euclid, Data, Prop. 38.) 4. Any straight line drawn || the parallel sides of a trapezium divides the non-parallel sides, or those sides produced proportionally. • 5. In the figures to the proposition, if DE be || BG, prove BA : AD — GA '. AjS, and conversely. 6. ABG is any angle, and P a given point within it ; draw through P a straight line terminated by BA, BG, and bisected at P. 7. In the base BG of a ABC any point D is taken, and DE, DF, drawn || AB, AG respectively, meet the other sides at E, F: prove A AFE a mean proportional between as FBD, EDO. Examine the case when D is taken in BG produced. 8. ABG, DBG are two triangles either on the same side, or on opposite sides of a common base BG ; from any point E in BG there are drawn EF, BO respectively || BA, BD, and meeting the other sides m F, G. Prove FO || AD. Examine the case when E is taken in BG produced. 9. ABG is any triangle ; D and E are points on AB and ^C such that DE is || BG ; BE and GD intersect at F. Prove that A ADF = A AEF, and that AF produced bisects BG. Examine also the cases when D and i^ are on A B a,nd AG produced. ProVe the following construction for trisectmg a straight line ABinG and H: On ^5 as diagonal construct a ||«» AGBD; bisect AG, BD in E and F. Join DE, FG cuttmg AB inG and H. 11. AB is a straight line, and G is any point in it ; find in AB produced a point D such that AD : DB = AG : GB. 10 m BUCLID^S ELEMENTS. [Book Vt PROPOSITION 3. Theorems. If the vertical angle of a triangle be bisected by a straight line which also cats the base, the internal segments of the base shall have to one another the same ratio as the other sides of the triangle have. Conversely : If the intei'nal segments of the base have to one another the same ratio as the other sides of the tnangle have, the straight line drawn from the vertex to the point of section shall bisect the vertical angle. (1) Let the vertical l BAC oi the A i4J5(7 be bisected by AD, which meets the base at D : it is required to prove that BD : DG = BA :AC, Through G draw GE \\ DA, and let GE meet BA produced sX E. Because DA and GE are parallel, . .-. L BAD = L AEG, and l DAG = u ACE. L DAG; L AGE; AE. GE, a side of the A BCE, But L BAD = .-. L AEG = AG = Because DA is I /. 31 7.29 Hyp. 7.6 BD:DG= BA'.AE; BD'.DG= BA.AG. VI. 2 V. 7 [Book Vt PROPOSITION 3. straight ments of lo as the ve to owe I tnangle X to the bisected /. 31 7.29 Hyp. 7.6 F7. 2 V. 7 Book Vi.} (2) Let BD:DG= BA : AC, and AD be joined : it is required to prove l BAD = l DAC. Through C draw CE \\ DA, and let CE meet BA produced at E. Because DA is |( CE, a side of the A BCE, .'. BD.DC = BA.AE. B\itBD:DC= BA'.AC; .'. BA.AE = BA: AC; AE = AC, and L AEC = L ACE. But because DA and CE are parallel, .-. L AEC = L BAD, and l ACE = l DAC; .'. L BAD = L DAC. 2dd 7 31 VI. 2 r. 11 V. 9 7 5 7 29 1. With the same figure and construction as in I. 10, prove that ^ JS is bisected. 2. If a straight line bisect both the base and the vertical angle of % triangle, the triangle must be isosceles. a The bisector of an angle of a triangle divides the triangle into two others, which are proportional to the sides of the bisected angle. 4. ABC \9 a. triangle whose base BG is bisected at Z>; z s ADB, ADC are bisected by DE, DF meeting AB, AG at E F Prove EF || BG. 6. Trisect a given straight hne. 6. Divide a given straight line mto parts which shall be to one another as 3 to 2. 7. Divide a given straight line into n equal parts. 8. The bisectors of the angles of a triangle are concurrent. 9. Express BD and DG (fig. to the proposition) in terms of o, b, c, the three sides of the triangle. 10. AB\Bti, diameter of a circle, GD a chord at right angles to It, and E any point in GD ; AE, BE produced cut the circle ' at i^and O. Prove that the quadrilateral GFDO has any two of its adjacent sides in the same ratio as the other two. 11. H is the middle point of BG (fig. to the proposition) : prove HG'.HD = BA+AG'.BA-AO. 12. The straight hues which trisect an angle of a triangle do not trisect the opposite side. 3M mrotiD^s mMumfB. riook VI I PROPOSITION A.* Thboremp. If the exterior vertical angle of a triangle be bisected by a straight line which also cuts the base produced, the external segments of the base shall have to one another the same ratio as the other sides of the triangle have. Conversely : If the external segments of the base have to one another the same ratio as the other sides of the triangle have, the straight line drawn from the vertex to tha point of section shall bisect the exterior vertical angle. (1) Let the exterior vertical l CAF of the A ABC be bisected by AD, which meets the base produced at D: it is required to prove that BD : DO = BA : AC. Through C draw CE \\ DA, and let CE meet BA at E. Because DA and CE are parallel, ■ .-. L FAD = L AEC, and l DAC = L ACE. But L FAD = L DAC; .'. L AEC = L ACE; AC = AE. Because DA is || CE, a side of the A BCE, /. 31 I. 29 Hyp. J. 6 ! 1 * Assumed in PappuSi VII. 39, second proof. ■«* TI.] PROPOSITION A. BD:DO= BA: AE; BD '. DG = BA '. AC. (2) Let BD '. DC = BA '. AC, and AD be joined : it is required to prove l FAD = l DAC. Through C draw CE || DA, and let CE meet J9il at E. Because DA is || CE, a side of the A ^GS; .-. BDiDC = BA :AE. But BD: DC = BA.AC; .-. BA:AE = BA: AC; AE = ^C and L AEC = z. ACE. But because Z)i4 and CE are parallel, .*. -L AEC = £. ^^Z>, and l ACE = l DAC; .'. L FAD = L DAC. 295 VI. 2 F. 7 /.81 VI. 2 Hyp. V. 11 V. 9 /. 6 7.29 i. What does the proposition become when the triangle is isosceles ? 2. The bisector of the vertical angle of a triangle, and the bisectors of the exterior angles below the base, are concurrent. 3. Express BD and DC (fig. to the proposition) in terms of a, b, c, the three sides of the triangle. 4. Prove the tenth deduction from VI. 3 when JS is taken in CD produced. 6. P is any point in the 0«» of the circle of which AB ia a, diameter; PO, PD drawn on opposite sides of AP, and making equal angles with it, meet AB sA, G and D. Prove AG:GB = AD:DB. »). ^5 is a straight line, and G is any point in it; find in ^5 produced a point D such that AD : DB = AG : GB. 1. Prove the proposition by cutting oflF from BA produced, AM — AG, and joining DE. 8. If in any a ABG there be inscribed a t\XYZ {X being on BG, Y onCA, Z on AB), such that every two of its sides make equal angles with that side of t, ABG on which they meet, th«n AX, BY, GZ are respectively ± BG, GA, AB. Examine the case when X and Y are on 5(7 and .4 C produced. 296 BUOLIDS ELEMENTS. Oxdi VL • PROPOSITION 4. Theorem. If two triangles he mutually equiangulaVy they shall be similar , those sides being homologous which are opposite to e^ual angles.* In As ABC, DCE, let l ABC = l DCE, l BCA L CED, L BAC= L CDE: it is required to prove As ABC, DCE similar. Place A DCE so that CE may be contiguous and in the same straight line with it. Because l ABC + l ACB is less than 2 rt. and L ACB = l DEC; L ABC + L DEC is less than 2 rt. . •. BA and ED if produced will meet. Let them be produced and meet at F. Because l DCE = l ABC, .'. BFi%\\ CD; and because l BCA = l CED, .-. AC\b\\FE; .-. i^^CZ^isaJI-"; ••. AF = CD, and AC = FD. I 34 * This theorem is usuallj attributed to Thales (640-546 5.C.). to BC, I. 22 La; /. 17 Hyp. La; L 29, Cor. Hyp. I. 28 Hyp. i.28 Book VI.] PROPOSITION 4. 297 Now because i4C is || FE, a side of the A FBE, .♦. BA-.AF = BC: CE; VI 2 .-. BA : CD = BG : CE; V. 7 .-. BA.BC = CD: CE, by altornation. V. 16 Again, because CZ> is || BF, a side of the A i^5^, .-. i5C' : CE = FD : Z)^; , VI. 2 .-. BC.CE = AC'.DE; V. ? . .-. BC : CA = CE : DE, by alternation. F. 16 Lastly, because AB, BC, CA are three magnitudes, and DC, CE, ED other three ; and since it has been proved that AB : BO = DC : CE, and BC.CA = CEiED; .'. ABiAC = DC: DE, by direct equality. V. 22 Hence As ABC, DCE are similar. VI. Def. 1 1. From a given triangle another is cut ofiF by a parallel to the base ; prove the two triangles similar. 2. Two right-angled triangles are similar if an acute angle of the one be equal to an ao ite angle of the other. 3. Two isosceles triangles are similar if their vertical angles are equal. 4 ABCD is a rhombus ; through D a straight line is drawn so as to cut BA and BC produced at E and F. Prove L& BAD, DGF similar. A, Two chords AG, BD of a circle ABC intersect at E, either within or without the circle ; prove as AEB, CED similar, and also as A ED, BEC. 6. The straight line which joins the middle points of two sides of a triangle is half of the third side. 7. A straight line which is || one of the sides of a triangle and = half of it must bisect each of the other sides. 8. If one of the two j)aiallel si*les of a trapezium be double of the other, the diagonals intersect at a point of trisection. 9. In mutually equiangular triangles the perpendiculars drawn from corresponding vertices to the opposite sides are proportional to those sides. 10, The median to the base of a triangle bisects all the parallelfl to the base intercepted by the sides. ■: 398 BUOLID 8 BLBMBNTS. [Book ft 11. Three straight lines AB, AO^ AD Me drawn through one point At and are out by two parallels at the points E, F, Q and, B, C, D respectively : prove BC : CD = EF : FO. 12. Hence devise a method of dividing a given straight line into any number of equal parts. 13. Prove the proposition from VI. 2, by auperposing the one triangle on the other. PROPOSITION 5. Theorem. ^ two triangles have the sides taken in order about tuch of their angles proportional^ they shall he similar, those angles being equal which are opposite to the homologous tides. D \ / V G In As ABC, DEF, let AB . BC ^ DE -. EF, BC : CA .= EF : FD, QXidi BA: AC = ED : DF : it is required to prove As ABC, DEF similar. At E make l FEG = l ABC >A at F make l EFG = L ACB. Then l G = L A, and A ABC is equiangular to A GEF , .'