m ''■r-r: '.fe*::^ M w rS^ ?^*';.'?ii my^ ■•'it "r^- ■ L/Cl^iyC C^ . /(Ji^^.-'^C-v^.a^-#--»-«--<^L_, -f- "/T THE ELEMENTS OF EUCLID, VIZ. THE FIRST SIX BOOKS, TOGETHER WITH THE ELEVENTH AND TWELFTH THE ERRORS, »Y WHICH THEON, OR OTHERS, HAVE LONG AGO VITIATED THESE BOOKS, ARE CORRECTED, AND SOME OF EUCLID'S DEMONSTRATIONS ARE RESTORED. ALSO, THE BOOK OF EUCLID'S DATA, IN LIKE MANNER CORRECTED. BY ROBERT SIMSON, M. D. EMERITUS PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF GLASGOW. TO THIS EDITION ARE ALSO ANNE.XED, ELfiMENTS OF PLANE AND SPHERICAL TRIGONOMETRY. PHILADELPHIA : PUBLISHED BY MATHEW CAREY, AND SOLD BY J. CONRAD & CO. S. F. BRADFORD, BIRCH & SMALL, AND SAMUEL ETHERIDGE. yRINTED BY T. ISf G. PALMER, 116, HIGH-STR»«T. 1806. TO THE KING, THIS EDITION OF THE PRINCIPAL BOOKS or THE ELEMENTS OF EUCLID, AND OF THE BOOK OF HIS DATAy IS MOST JIUMBLY DEDICATED BY HIS MAJESTY'S MOST DUTIFUL, AND MOST DEVOTED SUBJECT AND SERVANT ROBERT SIMSON. PREFACE. The opinions of the modems concerning the author of the Elements of Geometry, which go under Euclid's name, are very different and contrary to one another. Peter Ramus ascribes the propositions, as well as their demonstrations, to Theon ; others think the propositions to be Euclid's, but that the demonstrations are Theon's ; and others maintain that all the propositions and their demonstrations are Euclid's own. John Buteo and Sir Henry Savile are the authors of greatest note who assert this last, and the greater part of geometers have ever since been of this opinion, as they thought it the most probable. Sir Henry Savile, after the several arguments he brings to prove it, makes this conclusion (page 1 3 Praelect.), " That, excepting a very few interpolations, explications, and " additions, Theon altered nothing in Euclid." But, by often, considering and comparing together the definitions and de- monstrations as they are in the Greek editions we now have, I found that Theon, or whoever was the editor of the present Greek text, by adding some things, suppressing others, and mixing his own with Euclid's demonstrations, had changed more things to the worse than is commonly supposed, and those not of small moment, especially in the fifth and eleventh books of the Elements, which this editor has greatly vitiated ; for instance, by substituting a shorter, but insufficient demon- stration of the 18th prop, of the 5th book, in place of the legiti- mate one which Euclid had given ; and by taking out of this book, besides other things, the good definition which Eudoxus or Euclid had given of compound ratio, and given an absurd one in place of it in the 5th definition of the 6th book, which nei- ther Euclid, Archimedes, Appolonius, nor any geometer be- fore Theon's time, ever made use of, and of which there is not to be found the least appearance in any of their writings ; THE ELEMENTS OF EUCLID. BOOK I. / DEFINITIONS. ^ ' I. Book I. A POINT is that which hath no parts, or which hath no mag- 1 ^^ nitude. N?tes. 11. A line is length without breadth. III. The extremities of a line are points. IV. A straight line is that which lies evenly between its extreme points. V. A supei'ficies is that which hath only length and breadth. VI. The extremities of a superficies are lines. VII. A plane superficies is that in which any two points being taken, See N. the straight line between them lies wholly in that superficies. VIII. " A plane angle is the inclination of two lines to one another See N, " in a plane, which meet together, but are not in the same " direction." IX. A plane rectilineal angle is the inclination of two straight lines to one another, which meet together, but are not in the same straight line. B 10 Book I. THE ELEMENTS B N. B. ' When several angles are at .one point B, any one of * them is expressed by three letters, of which the letter that is * at the vertex of the angle, that is, at the point in which the ' straight lines that contain the angle meet one another, is put * between the other two letters, and one of these two is some- ' where upon one of those straight lines, and the other upon 'the other line: thus the angle which is contained by the ' straight lines AB, CB is named, the angle ABC, or CBA ; that ' vv'hich is contained by AB, DB is named the angle ADB, or ' DBA ; and that which is contained by DB, CB is called the * angle DBC, or CBD ; but, if there be only one angle at a ' point, it may be expressed by a letter placed at that point ; as < the angle at E.' X. When a straight line standing on ano- ther straight lirhe makes the adjacent angles equal to one another, each of the angles is called a right angle'; and the straight line which stands on the other is called a perpendicular to it. XI. ^n obtuse angle is that v/hich is greater than a right angle. XII. An acute angle is that which is less than a right anp-le. XIII. " A term or boundary is the extremity of any thing." XIV. A figure is th^it vvhich is enclosed by onQ or more boundaries. OF EUCLID. 11 XV. Book h 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 circumferenccj are equal to one another : XVI. And this point is called the centre of the circle. XVII. A diameter of a circle is a straight line drawn through the cen- tre, and terminated both wdys by the circumference. XVIII. A semicircle is the figure contained by a diameter and the part of the circuniference cut off by the diameter. XIX. " A segment of a circle is the figure contained by a straight line " and the circumference it cuts off." XX. Rectilineal figures are those which are contained by straight lines. XXI. Trilateral figures, or triangles, by three straight lines. XXII. Quadrilateral, by four straight lines. XXIII. Multilateral figures, or polygons, by more than four straight lines. XXIV. Of three sided figures, an equilatei'al triangle is that which has three equal sides. XXV. An isosceles triangle is that which has only two sides equal. THE ELEMENTS XXVI. A scalene triangle is that which has three unequal sides. XXVII. A right angled triangle is that which has a right angle. XXVIII. An obtuse angled triangle is that which has an obtuse angle. XXIX. An acute angled triangle is that which has three acute angles. XXX. Of four sided figures, a square is that which has all its sides tqual, and all its angles right angles. XXXI. An oblong is that which has all its angles right angles, but has not all its sides equal. XXXII. A rhombus is that which has all its sides equal, but its angles are not right angles. XXXIII. A rhomboid is that which has its opposite sides equal to one another,''but all its sides are not equal, nor its angles right angles. OF EUCLID* XXXIV. All other four sided figures, besides these, are called trape- ziums. XXXV. Parallel straight lines are such as are in the same plane, and which being produced ever so far both ways, do not meet. POSTULATES. I. LET it be granted that a straight line may be drawn from any one point to any other point. II. That a terminated straight line may be produced to any length in a straight line. III. And that a circle may be described from any centre, at any distance from that centre. AXIOMS. I. THINGS which are equal to the same are equal to one an- other. II. If equals be added to equals, the wholes are equal. III. If equals be taken from equals, the remainders are equal. IV. If equals be added to unequals, the wholes are unequal. V. If equals be taken from unequals, the remainders are unequal. VL Things which are double of the same are equal to one another. VII. Things which are halves of the same are equal to one another. VIII. Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another. 14 THE ELEMENTS Book I. IX. ^— v*^ The whole is greater than its part. X. Two straight lines cannot enclose a space. XL All right angles are equal to one another. XII. " If a straight line meets two straight lines, so as to make the " two interior angles on the same side of it taken together " less than two right angles, these straight lines being con- " tinually produced, shall at length meet upon that side on " which are the angles which are less than two right angles. " See the notes on prop. 29. of book I." OF EUCLID. PROPOSITION I. PROBLEM. TO describe an equilateral triangle upon a given finite straight line. Let AB be the given straight line ; it is required to describe an equilateral triangle upon it. C From the centre A, at the dis- tance AB, describe * the circle . y"^ /^)^\ ^^N. ^ ^- ^°^' BCD, and from the centre B, at / // \\ \ tulate. the distance BA, describe the cir- cle ACE ; and from the point C, in which the circles cut one ano- ther, draw the straight lines '^ C A, \ \ /I bl.Post. CB to the points A, B ; ABC shall be an equilateral triangle. Because the point A is the centre of the circle BCD, AC is equal <= to AB ; and because the point B is the centre of thee 15. De- circle ACF, BC is equal to BA : but it has been proved that CA finition. is equal to AB ; therefore CA, CB are each of them equal to AB ; but things which are equal to the same are equal to one another d ; therefore CA is equal to CB ; wherefore CA, AB, BC d 1st Ax- are equal to one another ; and the triangle ABC is therefore ^°'"* equilateral, and it is described upon the given straight line AB. Which was required to be done. PROP. II. PROB. FROM a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line ; it is required to draw from the point A a straight line equal to BC. From the point A to B draw* the straight line AB ; and upon it describe ^ the equilateral triangle DAB, and produce « the straight lines D A, DB to E and F ; from' the centre B, at the distance BC, describe "^ the circle CGH, and from the centre D, at the distance DG, describe the circle GKL. AL chall be equal to BC. a 1. Post. bl. 1. . c 2. Post, d 3. Post, 16 THE ELEMENTS Book I. Because the point B is the centi-e of the circle CGH, BC is *— V— ' equal e to BGj and because D is the centre of the ciixle GKL, elS.Def. DL is equal to DG, and DA, DB, parts of them, are equal; i 3. Ax. therefore the remainder AL is equal to the remainder*" BG: but it has been shown, that BC is equal to BG ; wherefore AL and BC are each of them equal to BG ; and things that are equal to the same are equal to one another ; therefore the straight line AL is equal to BC. Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC. Which was to be done. PROP. in. PROB. FROM the greater of two given straight lines to cut off a part equal to the less. Let AB and C be the two given straight lines, whereof AB is the greater. It is required to cutoff from AB, the greater, a partj equal to C, the less. a 2. 1. From the point A draw ^ the straight line AD equal to C ; and from the centre A, and at the dis- b 3. Post, tance AD, describe ^ the circle DEF ; and because A is the cen- F tre of the circle DEF, AE shall be equal to AD ; but the straight line C is likewise equal to AD ; whence AE and C are each of them equal to AD: wherefore the straight line AE is equal to*^ C, and from AB, the greater of two straight lines, a part AE has been cut off equal to C the less. Which was to he done. PROP. IV. THEOREM. IF two triangles have two sides of the one equal to two sides of the other, each to each ; and have likewise the angles contained by those sides equal to one another ; they shall likewise have their bases, or third sides, equal ; and the two triangles shall be equal ; and their other angles shall be equal, each to each, viz. those to which the equal sides are opposite. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. c 1. Ax. OF EUCLID. 17 AB to DE, and AC to DF ; A D Book I. and the angle BAG equal to the angle EDF, the base BC shall be equal to the base EF ; and the triangle ABC to the triangle DEF ; and the other angles, to which the equal sides are opposite, shall be equal, each to each, viz. the angle ABC to the B angle DEF, and the angle ACB to DFE. For, if the triangle ABC be applied to DEF, so that the point A may be on D, and the straight line AB upon DE ; the point B shall coincide with the point E, because AB is equaj to DE ; and AB coinciding with DE, AC shall coincide with DF, be- cause the angle BAC is equal to the angle EDF; wherefore also the point C shall coincide with the point F, because the straight line AC is equal to DF : but the point B coincides with the point E; wherefore the base BC shall coincide with the base EF, because the point B coinciding with E, and C Avith F, if the base BC does not coincide with the base EF, two straight lines would inclose a space, which is impossible^. Therefore a 10. Axi the base BC shall coincide with the base EF, and be equal to it. Wherefore the whole triangle ABC shall coincide with the whole triangle DEF, and be equal to it ; and the other angles of the one shall coincide with the remaining angles of the other, and be equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to DFE. Therefore, if two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to one another, their bases shall likewise be equal, and the triangles be equal, and their other angles to which the equal sides are opposite shall be equal, each to each. Which was to be demonstrated. PROP. V. THEOR. THE angles at the base of an isosceles triangle are equal to one another ; and, if the equal sides be pro- duced, the angles upon the other side of the base shall be equal. Let ABC be an ir,osceIes triangle, of which the side AB ip C 18 THE ELEMENTS Book I. equal to AC, and let the straight lines AB, AC be produced to ^ — y*— ■>. D and E ; the angle ABC shall be equal to the angle ACB, and the ongle CBD to the angle BCE. In BD take any point F, and ffom AE the greater, cut off AG a 3. 1. equals to AF, the less, and join FC, GB. Because AF is equal to AG, and AB to AC, the two sides FA, AC are equal to the two GA, AB, each to each; and they contain the angle FAG conj- mon to the two triangles AFC, AGB ; therefore the base FC is b 4. 1. equal ^ to the base GB, and the tri- angle AFC to the triangle AGB; and the remaining angles of the one are equal •^ to the remaining angles of the other, each to each, to which the equal sides are opposite ; viz. the angle ACF to the angle ABG, and the angle AFC to the angle AGB : and because the whole AF is equal to the whole AG, of which the parts AB, AC are equal ; the c 3. Ax. remainder BF shall be equal = to the remainder CG ; and FC Avas proved to be equal to GB ; therefore the two sides BF, FC are equal to the two CG, GB, each to each ; and the angle BFC is equal to the ?ngle CGB, and the base BC is common to the two triangles BFC, CGB ; wherefore the triangles are equal W and their remaining angles, each to each, to which the equal sides are opposite ; therefore tlie angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG : and, since it has been demonstrated that the whole angle ABG is equal to the whole ACF, the parts of which, the angles CBG, BCF, are also equal ; the remaining angle ABC is therefore equal to the re- nnaining angle ACB, v/hicii are the angles at the base of the triangle ABC : and it has also been proved that the angle FBC is equal'to the anL;le GCB, wliich are the angles upon the other side of the base. Therefore the angles at the base. Sec. Q. E. D. CoROLLARv. Hence every equilateral triangle is also equi- angular. PROP. VI. THEOR. IF two angles of a triangle be equal to one another, the sides also which subtend, or are opposite tOy the equal angles shall be equal to one another. OF EUCLID. 19 Let ABC be a triangle haviiicc the angle ABC equal to the Book I. angle ACB ; the side AB is also equal to the side AC. ^.— y.^^ For if AB be not equal to AC, one of them is greater than the other; let AB be the greatei', and from it cut » off DB a 3. 1. equal to AC, the less, and join DC ; there- A fore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides DB, BC are equal to to the two AC, CB, each to each ; and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC is equal to the triangle'' ACB, the less to the great- / \\ b^.!. er ; which is absurd. Therefore AB is not unequal to AC, that is, it is equal to it. Wherefore, if two angles, 8cc. Q. E. D. ^ CoR. Hence every equiangular triangle is also equilateral. PROP. Vn. THEOR. UPON the same base, and on the same side of it, See n. there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity. If it be possible, let there be two triangles ACB, ADB, up- on the same base AB, and upon the same side of it, which have their sides CA, DA terminated in the extremity A of the base equal to one another, and likewise q q their sides CB, DB that are termi- nated in B. Join CD ; then, in the case in which the vertex of each of the tri- angles is without the other triangle, because AC is equal to AD, the angle ACD is equal * to the angle ADC : but the angle ACD is greater than the angle BCD ; therefore the angle ADC is greater also than BCD ; ^ much more then is the angle BDC greater than the angle BCD. Again, because CB is equal to DB, the angle BDC is equal a to a 5. 1. the angle BCD ; but it has been demonstrated to be greater than it, Avhich is impossible. 20 THE' ELEMENTS Book I. Bui ii'one of the vertices, as D, be within the other triangle *— v— ' ACB ; produce AC, AD to E, F ; there- E fore, because AC is equal to AD in the triangle ACD, the angles ECD, FDC upon the other side of the base CD are a 5. 1. equal a to one another, but the angle ECD is greater than the angle BCD ; wherefore the angle FDC is likewise greater than BCD ; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal to DB, the angle BDC is equal ^ to the angle- BCD ; but BDC has been proved to be A B greater than the same BCD; which is impossible. The case in which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore upon the same base, and on the- same side' of it, there cannot be two triangles that have their sides which are terminated in one extremity of the l)ase equal to one another, and likewise those winch are terminated in the other extremity* Q. E. D. PROP. VIII. THEOR. IF two triangles have two sides of the one equal to two sides of the other, each to each, and have like- wise their bases equal ; the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides equal to them, of the other. liCt ABC, DEF be tAvw triangles, having the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB to 1)E, and AC to A D G DF ; and also the base BC equal to the base EF. The angle BAC is e- qual to the angle EDF. For, if the tri- angle ABC be ap- plied to DEI', so B ' C E F that the point B be on E, and the straight line BC upon EF : the point C shall also coincide with the point F. Because OF EUCLID. 21 BC is equal to EF ; therefore BC coinciding with EF, BA and Book I. AC shall coincide with ED and DF ; for, if the base BC coin- y-m- y mmJ cides with the base EF, but the sides BA, CA do not coincide with the sides ED, FD, but have a diiferent situation, as EG, FG ; then, upon the same base EF, and upon the same side of it, there can be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise their sides terminated in the other extremity ; but this is impossible^; therefore, if the base BC coincides with the a 7- 1. base EF, the sides BA, AC cannot but coincide with the sides ED, DF ; wherefore, likewise, the angle BAC coincides with the angle EDF, and is equal ^ to it. Therefore, if two triangles, b 8. Ax. kc. Q. E. D. PROP. IX. PROB. TO bisect a given rectilineal angle, that is, to di- vide it into two equal angles. Let BAC be the given rectilineal angle ; it is required to bi- sect it. Take any point D in AB, and from AC cut-off AE equal to a 3. 1. AD ; join DE, and upon it describe ^ A b 1. L an equilateral triangle DEF ; then join AF ; the straight line AF bisects the angle BAC. Because AD is equal to AE, and AF is common to the two triangles DAF, EAF ; the two sides DA, AF are equal to the two sides EA, AF, each to each ; and the base DF is equal to the base EF ; therefore the angle DAF is equal ^ to the angle B' C c 8. 1. EAF; wherefore the given rectilineal angle BAC is bisected by the straight line AF. Which was to be done. PROP. X. PROB. TO bisect a given finite straight line, that is, to divide it into two equal parts. Let AB be the given straight line ; it is required to divide it into two equal parts. Describe ^ upon it an equilateral triangle ABC, and biscc'. ^ a 1. 1. the angle ACB by the straight line CD. AB is cut into two b 9. 1- equal parts in the point D. 22 THE ELEMENT Book I. Because AC is equal to CB, and CD ^■-y*^ common to the two triangles ACD, BCD ; the two sides AC, CD are equal to BC, CD, each to each ; and the angle ACD is equal to the angle BCD ; there- c 4, 1. fore the base AD is equal to the base ^ DB, and the straight line AB is divided into two equal parts in the point D. WJjiich was to be done. PROP. XI. FROB. SeeN. a 3,1. t> 1. 1. TO draw a straight line at right angles to a given straight line, from a given point in the same. Let AB be a given straight line, and C a point given in it ; it is required to draw a straight line from the point C at right an- gles to AB. Take any point D in AC, and » make CE equal to CD, and upon DE describe '' the equi- lateral triangle DFE, and join FC ; the straight line FC drawn from the given point C is at right angles to the given straight line AB. Because DC is equal to CE, and FC common to the tvi^o ? triangles DCF, ECF; the two A D C E B sides DC, CF are equal to the two EC, CF, each to each ; and the base DF is equal to the base EF; therefore the angle DCF G 8. 1. is equal <: to the angle ECF ; and they are adjacent angles. But, when the adjacent angles whi[:h one straight line m-ikes with another straight line are equal to one another, each of them <110.Def. is called a right *! angle ; therefore each of the angles DCF, 3- ECF is a right angle. Wherefore, from the given point C, in the given straigitt line AB, FC iias been drawn at I'ight angles to All. Which was to be done. Cor. By help of this ])iob!eni, it may be demonstrated, that two straight lines cannot have a common segment. If it be possible, let the two straight lines ABC, ABD have the segment AB common to both of them. From the point B draw BE at right angles to- AB ; and because ABC is a straight OF EUCLID. 23 line, the angle CBE is equal -"^ to the angle EBA ; in the same manner, because ABD is a straight line, the angle DBE is equal to the angle EBA; where- fore tlie angle DBE is equal to the angle CBE, the less to the greater, which is impossible ; therefore two straight lines can- not have a common segment. E Book I. D A B C b 3- Post. 10. 1. PROP. XII. PROB. TO draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it. Let AB be the given straight line, which may be produced to any length both ways, and let C be a point without it ; it is re- quired to drav/ a straight hne C perpendicular to AB from the point C. Take any point D upon the other side of AB, and from the centre C, at the distance CD, describe "^ the circle EGF meeting AB in F. G ; and bi- sect <= EG in H, and join CF, CH, CG ; the straight line CH, drawn from the given point C, is perpendicular to the given straight line AB. Because EH is equal to HG, and IIC common to the two triangles EHC, GHC, the two sides EH, HC are equal to the two GH, HC, each to each ; and the base CF is equal d to thedl5.Def. base CG ; therefore the angle CHF is equal e to the angle CHG ; 1- and they are adjacent angles ; but wh.?n a straight line standing e 8. 1. on a straight line makes the adjacent angles equal to one ano- ther, each of them is a right angle, and the straight line which stands upon the other is called a perpendicular to it ; therefore from the given point C a perpendicular CH has been drav/n to the given straight line AB. Which v/as to be done. PROP. XIII. THEOR. THE angles which one straight line makes with another upon the one side of it, are either two right angles, or are together equal to two right angles. 24 Book I. THE ELEMENTS Let the straight line AB make with CD, upon one side of it, the angles CBA, ABD ; these are either two right angles, or are together equal tu two right angles. For, if the angle CBx\ be equal to ABD each of them is a aDef.lO. right » angle; but if not, from the point B draw* BE at right A E A D B C D B C c 2. Ax. b 11. 1. angles ^ to CD ; therefore the angles CBE, EBD are two right angles^ ; and because CBE is equal to the two angles CBA, ABE together, add the angle EBD to each of these equals ; there- fore the angles CBE, EBD are equal «= to the three angles CBA, ABE, EBD. Again, because the angle DBA is equal to the two angles DBE, EBA, add to these equals the angle ABC ; tiierefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC ; but the angles CBE, EBD have been de- monstrated to be equal to the same three angles ; and things (1 1. Ax. that are equal to the same are equal ^ to one another ; therefore the angles CBE, EBD are equal to, the angles DBA, ABC; but CBE, EBD are two right angles ; therefore DBA, ABC are to- gether equal to two right angles. Wherefore, when a straight line, &c. Q. E. D. PROP. XIV. THEOPv. IF, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line. At the point B, in the straight -^ line AB, let the two straight lines BC, BD, upon the opposite sides of AB, make the adjacent angles ABC, ABD equal together to two right angles; BD is in the same straight line with CB. For, if BD be not in the same' straight lii^e with CB, let BE be C B D OF EUCLID. 33 in the same straight line with it; therefore, because the straight Book I. line AB makes angles with the straight line CBE, upon one side v. ^. y —^ of it, the angles ABC, ABE are together equal » to two right a 13. 1. angles ; but the angles ABC, ABD are likewise together equal to two right angles ; therefore the angles CBA, ABE are equal to the angles CBA, ABD : take away the common angle ABC, the remaining angle ABE is equal ^ to the remaining angle b 3. Ax, ABD, the less to the greater, which is impossible ; therefore BE is not in the same straight line with BC. And, in like manner, it may be demonstrated, that no other can be in the same straight line with it but BD, which therefore is in the same straight line with CB. Wherefore, if at a point, Sec, Q. E. D. PROP. XV. THEOR. IF two straight lines cut one another, the vertical, or opposite, angles shall be equal. Let the two straight lines AB, CD cut one another in the point E ; the angle AEC shall be equal to the angle DEB, and CEB to AED. Because the straight line AE makes with CD the angles CEA, AED, these angles are together equal » to two right angles. ^v^ ^ * ^^' ^' Again, because the straight line DE makes with AB the angles AED, DEB, these also are to- gether equal * to two right angles ; and CEA, AED have been de- monstrated to be equal to two right angles ; wherefore the angles CEA, AED are equal to the angles AED, DEB. Take away the common angle AED, and the remaing angle CEA is equal '^ b 3. Ax, to the remaining angle DEB. In the same manner it can be de- monstrated that the angles CEB, AED are equal. Therefore, if two straight lines, 8cc. Q. E. D. CoR. 1. From this it is manifest, that, if two straight lines cut one another, the angles they make at the point where they cut are together equal to four right angles. Cor. 2. And, consequently, that all the angles made by any number of lines meeting in one point, are together equal to four right angles. D 26 THE ELEMENTS a 10. 1. Book I. PROP. XVI. THEOR. IF one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. Let ABC be a triangle, and let its side BC be produced to D ; the exterior angle ACD is greater than either of the interior op- posite angles CBA, BAG. A Bisect a AC in E, join BE and produce it to F, and make EF equal to BE ; join also FC, and produce AC to G. Because AE is equal to EC, and BE to EF ; AE, EB are equal to CE, EF, each to each ; and the angle AEB is equal •» to the angle CEF, because they are oppo- site vertical angles; therefore the base AB is equal ^ to the base CF, and the triangle AEB to the triangle CEF, and the remaining angles to the re- maining angles, each to each, to which the equal sides are oppo- site ; wherefore the angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle ECF ; therefore the angle ACD is greater than BAE : in the same manner, if the side BC be bisected, it may be demonstrated that the angle BCG, d 15. 1. that is ^, the angle ACD, is greater than the angle ABC. There- fore, if one side, &c. Q. E. D. b 15 1. c4. 1. PROP. XVn. THEOR. ANY two angles of a triangle are together less than two right angles. al6. 1. Let ABC be any triangle ; any two of its angles together are less than two right angles. Produce BC to D ; and be- cause ACD is the exterior angle of the triangle ABC, ACD" is greater * than the interior and opposite angle ABC ; to each of OF EUCLID. 87 these add the angle ACB ; therefore the angles ACD, ACB are Book L greater than the angles ABC, ACB ; but ACD, ACB are to- * — v— ^ gether equal ^ to two right angles ; the-refore the angles ABC, b ^3. 1. BCA are less than two right angles. In Hke manner, it may be demonstrated, that BAC, ACB, as also CAB, ABC, are less than two right angles. Therefore, ajiy two angles, Sec. Q. E. D. PROP. XVIII. THEOR. THE greater side of every triangle is opposite to 2 greater anele. o the greater angle. Let ABC be a triangle, of wlUeh the side AC is greater than the side AB ; the angle ABC is also greater than the angle BCA. Because AC is greater than AB, make^ AD equal to AB, and join / ^ — — "■ \ a 3. 1. BD ; and because ADB is the ex- p terior angle of the triangle BDC, " it is greater i» than the interior and opposite angle DCB ; butb 16. 1. ADB is equal <= to ADB, because the side AB is equal to the side c 5. 1. AD ; therefore the angle ABD is likewise greater than the angle ACB ; wherefore much more is the angle ABC greater than ACB. Therefore, the greater side, Sec. Q. E. D. PROP. XIX. THEOR. THE greater angle of every triangle is subtended by the greater side, or bas the greater side opposite to it. Let ABC be a triangle, of which the angle ABC is greater than the angle BCA; the side AC is likewise greater than the side AB. For, if it be not greater, AC must either be equal to AB, or less than it ; it is not equal, because then the angle ABC would be equal » to the angle ACB ; but it is not ; therefore AC is not equal to AB ; neither is it less ; because then the angle ABC would be lessi> than the angle ACB ; B C b 18. 1, a5. 1. 28 THE ELEMENTS Book I. but it is not ; therefore the side AC is not less than AB ; and it •— "v^^ has been shown that it is not equal to AB ; therefore AC is greater than AB. Wherefore, the greater angle, See. Q. E. D. PROP. XX. THEOR. See N. ANY two sidcs of a triangle are together greater than the third side. Let ABC be a triangle ; any two sides of it together are greater than the third side, viz. the sides BA, AC greater than the side BC ; and AB, BC greater than AC ; and BC, CA great- er than AB. Produce BA to the point D, and a 3. 1. make ^ AD equal to AC ; and join DC. Because DA is equal to AC, the b 5. 1. angle ADC is likewise equal '' to ACD ; but the angle BCD is great- er than the angle ACD ; therefore the angle BCD is greater than the B angle ADC ; and because the angle BCD of the triangle DCB c 19. 1. is greater than its angle BDC, and that the greater <= side is op- posite to the greater angle ; therefore the side DB is greater than the side BC ; but DB is equal to BA and AC ; therefore the sides BA, AC are greater than BC. In the same manner it may be demonstrated, that the sides AB, BC are greater than CA, and BC, CA greater than AB. Therefore, any two sides, Sec. Q. E. D. PROP. XXL THEOR. See N. IF, from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the tri- angle, but shall contain a greater angle. Let the two straight lines BD, CD be drawn from B, C, the ends of the side BC of the triangle ABC, to the point D within it ; BD and DC are less than the other two sides BA, AC of the triangle, but contain an angle BDC greater than the an- gle BAC. Produce BD to E ; and because two sides of a triangle are greater than the third side, the two sides BA, AE of the tri- OF EUCLID. 29 angle ABE are greater than BE. To each of these add EC ; Book I. therefore the sides BA, AC are A ^'--r-^ greater than BE, EC : again, because the two sides CE, ED of the triangle CED are great- er than CD, add DB to each of these ; therefore the sides CE, EB are greater than CD, DB ; but it has been shown that BA, AC are greater than BE, EC ; xnuch more then are BA, AC greater than BD, DC. Again, because the exterior angle of a triangle is greater than the interior and opposite angle, the exterior angle BDC of the triangle CDE is greater than CED ; for the same reason, the exterior angle CEB of the triangle ABE is greater than BAC; and it has been demonstrated that the angle BDC is greater than the angle CEB ; much more then is the angle BDC greater than the angle BAC. Therefore, if from the ends of, &c. Q. E. D* PROP. XXII. PROB. TO make a triangle of which the sides shall be See n. equal to three given straight lines, but any two what- ever of these must be greater than the third ^ a20. i Let A, B, C be the three given straight lines, of which any two whatever are greater than the third, viz. A and B greater than C ; A and C greater than B ; and B and C than A. It is required to make a triangle of which the sides shall be equal to A, B, C, each to each. Take a straight line DE terminated at the point D, but un- limited towards E, and make ^ DF equal to A, FG to B, and GH equal to C ; and from the centre F, at the distance FD, de- scribe '' the circle DKL ; and from the centre G, at the distance GH, describe ^ another circle HLK ; and join KF, KG; the triangle KFG has its sides equal to the three straight lines A, B, C. Because the point F is the centre of the circle DKL, FD a 3.1. .Post. 30 THE ELEMENTS Book I. equal ^ to FK ; but FD is equal to the straight line A ; there- ^— >r— ' fore FK is equal to A : again, because G is the centre of the cl5.Def. circle LKH, GH is equal « to GK ; but GH is equal to C; therefore also GK is equal to C ; and FG is equal to B ; there- fore the three straight lines KF, FG, GK are equal to the three A, B, C : and therefore the triangle KFG has its three sides KF, FG, GK equal to the three given straight lines A, B, C. Which was to be done. a 22. 1. b 8. 1. PROP. XXIII. PROB. AT a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle : it is required to make an angle at the given point A in the given straight line AB, that shall be equal to the given rectilineal an- gle DCE. Take in CD, CE any points D, E, and join DE ; and make ^ the triangle AFG, the sides of vi'hich shall be equal to the three straight lines CD, DE, CE, so that CD be equal to AF, CE to AG, and DE to FG, and because DC, CE are equal to FA, AG, each to each, and the base DE to the base FG ; the angle DCE is equal ^ to the angle FAG. Therefore, at the given point A, in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Which was to be done. PROP. XXIV. THEOR. See N. IF two tnanglcs have two sides of the one equal to two sides of the^other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them, of the other ; the base of that which has the greater angle shall be greater than the base of the other. OF EUCLID. SI Let ABC, DEF be two triangles which have the two sides Book I, AB, AC equal to the two DE, DF, each to each, viz. AB equal '— v— ' to DE, and AC to DF ; but the angle BAC greater than the an- gle EDF ; the base BC is also greater than the base EF. Of the two sides DE, DF, let DE be the side which is not greater than the other, and at the point D, in the straight line DE, make » the angle EDG equal to the angle BAC ; and make a 23. 1. DG equal b to AC or DF, and join EG, GF. b 3. 1. Because AB is equal to DE, and AC to DG, the two sides BA, AC are equal to the two ED, DG, each to each, and the angle BAC is equal A D to the angle EDG ; therefore the base BC is equal = to the base EG; and because DG is equal to DF, the angle DFG is equal"^ to the angle DGF; but the angle DGF is greater than the an- gle EGF ; therefore the angle DFG is greater than EGF ; and much more is the angle EFG greater than the angle EGF ; and because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater e side is opposite to the greater angle; the side EG^ ^^' ^' is therefore greater than the side EF; but EG is equal to BC ; and therefore also BC is greater than EF. Therefore, if two triangles, Sec. Q. E. D. c4. 1. d5. 1. PROP. XXV. THEOR. IF two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other ; the angle also contained by the sides of that which has the greater base, shall be greater than the angle contained by the sides equal to them, of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF ; but the base CB is greater than the base EB' ; the angle BAC is likewise greater than the anele EDF. 32 THE ELEMENTS Book I. For, if it be not greater, it must either be equal to it, or less; ^— v "-~^ but the angle BAG is not equal to the angle EDF, because then the base BC would a4fl- beequalatoEF; but A D it is not ; therefore the angle BAG is not equal to the angle EDF; neither is it less ; because then the base BG would b 24. 1. be less ^ than the base EF ; but it is not ; therefore the angle BAG is not less than the angle EDF ; and it was shown that it is not equal to it ; therefore the angle BAG is greater than the angle EDF. Wherefore, if two triangles, &c. Q. E. D. PROP. XXVI. THEOR. IF two triangles have two angles of one equal to two angles of the other, each to each ; and one side equal to one side, viz. either the sides adjacent to the equal angles, or the sides opposite to equal angles in each ; then shall the other sides be equal, each to each; angle of the other. and also the third angle of the one to the third Let ABG, DEF be two triangles which have the angles ABG, BGA equal to the angles DEF, EFD, viz. ABG to DEF, and BGA to EFD ; also one side equal to one side ; and first let those sides be equal which are adjacent to the angles that are equal in the two tri- A D angles, viz. BG to EF ; the other sides q. shall be equal, each to each, viz. AB to DE, and AG to DF ; and the third angle BAG to the third an- gle EDF. For, if AB be not B G E F equal to DE, one of them must be the greater. Let AB be the greater of the two, and make BG equal to DE, and join GC ; therefore, because BG is equal to DE, and BG to EF, the two OF EUCLID. 3S sides GB, BC are equal to the two DE, EF, each to each ; and Book I. the angle GBC is equal to the angle DEF ; therefore the base ^■— v— > GC is equal ^ to the base DF, and the triangle GBC to the tri- a 4. 1. angle DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite ; therefore the angle GCB is equal to the angle DFE ; but DFE is, by the hypothe- sis, equal to the angle BCA ; wherefore also the angle BCG is equal to the angle BCA, the less to the greater, which is im- possible ; therefore AB is not unequal to DE, that is, it is equal to it ; and BC is equal to EF ; therefore the two AB, BC are equal to the two DE, EF, each to each ; and the angle ABC is equal to the angle DEF ; the base therefore AC is equal » to the base DF, and the third angle BAG to the third angle EDF. Next, let the sides which are opposite to A D equal angles in each triangle be equal to one another, viz. AB to DE ; likewise in this case, the other sides shall be equal, AC to DF, and BC to EF : and also the third angle BAG to the third EDF. For, if BC be not equal to EF, let BC be the greater of them, and make BH equal to EF, and join AH ; and because BH is equal to EF, and AB to DE ; the two AB, BH are equal to the two DE, EF, each to each ; and they contain equal angles ; therefore the base AH is equal to the base DF, and the triangle ABH to the triangle DEF, and the other angles shall be equal, each to each, to which the equal sides are opposite ; therefore the angle BHA is equal to the angle EFD ; but EFD is equal to the angle BCA ; therefore also the angle BHA is equal to the angle BCA, that is, the exterior angle BHA of the triangle AHC is equal to its interior and opposite angle BCA; which is impossible'*; wherefore BC is not unequal to EF, that is, it isbl6. 1. equal to it; and AB is equal to DE ; therefore the two AB, BC are equal to the two DE, EF, each to each ; and they con- tain equal angles ; wherefore the base AC is equal to the base DF, and the third angle BAG to the third angle EDF. There- fore, if two triangles, &c. Q. E. D. 34 THE ELEMENTS Book I. PROP. XXVII. THEOR. IF a straight line falling upon two other straight lines makes the alternate angles equal to one another, these two straight lines shall be parallel. . Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD equal to one another ; AB is parallel to CD. For, if it be not parallel, AB and CD being produced shall meet either towards B, D, or towards A, C ; let them be pro- duced and meet towards B, D, in the point G ; therefore GEF a 16. 1. is a triangle, and its exterior angle AEF is greater » than the in- terior and opposite angle EFG ; but it is also equal ^ / to it, which is impossible; . ^f ^ therefore AB and CD be- ing produced do not meet towards B, D. In like man- ner it may be demonstrat- ed, that they do not meet towards A, C ; but those straight lines which meet neither way, though produced ever s» b35.Def. far, are parallel ^ to one another. AB therefore is^parallel to CD. Wherefore, if a straight line, Sec Q. E. D. PROP. XXVIII. THEOR. IF a straight line falling upon two other straight lines makes the exterior angle equal to the interior and opposite upon the same side of the line ; or makes the interior angles upon the same side together equal to two right angles ; the two straight lines shall be parallel to one another. Let the straight line EF, which E ,falls upon the two straight lines AB, CD, make the exterior angle EGB equal to the interior and op- A ■ posite angle GHD upon the same side ; or make the interior angles on the same side BGH, GHD toge- ther equal to two right angles ; AB is parallel to CD. 13ecause the angle EGB is equal to the angle GHD, and the angle OF EUCLID. 35 EGB equal * to the angle AGH, the angle AGH is equal Book I. to the angle GHD ; and they are the alternate angles ; therefore *>-— v— -^ AB is parallel ^ to CD. Again, because the angles BGH, GHD a 15. 1. are equal <= to two right angles ; and that AGH, BGH are also b 27. 1, equal d to two i^ight angles ; the angles AGH, BGH are equal cBy hyp. to the angles BGH, GHD: take away the common angle d 13. 1 BGH ; therefore the remaining angle AGH is equal to the re- maining angle GHD ; and they are alternate angles ; therefore AB is parallel to CD. Wherefore, if a straight line, Sec. Q. E. D. PROP. XXIX. THEOR. IF a straight line fall upon two parallel straight See the lines, it makes the alternate angles equal to one ano- "iJJs"°" ther ; and the exterior angle equal to the interior and position opposite upon the same side ; and likewise the two interior angles upon the same side together equal to two right angles. Let the straight line EF fall upon the parallel straight lines AB, CD ; the alternate angles, AGH, GHD are equal to one another; and the exterior angle EGB is equal to the interior and opposite, upon the same side E GHD ; and the two interior an- gles BGH, GHD upon the same . . side are together equal to two right angles. For, if AGH be not equal to ^ GHD, one of them must be great- er than the other ; let AGH be the greater ; and because the an- gle AGH is greater than the an- gle GHD, add to each of them the angle BGH ; therefore the angles AGH, BGH are greater than the angles BGH, GHD ; but the angles AGH, BGH are equal » to two right angles ; therefore a 13. 1. the angles BGH, GHD are less than two right angles ; but those straight lines which, with another straight line falling upon them, make the interior angles on the same side less than two right angles, do meet* together if continually produced ; therefore » 12. ax. the straight lines AB, CD, if produced far enough, shall meet ; See the but they never meet, since they are parallel by the hypothesis ; "o'es on therefore the angle AGH is not unequal to the angle GHD, that *^'^. P'"' is, it is equal to it ; but the angle AGH is equal ^ to the angle n'*'7' EGB; therefore likewise EGB is equal to GHD ; add to each 36 THE ELEMENTS Book I. of these the angle BGH ; therefore the angles EGB, BGH are ^— V— ' equal to the angles BGH, GHD ; but EGB, BGH are equal c~ c 13. 1. to two right angles ; therefore also BGH, GHD are equal to two right angles. Wherefore, if a straight line, Sec. Q. E. D. PROP. XXX. THEOR. STRAIGHT lines which are parallel to the same straight line are parallel to one another. Let AB, CD be each of them parallel to EF ; AB is also pa- rallel to CD. Let the straight line GHK cut AB, EF, CD ; and because GHK cuts the parallel straight lines AB, EF, the angle AGH is a 29. 1. equal ^ to the angle GHF. Again, because the straight line GK cuts the parallel straight lines EF, CD, the an^le GHF is equal to the angle GKD ^ ; and it was shown that tlie angle AGK is equal to the angle GHF ; there- fore also AGK is equal to GKD ; and they are alternate angles ; b 27. 1. therefore AB is parallel ^ to CD. Wherefore, straight lines, &c. Q. E. D. PROP. XXXL PROB. TO draw a straight line through a given point pa- rallel to a given straight line. Let A be the given point, and BC the given straight line ; it is required to draw a straight line through the point A, parallel to the E A F straight line BC. In BC take any point D, and join AD ; and at the point A, in the A 23. 1. straight line AD, make * the angle DAE equal to the angle ADC ; and B D C produce the straight line EA to F. Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal to b 27. 1. one another, EF is parallel ^ to BC Therefore the straight line OF EUCLID. 37 EAF is drawn through the given point A parallel to the given Book I. straisrht line BC. Which was to be done. V-i-Y^i^* PROP. XXXII. THEOR, IF a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles ; and the three interior angles of every triangle are equal to two right angles. Let ABC be a triangle, and let one of its sides BC be produ- ced to D ; the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC ; and the three interior angles of the triangle, viz. ABC, BCA, CAB, are together equal to two right angles. Through the point C A draw CE parallel » to the y\ ^E a 31. 1. straight line AB ; and be- cause AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are equal b. Again, B v^ ^ b 29. 1. because AB is parallel to CE, and BD falls upon them, the ex- terior angle ECD is equal to the interior and opposite angle ABC ; but the angle ACE was shown to be equal to the angle BAC ; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC ; to these equals add the angle ACB, and the angles ACD, ACB are equal to the three angles CBA, BAC, ACB ; but the angles ACD, ACB are equal c to two right angles : therefore also the angles CB A, c 13. 1. BAC, ACB are equal to two right angles. Wherefore, if a side of a triangle, Sec. Q. E. D. Cor. 1. AH the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. And, by 38 THE ELEMENTS Book I. ^^^ preceding proposition, all the angles of these triangles are \,^.^^mmJ equal to twice as many right anglts as there are triangles, that is, as there are sides of the figure ; and the same angles are equal to the angles of the figure, together with the angles at the point a 2. Cor. F, which is the common vertex of the triangles ; that is ^, to- 15. 1. gether with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. CoR. 2, All the exterior angles of any rectilineal figure, are together equal to four right angles. Because every interior angle ABC, with its adjacent exterior I>13. 1. ABD, is equal •> to two right angles ; therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure : D that is, by the foregoing corol- lary, they are equal to all the interior angles of the figure, to- gether with four right angles ; therefore all the exteripr an- gles are equal to four right angles. PROP. XXXIII. THEOR. THE Straight lines wliich join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel. Let AB, CD be equal and pa- rallel straight lines, and joined to- wards the same parts by the straight lines AC, BD ; AC, BD arc also equal and parallel. Join BC ; and because AB is parallel to CD, and IIC meets C D 91 29. 1. them, the alternate angles ABC, BCD are equal a ; and because AB is equal to CD, and BC common to the two triangles ABC, DCB, the two sides AB, BC are equal to the two DC, CB ; and the angle ABC is equal to the angle BCD ; therefore the b 4 1 base AC is equal •» to the base BD, and ihe triangle ABC to the triangle BCD, and the other angles to the other angles ^^ each to each, to which the equal sides are opposite : therefore th? OF EUCLID. 3S angle ACB is equal to the angle CBD ; and because the straight Book I. line BC meets the two straight lines AC, BD, and makes the al- ^■— y— ' ternate angles ACB, CBD equal to one another, AC is parallel ^ to BD ; and it was shown to be equal to it. Therefore, straight c 27. 1. lines, &c. Q. £. D. PROP. XXXIV. THEOR. THE opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them, that is, divides them into two equal parts. N. B. A parallelogram is a four sided figure^ of which the opposite sides are parallel ; and the diameter is the. straight line joining tvjo of its opposite angles. Let ACDB be a parallelogram, of which BC1s a diameter; the opposite sides and angles of the figure are equal to one an- other ; and the diameter BC bisects it. Because AB is parallel to CD, A B and BC meets them, the alter- nate angles ABC, BCD are equal =* to one another ; and because \ y^ \ a 29. AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal ^ to one C D another; wherefore the two triangles ABC, CBD have two angles ABC, BCA in one, equal to two angles BCD, CBD in the other, each to each, and one side BC common to the two triangles, which is adjacent to their equal angles ; therefore their other sides shall be equal, each to each, and the third angle of the one to the third angle of the other ^, viz. the b 25. 1. side AB to the side CD, and AC to BD, and the angle BAC equal to the angle BDC ; and beciuse the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD : and the angle BAC has been shown to be equal to the angle BDC ; therefore the opposite sides and angles of parallelograms are equal to one another ; also, their diameter bisects them ; for AB being equal to CD, and BC common, the two AB, BC are equal to the two DC, CB, each to each ; and the angle ABC is 40 THE ELEMENTS Book I. equal to the angle BCD ; therefore the triangle ABC is equal *'''""' <= to the triangle BCD, and the diameter BC divides the parallelo- c 4. 1. gram ACDB into two equal parts. Q. E. D. PROP. XXXV. THEOR. SeeN. PARALLELOGRAMS upon the same base, and between the same parallels, are equal to one another. figures. a 34. 1. See the^ Let the parallelograms ABCD, EBCF be upon the same base -^^J^'^^'^BC, and between the same parallels AF, BC ; the parallelogram ABCD shall be equal to the parallelogram EBCF. If the sides AD, DF of the paral- lelograms ABCD, DBCF opposite to the base BC be terminated in the same point D, it is plain that each of the parallelograms is double ^ of the trian- gle BDC ; and they are therefore equal to one another. But, if the sides AD, EF, opposite to the base BC of the parallelograms ABCD, EBCF, be not terminated in the same point, then, be- cause ABCD is a parallelogram, AD is equal ^ to BC ; for the same reason, EF is equal to BC ; wherefore AD is equal ^ to EF, and DE is common ; therefore the whole, or the remain- c 2. or 3. der AE, is equal <= to the whole, or the remainder DF ; AB also Ax. is equal to DC ; and the two EA, AB are therefore equal to the b 1. A> d 19. 1. 4. 1. Ax. two FD, DC, each to each ; and the exterior angle FDC is equal ^ to the interior EAB ; therefore the base EB is equal to the base FC, and the triangle EAB equal <= to the triangle FDC ; take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB; the remainders there- fore are equal f, that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore, parallelograms upon the same base, &c. -Q. E. D. ^ OF EUCLID. PROP. XXXVI. THEOR. PARALLELOGRAMS upon equal bases, and between the same parallels, are equal to one another. Let ABCD, EFGH be parallelograms upon equal bases BC, FG, and be- tween the same parallels AH, BG ; the parallelo- gram ABCD is equal to EFGH. Join BE, CH; and be- B C F G cause BC is equal to FG, and FG to » EH, BC is equal to EH ; a 34 1, and they are parallels, and joined towards the same parts by the straight lines BE, CH : but straight lines which join equal and parallel straight lines towards the same parts, are themselves equal and parallel ^ ; therefore EB, CH are both equal and pa- b 33. 1- rallel, and EBCH is a parallelogram ; and it is equal ^ to ABCD, c 35. 1. because it is upon the same base BC, and between the same parallels BC, AD : for the like reason, the parallelogram EFGH is equal to the same EBCH: therefore also the parallelogram ABCD is equal to EFGH. Wherefore, parallelograms, &c. Q. E. D. PROP. XXXVII. THEOR. TRIANGLES upon the same base, and between the same parallels, are equal to one another. Let the triangles ABC, DBC be upon the same base BC and between the same parallels ^^ An v AD, BC : the triangle ABC ^ ^ " is equal to the triangle DBC. Produce AD both ways to the points E, F, and through B draw » BE. parallel to CA ; and through C draw CF pa- rallel to BD : therefore each of the figures EBCA, DBCF is a parallelogram ; and EBCA is equal ^ to DBCF, because b 35. 1. they are upon the same base BC, and between the same parallels BC, EF ; and the triangle ABC is the half of the parallelo- F 31. 1 42 THE ELEMENTS Book I. gram EBCA, because the diameter AB bisects « it ; and the tri- angle DBC is the half of the parallelogram DBCF, because the diameter DC bisects it: but the halves of equal things are equal d; therefore the triangle ABC is equal to the triangle DBC. Wherefore, triangles, Sec Q. E. D. PROP. XXXVIIL THEOR. TRIANGLES upon equal bases, and between the same parallels, are equal to one another. Let the triangles ABC, DEF be vjpon equal bases BC, EF, and between the same parallels BF, AD : the triangle ABC is equal to the triangle DEF. Produce AD both ways to the points G, H, and through B a SI. 1- draw BG parallel » to CA, and through F draw FH parallel to ED : then each of p . p. tt the figures GBCA, ^ A u ti DEFH i5 a parallel- ogram, and they b 36. 1. are equal ^ to one another,because they are upon equal bases BC, EF, and be- tween the same pa- c34. 1. rallels BF, GH ; and the triangle ABC is the half <= of the pa- rallelogram GBCA, because the diameter AB bisects it ; and the triangle DEF is the half ^ of the parallelogram DEFH, because the diameter DF bisects it : but the halves of equal things are d 7. Ax. equal ^ ; therefore the triangle ABC is equal to the triangle DEF. Whprefore, triangles, 8cc. Q, E. D. PROP. XXXIX. THEOR. EQUAL triangles upon the same base, and upon the same side of it, are between the same parallels. Let the equal triangles ABC, DBC be upon the same base BC, and upon the same side of it ; they are between the same parallels. Join AD ; AD is parallel to BC ; for, if it is not, through the » SI. 1- point A draw ^ AE parallel to BC, and join EC : the triangle OF EUCLID. 43 ABC is equal •» to the triangle EBC, because it is upon the same Book I. base BC, and between the same paral- lels BC, AE: but the triangle ABC is equal to the triangle BDC ; therefore also the triangle BDC is equal to the triangle EBC, the greater to the less, which is impossible: therefore AE is not parallel to BC. In the same man- ner, it can be demonstrated that no other line but AD is parallel to BC ; AD is therefore parallel to it. Q. E. D. B C Wherefore, equal triangles, &c. PROP. XL. THEOR. EQUAL triangles upon equal bases, in the same straight line, and towards the same parts, are between the same parallels. Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the same straight line BF, and towards the same parts ; they are be- tween the same parallels. Join AD ; AD is paral- lel to BC : for, if it is not, through A draw ^ AG pa- rallel to BF, and join GF : „ r v t? the triangle ABC is equal »> ^ *- £. ^ b 38. 1, to the triangle GEF, because they are upon equal bases BC, EF, and between the same parallels BF, AG : but the triangle ABC is equal to the triangle DEF ; therefore also the triangle DEF is equal to the triangle GEF, the greater to the less, which is impossible : therefore AG is not parallel to BF : and in the same manner it can be demonstrated that there is no other parallel to it but AD ; AD is therefore parallel to BF. Wherefore, equal triangles, &c. Q. E. D. a 31. 1. PROP. XLI. THEOR. IF a parallelogram and triangle be upon the same base, and between the same parallels ; the parallelo- gram shall be double of the triangle. 44 THE ELEMENTS Book I. Let the parallelogram ABCD and the triangle EBC be «— V— ' the same base BC, and between the same parallels BC, AE parallelogram ABCD is double of the triangle EBC. Join AC ; then the triangle ABC a 37. 1. is equal * to the triangle EBC, because they are upon the same base BC, and between the same parallels BC, AE. b 34. 1. But the parallelogram ABCD is double ^ of the triangle ABC, because the diame- ter AC divides it into two equal parts ; wherefore ABCD is also double of the triangle EBC. Therefore, if a parallelo- gram, Sec. Q. E. D. upon ;the PROP. XLIL PROB. A F TO describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let ABC be the given triangle, and D the given rectilineal angle. It is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D. a 10. 1. Bisect a BC in E, join AE, and at the point E in the straight b 23. 1. line EC make ^ the angle CEF equal to D ; and through A draw c 31. 1. ^ AG parallel to EC, and through C draw CG «= parallel to EF : therefore FECG is a parallelo- gram : and because BE is equal to EC, the triangle ABE is like- d 38. 1. wise equal ^ to the triangle AEC, since they are upon equal bases BE, EC, and between the same parallels BC, AG : therefore the triangle ABC is double of the triangle AEC: and the paral- e4l. 1. lelogram FECG is likewise double e of the triangle AEC, be- cause it is upon the same base, and belv/een the same parallels : therefore the parallelogram FECG is equal to the triangle ABC, and it has one of its angles CEF equal to the given angle D. Wherefore there has been described a parallelogram OF EUCLID. 45 FECG equal to a given triangle ABC, having one of its angles Book I. CEF equal to the given angle D. Which was to be done. ^— v-*' PROP. XLIII. THEOR. THE complements of the parallelograms which are about the diameter of any parallelogram, are equal to one another. Let ABCD be a parallelogram, of which the diameter is AC and EH, FG the parallelograms about AC, that is, through which AC passes, and BK, KD the other parallelograms which make up the whole figure ABCD, which aTe therefore called the complements ; the complement BK is equal to the complement KD. Because ABCD is a paral- lelogram, and AC its diame- ter, the triangle ABC is equal * to the triangle ADC : and a 34< !♦ because EKHA is a parallelogram, the diameter of which is AK, the triangle AEK is equal to the triangle AHK : by the same reason, the triangle KGC is equal to the triangle KFC : then, because the triangle AEK is equal to the triangle AHK, and the triangle KGC to KFC; the triangle AEK together with the triangle KGC is equal to the triangle AHK together with the triangle KFC : but the whole triangle ABC is equal to the whole ADC ; therefore the remaining complement BK is equal to the remaining complement KD. Wherefore, the complements, &c. Q. E. D. PROP. XLIV. PROB. TO a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle. It is required to apply to the straight line AB a parallelogram equal to the triangle C, and having an angle equal to D. 46 THE ELEMENTS Book I. Make » the ^"--^-^^ parallelogr«im a 42.1, BEFG equal to the triangle C, and having the angle EBG equal to the angje D, so that BE be in thC' same straight line b 31. 1. with AB, and produce EG to H ; and through A draw •' AH pa- rallel to BG or EF, and join HB. Then, because the straight line HF falls upon the parallels AH, EF, the angles AHF, HFE c 29. 1. are together equal *= to two right angles; wherefore the angles BHF, HFE are less than two right angles : but straight lines which with another straight line make the interior angles upon d 12. Ax. the same side less than two right angles do meet •*, if produced far enough : therefore HB, FE shall meet, if produced ; let them meet in K, and through K draw KL parallel to EA or FH, and produce HA, GB to the points L, M : then HLKF is a paral- lelogram, of which the diameter is HK, and AG, ME are the parallelograms about HK ; and LB, BF are the complements ; therefore LB is equal e to BF ; but BF is equal to the triangle C ; wherefore LB is equal to the triangle C : and because the angle GBE is equal f to the angle ABM, and likewise to the angle D ; the angle ABM is equal to the angle D : therefore the parallelogram LB is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D. Which was to be done. e 43. i: f 15. 1. PROP. XLV. PROB. TO describe a parallelogram equal to a given rec- tilineal figure, and having an angle equal to a given rectilineal angle. a42. 1. b44. 1. Let ABCD be the given rectilineal figure, and E the given rectilineal angle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E. Join DB, and describe » the parallelogram FH equal to the triangle ADB, and having the angle HKF equal to the angle E; and to the straight line GH apply ^ the parallelogram GM equal OF EUCLID. 47 to the triangle DBC, having the angle GHM equal to the angle Book I. E ; and because the angle E is equal to each of the angles FKH, *— v— ' GHM, the angle FKH is equal to GHM ; add to each of these the angle KHG ; therefore the angles FKH, KHG are equal to the angles KHG, GHM ; but FKH, A ^ F G L KHG are equal c to two right an- \ / \ I t / / / c 29. 1. gles ; therefore also KHG, GHM are equal to two right angles ; and because at the point H in the straight line GH the two straight lines KH, H!M, upon the opposite sides of it, make the adjacent angles equal to two right angles, KH is in the same straight line ^ with HM ; and because the straight line d 14. 1. HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal « : add to each of these the angle HGL : there- fore the angles MHG, HGL are equal to the angles HGF, HGL : but the angles MHG, HGL are equal <= to two right an- gles ; wherefore also the angles HGF, HGL are equal to two right angles, and FG is therefore in the same straight line with GL : and because KF is parallel to HG, and HG to ML ; KF is parallel* to ML : and KM, FL are parallels ; wherefore KFLM e 30. 1. is a parallelogram ; and because the triangle ABD is equal to the parallelogram HF, and the triangle DBC to the parallelo- gram GM ; the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM ; therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD having the angle FKM equal to the given angle E. Which was to be done. Cor. From this it is manifest how to a given straight line to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure, viz. by applying ^ to the given straight line a parallelogram b 44. 1. equal to the first triangle ABD, and having an angle equal to the given angle. *8 THE ELEMENTS Book I. PROP. XLVI. PROB. TO describe a square upon a given straight line. Let AB be the given straight line ; it is required to describe a square upon AB. a 11. 1. From the point A draw a AC ^t right angles to AB ; and b 3. 1. make i" AD equal to AB, and through the point D draw DE c 31. 1. parallel ^ to AB, and through B draw BE parallel to AD ; there- d34. 1. fore ADEB is a parallelogram: whence AB is equal d to DE, and AD to BE : but BA is equal to C AD ; therefore the four straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADiiB is equilateral, likewise all its angles are right angles ; because the straight line AD meeting the parallels AB, DE, the angles BAD, ADE are equal e 29. 1. *= to two right angles ; but BAD is a right angle; therefore also ADE is a right angle ; but the opposite angles of parallelograms are equal ^ ; tliere- A fore each of the opposite angles ABE, BED is a right angle ; wherefore the figure ADEB is rectan- gular, and it has been demonstrated that it is equilateral ; it is therefore a square, and it is described upon the given straight line AB. Which was to be done. CoR. Hence every parallelogram that has one right angle has all its angles right angles. B PROP. XLVn. THEOR. IN any right angled triangle, the square which is described upon the side subtending the right angle is equal to the squares described upon the sides which contain the right angle. Let ABC be a right angled triangle, having the right angle BAC ; the square described upon the side BC is equal to the squares described upon BA, AC. ^46.1. On BC describe » the square BDEC, and on BA, AC the OF EUCLID. 49 i 14. 1. squares GB, HC ; and through A draw ^ AL parallel to BD or Book 1. CE, and join AD, FC ; then, because each of the angles BAG. ^-— y««> BAG is a right angle <=, the G b 31. 1. two straight lines AC, AG, X\ ». «30.Def. cpon the opposite sides of AB, make with it at the point A the adjacent angles equal F to two right angles ; therefore CA is in the same straight line ^ Avith AG ; for the same reason, AB and AH are in the same straight line ; and because the angle DBC is e- qual to the angle FBA, each of them being a right angle, add to each the angle ABC, and the whole angle DBA is equal ^ lo the whole FBC ; and e 2, Ax. because the two sides AB, BD are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC ; therefore the base AD is equal f to the base FC, and the trian- f 4. 1. gle ABD to the triangle FBC : now the parallelogram BL is double s of the triangle ABD, because they are upon the same g 41. 1. base BD, and between the same parallels, BD, AL ; and the square GB is double of the triangle FBC, because these also are upon the same base FB, and between the same parallels FB, GC. But the doubles of eqvials are equal ^ to one another : h 6. Ax. therefore the parallelogram BL is equal to the square GB : and in the same manner, by joining AE, BK, it is demonstrated that the parallelogram CL is equal to the square HC : therefore the whole square BDEC is equal to the two squares GB, HC ; and the square BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC : wherefore the square upon the side BC is equal to the squares upon the sides BA, AC« Therefore, in any right angled triangle, &c. Q. E. D. PROP. XLVin. THEOR. IF the square described upon one of the sides of a triangle be equal to the squares described upon the other two sides of it ; the angle contained by these two sides is a right angle. G 5.0 THE ELEMENTS, £cc. Book I. If the square described upon BC, one of the sides of the tw- >-- v*-^ angle ABC, be equal to the squares upon the other sides BA, AC, the angle BAC is a right angle, a 11. 1. From the point A draw » AD at right angles to AC, and make AD equal to BA, and join DC : then, because DA is equal to AB, the square of DA is equal to D the square of AB : to each of these add the square of AC ; therefore the squares of DA, AC are equal to the squares of b 47. 1. BA, AC : but the square of DC is equal ^ to the squares of DA, AC, because DAC is a right angle ; and the square of BC, by hypothesis, is equal to the squares of BA, AC ; therefore the square of DC is equal to the square of BC ; and therefore also ^ C the side DC is equal to the side BC. And because the side DA is equal to AB, and AC common to the two triangles DAC, BAC, the two DA, AC are equal to the two BA, AC ; and the base DC is equal to the base BC ; therefore the angle DAC is c 8. 1. equal c to the angle BAC : but DAC is a right angle ; therefore also BAC is a right angle. Therefore, if the square, &c» Q. E. D. THE ELEMENTS OF EUCLID. BOOK IL DEFINITIONS. I. Every right angled parallelogram is said to be contained by Book II. any two of the straight lines which contain one of the right Vi^v" ^ angles. II. In every parallelogram, any of the parallelograms about a dia- meter, together with the P two complements, is called A , , . D a gnomon. ' Thus the pa- ' rallelogram HG, together ' with the complements AF, ' ' FC, is the gnomon, which * is more briefly expressed H -~y^ — jj^ ' by the letters AGK, or ' EHC, which are at the B * opposite angles of the pa- ' rallelograms v/hich make the gnomon.' F PROP. I. THEOR. IF there be two straight lines, one of which is divided into any number of parts ; the rectangle con- tained by the two straight lines, is equal to the rect- angles contained by the undivided line and the several parts of the divided line. 52 THE ELEMENTS Book II. Let A and BC be two straight lines ; and let BC be divided *— V— ^ into any parts in the points D, E ; the rectangle contained by the straight lines A, BC is equal to the rectangle contained by A, BD, together with that contained by A, DE, and that contained by A, EC. a 11.1. From the point B draw ^ BF at right angles to BC, and make G b 3. 1. BG equal'' to A; and through c31. 1. G drawc GH parallel to BC ; and through D, E, C draw ^ DK, F EL, CH parallel to BG ; then the rectangle BH is equal to the i'ectangles BK, DL, EH ; and BH is contained by A, BC, for it is contained by GB, BC, and GB is equal to A ; and BK is contained by A, BD, for it is contained by GB. BD, of which GB is equal to A ; and DL is d 34. 1. contained by A, DE, because DK, that is 'i, BG, is equal to A ; and in like manner the rectangle EH is contained by A, EC : therefore the rectangle contained by A, BC is equal to the se- veral rectangles contained by A, BD, and by A, DE ; and also by A, EC. Wherefore, if there be two straight lines, Sec. Q. E. D. PROP. n. THEOR. IF a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts are together equal to the square of the whole line. •i. 46. 1. bSl.l. Let the straight line AB be divided into any two pans in the point C ; the rect- angle contained by AB, BC, together with the rectangle * AB, AC, shall be equal to the square of AB. Upon AB describe » the square ADEB, and through C draw ^ CF, parallel to AD or BE ; then AE is equal to the rectangles AF, CE ; and AK is the square of AB ; and AF is the rectangle contained by BA, * N. B. To avoid repeating the word contained too frequently, the rectangle contaiiipcl by tv/o straiglit lines AB, AC is sometimes simply called the rect- angle AB, AC. OF EUCLID. 5«5 AC ; for it is contained by DA, AC, of which AD is equal to Book II. AB ; and CE is contained by AB, BC, for BE is equal to AB; ^— •v^»"' therefore the rectangle contained by AB, AC, together Avith the rectangle AB, BC, is equal to the square of AB. If therefore a straight line, &c. Q. E. D. PROP. III. THEOR. IF a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts is equal to the rectangle contained by the two parts, together with the square of the foresaid part. Let the straight line AB be divided into any two parts in the point C ; the rectangle AB, BC is equal to the rectangle AC, CB, together with the square of BC. Upon BC describe ^ the square CDEB, and produce ED to F, and through A draw •» AF parallel to CD or BE ; then the rectangle AE is equal to the rectangles AD, CE : and AE is the rectangle contained by AB, BC, for it is contained by AB, BE, of which BE is equal to BC ; and AD is contained by AC, CB, for CD is equal to CB ; and DB is the square of BC ; therefore the rectangle AB, BC is equal to the rectangle AC, CB, together with the square of BC. If there- fore a straight line, See. Q. E. D. B a 46. 1. b 31. 1. PROP. IV. THEOR. IF a Straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts. Let the straight line AB be divided into any two parts in C ; the square of AB is equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB. 54 THE ELEMENTS Book II. Upon AB describe * the square ADEB, and join BD, and through C draw ^ CGF parallel to AD or BE, and through G draw HK parallel to AB or DE : and because CF is parallel to AD, and BD falls upon them, the exterior angle BGC is equal c to the interior and opposite angle ADB ; but ADB is equal ^ to the angle ABD, because BA is equal to AD, being sides of a square ; wherefore the angle CGB A C B is equal to the angle GBC ; and there- fore the side BC is equal ^ to the side CG : but CB is equal *^ also to GK, and CG to BK ; wherefore the figure H CGKB is equilateral : it is likewise rectangular ; for CG is parallel to BK, and CB meets them ; the angles KBC, GCB are therefore equal to two right an.s:les ; and KBC is a right angle ; e 6.1. £34.1 G ^ K E D F wherefore GCB is a right angle ; and therefore also the angles f CGK, GKB, opposite to these, are right angles, and CGKB is rectangular: but it is also equilateral, as was demonstrated; wherefore it is a square, and it is upon the side CB : for the same reason HF also is a square, and it is upon the side HG, which is equal to AC : therefore HF, CK are the squares of g 43. 1. AC, CB ; and because the complement AG is equal s to the complement GE, and that AG is the rectangle contained by AC, CB, for GC is equal to CB ; therefore GE is also equal to the rectangle AC, CB : wherefore AG, GE are equal to twice the rectangle AC, CB : and HF, CK are the squares of AC, CB ; wherefore the four figures HF, CK, AG, GE are equal to the sqiiares of AC, CB, and to twice the rectangle AC, CB ; but HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB : therefore the square of AB is equal 40 the squares of AC, CB, and twice the rectangle ACj CB. Wherefore, if a straight line, &c. Q. E. D. Cor. From the demonstration, it is manifest, that the paral- lelograms about the diameter of a square are likewise squares. OF EUCLID. PROP. V. THEOR. IF a straight line be divided into two equal parts, and also into two unequal parts; the rectangle con- tained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line. Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts at the point D ; the rectangle AD, DB, together with the square of CD, is equal to the square of CB. Upon CB describe » the square CEFB, join BE, and through ^ 46. 1. D draw b DHG parallel to CE or BF ; and through H draw b 31. 1. KLM parallel to CB or EF ; and also through A draw AK pa- - rallel to CL or BM : and because the complement CH is equal = to the complement HF, to each of these add DM ; ^ 43. 1. therefore the whole CM is equal to the whole DF ; A C D B but CM is equal ^ to AL, ( i 71 d 36. 1. because AC is equal to CB ; therefore also AL is equal to DF. To each of these add CH, and the whole AH is equal to DF and CH : but AH is the rectangle contained by AD, DB, for DH is equal e to DB ; and DF together with CH is the eCor,4.2. gnomon CMG ; therefore the gnomon CMG is equal to the rectangle AD, DB : to each of these add LG, which is equal « to the square of CD ; therefore the gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the square of CD: but the gnomon CMG and LG make up the whole figure CEFB, which is the square of CB : therefore the rectangle AD, DB, together with the square of CD, is equal to the square of CB. Wherefore, if a straight line, Sec. Q. E. D. From this proposition it is manifest, that the difference of the squares of two unequal lines AC, CD, is equal to the rectangle contained by their sum and difference. THE ELEMENTS PROP. VI. THEOR. a 46. 1. bSl. 1. IF a straight line be bisected, and produced to any point, the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal to the square of the straight line which is made up of the half and the part produced. Let the straight line AB be bisected in C, and produced to the point D ; the rectangle AD, DB, together with the square of CB, is equal to the square of CD. Upon CD describe » the square CEFD, join DE, and through B draw b BHG parallel to CE or DF, and through H draw KLM parallel to AD or EF, and also through A draw AK parallel toCL B D L K / / M G F or DM : and because AC is equal to CB, the rectangle c 36. 1. AL is equal <=■ to CH ; but d 43. 1. CH is equal ^ to HF ; there- fore also AL is equal to HF : to each of these add CM ; therefore the whole AM is equal to the gnomon CMC : and AM is the rect- angle contained by AD, DB, eCor-4.2. for DM is equal ^ to DB : therefore the gnomon CMG is equal to the rectangle AD, DB : add to each of these LG, which is equal to the square of CB ; therefore the rectangle AD, DB, to- gether Avith the square of CB, is equal to the gnomon CMG and the figure LG : but the gnomon CMG and LG make up the whole figure CEFD, which is the square of CD ; therefore the rectangle AD, DB, together with the square of CB, is equal to the square of CD. Wherefore, if a straight line, &c. Q. E. D. PROP. Vn. THEOR. IF a straight line be divided into any two parts, the squares of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the straight line AB be divided into any two parts in OF EUCLID. 57 the ijoint C ; the squares of AB, BC are equal to twice the rect-Book II. angle AB, BC, together with the square of AC. v.^.^.— > Upon AB describe * the square ADEB, and construct the a 46. 1. figure as in the preceding propositions : and because AG is equal ^ to GE, add to each of them CK ; the whole AK is b 43- 1- therefore equal to the whole CE ; therefore AK, CE are double of AK : but AK, CE are the gnomon AKF together with the square CK ; therefore the gnomon AKF, toge- ther with the square CK, is double Hi -p^ |K of AK : but twice the rectangle AB, BC is double of AK, for BK is equal c to BC : therefore the gnomon AKF, together with the square CK, is equal to twice the rectangle AB, BC : to D F E each of these equals add HF, which is equal to the square of AC ; therefore the gnomon AKF, toge- ther with the squares CK, HF, is equal to twice the rectangle AB, BC, and the square of AC : but the gnomon AKF, together with the squares CK, HF, make up the whole figure ADEB and CK, which are the squares of AB and BC : therefore the squares of AB and BC are equal to twice ihe rectangle AB, BC, together with the square of AC. Wherefore, if a straight line, 8cc. Q. E. D, A C B G / X c cor. 4.2. PROP. VHI. THEOR. IF ia straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made up of the whole and that part. Let the straight line AB be divided into any two parts in the point C ; four times the rectangle AB, BC, together with the square of AC, is equal to the square of the straight line m;ide up of AB and BC together. Produce AB to D, so that BD be equal to CB, and upon AD describe the square AEFD ; and construct two figures such as in the preceding. Because CB is equal to BD, and that CB is equal * to GK, and BD to KN ; therefore GK is a 34 1. H SB THE ELEMENTS — — — — 1 G I / P / R X Book II. equal to KN ; for the same reason, PR is equal to RO ; and because CB is equal to BD, and GK to KN, the rectangle CK is equal •» to BN, and GR to RN : but CK is equal <= to RN,^ because they are the complements of the parallelogram. CO ; therefore also BN is equal to GR ; and the four rect- angles BN, CK, GR, RN are therefore equal to one another, and so are quadruple of one of them CK : again, because CB is equal to BD, and that BD is dCor.4.2. equal ^ to BK, that is, to CG ; C B and CB equal to GK, that ^ is, to A , 1 1 yt D GP ; therefore CG is equal to GP : and because CG is equal to M I ■ .. •{ ^ — \ N GP, and PR to RO, the rectangle AG is equal to MP, and PL to X 7f — ! 1 O e 43. 1. RF : but MP is equal « to PL, because they are the complements of the parallelogram ML ; where- fore AG is equal also to RF : ElZ^ :^ — J 1 F therefore the four rectangles " ■'-' AG, MP, PL, RF are equal to one another, and so are quadruple of one of them AG. And it was demonstrated, that the four CK, BN, GR, and RN are quad- ruple of CK : therefore the eight rectangles which contain the gnomon AOH are quadruple of AK : and because AK is the rectangle contained by AB, BC, for BK is equal to BC, four times the rectangle AB, BC is quadruple of AK ; but the gno- mon AOH was demonstrated to be quadruple of AK ; therefore four times the rectangle AB, BC is equal to the gnomon AOH. To each of these add XH, which is equal '^ to the square of AC : therefore four times the rectangle AB, BC, together with the square of AC, is equal to the gnomon AOH and the square XH : but the gnomon AOH and XH make up the figure AEFD, which is the square of AD : therefore four times the rectangle AB, BC, together with the square of AC, is equal to the square of AD, that is, of AB and BC added together in one straight line. Wherefore, if a straight line, &c. Q. E. D. OF EUCLID. PROP. IX. THEOR. IF a straight line be divided into two equal, and also into two unequal parts ; the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section. Let the straight line AB be divided at the point C into two equal, and at D imo two unequal parts : the squares of AD, DB are together double of the squares of AC, CD. From the point C draw * CE at right angles tb AB, and a 11. 1 make it equal to AC or CB, and join EA, EB ; through D draw *> DF parallel to CE, and through F draw FG parallel to AB ; b 31. 1. and join AF : then, because AC is equal to CE, the angle EAC is equal «= to the angle AEC ; and because the angle c 5. 1. ACE is a right angle, the two others, AEC, EAC together make one riglit angle ^ ; and they are equal to one another ; d 32. 1. each of them therefoce is half E of a right angle. For the same reason each of the angles CEB, EBC is half a right angle ; and therefore the whole AEB is a right angle : and because the an- gle GEF is half a right angle, and EGF a right angle, for it is A CD B equal e to the interior and oppo- e 29. 1. site angle ECB, the remaining angle EFG is half a right angle ; therefore the angle GEF is equal to the angle EFG, and the side EG equal ^ to the side GF: again, because the angle at Bf 6. 1. js half a right angle, and FDB a right angle, for it is equal ^ to the interior and opposite angle ECB, the remaining angle BFD is half a right angle ; therefore the angle at B is equal to the angle BFD, and the side DF to ^ the side DB : and be- cause AC is equal to CE, the square of AC is equal to the square of CE ; therefore the squares of AC, CE are double of the square of AC : but the square of EA is equal s to the g 47. 1. squares of AC, CE, because ACE is a right angle ; therefore the square of EA is double of the square of AC: again, be- cause EG is equal to GF, the square of EG is equal to the 2quare of GF ; therefore the squares of FG, GF are double of 60 THE ELEMENTS Book II. the square of GF ; but the square of EF is equal to the squares '■"-V— ^ of EG, GF ; therefore the square of EF is double of the square h 34. 1. GF ; and GF is equal ^ to CD ; therefore the square of EF is double of the square of CD : but the square of AE is likewise double of the square of AC ; therefore the squares of AE, EF are double of the squares of AC, CD : and the square of AF is i 47. 1. equal ' to the squares of AE, EF, because AEF is a right angle ; therefore the square of AF is double of the squares of AC, CD : but the squares of AD, DF are equal to the square of AF, because the angle ADF ia a right angle; therefore the squares of AD, DF are double of the squares of AC, CD : and DF is equal to DB; therefore the squares of AD, DB are double of the squares of AC, CD. If therefore a straight line, Sec. Q. E. D. prop: X. THEOR. IF a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced. Let the straight line AB be bisected in C, and produced to the point D ; the squares of AD, DB are double of the squares of AC, CD. all. 1. From the point C draw ^ CE at right angles to AB : and make it equal to AC or CB, and join AE, EB ; through E draw b31. 1. b EF parallel to AB, i-.nd through D draw DF parallel to CE : and because the straight line EF meets the parallels EC, FD, the c 29. 1. angles CEF, EFD are equal <= to two right angles ; and therefore the angles BEF, EFD are less than two right angles : but straight lines Avhich with another straight line make the interior angles d 12. Ax. upon the same side less than two right angles, do meet 'i if pro- duced far enough : therefore EB, FD shall meet, if produced, towards B, D : let them meet in G, and join AG : then, because e 5. 1. AC is equal to CE, the angle CEA is equal e to the angle EAC ; and the angle ACE is a right angle ; therefore each of the f 3S. 1. angles CEA, EAC is half a right angle <": for the same reason. OF EUCLID. 61- h 34. t. each of the angles CEB, EBC is half a right angle ; therefore Book II. AEB is a right angle : and because EBC is half a right angle, ^— -v— ^ DBG is also *■ half a right angle, for they are vertically oppo-fl5. 1. site; but BDG is a right angle, because it is equal ^ to the al-c29. 1. ternate angle DCE ; therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG ; wherefore also the side BD is equal s to the side DG : again, g 6. t. because EGF is half a right angle, and that the angle at F is a right angle, because it is e- qual ^ to the opposite angle ECD, the remain- ing angle FEG is half a right angle, and equal to the angle EGF ; wherefore also the side GF is equal s to the side FE. And because EC is equal to CA, the square of EC is equal to the square of CA ; therefore the squares of EC, CA are double of the square of CA : but the square of EA is equal » to the squares of EC, C A ; there- i 47. 1. fore the square of EA is double of the square of AC : again, because GF is equal to FE, the square of GF is equal to the square of FE ; and therefore the squares of GF, FE are dou- ble of the square of EF : but the square of EG is equal ' to the squares of GF, FE ; therefore the square of EG is double of the square of EF : and EF is equal to CD ; wherefore the square of EG is double of the square of CD : but it was demon- strated, that the square of EA is double of the square of AC ; therefore the squares of AE, EG are double of the squares of AC, CD : and the square of AG is equal ' to the squares of AE, EG ; therefore the square of AG is double of the squares of AC, CD : but the squares of AD, GD are equal '■ to the square of AG ; therefore the squares of AD, DG are double of the squares of AC, CD : but DG is equal to DB ; therefore the squares of AD, DB are double of the squares of AC, CD. Wherefore, if a straight Ihie, &c. Q. E. D. THE ELEMENTS PROP. XI. PROB, a 46. 