. AB:BC= GE: EF. But AB:BG= DE: EF; .'. DE.EF = GE: EF,' DE = GE. I. 23 /. 32, Car. 1 VI. 4 Hyp. V 11 I 1 r Book VI.] PROPOSITIONS 6, 6. 399 Similarly, DF = OF. Now As DEF, OFF have the throe sides of the one respectively equal to tho throe aides of the other; .-. they are mutually equiangular. ' /^ g But A ABC is equiangular to A OFF; .-. A ABC is equiangular to A DFF. Hence As ABC, DFF are similar. yi D^f i 1. What is the analogous propoaition in the Firrt Book provinc the equality of two triangles ? «■ o 2. The triangle formed by joining the middle points of the sides of another triangle is similar to that other. a Prove the proposition from the following construction • From ^^ cut offAO = DE, and through G draw GU || BC, meetimr AC aX H. " VI. 4 Hyp. V 11 v:9 PROPOSITION 6. Theorem. If two triangles have one angle of the one equal to one, angle of the other, and the ddea about these angles pro- pcrtional, they shall be similar, those angles being equal which are opposite to the homologous sides. = ^Ed'df^' ^^^' ^'* "■ ^^^ = ^ ^^^' ««d BA'.AG it is re. ( ED = GD Now in As EDF, GDF, ] D^' = DF ( L EDF = L GDF; .-. L E= L G, and l DFE = l DFG. But z. J? = z. Gf, and l AGB = l DFG; .'. L B = L E, and l AGB = ^ ZJjF^fi'. Hence As ABC, DEF oxe similar. VI. I. 32, Cor. J F/. 4 F. 11 K9 Const. LA Def.l m i 1. What is the analogous proposition in the First Book proving the equality of two triangles ? 2. Prove the proposition with the same construction as in the third deduction from VI. 6. 3. ABC is a triangle, and the perpendicular AD drawn from A to BC falls within the triangle. Prove that it AD ia a. mean proportional between BD and DC, L BAC is right, and that if ^£ is a mean proportional between BC and BD, L BAG is right. 4. ABi^a. straight line, D and E two points on it ; 2)/^ and EO are parallel, and proportional to ^JD and AE. Prove A, Ff and O to be in one straight line. 5. AB is divided internally at G and D so that AB :AG s= AG : AD. From A any other straight line AE is drawn bAC. Prove t,^ ABE, A ED similar, and that EC bieeoti I BMID, look vij PROPOSITIONS 6, 7. 301 PROPOSITION 7. Theorem. If two triangles have two sides of the one proportional to two ^des of the oth^r, and the angles opposite to one pair of homologous ddes equal, the angles opposite to the other pair of homologcms sides shall he either equal or supplementari/. . A A E F G In As ABC, DEF, I'di BA\AC= ED = L E: it is required to prove either L C = l F = 2 rt. L8. (1) L Aie either = z. Z), or not. If L A ^ L Df then since l B ~ l E .'. L C=lF, (2) If ^ ^ is not = L D, at D make l EDG = l A; and, if necessary, produce EF io meet DG. Because l B ^ l DEG, and L A =. L EDG; .'. A ABC is equianj^ular to A DEG • .'. BA'.AC = ED'.DG. But BA : AC = ED : DF ; .-. ED.DF= ED.DG; DF - DG ; E F : DF, and l B L DFG = lQ, or L G ■{■ L F Hyp. /. 32, Cor. 1 /. 23 Hyp, Const. I. 32, Car. 1 VI. 4 Hyp. V. 11 .K9 /.5 302 Euclid's elements. TBook YL /. 13 Now L DFE ia supplementary to l DFG; c DFE is supplementary to l G; L. Z>i?'J^ is supplementary to l G. Note. — See the note appended to I. A., p, 62. 1. What is the analogous proposition in the First Book proving, under certain conditions, the equality of two triangles ? 2. ABG is a triangle, and ^D is drawn A. BC. If £C : CA = AB : AD, then a ABG is right-angled. PROPOSITION 8. Theorem. In a right-angled triangle, if a perpendicular be drawn from the right angle to the hypotenuse, the triangles on each side of it are similar to the whole triangle and to one another. Let A ABC be right-angled at A, and let AD be drawn perpendicular to the hypotenuse BC: it is required to prove £\s DBA and D AC similar to A ABCy and to one another. ^ f L ADB = L CAB In As DBA, ABC, | l B = l B- .'. these triangles are mutually equiangular; 7. 32, Cor. 1 .'. they are similar. VI. 4 In the same way, As D^Cand ABC may he proved similar. Now since As DBA and DAC are similar to A ABC, they are similar to one another. Book VL] PROPOSITIONS 7, 8. 303 Cor.— From the similarity of As DBA, DAC \i foUows *hat BD:DA== AD: DG. (i) From the similarity of As ABC, DBA it foUows that CB:BA = AB: BD. (2) From the simUarity of As ABG, DAC it foUows that BC:CA = AC.CD, (31 and BC : BA ^ AC : AD. U) These results expressed m words are : (1) The perpendicular from the right angle on the hypo- tenuse is a mean proportional between the two segments into which it divides the hypotenuse. (2) and (3) Either of the sides is a mean proportional between the hypotenuse and its projection on the hypo- tenuse. ^ (4) The hypotenuse is to either side as'the other side is to the perpendicular. 1. If from any point in the O «• of a circle a perpendicular be drawn to any radius, and a tangent from the same point to meet the radius produced, the radius will be a mean proportional between the segments intercepted between the centre and the points of concourse. 2. That part of a tangent to a circle intercepted by tangents at the extremities of any diameter is divided at the point of contact so that the radius is a mean prop<,rtional between the sec- ments. ° & Prove BD:DG = dupUcate of BA-.AG. 4. ABC is a triangle ; AD and A E are drawn to the base j?(7 so as to make i a ADB, AEG each = the vertical ^ BAG- prove (1) BD'.AD = AE: GE, (2) CB -. BA =. AB ■ BD (3) BG :GA =AG:GE, (4) BG : BA ^ AG '. AE. Draw figures for the cases when i BAG ia acute and obtuse and deduce from this theorem the results given in the Cor. to the proposition. 6. Examine the converses of the results (1), (2), (3), (4) of the Cor. to the proposition, and of the preceding dtduction. II 304 EUCLID S ELEMENTS. [Book VL PROPOSITION 9. Problem. From a given straight line to cut off any aliquot part, o A F B Let AB be the given straight line : , it is required to cut off from AB any aliquot part. From A draw AC, making any angle with AB; in AC take any point D; and from AC cut off AE, containing AD as many times afl AB contains the part required. /. 3 Join EB, and through D draw DF \\ EB. I. 31 AF is the part required. Because DF is || EB, a side of A ABE, ED'.DA = BF.FA; EA :DA = BA: FA, by addition. But EA contains DA a certain n'umber of times ; .*. BA contains FA the same number of times. VI. 2 V. 18 V.C 1. Which proposition in the First Book is a particular case of this ? 2. Trisect a given straight Une. 3. Show how to find three-fifths of a given straight line. 4. From a given triangle or ||™ cut off any aliquot part. 6. Show how to find four-sevenths of a given ||°*. [Book VL part. times afl /. 3 /. 31 luired. F/. 2 V. 18 V.C se of this ? ^J i»ROP08ITIONS 9, 10. 90d PROPOSITION 10. Problem. ' lb divide a given straight line internally and externally a given ratio.* ^n ^ Let AB be the given straight line, K:L the given ratio : tt 18 required to divide AB internally and externally in m ratio K : L. Draw a straight line AE making an angle with AB ; . ut off AF = K, and FG, FH on opposite sides of F, each Join BG, BH; ttid through F draw FC \\ BG, and FD \\ BH, meeting AB Dioduced at D. C and D are the required points. Because FC is || BG, a side of the A ^^G^ ACiCB^ AFiFG, ' Again, because /7) is || BH, a side of the A ^^jy AD.DB^AF.FH, == K:L. VI. 2 r. 11 • • F/. 2 Kll 1. AB and ^C are two straight lines, and AG is divided inter- naJly at the points D and E. Divide AB similarly to AG. ♦This proposition has bMn ins«rt«d instead of Euclid's i.>tb «bi.K is given as tii« first dednolien. * f ' 306 euolid's elements. iBook VL 2. Make the figure and prove the proposition when K is less than L. What becomes of the external section when K — L- 3. Divide a given triangle or H'" into two parts which shall have to each other a given ratio. 4. Given two points on the C" of a circle, to find a third point im the o°* siich that the ratio of its distances from the two given points may be equal to a given ratio. PROPOSITION 11. Problem. To find a third proportional to two given straight lines. A B D Let AB, AChQ the two given straight lines : if is reqiiired to find a third proportional to AB, AG, Place AB, AC so as to contain any angle ; produce AB, AC, making BD = AC; I. S join BC, and through D draw DE \\ BC. L 31 CE is the third proportional. Because BC is || DE, a side of A^ADE^ AB:BD = AC'.CE; VL 2 AB'.AC = AC: CE, since BD = AC. V. 7 1. Does the magnitude of the third proportional to two straight lines depend on the order in which the straight lines are taken ? How many third proportionals can be found to two straight lines ? 2. To AB and AC obtain the third proportional measured from A, 3. By VI. 8, Cor., find a third proportional to two straight lines in VTTvr urviicx rratjsst •ttok VI.] PROPOSITIONS 11, 12. 307 4. ^5 and AC are two straight lines drawn from A. Produce " "^ttoXnt^^^^^^^^^^^ ^'- ^ *« '^' ^ third proportional 6. Use the fourth deduction from VI. 8 for the same pun)ose. PROPOSITION 12. Problem. Tojind a fourth proportional to three given straight lines. iF Let A, B, C be the three given straight lines • tt IS required h find a fourth proportional to A, ' B, C. Take two straight lines DE, DF containing Iny ande • W tl^se cut off DG = A,GE=. B, and DhIc^i 3 join GH, and through E draw EF \\ GH. ' /'^^ Because GE is „ EF, a sMfc^ A^j^' '"'"'^^^^• DG:GE=:DH:HF; * ^. ^ A:B = C:HF y ^^ 1. Which prarious proposition is a particular case of this' X R^ "*' '^■v,^.""*'.? *"" '"""^ proportional measured from S *. By a method similar to that of th« f™,-*^ j.j."" . ".?• _ find a foa^ proportional to three-g^vor;;;;:;^ IZ "' "' 308 Euclid's ELBMKNTa ' [Book VL 5. Given a triangle or ||°> ; construct another triangle or ||°» which shall have to it a given ratio. 