1. b 10. 1. c 3. 1. d6. 2. e 47. 1. TO divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square of the other part. Let AB be the given straight line ; it is required to divide it into tv/o parts, so that the rectangle contained by the whole and one of tlie parts shall be equal to the square of the other part. Upon AB describe ^ the square ABDC ; bisect b AC in E, and join BE; produce CA to F, and make «= EF equal to EB ; and upon AF describe » the square FGHA ; AB is divided in H, so that the rectangle AB, BH is equal to the square of AH. Produce GH to K : because the straight line AC is bisected in E, and produced to the point F, the rectangle CF, FA, to- gether Vv'ith the square of AE, is equal ^ to the square of EF : but EF is equal to EB ; therefore the rectangle CF, FA, toge- ther with the square of AE, is equal to the square of EB : and the squares of BA, AE are equal e to the F G square of EB, because the angle EAB is a right angle ; therefore the rect- angle CF, FA, together with the square of AE, is equal to the squares of BA, AE : take away the square of AE, "which is common to both, therefore the remaining rectangle CF, FA is equal to the square of AB : and the fi- gure FK is the rectangle contained by CF, FA, for AF is equal to FG ; and AD is the square of AB ; therefore FK is equal to AD : take away the common part AK, and the remainder FH is equal to the remainder HD : C K D and HD is the rectangle contained by AB, BH, for AB is equal to BD ; and FH is the square of AH : therefore the rectangle AB, BH is equal to the square of AH : wherefore the straight lirie AB is divided in H so, that the rectangle AB, BH is equal to the square of AH. Which was to be done. OF EUCLID. PROP. XII. THEOR. IN obtuse angled triangles, if a perpendicular be * drawn from any of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line inter- cepted without the triangle between the perpendicu- lar and the obtuse angle. Let ABC be an obtuse angled triangle, having the obtuse angle ACB, and from the point A let AD be drawn ^ perpendi- a 12. 1. cular to BC produced: the square of AB is greater than the squares of AC, CB by twice the rectangle BC, CD. Because the straight line BD is divided into two parts in the point C, the square of BD is equal ^ to the squares of BC, CD, and ^ A b 4. 2. twice the rectangle BC, CD : to each of these equals add the square of DA ; and the squares of BD, DA ai'e equal to the squares of BC, CD, DA, and twice the rectangle BC, CD : but the square of B A is equal <^ to the squares of BD, DA, be- y / | c 47. 1. cause the angle at D is a right -d'' angle ; and the square of C A is equal « to the squares of CD, DA : therefore the square of B A is equal to the squares of BC, CA, and twice the rectangle BC, CD ; that is, the square of BA is greater than the squares of BC, CA, by twice the rectangle BC, CD. Therefore, in obtuse angled triangles, &c. Q. E. D. 64 Book II. THE ELEMENTS PROP. XIII. THEOR. and because See N. IN every triangle, the square of the side subtend-- ing any of the acute angles is less than the squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle. Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the sides containing it, let fall the a 12. 1. perpendicular * AD from the opposite angle : the square of AC, opposite to the angle B, is less than the squares of CB, BA, by twice the rectangle CB, BD. First, Let AD fall within the triangle ABC; the straight line CB is divided into two parts in the point D, the squares of CB, BD are equal b 7. 12 ^ to twice the rectangle contain- ed by CB, BD, and the square of DC: to each of these equals add the square of AD ; therefore the squares of CB, BD, DA are equal to twice the rectangle CB, BD, and the squares of AD, DC : <; 47- 1 but the square of AB is equal = to B D the squares of BD, DA, because the angle BDA is a right angle ; and the square of AC is equal to the squares of AD, DC : therefore the squares of CB, BA are equal to the square of AC, and twice the rectangle CB, BD, that is, the square of AC alone is less than the squares of CB, BA by twice the rectangle CB, BD. Secondly, Let AD fall with- A out the triangle ABC : then, be- cause the angle at D is a right angle, the angle ACB is greater (1 16. 1. 'I ihan a right angle ; and there- e 12. 2. fore the square of AB is equal'^ to the squares of AC, CB, and twice the rectangle BC, CD : to these e- qualsaddthe squareofBC,andthe BCD OF EUCLID. 65 squares of AB, BC are equal to the square of AC, and twice Book ir, the square of BC, and twice the rectangle BC, CD : but be- ^■— v — ^ cause BD is divided into two parts in C, the rectangle DB, BC is equal f to the rectangle BC, CD and the square of BC : and f 3. 2. the doubles of these are equal : therefore the squares of AB, BC are equal to the square of AC, and twice the rectangle DB, BC : therefore the square of AC alone is less than the squares of AB, BC by twice the rectangle DB, BC. A Lastly, Let the side AC be perpendicular to BC ; then is BC the straight line between the perpendicular and the acute angle at B ; and it is manifest that the squares of AB, BC are equal K to the square of AC and twice the square of BC. Therefore, in every triangle, &:c. Q. E. D. . PROP. XIV. PROB. TO describe a square that shall be equal to a given See n. rectilineal figure. Let A be the given rectilineal figure ; it is required to describe a square that shall be equal to A. Describe * the rectangular parallelogram BCDE equal tt) the a 45. 1. rectilineal figure A. If, then, the sides of it BE, ED are equal to one another, it a ^ ^.^^^ jr is a square, and what was requir- ed is now done : but if they are not equal, produce one \ / •' B of them BE to F, and make EF e- qual to ED, and bisect BF in G ; and from the centre G, at the distance GB, or GF, describe the semicircle BHF, and produce DE to H, and join GH ; therefore, because the straight line BF is divided into two equal parts in the point G, and into two unequal at E, the rectangle BE, EF, together with the square of EG, is equal ^ to the square ofb 5, 2. GF : but GF is equal to GH ; therefore the rectangle BE, EF, I 66 THE ELEMENTS, Sec. Book II. together with the square of EG, is equal to the square of GH ; *--"v-— ^ but the squares of HE, EG are equal « to the square of GH : c 47. 1. therefore the rectangle BE, EF, together with the square of EG, is equal to the squares of HE, EG: take away the square of EG, which is common to both ; and the i^emaining rect- angle BE, EF is equal to the square of EH : but the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED ; therefore BD is equal to the square of EH ; but BD is equal to the rectilineal figure A ; therefore the rectilineal figure A is equal to the square of EH : wherefore a square has been made equal to the given rectilineal figure A, viz. the square described upon EH. Which was to be done. THE ELEMENTS OF EUCLID- BOOK III. DEFINITIONS. I. JljQUAL circles are those of which the diameters are equal, or Book III. from the centres of which the straight lines to the circumfer- s— ^-—^ ences are equal. ' This is not a definition but a theorem, the truth of which is ' evident ; for, if the circles be applied to one another, so that ' their centres coincide, the circles must likewise coincide, since * the straight lines from the centres are equal.' II. A straight line is said to touch a circle, when it meets the circle, and being produced does not cut it. III. Circles are said to touch one another, which meet, but do not cut one another. IV. Straight lines are said to be equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal. V. And the straight line on which the greater perpendicular falls, is said to be farther from the centre. 68 THE ELEMENTS Book III- VI. ^■■^■v^*^ A segment of a circle is the figure con- tained by a straight line and the cir- cumference it cuts off. VII. " The angle of a segment is that which is contained by the " straight line and the circumference." VIII. An angle in a segment is the angle con- tained by two straight lines drawn from any point in the circumference of the segment, to the extremities of the straight line which is the base of the segment. IX. And an angle is srid to insist or stand upon the circumference intercepted between the straight lines that contain the angle. X. The sector of a circle is the figure contain- ed by two straight lines drawn from the centre, and the circumference between them. XI. Similar segments of a circle are those in which the an- gles are equal, or which contain equal angles. PROP. I. PROB. See N. To find the centre of a given circle. a 10. 1. bll. 1. Let ABC be the given circle ; it is required to find its centre. Draw within it any straight line AB, and bisect » it in D ; from the point D draw ^ UC at rjgh.t angles to AB, and pro- duce it to E, and bisect CE in F : the point F is the centre of the circle ABC. OF EUCLID, 69 For, if it be not, let, if possible, G be the centre, and join B. Ill, GA, GD, GB : then, because DA is equal to DB, and DG *— v— common to the two triangles ADG, BDG, the two sides AD, DG are e- G qual to the two BD, DG, each to each ; and the base GA is equal to the base GB, because they are drawn from the centre G * : therefore the angle ADG is equal <= to the angle GDB : but when a straight line stand- ing upon another straight line makes the adjacent angles equal to one ano- ther, each of the angles is a right an- gle er, each of them is a right « an- ^ gle : therefore each of the angles AFE, BFE is a right angle ; wherefore the straight line CD, drawn through the cen- tre bisecting another AB that does not pass through the centre, cuts the same at right angle:;. But let CD cut AB at right angles j CD also bisects it, that is, AF is equal to FB. The same construction being made, because EA, EB from <\5A. the centre a'e equal to one another, the angle EAF is equal «l to the angle EBF ; and the right angle AFE is equal to the right angle BFE ; therefore, in the two triangles EAF, EBF, OF EUCLID. 7i there are two angles in one equal to two angles in the other, Book III, and the side EF, which is opposite to one of the equal angles v —^..^ in each, is common to both ; therefore the other sides are equal e ; e 26. 1. AF therefore is equal to FB. Wherefore, if a straight line, &c. Q. E. D. PROP. IV. THEOR. IF in a circle two straight lines cut one another which do not both pass through the centre, they do not bisect each other. Let ABCD be a circle, and AC, BD two straight lines in it •which cut one another in the point E, and do not both pass through the centre ; AC, BD do not bisect one another. For, if it is possible, let AE be equal to EC, and BE to ED : if one of the lines pass through the centre, it is plain that it cannot be bisected by the other which does not pass through the centre : but, if neither of them pass through the centre, take * F the centre of the circle, / p \D dii.3. and join EF : and because FE, a straight line through the centre, bisects -^ another AC which does not pass through the centre, it shall cut it at right •» angles ; wherefore FEA is a B "^ — — -^^ C b 3. 3, right angle : again, because the straight line FE bisects the straight line BD which does not pass through the centre, it shall cut it at right bangles: wherefore FEB is a right angle : and FEA was shown to be a right angle ; therefore FEA is" equal to the angle FEB, the less to the greater, which is impossible : there- fore AC, BD do not bisect one another. Wherefore, if in a cir- cle, &c. Q. E. D. PROP. V. THEOR. IF two circles cut one another, they shall not have the same centre. Let the two circles ABC, CDG cut one another in the points B, C ; they have not the same centre. 72 THE ELEMENTS Booklll. For, if it be possible, let E be their centre: join EC, and ^ "■ ^v *^ draw any straight line EFG meet- ing them in F and G ; and because C E is the centre of the circle ABC, CE is equal to EF : again, be- cause E is the centre of the circle CDG, CE is equal to EG: but ^ CE was shown to be equal to EF ; therefore EF is equal to EG, the less to the greater, which is impos- sible : therefore E is not the centre of the circles ABC, CDG. Where- fore, if two circles, &c. Q. E. D. PROP. VI. THEOR. IF two circles touch one another internally, they shall not have the same centre. Let the two circles ABC, CDE touch one another internally in in the point C ; they have not the same centre. For, if they can, let it be F ; join FC, and draw any straight line FEB meeting them in E and B; C and because F is the centre of the circle ABC, CF is equal to FB ; also, because F is the centre of the circle CDE, CF is equal to FE : and CF was shown equal to FB ; there- fore FE is equal to FB, the less to A the greater, which is impossible: wherefore F is not the centre of the circles ABC, CDE. Therefore, if two circles, Stc. Q. E. D. OF EUCLID. PROP. VII. THEOR, IF any poir>t be taken in the diameter of a circle, which is not the centre, of all the straight lines which can be drawn from it to the circumference, the greatest is that in which the centre is, and the other part of that diameter is the least ; and, of any others, that which is nearer to the line which passes through the centre is always greater than one more remote : and from the same point there can be drawn only two straight lines that are equal to one another, one upon each side of the shortest line. Let ABCD be a circle, and AD its diameter, in which let any point F be taken which is not the centre; let the centre be E ; of all the straight lines FB, FC, FG, Sec. that can be drawn from F to the circumference, FA is the greatest, and FD, the other part of the diameter AD, is the least: and of the others, FB is greater than FC, and FC than FG. Join BE, CE, GE ; and because two sides of a triangle are greater a than the third, BE, EF are greater than BF; but AE*20. !.. is ^equal to EB ; therefore AE, EF, that is, AF, is greater than BF : again, because BE is equal to CE, and FE common to the triangles BEF, CEF, the two sides BE, EF are equal to the two CE, EF; but the angle BEF is greater than the angle CEF ; therefore the base BF is ^^ greater »» than the base FC : for the \ //ivN^ / b24. 1, same reason, CF is greater than GF : again, because GF, FE are greater * than EG, and EG is equal to ED; GF, FE are greater than ED: take away the comTnon part FE, and the remainder GF is greater than the remainder FD : therefore FA is the greatest, and FD the least of all the straight lines from F to the circumference ; and BF is greater than CF, and CF than GF. Also there can be drawn only two equal straight lines from ihe point F to the circumference, one upon eaeh side of the K r* THE ELEMENTS Book in, shortest line FD : at the point E, in the straight line EF, make ^•^-v-*^ c the angle FEH equal to the angle GEF, and join FH : then c 23. 1. because GE is equal to EH, and EF common to the two tri- angles GEF, HEF ; the two sides GE, EF are equal to the two HE, EF ; and the angle GEF is equal to the angle HEF ; there- (j 4. 1, fore the base EG is equal ^ to the base FH : but, besides FH, no other straight line can be drawn from F to the circumference equal to FG : for, if there can, let it be FK ; and because FK is equal to FG, and FG to FH, FK is equal to FH ; that is, a line nearer to that which passes through the centre, is equal to one which is more remote; which is impossible. Therefore, if any point be taken, Sec. Q. E. D. PROP. VHI. THEOR. IF any point be taken without a circle, and straight lines be drawn from it to the circumference, where- of one passes through the centre, of those which fall upon the concave circumference, the greatest is that which passes through the centre, and, of the rest, that ^vhich is nearer to that through the centre is always greater than the more remote : but of those which fall upon the convex circumference, the least is that be- tween the point without the circle, and the diameter; and, of the rest, that which is nearer to the least is always less than the more remote : and only two equal straight lines can be drawn from the point unto the circumference, one upon each side of the least. Let ABC be a circle, and D any point without it, from which let the straight lines DA, DE, DF, DC be drawn to the cir- cumference, whereof DA passes through the centre. Of those which fall upon the concave part of the circumference AEFC, the greatest is AD which passes through the centre ; and the nearer to it is always greater than the more remote, viz. DE than DF, and DF than DC ; but of those which fall upon the convex circumference HLKG, the least is DG between the OF EUCLID. 75 d 4. Ax. point D and the diameter AG; and the nearer to it is always Book III less than the more remote, viz. DK than DL, and DL than *— y— ^ DH. Take » M the centre of the circle ABC, and join ME, MF, » 1. 3. MC, MK, ML, MH: and because AM is equal to ME, add MD to each, therefore AD is equal to EM, MD ; but EM, MD are greater^ than ED; therefore also AD is greater than ED : b 20. 1, again, because ME is equal to MF, and MD common to the triangles EMD, KMD ; EM, MD are equal to FM, MD ; but the angle EMD is greater than the angle FMD ; therefore the base ED is greater '^ than the base FD : /// \\ c24. 1, in like manner it may be shown that FD is greater than CD : therefore DA is the greCest : and DE greater than DF, and DF than DC: and because MK, KD are greater •> than MD, and MK is equal to MG, the remainder KD is greater '^ than the remainder GD, that is, GD is less than KD : and because MK, DK are drawn to the point K within the triangle MLD from M, D, the extremi- ties of its side MD ; MK, KD are less e than ML, LD, whereof MK E A e 21- 1. is equal to ML ; therefore the remainder DK is less than the re- mainder DL : in like manner it may be shown, that DL is less than DH : therefore DG is the least, and DK less than DL, and DL than DH : also there can be drawn only two equal straight lines from the point D to the circumference, one upon each side of the least : at the point M, in the straight line MD, make the angle DMB equal to the angle DMK, and join DB : and because MK is equal to MB, and MD common to the triangles KMD, BMD, the two sides KM, MD are equal to the two BM, MD ; and the angle KMD is equal to the angle BMD ; there- fore the base DK is equal f to the base DB : but, besides DB, f 4. 1. there can be no straight line drawn from D to the circumference equal to DK : for, if there can, let it be DN ; and because DK is equal to DN, and also to DB ; therefore DB is equal to DN, that is, the nearer to the least equal to the more remote, which is impossible. If, therefore, any point, &c. Q. E, D. THE ELEMENTS PROP. IX. PROB. a 7. 3. IF a point be taken within a circle, from which there fall more than two equal straight lines to the circumference, that point is the centre of the circle. Let the point D be taken within the circle ABC, from Avhich to the circumference there fall more than two equal straight lines, viz. DA, DB, DC ; the point D is the centre of the circle. For, if not, let E be the centre, join DE and produce it to the cir- cumference in F, G ; then FG is a diameter of the circle ABC : and because in FG, the diameter of the circle ABC, there is taken the point D which is not the centre, DG shall be the greatest line from it to the circumference, and DC greater a than DB, and DB than DA ; but they are likewise equal, which is impossible : therefore E is not the centre of the circle ABC : in like manner, it may be demon- strated, that no other point but D is the centre ; D therefore is the centre. Wherefore, if a point be taken, &c. Q. E. D. a 9. 3. PROP. X. THEOR. ONE circumference of a circle cannot cut another in more than two points. If it be possible, let the circumfe- rence FAB cut the circumference DEF in more than two points, viz. in B, G, F ; take the centre K of the circle ABC, and join KB, KG, KF: and because within the circle DEF there is taken the point K, from which to the circumference DEF fall more than two equal straight lines KB, KG, KF, the point K is» OF EUCLID. 77 the centre of the circle DEF : but K is also the centre of the Book III. circle ABC ; therefore the same point is the centre of two cir ^ ."r-mj cles that cut one another, which is impossible''. Therefore one b 5. 3. circumference of a circle cannot cut another in more than two points. Q. E. D. PROP. XL THEOR. IF two circles touch each other internally, the straight line which joins their centres being produced shall pass through the point of contact. Let the two circles ABC, ADE touch each other internally in the point A, and let F be the centre of the circle ABC, and G the centre of the circle ADE ; the ^ straight line which joins the centres F, G, being produced, passes through the point A. For, if not, let it fall otherwise, if pos^ sible, as FGDH, and join AF, AG : and because AG, GF are greater^ than |y ^ 7?\^ / i ^20.1- FA, that is, than FH, for FA is equal to FH, both being from the same centre ; take away the common part FG ; there- fore the remainder AG is greater than the remainder GH : but AG is equal to GD ; therefore GD is greater than GH, the less than the greater, which is impossible. Therefore the straight line which joins the points F, G cannot fall otherwise than upon the point A, that is, it must pass through it. Therefore, if two circles, 5cg. Q. E. D. PROP. XIL THEOR. IF two circles touch each other externally, the straight line which joins their centres shall pass through the point of contact. Let the two circles ABC, ADE touch each other externally in the point A ; and let F be the centre of the ciix;le ABC, and G the centre of ADE : the straight line which joins the points F, G shall pass through the point of contact A. Far, if not, let it pass otherwise, if possible, as FCDG, and 78 THE ELEMENTS Book III. join FA, AG: and because F is the centre of the circle ABC, *— Y'"-.' AF is equal to FC : also, ^ E because G is the centre of the circle ADE, AG is e- qual to GD : therefore FA, AG are equal to FC, DG ; wherefore the whole FG is greater than FA, a 20. 1. AG: but it is also less*; which is impossible : therefore the straight line which joins the points F, G shall not pass otherwise than through the point of contact A, that is, it must pass through it. Therefore, if two circles, Sec. Q. E. D. PROP. XIII. THEOR. See N. ONE circle cannot touch another in more points than one, whether it touches it on the inside or outside. For, if it be possible, let the circle EBF touch the circle ABC in more points than one, and first on the inside, in the points a 10111. B, D; join BD, and draw* GH bisecting BD at right angles: Therefore, because the points B, D are in the circumference of b 2. 3. " each of the circles, the straight line BD falls within each** of cCor. 1.3. them : and their centres are«= in the straight line GH which bi- sects BD at right angles; therefore GH passes through the point d 11. 3. of contact J ; but it does not pass through it, because the points li, D are without the straight line GH, which is absurd: there- fore one circle cannot touch another on the inside in more points than one. OF EUCLID. 79 Nor can two circles touch one another on the outside in Book III. more than one point: for, if it be possible, let the circle ACK <^- y mJ touch the circle ABC in the points A, C, and join AC : there- fore, because the two points A, C are in the circumference of the circle ACK, the straight line AC which joins them shall fall within *» the circle ACK: and the cir- cle ACK is without the circle ABC ; and therefore the straight line AC is without this last circle ; but, because the points A, C are in the circumference of the circle ABC, the straight line AC must be within^* the same circle, which is absurd : there- fore one circle cannot touch another on the outside in more than one point : and it has been shown, that they cannot touch on the inside in more points than one. There- fore, one circle, &c. Q. E. D. PROP. XIV. THEOR. EQUAL straight lines in a circle are equally dis- tant from the centre; and those which are equally distant from the centre, are equal to one another. Let the straight lines AB, CD, in the circle ABDC, be equal to one another ; they are equally distant from the centre. Take E the centre of the circle ABDC, and from it draw EF, EG perpendiculars to AB, CD : then, because the straight line EF, passing through the centre, cuts the straight line AB, which does not pass through the centre, at right angles, it also bisects » it : wherefore AF is equal toFB, and AB double of AF. For the same reason, CD is double of CG : and AB is equal to CD ; therefore, AF is equal to CG: and because AE is equal to EC, the square of AE is equal to the square of EC ; but the squares of AF, FE are equal b to the square of AE, because the angle AFE is a right angle ; and, for the like reason, the squares of EG, GC are equal to the square of EC : therefore the squares of AF, FE are equal to the squares of vG, GE, of which the square of AF is equal to the square of a 3. b 4r. 1. 80 THE ELEMENTS Book III. CG, because AF is equal to CG ; therefore the remaining square V— Y-^ of FE is equal to the remaining square of EG, and the straight line EF is therefore equal to EG : but straight lines in a circle are said to be equally distant from the centre, when the perpen- c4.def.3. diculars drawn to them from the centre are equal <= : therefoi'c AB, CD are equally distant from the centre. Next, if the straight lines AB, CD be equally distant from the centre, that is, if FE be equal to EG, AB is equal to CD: for, the same construction being made, it may, as before, be de- monstrated, that AB is double of AF, and CD double of CG, and that the squares of EF, FA are equal to the squares of EG, GC ; of which the square of FE is equal to the square of EG, because FE is equal to EG ; therefore the remaining square of AF is equal to the remaining square of CG ; and the straight line AF is therefore equal to CG : and AB is double of AF, and CD double of CG ; wherefore AB is equal to CD. Therefore, equal straight lines, &c. Q. E. D. PROP. XV. THEOR. See N THE diameter is the greatest straight line in a circle; and, of all others, that which is nearer to the centre is always greater than one more remote ; and the greater is nearer to the centre than the less. 20.1. Let ABCD be a circle, of which the diameter is AD, and the centre E; and let BC be nearer to the centre than FG ; AD is greater than any straight line BC which is not a diameter, and BC greater than FG. From the centre draw EH, EK per- pendiculars to BC, FG, and join EB, EC, EF; and because AE is equal to EB, and ED to EC, AD is equal to EB, EC : but EB, EC are greater » than BC; wherefore also AD is greater than BC And, because BC is nearer to the centre than FG, EH is OF EUCLID. 81 less** than EK ; but, as was demonstrated in the preceding, BC Booklll. is double of BH, and FG double of FK, and the squares of EH ^■— v— ^ HB are equal to the squares of EK, KF, of which the square ofb5.def.3. EH is less than the square of EK, because EH is less than EK ; therefore the square of BH is greater than the square of FK, and the straight line BH greater than FK ; and therefore BC is greater than FG. Next, Let BC be greater than FG ; BC is nearer to the cen- tre than FG, that is, the same construction being made, EH is less than EK : because BC is greater than FG, BH likewise is greater than KF : and the squares of BH, HE are equal to the squares of FK, KE, of which the square of BH is greater than the square of FK, because BH is greater than FK ; therefore the square of EH is less than the square of EK, and the straight line EH less than EK. Wherefore the diameter, &c. Q. E. D. PROP. XVL THEOR. THE straight line drawn at right angles to the dia- See n. meter of a circle, from the extremity of it, falls with- out the circle ; and no straight line can be drawn between that straight line and the circumference from, the extremity, so as not to cut the circle ; or, which is the same thing, no straight line can make so great an acute angle with the diameter at its extremity, or so small an angle with the straight line which is at right angles to it, as not to cut the circle. Let ABC be a circle, the centre of which is D, and the dia- meter AB ; the straight line drawn at right angles to AB from its extremity A, shall fall without the circle. For, if it does not, let it fall, if possible, within the circle, as AC, and draw DC to the point C where it meets the circumference : and because DA is equal to DC, the angle DAC is equal* to the angle ( y^ ~) "^ * ^- '^ ACD ; but DAC is a right angle, therefore ACD is a right angle, and the angles DAC, ACD are therefore equal to two right an- gJesj which is impossible'': therefor^ the straight line drawn b 17. 1. L 82 THE ELEMENTS Book III. from A at right angles to BA does not fall within the circle : in *— (T-*-' the same manner it may be demonstrated, that it does not fall upon the circumference ; therefore it must fall without the cir- cle, as AE. And between the straight line AE and the circumference no straight line can be drawn from the point A which does not cut the circle : for, if possible, let FA be between them, and cl2. 1. from the point D draw^ DG perpendicular to FA, and let it meet the circumference in H : and because AGD is a right d 19. 1. angle, and EAG less ^ than a right angle : DA is greater ^ than DG: but DA is equal to DH ; " therefore DH is greater than DG, the less than the greater, which is impossible : therefore no straight line can be drawn from the point A between AE and the circumfe- rence, which does not cut the cir- cle, or, which amounts to the same thing, however great an acute angle a straight line makes with the dia- meter at the point A, or however small an angle it makes with AE, the circumference passes between that straight line and the perpendicular AE. ' And this is •all ' that is to be understood, when, in the Greek text and transla- ' tions from it, the angle of the semicircle is said to be greater ' than any acute rectilineal angle, and the remaining angle less ' than any rectilineal angle.' Cor. From this it is manifest, that the straight line which is drawn at right angles to the diameter of a circle from the ex- tremity of it, touches the circle ; and that it touches it only in one point, because, if it did meet the circle in two, it would e 2. 3. fall within it <=. ' Also it is evident that there can be but one * straight line which touches the circle in the same point.' PROP. XVn. PROB. TO draw a straight line from a given point, either without or in the circumference, which shall touch a given circle. First, Let A be a given point without the given circle BCD ; OF EUCLID. it is required to draw a straight line from A whicli shall touch Book III. the circle. ^— -v"*.^ Find a the centre E of the circle, and join AE ; and from the a 1. 3. centre E, at the distance EA, describe the circle AFG ; from the point D draw ^ DF at right angles to EA, and join EBF, AB. b 11. 1. AB touches the circle BCD. Because E is the centre of the circles BCD, AFG, EA is equal to EF : and ED to EB ; therefore the two sides AE, EB are equal to the two FE, ED, and they contain the angle at E common to the two triangles AEB, FED ; there- fore the base DF is equal to the base AB, and the trian- gle FED to the triangle AEB, and the other angles to the other angles « : therefore the angle c 4. 1. EBA is equal to the angle EDF : but EDF is a right angle, wherefore EBA is a right angle : and EB is drawn from the *»> centre : but a straight line di'awn from the extremity of a diameter, at right angles to it, touches the circle^: therefore d Cor. 16 AB touches the circle ; and it is drawn from the given point A. 3. Which was to be done. But, if the given point be in the circumference of the circle, as the point D, draw DE to the centre E, and DF at right angles to DE ; DF touches the circle •*. PROP. XVIII. THEOR. IF a Straight line touches a circle, the straight line drawn from the centre to the point of contact shall be perpendicular to the line touching the circle. Let the straight line DE touch the circle ABC in the point C ; take the centre F, and draw the straight line FC : FC is per- pendicular to DE. For, if it be not, from the point F draw FBG perpendicular to DE ; and because FGC is a right angle, GCF is a an acute a 17. 1 angle ; and to the greater angle the greatest ^ side is opposite : b 19. 1 84 THE ELEMENTS Book III. therefore FC is greater than FG ; v.i-v-*^ but FC is equal to FB ; therefore FB is greater than FG, the less than the greater, which is impos- sible ; wherefore FG is not per- pendicular to DE : in the same manner it may be shown, that no other is perpendicular to it besides FC, that is, FC is perpendicular to DE. Therefore, if a straight line, &c. Q. E. D. a 18. 3. PROP. XIX. THEOR. IF a straight line touches a circle, and from the point of contact a straight line be drawn at right an- gles to the touching line, the centre of the circle shall be in that line. Let the straight line DE touch the circle ABC in C, and from C let CA be drawn at right angles to DE ; the centre of the cir- cle is in CA. For, if not, let F be the centre, if possible, and join CF : because DE touches the circle A ABC, and FC is drawn from the centre to the point of contact, FC is perpendicular » to DE ; there- fore FCE is a right angle: but ACE is also a right angle ; therefore the angle FCE is equal to the an- gle ACE, the less to the greater, which is impossible : wherefore F is not the centre of the circle ABC: in the same njanner it may be shown that no other point which is not in CA is the centre ; that is, D C the centre is in CA. There- fore, if a straight line, Sec. Q. E. D, PROP. XX. THEOR. See N. THE angle at the centre of a circle is double of the angle at the circumference, upon the same base, that is', upon the same part of the circumference. OF EUCLID. b32. 1 Let ABC be a circle, and BEC an angle at the centre, and Book III, BAC an angle at the circumference, which have the same cir- <^ v ^ cumference BC for their base ; the angle BEC is double of the angle BAC. First, let E the centre of the circle be within the angle BAC, and join AE, and produce it to F : because EA is equal to EB, the angle EAB is equal ^ to the angle EBA ; therefore the angles EAB, EBA are double of the angle EAB ; but the angle BEF is equal '' to the angles EAB, EBA ; therefore also the angle BEF is double of the angle EAB : for the same reason, the angle FEC is double of the angle EAC : therefore the whole angle BEC is double of the whole angle BAC. Again, Let E the centre of the A circle be without the angle BDC, and join DE, and produce it to G. It may be demonstrated, as in the first case, that the angle GEC is double of the angle GDC, and that GEB a part of the first is double of GDB a part of the other ; therefore the re- maining angle BEC is double of the remaining angle BDCr Therefore, the angle at the centre, 8cc, Q. E. D. PROP. XXL THEOR. THE angles in the same segment of a circle are Sec n equal to one another. Let ABCD be a circle, and BAD, BED angles in the same segment BAED : the angles BAD, BED are equal to one another. Take F the centre of the circle ABCD: and, first, let the segment BAED be greater than a semicircle, and join BF, FD : and because the angle BED is at the centre, and the angle BAD at the circumference, and that they have the same part of 86 THE ELEMENTS Booklll. the circumference, viz. BCD, for their base ; therefore the an- *>— -v— ^ gle BFB is double » of the angle BAD : for the same reason, a 20. 3. the angle BFD is double of the angle BED: therefore the angle BAD is equal to the angle BED. But, if the segment BAED be not greater than a semicircle, let BAD, BED be angles in it; these also are equal to one another: draw AF to the centre, and produce it to C, and join CE : therefore the seg- ment BADC is greater than a semi- circle ; and the angles in it BAG, BEC are equal, by the first case : for the same reason, because CBED is greater than a semicircle, the angles CAD, CED are equal : therefore the whole angle BAD is equal to the whole angle BED. Wherefore, the angles in the same segment, &:c. Q. E. D. PROP. XXII. THEOR. THE opposite angles of any quadrilateral figure de- scribed in a circle are together equal to two right angles. Let ABCD be a quadrilateral figure in the circle ABCD ; any two of its opposite angles are together equal to two right angles. Join AC, BD ; and because the three angles of every tri- a 32. 1. angle are equal* to two right angles, the three angles of the triangle CAB, viz. the angles CAB, ABC, BCA are equal to two right angles : but tlic angle CAB b21. 3. is equal'' to the angle CDB, because they are in the same segment BADC, and the angle ACB is equal to the angle ADB, because they are in the same segment ADCB: therefore the whole angle ADC is equal to the angles CAB, ACB : to each of these equals add the angle ABC ; therefore the angles ABC, CAB, BCA are equal to the angles AHC, ADC : but 7\BC, CAB, BCA are equal to two right angles; therefore also the angles ABC, ADC are equal to tv.o right angles: in the same manner, the angles OF EUCLID; 87 BAD, DCB may be shown to be equal to two right angles. Book III. Therefore, the opposite angles, £cc. Q. E. D. •— v— ' PROP. XXIII. THEOR. UPON the same straight line, and upon the same See n. side of it, there cannot be two similar segments of circles not coinciding with one another. If it be possible, let the two similar segm-ents of circles, viz. ■ ACB, ADB, be upon the same side of the same straight line AB, not coinciding with one another : then, because the cir- cle ACB cuts the circle ADB in the two points A, B, they cannot cut one another ^ in any other point ^: one of the segments / ^X \ ('^-^ ^ ^^' ^' must therefore fall within the other; let ACB fall within ADB, and draw the straight line BCD, and join C A, DA : and because the segment ACB is similar to the segment ADB, and that similar segments of circles contain ^ equal an- b 11. def. gles ; the angle ACB is equal to the angle ADB, the exterior to 3. the interior, which is impossible <=. Therefore, there cannot be c 16. 1. two similar segments of a circle upon the same side of the same line, which do not coincide. Q. E. D. PROP. XXIV. THEOR. SIMILAR segments of circles upon equal straight See n lines are equal to one another. Let AEB, CFD be similar segments of circles upon the equal straight lines AB, CD ; the segment AEB is equal to the seg- ment CFD. For, if the seg~ E F ment AEB be ap- plied to the seg- ment CFD, so as the point A be on C, and the straight line AB upon CD, the point B shall coincide with the point D, 88 THE ELEMENTS Book III. because AB is equal to CD : therefore, the straight line AB co- <^^^fmmJ inciding with CD, the segment AEB must * coincide with the % 23. 3. segment CFD, and therefore is equal to it. Wherefore, similar segments, &c. Q. E. D. PROP. XXV. PROB. See N. A SEGMENT of a circle being given, to describe the circle of which it is the segment. a 10. 1. b 11. 1. c6. 1. d9.3. Let ABC be the given segment of a circle ; it is required to describe the circle of which it is the segment. Bisect ^ AC in D, and from the point D draw ^ DB at right angles to AC, and join AB: first, let the angles ABD, BAD be equal to one another; then the straight line BD is equal '^ to DA, and therefore to DC ; and because the three straight lines DA, DB, DC are all equal, D is the centre of the cir- cle *!: from the centre D, at the distance of any of the three DA, DB, DC describe a circle ; this shall pass through the other points ; and the circle of which ABC is a segment is described : and because the centre D is in AC, the segment ABC is a se- micircle: but if the angles ABD, BAD are not equal to one e 23. 1. another, at the point A, in the straight line AB, make « the angle BAE equal to the angle ABD, and produce BD, if necessary, to E, and join EC : and because the angle ABE is equal to the an- gle BAE, the straight line BE is equal e to EA : and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each ; and the angle ADE is equal to the angle CDE, for each of them is a right angle; therefore the base AE is equal f 4. 1. f to the base EC : but AE was shown to be equal to EB ; where- fore also BE is equal to EC: and the three straight lines AE; OF EUCLID. 89 EB, EC are therefore equal to one another; Avherefore* E is Book III. the centre of the circle. From the centre E, at the distance of ^ - y ^ any of the three AE, EB, EC, describe a circle ; this shall pass d 9. 3. through the other points ; and the circle of which ABC is a seg- ment is described: and it is evident, that if the angle ABD be greater than the angle BAD, the centre E falls without the seg- ment ABC, which therefore is less than a semicircle : but if the angle ABD be less than BAD, the centre E falls within the seg- ment ABC, which is therefore greater than a semicircle : where- fore a segment of a circle being given, the circle is described of which it is a segment. Which was to be done. PROP. XXVI. THEOR. IN equal circles, equal angles stand upon equal circumferences, whether they be at the centres or circumferences. Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centres, and BAC, EDF at their circumferences : the circumference BKC is equal to the circumference ELF. Join BC, EF ; and because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal: there- fore the two sides BG, GC are equal to the two EH, HF; and the angle at G is equal to the angle at H ; therefore the base BC is equal ^ to the base EF : and because the angle at A is equal a 4. 1. to the angle at D, the segment BAC is similar •> to the segment b 11. def. EDF ; and they are upon equal straight lines BC, EF ; but simi- 3. lar segments of circles upon equal straight lines are equal <= to * one another : therefore the segment BAC is equal to the segment £. 94 " EDF : but the whole circle ABC is equal to the whole DEF ; " "^ M 90 THE ELEMENTS Booklll. therefore -the remaining segment BKC is equal to the remaining ^^yi^ segment ELF, and the circumference BKC to the circumference ELF. Wherefore, in equal circles, &;c. Q. E. D. PROP. XXVn. THEOR. IN equal circles, the angles which stand upon equal circumferences are equal to one another, whether they be at the centres or circumferences. a 20. 3. Let the angles BGC, EHF at the centres, and BAG, EDF at the circumferences of the equal circles ABC, DEF stand upon the equal circumferences BC, EF : the angle BGC is equal to the angle EHF, and the angle BAG to the angle EDF. If the angle BGC be equal to the 'angle EHF, it is manifest * that the angle BAG is also equal to EDF : but, if not, one of them is the greater : let BGC be the greater, and at the point b 23. 1. G, in the straight line BG, make^ the angle BGK equal to the c26. 3. angle EHF; but equal angles stand upon equal circumferences S when they are at the centre ; therefore the circumference BK is equal to the circumference EF : but EF is equal to BC ; there- fore also BK is equal to BC, the less to the greater, which is impossible : therefore the angle BGC is not unequal to the an- gle EHF; that is, it is equal to il : and the angle at A is half of the angle BGC, and the angle at D half of the angle EHF : there- fore the angle at A is equal to the angle at D. Wherefore, in equal circles, S^c Q- E. D. OF EUCLID. PROP. XXVIII. THEOR. IN equal circles, equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less. Let ABC, DEF be equal circles, and BC, EF equal straight lines in them, which cut off the two greater circumferences BAC, EDF, and the two less BGC, EHF; the greater BAC is equal to the greater EDF, and the less BGC to the less EHF. Take » K, L the centi-es of the circles, and join BK, KC, a- 1- 3- EL, LF : and because the circles are equal, the straight lines from their centres are equal; therefore BK, KC are equal to EL, LF ; and the base BC is equal to the base EF ; therefore the angle BKC is equal ^ to the angle ELF : but equal angles b 8. 1. stand upon equal <= circumferences, when they are at the cen-c26. 3. tres ; therefore the circumference BGC is equal to the circum- ference EHF. But the whole circle ABC is equal to the whole EDF ; the remaining part, therefore, of the circumference, viz. BAC, is equal to the remaining part EDF. Therefore, in equal circles, &:c. Q. E. D. PROP. XXIX. THEOR. IN equal circles, equal circumferences are subtended by equal straight lines. Let ABC, DEF be equal circles, and let the circumferences BGC, EHF also be equal ; and join BC, EF : the straight line BC is equal to the straight line EF. 92 THE ELEMENTS Book III. Take » K, L, the centres of the circles, and join BK, KC, v^.i'V—^ EL, LF : and because the circumference BGC is equal to the a 1. 3. A D G H b27. 3. circumference EHF, the angle BKC is equal ^ to the angle ELF : and because the circles ABC, DEF are equal, the straight lines from their centres are equal : therefore BK, KC are equal to EL, LF, and they contain equal angles : therefore the base c 4. 1. BC is equal <= to the base EF. Therefore, in equal circles, 8cc. Q. E. D. PROP. XXX. PROB. a 10. 1. TO bisect a given circumference, that is, to dhide it into two equal parts. Let ADB be the given circumference ; it is required to bisect it. Join AB, and bisect » it in C ; from the point C draw CD at right angles to AB, and join AD, DB : the circumference ADB is bisected in the point D. Because AC is equal to CB, and CD common to the triangles ACD, BCD, the two sides AC, CD are equal to the two BC, CD ; and the angle ACD is equal to the angle BCD, because each of them is a right angle ; therefore the base AD is equal ^ to the base BD : but equal straight lines cut off equal <= circumferences, the greater equal to the greater, and the less to the less, and AD, DB are each of them d Cor, 1. less than a semicircle ; because DC passes through the centre ^ : '• wherefore the circumference AD is equal to the circumference DB : therefore the given circumference is bisected in D. Which was to be done. b4. 1. c28.3 OF EUCLID. PROP. XXXI. THEOR. IN a circle, the angle in a semicircle is a right angle ; but the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle. Let ABCD be a circle, of which the diameter is BC, and centre E; and draw CA dividing the circle into the segments ABC, ADC, and join BA, AD, DC ; the angle in the semi- circle BAG is a right angle ; and the angle in the segment ABC, which is greater than a semicircle, is less than a right angle ; and the angle in the segment ADC, which is less than a semicircle, is greater than a right angle. Join AE, and produce BA to F ; and because BE is equal to EA, the angle EAB is equal ^ to EBA; also, because AE a_5. 1. is equal to EC, the angle EAC is equal to ECA; wherefore the whole angle BAC is equal to the two angles ABC, ACB ; but FAC, the exterior angle of the triangle ABC, is equal ^ to the two angles ABC, ACB; therefore the angle BAC is equal to the angle FAC, and each of them is therefore a rights angle : wherefore the angle BAC in a semicircle is a right angle. And because the two angles ABC, BAC of the triangle ABC are together less— v-^ fore the rectangle AE, EC, together with the square of EG, is a 3. 5. equal ^ to the square of AG : to each of these equals add the hS.2. square of GF; therefore the rectangle AE, EC, together with the squares of EG, GF, is equal to the squares of AG, GF : but the c 47, 1. squares of EG, GF are equal ^ to the square of EF ; and the squares of AGj GF are equal to the square of Af: therefore the rectangle AE, EC, together with the square of EF, is equal to the square of AF ; that is, to the square of FB : but the square of FB is equal "^ to the rectangle BE, ED, together with the square of EF : therefore the rectangle AE, EC, together with the square of EF, js equal to the rectangle BE, ED, toge- ther with the square of EF : take away the common square of EF, and the remaining rectangle AE, EC is therefore equal to the remaining rectangle BE, ED. Lastly, Let neither of the straight lines AC, BD pass through the centre : take the centre F, and through E, the intersection of the straight lines AC, DB, draw the dia- meter GEFH: and because the rect- angle AE, EC is equal, as has been shown, to the rectangle GE, EH ; and, for the same reason, the rectangle BE, ED is equal to the same rectangle GE, EH ; thficfore the rectangle AE, EC is equal to the rectangle BE, ED. Whers-fbre, if two straight lines, £^c. PROP. XXXVL THEOR. IF from any point without a circle two straight lines be drawn, one of which cuts the circle, and the Other touches it; the rectangle contained by the whole Une which cuts the circle, and the part of it without the circle, shall be equal to the square of the li.n^ which touches it. OF EUCLID* n c 47. J. Let D be any point without the circle ABC, and I>CA, DB Book III. two straight lines drawn from it, of which DCA cuts the circle, Vi^^— ^ and DB touches the same : the rectangle AD, DC is equal to the square of DB. Either DCA passes through the centre, or it does not ; first, let it pass through the centre E, and join EB ; therefore the an- gle EBD is a right * angle t and be- a 18. 3» cause the straight line AC is bisected in E, and produced to the point D, the rectangle AD, DC, together with the Square of EC, is equal ^ to the square / C b 6. 2. of ED, and CE is equal to EB : there- fore the rectangle AD, DC, together with the square of EB, is equal to the square of ED : but the square of ED is equals to the squares of EB, BD, be- cause EBD is a right angle : therefore the rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD : take away the common square of EB ; tljerefore the remain- ing rectangle AD, DC is equal to the square of the tangent DB. But if DCA does not pass through the centre of the circle ABC, take <* the centre E, and draw EF perpendicular* to AC, d 1,3. and join EB, EC, ED : and because the straight line EF, which c 12. 1. passes through the centre, cuts the straight lirie AC, which does not pass through the centre, at right ~ angles, it shall likewise bisect *^ it ; Jfi f3. 3. therefore AF is equal to PC s and be- cause the straight line AC is bisected in F, and produced to D, the rectangle AD, DC, together with the square of FC, is equal ^ to the square of FD : to each of these equals add the square of FE ; therefore the rectangle AD, DC, together with the squares of CF, FE, is equal to the squares of DF, FE ; but the square of ED is equal «= to the squares of DF, FE, because Et'D is a right an- gle ; and the square of EC is equal to the squares of CF, FE ; therefore the rectangle AD, DC» to- gether with the square of EC, is equal to the square of ED : and CE is equal to EB ; therefore the rectangle AD, DC, tO' gether with the square of EB, is equal to the square of ED : but the squares of EB, BD are equal to the square « of ED, be- 100 THE ELEMENTS Book III. cause EBD is a right angle ; therefore the rectangle AD, DC, U -.^— ^ together with the square of EB, is equal to the squares of EB, BD : take away the common square of EB ; therefore the re- maining rectangle AD, DC is equal to the square of DB. Wherefore, if from any point, &c. Q. E. D. Cor. If from any point without a A circle, there be drawn two straight lines cutting it, as AB, AC, the rect- angles contained by the whole lines and the parts of them without the circle, are equal to one another, viz. the rectangle B A, AE to the rectan- gle CA, AF : for each of them is equal to the square of the straight line AD which touches the circle. PROP. XXXVII. THEOR. See U. IF from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it ; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle be equal to the square of the line which meets it, the Jine which meets shall touch the circle. a 17. 3. b 18. 3. c 36. 3. Let any point D be taken without the circle ABC, and from it let two straight lines DCA and DB be drawn, of which DCA cuts the circle, and DB meets it ; if the rectangle AD, DC be equal to the square of DB, DB touches the circle. Draw a the straight line DE touching the circle ABC, find its centre F, and join FE, FB, FD ; then FED is a right •> an- gle : and because DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal ^ to the square of DE : but the rectangle AD, DC is, by hypothesis, equal to the square of DB : therefore the square of DE is equal to the square of DB ; And the straight line DE equal to the straight line DB ; and OF EUCLID. 101 FE is equal to FB, wherefore DE, EF are equal to DB, BFj Book III. and the base FD is common to the ~ two triangles DEF, DBF; therefore the angle DEF is equal ^ to the angle / 1 \ d 8. 1. DBF; but DEF is a right angle, therefore also DBF is a right angle : and FB, if produced, is a diameter, and the straight line which is drawn at right angles to a diameter, from the extremity of it, touches ^ the circle : i / ^i^ \ g 16. 3. therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E. D. fHE ELEMENTS OF EUCLID, BOOK IV. DEFINITIONS. SeeN. I. Book IV. ^ RECTILINEAL figure is said to be inscribed in another rectilineal figure, when all the angles of the inscribed figure are upon the sides of the figure in which it is inscribed, each upon each. n. In like manner, a figure is said to be described about another figure, when all the sides of the circumscribed figure pass through the an- gular points of the figure about which it is descr ibed, each through each* III. A rectilineal figure is said to be inscribed in a circle, when all the angles of the in- cribed figure are upon the circumference of the circle. IV. A rectilineal figure is said to be described about a circle, when each side of the circumscribed figure touches the circumference of the circle. V. In like manner, a circle is said to be inscrib- ed in a rectilineal figure, when the cir- cumference of the circle touches each side of the figure. OF EUCLID. 103 VI. A circle is said to be described about a rec- tilineal figure, when the circumference of the circle passes through all the angular points of the figure about which it is de- scribed. VII. A straight line is said to be placed in a circle, when the extremi- ties of it are in the circumference of the circle. fiooklV. PROP. I. PROB. IN a given circle to place a straight line, equal to a given straight line not greater than the diameter of the circle. Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle. Draw BC the diameter of the circle ABC ; then, if BC is equal to D, the thing required is done ; for in the circle ABC a straight line BC is placed equal to D ; but, if it is not, BC is greater than D ; make CE equal » to D, and from the cen- tre C, at the distance CE, de- scribe the circle AEF, and join CA ; therefore, because C is the centre of the circle AEF, CA is equal to CE ; but D is equal to CE; therefore D is equal to CA : wherefore, in the circle ABC, a straight line is placed equal to the given straight line D, which is not greater than the diameter of the circle. Which was to be done. a 3.1. PROP. II. PROB. IN a given circle to inscribe a triangle equiangular to a given triangle. 104 THE ELEMENTS Book IV. a 17. 3. b 23. 1. c 32. 3. Let ABC be the given circle, and DEF the given triangle ; it is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF. Draw a the straight line GAH touching the circle in the point A, and at the point A, in the straight line AH, make'' the angle HAC equal to the angle DEF ; and at the point A, in the straight line AG, make the an- gle GAB equal to the ^ """^""-^^ A angle DFE, and join BC : therefore because HyVG touches the cir- cle ABC, and AC is drawn from the point of contact, the angle PL\C is equal c to the angle ABC in the alter- nate segment of the cir- cle : but HAC is equal to the angle DEF ; therefore also the an- gle ABC is equal to DEF : for the same reason, the angle ACB is equal to the angle DFE ; therefore the remaining angle BAC is equal ^ to the remaining angle EDF : wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC. Which was to be done. PROP. HI. PROB. ABOUT a given circle to describe a triangle equi- angular to a given triangle. Let ABC be the given circle, and DEF the given triangle ; it is required to describe a triangle about the circle ABC equiangu- lar to the triangle DEF. Produce EF both ways to the points G, H, and find the centre K of the circle ABC, and from it draAV any straight line KB ; DG is equal to DF ; therefore the three straight lines DE, DF, c 26. 1. DG are equal to one another, and the circle described from the centre D, at the distance of any of them, shall pass through the extremities of the other two, and touch the straight lines AB, BC, CA, because the angles at the points E, F, G are right an- gles, and the straight line which is drawn from the extremity of d 16. 3. a diameter at right angles to it, touches 'l the circle: therefore the straight lines AB, BC, CA do each of them touch the circle, and the circle EFG is inscribed in the triangle ABC. Which was to be'done. PROP. V. PROB. See N. TO describe a circle about a given triangle. Let the given triangle be ABC ; it is required to describe a circle about ABC. a 10. 1. Bisect a AB, AC in the points D, E, and from these points b 11. 1. draw DF, EF at right angles'* to AB, AC; DF, EF produced V> C B c4. 1. meet one another: for, if they do not meet, they are parallel, wherefore AB, AC, Avhich are at right angles to them, are pa- rallel ; which is absurd : let them meet in F, and join FA ; also, if the point F be not in BC, join BF, CF : then, because AD is equal to DB, and DF common, and at right angles to AB, the base AF is ecjual^ to the base FB: in like manner, it may be shown, that CF is equal lo FA ; and therefore BF is equal to IC; and FA, FB, FC are equal to one anothei: ; OF EUCLID. lor wherefore the circle described from the centre F, at the distance Book IV. of one of them, shall pass through the extremities of the other *'■— r— ^ two, and be described about the triangle ABC. Which was to be done. Cor. And it is manifest, that when the centre of the circle falls within the triangle, each of its angles is less than a right angle, each of them being in a segment greater than a semicir- cle ; but, when the centre is in one of the sides of the triangle, the angle opposite to this side, being in a semicircle, is a right angle ; and, if the centre falls without the triangle, the angle op- posite to the side beyond which it is, being in a segment less than a semicircle, is greater than a right angle : wherefore, if the given triangle be acute angled, the centre of the circle falls within tt; if it be a right angled triangle, the centre is in the side oppo- site to the right angle ; and, if it be an obtuse angled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle. PROP. VI. PROB. TO inscribe a square in a given circle. ft 4.1. Let ABCD be the given circle ; it is required to inscribe a square in ABCD. Draw the diameters AC, BD at right angles to one another ; and join AB, BC, CD, DA ; because BE is equal to ED, for E is the centre, and that EA is common, A and at right angles to BD ; the base BA is equal ^ to the base AD ; and, for the same reason, BC, CD are each of them equal to BA or AD ; therefore the quadrilateral figure ABCD is equi- lateral. It is also rectangular ; for the straight line BD, being the diameter of the circle ABCD, BAD is a semi- circle ; wherefore the angle BAD is a right ^ angle ; for the same reason each of the angles ABC, BCD, b 31. 3. CDA is a right angle ; thei*efore the quadrilateral figure ABCD is rectangular, and it has been shown to be equilateral ; therefore it is a square ; and it is inscribed in the circle ABCD. Which was to be done. Book IV. THE ELEMENTS a 17. 3. b 18. 3. c 28. 1. d34.1. PROP. VII. PROB. TO describe a square about a given circle. Let ABCD be the given circle ; it is required to describe a square about it. Draw two diameters AC, BD of the circle ABCD, at right an- gles to one another, and through the points A, B, C, D draw * FG, GH, HK, KF touching the circle ; and because FG touches the circle ABCD, and EA is drawn from the centre E to th& point of contact A, the angles at A are right ^ angles ; for the same reason, the angles at the points B, C, D are right angles ; and because the angle AEB is a right G A F angle, as likewise is EBG, GH is pa- rallel "^ to AC ; for the same reason, AC is parallel to FK, and in like man- ner GF, HK may each of them be de- monstrated to be parallel to BED ; therefore the figures GK, GC, AK, FB, BK are parallelograms ; and GF is therefore equal ^ to HK, and GH to FK ; and because AC is equal to BD, H C K and that AC is equal to each of the two GH, FK ; and BD to each of the two GF, HK : GH, FK are each of them equal to GF or HK ; therefore the quadrilateral figure FGHK is equilate- ral. It is also rectangular ; for GBEA being a parallelogram, and AEB a right angle, AGB ^ is likewise a right angle : in the same manner, it may be shown that the angles at H, K, F are right angles ; therefore the quadrilateral figure FGHK is rect- angular, and it was demonstrated to be equilateral ; therefore it is a square ; and it is described about the circle ABCD. Which was to be done. B r. N J D PROP. VIII. PROB. TO inscribe a circle in a given square. I^et ABCD be the given square ; it is required to inscribe a circle in ABCD. a. 10. 1. Bisect ^ each of the sides AB, AD, in the points F, E, and b31 1. through E draw •» EH parallel to AB or DC, and through F OF EUCLID. 109 draw FK parallel to AD or BC ; therefore each of the figures Book IV. AK, KB, AH, HD, AG, GC, BG, GD is a parallelogram, and their ^— r— ' opposite sides are equal^ ; and because AD is equal to AB, and c 34. 1. that AE is the half of AD, and AF the half of AB, AE is equal to AF; wherefore the sides opposite A E D to these are equal, viz. FG to GE ; in the same manner, it may be demon- strated that GH, GK are each of them equal to FG or GE ; therefore the four straight lines GE, GF, GH, GK are equal to one another ; and the cir- cle described from the centre G, at the distance of one of them, shall pass through the extremities of the other three, and touch the straight lines AB, BC, CD, DA : because the angles at the points E, F, H, K are right "i angles; and that the straight line which is drawn from d the extremity of a diameter, at right angles to it, touches the circle « ; therefore each of the straight lines AB, BC, CD, DAe 16. 3. touches the circle, which therefore is inscribed in the square ABCD. Which was to be done. iJ B H C 29.1. PROP. IX. PROB. TO describe a circle about a given square. a 8.1. Let ABCD be the given square; it is required to describe a circle about it. Join AC, BD cutting one another in E ; and because DA is equal to AB, and AC common to the triangles DAC, BAC, the two sides DA, AC are equal to the two BA, AC ; and the base DC is equal ^ to the base BC ; wherefore the angle DAC is equal* to the angle BAC, andi the angle DAB is bisected by the straight] line AC : in the same manner, it may be demonstrated that the angles ABC, BCD, CDA are severally bisected by the straight ^^ lines BD, AC ; therefore, because the angle DAB is equal to the angle ABC, and that the angle EAB is the half of DAB, and EBA the half of ABC; the angle EAB is equal to the angle EBA; wherefore the side EA is equal *» to the side EB : in the same manner, it may be b 6.1. HO TITE ELEMENTS Book IV. demonstrated, that the strais^ht lines EC, ED are each of them *— "v^-* equal to EA or EB ; therefore the four straight lines EA, EB, EC, ED are equal to one another ; and the circle described from the centre E, at the distance of one of them, shall pass through the extremities of the other three, and be described about the square ABCD. Which was to be done. PROP. X. PROD. TO describe an isosceles triangle, having each of the angles at the base double of the third angle. all. 2. Take any straight line AB, and divide^ it in the point C, so that the rectangle AB, BC be equal to the square of CA ; and from the centre A, at the distance AB, describe the circle BDE, b 1. 4. in which place ^ the straight line BD equal to AC, which is not greater than the diameter of the circle BDE ; join DA, DC, and c 5. 4. about the triangle ADC describe = the circle ACD ; the triangle ABD is such as is required, that is, each of the angles ABD, ADB is double of the angle BAD. Because the rectangle AB, BC is equal to the square of AC, and that AC is equal to BD, the rectangle AB, BC is equal to the square of BD ; and because from the point B, without the circle ACD, two straight lines BCx\, BD are drawn to the cir- cumference, one of which cuts, and the other meets the circle, and that the rectangle AB, BC contained by tlie whole of the cutting line, and the part of it without the circle, is equal to the square of BD which meets it; d ^r. o. the straight line BD touches 'i the circle ACD ; and because BD touches the circle, and DC is drawn from the point of con- e 33. 3. tact D, the angle BDC is equals to the angle DAC in the alternate segment of the circle; to each of these add the angle CDA; therefore the whole angle BDA is equal to the two f .32. 1. angles CDA, DAC; but the exterior angle BCD is equal f to the angles CD.\, DAC ; therefore also BDA is equ&l to BCD ; OF EUCLID. Ill but BDA is equals to the aiif^le CBD, because the side AD Book IV. is equal to the side AB ; therefore CBD, or DBA is equal to *»— v— ' BCD ; and consequently thT three angles BDA, DBA, BCD g 5. 1. are equal to one another ; and because the angle DBC is equal to the angle BCD, the side BD is equal ^ to the side DC ; but h 6. 1 BD was made equal to CA ; therefore also CA is equal to CD, and the angle CDA equals to the angle DAC ; therefore the angles CD A, D AC together, are double of the angle DAC: but BCD is equal to the angles CDA, DAC ; therefore alsc^ BCD is double of DAC, and BCD is equal to each of the angles BDA, DBA ; each therefore of the angles BDA, DBA is double of the angle DAB; wherefore an isosceles triangle ABD is described, having each of the angles at the base double of the third angle. Which was to be done. PROP. XI. PROB. TO inscribe an equilateral and equiangular penta- gon in a given circle. Let ABCDE be the given circle ; it is required to inscribe an equilateral and equiangular pentagon in the circle ABCDE. Describe* an isosceles triangle FGH, having each of the a 10. 4 angles at G, H, double of the angle at F ; and in the circle ABCDE inscribe ^ the triangle ACD equiangular to the tri- b 2. 4 angle FGH, so that the angle A CAD be equal to the angle at F, and each of the angles ACD, CDA equal to the angle at G or H ; wherefore each of the angles ACD, CDA is double of the angle CAD. Bisect c the angles / \ \\ / ^ the circle; take the centre F, and join FB, FK, FC, FL, FD : and because the straight line KL touches the circle ABCDE in the point C, to which FC is drawn from the c 18. 3. centre F, FC is perpendicular = to KL ; therefore each of the angles at C is a right angle : for the same reason, the angles at the points B, D, are right angles : and because FCK is a d 47. 1. right angle, the square of FK is equal ^ to the squares of FC, CK : for the same reason, the square of FK is equal to the squares of FB, BK : therefore the squares of FC, CK are equal to the squares of FB, BK, of which the square of FC is equal to the squa:-:; of FB ; tl.c remaining square of CK is therefore equal to OF EUCLID. 113 the remaining square of BK, and the straight line CK equal to Book IV. BK : and because FB is equal to FC, and FK common to the ^— -y — ^ triangles BFK, CFK, the two BF, FK are equal to the two CF, FK ; and the base BK is equal to the base KC ; therefore the angle BFK is equal e to the angle KFC, and the angle BKF to e 8. 1. FKC ; wherefore the angle BFC is double of the angle KFC, and BKC double of FKC ; for the same reason, the angle CFD is doublp of the angle CFL, and CLD double of CLF : and be- cause the circumference BC is equal to the circumference CD, the angle BFC is equal f to the • f 27. 3. angle CFD; and BFC is dou- G ble of the angle KFC, and CFD double of CFL ; therefore the angle KFC is equal to the angle CFL ; and the right angle FCK is equal to the right angle FCL: therefore, in the two triangles FKC, FLC, there are two an- gles of one equal to two angles of the other, each to each, and the side FC, which is adjacent to the equal angles in each, is common to both ; therefore the other sides shall be equal s to the other sides, and the third angle S 26. 1 to the third angle : therefore the straight line KC is equal to CL, and the angle FKC to the angle FLC : and because KC is equal to CL, KL is double of KC : in the same manner, it may be shown that HK is double of EK : and because BK is equal to KC, as was demonstrated, and that KL is double of KC, and HK double of BK, HK shall be equal to KL : in like manner it may be shown that GH, GM, ML are each of them equal to HK or KL : there- fore the pentagon GHKLM is equilateral. It is also equiangular ; for, since the angle FKC is equal to the angle FLC, and that the angle HKL is double of the angle FKC, and KLM double of FLC, as was before demonstrated, the angle HKL is equal to KLM: and in like manner it may be shown, that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM : therefore the five angles GHK, HKL, KLM, LMG, MGH being equal to one another, the pentagon GHKLM is equiangular : and it is equila- teral, as was demonstrated ; and it is described about the circle ABCDE. Which Avas to be done. 114 Book IV. THE ELEMENTS PROP. XIII. PROB. A TO inscribe a circle in a given equilateral and equi- angular pentagon. Let ABCDE be the given equilateral and equiangular penta- gon ; it is required to inscribe a circle in the pentagon ABCDE. a 9.1. Bisect 2 -.he angles BCD, CDE by the straight lines CF, DF, and from the point F, in which they meet, draw the straight lines FB, FA, FE : therefore, since BC is equal to CD, and CF com- mon to the triangles BCF, DCF, the two sides BC, CF are equal to the two DC, CF ; and the angle BCF is equal to the angle b 4. 1. DCF ; therefore the base BF is equal ^ to the base FD, and the other angles to the other angles, to which the ecjual sides are op- posite ; therefore the angle CBF is equal to the angle CDF : and because tiie angle CDE is double of CDF, and that CDE is equal to CBA, and CDF to CBF ; CBA is also double of the angle CBF ; therefore the angle ABF is equal to the angle CBF; wherefore the angle ABC is bisected by the straight line BF : in the same manner, it may be demonstrated, that the angles BAE, AED are bisected by the slraigiit lines AF, c 12. 1. FE : froui the point F draw c FG, FH, FK, FL, FM perpen- diculars to the straight lines AB, BC, CD, DE, EA: and be- cause tile angle IICF is equal to KCF, and tiie right angle FHC equal to the right angle FKC ; in liie triangles FHC, FKC there are two angles of one equal to two anj^;les of the other, and the side FC, which is opposite to one of the equal angles in each, is common to both ; therefore d 26. 1. tlie otiier sides shall be eciual 'i, each to each; wherefore the perpendicular FII is equal to tiie perpendicular FK : in the same manner it may be demonstrated that FL, FM, FG are each pf them equal to FII or FK ; therefore the five straight lines FG, I'll, FK, FL, FM are equal to one another : wherefore the cir- cle described from the centre F, at the distance of one of these live, shall pass through the extremities of the other four, and OF EUCLID. 115 touch the straight lines AB, BC, CD, DE, EA, because the Book IV. angles at the points G, H, K, L, M are right angles ; and that ^— v— ' a straight line drawn from the extremity of the diameter of a circle at right angles to it, touches e the circle : therefore each e 15. 3- of the straight lines AB, BC, CD, DE, EA touches the circle ; wherefore it is inscribed in the pentagon ABCDE. Which was to be done. PROP. XIV. PROB. TO describe a circle about a given equilateral and equiangular pentagon. Let ABCDE be the given equilateral and equiangular penta- gon ; it is required to describe a circle about it. Bisect a the angles BCD, CDE by the straight lines CF, FD, a 9. 1. and from the point F, in which they meet, draw the straight lines FB, FA, FE to the points B, Aj E. It may be demonstrated, in the same manner as in the preceding proposition, that the angles CBA, BAE, AED are bisected by the straight lines FB, FA, FE : and because the angle BCD is equal to the angle CDE, and that FCD is the half of the angle BCD, and CDF the half of CDE ; the angle FCD is equal to FDC ; wherefore the side CF is equal *> to the side FD : in Hke manner it may be demon- b 6. l\ strated that FB, FA, FE are each of them equal to FC or FD : therefore the five straight lines FA, FB, FC, FD, FE arc equal to one another ; and the circle described from the centre F, at the distance of one of them, shall pass through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Which wa« to be done. THE ELEMENTS PROP. XV. PROB. SeeN. a 5. 1. b 32. 1. C 13. 1. d 15. 1. e26. f 29. .1. TO inscribe an equilateral and equiangular hexa- gon in a given circle. Let ABCDEF be the given circle ; it is required to inscribe im. ecfuilateral and equiangular hexagon in it. Find the centre G of the circle ABCDEF, and draw the dia- meter AGD ; and from D as a centre, at the distance DC, de- scribe the circle EGCH, join EG, CG, and produce them to the points B, F ; and join AB, BC, CD, DE, EF, FA : the hexagon ABCDEF is equilateral and equiangular. Because G is the centre of the circle ABCDEF, GE is equal to GD : and because D is the centre of the circle EGCH, DE is equal to DG ; wherefore GE is equal to ED, and the tri- angle EGD is equilateral; and therefore its three angles EGD, GDE, DEG are equal to one another, because the angles at the base of an isosceles triangle are equal *; and the three angles of a triangle are equal b to two right angles ; therefore the angle EGD is the third part of two right angles: in the same manner it may be demonstrated, that the angle DGC is also the third part of two right angles : and because the straight line GC makes with EB the adjacent angles EGC, CGB equal <= to two right angles ; the remaining angle CGB is the third part of two right angles ; therefore the angles EGD, EIGC, CGB are equal to one another: and to these arc equaH the vertical opposite angles BGA, AGF, FGE : therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another : but equal angles stand upon equal « circumfe- rences ; therefore the six circumfe- rences AB, BC, CD, DE, EF, FA are equal to one another : and equal circumferences are subtended by equal *" straight lines ; therefore the six straight lines are equal to one another, and the hexagon ABCDEF is equilateral. It is also equiangu- lar ; for, since the circumference AF is equal to ED, to each of thtfcC add the circumference ABCD : therefore the whole cir- cumference FABCD shall be equal to the whole EDCBA ; OF EUCLID. iir and the angle FED stands upon the circumftiience FABCD, and Book IV. the angle AFE upon EDCBA ; therefore the angle AFE is *— v—' equal to FED : in the same manner it may be demonstrated that the other angles of the hexagon ABCDEF are each of them equal to the angle AFE or FED ; therefore the hexagon is equiangular; and it is equilateral, as was shown; and it is in- scribed in the given circle ABCDEF. Which was to be done. Cor. From this it is manifest, that the side of the hexagon is equal to the straight line from the centre, that is, to the semi- diameter of the circle. And if through the points A, B, C, D, E, F there be drawn straight lines touching the circle, an equilateral and equiangular hexagon shall be described about it, which may be demonstrat- ed from what has been said of the pentagon ; and likewise a cir- cle may be inscribed in a given equilateral and equiangular hexa- gon, and circumscribed about it, by a method like to that used for the pentagon. PROP. XVI. PIIOB. TO inscribe an equilateral and equiangular quin- See n. decagon in a given circle. Let ABCD be the given circle ; it is required to inscribe an equilateral and equiangular quindecagon in the circle ABCD. Let AC be the side of an equilateral triangle inscribed ^ in a 2. 4. the circle, and AB the side of an equilateral and equiangular pentagon inscribed ^ in the same ; therefore, of such equal parts b 11. 4- as the whole circumference ABCDF contains fifteen, the cir- cumference ABC, being the third \ part of the whole, contains five ; and the circumference AB, which is the fifth part of the whole, contains three ; therefore BC their difference contains two of the same parts: bi- sect ^BC in E; therefore BE, EC are, each of them, the fifteenth part of the whole circumference ABCD : tlicrefore, if the straight lines BE, EC be drawn, and straight lines equal to them be placed d around in the whole circle, an equila-d L4 leral and equiangular quindecagon shall be inscribed in it- Which was to be done. cSO. jU THE ELEMENTS, &c. Book IV, And in the same manner as was done in the pentagon, if S^^mmJ through the points of division made by inscribing the quindeca- gon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon shall be described about it : and likewise, as in the pentagon, a circle may be inscribed in a given equilateral and equiangular quindecagon, and circumscribed about it. THE ELEMENTS OF EUCLID. BOOK V. DEFINITIONS. I. A LESS magnitude is said to be a part of a greater magnitude, Book V. when the less measures the greater, that is, ' when the less is ^•^f'^ ' contained a certain number of times exactly in the greater.' II. 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.* III. * Ratio is a mutual relation of two magnitudes of the same kind S«e N ' to one another, in respect of quantity.' IV. Magnitudes are said to have a ratio to one another, when the less can be multiplied so as to exceed the other. V. The first of four magnitudes is said to have the same ratio to the \ second, which the third has to the fourth, when any equimul- tiples whatsoever of the first and third being taken, and any equimultiples whatsoever of the second and fourth ; if tlie mul- tiple of the first be less than that of the second, the multiple of the third is also less than that of the fourth ; or, if the multiple of the first be equal to that of the second, the multiple of the third is also equal to that of the fourth; or, if the multiple of 120 THE ELEMENTS Book V. the first be greater than that of the second, the multiple of the •— v^— ' third is also greater than that of the fourth. VI. Magnitudes which have the same ratio are called propoi'tionals. N. B. ' When four magnitudes are proportionals, it is usually * expressed by saying, the first is to the second, as the third to ' the fourth.' VII. 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 multiple of the third is not greater than the multiple of the fourth ; then the first is said to have to the second a greater ratio than the third magni- tude has to the fourth ; and, on the contrary, the third is said to have to the fourth a less ratio than the first has to the second. VIII. " Analogy, or proportion, is the similitude of ratios." IX. Proportion consists in three terms at least. X. When three magnitudes are proportionals, the first is said to have to the third the duplicate ratio of that which it has to the second. XI. See N. When four magnitudes are continual proportionals, the first is said to have to the fourth the triplicate ratio of that which it has to the second, and so on, quadruplicate. Sec increasing the denomination still by unity, in any number of propor- tionals. Definition A, to wit, of compound ratio. When there are any number of magnitudes of the same kind, the first is said to have to the hist of tliem the ratio com- pounded of the ratio which the first has to the second, and of the ratio which the second has to the third, and of the ratio which the third has to the fourth, and so on unto the last magnitude. Tor example, if A, B, C, D be four magnitudes of the same kind, the first A is said to have to the hist D the ratio com- pounded of the ratio of A to B, anil of the ratio of B to C, and of the ratio of C to D ; or, the ratio of A to D is said to be compounded of the ratios of A to B, B to C, and C to D : OF EUCLID. 121 Afid if A has to B the same ratio which E has to F ; and B to C Book V. the same ratio that G has to H ; and C to D the same that K *— y —^ has to L ; then, by this definition, A is said to have to D the ratio compounded of ratios which are the same with the ratios of E to F, G to H, and K to L : and the same thing is to be un- derstood when it is more briefly expressed, by saying A has to D the ratio compounded of the ratios of E to F, G to H, and K to L. Jn like manner, the same things being supposed, if M has to N the same ratio which A has to D; then, for shortness' sake, M is said to have to N the ratio compounded of the ratios of E to F, G to H, and K to L. XII. In proportionals, the antecedent terms are called homologous to one another, as also the consequents to one another. * Geometers make use of the following technical words to sig- * nify certain ways of changing either the order or magnitude * of proportionals, so as that they continue still to be propor- * tionals.' XIII. Permutando, or alternando, by permutation, or alternately; this word is used when there are four proportionals, and it is See N- inferred, that the first has the same ratio to the third, which the second has to the fourth ; or that the first is to the third, as the second to the fourth : as is shown in the 16th prop, of this 5 til book. XIV. Invertendo, by inversion ; when there are four proportionals, and it is inferred, that the second is to the first as the fourth to the third. Prop. B, book 5. XV. Componendo, by composition ; when there are four proportionals, and it is inferred, that the first, together with the second, is to the second, as the third, together with the fourth, is to the fourth. 18 th prop, book 5. XVI. Dividendo, by division ; when there are four proportionals, and it is inferred, that the excess of the first above the second is to the second as the excess of the third above the fourth is to the fourth. 17th prop, book 5. XVII. Convertendo, by conversion ; when there are four proportion- als, and it is inferred, that the first is to its excess above the Q 123 THE ELEMENTS BookV. second, as the third to its excess above the fourth. Prop. E^ ^x-^^^ book 5. XVIII. Ex sequali (sc. distantia), or ex squo, from equality of distance ; when there is any number of magnitudes more than two, and as many others, so that they are proportionals when taken two and two of each rank, and it is inferred, 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 different order in which the magnitudes < are taken two and two.' XIX. Ex aquali, from equality ; this term is used simply by itself, when the first magnitude is to the second of the first rank» as the first to the second of the other rank ; and as the se- cond is to the third of the first rank, so is the second to the third of the other; and so on in order, and the inference is as mentioned in the preceding definition ; whence this is called ordinate proportion. It is demonstrated in 2 2d prop, book 5. XX. Ex aequali, in proportione perturbata, seu inordinata ; from equa- lity, in perturbate or disorderly proportion* ; this term is used 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 third from the last to the last but two of the second rank ; and so on in a cross order : and the inference is as in the 18th definition. It is demonstrated in the 23d prop, of book 5. AXIOMS. I. EQUIMULTIPLES of the same, or ot equal magnitudes, are equal to one another. * 4 Prop. lib. 2- Archimedis dc sphsera et cylkidro. OF EUCLID. II. Those magnitudes of which the same, or equal magnitudes, are equimultiples, are equal to one another. III. A multiple of a greater magnitude is greater than the same multiple of a less. IV. That magnitude of which a multiple is greater than the same multiple of another, is greater than that other magnitude. PROP. I. THEOR. IF any number of magnitudes be equimultiples of as many, each of each; what multiple soever any one of them is of its part, the same multiple shall all the first magnitudes be of all the other. Let any number of magnitudes, AB, CD be equimultiples of as may others E, F, each of each ; whatsoever multiple AB is of E, the same multiple shall AB and CD together be of E and F together. Because AB is the same multiple of E that CD is of F, as many magnitudes as are in AB, equal to E, so many are there in CD, equal to F. Divide AB into magni- tudes equal to E, viz. AG, GB ; and CD into A ( CH, HD, equal each of them to F: the num- j bcr therefore of the magnitudes CH, HD shall | be equal to the number of the others AG, GB : ^T and because AG is equal to E, and CH to j F, therefore AG and CH together are equal g I to a E and F together: for the same reason, ' a As. 2, because GB is equal to E, and FID to F; GB q j i 5 and HD together are equal to E and F together. 1 Wherefore, as many magnitudes as are in AB I I equal to E, so many are there in AB, CD to- H-L gether equal to E and F together. There- fore, whatsoever multiple AB is of E, the same multiple is AB and CD together of E and F ^ together. Therefore, if any magnitudes, how many soever, be equi- multiples of as many, each of each, whatsoever multiple any one of them is of its part, the same multiple shall all the first magnitudes be of all the other : ' For the same demonstration !24 THE ELEMENTS BookV. ' holds in any numbei- of magnitudes, which was here applied •^-nr*-' ' to two.