6. AB and AC are two straight lines, and Z) is a point between them. Draw through I) a atraij^'ht line such that the partt* of it intercepted between D and the two given straight liner may be in a given ratio. PEOPOSITION 13. Problem. To find a mean proportional between tico given straight lines. Let ABj BO be the two given straight lines : it is required to find a mean proportional between ABj BC. Place ABy BC in the same straight line, and on ^C describe the semicircle ADC; I. 10 from B draw BD ± AC. L 11 BD is the mean proportional. Join AD, CD. Then A ADC is right-angled, and AC is the hypoten- use ; ///. 31 . • . BD is a mean prop or lion ^ between AB,BC. VI. 8, Cor. 1. If the given straight lines were AG, BG, placed as in the figure to the proposition, show how to find a mean proportional between them. I 2. To find a mean proportional between AB, 5C placed as in the figure to the proposition. Describe any circle passing through A and C; join B to the centre 0, and diftW DUE L OB, meeting the o<=« at D and E. BD or BE\& the mean pro- portional. [Book VL T ||°» which nt between ; the partty raight liner ight lines. Book vil PROPOSITIONS 13, 14. 309 ^' '^''^LTT P~P«^«°»^ ^«*^««« ^C-. BC, placed aa in the hgure to the proposition. Describe any segment of a circle 'cD bntt^' ''^'' -''''' *"«^« ^" *^« segment. andlSn o^y. ox/ 18 the mean proportional. 4. Half the sum of two straight lines is greater than the mean proportional between them ^ ^^l^nf^" !S^'° ^^ theside^^of a rABCD; DEmeetn BO produced m F. Prove a ^^^ a mean proportional between a s AED and ^J?/' proportional ^ \hTf T °^ J^' P"'"'"' °^ '^^^« » «»«»« proportional, what numbers of mean proportionals could be found between two given s raight lines so as to form a continued prop^^on" nuXra'' ^ "'"^ expression which will include all theae AB, BC. I. 10 /. 11 itional. hypoten- ///. 31 7. 8, Cor. 1 the figure proportional id as in the :ng through BE L OB, mean pro- PROPOSITION 14. Theorems. Equal parallelograms, which have me angle of the one equal to me angle of the other, have their sides about the equal angles reciprocally proportional. Cmversely: Parallelograms which have one angle of the me equal to one angle of the other, and the sides about the equal angles redjprocally proportional, are equal A ^3?r>^* ^^ ^"^^ ^^ ^^ ^^^^ II" leaving L DBF = L QBE: a is required to prove that DB : BE = GB : BF. Plaee the jj- so that DB and BE may be in ob« «tmiaU hne. °~" 1 dlO I . '9 II euolid's blembniv. F [Book WL Then since l GBE = l DBF; .'. L QBE + L FBE = L DBF + l FBE, = 2 rt. z. s ; .*. GB and BF are in one straight line. Complete the |r FE. Because r AB = "'BC, ••• r AB • m FE == a 'BG a ' FE. But "» AB . m FE = DB BE, and "" BG m i!S = GB BF; • • • DB 5jB; = GB. BF. (2) Let L DBF = l GBE, and DB : BE it is required to prove |p AB = \^ BC. Make the same construction as before. Because DB and DB and GB rAB BE = BE = BF = II" FE = ir ^5 = GB: !«" ^5 : ir^c: ir 5G : ir BG. BF, FE, FE; FE; Hyp, I. 13 /. 14 Hyp. V.I. VI. 1 VI. 1 7.11 GB'.BF: Hyp. VL 1 VI. 1 F. 11 r. 9 1. Prove the proposition by joining EF and DO, and using the fifth deduction from VI. 2. 2. Prove AD, CO and the diagonal of the ||"" FE drawn through B concurrent. 3. Prove AG || EF. 4. Equal rectangles have their bases and altitudes reciprocal^ proportional, and conversely, f^ Equal II'"* that have their sidea reciprocally proportional ai« mutually equiangular. ^.1 FfiOPosinoNs 14. 16. 8U PROPOSITION 15. Thjoremb. Equal triangles which have one angle of the one eqttal to one angle of the other have their sides about the equal angles reciprocally proportional. Conversely : Triangles which have one angle of the me equa! to one angle of the others and their sides about th« equnX angles reciprocally proportional^ are equal (1) Let BACy DAE he equal triangles having L BAC = L DAE: it is required to prove that AC'. AD - AE : AB. Place the triangles so that AG and AD may be m one straight line. Then since l DAE = l BAC, .-. L DAE + L BAD = L BAC + l BAD, = 2 rt. z. s ; .•. EA and AB are in one straight line. Join BD. Because A BAG = A DAE, Hyp, I. 13 .\ A BAC : A BAD = A DAE But A BAG: A BAD = AG and A DAE : A BAD = AE 9 • AG AD = AE A BAD. AD, AB; AB. Hyp. V. y VI. 1 V. n 9ii SUCLIDS ELlfilUBNfS. jlOOB VI (2) L* L BAC = L DAE, and AG'. AD = AE.ABi it is required to prove A BAG = A DAE. Make the same construction as before. Because AG and AG and AE .-. A BAG AD = AE : AB, Hyp. AD = A BAC : A BAD, VI. 1 AB == A DAE : A BAD; VI. 1 A BAD = A DAE : A BAD; V. 11 A BAG = A i)il^. F. 9 1. Gould this proposition have been inferred from VI. 14 ? 2. Prove the proposition by joining GE, and using the fifth deduo- tion from VI. 2. 3. If in the figure to VL 14, AB and EO be joined, what modifica* tion of this proposition should we be enabled to prove ? 1. If A ABC is right-angled at B, and BD, the perpendicular on A C, is produced to ^ so that DE is a third proportional to . BD and DC, h ADE = t. BDp. 5. Equal triangles which have the sides about one pair of angles reciprocally proportional have those angles either equal or supplementary. 6. If, in the fig. to VI. 8, BE be drawn ± BA, and meet AD pro- duced at E, then A ABD = A EGD. '. Find a point in a side of a triangle, from which two straight lines drawn, one to the opposite angle, and the other || the base, shall cut off towards the vertex and towards the base, equal triangles. BUamine the cm9 for a point in a side produced. iMk VLl PROPOSITIONS 16, la M PROPOSITION 16. Thborems. If f^yur straight Im^a be proportional, the rectangle contained by the extrerrtes is equal to the rectangle contained by the means. Conversely .- If the rectangle contained by the extremes be efjual tu the rectangle contained by the means, the fintr 9tratght lines art proportional. E< — F Jb O Jd <1) Let AB'.CD^ EF: GH .- tt is required to prove AB • GH = CD . lilF, From A draw AK ± Ali, and = GH , from C draw CL A. CD, and = EF; and complete tlie rectangles KB, LD. Because AB : CD = EF' . GH and CL = EF, and AK - GH ; '•' AB : CD = CL : AK,' that IS, the sides about the equal angles of the ir KB LD are reciprocally proportional. ' *' ^^ ~ LD; • rrj j j .-. AB.AK= CD.CL; ,\ AB.GH= CDEF. (2) Let AB-GH = CD- EF: it is reauired in nrmie A « . nn _ ^v xvr. I. 11, 3 /. 11, 3 7.31 Hyp, Const. V. 7 su I - G. Euclid's elements. T [Book VK Lr T t Al- Ol- D Make the same construction as before. Because AB - GH = CD - EF, ^d AK = GH, and CL = EF ; AB • AK =^ CD . CL, that is, the \\"" KB, LD which have l A .'. AB:CD= CL'.AK; .-. AB'.CD^ EF.GH. Hyp, Cansi, L C; are equal. VI. 14 V.I ,. In the figure to VI. 8, prove (1) BD - DG = AD^, (2) CB • BD = AB^, i3) BG' CD = AG\ (4) BG • AD = BA - AG. 9. Using the results (2) and (S) of the preceding deduction, prove 1.47. 3. Show that these results are established in Euclid's proof of 1. 47. 4. Two chords AG, BD of a circle ABG intersect at E, either within or without the circle ; prove AE • EG = BE • ED, 6. In the figure to the fourth deduction from VI. 8, prove (I) BDGE=AD- AE, (2) GB . BD = AB?, (3) BG'CE=^ AG% (4) BG • AE = BA . AG. Using the results (2) and (3) oi the preceding deduction, show that when a BAG ia acute, AB^ + AG^ is greater than BC" by BG- DE; when i BAG is .obtuse, AB^ + AG^ is less thAvi BG"^ hy BG • DE. 7. What becomes of the rectangle BG • DE when L BAG Sb right? o. <*ive another proof of III. 35 and its Cor. 9 A square is inscribed in a right-angled triangle, one side of the square coinciding with the hypotenuse ; prove that the area of the square = the rectangle contained by tC = BD, (2) Let ^.(7 = B\- it is required to prove A :B = B'.O. Make the same construction as before. Because A - C = B^, A'C = B.D; A:B = D.G; A:B = BiG. .*. Byp VLU Hyp. VL 16 F. 7 1. Of which proposition is this merely a particalar case ? T-f 5*.^ '^'^''^^^ ^^'^'^ *^^^"^^^ "^ ^^t'-eme and mean ratio is divided m medial section, and conversely. 3. From 5, one of the vertices of |!>" ABCD, a straight line is drawn cuttinij the diagonal Jtl af a; nn of jt- j ^ r. . , , . ^ - — J — ""^ ^ » and »ix/ pruaucea Ai M6 Euclid's elements. rsook VI. PROPOSITION 18. Problem. On a (jiven strnight line to describe a rectilineal figure lohich shall be similar to a givev. rectilir^al figure * Let AB U the given straight line, CDEF the fjiven rectilineal iigure : it is required to describe en AB a rectilineal figure which shall be similar to CDEF. meet at O. Join 0^. Oi?; and let AG, BH drawn respectively || Ci^, i>^ meet OF, OE at C, ^. Join GH. 3. If on BA,BG BH, or on these lines produced, there be taken points L, M, N, such that BL . BA = BM : BG = BN - BH tgxxre^BAin!^^ '' ''"""^ ^""^ '''""^'^^ '^*"**'^ *'° *^^ 1 How cmUd a figure 5ZJM;i\r similar and oppositely situated tc the figure ^ J G^irbs obtained? ^^ J ^ 318 EUCUD S ELEMBNTS. [Book yi PROPOSITION 19. Theorem. Similar triangles are to one another in the duplicatb raid of their homologous sides. , E A Let ABC and DEF be similar triangles, having l B ^ L JEj and l G = l F, so that BC and FF are homologous sides : it is required to prove A ABC : A DFF = duplicate ^i BG : EF. Take BG a third proportional to BG and EF^ 80 that BC:EF = EF: BG; VL il and join AG. Because AB : BG = DE : EF, Hyn. AB.DE = BG: EF, by alternation, V. iC» = EF:BG; V. M that is, the sides of As ABG, DEF about their equal angles B and E are reciprocally proportiolial. .-. A ABG = A DEF. VI. 15 Again, because BG : EF = EF : BG, Const. BG : BG = duplicate of BG : EF. V. Def. 13, Gm, But BG: BG = A ABC : A ABG, VI 1 .•. A ABC: A ABG = duplicate of BG : EF; V. 1 1 .«. A il.BC : A DEF = duplicate of BG .EF. VI Book VI, lib raivi F mologous dicate A vm V. n iial angles VI. 15 Const. 13, Gw, VI 1 V. u V 7 Book VI.] PROPOSITIONS 19, 20. 