* Q. E. D. PROP. II. THEOR. IF the first magnitude be the same multiple of the second that the third is of the fourth, and the fifth the same multiple of the second that the sixth is of the fourth ; then shall the first together with the fifth be the same multiple of*^*lJ|e second, that the third together with the sixth is o^me-i -fourth. B Let AB the first, be the same multiple of C the second, that DE the third is of F the fourth ; and BG the fifth, the same multiple of C the second, that EH j) the sixth is of F the fourth : then ^ is AG the first, together with the fifth, the same multiple of C the second, that DH the third, together with the sixth, is of F the fourth. Because AB is the same multiple of C, that DE is of F ; there are as many magnitudes in AB equal to C, as there are in DE equal to F: in like G C H F manner, as many as there are in BG equal to C, so many are there in EH equal to F ; as many, then, as are in the whole AG equal to C, so many are there in the whole DH equal to ¥: therefore AG is the same multiple of C, that DH is of F; that is, AG the first and fifth together, is the same multiple of the second C, that D DH the third and sixth together is of the A fourth F. If, therefore, the first be the same multiple, &c. Q. E. D. CoR. ' From this it is plain, that, if any B ' number of magnitudes AB, BG, GH ' be multiples of another C, and as many ' DE, ER, K.L be the same multiples of G-- ' F, each of each, the whole of the first, * viz. AH, is the same multiple of C, ' that the whole of tlie last, viz. DL, is ' of F.' H C L F K OF EUCLID. 125 BookV. PROP. III. THEOR. IF the first be the same multiple of the second, which the third is of the fourth; and if of the first and third there be taken equimultiples, these shall be equimultiples, the one of the second, and the other of the fourth. H K • Let A the first, be the same multiple of B the second, that C the third is of D the fourth ; and of A, C let the equimultiples EF, GH be taken : then EF is the same multiple of B, that GH is of D. Because EF is the same multiple of A, that GH is of C, there are as many magnitudes in EF equal to A, as are in GH equal to C : let EF be divided into the magnitudes EK, KF, each equal to A, and GH into GL, LH, each equal to C : the number therefore of the magnitudes EK, KF shall be equal to the number of the others GL, LH : and be- cause A is the same multiple of B, that C is of D, and that EK is equal to A, and GL to C ; there- fore EK is the same multiple of B, that GL is of D ; for the same reason, KF is the same multiple of B, that LH is of D ; and so, if there be more parts in EF, GH equal to A, C : because, there- E A B fore, the first EK is the same multiple of the second B, which the third GL is of the fourth D, and that the fifth KF is the same multiple of the second B, ■which the sixth LH is of the fourth D ; EF the first, together ■with the fifth, is the same multiple » of the second B, which GH a 2. 5. the third, together with the sixth, is of the fourth D. If, there- fore, the first, &C. Q. E. D. C D THE ELEMENTS PROP. IV. THEOR. See N. lY the first of four magnitudes has the same ratio to the second which the third hath to the fourth, then any equimuhiples whatever of the first and third shall have the same ratio to any equimultiples of the second and fourth, viz. ' the equimultiple of the first shall * have the same ratio to that of the second, which the * equimultiple of the third has to that of the fourth.' Let A the first have to B the second the same ratio which the third C has to the fourth D; and of A and C let there be taken any equimultiples whatever E, F : and of B and D any equi- multiples whatever G, H : then E has the same ratio to G, which F has to H. Take of E and F any equimul- tiples whatever K, L, and of G, H, any equimultiples whatever M, N: then, because E is the same multiple of A, that F is of C; and of E and F have been taken equimultiples K, L ; therefore K is the same multiple of A, that L a 3. 5, isofC*: for the same reason M is the same multiple of B, that N is of D: and because as A is to b Hyp. E, so is C to D^, and of A and C have been taken certain equi- multiples K, L; and of B and D have been taken certain equimul- tiples M, N ; if, ihereforc, K be greater than M, L is greater than N; and if equal, equal; if less, c 5 def.5. less«^. And K, L are any equi- multiples whatever of E, F ; and M, N any whatever of G, H : as therefore E is to G, so is'= F to il. Therefore, if the first, Sec. Q. E. D. See N. Coil. Likewise, if the first has th.e same ratio to the second, which the third has to (he fourth, then also any equimultiples K E F A C 13 D G H M N OF EUCLID. i2r whatever of the first and third have the same ratio to the se- BookV. cond and fourth : and in like manner, the first and the third ' — v— ' have the same ratio to any equimultiples whatever of the second and fourth. Let A the first have to B the second the same ratio which the third C has to the fourth D, and of A and C let E and Y he any equimultiples whatever ; then E is to B, cvs F to D. Take of E, F any equimultiples whatever K, L, and of B, D any equimultiples whatever G, H ; then it may be demonstrated, as before, that K is the same multiple of A, that L is of C ; and because A is to B, as C is to D, and of A and C certain equi- multiples have been taken, viz. K and I^ ; and of B and D cer- tain equimultiples G, H ; therefore, if K be greater than G, L is greater than H ; and if equal, equal; if less, less<^: and KcSdef.S L are any equimultiples of E, F, and G, H any whatever of B, D ; as therefore E is to B, so is F to D : and in the same way the other case is demonstrated. PROP. V. THEOR* IF one magnitude be the same multiple of another, See n. which a magnitude taken from the first is of a mag- nitude taken from the other ; the remainder shall be the same multiple of the remainder, that the whole is of the whole. Let the magnitude AB be the same multiple G of CD, that AE taken from the first is of CF taken from the other; the remainder EB shall be the same multiple of the remainder FD, that A the whole AB is of the whole CD. Take AG the same multiple of FD, that AE is of CF : therefore AE is ^ the same mul- tiple of CF, that EG is of CD : but AE, by the hypothesis, is the same multiple of CF that AB is of CD ; therefore EG is the same mul- tiple of CD that AB is of CD ; wherefore EG is equal to ABb. Take from them the common magnitude AE ; the remainder AG is equal to the remainder EB. Wherefore, since AE is the sartie multiple of CF, that AG is of FD, and that AG is equal to EB ; therefore AE is the same multiple of CF that EB is of FD : but AE is ti)e same multiple of CF E— B al. 5. I bl.Ax.5. D 12B THE ELEMENTS BookV. that AB is of CD ; therefore EB is the same multiple of FD *— y-'*-' that AB is of CD. Therefore, if any magnitude, Sec Q. E. D. PROP. VI. THEOR. A K C± G~ B H- see N. IF two magnitudes be equimultiples of two others, and if equimultiples of these be taken from the first two, the remainders are either equal to these others, or equimultiples of them. Let the two magnitudes AB, CD be equimultiples of the two E, F, and AG, CH taken from the first two be equimultiples of the same E, F ; the remainders GB, HD are either equal to E, F, or equimultiples of them. First, Let GB be equal to E ; HD is equal to F : make CK equal to F ; and because AG is the same multiple of E, that CH is of F, and that GB is equal to E, and CK to F ; therefore AB is the same multiple of E, that KH is of F. But AB, by the hypothesis, is the same multiple of E that CD is of F ; there- fore KH is the same multiple of F, that CD is of F ; wherefore KH is equal to al.Ax.5. CD a : take away the common magni- tude CH, then the remainder KG is equal to the remainder HD : but KC is equal to F ; HD therefore is equal to F. But, let GB be a multiple of E ; then HD is the same multiple of P': makeCK the same multiple of F that GB is of E: and because AG is the same multiple of E that CH is of F ; and GB the same multiple of E that CK is of F: therefore AB is the same multiple of E that KH b2. 5. isofF^: but AB is the same multiple of E that CD is of F ; therefore KH is the same multiple of F that CD is of it ; wherefore KH is equal to CD^: take away CH from both ; therefore the re- mainder KC is equal to the remainder HD : and because GB is the same multiple of E, that KC is of F. and that KC is equal to HD; therefore HD is the same multiple of F, that GB is of E. If, therefore, two magnitudes, 8cc. Q. E. D» D E F G- K C— H- B D E F OF EUCLID. PROP. A. THEOR. IF the first of four magnitudes has to the second the See n. same ratio which the third has to the fourth ; then, if the first be greater than the second, the third is also greater than the fourth; and, if equal, equal; if less, less. Take any equimultiples of each of them, as the doubles of each; then, by def. ^th of this book, if the double of the first be greater than the double of the second, the double of the third is greater than the double of the fourth ; but, if the first be greater than the second, the double of the first is greater than the double of the second ; wherefore also the double of the third is greater than the double of the fourth ; therefore the third is greater than the fourth : in like manner, if the first be equal to the second, or less than it, the third can be proved to be equal to the fourth, or less than it. Therefore, if the first, &c. Q. E. D. PROP. B. THEOR. IF four magnitudes are proportionals, they are Se? i^. proportionals also when taken inversely. If the magnitude A be to B as C is to D, then also inversely B is to A as D to C. Take of B and D any equimultiples whatever E and F; and of A and C any equimultiples whatever G and H. First, Let E be greater than G, then G is less than E ; and, because A is to B as C is to D, and of A and C, the first and third, G and H are equimultiples ; and of B and D, the se- cond and fourth, E and F are equimulti- ples ; and that G is less than E, H is also G A B E ^ less than F ; that is, F is greater than H ; a 5. defs if therefore E be greater than G, F is great- H C D F 5. er than H : in like manner, if E be equal to G, F may be shown to be equal to H ; and if less, less ; and E, F are any equi- multiples whatever of B and D, and G, II any whatever of A and C ; therefore, as B R 130 BookV. is to A, so is D to C. <— v—' Q. E. D. THE ELEMENTS If, then, four magnitudes, &c. See N. PROP. C. THEOR. IF the first be the same muUiple of the second, or the same part of it, that the third is of the fourth ; the first is to the second as the third is to the fourth. A B C D E G F H Let the first A be the same multiple of B the second, that C the third is of the fourth D : A is to B as C is to D. Take of A and C any equimultiples what- ever E and F ; and of B and U any equi- multiples whatever G and H : then because A is the same multiple of B that C is of D ; and that E is the same multiple of A that F is of C ; E is the same multiple of B that a 3. 5. F is of D » ; therefore E and F are the same multiples of B and D : but G and H are equi- multiples of B and D ; therefore, if E be a greater multiple of B than G is, F is a great- er multiple of D than H is of D ; that is, if E be greater than G, F is greater than H : in like manner, if E be equal to G, or less, F is equal to H, or less than it. But E, F are equimultiples, any whatever, of A, C, and G, H any equimultiples whatever of B, b S.dci.S. D. Therefore A is to B as C is to D t. Next, Let the first A be the same part of the second B, that the third C is of the fourth D : A is to B as C is to D : for B is the same multiple of A, that D is of C : wherefore, by the preceding case, B is to A as D is to C ; and in- c B 5. verselyc, A is to B as C is to D. There- fore, if the first be the same multiple, 8cc. Q. E. D. A B C D OF EUCLID. PROP. D. THEOR. IF the first be to the second as to the third to the See n. fourth, and if the first be a multiple, or part of the second ; the third is the same multiple, or the same part of the fourth. Let A be to B as C is to D ; and first let A be a multiple B ; C is the same multiple of D. Take E equal to A, and whatever mul- tiple A or E is of B, make F the same mul- tiple of D : then, because A is to B as C is to D ; and of B the second and D the four h equimultiples have been taken E and F ; A is to E as C to F a : but A is equal to E, therefore C is equal to F •» : and F is the same multiple of D that A is of B. A B C D Wherefore C is the same multiple of D that A is of B. Next, Let the first A be a part of the se- cond B ; C the third is the same part of the fourth D. Because A is to B as C is to D ; then, inversely, B is « to A as D to C: but A is 'a part of B, therefore B is a multiple of A; and, by the preceding case, D is the same multiple of C, that is, C is the same part of D, that A is of B. Therefore, if the first, &c. Q. E. D. of a Cor .4.5. b A.5. See the figure at the foot of the preced- ing pag<^ cB. 5. PROP. Vn. THEOR. EQUAL magnitudes have the same ratio to thje same magnitude ; and the same has the same ratio to equal magnitudes. Let A and B be equal magnitudes, and C any other. A and B have each of them the same ratio to C, and C has the same ratio to each of the magnitudes A and B. Take of A and B any equimultiples whatever D and E, and 132 THE ELEMENTS BookV of C any multiple whatever F: then, because D is the same ''—V—' n)ultiple of A that E is of B, and that A is a l.Ax.5. equal to B ; D is ^ equal to E : therefore, if D be greater than F, E is greater than F ; and if equal, equal ; if less, less : and D, E are any equimultiples of A, B, and F is any b5.def.5. multiple of C. Therefore b, as A is to C, so is B to C. Likewise C has the same ratio to A, that it has to B : for, having made the same con- struction, D may in like manner be shown equal to E : therefore, if F be greater than D, it is likewise greater than E ; and if equal, equal ; if less, less : and F is any multiple whatever of C, and D, E are any equimulti- ples whatever of A, B. Therefore C is to A as C is to B''. Therefore, equal magnitudes, &c. Q. E. D. See N. D E A B PROP. Vin. THEOR. OF unequal magnitudes, the greater has a greater ratio to the same than the less has ; and the same magnitude has a greater ratio to the less, than it has to the greater. Let AB, BC be unequal magnitudes, of which AB is the greater, and let D be any magnitude whatever : AB has a greater ratio to D Fig. 1. than BC to D: and D has a greater ra- tio to BC than unto AB. If the magnitude which is not the greater of the two AC, CB, be not less A than D, lake EF, EG, the doubles of AC, CB, as in Fig. 1. But, if that which is not the greater of the two AC, CB be less than D (as in Fig. 2. and 3.) this magnitude can be multiplied, so as to G B become greater than D, whether it be AC, or CB. Let it be multiplied until L K H D it become greater than D, and let the other be multiplied as often; and Itt EF be the multiple thus taken of AC, and FG the same multiple of CB : therefore EF and FG are each of them greater than F — OF EUCLID. 1J3 D : and in every one of the cases, take H the double of D, K BookV. its triple, and so on, till the multiple of D be that which first •*— #^-^ becomes greater than FG : let L be that multiple of D which is first greater than FG, and K the multiple of D which is next less than L. Then, because L is the multiple of D, which is the first that becomes greater than FG, the next preceding multiple K is not greater than FG ; that is, FG is not less than K : arwl since EF is the same multiple of AC, that FG is pf CB ; FG is the same multiple of CB, that EG is of AB^; wherefore EG and»l*5- FG are equimultiples of AB and CB ; and it was shown, that FG was not less than Fig. 2, Fig. 3. G A C- B F— G K, and, by the con- struction, EF is great- er than D ; therefore the whole EG is great- er than K and D toge- ther : but K, together with D, is equal to L ; therefore EG is great- er than L ; but FG is not greater than L ; and EG, FG are equi- multiples of AB, BC, and L is a multiple of D ; therefore^ AB has to D a greater ratio than BC has to D. Also, D has to BC a greater ratio than it has to AB : for, hav- ing made the same construction, it may be shown, in like man- ner, that L is greater than FG, but that it is not greater than EG : and L is a multiple ofD; and FG, EG are equimultiples of CB, AB : therefore D has to CB a greater ratio'' than it has to AB. Wherefore, of unequal magnitudes, &c. Q. E. D. K H D C-- B K D b 7. def. 5. 134 THE ELEMENTS Book V. PROP. IX. THEOR. See N. MAGNITUDES which have the same ratio to the same magnitude are equal to one another; and those to which the same magnitude has the same ratio are equal to one another. Let A, B have each of them the same ratio to C: A is equal to B: for, if they are not equal, one of them is greater than the other; let A be the greater; then, by what was shown in the preceding proposition, there are some equimultiples of A and B, and some multiple of C such, that the multiple of A is greater than the multiple of C, but the multiple of B is not greater than that of C. Let such multiples be taken, and let 13, E be the equimultiples of A, B, and F the multiple of C, so that D may be greater than F, and E not greater than F : but, because A is to C as B is to C, and of A, B are taken equimultiples D, E, and of C is taken a multiple F ; and that D is greater than F ; a5. def. E shall also be greater than F^; but E is D ^' not greater than F, which is impossible; A therefore and B are not unequal ; that is, they are equal. Next, Let C have the same ratio to each of the magnitudes A and B ; A is equal to B : for, if they are not, one of them is greater than the other; let A be the greater; therefore, as was shown in Prop. 8lh, there is some multiple F of C, and some equimultiples E and D, of B and A such, that F is greater than E, and not greater than D ; but be- cause C is to B, as C is to A, and that F, the multiple of the first, is greater than E, the multiple of the second ; F, the mul- tiple of the third, is greater than D, the multiple of the fourth*: but F is not greater than D, which is impossible. Therefore A is equal to B. Wherefore, magnitudes which, &c. Q. E. D. A B OF EUCLID. PROP. X. THEOR. THAT magnitude which has a greater ratio than See N. another has unto the same magnitude is the greater of the two : and that magnitude, to which the same has a greater ratio than it has unto another magnitude, is the lesser of the two. Let A have to C a greater ratio than B has to C : A is great- er than B: for, because A has a greater ratio to C than B has to C, there are * some equimultiples of A and B, and a 7- def some multiple of C such that the multiple of A is greater than 5. the multiple of C, but the multiple of B is not greater than it: let them be taken, and let D, E be equi- multiples of A, B, and F a multiple of C such that D is greater than F, but E is not greater than F : therefore D is greater than E: and, because D and E are equi- A D multiples of A and B, and D is greater than E ; therefore A is *» greater than B. I l*' b 4. Ak. Next, Let C have a greater ratio to B CI 5. than it has to A ; B is less than A: for* there is some multiple F of C, and some equimultiples E and D of B and A such B that F is greater than E, but is not greater than D : E therefore is less than D ; and because E and D are equimultiples of B and A, therefore B is '' less than A. That magnitude, therefore, &;c. Q. E. D. PROP. XL THEOR. RATIOS that are the same to the same ratio are the same to one another. Let A be to B as C is to D ; and, as C to D, so let E be to F; A is to B as E to F. Take of A, C, E any equimultiples whatever G, H, K ; and of B, D, F any equimultiples whatever L, M, N. Therefore, since A is to B as C to D, and G, H are taken equimultiples of 136 THE ELEMENTS Book V. A, C, and L, M of B, D ; if G be greater than L, H is greater than M ; and if equal, equal ; and if less, less *. Again, be- cause C is to D as E is to F, and H, K are taken equimultiples of C, E ; and M, N of D, F : if H be greater than M, K is greater than N j and if equal, equal ; and if less, less: but if G H- K- C E- B D- L M- be greater than L, it has been shown that H is greater than M ; and if equal, equal; and if less, less; therefore, if G be greater than L, K is greater than N; and if equal, equal; and if less, less: and G, K are any equimultiples whatever of A, E; and L, N any whatever of B, F : therefore, as A is to B so is E to F». Wherefore, ratios that, &c. Q. E. D. PROP. XII. THEOR. IF any number of magnitudes be proportionals, as one of the antecedents is to its consequent, so shall all the antecedents taken together be to all the conse- quents. Eet any number of magnitudes A, B, C, D, E, F be propor- tionals ; that is, as A is to B so C to D, and E to F : as A is to B, so shall A, C, E together be to B, D, F together. Take of A, C, E any equimultiples whatever G, H, K ; H K- A C E- B D F- •M N- and of B, D, F any equimultiples whatever L, M, N: then, because A is to B as C is to D, and as E to F ; and that G, H, OF EUCLID. 13? K are equimultiples of A, C, E, and L, M, N equimultiples of BookV. B, D, F ; if G be greater than L, H is greater than M, and K ^ — v».»* greater than N ; and if equal, equal ; and if less, less*. Where-a5. def.5. fore, if G be greater than L, then G, H, K together are greater than L, M, N together ; and if equal, equal ; and if less, less. And G, and G, H, K together are any equimultiples of A, and A, C, E together; because, if there be any number of magni- * tudes equimultiples of as many, each of each, whatever multi- ple one of them is of its part, the same multiple is the whole of the whole •> : for the same reason L, and L, M, N are any equi- b 1. 5. multiples of B, and B, D, F : as therefore A is to B, so are A, C, E together to B, D, F together. Wherefore, if any number, &c. Q. E. D. PROP. XIII. THEOR. IF the first has to the second the same ratio which See N. the third has to the fourth, but the third to the fourth a greater ratio than the fifth has to the sixth; the first shall also have to the second a greater ratio than the fifth has to the sixth. Let A the first have the same ratio to B the second, which C the third has to D the fourth, but C the third to D the fourth, a greater ratio than E the fifth to F the sixth : also, the first A shall have to the second B, a greater ratio than the fifth E to the sixth F. Because C has a greater ratio to D, than E to F, there are some equimultiples of C and E, and some of D and F such, that the multiple of C is greater than the multiple of D, but M. . G H A . C— E B D F N K L the multiple of E is not greater than the multiple of F^: Ieta7.def.5. such be taken, and of C, E let G, H be equimultiples, and K, L equimultiples of D, F, so that G be greater than K, but H not greater than L ; and whatever multiple G is of C, take M the same multiple of A ; and whatever multiple K is of D, take N the same multiple of B : then, because A is to B, as C to D, and S 138 THE ELEMENTS Book V. of A and C, M and G are equimultiples ; and of B and D, N ^-"ymmJ and K are equimultiples ; if M be greater than N, G is greater b 5.def.5. than K; and if equal, equal; and if less, less^; but G is greater than K, therefore M is greater than N : but H is not greater than L ; and M, H are equimultiples of A, E ; and N, L equimultiples c7.def.5. of B, F : therefore A has a greater ratio to B than E has to F<=. Wherefore, if the first, 8cc. Q. E. D. Cor. And if the first has a greater ratio to the second, than the third has to the fourth, but the third the same ratio to the fourth, which the fifth has to the sixth ; it may be demonstrated, in like manner, that the first has a greater ratio to the second, than the fifth has to the sixth. PROP. XIV. THEOR. SeeN. A 8.5. b 13. 5. c 10. 5. d9.5. IF the first has to the second the same ratio which the third has to the fourth; then, if the first be great- er than the third, the second shall be greater than the fourth; and if equal, equal; and if less, less. Let the first A have to the second B, the same ratio which the third C has to the fourth D 5 if A be greater than C, B is greater than D. Because A is greater than C, and B is any other magnitude, A has to B a greater ratio than C to B*: but, as A is to B, so A B C D A B C D A B C D is C to D ; therefore also C has to D a greater ratio than C has to B'': but of two magnitudes, that to which the same has the greater ratio is the lesser <= : wherefore D is less than B ; that is, B is greater than D. Secondly, If A be equal to C, B is equal to D : for A is to B, as C, that is. A, to D ; B therefore is equal to D^. Thirdly, If A be less than C, B shall be less than D : for C is. greater than A, and because C is to D, as A is to B, D is greater than B, by the first case ; wherefore B is less than D. There- fore, if the first, &c. Q. E. D. OF EUCLID. PROP. XV. THEOR. MAGNITUDES have* the same ratio to one an- other which their equimultiples have. Let AB be the same multiple of C that DE is of F : C is to F as AB to DE. Because AB is the same multiple of C that DE is of F, there are as many magnitudes in AB equal to C ^ as there are in DE equal to F : let AB be divided into magnitudes, each equal to C, viz. AG, GH, HB ; and DE into magni- tudes, each equal to F, viz. DK, KL, LE ; G then the number of the first AG, GH, HB shall be equal to the number of the last DK, KL, LE : and because AG, GH, HB are H ail equal, and that DK, KL, LE are also equal to one another : therefore AG is to DK as GH to KL, and as HB to LE=^: I ^7.5. and as one of the antecedents to its conse- B C E F quent, so are all the antecedents together to all the consequents together'^; Avherefore, as AG is to DK so is AB to DE : but b 12 AG is equal to C, and DK to F : therefore, as C is to F so is AB to DE. Therefore, magnitudes, &c. Q. E. D. D K— PROP. XVL THEOR. IF four magnitudes of the same kind be propor- tionals, they shall also be proportionals when taken alternately. Let the four magnitudes A, B, C, D be proportionals, viz. as A to B so C to D ; they shall also be proportionals Avhen taken alternately, that is, A is to C as B to D. Take of A and B any equimultiples whatever E and F ; and of C and D take any equimultiples whatever G and H: and 140 THE ELEMENTS Book V. because E is the same multiple of A that F is of B, and that •"— v-—* magnitudes have the same ratio to one another which their a 15. 5. equimultiples have a; therefore A is to B, as E is to F : but as A is to B, so is C to D : wherefore, as C is to D, E G b 11. 5. so t* is E to F: again, be- cause G, H are equimul- A C — ■ tiples of C, D, as C is to D, so is G to II a ; but B- D as C is to D, so is E to F. Wherefore, as E is F — H to F, so is G to HW But, when four magnitudes are proportionals, if the first be greater than the third, the second shall be greater than the c 14. 5. fourth; and if equal, equal ; if less, less c. Wherefore, if E be greater than G, F likewise is greater than H ; and if equal, equal ; if less, less : and E, F are any equimultiples whatever of A, B ; and d5.def.5. G, H any whatever of C, D. Therefore, A is to C, as B to D^ If then four magnitudes, 8cc. Q. E. D, PROP. XVII. THEOR. SeeN. IF magnitudes, taken jointly, be proportionals, they shall also be proportionals when taken sepa- rately; that is, if two magnitudes together have to one of them the same ratio which two others have to one of these, the remaining one of the first two shall have to the other the same ratio which the remaining one of the last two has to the other of these. Let AB, BE, CD, DF be the magnitudes taken jointly which are proporitonals ; that is, as AB to BE, so is CD to DF ; they shall also be proportionals taken separately, viz. as AE to EB, so CF to FD. Take of AE, EB, CF, FD any equimultiples whatever GH, IlK, LM, MN ; and again, of EB, FD take any e : but it was shown that LM is the same multiple of CD, that GK is of AE ; therefore GK is the same multiple of AE, that LN is of CF ; that is, GK, E LN are equimultiples of AE, CF : and because KO, NP are equimul- tiples of BE, DF, if from KO, NP G ti.ere be taken KH, NM, which are likewise equimultiples of BE, DF, the remainders HO, MP are either equal to BE, r f>. 5. i^F) or equimultiples of lhem<:. First, Let HO, MP be L-qual to BI;. DF ; and because AE is lo EB^ as. CF to FD, and B D C L OF EUCLID. 143 that GK, LN are equimultiples of AE, CF ; GK shall be to BookV EB, as LN to FD d : but HO is equal to EB, and MP to FD ; <>.-v^ wherefore GK is to HO as LN to MP. If, therefore, GK bedCor.4.5. greater than HO, LN is greater than MP ; and if equal, equal; and if less «, less. e A. 5. But let HO, MP be equimultiples of EB, FD ; and because AE is to EB as CF to FD, and that of AE, CF are taken equi- multiples GK, LN ; and of EB, FD, the equimultiples HO, MP ; if GK be greater than HO, LN is greater than MP ; and if equal, equal ; and if less, less ^ ; which was f 5.def.5. likewise shown in the preceding case. If, therefore, GH be greater than KO, taking KH from both, GK is greater than HO ; wherefore also LN is greater than MP ; and, conse- quently, adding NM to both, LM is greater than NP : therefore, if GH be greater than KO, LM is great- er than NP. In like manner it may be shown, that if GH be equal to KO, LM is equal to NP ; and if less, less. And in the case in which KO is not greater than KH, it has been shown that GH is always greater than KO, and likewise LM than NP : but GH, LM are any equimultiples of AB, CD, and KO, NP are any whatever of BE, DF ; therefore f, as AB is to BE so is CD to DF. If then magnitudes, &c. Q. E. D. O H— K — M- N— B E— D F-- PROP. XIX. THEOR. IF a whole magnitude be to a whole, as a magnitude See n. taken from the first is to a magnitude taken from the other ; the remainder shall be to the remainder, as the whole to the whole. Let the whole AB be to the whole CD as AE, a magnitude taken from AB, to CF, a magnitude taken from CD ; the re- manider EB shall be to the remainder FD as the whole AB to the whole CD. Because AB is to CD as AE to CF ; likewise, alternately », a 16. S. 144 THE ELEMENTS E— BookV. BA is to AE as DC to CF : and because, if mag- v,..y...^ nitudes, taken jointly, be proportionals, they are b 17. 5. also proportionals ^ when taken separately ; there- fore, as BE is to DF so is EA to FC ; and alter- nately, as BE is to EA, so is DF to FC : but, as AE to CF, so by the hypothesis is AB to CD ; therefore also BE, the remainder, shall be to the remainder DF, as the whole AB to the whole CD. Wherefore, if the whole, Sec. Q. E. D. Cor. If the whole be to the whole, as a mag- " nitude taken from the first is to a magnitude taken from the other ; the remainder likewise is to the remainder, as the magnitude taken from the first to that taken from the other ; the demonstration is contained in the pi*eceding. B D PROP. E. THEOR. IF four magnitudes be proportionals, they are also proportionals by conversion, that is, the first is to its excess above the second, as the third to its excess above the fourth. a 17. 5. bB. 5. c 18. 5. Let AB be to BE as CD to DF ; then BA is to AE as DC to CF. Because AB is to BE as CD to DF, by divi- sion a, AE is to EB as CF to FD ; and by inver- sion b, BE is to E A as DF to FC. Wherefore, by composition S BA is to AE, as DC is to CF. therefore, four, Sec. Q. E. D. If, B F— D PROP. XX. THEOR. SeeN. IF there be three magnitudes, and other three, which, taken two and two, have the same ratio ; if the first be greater than the third, the fourth shall be greater than the sixth ; and if equal, equal j and if less, less. OF EUCLID. f45 Let A, B, C be three magnitudes, and D, E, F other three, Book V. which, taken two and two, have the same ratio, viz. as A is to ^■- y «n^ B, so is D to E ; and as B to C, so is E to F. If A be greater than C, D shall be greater than F; and if equal, equal; and if less, less. Because A is greater than C, and B is any other magnitude, and that the greater has to the same magnitude a greater ratio than the less has to it*, therefore A has to B a greater ratio than C has to B : but as D is to E, so is A to B ; therefore ^ D has to E a greater ratio than C to B ; and because B is to C, as E to F, by inver- sion, C is to B, as F is to E ; and D was shown to have to E a greater ratio than C to B ; there- fore D has to E a greater ratio than F toE^ : but the magnitude which has a greater ratio than another to the same magnitude, is the greater of the two"*: D is therefore greater than F. Secondly, Let A be equal to C ; D shall be equal to F cause A and C are equal to one an- A D C F a 8. 5. b 13. 5. c Cor. 13.5. d 10. 5, be- other, A is to B, as C is to B^: but A is to B, as D to E ; and C is to B, as F to E ; wherefore D is to E, as F to Ef; and therefore D is equal to F s. Next, Let A be less than C ; D shall be less than F : for C is great- er than A, and, as was shown in the first case, C is to B, as F to E, and in like manner B is to A, as E to D ; therefore F is greater than D, by the first case ; and therefore D is less than F. Therefore, if there be three, &c. Q. E. D. A D B E C F A D B E t7.S. fll.5. g9. 5. PROP. XXI. THEOR. IF there be three magnitudes, and other three, See n. which have the same ratio taken two and two, but in a cross order ; if the first magnitude be greater than the third, the fourth shall be greater than the sixth; and if equal, equal; and if less, less. T u& THE ELEMENTS Book V, h 13. 5. c Cor. 13. 5. d 10. S. f 11. 5. S9 5. A D B E Let A, B, C be three magnitudes, and D, E, F other three, ' which have the same ratio, taken two and two, but in a cross order, viz. as A is to B, so is E to F, and as B is to C, so is D to E. If A be greater than C, X) shall be greater than F ; and if equal, equal ; and if less, less. Because A is greater than C, and B is any other magnitude, A has to B a greater ratio* than C has to B : but as E to F, so is A to B ; therefore ^ E has to F a greater ratio than C to B : and because B is to C, as D to E, by inver- sion, C is to B, as E to D : and E was shown to have to F a greater ratio than C to B ; there- fore E has to F a greater ratio than E to D <= ; but the magnitude to which the same has a greater ratio than it has to another, is the lesser of the two'i ; F therefore is less than D ; that is, U is greater than F. Secondly, Let A be equal to C ; D shall be equal to F. Be- cause A and C are equal, A is e to B, as C is to B : but A is to B, as E to F ; and C is to B as E to D ; wherefore E is to F as E to Df; and therefore D is equal to Fs. Next, Let A be less than C ; D shall be less than F : for C is greater than A, and, as was shown, C is to B, as E to D, and in like manner B is to A, as F to E ; therefore F is great- er than D, by case first; and therefore D is less than F. Therefore, if there be three. Sec. Q. E. D. A D B E A D B E C F PROP. XXn. THEOR. SseN. IF there be any number of magnitudes, and as many others, which, taken two and two in order, have the same ratio; the first shall have to the last of the first magnitudes the same ratio which the first of the others has to the last. N. B. This is usually died by the words *' ex aqualiy^'' or *' ex aquo.''^ OF EUCLID. 14? c M D H E L F N 3.4.5-.. First, Let there be three magnitudes A, B, G, and as many BookV. others D, E, F, which, taken two and two, have the same ratio, ^m- ^^ mmj that is, such that A is to B as D to E ; and as B is to C, so is E to F ; A shall be to C, as D to F. Take of A and D any equimultiples whatever G and H;- and of B and E any equimultiples whatever K and L ; and of C and F any whatever M and N : then, because A is to B, as D to E, and that G, H are equimultiples of A, D, and K, L equimultiples of B, A B E ; as G is to K, so is » H to L. G K For the same reason, K is to M, as L to N : and because there are three magnitudes G, K, M, and other three H, L, N, which, two and two, have the same ratio ; if G be greater than M, H is great- er than N ; and if equal, equal ; and if less, less^; and G, Hare b 20. §. any equimultiples whatever of A, D, arid M, N are any equimul- tiples whatever of C, F. Therefore c, as A is to C, so is D c 5 de£ toF. ,5. Next, let there be four magnitudes A, B, C, D, and other four E, F, G, H, which two and two have the same ratio, viz. as A is to B, so is E to F, and A. B. C. D. as B to C, so F to G ; and as C to D, so G to E. F. G. H. H : A shall be to D, as E to H. 1 Because A, B, C are three magnitudes, and E, F, G other three, which, taken two and two, have the same ratio ; by the foregoing case A is to C, as E to G. But C is to D, as G is to H ; wherefore again, by the first case, A is to D, as E to H : and so on, whatever be the number of magnitudes. Therefore, if there be any Humber, &c. Q. E. D. 148 Book V. THE ELEMENTS PROP. XXIII. THEOR. SeeN. IF there be any number of magnitudes, and as many others, which, taken two and two, in a cross order, have the same ratio, the first shall have to the last of the first magnitudes the same ratio which the first of the others has to the last. N. B. This is usu- ally cited by the words " ex tequali in proportione per- *' turbata ;^^ or, " ex aquo perturbate.'*'* a 15. 5. b 11 5. c4. 5. d 21. 5. First, Let there be three magnitudes A, B, C, and other three D, E, F, which, taken two and two, in a cross order, have the same ratio, that is, such that A is to B, as E to F ; and as B is to C, so is D to E : A is to C, as D to F. Take of A. B, D any equimultiples whatever G, H, K ; and of C, E, F any equimultiples whatever L, M, N ; and because G, H are equimultiples of A, B, and that magnitudes have the same ratio which their equimul- tiples have»; as A is to B, so is G to H. And, for the same rea- son, as E is to F, so is M to N : but as A is to B, so is E to F; A B C D E F as therefore G is to H, so is M to G H L KM N N''. And because as B is to C, so is D to E, and that H, K are equimultiples of B, D, and L, M of C, E ; as H is to L, so is ^ K to M : and it has been shown, that G is to H, as M to N : then, because there are three magni- tudes G, H, L, and other three K, M, N, which have the same ratio taken two and two in a cross order; if G be greater than L, K is greater than N; and if equal, equal; and if less, less**; and G, K are any equimultiples whatever of A, D ; and L, N any whatever of C, F; as, therefore, A is to C, so is D to F. OF EUCLID. 140 A. B. C. D. E. F. G. H. Next, Let there be four magnitudes, A, B, C, D, and other Book V. four E, F, G, H, which taken two and two in a cross order have the same ratio, viz. A to B, as G to H ; B to C, as F to G ; and C to D, as E to F : A is to D as E toH. Because A, B, C are three magnitudes, and F, G, H other three, which, taken two and two in a cross order, have the same ratio ; by the first case, A is to C, as F to H : but C is to D, as E is to F ; wherefore agani, by the first case, A is to D, as E to H : and so on, whatever be the number of magnitudes. Therefore, if there be any number, &c. Q. E. D. PROP. XXIV. THEOR. IF the first has to the second the same ratio which See n. the third has to the fourth ; and the fifth to the se- cond the same ratio which the sixth has to the fourth ; the first and fifth together shall have to the second the same ratio which the third and sixth together have to the fourth. the same ratio which let BG the fifth have B— H E— Let AB the first have to C the second DE the third has to F the fourth ; and to C the second the same ratio which EH the sixth has to F the fourth : AG, the first and fifth together, shall have to C the second the same ratio which DH, the third and sixth together, has to F the fourth. Because BG is to C, as EH to F ; by inversion, C is to BG, as F to EH : and because, as AB is to C, so is DE to F ; and as C to BG, so F to EH ; ex aquali », AB is to BG, as DE to EH : and because these magnitudes are proportionals, they shall likewise be proportionals when taken jointly *> : as, therefore, AG is to G3, so is DH to HE ; but as GB to C, so is HE to F. Tiierefore, ex xquali^^ as AG is to C, so is DH to F. Wherefore, if the first, &c. Q. E. D. Cor. 1. If the same hypothesis be made as in the proposi- tbn, the excess of the first and fifth shall be to the second, as A C D F a 22. 5. bl&5. 150 THE ELEMENTS BooRV. the excess of the third and sixth to the fourth. The demonstra- ^-^v-^ii^ tion of this is the same with that of the proposition, if division be used instead of composition. CoR. 2. The proposition holds true of tv/o ranks of magni- tudes, whatever be their number, of which each of the first rank has to the second magnitude the same ratio that the corres- ponding one of the second rank has to a fourth magnitude ; as is manifest. PROP. XXV. THEOR. IF four magnitudes of the same kind are propor- tionals, the greatest and least of them together are greater than the other two together. a A. 8t 14.5. b 19. 5. c A. 5. B Let the four magnitudes AB, CD, E, F be proportionals, viz. AB to CD, as E to F ; and let AB be the greatest of them, and consequently F the least ». AB, together with F, are greater than CD, together with E. Take AG equal to E, and CH equal to F : then, because as AB is to CD, so is E to F, and that AG is equal to E, and CH equal to F ; AB is to CD, as AG to CH. And because AB the whole is to the whole CD, as AG is to CH, likewise the remainder GB shall be to the remainder HD, as the whole AB is to the whole ^ CD : but AB is greater than CD, there- fore «= GB is greater than HD : and be- cause AG is equal to E, and CH to F ; AG and F together are equal to CH and E together. If, therefore, to the unequal magnitudes GB, HD, of which GB is the greater, there be added equal magnitudes, viz. to GB the two AG and F, and CH and E to HD ; AB and F together are greater than CD and E. Therefore, if four magnitude?, 8cc. Q. i:. D. G- D H— A C E F PROP. F. THEOR. See N. RATIOS which are compounded of the same r^- tios, are the same with one another. OF EUCLID. 151 A. B. C. D. E. F. Let A be to B as D to E ; and B to C as E to F : the ratio which BookV. is compounded .of the ratio of A to B, and B to C, which, by the definition of compound ratio, is the ratio of A to C, is the same with the ratio of D to F, which, by the same definition, is compounded of the ratios of D to E, and E to F. Because there are three magnitudes A, B, C, and three others D, E, F, which, taken two and two in order, have the same ra- tio ; ex xquali, A is to C as D to F 3. a 22. 5. Next, Let A be to B as E to F, and B to C as D to E ; there- fore, ex xqtiali in profiortione fierturbata ^, A is to C as D to F ; that is, the ratio of A to C, which is compounded of the ratios of A to B, and B to C, is the same with the ratio of D to F, which is compounded of the ratios of D to E, and E to F : and in like manner the proposition may be demons strated, whatever be the nuciber of ratios in either case. A. B. C. D. E. F. b23-5. PROP. G. THEOR. IF several ratios be the same with several ratios, see ^f. each to each ; the ratio which is compounded of ra- tios which are the same with the first ratios, each to each, is the same with the ratio compounded of ratios which are the same with the other ratios, each to each. A. B. C. D. E. F. G. H. K. L. M. N. O. P. Let A be to B as E to F ; and C to D as G to H : and let be to B as K to L ; and C to D as L to M : then the ratio of to M, by the definition of com- pound ratio, is compounded of the ratios of K to L, and L to M, which are the same with the ra- tios of A to B, and C to D : and as E to F, so let N be to O ; and as G to H, so let O be to P ; then the ratio of N to P is compounded of the ratios of N to O, and O to P, which are the same with the ratios of E to F, and G to H: and it is to be shown that the ratio of K to M is the same with the ratio of N to P, or that K is to M as N to P. Because K is to L as (A to B, that is, as E to F, that is, as) N t» O ; and as L to M, so is (C to D, and so is G to H, and so is) 152 THE ELEMENTS Book V. O to P : ex isqnali ^, K is to M as N to P. Therefore, if several ^■i-v^— ; ratios, Sec. Q. E. D. a 22. 5. PROP. H. THEOR. See N. IF a rp.tio compounded of several ratios be the same with a ratio compounded of any other ratios, and if one of the first ratios, or a ratio compounded of any of the first, be the same witli one of the last ratios, or with the ratio compounded of any of the last ; then the ratio compounded of the remaining ratios of the first, or the remaining ratio of the first, if but one remain, is the same with the ratio compounded of those remaining of the last, or with the remaining ratio of the last. Let the first ratios be those of A to B, B to C, C to D, D to E, and E to F ; and let the other ratios be those of G to H, H to K, K to L, and L to M : also, let the ratio of A to F, which is com- a Defini- pou"fl'-'fl of* the first ratios, be the tion of same Avith the ratio of G to M, which com- is compounded of the other ra- pounded jj^g . ^^^ besides, let the ratio of A to D, which is compounded of the ratio. A. B. C. D. E. F. G. H. K. L. M. ratios of A to B, B to C, C to D, be the same with the ratio of G to K, which is compounded of the ratios of G to H, and H to K : then the ratio compounded of the remaining first ratios, to wit, of the ratios of D to E, and E to F, which compounded ratio is the ratio of D to F, is the same with the ratio of K to M, which is compounded of the remaining ratios of K to L, and L to M of the other ratios, b B. 5. Because, by the hypothesis, A is to D as G to K, by inversion'', c 22. 5. D is to A as K to G ; and as A is to F, so is G to M ; therefore *=, ex squally D is to F as K to iM. If. therefore, a ratio which is, &c. Q. E. D. OF EUCLID. PROP. K. THEOR. IF there be any number of ratios, and any number See n. of other ratios such, that the ratio compounded of ratios which are the same with the first ratios, each to each, is the same with the ratio compounded of ratios which are the same, each to each, with the last ratios; and if one of the first ratios, or the ratio which is compounded of ratios which are the same with several of the first ratios, each to each, be the same with one of the last ratios, or with the ratio compounded of ratios which are the same, each to each, with several of the last ratios: then the ratio compounded of ratios which are the same with the remaining ratios of the first, each to each, or the re- maining ratio of the firtst, if but one remain; is the same with the ratio compounded of ratios which are the same with those remaining of the last, each to each, or with the remaining ratio of the last. Let the ratios of A to B, C to D, E to F, be the first ratios ; and the ratios of G to H, K to L, M to N, O to P, Q to R, be the other ratios : and let A be to B, as S to T ; and C to D, as T to V; and E to F, as V to X : therefore, by the de- finition, of compound ratio, the ratio of S to X is compounded h, k,I. A, B ; C, D ; E, F. G, H ; K, L ; M, N ; O, P ; Q, R. e, f, g. m, n, 0, p. S, T, V, X, Y, Z, a, b, c, d. of the ratios uf S to T, T to V, and V to X, which are the same with the ratios of A to B, C to D, E to F, each to each ; also, as G to H, so let Y be to Z ; and K to L, as Z to a ; M to N, as a to b, O to P, as b to c ; and Q to R, as c to d : therefore, by the same definition, the ratio of Y to d is com- pounded of the ratios of Y to Z, Z to a, a to. b. b to c, and U 154 THE ELEMENTS, &c. Book V. c to d, which are the same, each to each, with the ratios of G *--v-— ^ to H, K to L, M to N, O to P, and Q to R : therefore, by the hypothesis, S is to X, as Y to d : also, let the ratio of A to B, that is, the ratio of S to T, which is one of the first ratios, be the same with the ratio of e to g, which is compounded of the ratios of e to f, and f to g, which, by the hypothesis, are the same with the ratios of G to H, and K to L, two of the other ratios ; and let the ratio of h to 1 be that which is com- pounded of the ratios of h to k, and k to 1, which are the same with the remaining first ratios, viz. of C to D, and E to F ; also, let the i-atio of m to p be that which is compounded of the ratios of m to n, n to o, and o to p, which are the same, each to each, Avith the remaining other ratios, viz. of M to N, O to P, and Q to R : then the ratio of h to 1 is the same with the ratio of m to p, or h is to 1, as m to p. h, k, 1. A, B ; C, D ; E, F. S, T, V, X. G, H;K, L; M, N ; O, P ; Q, R. Y, Z, a, b, c, d. e, f, g. m, n, o, p. a 11. 5. Because e is to f, as (G to H, that is, as) Y to Z ; and f is to g, as (K to L, that is, as) Z to a ; therefore, ex xquali^ e is to g, as Y to a: and by the hypothesis, A is to B, that is, S to T, as e to g ; wherefore S is to T, as Y to a ; and, by inversion, T is to S, as a to Y ; and S is to X, as Y to d ; therefore, ex (zquali^ T is to X, as a to d : also, because h is to k, as (C to D, that is, as) T to V ; and k is to 1, as (E to F, that is, as) V to X ; therefore, ex aquali^ h is to 1, as T to X : in like manner, it may be demon- strated, that m is to p, as a to d : and it has been shown, that T is to X, as a to d ; therefore » h is to 1, as m to p. Q. E. D. The propositions G and K are usually, for the sake of brevity, expressed in the same terms with propositions F and H : and therefore it was proper to show the true meaning of them when they are so expressed ; especially since they are very frequently njade use of by geometers. THE ELEMENTS OF EUCLID. BOOK VI. DEFINITIONS. I. Similar rectilineal figures y\ BookVI. are those which have their se- veral angles equal, each to each, and the sides about the equal angles proportionals. II. " Reciprocal figures, viz. triangles and parallelograms, are such See IJ: " as have their sides about two of their angles proportionals " in such manner, that a side of the first figure is to a side " of the other, as the remaining side of this other is to the re- " maining side of the first." III. A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment, as the greater segment is to the less. IV. The altitude of any figure is the straight line drawn from its vertex perpendicular to the base* 156 THE ELEMENTS Book VI, PROP. I. THEOR. See N. TRIANGLES and parallelograms of the same al- titude are one to another as their bases. Let the triangles ABC, ACD, and the parallelograms EC, CF, have the same altitude, viz. the .perpendicular drawn from the point A to BD : then, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD, and the parallelogram EC to the parallelogram CF. Produce BD both ways to the points H, L, and take any num- ber of straight lines BG, GH, each equal to the base BC ; and DK, KL, any number of them, each equal to the base CD ; and join AG, AH, AK, AL : then, because CB, BG, GH are all equal, 4 38. 1. the triangles AHG, AGB, ABC are all equal a; therefore, what- ever multiple the base HC is of the base BC, the same multiple is the triangle AHC of the triangle ABC: for the same reason, ■whatever multiple the base E A F LC is of the base CD, the same multiple is the trian- gle ALC of the triangle ADC : and if the base HC be equal to the base CL, the triangle AHC is also equal to the triangle ALC a ; and if the base H G B C D K L HC be greater than the base CL, likewise the triangle AHC is greater than the triangle ALC ; and if less, less : therefore, since there are four magnitudes, viz. the two bases BC, CD, and the two triangles ABC, ACD ; and of the base BC and the triangle ABC, the first and third, any equimultiples whatever have been taken, viz. the base HC and triangle AHC ; and of the base CD and triangle ACD, the second and fourth, have been taken any equimultiples whatever, viz. the base CL and triangle ALC ; and that it has been shown, that, if the base HC be greater than the base CL, the triangle AHC is greater than the triangle ALC ; hS.delS. and if equal, equal ; and if less, less : therefoie •>, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD. And because the parallelogram CE is double of the triangle OF EUCLID. 157 ABC', and the parallelogram CF double of the triangle ACD, Book VI. and that magnitudes have the same ratio which their equimul- ^-— v— -^ tiples haved; as the triangle ABC is to the triangle ACD, so is c 41. 1. the parallelogram EC to the parallelogram CF : and because it d 15. 5. has been shown, that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD ; and as the triangle ABC to the triangle ACD, so is the parallelogram EC to the parallelo- gram CF; therefore, as the base BC is to the base CD, so is^e 11. 5. the parallelogram EC to the parallelogram CF. Wherefore, tri- angles, &c. Q. E. D. Cor. From this it is plain, that triangles and parallelograms that have equal altitudes, are one to another as tht-ir bases. Let the figures be placed so as to have their bases in the same straight line ; and having drawn perpendiculars from the vertices of the triangles to the bases, the straight line which joins the vertices is parallel to that in which their bases are ^, because the f 33. 1. perpendiculars are both equal and parallel to one another: then, if the same construction be made as in the proposition, the de- monstration will be the same. PROP. n. THEOR. IF a straight line be drawn parallel to one of the See n. sides of a triangle, it shall cut the other sides, or those produced, proportionally : and if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section shall be paral- lel to the remaining side of the triangle. Let DE be drawn parallel to BC, one of the sides of the trian- gle ABC : BD is to DA, as CE to EA. Join BE, CD; then the triangle BDE is equal to the triangle CDEa, because they are on the same base DE, and between the a 37. 1. same parallels DE, BC: ADE is another triangle, and equal magnitudes have to the same the same ratio^; therefore, as the b 7. 5. triangle BDE to the triangle ADE, so is the triangle CDE to the triangle ADE; but as the triangle BDE to the triangle ADE, so isc BD to DA, because haviug the same altitude, viz. c 1. 6. the perpendicular drawn from the point E to AB, they are to one another as their bases; and for the same reason, as ihe triangle 158 THE ELEMENTS BookVI.CDE to the triangle ADE, so is E to EA. Therefore, as BD »— V— ' to DA, so is CE to EAd. d 11.5. Next, Let the sides AB, AC of the triangle ABC, or these el. 6. f9. 5. g 39. 1. produced, be cut proportionally in the points D, E, that is, so that BD be to DA, as CE to EA, and join DE ; DE is parallel to BC. The same construction being made, because as BD to DA, so is CE to EA ; and as BD to DA, so is the triangle BDE to the triangle ADE*; and as CE to EA, so is, the triangle CDE to the triangle ADE ; therefore the triangle BDE is to the triangle ADE, as the triangle CDE to the triangle ADE ; that is, the triangles BDE, CDE have the same ratio to the tri- angle ADE ; and therefore f the triangle BDE is equal to the tri- angle CDE : and they are on the same base DE ; but equal tri- angles on the same base are between the same parallels s ; there- fore DE is parallel to BC. Wherefore, if a straight line. Sec. Q. E. D. PROP. in. THEOR. See N IF the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; the segments of the base shall have the same ratio which the other sides of the triangle have to one another : and if the segments of the base have the same ratio which the other sides of the triangle have to one ano- ther, the straight line drawn from the vertex to the point of section divides the vertical angle into two equal angles. Let the angle BAC of any triangle ABC be divided into two equal angles by the straight line AD : BD is to DC, as B A to \C. OF EUCLID. 15^ Through the point C draw CE parallel * to DA, and let BA Book VI- produced meet CE in E. Because the straight line AC meets ^— v— ' the parallels AD, EC, the angle ACE is equal to the alternate a 31. 1. angle CAD ^ : but CAD, by the hypothesis, is equal to the angle b 29. 1- BAD ; wherefore BAD is equal to the angle ACE. Again, be- cause the straight line BAE E meets the parallels AD, EC, the outward angle BAD is equal to ^ the inward and opposite angle AEC : but the angle ACE has been proved equal to the angle BAD ; therefore also ACE is equal to the angle AEC, and consequently the side AE is equal to the side <=■ AC ; and •" xy v. c 6. 1. because AD is drawn parallel to one of the sides of the triangle BCE, viz. to EC, BD is to DC as BA to AEd ; but AE is equal d 2. 6. to AC ; therefore, as BD to DC, so is BA to AC «. e 7. 5. Let now BD be to DC as BA to AC, and join AD ; the angle BAC is divided into two equal angles by the straight line AD. The same construction being made ; because, as BD to DC, so is BA to AC ; and as BD to DC, so is BA to AE ^, because AD is parallel to EC ; therefore BA is to AC as BA to AE f : conse-f 11.3; quently AC is equal to AE s, and the angle AEC is therefore g 9. 5. equal to the angle ACE ^ : but the angle AEC is equal to the out- h 5. 1, ward and opposite angle BAD : and the angle ACE is equal to the alternate angle CAD'^: wherefore also the angle BAD is equal to the angle CAD : therefore the angle BAC is cut into two equal angles by the straight line AD. Therefore, if the an- gle, 8cc. Q. E. D. THE ELEMENTS PROP. A. THEOR. IF the outward angle of a triangle, made by produ- cing one of its sides, be divided into two equal angles, by a straight line which also cuts the base produced; the segments between the dividing line and the ex- tremities of the base have the same ratio which the other sides of the triangle have to one another : and if the segments of the base produced, have the same ratio which the other sides of the triangle have, the straight line drawn from the vertex to the point of section divides the outward angle of the triangle into two equal angles. Let the outward angle CAE of any triangle ABC be divided into two equal angles by the straight line AD* which meets the base produced in I) : BD is to DC as BA to AC. a 31. 1. Throu,i^h C draw CF parallel to AD » : and because the straight line AC meets the parallels AD, FC, the angle ACF is equal to b 29. 1. the alternate angle CAD**: but CAD is equal to the angle c Hyp. DAE c ; therefore also DAE is equal to the angle ACF. Again, because the straight line FAE meets the parallels AD, FC, the outward angle DAE is equal ~ to the inward and opposite angle CFA: but the angle ACF has been proved equal to the angle DAE; therefore also the angle ACF is equal to the angle CP'A, and conse- quently the side AF is equal d 6. 1. to the side AC ^ : and because AD is parallel to FC, a side of the triangle BCF, BD is to DC as e 2. 6. BA to AF^ ; but AF is equal to AC ; as therefore BD is to DC, so is BA to AC. Let now BD be to DC as BA to AC, and join AD ; the angle CAD is equal to the angle DAE. The same construction being made, because BD is to DC fU. 5. as BA to AC; and that BD is also to DC as BA to AF f ; g 9. Tf. tiierefore BA is to AC as BA to AF e ; wherefore AC is equal hS.l. to AFh, and the angle AFC equal *» to the angle ACF: but OF EUCLID. IS I the angle AFC is equal to the outward angle EAD, and the Book VI. angle ACF to the alternate angle CAD ; therefore also EAD is *— y^ equal to the angle CAD. Wherefore, if the outward, &c. Q. E. D. PROP. IV. THEOR. THE sides about the equal angles of equiangular triangles are proportionals ; and those which are op- posite to the equal angles are homologous sides, that is, are the antecedents or consequents of the ratios. Let ABC, DCE be equiangular triangles, having the angle ABC equal to the angle DCE, and the angle ACB to the angle DEC, and consequently » the angle BAC equal to the angle a 32. 1. CDE. The sides about the equal angles of the triangles ABC, DCE are proportionals ; and those are the homologous sides which are opposite to the equal angles. Let the triangle DCE be placed, so that its side CE may be contiguous to BC, and in the same straight line with it : and because the angles ABC, ACB are together less than two right angles b, ABC, and DEC, which is F Jn^ b 17.1. equal to ACB, are also less than two right angles ; wherefore BA, ED produced shall meet <= ; let them be produced and meet in the point F : and because the angle ABC is equal to the angle DCE, BF is parallel ^ \ ^\.,^ \ \^ d28. 1, to CD. Again, because the angle ACB is equal to the angle DEC, AC is parallel to FE d : therefore FACD is a parallelogram ; and consequently AF is equal to CD, and AC to FD « : and because AC is parallel to FE, one of e 34. 1. the sides of the triangle FBE, BA is to AF, as BC to CE f : but f 2. 6. AF is equal to CD ; therefore s, as BA to CD, so is BC to CE ; g 7.5. and alternately, as AB to BC, so is DC to CE* : again, because CD is parallel to BF, as BC to CE, so is FD to DE^ : but FD is equal to AC ; therefore, as BC to CE, so is AC to DE : and al- ternately, as BC to CA, so CE to ED : therefore, because it has been proved that AB is to BC, as DC to CE, and as BC to CA, so CE to ED, ex aquuli\ B A i.s to AC, as CD to DE. Therefore, h 22. 5.. the sides, &:c. Q. E. D. X 162 THE ELEMENTS Book VI. PROP. V. THEOR. IF the sides of two triangles, about each of theii* angles, be proportionals, the triangles shall be equi- angular, and have their equal angles opposite to the homologous sides. Let the triangles ABC, DEF have their sides proportionals, so that A B is to BC, as DE ^o EF; and BC to CA, as EF to FD ; and consequently, ex (equali, BA to AC, as ED to DF ; the triangle ABC is equiangular to the triangle DEF, and their equal angles are opposite to the homologous sides, viz. the angle ABC equal to the angle DEF, and BCA to EFD, and also BAC to EDF. a 23. 1. At the points E, F, in the straight line EF, make » the angle FEG equal to the angle ABC, and the angle EFG equal to BCA ; wherefore the remain- ~ ing angle BAC is equal to the b32. 1. remaining angle EGF^, and the triangle ABC is therefore equiangular to the triangle GEF ; ana consequently they have their sides opposite to the equal angles proportion- ed 6. alsc. Wherefore, as AB to BC, so is GE to EF ; but as AB to BC, so is DE to EF ; there- d 11. 5. fore as DE to EF, so^ GE to EF : therefore DE and GE have e 9. 5. the same ratio to EF, and consequently are equal e : for the same reason, DF is equal to FG : and because in the triangles DEF, GEF, DE is equal to EG, and EF common, the two sides DE, EF are equal to the two GE, EF, and the base DF is equal to the f 8. 1. bass GF ; therefore the angle DEF is equal f to the angle GEF, and the other angles to the other angles which are subtended by S *• ^- the equal sides e. Wherefore the angle DFE is equal to the an- gle GFE,and FDF to EGF : and because the angle DEF is equal to the angle GEF, and GEF to the angle ABC ; therefore the angle ABC is equal to the angle DEF : for the same reason, the angle ACB is equal to the angle DFE, and the angle at A to the angle at D. Thert^fore the triangle ABC is equiangular to the triangle DEF. Wherefore, if the sides, &c. Q. E. D. OF EUCLID. PROP. VI. THEOR. IF two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equiangu- lar, and shall have those angles equal which are oppo- site to the homologous sides. Let the triangles ABC, DEF have the angle BAC in the one equal to the angle EDF in the other, and the sides about those angles proportionals ; that is, BA to AC, as ED to DF ; the triangles ABC, DEF are equiangular, and have the angle ABC equal to the angle DEF, and ACB to DFE. At the points D, F, in the straight line DF, make » the angle a 23. 1. FDG equal to either of the angles BAC, EDF ; and the angle DFG equal to the angle ACB ; A wherefore the remaining an- \ j) gle at B is equal to the re- I \ /V— — — _' G maining one at G •>, and con- | \ / \^ ' 1 b32. 1. scquently the triangle ABC is equiangular to the triangle DGF ; and therefore as BA to AC, so is« GD to DF: hnt,/ \ I N^ c4.6. by the hypothesis, as BA to B C E F AC, so is ED to DF ; as therefore ED to DF, so is 4 GD to DF ; «1 11- ?• wherefore ED is equal e to DG ; and DF is common to the two e 9. 5. triangles EDF, GDF : therefore the two sides ED, DF are equal to the two sides GD, DF ; and the angle EDF is equal to the angle GDF ; wherefore the base EF is equal to the base FGf, **-^- and the triangle EDF to the triangle GDF, and the remaining angles to the remaining angles, each to each, which are subtend- ed by the equal sides : therefore the angle DFG is equal to the angle DFE, and the angle at G to the angle at E : but the angle DFG is equal to the angle ACB ; therefore the angle ACB is equal to the angle DFE : and the angle BAC is equal to the an- gle EDF g ; wherefore also the remaining angle at B is equal to S Hyp- the remaining angle at E. Therefore the triangle ABC is equi- angular to the triangle DEF. Wherefore, if two triangles, 8cq. Q. E. D. J.64 Book VI. THE ELEMENTS PRO?. VII. THEOR. See N. a 23. 1. b52.1. c4. 6. d 11. 5. e9. 5. f5. 1. g 13. 1. IF two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals, then, if each of the re- maining angles be either less, or not less, than a right angle; or if one of them be a right angle: the trian- gles shall be equiangular, and have those angles equal about which the sides are proportionals. Let the two triangles ABC, DEF have one angle in the one equal to one angle in the other, viz. the angle BAC to the angle EDF, and the sides about two other angles ABC, DEF propor- tionals, so that AB is to BC, as DE to EF ; and, in the first case, let each of tlie remaining angles at C, F be less than a right angle. The triangle ABC is equiangular to the triangle DEF, viz. the angle ABC is equal to the angle DEF, and the remain- ing angle at C to the remaining angle at F. For, if the angles ABC, DEF be not equal, one of them is greater than the other: let ABC be the greater, and at the point B, in the straight line AB, make the angle ABG equal to the angle ^ DEF : and because the angle at A is equal to the angle at D, and the angle ABG to the angle DEF ; the remaining angle AGB is equal ^ to the remaining angle DFE : therefore the triangle ABG is equi- angular to the triangle DEF ; wherefore <= as AB is to BG, so is DE to EF ; but as DE to EF, so, by hypothesis, is AB to BC ; therefore as AB to BC, so is AB to BG d ; and because AB has the same ratio to each of the lines BC, BG ; BC is equal * to BG, and therefore tlie angle BGC is equal to the angle BCGf: but the angle BCG is, by hypothesis, less than a right angle ; therefore also the angle BGC is less than a right angle, and the adjacent angle AGB must be greater than a right angled. But it was proved that the angle AGB is equal to the angle at F ; therefore the angle at F is greater than a right angle : but by the hypothesis, it is less than a right angle ; which is absurd. There- OF EUCLID. 165 fore the angles ABC, DEF are not unequal, that is, they are Book VI. equal : and the angle at A is equal to the angle at D ; where- ^-^v— ^ fore the remaining angle at C is equal to the remaining angle at F : therefore the triangle ABC is equiangular to the triangle DEF. Next, Let each of the angles at C, F, be not less than a right angle : the triangle ABC is also in this case equiangular to the triangle DEF. The same construction being A made, it may be proved in like a. D manner that BC is equal to BG, and the angle at C equal to the angle BGC : but the angle at C is not less than a right angle ; therefore the angle BGC is not less than a right angle : wherefore two angles of the triangle BGC are together not less than two righl angles, which is im- possible*'; and therefore the triangle ABC may be proved to behl71- equiangular to the triangle DEF, as in the first case. Lastly, Let one of the angles at C, F, viz. the angle at C, be a right angle ; in this case likewise the triangle ABC is equiangu- lar to the triangle DEF. For, if they be not equiangu- A lar, make, at the point B of the straight line AB, the angle ABG equal to the angle DEF ; then it may be proved, as in the first case, that BG is equal to BC: but the angle ECG is a right angle, therefore' the angle BGC is also a right angle ; whence two of the angles of the trian- gle BGC are together not less than two right angles, which is )mpossiblel»: therefore the tri- B angle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, &c. Q. E. D. i 5. 1. THE ELEMENTS PROP. VIII. THEOR. SeeN IN a right angled triangle, if a perpendicular be- drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle, and to one another. a 32.1 b4. 6. Let ABC be a right angled triangle, having the right angle BAC ; and from the point A let AH be drawn perpendicular to the base BC : the triangles ABD, ADC are similar to the whole triangle ABC, and to one another. Because the angle BAC is equal to the angle ADB, each of them being a right angle, and that the angle at B is common to the two triangles ABC, ABD; , the remaining angle ACB is equal to the remaining angle BAD*: therefore the triangle ABC is equi- angular to the triangle ABD, and the sides about their equal angles are proportionals''; wherefore the cl.def.6. trjjtngles are similar^: in the like manner it may be demonstrated, that the triangle ADC is equiangular and similar to' the triangle ABC; and the triangles ABD, ADC, being both equiangular and similar to ABC, are equiangular and similar to each other. Therefore, in a right angled. Sec. Q. E. D. CoH. From this it is manifest, that the perpendicular drawn from the right angle of a right angled triangle to the base, i,s a mean proportional between the segments of the base : and also that each of the sides is a mean proportional between the base, and Its segment adjacent to that side: because in the triangles BDA, ADC, BD is to DA, as DA to DC •» ; and in the triangles ABC, DBA, BC is to BA, as BA to BD»>; and in the triangle?. ABC, ACDj BC is to CA^ as CA to CDi>. OF EUCLID. PROP. IX. PROB. FROM a given straight line to cut off any part See n. required. Let AB be the given straight line ; it is required to cut off any part from it. From the point A draw a straight line AC making any angle with AB ; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to be cut off from it ; join BC, and draw A DE parallel to It : then AE is the part requir- ed to be cut off. Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, as CD is to DA, so is * BE to EA ; and, by composition t, CA is to AD as BA to AE : but CA is a mul- tiple of AD ; therefore c BA is the same mul- tiple of AE : whatever part therefore AD is of AC, AE is the same part of AB : where- fore, from the straight line AB the part re- quired is cut off. Which was to be done. PROP. X. PROB. TO divide a given straight line similarly to a given divided straight line, that is, into parts that shall have the same ratios to one another which the parts of the divided given straight line have. Let AB be the straight line given to be divided, and AC the divided line ; it is required to divide AB similarly to AC. Let AC be divided in the points D, E ; and let AB, AC be placed so as to contain any angle, and join BC, and through the points D, E draw » DP, EG parallels to it ; and through D a 31. 1. draw DHK parallel to AB : therefore each of the figures FH, IIB, is a parallelogram; wherefore DH is equal ^ to FO, andbSi. 1 168 THE ELEMENTS Book VI. HK to GB: and because HE is paral- * — ■V-— ^ lei to KC, one of the sides of the trian- c 2. 6. gle DKC, as CE to ED, so is c KH to HD : but KH is equal to BG, and HD to GF ; therefore as CE to ED, so is BG to GF : again, because FD is pa- rallel to EG, one of the sides of the tri- angle AGE, as ED to DA, so is GF to FA ; but it has been proved that CE is to ED as BG to GF; and as ED to DA, so GF to FA : therefore the given straight line AB is divided si- milarly to AC. Which was to be done. PROP. XI. PROB. TO find a third proportional to two given straight lines* a31. 1. b 2. 6. Let AB, AC be the two given straight lines, and let them be placed so as to contain any angle ; it is requir- ed to find a third proportional to AB, AC. Produce AB, AC to the points D, E ; and make BD equal to AC ; and having joined BC, through D draw DE parallel to it». Because BC is parallel to DE, a side of the triangle ADE, AB is ^ to BD as AC to CE : but BD is equal to AC; as therefore AB to AC, so is AC to CE. Wherefore, to the two given straight lines AB, AC a third proportional CE is found. Which was to be done. PROP. Xn. PROB. TO find a fourth proportional to three given straight lines. Let A, B, C be the three given straight lines ; it is required to find a fourth j)roportional to A, B, C OF EUCLID. 169 Take two straight lines DE, DF, containing any angle EDF ; Book VI and upon these make DG equal to A, GE equal to B, and DH equal to C ; and havint^ joined GH, draw EF parallel » to it through the point E : and be- cause GH is parallel to EF, one of the sides of the triangle DEF, DG is to GE, as DH to HFb; but DG is equal to A, GE to B, and DH to C ; there- fore, as A is to B, so is C to HF. Wherefore to the three given straight lines A, B, C a fourth proportional HF is found. Which was to be done. b%G, PROP. XIII. PROB. TO find a mean proportional between two given straight lines. Let AB, BC be the two given straight lines ; it is required to find a mean proportional between them. Place AB, BC in a straight line, and upon AC descrih'^ the semicircle ADC, and from the point B draw * BD at right an- gles to AC, and join AD, DC. Because the angle ADC in a semicircle is a right angle *>, and because in the right angled tri- angle ADC, DB is drawn from the right angle perpendicular to the base, DB is a mean propor- tional between AB, BC, the segments of the base <= : therefore be- c Cor. 8, tween the two given straight lines AB, BC a mean proportional ^ DB is found. Which was to be done. a 11. 1. b 31. 3, THE ELEMENTS PROP. XIV. THEOR. EQUAL parallelograms which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportion- al : and parallelograms that have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another. Let AB, BC be equal parallelosframs, which have the angles at B equal, and lee the sides DB, BE be placed in the same a 14. 1* straight line ; wherefore also FB, BG are in one straight line » : the sides of the parallelograms AB, BC about the equal angles, are reciprocally proportional ; that is, DB is to BE, as GB to BF. Complete the parallelogram FE ; and because the parallelo- gram AB is equal to BC, and that A F FE is another parallelogram, AB ^7.5. is to FE, as BC to FE^: but as AB to FE, so is the base DB to c 1. 6. BE c ; and, as BC to FE, so is the base GB to BF ; therefore, as DB •i 11- ^' to BE, so is GB to BF d. Where- fore the sides of the parallelograms AB, BC about their equal angles are reciprocally proportional. But, let the sides about the equal angles be reciprocally pro- portional, viz. as DB to BE, so GB to BF ; the parallelogram AB is equal to the parallelogram BC. Because, as DBto BE, so is GB to BF ; and as DB to BE, so is the parallelogram AB to the parallelogram FE ; and as GB to BF, so is the parallelogram BC to the parallelogram FE ; there- fore as AB to FE, so BC to FE ^ : wherefore the parallelogram * ^- ■^' AB is equal « to the parallelogram BC. Therefore, equal paral. lelogramsj Sec. Q. E. D. OF EUCLID. PROP. XV. THEOR. EQUAL triangles which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional : and trian- gles which have one angle in the one equal to one an- gle in the other, and their sides about the equal angles reciprocally proportional, are equal to one another. Let ABC, ADE be equal triangles, which have the angle BAC equal to the angle DAE ; the sides about the equal angles of the triangles are reciprocally proportional ; that is, CA is to AD, as EA to AB. Let the triangles be placed so that their sides CA, AD be in one straight line ; wherefore also EA and AB are in one straight line a ; and join BD. Because the triangle ABC is equal to the a 14. 1. triangle ADE, and that ABD is another triangle ; therefore as the triangle CAB is to the trian- gle BAD, so is triangle EAD to triangle DABb : but as triangle / ^'^S. \ b 7- 5- CAB to triangle BAD, so is the base C A to AD <=■ ; and as trian- / y^ ^V^ \ c 1. & gle EAD to triangle DAB, so is the base EA to AB^ ; as there- C E fore C A to AD, so is E A to AB ^: d 11. 5. wherefore the sides of the triangles ABC, ADE abaut the equal ^ angles are reciprocally proportional. But let the sides of the triangles ABC, ADE about the equal angles be reciprocally proportional, viz. C A to AD, as EA to AB ; the triangle ABC is equal to the triangle ADB. Having joined BD as before ; because as CA to AD, so is EA to AB ; and as CA to AD, so is triangle BAC to triangle ^AD^; and as EA to AB, so is triangle EAD to triangle BAD = ; there- fore d as triangle BAC to triangle BAD, so is triangle EAD to triangle BAD ; that is, the triangles BAC, EAD have the same ratio to the triangle BAD : wherefore the triangle ABC is equal « g 9. 5, to the triangle ADE. Therefore, equal triangles, &c. Q. E. D. 172 THE ELEMENTS Book VI. PROP. XVI. THEOR. IF four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means : and if the rectangle contain- ed by the extremes be equal to the rectangle contained by the means, the four straight lines are proportionals. Let the four straight lines AB, CD, E, F be proportionals, viz. as AB to CD, so E to F ; the rectangle contained by AB, F is equal to the rectangle contained by CD, E. * 11. 1. From the points A, C draw a AG, CH at right angles to AB, CD ; and make AG equal to F, and CH equal to E, and com- plete the parallelograms BG, DH : because as AB to CD, so is h7.5. E to F ; and that E is equal to CH, and F to AG ; AB is •> to CD, as CH to AG : therefore the sides of the parallelograms BG. DH about the e(|ual angles are reciprocally proportional ; but parallelograms which have their sides about equal anglesr cl4. 6. reciprocally proportional, are equal to one another*: ; therefore the parallelogram BG is equal to the parallelogram DH : and the parallelogram BG is contained p i' by the straight lines AB, F, be- jj cause AG is equal to F ; and the parallelogram DH is con- tained by CD and E, because CH is equal to E: therefore the rectangle contained by the straight lines AB, F is equal to that which is contained by CD and E. And if the rectangle contain- A BCD ed by*the straight lines AB, F be equal to that which is contained by CD, E ; these four lines are proportionals, viz. AB is to CD, as E to F. The same construction being made, because the rectangle contained by the straight lines AB, F is equal to that which is contained by CD, E, and that the rectangle BG is contained by Ali, F, because AG is equal to F; and the rectangle DH by CD, E, because CH is equal to E ; therefore the parallelogram BG is equal to the parallelogram DH ; and they are equiangu- OF EUCLID. ifS lar: but the sides about the equal angles of equal parallelograms Book VI. are reciprocally pro,portionalc : wherefore, as A B to CD, so isCH ^■— ■/•"ii^ to AG ; and CH is equal to E, and AG to F : as therefore AB is c 14. 6. to CD, so E to F. Wherefore, if four, Sec Q. E. D. PROP. XVII. THEOR. IF three straight lines be proportionals, the rectan- gle contained by the extremes is equal to the square of the mean: and if the rectangle contained by the extremes be equal to the square of the mean, the three straight lines are proportionals. Let the three straight lines A, B, C be proportionals, viz. as A to B, so B to C ; the rectangle contained by A, C is equal to the square of B. Take D equal to B ; and because as A to B, so B to C, and that B is equal to D ; A is^ to B, as D to C : but if four straight lines a 7. 5. be proportionals, the rect- angle contained by the A extremes is equal to that B which is contained by the D means ^ : therefore the C j | 5 15. 6. rectangle contained by A, C is equal to that con- | "I C D tained by B, D. But the rectangle contained by B, D is the square of B ; be- A B cause B is equal to D : therefore the rectangle contained by A, C is equal to the square of B. And if the rectangle contained by A, C be equal to the square of B ; A is to B, as B to C. The same construction being made, because the rectangle contained by A, C is equal to the square of B, and the square of B is equal to the rectangle contained by B, D, because B is equal to D ; therefore the rectangle contained by A, C is equal to that contained by B, D : but if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportionals b; therefore A is to B, as D to [ C . I 174 THE ELEMENTS Book VI. C; but B is equal to D; wherefore as A to B, so B to C. <— >— ' fore, if three sti-aight lines, Sec. Q. E. D. There. PROP. XVIII. PROB. See N. UPON a given straight line to describe a rectili- neal figure similar, and similarly situated to a given rectilineal figure. a 23. 1. b i2. 1. 4.6. d 22. 5. Let AB be the sjiven straight line, and CDEF the given recti- lineal figure of four sides; it is required upon the given straight line AB to describe a rectilineal figure similar, and similarly situ- ated to CDEF. Join DF, and at the points A, B, in the straight line AB, make^ the angle BAG equal to the angle at C, and the angle ABG equal to the angle CDF ; therefore the remaining angle CFD is equal to the remaining angle AGB^*: wherefore the triangle FCD is e- quiangular to the triangle GAB : a- gain, at the points G, B, in the straight line GB, make^ the angle BGH equal to the angle DFE, and the angle GBH e- qual to FDE; there- fore the remaining angle FED is equal to the remaining angle GHB, and the triangle FDE equiangular to the triangle GBH : then, because the angle AGB is equal to the angle CFD, and BGH to DFE, the whole angle AGH is equal to the whole 'CFE: for the same reason, the angle ABH is equal to the angle CDE ; also the angle at A i-s equal to the angle at C, and the angle GHB to FED : therefore the rectilineal figure ABHG is equiangular to CDEF : but likewise these figures have their sides about the equal angles proportionals: because the triangles GAB, FCD being equiangular, BA is^ to AG, as DC to CF ; and because AG is to GB, as CF to FD ; and as GB to GH, so, by reason of the equiangular triangles BGH, DFE, is FD to FE ; therefore, ex xrjuali'^, AG is to Gil, as CF to FE : in the same manner it may be proved that AB is to BH, as CD to DE : and GH is to liB, as FE to ED c. Wherefore, becausy OF EUCLID. ^^5 the rectilineal figures ABHG, CDEF are equiangular, and have ^^"okVI. their sides about the equal angles proportionals, they are similar '^"^ to one another e. ^ l.def.e. Next, Let it be required to describe upon a given straight line AB, a rectilineal figure similar, and similarly situated to the rec-* tilineal figure CDKEF. Join DE, and upon the given straight line AB describe the rectilineal figure ABHG similar, and similarly situated to the quadrilateral figure CDEF, by the former case; and at the points B, H, in the straight line BH, make the angle HBL equal to the angle EDK, and the angle BHL equal to the angle DEK ; therefore the remaining angle at K is equal to the re- maining angle at L : and because the figures ABHG, CDEF are similar, the angle GHB is equal to the angle FED, and BHL is equal to DEK ; wherefore the whole angle GHL is equal to the whole angle FEK : for the same reason the angle ABL is equal to the angle CDK : therefore the five sided figures AGHLB, CFEKD are equiangular; and because the figures AGHB, CFED are similar, GH is to HB, as FE to ED ; and as HB to HL, so is ED to EK^ ; therefore, e:r cEcjuali^f ^ GH is to HL, as FE to EK : for the same reason, AB is to \iL, ^ ^^' ^' as CD to DK: and BL is to LH, as^ DK to KE, because the triangles BLH, DKE are equiangular ; therefore, because the five sided figures AGHLB, CFEKD are equiangular, and have their sides about the equal angles proportionals, they are similar to one another : and in the same manner a rectilineal figure of six or more sides may be described upon a given straight line similar to one given, and so on. Which was to be done. PROP. XIX. THEOR. SIMILAR triangles are to one another in the du- plicate ratio of their homologous sides. Let ABC, DEF be similar triangles, having the angle B equal to the angle E, and let AB be to BC, as DE to EF, so that the side BC is homologous to EF « ; the triangle ABC has to the *12.def.5. triangle DEF the duplicate ratio of that which BC has to EF. Take BG a third proportional to BC, EF^ so that BC is to b 11. 6 EF, as EF to BG, and join GA ; then, because as AB to BC, 3© DE to EF ; alternately S AB is to DE, as BC to EF : but c 16- 5. 176 THE, ELEMENTS Book VI. as BC to EF, so is EF to BG ; therefore d, as AB to DE, so is *^-^— ^ EF to EG : wherefore the sides of the triangles ABG, DEF, d 11. 5. which are about the equal angles, are reciprocally proportional : but triangles which have the sides about two equal angles reci- procally proportional, are equal to one an- A e 15. 6. other <= : therefore the triangle ABG is equal to the triangle DEF : and because as BC is to EF, so EF to BG ; and that if three straight lines be pro- portionals, the first is flO.def.5. saidf to have to the third the duplicate ratio of that which it has to the second ; BC therefore has to BG the duplicate ratio of 1.6. t^^^^ which BC has to EF : but as BC to BG, so is e the trian- * " - gle ABC to the triangle ABG. Therefore the triangle ABC has to the triangle ABG the duplicate ratio of that which BC has to EF : but the triangle ABG is equal to the triangle DEF ; wherefore also the triangle ABC has to the triangle DEF the du- plicate ratio of that which BC has to FF. Therefore, similar tri- angles, &c. Q. E. D. Cor. From this it is manifest, that if three straight lines be proportionals, as the first is to the third, so is any triangle upon the first to a similar, and similarly described triangle upon the second. PROP, XX. THEOR. 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 have to one another the duplicate ratio of that vi^hich their homologous sides have. Let ABCDE, FGHKL be similar polygons, and let AB be the homologous side to FG : the polygons ABCDE, FGHKL may be divided into the same number of similar triangles, whereof each to each has the same ratio which the polygons have; and the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that wliich the side AB has to the side FG. Join BE. EC, GL, LH : and because the polygon ABCDE is OF EUCLID. 