319 1. If three straight lines be proportional, as the first is to the third 80 is any triangle described on the first to a similar and similarly described triangle on the second. Prove the proposition with either of the following constructions : 2. Take EO, measured along EF produced, a third proportional to EF and BC, and join DG. 3. From BG cut off BG = EF; join AG, and through G draw GHWAC. ^ 4. Similar triangles are to one another in the duplicate ratio of (1) their corresponding medians, (2) their corresponding alti- tudes, (3) the radii of their inscribed circles, (4) the radii of their circumscribed circles. (Assume, what can be easily proved from V. 23, Cor., that if two ratios be equal, their duplicates are equal.) PROPOSITION 20. Theorem. Similar polygons may be divided into the same number of similar triangles, having the same ratio to one another that the polygons have ; and the polygons are to one another in the duplicate ratio of their homologous sides. M Let ABODE, FGHKL be similar polygons, and let AL and FG be homologous sides : it is required to p>rove that ABODE and FGHKL m^ay be divided into the same number of similar triangles ; that these triangles have each to each the same ratio which the polygons have ; and that the polygons are to me another in tn€ duplicate ratio of their homologous sides. 320 Euclid's elements. [Book Vt V Bf Join BE, EC, GL, LH. Because the polygon ABODE is similar to the polygon FGHKL, Hyp. .'. L A = L F, and BA : AE = OF.FL; VI. Def. ] .-. A ABE is similar to A FGL. VI. 6 Because the polygons are similar, .-. L ABO = L FGH; VI Def. 1 and because As ABE, FGL are similar, .-. L ABE = L FGL; VI Def. 1 .*. the remainder, l EBO = remainder, l LGH. And because As ABE, FGL are similar, .-. EB:BA = LG: GF ; VL Def 1 and because the polygons are similar, .-. BA : B0= GF: GH. VI Def 1 .'.EB'.BO^LG: GH, by direct equality ; V 22 that is, the sides about the equal z. s EBO, LGH are pro* portional ; .\ A EBO is similar to A LGH. VI Q For the same reason, A EDO is similar to A LKH. Because A ABE is similar to A FGL, .-.A ABE : A FGL = duplicate of BE : GL. VI 19 Similarly, A EBO : A LGH = duplicate of BE : GL; .'. A ABE : A FGL = A EBO : A LGH. . ^- H Because A EBO is similar to A LGH, . •, A EBO • A LGH = duplicate of EO : LH. VI. 19 Book VI] PROPOSITION 20. 321 Similarly, A ECD : A LHK = duplicate of EC : LH ; .-. A EBC : A LGII = A ECD : A LHK. V. 11 Hence A ABE : A FGL = A EBC : A LGH = A ECD : A LHK; .'. A ABE : A i^(?L - A ABE + A EBC + A ^(7Z) : A FGL + A ZG^// + A LHK, V. 12 = polygon ABCDE : polygon FGHKL. Lastly, A ^i?^ : A FGL = duplicate of AB : i^G«; F7. 19 ABCDE: FGHKL =^u^\{cQ,iQoiAB: EG. V. 11 CoR. — If three straight lines be proportional, as the first is to the third, so is any rectilineal figure described on the first to the similar and similarly described rectilineal figure on the second. For, take M a third proportional to AB and EG. VL 1 1 Then since AB : EG = EG : M, Const .'. AB:M= duplicate of AB : EG, V. Def. 13, Cor. But ABCDE : FGHKL = duplicate of AB : EG, VL 20 ,'. AB.M^ ABCDE : FGHKL. V. 1 1 1. Squares are to one another in the duplicate ratio of their sides. 2. Similar polygons are to one another as the squares on their homologous sides, or homologous diagonals. 3. The perimeters of similar polygons are to one another as the homologous sides. 4. Polygons are similar which can be divided into tJ j same number of similar and similarly situated triangles. 5. Prove that similar polygons may be divided into the same number of similar triangles having their vertices at points situated within the polygons. (Such points are called homo- logous points with reference to the polygons.) 6. Could homologous points with reference to similar polygons be situated outside the polygons, or on their sides ? 7. If two polygons be similar and similarly situated, the straight lines joining their corresponding vertices are concurrent. Examine the case when the polygons are similar and oppositely situated. 322 Euclid's elements. TBook VL 8. Use the preceding theorem to inscribe a square in a given triangle. How many squares can be inscribed in a triangle ? 9. In a given triangle inscribe a rectangle similar to a givei rectangle. How many such rectangles can be inscribed ? PROPOSITION 21. Theorem. Polygons which are similar and equal have their homologous &ides equal.* bV ic r^ 'G K L Let ABCD, EFGH be two similar and equal polygon^ having BG and FG homologous sides : it is required to prove EC = FG. Take KL a third proportional to BC and FG. VL 11 Because BC : FG = FG : KL, Const. BC: KL = ABCD : EFGH. VL 20, Cor. But ABCD = EFGH; .'. BC = KL. V. 14 Again, since BC : FG = FG : KL, Const. BC-KL = FG^ ^ VL17 But BC' KL = BC^, since BC = KL; BC^ = FG\ and MC = FG. Prove the proposition indirectly. * Euclid's 21st proposition is ' Rectilineal figures which are similar to the same rectilineal figure are similar to each other,' a theorem which may be regarded as self-evident. In place of it there has been substituted the lemma which occurs after the 22d proposition, and which is assumed in the proof of it. The demonstration of this lemma given in the text is due to C!onuaandine {Eudidis Elementorum Libri XV., 1572). Book VLJ PROPOSITIONS 21, 22. 328 PROPOSITION 22. Theorems. Tf four straight lines be projMrtwnal, and there he similarly described on the first and second antj tioo similar poly- yons, and on the third and fourth any two similar polygons, the polygons shall be proportional. Conversely .• If there be similarly described on the first and second of four straight lines two similar polygons, and two similar polygon.^ on the third and fourth, and if the polygons be proportional, the four straight lines shall be proportional. L C D G^ iH ,St- O p\. Vr (1) Let AB: CD = EF.GH, and let there be similarly described on AB and CD the similar polygons KAB, LCD, and on EF and GH the similar polygons MF, NH : it is reqiiired to prove KAB : LCD = MF : NH. Take X a third proportional to AB and CD, and O a third proportional to FF and GH. Because AB : CD = EF : GH, and AB:CD= CD: X, and EF-.GH =^ GH:0; X = GH : O, CD VL 11 VL 11 Hyp. Const. Const. V. 11 324 Euclid's elements. L D [Book VL ) Af gV— Ah -?- V' -Vb Now since AB, CD, X are three magnitudes, and KF, GH, O other three ; and since AB : CD = EF : GH, and CD.X = GH.O; = EF'.O, by direct equality. V. 22 = KAB : LCD, VL 20, Cor. = MF:NH; VI. 20, Cor. But and AB:X AB'.X EF.O KAB '.LCD = MF-.NIL v.n (2) Let KAB : LCD = MF.NH: it is required to prove AB : CD = EF : GH. Take P22 a fourth proportional to AB, CD, EF, VI 12 and on PR let a polygon SB be similar and similarly described to the polygons MF, Nil. Because AB : CD = EF : PR, KAB:LCD = MF:SR. But KAB : LCD = iJ/i^ : NH; MF.SR = MF:NH; SR=^NH. Hence PR = (^^, since ^/2 and NHaie similar. liiow AB: CD = EF: PR; .'. AB:CD ^ EF:GH. l.liAB:CD = EF : (?//, then A B^ : CD^ = EF^ : C/f*. 2. M two ratios be equal, their duplicates are equal VL 18 Cmi'it: VL 22 Hyp. V 11 F. 9 VL 21 Const: V.I Book VI.] PROPOSITIONS 22, 23. 321 PROPOSITION 23. Theorem. Mutually oqidanijnhir pamlk'hxjrams have to one another the ratio ichich is compounded of the ratios of their A aides.* Let II" AB be equiangular to ||"' BC, having z. DBF = L GBE: it is required to prove ||™ AB : II'" BC = / ^^ ' ^^ I " \FB :BG )' Place the ||™» so that DB and BE may be in one straight line ; then GB and BF are in one straight line. VI. 14 Complete the Ij-" FF. Then and |« jr^ : l"^ FF = I™ BC = ( ir AB '.\\^FF) ( \r FF:\r BCJ ~ \ But ir FF : IP i^C ^i5 : |h i?C = _ / AB BC - DB : FB: DB: FB: II'" AB : I II'" Z:^ : I DB: FB : BE, VI. 1 BG; VI. 1 BE) BGj FE\ j^^]; V.DefU V. 23, Cor. BE) BG)' V. 11 1. Triangles which have one angle of the one equal or supplemen- tary to one angle of the other are to one another in the ratio compounded of the ratios of the sides about those angles. *The proof in the text, due to Franciscus Flussas Candalla, 1566, i« 3umcwhut shorter than Jiuoiid's. S26 BUOLID's BTJntfBNTS. [Book VL 2. Show that VT. 14 ia a particular case of the propoflition. 3. Show that j ^^ ; 1^ j = Z)Z? . fi/' : ^5 . BG. 4. Hence enunciate differently the proposition and tlie Hrst de- duction. 6. Prove the proposition from the accompanying figure. 6. Deduce VI. 19 from the first de- B duotion. 7 HK F D PROPOSITION 24. Theorem. Parallelograms about a diagonal of any parallelogram are similar ta the whole parallelogram, and to one another. Let ABGD be a ||™, AG one of its diagonals, and let EQ, HK be ir about AC : it is required to prove that ||"" EG, HK are similar to 11"" ABGD, and to one another. >■ Because DG is, \\ GF, .*. L ADG = l AGF ; I. 29 and because BG is || EF, .'. l ABG = l AEF. I. 29 And L s BGD, EFG are each = z. BAD; I. 34 .-. ir ABGD is equiangular to ||"' AEFG. Again, because l ABG = l AEF, I. 29 and L 5^(7 is common ; .*. As ABG, AEF sue mutually equiangular ; /. 32, Gor. 1 .-. AB:BG= AE: EF. VL 4 Book VI] PROPOSITIONS 24, 25. But since the opposite sides of H""* are equal, .-. AB:AD = AEiAG, and CD.BC = FO : EF, and CD '. DA = FG \ OA ; 827 /. 34 V. 7 V. 7 V. 7 that is, the sides of the ||™ ABCD, AEFG about iheir equal angles are proportional. .-. r ABCD is similar to |r AEFG. VI. Def. 1 Hence also, H"* ABCD is similar to ||™ FHCK ; .'. ir AEFG is similar to H" FHCK. 1. From this proposition and VI. 14, deduce I. 43. 2. Prove r EG : ||'» BF = r FD : r //A". 3. Prove that EG, BD, and HK are paraUeL PROPOSITION 25. Problem. 7b describe a rectilineal figure which shall be mnilar to one and equal to another given rectilineal figure. Let ABC be the one, and D the other given rectilineal figure : it is required to describe a figure similar to ABC, and = D. On BC describe any ||" BE = the figure ^^C, /. 