177 similar to the polygon FGHKL, the angle BAE is equal to the Book VL angle GFL a, and BA is to AE as GF to FL a : wherefore, be- ^— ^— i^^ cause the triangles ABE, FGL have an angle in one equal to a Idef. 6. an angle in the other, and their sides about these equal angles proportionals, the triangle ABE is equiangular •», and therefore b 6. 6. similar to the triangle FGL <= ; wherefore the angle ABE is c 4. 6, equal to the angle FGL : and, because the polygons are simi- lar, the whole angle ABC is equal » to the whole angle FGH ; therefore the remaining angle EBC is equal to the remaining angle LGH : and because the triangles ABE, FGL are similar, EB is to BA, as LG to GF * ; and also, because the polygons are similar, AB is to BC, as FG to GH ^; therefore, ex aqualid^d 22. $. EB is to BC, as LG to GH ; that is, the sides about the equal an- gles EBC, LGH are proportionals; therefore'' the triangle EBC is equiangular to the triangle LGH, and si- milar to it c. For the same reason, the tri- angle ECD likewise is si- milar to the tri- angle LHK : therefore the similar polygons ABCDE, FGHKL are divided into the same number of similar triangles. Also these triangles have, each to each, the same ratio which the polygons have to one another, the antecedents being ABE, ^ EBC, ECD, and the consequents FGL, LGH, LHK : and tiie po- lygon ABCDE has to the polygon FGHKL the duplicate ratio of that which the side AB has to the homologous side FG. Because the triangle ABE is similar to the triangle FGL, ABE has to FGL the duplicate ratio* of that which the side BE « 19. 6, has to the side GL : for the same reason, the triangle BEC has to GLH the duplicate ratio of that which BE has to GL : there- fore, as the triangle ABE to the triangle FGL, so^ is the trian-f 11. 5. gle BEC to the triangle GLH. Again, because the triangle EBC is similar to the triangle LGH, EBC has to LGH the duplicate ratio of that which the side EC has to the side LH : for the same reason, the triangle ECD has to the triangle LHK the duplicate ratio of that which EC has to LH : as therefore the triangle EBC to the triangle LGH, so is *' the triangle ECD to the triangle LHK : but it has been proved that the triangle EBC is likewise to the triangle LGH, as the triangle ABE to the triangle FGL. ^ Therefore, as the triangle ABE is to the triangle FGL, so is tri- angle EBC to triangle LGH, and triangle ECD to triangle LHK : and, therefore, as one of the antecedents to one of the consequents, Z irg THE ELEMENTS Book VI. SO are all the antecedents to all the consequents s. Where-* *«— v-"-' fore, as the triangle ABE to the triangle FGL, so is the polygon g 12. 5. ABCDE to the polygon FGHKL : but the triangle ABE has to the triangle FGL the duplicate ratio of that which the side AB has to the homologous side FG. Therefore also the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that ■which AB has to the homologous side FG. Wherefore, similar polygons, Sec. Q. E. D. CoR. 1. In like manner, it may be proved, that similar four sided figures, or of any number of sides, are one to another in the duplicate ratio of their homologous sides, and it has already been proved in triangles. Therefore, universally, similar I'ectilineal figures are to one another in the duplicate ratio of their homolo- gous sides. Cor. 2. And if to AB, FG, two of the homologous sides, a h 10. def. third proportional M be taken, AB has *» to M the duplicate ratio 5- of that which AB has to FG : but the four sided figure or poly- gon upon AB has to the four sided figure or polygon upon FG likewise the duplicate ratio of that which AB has to FG ; there- fore, as AB is to M, so is the figure upon ABtothe figure upon FG, i Cor. 19. which was also proved in triangles'. Therefore, universally, it 6. is manifest, that if three straight lines be proportionals, as the first is to the third, so is any rectilineal figure upon the first, to a similar and similarly described rectilineal figure upon the second* OF EUCLID. PROP. XXI. THEOR. . RECTILINEAL figures which are similar to the same rectilineal figure, are also similar to one ano- ther. Let each of the rectilineal figures A, B be similar to the recti- lineal figure C : the figure A is similar to the figure B. Because A is similar to C, they are equiangular, and also have their sides about the equal angles proportionals*. Again, be- al.def.6., cause B is similar to C,they are equiangu- lar, and have their sides about the equal angles proportion- als * : therefore the figures A, B are each of them equiangular to C, and have the sides about the equal an- gles of each qf them and of C proportionals. Wherefore the rectilineal figures A and B are equiangular *>, and have their sides b 1 Ax. about the equal angles proportionals <=. Therefore A is similar * ^• to B. Q. E. D. c 11. 5. PROP. XXII. THEOR. IF four straight lines be proportionals, the similar rectilineal figures similarly described upon them shall also be proportionals ; and if the similar rectilineal figures similarly described upon four straight lines be proportionals, those straight lines shall be propor- tionals. Let the four straight lines AB, CD, EF, GH be proportionals, viz. AB to CD, as EF to GH, and upon AB, CD let the similar rectilineal figures KAB, LCD be similarly described ; and upon EF, GH the similar rectilineal figures MF, NH in like manner : the rectilineal figure KAB is to LCD, as MF to NH. To AB, CD take a third proportional » X ; and to EF, GH a 11. 6. a third proportional O: and because AB is to CD, as EF to b 11. 5. GH, and that CD is ^ to X, as GH to O ; wherefore, ejc ac/iialic, c 22. 5. as AB to X, so EF to O : but as AB to X, so is ^ the rectilineal d 2. Cor. 20. 6. 1«© THE ELEMENTS Book VI. KAB to the rectilineal LCD, and as EF to O, so is» is MF to NH. And if the rectilineal KAB he to LCD, as MF to NH j the straight line AB is to CD, as EF to GH. Make « as AB to CD, so EF to PR, and upon PR describe f the rectilineal figure SR similar and similarly situated to either > M S E F G H PR of the figures MF, NH : then because as AB to CD, so is EF to PR, and that upon AB, CD are described the similar and si- milarly situated rectilineals KAB, LCD, and upon EF, PR, in like manner, the similar rectilineals MF, SR ; KAB is to LCD, as MF to SR ; but, by the hypothesis, KAB is to LCD, as MF to NH ; and therefore the rectilineal MF having the same ratio S^.S. to each of the two NH, SR, these are equals to one another: they are also similar, and similarly situated ; therefore GH is equal to PR: and because as AB to CD, so is EF to PR, and that PR is equal to GH ; AB is to CD, as EF to GH. If, there- fore, four straight lines, Sec. Q. E. D. PROP. XXHL THEOR. Se,eN. EQUIANGULAR parallelograms have to one another the ratio which is compounded of the ratios of their sides. Let AC CF be equiangular parallelograms, having the angle BCD equal to the angle ECG : the ratio of the parallelogram AC to the parallelogram CF is the same with the ratio which is com])ounded of the ratios of their sides. OF EUCLID. 181 Let BC, CG be placed in a straight line ; therefore DC and Book VI. CE are also in a straight line* ; and complete the parallelogram ^— v— ^ DG ; and, taking any straight line K, make^ as BC to CG, a 14. 1. so K to L, and as DC to CE, so make ^ L to M : therefore b 12. 6. the ratios of K to L, and L to M, are the same with the ratios of the sides, viz. of BC to CG, and DC to CE. But the ra- tio of K to M is that which is said to be compounded c of thecA.dcf.5. ratios of K to L, and L to M : wherefore also K has to M the ratio compounded of the ratios of the A D H sides ; and because as BC to CG, so is the parallelogram AC to the parallelo- gram CH ^ ; but as BC to CG, so is K to L ; therefore K is ^ to L, as the pa- rallelogram AC to the parallelogram CH : again, because as DC to CE, so is the parallelogram CH to the paral- lelogram CF ; but as DC to CE, so is L to M ; wherefore L is e to M, as the parallelogram CH to the parallelogram CF : therefore, since it has been proved that as K to L, so is the parallelogram AC to the parallelogram CH ; and as L to M, so the parallelogram CH to the parallelo- ,gram CF ; ex aquali f, K is to M, as the parallelogram AC to the f 22. 4, parallelogram CF : but K has to M the ratio which is compound- ed of the ratios of the sides ; therefore also the parallelogram AC has to the parallelogram CF the ratio which is compounded of the ratios of the sides. Wherefore, equiangular parallelo- grams, &c. Q. E. D. PROP. XXIV. THEOR* THE parallelograms about the diameter of any see n. parallelogram, are similar to the whole, and to one another. Let ABCD be a parallelog^ram, of which the diameter is AC ; and EG, HK the parallelograms about the diameter: the paral- lelograms EG, HK are similar both to the whole parallelogram ABCD, and to one another. Because DC, GF are parallels, the angle ADC is equal » to a 29. 1. the angle AGF : for the same reason, because BC, EF are pa- 182 THE ELEMENTS BookVI. rallels, the angle ABC is equal to the angle AEF : and each ^— -V— -^ ol" the angles BCD, EFG is equal to the opposite angle DAB •>, b 34. 1. and therefore are equal to one another ; wherefore the paral- lelograms ABCD, AEFG are equiangular : and because the angle ABC is equal to the angle AEF, and the angle BAC common to the two triangles BAC, EAF, they are equiangu- C.4 6. lar to one another; therefore <= as AB to BC, so is AE to EF : and because the opposite sides of parallelograms d 7.. 5- are equal to one another '', AB is ^ to AD, as AE to AG ; and DC to CB, as GF to FE ; and also CD to DA, as FG to GA : therefore the sides of the parallelograms ABCD, AEFG about the equal angles are proportion- als ; and they are therefore similar t© el.def.6. one another <= : for the same reason, the parallelogram ABCD is similar to the parallelogram FHCK. Wherefore each of the parallelograms GE, KH is similar to DB : but rectilineal figures which are similar to the same rectilineal figure, ar^ also similar £21.6. to one another f ; therefore the parallelogram GE is similar to KH. Wherefore, the parallelograms, Stc Q. E. D. PROP. XXV. PROB. See N. 'YO describe a rectilineal figure which shall be si- milar to one, and equal to another given rectilineal figure. Let ABC be the given rectilineal figure, to which the figure to be described is required to be similar, and D that to which it must be ecjual. It is required to describe a rectilineal figure similar to AIJC, and equal to D. a Cor. 45. Upon the straight line BC describe^ the parallelogram BE ^- equal to the figure ABC ; also upon CE describe * the paral- lelogram CM equal to D, and having the angle FCE equal ,,^29. Lio the angle CBL : therefore BC and CF are in a straight l^"*- 1- line I', as also LE and EM : between BC and CF find <= a mean c 13. 6. proportional GH, and upon CiH describe 'l the rectilineal fi- d 18. 6. gure KGII similar and similarly situated to the figUVe ABC : e 2. Cor. and because BC is GH as GH to CF, and if three straight 20. (i. lines be proportionals, as the first is to the third, so is « the OF EUCLID. 18; figure upon the first to the similar and similarly described figure Book Vt upon the second ; therefore as BC to CF, so is the rectilineal ^ — v— ^ figure ABC to KGH : but as BC to CF, so is f the parallelogram f i. 6. BE to the parallelogram EF : therefore as the rectilineal figure ABC is to KGH, so is the parallelogi'am BE to the parallelogram EF g : and the rectilineal figure ABC is equal to the parallelo- g 11. 5. gram BE ; therefore the rectilineal figure KGH is equal ^ to the h 14 5. parallelogram EF : but EF is equal to the figure D ; wherefore also KGH is equal to D ; and it is similar to ABC. Therefore the rectilineal figure KGH has been described similar to the figure ABC, and equal to D. Which was to be done. • PROP. XXVI. THEOR. IF two similar parallelograms have a common an- gle, and be similarly situated; they are about the same diameter. Let the parallelograms ABCD, AEFG be similar and simi- larly situated and have the angle DAB common. ABCD and AEFG are about the same diameter. For, if not, let, if possible, the parallelogram BD have its diame- ter AHC in a different straight line from AF, the diameter of the parallelogram EG, and let GF meet AHC in H ; and through H draw HK parallel to AD or BC : therefore the parallelograms ABCD, AKHG being about the same diameter, they are similar to one another, »: wherefore, as a 24. 6. DA to AB, so isb GA to AK: but because ABCD and AEFG are similar parallelograms, as DA is to AB, so is GA to AE;bl,def.6, 184 THE ELEMENTS Book VI. therefore ^ as GA to AE, so GA to AK ; wherefore GA has the same ratio to each of the straight lines AE, AK ; and conse- quently AK is equal ^ to AE, the less to the greater, which is impossible : therefore ABCD and AKHG are not about the same diameter; wherefore ABCD and AEFG must be about the same diameter. Therefore, if two similar, See. Q. E. D. * To understand the three following propositions more easily, ' it is to be observed, ' 1. That a parallelogram is said to be applied to a straight ' line, when it is described upon it as one of its sides. JSx. gr. ' the parallelogram AC is said to be applied to the straight line* » AB. ' 2. But a parallelogram AE is said to be applied to a straight ' line AB, deficient by a parallelogram, when AD the base of * AE is less than AB, and therefore ' AE is less than the parallelogram * AC described upon AB in the same ' angle, and between the same pa- ' rallels, by the parallelogram DC ; * and DC is therefore called the de- ' feet of AE. * 3. And a parallelogram AG is ' said to be applied to a straight line AB, exceeding by a parallel- ' ogram, when AF the base of AG is greater than AB, and there- ' fore AG exceeds AC the parallelogram described upon AB in ' the same angle, and between the same parallels, by the paral- ' lelogram BG.' PROP. XXVII. THEOR. SeeN. OF all parallelograms applied to the same straight line, and deficient by parallelograms, similar and si- milarly situated to that which is described upon the lialf of the line; that which is applied to the half, and is similar to its defect, is the greatest. Let AB be a straight line divided into two equal parts in C ; and let the parallelogram AD be applied to the half AC ; which is therefore deficient from the parallelogram upon the whole line AB by the parallelogram CE upon the other half CB : of all Uie parallelograms applied to any other parts of OF EUCLID. 185 AB, and deficient by parallelograms that are similar, and simi-BookVI. larly situated to CE, AD is the greatest. ^— 'v— -^ Let AF be any parallelogram applied to AK, any other part of AB than the half, so as to be deficient from the parallelogram upon the whole line AB by the parallelogram KH similar, and similar- ly situated to CE ; AD is greater than AF. First, let AK the base of AF, be greater than AC the half of AB ; and because CE is similar to the D L E parallelogram KH, they are about the I '^K 1 J same diameter 3 : draw their diameter I j\|p I 8,26.6. DB, and complete the scheme : be- p / J \i | • cause the parallelogram CF is equal *> i " j K I H b 43. X. to FE, add KH to ijoth, therefore the / / I \ 1 whole CH is equal to the whole KE : / / / \ I but CH is equal c to CG, because the | [_1__\ ^ "^^^ *• base AC is equal to the base CB j a C K TK therefore CG is equal to KE : to each of these add CF ; then the whole AF is equal to the gnomon CHL : therefore CE, or the parallelogram AD, is greater than the parallelogram AF. Next, let AK the base of AF, be less than AC, and, the same construction be- ing made, the parallelogram DH is equal to DG c, for HM is equal to MG d, be- cause BC is equal to CA; wherefore DH is greater than LG : but DH is equal ^ to DK ; therefore DK is greater than LG : to each of these add AL ; then the whole AD is greater than the whole AF. Therefore, of all parallelograms applied, £cc. Q. E. D. A K C 2 A THE ELEMENTS PROP. XXVIII. PROS. SeeN. a 10. 1. b 18. 6. c 25. 6. d 21. 6. TO a given straight line to apply a parallelogram equal to a given rectilineal figure, and deficient by a parallelogram similar to a given parallelogram : but the given rectilineal figure to which the parallelogram to be applied is to be equal, must not be greater than the parallelogram applied to half of the given line, having its defect similar to the defect of that which is to be applied ; that is, to the given parallelogram. Let AB be the given straight line, and C the given rectilineal figure, to which the parallelogram to be applied is required to be equal, which figure must not be greater than the parallelo- gram applied to the half of the line having its defect from that upon the whole line similar to the defect of that which is to be applied ; and let D be the parallelogram to which this defect is required to be similar. It is required to apply a parallelogram to the straight line AB, which shall be equal to the figure C, and be deficient from the pa- rallelogram upon the whole line by a parallelogram simi- lar to D. ^ Divide AB into two equal parts * in t'le point E, and upon EB describe the paral- lelogram EBFG similar b and similarly situated to D, and complete the parallelogram AG, which must either be equal to C, or greater, than it. by the determination : and if AG be equal to C, then what was required is already done ; for, upon the straight line AB, the parallelogram AG is applied equal to tl: ■ figure C, and deficient by the parallelogram EF similar to D : but, if AG be net equal to C, it is greater than it ; and EF is equal to AG ; therefore EF also is greater than C. Make <= the parallelogr^im KLMN equal to the excess of EF above C, and similar and similarly situated to D ; but D is similar to EF, therefore (j figure upon the second^: ''^ " therefore, as CB to BD, so is the figure upon CB to the similar and similarly described figure upon BA : and, inversely ^, as DB to BC, so is the figure upon BA to that upon BC ; for the same rea- son, as DC to CB, so is the figure uponCA to that upon CB. Where- fore, as BD and DC together to BC, so are the figures upon BA, AC to that upon BC^: but BD and DC together are equal to BC. Therefore the figure described on BC is equal ^ to the si- milar and similarly described figures on BA, AC. Wherefore, in right angled triangles, Sec. Q. E. D. e24. 5 f A. 5. PROP. XXXII. THEOR. See N. IF two triangles which have two sides of the one proportional to two sides of the other be joined at one angle, so as to have their homologous sides paral- lel to one another, the remaining sides shall be in a straight line. a 29. 1. Let ABC, DCE be two triangles which have the two sides BA, AC proportional to the two CD, DE, viz. BA to AC as CD to DE ; and let AB be parallel to DC, and AC to DE. BC aiid CE arc in a straight line. Because AB is pirallel to A^ DC, and the straight line AC meets them, the alternate an- gles BAC, ACD are equal » ; for the same reason, the angle CDE is equal to the angle ACD ; wherefore also BAC is equal to CDE : and because OF EUCLID. 191 tho, triangles ABC, DCE have one angle at A equal to one at D, Book VI. and the sides about these angles proportionals, viz. BA to '^— v— ^ AC as CD to DE, the triangle' ABC is equiangular b to DCE : b 6. 6. therefore the angle ABC is equal to the angle DCE: and the angle BAC was proved to be equal to ACD : therefore the whole angle ACE is equal to the two angles ABC, BAC ; add the common angle ACB, tiien the angles ACE, ACB are equal to the angles ABC, BAC, ACB : but ABC, BAC, ACB are equal to two right angles c; therefore also the angles ACE, ACB are c 32. l-. equal to two right angles : and since at tiie point C, in the straight line AC, the two straight lines BC, CE, which are on the oppo- site sides of it, make the adjacent angles ACE, ACB equal to two right angles; therefore '^ BC and CK are in a sl;'aight line, d 14. l*. Wherefore, if two triangles, &c. Q. E. D. PROP. XXXIII. THEOR. IN equal circles, angles, whether at the centres or See Nr. circumferences, have the same ratio which the cir- cumferences on which they stand have to one another : so also have the sectors. Let ABC, DEF be equal circles ; and at their centres the angles BGC, EHF, and 'the angles BAC, EDF at their cir- cumferences ; as the circumference BC to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF ; and also the sector BGC to the sector EHF. I'ake any number of circumferences CK, KL, each equal to BC, and any number whatever FM, MN each equal to EF : and join GK, GL, HM, HN. Because the circumferences BC, CK, KL are all equal, the angles BGC, CGK, KGL are also all equal*: therefore what multiple soever the circum- a 27. ference BL is of the circumference BC, the same multiple is the angle BGL of the angle BGC : for the same reason, what- ever multiple the circumference EN is of the circumference EF, the same multiple is the angle EHN of the angle EHF; 193 THE ELEMENTS Book VI. and if the circumference BL be equal to the circumferejjce C^y-»> EN, the angle BGL is also equal =» to the angle EHN ; and a 27. 3. if the circumference BL be greater than EN, likewise the angle BGL is greater than EHN ; and if less, less : there being then four magnitudes, the two circumferences BC, EF, and the two angles BGC, EHF ; of the circumference BC, and of the angle BGC, have been taken any equimultiples whatever, viz. the circumference BL, and the angle BGL ; and of the circum- ference EF, and of the angle EHF, any equimultiples what-^ E r ever, viz. the circumference EN, and the angle EHN : and it has been proved, that, if the circumference BL be greater than EN, the angle BGL is greater than EHN ; and if equal, equal ; and if less, less : as therefore the circumference b 5.def.5. BC to the circumference EF, so ^ is the angle BGC to the angle EHF : but as the angle BGC is to the angle EHF, so is c 15. 5. ' the angle BAC to the angle EDF, for each is double of d 20. 3. each to the angle ECA in a semicircle, and the angle ABD to the B angle AEC in the same segment <= ; the triangles ABD, AEC are equiangular; therefore, as^ BA to AD, so is EA to AC ; and consequently the rectangle BA, AC is equal e to the rectangle EA, AD. If, therefore, from an angle, &c. Q. E. D. e 16. 6. PROP. D. THEOR. THE rectangle contained by the diagonals of aSecN. quadrilateral inscribed in a circle is equal to both the rectangles contained by its opposite sides. Let ABCD be any quadrilateral inscribed in a circle, and join AC, BD ; the rectangle contained by AC, BD is equal to the two rectangles contained by AB, CD, and by AD, -BC*. Make the angle ABE equal to the angle DBC ; add to each of these the common angle EBD, then the angle ABD is equal to the angle EBC : and the angle BDA is equal » to the an- a 21. 3. gle BCE, because they are in the same segment ; therefore the triangle ABD is equiangular h ^'•'^ "^"•^^ to the triangle BCE : wherefore ^ as / ^^ ^ ^* ^' BC is to CE so is BD to DA ; and consequently the rectangle BC, AD is equal c to the rectangle BD, CE : again, because the angle ABE is e- qual to the angle DBC, and the an- gle a BAE to the angle BDC, the tri- angle ABE is equiangular to the tri- angle BCD : as therefore BA to AE, so is BD to DC ; wherefore the rect- angle BA, DC is equal to the rect- angle BD, AE : but the rectangle BC, AD has been shown equal to the rectangle BD, CE ; therefore the whole rectangle AC, BD^ d 1. 2. is equal to the rectangle AB, DC, together with the rectangle AD, BC. Therefor^, the rectangle, &c. Q. E. D. c 16. 6. • This is a lemma of CI. Ptolomjcus, in page 9. of his KryttM o-u»t*|<5- THE ELEMENTS OF EUCLID. BOOK XL DEFINITIONS. I. Book XI. A SOLID is that which hath length, breadth, and thickness. '^"■^^^ II. That which bounds a soUd is a superficies. III. A sti'aight line is perpendicular or at right angles to a plane, when it makes right angles with every straight line meeting it in that plane. IV. A plane is perpendicular to a plane when the straight lines drawn in one of the planes perpendicularly to the common section of the two planes are perpendicular to the other plane. V. The inclination of a straight line to a plane is the acute angle contained by that straight line and another drawn from the point in which the first line meets the plane, to the point in wliich a perpendicular to the plane drawn from any point of the first line above the plane, meets the same plane. VI. The inclination of a plane to a plane is the acute angle contained by two straight Ijnes drawn from any the same point of their common section at right angles to it, one upon one plane, and the other upon the other plane. OF EUCLID. 197 VII. Book XI, Two planes are said to have the same, or a like inclination to one ^-—v*^ another, which two other planes have, when the said angles of inclination are equal to one another. VIII. Parallel planes are such which do not meet one another though produced. IX. A solid angle is that which is made by the meeting of more than See N. two plane angles, which are not in the same plane, in one point. X. ' The tenth definition is omitted for reasons given in the notes.' See N. Similar solid figures are such as have all their solid angles equal, each to each, and which are contained by the same number of similar planes. XII. A pyramid is a solid figure contained by planes that are consti- tuted betwixt one plane and one point above it in which they meet. XIII. A prism is a solid figure contained by plane figures of which two that are opposite are equal, similar, and parallel to one ano- ther ; and the others parallelograms. XIV. A sphere is a solid figure described by the revolution of a semi- circle about its diameter, which remains unmoved. XV. The axis of a sphere is the fixed straight line about which the semicircle revolves. XVI. The centre of a sphere is the same with that of the semicircle. XVII. The diameter of a sphere is any straight line which passes through the centre, and is terminated both ways by the super- ficies of the sphere. XVIII. A cone is a solid figure described by the revolution of a right angled triangle about one of the sides containing the right angle, which side remains fixed. If the fixed side be equal to the other side containing the right *' angle, the cone is called a right angled cone ; if it be less than the other side, an obtuse angled, and if greater, an acute angled cone. THE ELEMENTS XIX. The axis of a cone is the fixed straight line about which the tri- angle revolves. XX. The base of a cone is the circle described by that side containing the right angle, which revolves. XXI. A cylinder is a solid figure described by the revolution of a right angted parallelogram about one of its sides, which remains fixed. XXII. The axis of a cylinder is the fixed straight line about which the pai'allelogram revolves. XXIII. The bases of a cylinder are the circles described by the two re- volving opposite sides of the parallelogram. XXIV. Similar cones and cylinders are those which have their axes and the diameters of their bases p: oportionals. XXV. A cube is a solid figure contained by six equal squares. XXVI. A tetrahedron is a solid figure contained by four equal and equi- lateral triangles. XXVII. An octahedron is a solid figure contained by eight equal and equi- lateral triangles. XXVIII. A dodecahedron is a solid figure contained by twelve equal penta- gons which are equilateral and equiangular. XXIX. An icosahedron is a solid figure contained by twenty equal and equilateral triangles. DEF. A. A parallelepiped is a solid figure contained by six quadrilateral figures, whereof every opposite two are parallel. OF EUCLID* PROP. I. THEOR. ONE part of a straight line cannot be in a plane sec n. and another part above it. If it be possible, let AB, part of the straight line ABC, be in the plane, and the part BC above it : and since the straight line AB is in the plane, it can be pro- duced in that plane : let it be pro- duced to D: and let any plane pass through the straight line AD, and be turned about it until it pass through the point C ; and because the points B, C are in this plane, the straight line BC is in it » : therefore there are two a7.def.,l. straight lines ABC, ABD in the same plane that have a common segment AB, which is impossible''. Therefore, one part, S^cbCor.ll. Q. E. D. 1. PROP. II. THEOR. TWO Straight lines which cut one another are in see N. one plane, and three straight lines which meet one another are in one plane. Let two straight lines AB, CD cut one another in E ; AB, CD are one plane : and three straight lines EC, CB, BE which meet one another, are in one plane. Let any plane pass through the straight line EB, and let the plane be turned about EB, produced, if necessary, until it pass through the point C : then because the points E, C are in this plane, the straight line EC is in it^ : for the same reason, the / \ a7.def,l. straight line BC is in the same ; and, by the hypothesis, EB is in it: therefore the three straight lines EC, CB, BE are in one plane : but in the plane in which EC, EB are, in the same are ^ CD, AB : therefore, AB, CD are in one plane. Wherefore, two straight lines. Sec. Q. E. D. THE ELEMENTS PROP. III. THEOR. See N. IF two planes cut one another, their common sec tion is a straight line. Let two planes AB, BC, cut one another, and let the line DB be their common section : DB is a straight line : if it be not, from the point D to B draw, in the plane AB, the straight line DEB, and in the plane BC the straight line DFB: then two straight lines DEB, DFB have the same extremities, and there- fore include a space betwixt them ; which a 10. Ax. is impossible * : therefore BD the common 1- section of the planes AB. BC cannot but be a straight line. Wherefore, if two planes, he. Q. E. D. PROP. IV. THEOR. See N. 5ll5. 1. b 4. 3. c26. 1. IF a straight line stand at right angles to each of two straight lines in the point of their intersection, it shall also be at right angles to the plane which passes through them, that is, to the plane in which they are. Let the straight line EF stand at right angles to each of the straight lines AB, CD in E, the point of their intersection: EF is also at right angles to the plane passing through AB, CD. Take the straight lines AE, EB, CE, ED all equal to one an- other ; and through E draw, in the plane in which are AB, CD, any straight line GEFI ; and join AD, CB ; then, from any point F in EF', draw FA, FG, FD, EC, FH, FB : and because the two straight lines AE, ED are equal to the two BE, EC, and that they contain equal angles ^ AED, BEC, the base AD is equal ^ to the base BC, and the angle DAE to the angle EBC : and the angle AEG is equal to the angle BEH^; therefore the triangles AEG, BEH have two angles of one equal to two angles of the other, each to each, and the sides AE, EB, adjacent to the equal angles, equal to one another ; where- fore they shall have their other sides equal «= : GE is therefore OF EUCLID. 301 equal to EH, and AG to BH : and because AE is equal to EB, Book XI. and FE common and at right angles to them, the base AF is v — y-— ; equal ^ to the base FB ; for the same reason, CF is equal to FD : b 4. 1. and because AD is equal to BC, and AF to FB, the two sides FA, AD are equal to the two FB, BC, each to each ; and the base DF was proved equal to the base FC ; therefore the angle FAD is equal <* to the angle /// \\\ d 8. 1. FBC : again, it was proved that GA is equal to BH, and also AF to FB ; FA, then, and AG are equal to FB and BH, and the angle FAG has been proved equal to the angle FBH ; therefore the base GF is equal '' to the base FH: again, because it was proved, that GE is equal to EH, and EF is common ; GE, EF are equal to HE, EF ; and the base GF is equal to the base FH ; therefore the an- gle GEF is equal ^ to the angle HEF ; and consequently each of these angles is a right ^ angle. Therefore tE makes right an- e 10. def . gles with GH, that is, with any straight line drawn through E in 1 the plane passing through AB, CD. In like manner, it may be proved, that FE makes right angles with every straight line which meets it in that plane. But a straight line is at right an- gles to a plane when it makes right angles with every straight line which meets it in that plane f; therefore EF is at right an- f 3. def. gles to the plane in which are AB, CD. Wherefore, if a straight 11 line, &c. Q, E. D. PROP. V. THEOR. IF three straight lines meet all in one point, and a See n. straight line stands at right angles to each of them in that point ; these three straight lines are in one and the same plane. . Let the straight line AB stand at right angles to each of the straight lines BC, BD, BE, in B the point where they meet ; BC, BD, BE are in one and the same plane. If not, let, if it be possible, BD and BE be in one plane, and BC be above it ; and let a plane pass through AB, BC, the com- mon flection of which with the plane, in which BD and BE are, 202 THE ELEMENTS Book XL shall be a straight » line ; let this be BF : therefore the three > — ^-mJ straight lines AB, BC, BF are all in one plane, viz. that which a 3. 11. passes through AB, BC ; and because AB stands at right angles to each of the straight lines BD, BE, it is also at right angles b 4. 11. ''to the plane passing through them ; and therefore makes right c f?. def. angles = with every straight line meet- 11. ing it in that plane ; but BF which is in that plane meets it: therefore the angle ABF is a right angle ; but the angle ABC, by the hypothesis, is also a right angle ; therefore the angle ABF is equal to the angle ABC, and they are both in the same plane, which is impossible : therefore the straight line BC is not above the plane in which are BD and BE : wherefore the three straight lines BC, BD, BE are in one and the same plane. fore, if three straight lines, &;c. Q. E. D. There- PROP. VI. THEOR. IF two straight lines be at right angles to the same plane, they shall be parallel to one another. n. Let the straight lines AB, CD be at right angles to the same plane ; AB is parallel to CD. Let them meet the plane in the points B, D, and draw the straight line BD, to which draw DE at right angles, in the same plane ; and make DE equal to AB, and join BE, AE, AD. Then, because AB is per- a 3. def. pendicular to the plane, it shall make right » angles with every straight line which meets it, and is in that plane : but BD, BE, which are in that plane, do each of them meet AB. Therefore each of the angles ABD, ABE is a rigiit angle: for the same reason, each of the angles CDB, CDE is a right angle : and because AB is equal to DE, and BD common, the two sides AB, BD are equal to the two ED, DB ; and they contain right angles ; therefore the base AD is equal ^ to the base BE : again, because AB is equal to UE, and BE to AD ; AB, BE are equal b4. 1. OF EUCLia 203 to ED, DA ; and, in the triangles ABE, EDA, tlfc base AE is Book XI, common ; therefore the angle ABE is equal «= to the angle EDA : *>— v— ♦ but ABE is a right angle; therefore EDA is also a right angle, c 8. 1. and ED perpendicular to DA : but it is also perpendicular to eaqh of the two BD, DC : wherefore ED is at right angles to each of the three straight lines BD, DA, DC in the point in which they meet : therefore these three straight lines are all in the same plane **: but AB is in the plane in which are BD, DA, because d 5. 11, any three straight lines which meet one another are in one plane e : e 2. 11. therefore AB, BD, DC are in one plane : and each of the angles ABD, BDC is a right angle; therefore AB is parallel f to CD. £28,1. Wher«fore, if two straight lines. Sec. Q. E. D. PROP. VII. THEOR. IF two straight lines be parallel, the straight line See n. drawn from any point in the one to any point in the other is in the same plane with the parallels. Let AB, CD be parallel straight lines, and take any point E in the one, and the point F in the other: the straight line which joins E and F is in the same plane with the parallels. If not, let it be, if possible, above the plane, as EGF ; and in the plane ABCD in which the paral- a F r lels are, draw the straight line EHF from E to F; and since EGF also is a straight line, the two straight lines EHF, EGF include a space be- ^ tween them, which is impossible a, \\ a 10. Therefore the straight line joining - — ■ ^1 A*!,. the points E, F is not above the C F D plane in which the parallels AB, CD are, and is therefore in that plane. Wherefore, if two straight lines, 8cc. Q. E. D. PROP. VIII. THEOR. IF two straight lines be parallel, and one of them sce n, is at right angles to a plane, the other also shall be at right angles to the same plane. '504 THE ELEMENl S Book XI gr.ii. a 3. def. 11. b 29. 1. c41. d8. 1. e 4. 11. f3. def. 11 Let AB, CD be two parallel straight lines, and let one of them AB be at right angles to a plane ; the other CD is at right angles to the same plane. Let AB, CD meet the plane in the points B, D, and join BD : therefore e AB, CD, BD are in one plane. In the plane to which AB is at right angles, draw DE at right angles to BD, and make DE equal to AB, and join BE, AE, AD. And because AB is per- pendicular to the plane, it is perpendicular to every straight line which meets it, and is in that plane » : therefore each of the angles ABDj ABE is a right angle : and because the straight line BD meets the parallel straight lines AB, CD, the angles ABD, CDB are together equal i> to two right angles: and ABD is a right an- gle ; therefore also CDB is a right angle, and CD perpendicular to BD : and because AB is equal to DE, and BD common, the two AB, BD, are equal to the two ED, DB, and the angle ABD is equal to the a angle EDB, because each of them is a * right angle ; therefore the base AD is equal = to the base BE : again, because AB is equal to DE, and BE to AD ; the two AB, BE are equal to the two ED, DA; and the base AE is common to the trian- g gles ABE, EDA ; wherefore the angle ABE is equal d to the angle EDA: and ABE is a right angle ; and therefore EDA is a right angle, and ED perpendicular to DA: but it is also perpendicular to BD ; therefore ED is perpendicular e to the plane which passes through BD, DA, and shall f make right an- gles with every straight line meeting it in that plane : but DC is in the plane passing through BD, DA, because all three are in the plane in which are the parallels AB, CD: wherefore ED is at right angles to DC ; and therefore CD is at right angles to DE: but CD is also at right angles to DB ; CD then is at right angles to the two straight lines DE, DB in the point of their in- tersection D ; and therefore is at right angles * to the plane pass- ing through DE, DB, which is the same plane to which AB is at right angles. Therefore, if two straight lines, &c. Q. E. D. OF EUCLID. PROP. IX. THEOR. TWO straight lines which are each of them paral- lel to the same straight line, and not in the same plane with it, are parallel to one another. Let AB, CD be each of them parallel to EF, and not in the same plane with it ; AB shall be parallel to CD. In EF take any point G, from which draw, in the plane passing through EF, AB, the straight line GH at right angles to EF ; and in the plane passing through EF, CD, draw GK at right an- gles to the same EF. And be- cause EF is perpendicular both to GH and GK, EF is perpendicu- lar » to the plane HGK passing \ a 4. 11. through them : and EF is parallel to AB ; therefore AB is at right angles •> to the plane HGK. For the same reason, CD is likewise at right angles to the plane HGK. Therefore AB, CD are each of them at right angles to the plane HGK. But if two straight lines be at right angles to the same plane, they shall be parallel <= to one another. Therefore AB isc 6. It. parallel to CD. Wherefore, two straight lines, &c. Q. E. D. PROP. X. THEOR. IF two straight lines meeting one another be paral- lel to two others that meet one another, and are not in the same plane with the first two, the first two and the other two shall contain equal angles. Let the two straight lines AB, BC which meet one another be parallel to the two straight lines DE, EF that meet one another, and are not in the same plane with AB, BC. The angle ABC is equal to the angle DEF. Take BA, BC, ED, EF all equal to one another ; and join AD, CF, BE, AC, DF : because B A is equal and parallel to ED, there- 206 THE ELEMENTS Book XI. fore AD is * both equal and parallel to BE. ^■— v—^ For the same reason CF is equal and pa- a 33. 1, rallel to BE. Therefore AD and CF are each of them equal and parallel to BE. But straight lines that are parallel to the same straight line, and not in the same b 9. 11. plane with it, are parallel '' to one ano- ther. Therefore AD is parallel to CF ; el.Ax.l. and it is equal ^ to it, and AC, DF join them towards the same parts ; and there- fore a AC is equal and parallel to DF. And because AB, BC are equal to DE, EF, and the base AC to the base DF ; the angle ABC is equal «! to the angle DEF. straight lines, &c. Q. E. D. d8. 1. Therefore, if two PROP. XI. PROB. TO draw a straight line perpendicular to a plane, from a given point above it. a 12. 1. b 11. 1. c31. 1. d 4. 11. e 8. 11. P3. def. 11. Let A be the given point above the plane BH ; it is required to draw from the point A a straight line perpendicular to the plane BH. In the plane draw any straight line BC, and from the point A draw a AD perpendicular to BC. If then AD be also perpendi- cular to the plane BH, the thing required is already done ; but if it be not, from the point D draw*", in the plane BH, the straight line DE at right angles to BC : and from the point A draw AF perpendicular to DE > and through F draw <= GH parallel to BC : and because BC is at right angles to ED and DA, BC is at right an- gles ^ to the plane passing through ED, DA. And GH is parallel to BC ; but, if two straight lines be parallel, tJne of which is at right angles to a plane, the other shall be at right e angles to the same plane; wherefore GH is at right angles to the plane through ED, DA, and is perpendicular ^ to every straight line meeting it in that plane. But AF, which is in the plane through ED, DA, meets it: therefore GH is pgK- OF EUCLID. 207 pendicular to AF ; and consequently AF is perpendicular to GH ; Book XL and AF is perpendicular to DE: therefore AF is perpendicular *-.,-,^v_f to each of the straight lines GH, DE. But if a straight jine stands at right angles to each of two straight lines in the point of their intersection, it shall also be at right angles to the plane passing through them. But the plane passing through ED, GH is the plane BH ; therefore AF is perpendicular to the plane BH ; there- fore, from the given point A, above the plane BH, the straight line AF is drawn perpendicular to that plane. Which was to be doneu PROP. XH. PROB. TO erect a straight line at right angles to a given plane, from a point given in the plane. D B Let A be the point given in the plane ; it is required to erect a straight line from the point A at right angles to the plane. From any point B above the plane draw » BC perpendicular to it ; andfrora A draw*" AD parallel to BC. Because, tiierefore, AD, CB are two parallel straight lines, r and one of them BC is at right angles to / the given plane, the other AD is also at / right angles to it <^. Therefore a straight ^ line has been erected at right angles to a given plane from a point given in it. Which was to be done. a 11. 