45 and on CE describe the ll"* CM = the figure D nri'l Viain-r.r, L. FCE = L CBL. L 45 328 Euclid's elements. [Book VL Between ^Cand (TFfind a mean proportional GH; VI. 13 and on GH construct the figure KGII similar and similarly desciibed to the figure ABC. VL 18 KGH io the figure required. It may be proved as in I. 45 that BG and GF form one straight line, and also LE and EM ; . •. BG: GF = ir BE : ||™ GM, VI. 1 = ABG'.D. V. 11 But because BG\OH=GH: GF, Gonsf. and because ABG, KGH are similar and similarly descril)ed j BG : GF = ABG : A'C?//. VL 20, (7on Hence ^5(7 : D = ^5(7 : KGH; V. 11 A'G^^ = Z>. V,d 1. Construct an equilateral triangle = a given square. 2. Construct a square = a given equilateral triangle. 3. Construct a scjuare = a given regular pentagon. 4. Construct a regular pentagon = a given square. 6. Construct an equilateral trianj^jle = a given regular hexagon. ^ 6. Construct a regular hexagon = a given equilateral triangle. 7. Construct a polygon similar to a given polygon, and having a given perimeter. 8. Construct a polygon similar to a given polygon, and having a given ratio to it. 9. Through a given point inside a circle draw a chord so that it shall be divided at the point in a given ratio. Book VI] PAOPoyiTiONS 2d| 26. 329 PROPOSITION 26. Theorem. JJ two mnrlar parallelograms have, a common angle and he similarlij situated^ they are about the same diagonal. ^ le^ the ir ABCD, AEFG bo similar and similarly situate**!, aint have the common angle BA D : it is required to prove that they are about the same diagonal Join AG, AP, Because H™ ABCD, AEFG are similar and similarly situated, .*. -4Cand ^i'' will divfde them into similar triangles; VI. 20 0-. A ABO is similar to A AEF; . •. L BAO^ L EAF; yi. Def. 1 .'. ^Z' falls along ^ a * Note— This proposition is the converse of VI. 24, and should be read immediately after it. 1. Prove the promisition by supposing AC to cut EF at //, and drawing HK \\ EA to meet AG ax; K. 2. Extend the proposition to the case of two similar and oppositely situated \\^\ 3. From a given ||°» cut off a. sunilar |J" having a given ratio to the 330 EUCLID S ELEMENTS. [Book VL 1 1 1 1 I: u PROPOSITION 27. Theorem. Of all the parallelograms inscribed in a trianrjle so as to have one of the amjles at the base common to them ally the greatest is that which is described on half the base,"^ B D G O Let ABC he a triangle, having its base BG bisected at D. Let BE and BH be H"*" inscribed in it so as to have l B of the triangle common to both : it is required to prove H"* BE greater than ||"" BH. Complete the H'" FBGL, and produce GH, KH to M andiV. /. 34 /. 36 7.43 4 \ ! t Because BD = DG, .'. FE = EL; ir KE = ir EN. Again, ir MN = ^ DH. But 11" EN is greater than ||™ MN ; IP KE is greater than ij'" DH. Add to each of these unequals |p KD ; then ir BE is greater than ||™ BU. 1. Make the construction and prove the proposition when G lies between B and D. 2. When AB =^ BG and L B v& right, what does the proposition become ? * The enunciation of this proposition is different from tnat given by Euclid, but the proposition itself is substantially the same. The proof has been somewhat modified. Book VL] PROPOSITIONS 27, 28. 331 PROPOSITION 28.* Problem. To divide a given straight line internally so that the rect- angle contained by its segments may he equal to a given rectangle. K- Let AB be the given straight line, K and L the sides of the given rectangle : it is required to divide AB internally so that the rectangle contained by the segments may be — K ' L. Draw AG X AB, and = K, /. 11, 3 and on the same side of AB draw BD J. AB, and ^L. 7.11,3 Join CD, and on it as diameter describe the semicircle CED cutting AB at E. AE - EB shall be = /iT • L. Join CE, ED. Because l CED is right, ///. 31 .*. L AEG = complement of l BED, /. 13 = L BDE; L 32 and L GAE = l EBD. * Some editors of Euclid omit this and the following proposition. In the form in which Euclid presents them, they are difficult to understand and apply. The problems in the text are particular cases of Euclid's propositions^and the solutions given are to be found in Willehrord Snail's Apoiloidus Jjutavu«, or Edmund Halley^s ApoUonii Pergcei Conica (1710), Book VIIL Prop. X8, Scholion. 832 Euclid's elements. [Book 7L .-. As AEGy BDE are mutually equiangular ; /. 32, Cor. 1 .'.AE.AG^BD'.BE; F/. 4 ,'.AE^EB=AG'BD, F/. 16 = KL. 1. If E' be the other point in which the semicircle cuts AB, prove AE' • ^'5 = K'L: 2. Prove ^J^' = £A' and E'B = AE. 3. What limits are there to the size of the rectangle K -L? 4. Solve the problem otherwise by converting the rectangle K • L into a sq^uare. PROPOSITION 29. Problem. To divide a given straight line externally so that the rect- angle contained by its segments may be eqiml to a given rectangle. K- L- let AB be the given straight line, K and L the sides of the given rectangle : it is required to divide AB externally so that the rectangle contained by the segments may be - K • L. Draw AG ^. AB, and = K, ana on the opposite side of AB drav /. 11, 3 1,.qv' Tiri I AH flTlfl /. 11. 3 Book VI.] PROPOSITIONS 28, 29, 30. 333 Join CD, and on it as diameter describe the semicircle CED^ cutting AB produced at E. AE • EB shall be = ^ • £• Join GE, ED. Because l CED is right, ///; 3j .'. L AEG = complement of l BED, = L BDE; /. 32 and L GAE = l EBD. .'. As AEG, BDE are mutually equiangular; /. 32, Cor. 1 .-. AE'.AG = BD: BE; yi. 4 . •. AE.EB == AG' BD, yi. 16 ^ K ' L. 1. If E' be the point in which the semicircle described on the other side of CD c-t3 AB produced, prove AE' • E'B = K'L, 2. Prove AE'=BE ana E'B = AE. 3. What limits are there to the size of the rectangle K - L? •*. Solve the problem otherwise by converting the rectangle K • L into a square. PROPOSITION 30. Problem. To divide a given straight line in extreme and mean ratio. O Let AB be the given straight line : it is required to divide it in extreme and mean ratio. Divide AB internally at G so that AB - BC = AC^. II. 11 Because AB . BG =^ AG\' Const. AB : AG = AG: BG. VL IV 1. If in the figure to VI. 8, BO be divided in extreme and mean ratio at Z), then AG = BD ; and conversely. 2. AB and DE are two straight lines divided internally at Cand F no that AC : CB = DF : FE ; if AB • BC = AC^, prove DE.EF=DF^ 334 EUGUD 8 ELEMENTS. [Book VL 1' PROPOSITION 31. Theorem. Any rectilineal figure described on the hypotenuse of a right- angled triangle is equal to the similar and similarly described fi^gures on the other two sides. Let A ABC be right-angled at A, and let X, Y, Z he tectilineal figures, similar and similarly described on BG, AB,AG: it is required to prove X = Y + Z, Draw AD ± BG. I. 12 Then GB : BA = AB : BD; VL 8, Gor. GB:BD= X : Y, VL 20, Gor. and BD.GB = F : ^, by inversion. V. A. Similarly, DG : GB = Z : X; .-. BD + DG:CB = Y + Z : X. F. 24 But BD + DG = GB; .'. Y + Z = X. 1. From this proposition deduce I. 47. 2. Has I. 47 ever been used in any of the propositions which help to prove VI. 31 ? 3. Prove VI. 31 from VI. 22 and I. 47. 4. If on AB, AG, BG semicircles are described, those on AB and AG being exterior to the triangle, that on BC not being so, the sum of the areas of the two crescent-shaped figures will = A ABO. Assume that semicircles are similar figures. T'U.n. y%M:nMA«^*^f nVnv^'A^ -I? ry II MAO o t»\ AB-.AD^ AE'.AG; ' ^Z . ^ED.AD-AD\ // 3 ^ BD.DC ~ AD^/ III 35 r' ^ .-. ^i)2 » i?i> . i)C7 - ^i? . AC ' ' 2. In that case prove AD^- = AB . AG - BD . DC, if ^2) be any straight line drawn to the base BC. 3. Could the bisector of the exterior vertical angle of a triangle be a diameter of the circle circumscribed about the triangle? 4. Prove AE • BD = BB^ or CA^a. ** fi. If a straight line be cut internally and externally in the same ratio, the square on the segment between the points of section = the difference between the rectangle contained by the external segments, and the rectangle contained by the internal segments. 6. P^«J^t^*t the converse of the proposition is true except when 7. Express in terms of a ft. c, the sides of a triangle, the bisectors ot the interior and the exterior vertical angles 8. Construct a triangle having given two sides and (1) the bisector of the angle included by them, (2) the bisector of the angle adjacent to thit included bv them. ' I I 340 buoud's elements. [Book VX. ' PROPOSITION C.» TnEORKM. If from the vertical angle of a triangle a perpendicular he (Irami to the bane, the rectangle contained by the sides of the triangle is espial to the rectangle contained by the perpcjidicular and the diameter of the circle circumscribed about the triangle. A Let ABCh^Q. triangle, AD the perpendicular from A on the base BG, and AE a diameter of the circle circumscribed about ABC- it is required to prove AB > AC = AD • AE. Join EO. c lADB= lACE ///. 31 liiA9ABD,AEC,^ ^j^lij)^ LAEG;IIL21,or22,Cor. .-. these triangles are mutually equiangular. /. 32, Cor. 1 .'. AB:AD = AE:AC; VI. 4: .'.AB-AC ^AD.AE. , VL 16 1. Conversely, if ABG be a triangle, AE the diameter of the circumscribed circle, and if AD be drawn to BG so that ADAE^ AB' AG, then AT> is ± BG. 2. Construct a triangle, having given +,he base, the vertical angle, and the rectangle contained by the sides. 3. If a circle be circumscribed about a triangle, and two straight lines be drawn from the vertex making equal angles with the sides, one of the straight lines meeting the base, or the base * Given by Brahmegupta, an Indian mathematician (bom 598 a.d.)« Book n] PROPOSITIONS C, D. 341 produced, and the other the o««. the rectangle containeerpen- diculars on the third sides. 7. If in the figure to VI. D the diagonals intersect at F, prove BA.BG:CB.GD = BF:CF,&adcouyeneiy. 8. In the same figure prove AJi ' AD + CB . CD :BA . BG + DA • DO =r. AC: BD. PROPOSITION D.* Theorem. The rectangle contained by the diagmah of a quadnlateral inscribed in a circle is equal to the mm of tlie two rect- angles contained by its opposite sides. A Let ABCD he a quadrilateral inscribed in a circle, and AO, BD its two diagonals : it is required to prove AO . BD = AB • CD + AD - BC. Make l BAE = z. DAO. j 23 n J, '^'"' k'^'^.T '' ''^^'' '*"'** Ptolemy's (about 140 A.D.) because it occurs m his Almagest, L 9. U2 t!UCLID*S ELEMENT^. A [^kVt To each of these equals add l EAQ; .-. L BAG = L EAD. Ir^ A. Ann ATPT^ i ^ B^^ = ^ EAD ,'. these triangles are mutually equiangular. I. .\BC:CA =ED'.DA; .'.AD' BO = AC 'ED. In As ABE, ACD, { "• ^^^ = ^ ^^^ M ^ ABE = I. ACD; .'. these triangles are mutually equiangular. /. .'.AB-.BE = AG:CD; .'.AB'CD = Aa-BE. Hence, AB - CD + AD . BC = AC BE + AC - ==ACBD. TIL 21 32, Cor. 1 VL 4 VI. 16 Const. III. 21 32, Cor. 1 VI. 4: VL 16 ED, II. 1 1. An equilateral triangle is inscribed in a circle, and from any point on tha o"- straight lines are drawn to the vertices; prove that one of these is equal to tl. i sum of tho other two. 2. In all quadrilaterals that cannot be inscribed in a circle, the rectangle contained by the diagonals is leas than the sum of the two rectangles contained by the opposite sides. 3. Prove the converse of the proposition. 4. ABC ia a triangle inscribed in a circle ; D, E are taken on AB, AG BO that B, D, E, G are concyclic; the circle A BE cuts the former in F. Prove that F£ + FB : FG ■{• FD = AB'.AG. (R. Tucker.) Book VL] APPENDIX VI. 343 APPENDIX VI. TRANSVERSALS. DBF l._When a straight line intersects a system of straight lines. It IS called a transversaL This definition of a transversal is not the most general (thlt is comprehensive) one, but it will suffice for our present purpose. ' Proposition 1. V<^*ran^rsalmt iU aides, or the sides produced, of a tHangU, the product of three alternate s^grmnU taken cyclically is egmlto the product of the other three, and conversely* J^t ABO be a tnangle, and let a transversal cut BG, CA AR or these sides produced at D, E, F respectively : It is required to prove AF - BD . GE = FB - DC ■ EA. Draw AG \\ BG, and meeting the transversal at G. Then AaAFG, BFD are nrntually equiangular • '.AF:AG = BF:BD; ^ ' '.AF.BD^AOBF. (1) 7.31 7.29 F7. 4 F7. 16 Given m the third book of the Spherics of Menelaus, who Uvod at ^exandna towards the close of the first century a.d. Fo; a fuU Hunt ■^ .u^ .ueorcm, see cimHles' Apergu Bistorique sur Vorigine et U d4v€U)ppment des M6thodes en Geom^trie, p. 291. ; 344 EUCLID'S ELEMENTS. [Book VL Again, A s AEO, GED are mutually equiangular ; /. 29 .'.AG:AE=GDi GE; VI. 4 .'.AO'GE = GD'AE. (2) VI. m Multiply equations (1) and (2) together, and strike out the common factor AG; then AF- BD ■ GE = FB • DG • EA. Cor. 1.— The equation AF BD - GE = FB - DG - EA may be put in any of the foUoAving four useful forms : AF BD GE FB= DG DG = EA EA = FB AF BD FB ' DG EA : FB : DG: GE^ ' EA BD- GE ' AF- = 1. GE, AF, BD, Cor. 2. — Consider ABG as the triangle, DEF aa th-: transversal: then AF'BD-GE=FB'MV-EA. (1) Consider* -4 i^'i^ as the triangle, BGD as the transversal ; then AB . FD ' EG = BF ' DE ■ GA. (2) Consider BDF as the triangle, AEG as the transversal ; then BG'DE'FA =GD-EF' AB. (3) Consider GED as the triangle, A FB as the transversal ; then GB-DFEA=BD-FE- AG. (4) Any one of these four equations may be deduced from the other three by multiplying them together and striking out the factors common to both sides. The converse of the theorem (which may be proved indirectly) is, If two points be taken in the sides of a triangle, and a third point in the third side produced, or if three points be taken in the three sides produced of a triangle, such that the product of three alternate the three points are collinear. Book VI.] APPENDIX VL 345 Proposition 2. If three concurrent straight lines be drawn from the vertices of a triangle to meet the opposite sides, or two of those sides produced, the product of three alternate segments oftlie sides taken cyclically w equal to the product of the other three; and conversely.* DC B CD Let ABC be a triangle, and let AD, BE, OF, which i)ass thrpugh any point 0, meet the opposite sides in Z>, E, F: it is required to prove AF • BD • GE = FB • DC • EA. Consider ABD as a triangle cut by the transversal COF • then AF • BC - DO = FB -CD . OA. (1) App. VL 1 Consider ADC as a triangle cut by the transversal BOE; then AO'DB'CE=OD.BC. EA. (2) App. VI. 1 Multiply equations (1) and (2) together, and strike out the common factors AO, DO, BC ; ikeri AF ' BD ' CE = FB • DC ■ EA. Cob.— Repeat Cor. 1 to the preceding theorem. The converse of the theorem (which may be proved indirectly) is, If three straight lines be drawn from the vertices of a triangle to meet the opposite sides, or two of those sides produced, so that the product of three alternate segments of the sides taken cyclically is equal to the product of the other three, the three straight lines are concurrent. * This theorem is first found in a work of the Marquis Giovanni Cava, De lineis rectis se invicem aecantibus, statica constructio (1678), Book L, Prop. 10. The proof given in the text is due to Camot, the founder of the Theory of Transversals. See his Essai sur la Thiorie de* Tranwer- sales (1806), p. 74. S46 Euclid's elements. [Book yi Noix-To distinguish readily between the converse of Menelaus's theorem and that of Ceva's, it should be observed that in the firet ease an even number of the points D, E, F are situated on the sides, and an odd number on the sides produced; in the second case mrttcrs are rsversed. Proposition 3. /f two triangles be situated so that ihe straight lines joining cor- responding vertices are concuirent, the points of intersectum of corresponding sides are collinear; and conversely* Let ABC, A'B'C be two triangles such that A A', Bff, GO' are Z7c:i'a\t'M^t%Tsr'''' -''- ^'- ^^' --' » it is required to prove L, M, A" coUinear. Consider AOB as a triangle cut by the transversal A'B'N • then AN ■ BB' . OA' = JVB • B'O • A'A. (1) f' yj , Consider AOG as a triangle cut by the transversal A'O'M then A A' • OC ■ CM = A'O ■ CO • MA. (2) Atm VI 1 Consider BOG as a triangle cut by the transversal ROL • then RO . G'G • LB = BE . OG' • GL. (3) App. VI. 1 * Due to Girard Desargues, an architect of Lyon, who was hnm ikq^ and died 1862. See Poudra's (Euvres dc Desargu^ tomell ^3, m Book ^ Multip com mo then A The If two corresp spondin Dep. same ra section harmon: Thus, same ra A, 0, B, Def, I Jugate t l)oints A G6omHri Since a in any r; ways. The ar to be in h d>fFerence the secon externall}) Hence, if . be seen th to the defi * i*ythag III., section Book VI] APPENDIX VI. 54 i (3) together, aad strike out ths MA; Multiply the equations (1), (2), common factors ; then ANBLCM = I^B- LG ''• L,M, N SLTecoWineax. A vi The converse of the theorem (which may be proved indir^cilv^V If two triangles be situated so that the points of intersection ol corresponding sides are coUinear, the straight lines joTnircorre spondmg vertices are concurrent. J"i"xng corre- HARMONICAL PROGRESSION. «ol . f ;, cut internally at C, and externally at D in the same ratio^ AB is said to be cut harmonically; and the Luts A, G, B, D are said to form a harmonic range. ^ i„ J^tl' f ~^^v! P°i''*' ^ *°^ ^ *^" '^^^ *« ^« harmonlcauy con- points ^ and ^ The segments AB, CD are sometimes (Chasles' Geometne SupSrieure, § 58) called harmonic conjugates. Since a straight line can be cut internally, and therefore «f^m»ii m any ratio, it may be cut harmonical/ln an inTn^r^utbT'l^ The ancient Greek mathematicians * defined three maanitnH.c to be m harmonical progression when the first is to the tSt tt externally at D in the sameT^tb, '"* "*""'"^ "* ^ »-^ , AD:DB=AG:GB; AD:AG==DB: GB by alternation, V 16 TT .... -^AJ^-AB:AB-Aa ^ Hence, if ^i> ^5, ^C be regarded as the three magnitudes it will to%rdetiti'r '-' "^ '^^'^^"^^^^ '-^'--- «^-' tw confrri: III.,!Su '"'*'^^'"' ^" *'>^ ^^^--* Progressions,see Pappus, W / 348 EUCLID'S ELEMENTS. [Book Yt I i Proposition 4. If G and D are harmonic conjugates with respect to A and B, then A and B are harmmic conjugates with respect to G and D. B D Since Cand D are harmonic conjugates with respect to A and B .-. AB 18 cut internally at G and externally at D in the same ratio • .'. AD '. DB = AG : GB; j^p^ yj j^ f X •'. AD: AG = DB :GB,hya.lterna.tioii, ' y'lQ that is, GD is cut externally at A and internally at B in the s^e ratio ; .-. A and B are harmonic conjugates with respect to G and D. Cor. 1. -Hence, if A, G, B, D form a harmonic range, not only are AD, AB, AG^in harmonic progression, but also AD, GD, BD. Coii. 2.— The points which are harmonic conjugates to two given p^mts are always situated on the same side of the middle of the line joming the two given points. (1) (2) O — ♦— c B -♦- O — ►- (3) D A C o B C A -4- (4) D O B Suppose A and B the given points, the middle of AB. Since G and D are harmonic conjugates with respect to ^ and B ^••^f = ^,^-/f-^^- App.VLDe/.S Now if i) be situated (as in figs. 1 and 2) to- the right of then ^Z) must be greater tlian Z)^; ' ,-. AG must be greater than GB, that is, G also is situated to the right of 0. If i) be situated (as in figs. 3 and 4) to the left of 0, then AD must be less than DB ; .'.AG must be less than GB, , ^ tlwt is, a also is situated to the left uf 0. Book TL] APPENDIX VI. 349 given, the ^ourtl may be determined ^ ^ OB D Pour cases are all that can arise, namely, when A, a B or D is to be found. > » f ^^ *o (1) If A, G, B are given, D can be found by dividing AB exter- nally m the ratio ^ C : CB. ^ "^"^ n If (7, 5, D are given, A can be found by dividing DC exter- naliy in the ratio DB.BC. ^ (3) If ^, 5, i) are given, G can be found by dividing AB inter- nally m the ratio AD . DB. (4) If A, (7, /) are given, B can be founcT by dividing DC inter- nally m the ratio DA .AC. Proposition 5. ^^A AB, AG are in harmonical progression, and the mean AB is oisected at O tlien OD, OB, OG are in geometrical progression: ana conversely.* D Since AD, AB, AG are in harmonical progression AD:DB = AG:GB; Avd Vf n.f 9 .-. OD+OB:OD - OB =. OB + OG.OB - OG ^ OD:OB=.OB:Oa Converse <^ V. l> Cor 1 -Since OD : OB = OB : OG, .-. OB^ = OG . OD. VI 17 Now If A and B are fixed points, OB^ is constant : .-. OG ■ OD is constant. Hence if OC diminishes, OD increases, that is, if G moves nearer to 0, i> moves farther away ; and if OC increases, 02) diminishes, that is, li G moves away from O, D moves nearer to 0. In other words, if C and D move in such a manner as always to remain harmonic conjugates with respect to the fixed points A and B thev Til TH''' 7^"^^ directions. Al.o, the nearer C approaches to O, the farther does D recede from it ; and when G coincides with O ^ must be mfimtely distant from it, or as it is ofte. expressed, at * Pappus, VII. 160. [BOOkfl. 850 Euclid's elbhents. Cor. 2. OD.OD = OBfi\ AG-CB. Cor. 3. OC .00 = AG^ : AD\ [Corn 2, 3 are given in De La Hire's Sectionea Conicce, 1685, p. 3.] Proposition 6. //AD, AB, AC are in harmonical progrrasicm, and the mean AB is bisected at 0, then AD, ODy CD, BD are j/roporlionals ; and conversely.* O — »- B D For AD'DB= {CD + OB) • (OD - OB), = 0/>3 - OB^, = 0Z>2 - OD . OG, ■ = OD'CD; : AD:OD=CDi BD. II. 5, G&r, App. VI. 5 ILZ VI. IQ Cor. l.—Since ODCD= AD - DB ; 2 0D'CD=2AD.DB; (AD + DB)'GD=2AD. DB, a result which, considering AD, GD, BD as the terms in hanuonioal progression, may be stated thus : The rectangle under the harmonic mean and the sum of the extremes is equal to twice the rectangle under the extremes. Cor. 2. — ^The geometric mean between two straight lines is a geometric mean between the arithmetic and the harmonic means of the same straight lines. [The arithmetic mean between two magni- tudes is half their sum.] Denote the arithmetic, geometric, and^harmonic means between AD and DB by a, g, h respectively ; then a = i {AD + DB) = OD, g"^ Now since AD • DB = OD - GD, .-. aig — g-.h. AD'DB, h==GD. g^ =:a'h; * Pappus, VIL 16a Book VL] APPENDIX VI. 351 Proposition 7. If AD, AB, AC are in harmonkal progression, and the mean AB U bisecUid at O, then CB : GD =. CO : CA ; and conversely* O — »- c — »- B -4- For AC-CB=: (OB + OC) • (OB - OC), = 0^ - 0C\ ji, 6, c.«<..■, fjAliy and externally in the ratio of the radii, the jwiuts of section aie uiOled the Internal and external centres of similitude of the two circles. (The phrase ' centre of similitude ' is due to Euler, 1777. See iN^ov. Act. Petrop., ix. 154.) Def. 6. — The figiu^ which results from producing all the sides of any ordinary quadrilafct./u lill they intersect is called a complete quadrilateral; and the straight line joining the intersections of pairs of opposite sides is called the third diagonal. (Caruot, Eaaai mr la Thiorie des Transversales, p, 69.) To the notation adopted for points and lines connected with the triangle ABG on pp. 98-100, 252, 253, should be added the following : Ny P, Q denote the points where the bisectors of the interior IB A, B, O meet the opposite sides. N't P', Q denote the points where the bisectors of the exterior IB A, B,G meet the opposite sides. A by itself denotes the area of a ABG. p denotes the radius of the circle inscribed in the orthocentrie DEDUCTIONS. i. <7and D are two points both in AB, or both in ^^ produced: show that AG : GB is not = AD : DB. 2. Find the geometric mean between the greatest and the least straight lines that can be drawn to the O "^ of a circle from a point (1) within, (2) without the circle. 3. In the figure to IV. 10, as ABD, AGD, DGB are in geometrical progression. 4. Construct a right-angled triangle whose sides shall be in geo- metrical progression. 5. If a straight line be a common tangent to two circles which touch each other externally, that part of the tangent between the points of contact is a geometric mean between the diameters of the circles. flL Any regular polygon inscribed in a circle is a geometric mean between the inscribed and circumscribed regular polygons of half the number of sides. iDOkTI.] APPENDIX VI. 353 7. To find a mean proportional between A B and BG, G boing situated between A and B. Produce AB to E, making BJiJ = AG ; with A and E as centres and AB as radius, describe arcs cutting in D ; join BD. BD is the mean pro- portional. (See Wallis's Algebra, Additions and Emendations, 1685, p. 164.) Of three straight lines in geometrical progression : 8. Given the mean and the sum of the extremes, to find the extremes. ft Given the mean and the difference of the extremes, to fhid the extremes. 10. Given one extreme and the sum of the mean and the other extreme, to find the mean and the other extreme. 11. Given one extreme and the difference of the mean and the other extreme, to find the mean and the other extreme. 12. Find two straight lines from any two of the six following data : their sum, their difference, the sum of their squares, the differ- ence of their squares, their rectangle, their ratio. 13. If two triangles have. two angles supplementary and other two angles equal, the sides about their third angles are propor- tional. 14. Divide a straight line into two parts, the squares on which i^hft H have a given ratio. 16. Describe a square which shall have a given ratio to a given polygon. 16. Cut off from a given triangle another similar to it, and in a given ratio to it. 17. Cut off from a given angle a triangle = a given space, and such that the sides about that angle shall have a given ratio. 18. AGB is a semicircle whose diameter is AB, and on AB \r described a rectangle A DEB, whose altitude = the chord of half the semicircle ; from G, any point in the 0<*, GD, GE are drawn cutting AB sX, F and G. Prove AQ^ + BF^ =^ AB^, (Due to Fermat, 1658. See Wallis's Opera MathenuUica, 1695, vol. i. p. 858.) 19. If two chords AB, GD intersect each other at a point E inside fk circle, the straight lines AD, BG cut off equal segments from the chord which passes through E and is there bisected. 20. Enunciate and prove the preceding theorem when the chorda ABf GD intersect each other outside the circle. n4 Euclid's ei-ement8. [Book .1 c,. Prove the following projKjrties of a ABG: 2Y. >«(« - a) : A = A ; (j» - 6) (« - c). 22. « : a - a - (s - A) (» - c) : r" r= r{^ :{s - b) (a 2." J. rrir,r3 = A ". 24. «■■» = nrj + r,r8 +■ r^ri. 2"». «i = ri(r, + rs), sb = rg {r^ + r,), sc -^ r, (»•, + r^'. 2()'. lir^ + r^ + r.^)^ AF' FB + i/Z) . DC + CA^ . AVI 27. A = i/^(Xr+ r^ + ifX). 28. 2« rJC }' + rz + j^A' = i? : n 29. A ABC : A A' YZ= R: p, 2Rp = AO • OX = BO ' OY = CO ' OZ. a2 + 63 + ,.2 ^ 8A'- + 4i?/3. 30. 32. 5/2 = /?(/? - 2»). 33. 34. 35. 36. 37. 38. 39. 5/,a = li{n + 2ri). -S-Za^ = R{R + 2r3), ^/a" = /?(/? + 2r3). .S72 + SI{^ + .S'/a'-J + 5/32 = 12i?2. a2 + 62 + c-' 4! r2 + ri2 + rgS + r32 = \QR"-. 11,^ + 11,^ + 7/32 + IJ,' + IJ.,^ + /3A2 = 48i?2. //i>2= HD,^ = //X . HN; HD.^= HlX? = //A' HX 'ND = HD ■ i)X ; ^X • iVA = HD. IIX . iVT'i)., = JID,.D,X ; HX • N'D-, = II D, 40. //iV 'NX =DN ■ NDi; HN' • NX = D.N' HN'. D,X. DuX. N'D,. [Regarding theorem 21, see p. 145. It has, however, beea con- jectured, and with probability, that the treatise in which it cjcurs is a work of Heron the younger, ami therefore long subsequent to the date of the elder Heron. The theorem was known to Brahmegupta, 628 A.D. For theorems 22, 36, 25, 26, see Davies in LaiivJ Diary, 1835, pp. 56, 59 ; 1836, p. 60 ; and Philosophical Magazine for June 1827, p. 28. For 23 and 24, see Lhuilier, Elemem d' Analyse, p. 224. For 27, 28, 29, 30, 31, 34, 35, see Feuerbach, Eigenschaften, &c., section vi., theorems 3, 4, 5, 6, 7 ; section iv., § 50 ; section ii., § Ja Theorem 32 is usually attributed to Euler, who gave it in 1765. It occurs, however, in vol. i. page 123, by William Chappie, of the Miscellanea Guriosa Matliematica, and probably appeared about 1746. Theorem 33 is given in John Landen's Mathematical Liicubrations, 1755, p. 8. Some of the proiierties 37-40 are well known ; but I cannot trace them to their sources. Hundreds of other beautiful properties of the triangle may be found in Thomas Weddle's papers in the Lady's and Gentleman's Diary for 1843, 1845. iS4a] Book VI.] APPENDIX VI. 365 41 T 4ii .tract a triangle, haviug giv m : ■^rtical angle, the ratio of the sidea containing it, and the » 8, . (Pappus, VII. 165.) I'l. ..rtical angle, the ratio of the sides containing it, and the ...ineter of the circumscribed circle. Tiio vertical angle, the median from it, and the angle which i>he mediau makes with the base. 44. The vertical angle, the perpendicular from it to the base, and the ratio of the segments of the base made by the perpen- dicular. 45. The vertical angle, the perpendicular from it to the base, and the sum or difference of the other two sides. 46. The base, the perpendicular from the vertex to the base, and the ratio of the other two tides. 47. The base, the perpendicular from the vertex to the base, and the rectangle contained by the other two sides. 48. The segments into which the perpendicular from the vertex divides the base, and the ratio of the other two sides. 49. The perpendiculars from the vertices to the opposite sides. 50. The Pides containing the vertical angle, and the distance of the vertex from the centre of the inscribed circle. TRANSVERSALS. The following five triads of straight lines are concurrent : 1. The medians of a triangle. 2. The bisectors of the angles of a triangle. 3. The bisector of any angle of a triangle and the bisectors of the two exterior opposite angles. 4. The perpendiculars from the vertices of a triangle on the opposite sid^s. 6. AL, BK, CF in the figure to I. 47. 6. If two sides of a triangle be cut proportionally (as in VI. 2), the straight lines drawn from the points of section to the opposite vertices will intersect on the median from the third vertex ; and conversely. 7. The points in which the bisectors of any two angles of a triangle and the bisector of the exterior third angle cut the opposite sides are coUinear. i. The points in which the bisectors of the three exterior angles 6f a triangle naeet the opposite sides are coUinear. ^•S S66 btjclid's elements. [Book Vt 11 12 13 14 9. If a circle be circumscribed about a triangle, the points in which tangents at the vertices meet the opposite sides are collmear. 10. The perpendiculars to the bisectors of the angles of a triangle at their middle points meet the sides opposite those angles in three points which are coUinear. (G. de Longchamps.) OA, O'A', 0"A'' are three parallel straight lines; 00', A A' meet at B" ; CO", A'A" at B; 0'V,A"A at B'. Prove B, B',B" coUinear. If a transversal cut the sides, or the sides produced, of any polygon, the product of one set of alternate segments taken cyclically is equal to the product of the other set. (Camot'a Essai sur la TMorie des Tramversales, p. 70.) If a hexagon be inscribed in a circle, and the opposite sides be produced to meet, the three points of intersection are coUinear. (Particular case of Pascal's theorem.) Prove with reference to fig. on p. 346. i A0-BO-C0:D0-E0.FO = AB.BOOAiAF-BD. CE. (Davies's edition of HuttorCa Mathematics, 1843. vol. ii p. 219.) 15. If a point A be joined with three coUinear points B, C, D, then will AC'^.BD ±AB^.CD = AD^ ■BC±BD-DG' BC, the upper sign being taken when D lies between B and C, and the lower when it does not. (Matthew Stewart's Sonie General Theorems of considerable use in the higher parts of Mathematics, 1746, Prop. II.) Deduce from the preceding theorem, App. II. 1 ; deduction 1 on p. 151 ; VI. B • and App. VI. 8. F , , IG. If the o"' of a circle cut the sides BC, CA, AB, or those sides produced, of ^ ABC at the points .D,D', E,E', F,F', then will AF.AF'-BD.BD',CE-GE'=FBrF'B.DC-D'G.EA.EA. (Camot's Essai, &c,, p. 72.) 17. Prove with reference to fig. on p. 251. AI-BI.CI.AB.BC.CA=AB-BC.CA:AI,.BI,.CI,. (C. Adams's Die merkwiirdir/sten Eigenschaften des gerad- linigen Dreiecks, 1846, p. 20.) 18. Prove the following triads of straight lines connected with A ABC concurrent : Book Vt] APPENDIX VI. 367 (1) AD, BE, OF (7) AD,,BF; CF^ (13) BO, E,F„ E.F. (2) AD,, BE,, CF, (8) AD,, BE,, OF (14) GA, F,D„F,D, (3) AD,, BE,, CF, (9) AB, DE, D^E, (16) AB, NP, 1,1, (4) AD,, BE,, OF^ (10) BC, EF, E^F, (16) BC, PQ, /,/, (5) AD„ BE,, OF, (11) GA, FD, F,D, (17) C^, QN, 1,1, (6) ^A ^^3. GF, (12) ^A A^2.i>a4 ' '^ ' « ^ 19, If the triads (6), (6), (7), (8) meet at the points /', //, I,', I^ respectively, prove that these four points are the inscribed and escribed centres of the triangle formed by drawing through A,B,G parallels to the opposite sides. 20. If the triads (9), (10), (11) meet at the points Q,, N„ P,, (12), (13), (14) n „ q2,N.,P„ (15), (16). (17) „ „ Q„Ns,P,; then Qi, N„ P, will lie on one straight line n, Q» -"8» ■* 2 l» II p, Qat -"8» °3 " " and E. Prove that the rectangles BD • DC, BA AC, BE • EC are in arithmetical progression when the difference of the base angles is equal to a right angle, in geometrical progression when one of the base angles is right, and in harmonical progression when the vertical angle is right (Lardner's Elements of Euclid, 1843, p. 206.) 20. If K and L represent two regular polygons of the same a umber of sides, the one inscribed in, and the other circumscribed about, the same circle, and if M and M represent the inscribed and circumscribed polygons of twice the number of aides ; M shall be a geometric mean between K and L, and iV" shaU be a harmonic mean between L and M. (Libraiy of Useful Knowledge, Geometry, 1847, p. 96.) CENTRBS OP SIMILITUDB. \. When is the internal centre of similitude situated on botK circles ? How, in that case, is the external centre situated ? 2. When is the external centre of similitude situated on both circles ? How, in that case, is the internal centre situated ? 3. When are both centres of similitude outside both circlts, and when inside both circles ? 4. When is the internal centre of similitude inside both circles, and the external centre outside both ? 5. When two circles intersect, the straight line joining either point of intersecticn to the internal centre of similitnde bisects tne an,-- of » sid^ "'' ""' '"^'"' "' *'"' ""«'■= * '■"^-'"'^ ««• femnine the case of the bisector of the exterior Migle at R 362 buolid's blbmbnts. [Book VL 12. ^5 18 a diameter of a circle. A right angle, whose vertex is at A, revolves round A, and its sides intersect the tangent at B m the points C and D ; find the locus of the intersection of the tangents drawn to the circle from the points G and D. 13. XYand X'Y' are two parallel straight lines, and A, B, G three fixed coUinear points. A straight line revolvef round A and meets Xr and X'Y' at D and U ; find the loci of the inter- sections of BE and GD, and of BD and GJS. 14. XY and X'Y' are two parallel straight lines, and is a point midway between them. Through straight lines are drawn terminated by XF and XT, and equilateral triangles are described on these straight lines ; find the locus of the third vertices of the triangles. 15. Jf r and X'Y' are two paraUel straight lines, and is a fixed point. Through O straight lines are drawn to XY&ndXi , and on the segments intercepted between XY and X'Y' simUar triangles are described ; find the locus of the third vertices of the triangles. (The last three examples are taken from Vuiuert's Journal de MatMmatiques EUmentaires 1" Annee, pp. 13, 20 ; 3" Anu^e, p. 5.) ' MISCELLANEOUS. ;. Show that the perpendiculars of a triangle arp concurrent, by a method which will prove at the same time that the circuw- scribed centre, the centroid, and the orfchocentre are coUinear and that their distances from each other are in a constant ratio. 2. The circumscribed centre, the centroid, the medioscribed centre and the orthocentre form a harmonic range ; and the centroid and the orthocentre are the internal and external centres of similitude of the circumscribed and medioscribed circles. 3. All straight lines drawn from the orthocentre to the O'^ of the circumscribed circle are bisected by the o^" of the medio- scribed circle. 4. What is the analogous property for the straight lines drawn from the centroid to the o«« of the circumscribed circle ? 5. The inscribed centre, the centroid, and the point /' (see the 1 9th deduction on p. 357) are coUinear, and their distances from •ach other are in a constant ratio. [Book VL vertex is at angent at B action of the dD. B, G three )und A and I the inter- Hs a point I are drawn iangies are f the third • ' is a fixed 'andX '/ , and X'Y' I the third 3 are taken ntaires, 1™ rrent, by a he circum. 3 colUnear, Et constant ►ed centre, le centroid centres of rcles. C^ of the he naedio- les drawn ircle ? 3 the 19th nces from 3ook Y1.1 APPENDIX VI. 363 9. 10. II 6. The middle point J of /'/ is the centre of the circle inwribed in the centroidal a HKL, inwjnoea m 7. The points H J and the middle point of AI' are coffinear. & The pointe / y. Q I form a harmonic range ; and G and /' ai^ the mternal and external centres of «Stude of the c JH mscnbed in a s ABC, HKL. «*-*"«'««e oi we circles The inscribed circle of a HKL i> •!.« *i. • -. , . ^< iu X • 1 , «A/y u aiBo the inscribed cirelA Al^Bl^'^ '^""^ '^ ^""^"^^ *^« middle ^r!j Deduce the properties corresponding to those in the last five deductions for the escribed centres, the centroll and^the points /,'. /,', f,', cho d 12 "iT circle at^ and B. In this circle place the ^5 or ^i) will be the radius of the given circle 13 H Swale, ^^e PhiloaopMcal Magazine, IS^l, 'p u\)' ' ^^' "^lyBD^tX^^"' f^^*r«^^.^ ** ^- ^"gl« ^* - bisected 5^^'r7 ™®®** ^C at i) ; prove 2 BC^ : ^C^ - (7Z)2 = ^A:CD. (John Pell, 1644. This theorem is susceptible of a good many proofs.) «'«'pwuie oi IS. Of the four trkngles formed bv /, /„ /„ f, («ee fig. on p. 251) the centroid of any one is tne orthocentr« of%he tnangli formed by the centroids of the other three 14. The middle points of the three diagonals of a complete quadri- three diagonals as diameters have the same radical axis ; t^ rad,cal axis is perpendicular to t^ straight Ime through the midde points of the dia^nals, and it contfins the ortStres of the four triangles formed by taking the sides of the quadri! Uteral three and three. (The first part of this theorem Ts ascribed to Gauss. 1810 ; the last part is due to CnTr See his QemmmeUe Werke, vol. i. p. 128?) 15. In a given circle to inscribe a triangle (a) whose thi^e sides shall be parallel to three given straight lines (6) two of whose s,des shall be parallel to Lo given^str^Tght lines, and the third shall pass through a given ix,inr 364 Euclid's elements. [Book VL (c) two of whose sides shall pass through two given points, and the third shall be parallel to a given straight line. C<0 whoee three sides shall pass through three given points. [The last of these problems is often called Castillon's, whose solution was published in 1776. A very full history, by T. S. Davies, both of it and of the more general problems to which it gave rise, will be found in The Mathematician, vol. iil (1866), pp. 76-87, 140-154, 226-233, 311-322. It may be interesting to compare alsc? Pappofl, VIL, 105, 107, 108, 109, 117.] IBM END I k : , [Book Vt points, and ine. points. a's, whose , by T. S. lich it gave pp. 76-87, mpare ainq