11. b 31. 1. o 8. lU PROP. XHL THEOR. FROM the same point in a given plane, there can- not be two straight lines at right angles to the plane, upon the same side of it ; and there can be but one perpendicular to a plane from a point above the plane. For, if it be possible, let the two straight lines AC, AB be at right angles to a given plane from the same point A in the plane, and upon the same side of it ; and let a plane pass through BA, 208 THE ELEMENTS Book XI. AC ; the common section of this with the given plane is a straight* v.-^^^ line passing through A : let DAE be their common section : there- a 3. 11. fore the straight lines AB, AC, DAE are in one plane : and be- cause CA is at right angles to the given plane, it shall make right angles with every straight line meeting it in that plane. But DAE, which is in that plane, meets CA; therefore CAE is a right angle. For the same reason BAE is a right angle. Wherefore B C " the angle CAE is equal to the angle BAE; and they are in one plane, which is impossible. Also, from a point above a plane, there can be but one perpendicular to that plane ; for, if there could be two, they would & v b6. 11. be parallel ^ to one another, which ^ A li. is absurd. Therefore, from the same point, &c. Q. E. D. PROP. XIV. THEOR. PLANES to which the same straight line is per- pendicular, are parallel to one another. a 3. def. 11. b 17. 1. c 8. def. 11. Let the straight line AB be perpendicular to each of the planes CD, EF ; these planes are parallel to one another. If not, they shall meet one another when produced ; let them meet ; their common section shall be a straight line GH, in which take any point K, and join AK, BK : then, be- cause AB is perpendicular to the plane EF, it is perpendicular » to the straight line BK which is in that plane. There- fore ABK is a right angle. For the same reason, BAK is a right angle ; wherefore the two angles ABK, BAK of the triangle ABK are equal to two right angles, which is impossible *» : therefore the planes CD, EF, though produced, do not meet one another ; that is, they are parallel ^. Therefore, planes, &c. Q. E. D. OF EUCLIDu PROP. XV. THEOR. IF two straight lines meeting one another, be pa- See N. rallel to two straight lines which meet one another, but are not in the same plane with the first two ; the plane which passes through these is parallel to the plane passing through the others. Let AB, BC, two straight lines meeting one another, be pa- rallel to DE, EF, that meet one another, but are not in the same plane with AB, BC : the planes through AB, BC, and DE, EF shall not meet, though produced. From the point B draw BG perpendicular » to the plane a 11. 11. which passes through DE, EF, and let it meet that plane in G ; and through G draw GH parallel »> to ED, and GK pa- b 31. 1. rallel to EF : and because BG is perpendicular to the plane through DE, EF, it shall E make right angles with every straight line meeting it in that plane *=. But the straight lines GH, GK in that plane meet it : therefore each of the an- gles BGH, BGK is a right an- gle : and because BA is pa- rallel ^ to GH (for each of them is parallel to DE, and they are not both in the same plane with it) the angles GBA, BGH are together equal* to two right angles: and BGH is a c 29. 1. right angle; therefore also GBA is a right angle, and GB per- pendicular to BA : for the same reason, GB is perpendicular to BC : since therefore the straight hne GB stands at right angles to the two straight lines BA, BC, that cut one another in B, Gfi is perpendicular f to the plane through BA, BC ; and it is f 4. 11. perpendicular to the plane through DE, EF : therefore BG is perpendicular to each of the planes through AB, BC, and DE, EF : but planes to which the same straight line is perpendicular, are parallel e to one another : therefore the plane through AB, g 14. 11. BC is parallel to the plane through DE,J£F. Wherefore, if two straight lines, Sec. Q. E. D. (T^ F K \ C ^^ c 3. def. 11. ^...^^ D d 9. 11.. 3D 210 Book XI. THE ELEMENTS SceN. PROP. XVI. THEOR. IF two parallel planes be cut by another plane, their common sections with it are parallels. Let the parallel planes AB, CD be cut by the plane EFHG, and let their common sections with it be EF, GH : EF is paral- lel to GH. For, if it is not, EF, GH shall meet, if produced, either on the side of FH, or EG : first, let them be produced on the side of FH, and meet in the point K ; therefore, since EFK is in the plane AB, every point in K EFK is in that plane : and K is a point in EFK ; therefore K is in the plane AB : for the same reason K is also in the plane CD : wherefore the planes AB, CD produced ^^ "*1 D meet one another ; but they do not meet since they are parallel by the hypothesis : therefore tke straight lines EF, GH do not meet when produced on the side of FH ; in the same manner it may be .proved, that EF, GH do not meet when produced on the side of EG : but straight lines which are in the same plane and do not meet, though produced either way, are parallel : therefore EF is parallel to GH. Wherefore, if two parallel planes, &c. Q. E. D. h' / fA ^ B k. C L^ E '^ G H PROP. XVH. THEOR. IF two Straight lines be cut by parallel planes, thev shall be cut in the same ratio. Let the straight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B ; C, F, D : as AE is to EB, so is CF to FD. Join AC, BD, AD, and let AD meet the plane KL in the point X ; and join EX, XF : because the two parallel planes KL, MN are cut by the plane EBDX, the common sections OF EUCLID. $11 b 2. 6. EX, Bl) are parallel". For the same reasorii because the two Book XI. parallel planes GH, KL are cut ^ -v ^ by the plane AXFC, the com- / _- — "!>. / H a 16. 11. men sections AC, XF are paral- lel: and because EX is paral- lel to BD, a side of the triangle ABD, as AE to EB so is ^ AX to XD. Again, because XF is parallel to AC, a side of the triangle ADC, as AX to XD, so is CF to FD : and it was proved that AX is to XD as AE to EB : therefore <=, as AE to EB so is CF to FD. Wherefore, if two straight lines, kc, Q. E. D. c 11. S; PROP. XVIII. THEOR. IF a straight line be at right angles to a plane, eve- ry plane which passes through it shall be at right an- gles to that plane. Let the straight line AB be at right angles to a plane CK ; every plane which passes through AB shall be at right angles to, the plane CK. Let any plane DE pass through AB, and let CE be the com- mon section of the planes DE, CK ; take any point F in CE, from which draw FG in the plane DE at right angles to CE : and because AB is per- pendicular to the plane CK, therefore it is also perpendi- cular to every straight line in that plane meeting it » ; and consequently it is perpendicu- lar to CE : wherefore ABF is a right angle ; but GFB is likewise a right angle: there- D G A H ' ' K \ \ a 3. de£. 11. B E fore AB is parallel ^ to FG. And AB is at right angles to the b 28. t plane CK ; therefore FG is also at right angles to the same plane «. But one plane is at right angles to another plane when c 8. 11. the straight lines drawn in one of the planes, at right angles to their common section, are also at right angles to the other 212 THE ELEMENTS Book XI. plane d : and any straight line FG in the plane DC, which is y^'-Y'mJ at right angles to CE the common section of the planes, has d4. def. been proved to be perpendicular to the other plane CK ; there- II- fore the plane DE is at rigist angles to the plane CK. In like manner it may be proved that all the planes which pass through AB are at right angles to the plane CK. Therefore, if a straight line, £cc. Q. E. D. PROP. XIX. THEOR. IF two planes cutting one another be each of them perpendicular to a third plane, their common section shall be perpendicular to the same plane. a 4 def. 11. b 13. 11 Let the two planes AB, BC be each of them perpendicular to a third plane, and let BD be the common section of the first two ; BD is perpendicular to the third plane. If it be not, from the point D draw, in the plane AB, the straight line DE at right angles to AD the common section of the plane AB with the third plane ; and in the plane BC draw DF at right angles to CD the common section of the plane BC with the third plane. And because the plane AB is perpendicular to the third plane, and DE is drawn in the plane AB at right angles to AD their common sec- tion, DE is perpendicular to the third plane a. In the same manner it may be proved that DF is perpendicular to the third plane. Wherefore, from the point D two straight lines stand at right angles to the third plane, upon the same side of it, which is impossible^: therefore from the point D there cannot be any straight line at right angles to the third plane, except BD the common section of the ^ ^ planes AB, BC BD therefore is perpendicular to the third plane. Wherefore, if two planes, Sec. Q. E. D. OF EUCLID. PROP. XX. THEOR. IF a solid angle be contained by three plane angles, See N. any two of them are greater than the third. Let the solid angle at A be contained by the three plane an- gles BAG, CAD, DAB. Any two of them are greater than the third. If the angles BAC, CAD, DAB be all equal, it is evident that any two of them are greater than the third. But if they are not, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB ; and at the point A, in the straight line AB, make, in the plane which passes through BA, AC, the angle BAE equal » to the angle DAB ; and make a 23.1. AE equal to AD, and through E draw BEC cutting AB, AC in the points B, C, and join DB, DC. And because DA is equal to AE, and AB is common, the two DA, AB are equal to the two EA, AB, and D the angle DAB is equal to the angle /k EAB : therefore the base DB is equal'' / 1 >< b 4, 1. to the base BE. And because BD, DC /I \^ are greater c than CB, and one of them / ^ | \^ ^ ^"* *' BD has been proved equal to BE a part of CB, therefore the other DC is great- ^^ er than the remaining part EC. And g ^ because DA is equal to AE, and AC common, but the base DC greater than the base EC : therefore the angle D AC is greater ^ than the angle EAC ; and, by the d 25. 1. construction, the angle DAB is equal to the angle BAE ; where- fore the angles DAB, DAC are together greater than BAE, EAC, that is, than the angle BAC. But BAC is not less than either of the angles DAB, DAC : therefore BAC, with either of them, is greater than the other. Wherefore, if a solid angle, 8cc. Q. E. D. PROP. XXI. THEOR. EVERY solid angle is contained by plane angles which together are less than four right angles. First, let the solid angle at A be contained by three plane angles BAC, CAD, DAB. These three together are less than four right angles. 21* THE ELEMENTS Book XI. Take in each of the straight lines AB, AC, AD any points B, *«— v^"-* C, D, and join BC, CD, DB : then because the solid angle at B is contained by the three plane angles CBA, ABD, DBC, any a 20. 11. two of them are greater » than the third ; therefore the angles CBA, ABD are greater than the angle DBC : for the same rea- son, the angles BCA, ACD are greater than the angle DCB ; and the angles CDA, ADB greater than BDC : wherefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than the three angles DBC, BCD, CDB : but the three angles DBC, BCD, CDB are b32. 1. equal to two right angles *> : therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than two right angles : and because the three angles of each of the triangles ABC, ACD, ADB are equal to two right angles, therefore the nine angles of these three triangles, viz. the angles CBA, BAC, ACB, ACD, CDA, DAC, ADB, DBA, BAD are equal to six right angles : of these the six angles CBA, ACB, ACD, CDA, ADB, DBA are greater than two right angles : therefore the remaining three angles BAC, DAC, BAD, which contain the solid angle at A, are less than four right angles. Next, let the solid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB ; these together are less than four right angles. Let the planes in which the angles are be cut by a plane, and let the common section of it Avith those planes be BC, CD, DE, EF, FB : and because the solid angle at B is contain- ed by three plane angles CBA, ABF, FBC, of which any two are greater » than the third, the angles CBA, ABF are greater than the angle FBC: for the same reason, the two plane angles at each of the points C, D, E, F, viz. the angles which are at the bases of the triangles, having the common vertex A, are greater than the third angle at the same point, which is one of the an- gles of the polygon BCDEF : there- fore all the angles at the bases of the triangles are together OF EUCLID. ^]s greater than all the angles of the polygon : and because all the Book XL angles of the triangles are together equal to twice as many right v -y^ angles as there are triangles ^ ; that is, as there arc sides in the b 32. 1. polygon BCDEF : and that all the angles of the polygon, toge- ther with four right angles, are likewise equal to twice as many right angles as there are sides in the polygon <= ; therefore all c 1. Coi;^ the angles of the triangles are equal to all the angles of the po- 32. 1. lygon together with four right angles. But all the angles at the bases of the triangles are greater than all the angles of the poly- gon, as has been proved. Wherefore the remaining angles of the triangles, viz. those at the vertex, which contain the solid angle at A, are less than four right angles. Therefore, every solid angle, &c. Q. £. D. PROP. XXII. THEOH. IF every two of three plane angles be greater than see ^^ the third, and if the straight lines which contain them be all equal ; a triangle may be made of the straight lines that join the extremities of those equal straight lines. Let ABC, DEF, GHK be three plane angles, whereof eveiy two are greater than the third, and are contained by the equal straight lines AB, BC, DE, EF, GH, HK ; if their extremities be joined by the straight lines AC, DF, GK, a triangle may be made of three straight lines equal to AC, DF, GK ; that is, eve- ry two of them are together greater than the third. If the angles at B, E, H are equal, AC, DF, GK are also equal a, and any two of them greater than the third: but if a 41. the angles are not all equal, let the angle ABC be not less than cither of the two at E, H; therefore the straight line AC is not less than either of the other two DF, GK '> ; and it is b 4. or plain that AC, together with either of the other two, must be 24. 1. greater than the third : also DF with GK are greater than AC: for, at the point B in the straight line AB make <^ the c 23. 1. 316 THE ELEMENTS Book XI. angle ABL equal to the an^le GHK, and make BL equal to **— V— ^ one of the straight lines AB, BC, DE, EF, GH, HK, and join AL, LC ; then, because AB, BL are equal to GH, HK, and the angle ABL to the angle GHK, the base AL is equal to the base GK : and because the angles at E, H are greater than the angle ABC, of which the angle at H is equal to ABL ; therefore the remaining angle at E is greater than the angle LBC : and be- F G d 24. 1. e20. 1. £23. 1. cause the two sides LB, BC are equal to the two DE, EF, and that the angle DEF is greater than the angle LBC, the base DF is greater ^ than the base LC : and it has been proved that GK is equal to AL ; therefore DF and GK are greater than AL and LC : but AL and LC are greater ^ than AC : much more then are DF and GK greater than AC. Wherefore every two of these straight lines AC, DF, GK are greater than the third ; and, therefore, a triangle may be made f, the sides of which shall be, equal to AC, DF, GK. Q. E. D. PROP. XXHL PROB. See N TO make a solid angle which shall be contained by three given plane angles, any two of them being greater than the third, and all three together less than four right angles. Let the three given plane angles be ABC, DEF, GHK, any two of which are greater than the third, and all of them toge- ther less than four right angles. It is required to make a solid angle contained by three plane angles equal to ABC, DEF, GHK, each to each. OF EUCLID. 217 From the straight lines containing the angles, cut off AB, BC, BookXI. DE, EF, GH, HK, all equal to one another; and join AC, DF, ' — r— ' GK : then a triangle may be made » of three straight lines ec^ual a 22. 11. to AC, DF, GK. Let this be the triangle LMN^, so that AC b 22. t be equal to LM, DF to MN, and GK to LN ; and about the tri- angle LMN describe *= a circle, and find its centre X, which will c 5. 4' cither be within the triangle, or in one of its sides, or without it. First, let the centre X be within the triangle, and join LX, MX, NX : AB is greater than LX : if not, AB must ed that it is greater than DEF, which is absurd. Therefore AB d 8. 1. is not equal to LX. Nor yet is it less; for then, as has been proved in the first case, the angle ABC is greater than the angle MXL, and the angle GHK greater than the angle LXN. At the point B, in the straight line CB, make the angle CBP equal to the angle GHK, and make BP equal to HK, and join CP, AP. And because CB is equal to GH ; CB, BP are equal to GH, HK, each to each, and they contain equal angles ; wherefore the base CP is equal to the base GK, that is, to LN. And in the isosceles triangles ABC, MXL, because the angle ABC is great- er than the angle MXL, therefore the angle MLX at the base is greater s than the angle ACB at the base. For the same rea- g 32. >. son, because the angle GHK, or CBP, is greater than the angle R LXN, the angle XLN is greater than the angle BCP. Therefore the whole angle MLN is greater than the whole angle ACP. And because ML, LN are equal to AC, CP, each to each, but the angle MLN is greater than the angle ACP, the base MN is greater ^ than the base AP. And MN is equal to DF ; there- fore also DF is greater than AP. Again, because DE,EF are equal to AB, BP, but the base DF greater than the base AP, the angle DEF is greater ^ than the 1- 35. l angle ABP. And ABP is equal to the two angles ABC, CBP, that is, to the two angles ABC, GHK ; therefore the angle DEF is greater than the two angles ABC, GHK ; but it is also less h 24. 1. 220 THE ELEMENTS b 3. def. 11. Book XI. than these, which is impossible. Therefore AB is not less than Ci-y-.^ LX, and it has been proved that it is not equal to it ; therefore AB is greater than LX. a 12. 11. From the point X erect a XR at right angles to the plane of the circle LMN. And because it has been proved in all the cases that AB is greater than LX, find a square equal to the excess of the square of AB above the square of LX; and make RX equal to its side, and join RL, RM, RN. Be- cause RX is perpendicular to the plane of the circle LMN, it is •> per- pendicular to each of the straight lines LX, MX, NX. And because LX is equal to MX, and XR com- mon, and at right angles to each of them, the base RL is equal to the base RM. For the same reason, RN is equal to each of the two RL, RM. Therefore the three straight lines RL, RM, RN are all equal. And because the square of XR is equal to the excess of the square of AB above the square of LX ; therefore the square of AB is equal to the squares of LX, XR. c4r. 1. But the square of RL is equal <= to the same squares, because LXR is a right angle. Therefore the square of AB is equal to the square of RL, and the straight line AB to RL. But each of the straight lines BC, DE, EF, GH, HK is equal to AB, and each of the two RM, RN is equal to RL. Wherefore AB, BC, DE, EF, GH, HK are each of them equal to each of the straight lines RL, RM, RN. And because RL, RM are equal to AB, BC, and the base LM to the base AC ; the angle LRM is equal d 8. 1. ^ to the angle ABC. For the same reason, the angle MRN is equal to the angle DEF, and NRL to GHK. Therefore there is made a solid angle at R, which is contained by three plane an- gles LRM, MRN, NRL, which are equal to the three given plane angles ABC, DEF, GHK, each to each. Which was to be done. OF EUCLID* PROP. A. THEOR. IF each of two solid angles be contained by three see n. plane angles equal to one another, each to each; the planes in which the equal angles are, have the same inclination to one another. a 6. def. U Let there be two solid angles at the points A, B ; and let the angle at A be contained by the three plane angles CAD, CAE, EAD ; and the angle at B by the three plane angles FBG, FBH, HBG; of which the angle CAD is equal to the angle FBG, and CAE to FBH, and EAD to HBG : the planes in which the equal angles are have the same inclination to one another. In the straight line AC take any point K, and in the plane CAD from K draw the straight line KD at right angles to AC, and in the plane CAE the straight line KL at right angles to the same AC : therefore the angle DKL is the inclina- tion a of the plane CAD to the plane CAE. In BF take BM equal to AK, and from the point M draw, in the planes FBG, FBH, the straight lines MG, MN at right angles to BF ; therefore the angle GMN is the inclination a of the plane FBG to the plane FBH : join LD, NG; and because in the triangles KAD, MBG, the angles KAD, MBG are equal, as also the right angles AKD, BMG, and that the sides AK, BM, adjacent to the equal angles, are equal to one another; therefore KD is equal ^ to b 26. 1 MG, and AD to BG : for the same reason, in the triangles KAL, MBN, KL is equal to MN, and AL to BN : and in the triangles LAD, NBG, LA, AD are equal to NB, BG, and they contam equal angles; therefore the base LD is equal <= to the c 4. 1 base NG. Lastly, in the triangles KLD, MNG, the sides DK, KL are equal to GM, MN, and the base LD to the base NG; therefore the angle DKL is equal d to the angle GMN : but the d 8. 1. angle DKL is the inclination of the plane CAD to the plane CAE, and the angle GMN is the inclination of the plane FBG 222 THE ELEMENTS Book XI. to the plane FBH, which planes have therefore the same incli- *— V— ^ nation » to one another : and in the same manner it may be de- a 7. monstrated, that the other planes in which the equal angles are dof. 11. Yiave the same inclination to one another. Therefore, if two so- lid angles, See. Q. E. D. PROP. B. THEOR. See N, IF two solid angles be contained, each by three plane angles which are equal to one another, each to each, and alike situated; these solid angles are equal to one another. Let there be two solid angles at A and B, of which the solid angle at A is contained by the three plane angles CAD, CAE, EAD ; and that at B by the three plane angles FBG, FBH, HBG ; of which CAD is equal to FBG ; CAE to FBH ; and EAD to HBG : the solid angle at A is equal to the solid angle at B. Let the solid angle at A be applied to the solid angle at B ; and, first, the plane angle CAD being appHed to the plane angle FBG, so as the point A may coincide with the point B, and the straight line AC with BF ; then AD coincides wiih BG, because the angle CAD is equal to the angle FBG : and because the in- clination of the plane CAE to the a A. 11. plane CAD is equal » to the in- clination of the plane FBH to the plane FBG, the plane CAE coin- cides with the plane FBH, be- cause the planes CAD, FBG co- C incide with one another : and because the straight lines AC, BF coincide, and that the angle CAE is equal to the angle FBH ; therefore AE coincides with BH, and AD coincides with BG ; wherefore the plane EAD coincides with the plane HBG : there- fore the scHd angle A coincides with the solid angle B, and con- b 8. A. 1. sequently they are equal *> to one another. Q. E. D. OF EUCLID, PROP. C. THEOR. ' SOLID figures contained by the same number of See n. equal and similar planes alike situated, and having none of their solid angles contained by more than three plane angles, are equal and similar to one an- other. Let AG, KQ be two solid figures contained by the same num- ber of similar and equal planes, alike situated, viz. let the plane AC be similar and equal to the plane KM, the plane AF to KP; BG to LQ ; GD to QN ; DE to NO ; and lastly, FH similar and equal to PR : the solid figure AG is equal and similar to the so- lid figure KQ. Because the solid angle at A is contained by the three plane angles BAD, BAE, EAD, which, by the hj^pothesis, are equal to the plane angles LKN, LKO, OKN, which contain the solid angle at K, each to each ; therefore the solid angle at A is equal ^ a B. 11. to the solid angle at K : in the same manner, the other solid an- gles of the figures are equal to one another. If, then,, the solid figure AG be applied to the solid figure KQ, first, the plane figure AC being ap- H G . R \ E \ C \ \ O Q Mi A B K plied to the plane figure KM ; the straight line AB co- inciding with KL, the figure AC must ^ coincide with the figure KM, because they are equal and similar : therefore the straight lines AD, DC, CB coincide with KN, NM, ML, each with each ; and the points A, D, C, B, with the points K, N, M, L : and the solid angle at A coincides with* the solid angle at K ; wherefore the plane AF coincides with the plane KP, and the figure AF with the figure KP, because they are equal and similar to one another: therefore the straight lines AE, EF, FB coincide with KO, OP, PL ; and the points E, F with the points O, P. In the same manner, the figure AH coincides with the figure KR, and the straight line DH with NR, and the point H with the point R : and because the solid angle at B is equal to the solid angle at L, it may be proved, in the sanae manner, that the figure BG coincides with the figure 224 THE ELEMENTS Book XL LQ, and the straight line CG with MQ, and the point G with ^-— v-i-^ the point Q: since, therefore, all the planes and sides of the so- lid figure AG coincide with the planes and sides of the solid fi- gure KQ, AG is equal and similar to KQ : and, in the same man- ner, any other solid figures whatever contained by the same num- ber of equal and similar planes, alike situated, and having none of their solid angles contained by more than three plane angles, may be proved to be equal and similar to one another. Q. E. D. PROP. XXIV. THEOR. SeeN. IF a solid be contained by six planes, two and two of which are parallel ; the opposite planes are similar and equal parallelograms. Let the solid CDGH be contained by the parallel planes AC, GF; BG, CE ; FB, AE : its opposite planes are similar and equal parallelograms. Because the two parallel planes BG, CE are cut by the plane a 16. 11. AC, their common sections AB, CD are parallel ». Again, be- cause the two parallel planes BF, AE are cut by the plane AC, their common sections AD, BC are parallel*: and AB is paral- lel to CD ; therefore AC is a parallelogram. In like manner, it may be proved that each of the figures CE, FG, GB, BF, AE is a parallelo- B H gram : join AH, DF ; and because AB is parallel to DC, and BH, to CF; the two straight lines AB, BH which A meet one another, are parallel to DC and CF which meet one another, and are not in the same plane with the other two; wherefore they contain b 10. 11. equal angles ^ ; the angle ABH is D therefore equal to the angle DCF : and because AB, BH are equal to DC, CF, and the angle ABH c 4. 1. equal to the angle DCF ; therefore the base AH is equal <= to the base DF, and the triangle ABH to the triangle DCF : and the d34. 1. parallelogram BG is double ^ of the triangle ABH, and the pa- rallelogram CE double of the triangle DCF; therefore the paral- lelogram BG is equal and similar to the parallelogram CE. In the same manner it may be proved, that the parallelogram AC ^^--^ c G ^ ^?^ E GF EUCLID. 225 'is equal and similar to the parallelogram GF, and the parallelo-BookXI. gram AE to BF. Therefore, if a solid, 8cc. Q. E. D. *— v— ' PROP. XXV. THEOR. IF a solid parallelepiped be cut by a plane parallel See n. to two of its opposite planes, it divides the whole into two solids, the base of one of which shall be to the ^ base of the other as the one solid is to the other. Let the solid parallelepiped ABCD be cut by the plane EV, which is parallel to the opposite planes AR, HD, and divides the whole into the two solids ABFV, EGCD ; as the base AEFY of the first is to the base EHCF of the other, so is the solid ABFV to the solid EGCD. Produce AH both ways, and take any number of straight lines HM, MN, each equal to EH, and any number AK, KL, each equal to EA, and complete the parallelograms LO, KY, HQ, MS, and the solids LP, KR, HU, MT; then, because the straight lines LK, KA, AE are all equal, the parallelograms LO, KY, AF are equal a : and likewise the parallelog»^inis KX, KB, AG » ; a 36. 1. as also ^ the parallelograms LZ, KP, A^i because they are op- b 24. 11. posite planes : for the same reason, t.ie parallelograms EC, HQ, MS are equal = ; and the parallelr&i'ams HG, HI, IN, as also^ HD, MU, NT : therefore three planes of the solid LP are equal and similar to three planes of «:he solid KR, as also to three planes of the solid A V : but the -'hree planes opposite to these three are equal and similar *> tJ them in the several solids, and none of their solid angles a-'-e contained by more than three plane an- gles : therefore the three solids LP, KR, A.V are equal c to one c C. 11, another: iov the same reason, the three solids ED, HU, MT are equal to one another : therefore, what multiple soever the 2 F 226 THE ELEMENTS Book XL base LF is of the base AF, the same, multiple is the solid LV of *--v— ^ the solid AV: for the same reason, whatever multiple the base NF is of the base HF, the same multiple is the solid NV of the solid ED : and if the base LF be equal to the base NF, the solid cC. 11 LV is equal = to the solid NV; and if the base LF be greater than the base NF, the solid LV is greater than the solid NV ; and if less, less : since then there are four magnitudes, viz. the tv(^o bases AF, FH, and the two solids AV, ED, and of the X B G I d 5. def. 5. R O Y F C Q S base AF and solid AV, the base LF and solid LV are any equi- multiples whatever; and of the base FH and solid ED, the base FN and solid N \^ are any equimultiples whatever ; and it has been proved, that if the base LF is greater than the base FN, the solid LV is greater than the solid NV ; and if equal, equal ; and if less, less. Therefore ^ as the base AF is to the base FH, so is the solid AV to the solid ED. Wherefore, if a solid, &:c. Q. E. D. PROP. XXVL PROB. SeeN. AT a given point in a given straight line, to make a solid angle equal to a given solid angle contained by three plane angles. Let AB be a given straight line, A a given point in it, and D a given solid angle contained b; the three plane angles EDC, EDF, FDC : it is required to make at the point A, in "the straight Ime AB, a solid angle equal to the soh-\ anglo D. In the straight line DF take any point F, from which draw a 11. 11. a FG perpendicular to the plane ED^,, meeting that plane in G ; join DG, and at the point A, in the straight line AB, b 23. 1. make ^ the angle BAL equal to the angle EDC, and in the plane BAL make the angle BAK equal to the angle EDG; c 12. 11. then make AK equal to DG, and from the point K erect ^ KH OF EUCLID. 227 at right angles to the plane BAL ; and make KH equal to Book XI. GF, and join AH : then the solid angle at A, which is contain- v— y^«ii ed by the three plane angles BAL, BAH, HAL, is equal to the solid angle at D contained by the three plane angles EDC, EDF, FDC. Take the equal straight lines AB, DE, and join HB, KB, FE, GE : and because FG is perpendicular to the plane EDC, it makes right angles ^ with every straight line meeting it in d 3. def. that plane : therefore each of the angles FGD, TGE is a right ^^^ angle : for the same reason, HKA, HKB are rTght angles : and because KA, AB are equal to GD, DE, each to each, and contain equal angles, therefore the base PK. is equal « to the e 4. 1. base EG : and KH is equal to GF, and HKB, FGE are right angles, therefore HB is equal e to FF: again, because AK, KH are • equal to DG, GF, and contain right angles, the base AH is equal to the base DF : and AB is equal to DE ; therefore HA, AB are equal to FD? DE, and the base HB is e- qual to tlxe base FE, ~ therefore the angle BAH is equal f to //1\ //\\ £8.1, the angle EDF: for th^ same reason, the angle HAL is equal to tht angle FDC. Because if AL and DC be made equal, and KL, HL, GC, FC be joined, since whole EDC, and the parts of the construction, equal ; therefore is equal to the remaining angle B the whole parts angle them the by BAL is equal to BAK, EDG are, the remaining angle KAL GDC : and because KA, AL are equal to GD, DC, and contain equal angles, the base KL is equal e to the base GC : and KH is equal to GF, so that LK, KH are equal to CG, GF, and they contain right angles ; therefore the base HL is equal to the base FC : again, because HA, AL are equal to FD, DC, and the base HL to the base FC, the angle HAL is equal f to the angle FDC : therefore, because the three plane angles BAL, BAH, HAL, which contain the solid angle at A, are equal to the three plane angles EDC, EDF, FDC, which contain the solid angle at D, each to each, and are situated in the same order, the solid angle at A is e- qual e to the solid angle at D. Therefore, at a given point in g B. Jl a given straight line, a solid angle has been made equal to a gi- ven solid angle contained by three plane angles. Which Avas tp be done. THE ELEMENTS PROP. XXVII. PROB. Nl D TO describe froril a given straight line a solid pa- rallelepiped similar and similarly situated to one given. Let AB be the g.ven straight line, and CD the given solid pa- rallelepiped ; it is rtquired from AB to describe a solid paral- lelepiped similar and sonilarly situated to CD. a 26. 11. At the point A of th«t given sti-aight line AB make» a solid angle equal to the solid a^gle at C, and let BAK, KAH, HAB be the three plane angles '^hich contain it, so that BAK be e- qual to the angle ECG, aiid KAH to GCF, and HAB to FCE : and as EC to CG, so mcAce ^ BA to AK ; and as GC to b 12. 6. CF, so make ^ K A to AH ; wherefore, ex (cquali c, as EC to c 22. 5. CF, so is BA to AH: complete the parallelogram BH, and the solid AL: and L because as EC to CG, so BA to AK, the sides about the equal angles ECG, BAK are propor- tionals : therefore the parallelogram BK is similar to EG. For the same reason, the parallelogram KH is similar to GF, and HB to FE. Wherefore, three parallelograms of the solid AL are similar to three of (he solid CD ; and the three opposite ones in eath solid d 24. 11. are equaH and similar to these, each to each. Also, because the plane angles which contain the solid angles of the figures are equal, each to each, and situated in the same order, the solid e B. 11. angles are equal *, each to each. Therefore the solid AL is f 11. def. similar f to the solid CD. Wherefore from a given straight 11- line AB a solid parallelepiped AL has been described, similar and similarly situated to the given one CD. Which was to be done. K«^ \ H \ \ \ \ ^ — K A B OF EUCLID. PROP. XXVIII. THEOR. IF a solid parallelepiped be cut by a plane passing See n> through the diagonals of two of the opposite planes ; it shall be cut in two equal parts. Let AB be a solid parallelepiped, and DE, CF the diagonals of the opposite parallelograms AH, GB, viz. those which are drawn betwixt the equal angles in each : and because CD, FE are each of them parallel to GA, and not in the same plane with it, CD, FE are parallel^; wherefore the diagonals CF, DE are a 9. 11. in the plane in which the parallels are, and are themselves parallels^: and the \^ ~ b 16. 11. plane CDEF shall cut the solid AB in- to tv/o equal parts. Because the triangle CGF is equal <^ 1 "" c 34. 1- to the triangle CBF, and the triangle DAE to DHE: and that the paral- lelogram CA is equaH and similar to L^Jtr— ■ !■ — JH d 24. 11. the opposite one BE ; and the paral- lelogram GE to CH : therefore the prism contained by the two triangles CGF, DAE, and the three parallelograms CA, GE, EC, is equal e e C 11. to the prism contained by the two triangles CBF, DHE, and the three parallelograms BE, CH, EC, because they are contained by the same number of equal and similar planes, alike situated, and none of their solid angles are contained by more than three plane angles. Therefore the solid AB is cut into two equal parti by the plane CDEF. Q. E. D. ' N. B. The insisting straight lines of a parallelepiped, men- ' tioned in the next and some folio wingpropositions, are the sides ' of the parallelograms betwixt the base and the opposite plane ' parallel to it.' c B o^ D F A / PROP. XXIX. THEOR. SOLID parallelepipeds upon the same base, and of See x the same altitude, the insisting straight lines of which are terminated in the same straight lines in the plane opposite to the base, are equal to one another. 230 THE ELEMENTS See the figures below. Book XI. Let the solid parallelepipeds AH, AK be upon the same base AB, and of the same altitude, and let their insisting straight lines AP\ AG, LM, LN be terminated in the same straight line FN, and CD, CE, BH, BK be terminated in the same straight line DK ; the solid AH is equal to the solid AK. First, let the parallelograms DG, UN, which are opposite to the base y\.B, have a common side HG : then, because the solid AH is cut by the plane AGHC passing tn rough the diagonals AG, CH of the opposite planes ALGF, CB'ilD, AH is cut into a 28. 11. two equal parts » by the plane AGHC : therefore the solid AH is double of the prism which is contained betwixt the triangles ALG, CBH: for the same rea- son, because the solid AK is cutby the plane LGHB through ^ the diagonals LG, BH of the op- ^ ^ ^ I ^ ^^ ^ N posite planes ALNG, CBKH, the solid AK is double of the same prism which is contain- ed betwixt the triangles ALCi, CBH. Therefore the solid AH ^ L is equal to the solid AK. But, let the parallelograms DM, EN opposite to the base, have no common side : then, because CH, CK are parallelo- grams, CB is equal i> to each of the opposite sides DH, EK ; wherefore DH is equal to EK : add, or take away the common part HE; then DE is equal to HK : wherefore also the tri- angle CDE is equal c to the triangle BHK : and the parallelo- gram DG is equal ti to the parallelogram HN : for the same reason, the triangle AFG is equal to the triangle LMX, and the parallelogram CF is equal e to the parallelogram BM, and D H K \F ^fV!/" r / 7' / ^ ^ / \ / 1 / b 34. 1. c 38. 1. d 36. 1. e 24- 11 D CG to BN ; for they are opposite, Therefore the prism which is contained by the two triangles AFG, CDE, and the three parallelograms AD, DG, GC, is equal f to the prism contain- t C 11. <-'d by tlie two triangles LMN% BHK, and the three parallelo- grams BM, IMK, KL. If therefore the prism Lr.INBHK bp OF EUCLID. 231 taken from the solid of which the base is the parallelogram AB, Book XI. and in which FDKN is the one opposite to it ; and if from this *«— v-^* same solid there be taken the prism AFGCDE, the remaining solid, viz. the parallelepiped AH, is equal to the remaining pa- rallelepiped AK. Therefore, solid parallelepipeds, 8cc. Q. E. D. PROF. XXX. THEOR. SOLID parallelepipeds upon the same base, and See n of the same altitude, the insisting straight lines of which are not terminated in the same straight lines in the plane opposite to the base, are equal to one another. Let the parallelepipeds CM, CN be upon the same base AB, and of the same altitude, but their insisting straight lines AF, AG, LM, LN, CD, CE, BH, BK not terminated in the same straight lines : the solids CM, CN are equal to one another. Produce FD, MH, and NG, KE, and let them meet one ano- ther in the points O, P, Q, R ; and join AO, LP, BQ, CR : and because the plane LBHM is parallel to the opposite plane ACDF, N K M H and that the plane LBHM is that in which are the parallels LB, MHPQ, in which also is the figure BLPQ ; and the plane ACDF is that in which are the parallels AC, FDOR, in which also is the figure C AOR ; therefore the figures BLPQ, C AOR are in parallel planes : in like manner, because the plane ALNG is pa- rallel CO the opposite plane CBKE, and that the plane ALNG is that in which are the parallels AL, OPGN, in which also is the 9,32 THE ELEMENTS Book XI. figure ALPO ; and the plane CBKE is that in which are the pa- '^— V— ' rallels CB, RQEK, in which also is the figure CBQR ; therefore the figures ALPO, CBQR are in parallel planes: and the planes ACBL, ORQP are parallel ; therefore the solid CP is a paralle- lepiped : but the solid CM, of which the base is ACBL, to which a 29. 11. FDHM is the opposite parallelogram, is equal ^ to the solid CP, of which the base is the parallelogram ACBL, to which ORQP A C is the one opposite, because they are apon the same base, and their insisting straight lines AF, AO, CD, CR ; LM, LP, BH, BQ are in the same straight lines FR, MQ : and the solid CP is equal » to the solid CN ; for they are upon the same base ACBL, and their insisting straight lines AO, AG, LP, LN ; CR, CE, BQ, BK are in the same straight lines OL, RK; therefore the solid CM is equal to the solid CN. Wherefore, solid parallelepi- peds, &c. Q. E. D. PROP. XXXL THEOR. SeeN. SOLID parallelepipeds which are upon equal bases, and of the same altitude, are equal to one another. Let the solid parallelepipeds AE, CF be upon equal bases AB, CD, and be of the same altitude ; the solid AE is equal to the so- lid CF. First, Let the insisting straight lines be at right angles to the bases AB, CD, and let the bases be placed in the same plane, OF EUCLID. 333 and so as that the sides CL, LB be in a straight line ; therefore Book XL the straight line LM, which is at right angles to the plane in ' — r— ' which the bases are, in the point L, is common » to the two so- a 13. 11. lids AE, CF : let the other insisting lines of the solids be AG, HK, BE; DF, OP, CN : and first let the angle ALB be equal to the angle CLD ; then AL, LD are in a straight linet. Pro- b 14.1. duce OD, HB, and let them meet in Q, and complete the solid parallelepiped LR, the base of which is the parallelogram LQ, and of which LM is one of its insisting straight lines : therefore, because the parallelogram AB is equal to CD, as the base AB is to the base LQ, so is <= the base CD to the same LQ : and be- c 7. 5. cause the solid parallelepiped AR is cut by the plane LMEB, which is parallel to the opposite planes AK, DR ; as the base AB is to the base LQ, so is d the solid AE to the solid LR : d 25. 11. for the same reason, because the solid parallelepiped CR is cut by the plane LMFD, which is parallel to the opposite planes R O "" NT \iVl \ E ^ ■v^V 7 ^ p-^ G K B K \ \ c L ^ h-, ^ CP, BR ; as the base CD to the base LQ, so is the solid CF to the solid LR : x r-^j | "^^^^T^ X but as the base AB '^ to the base LQ, so the base CD to the base LQ, as before was proved: there- fore as the solid AE a c H T to the solid LR, so ^ ^ " ^ is the solid CF to the solid LR ; and therefore the solid AE is equal e to the solid CF. * 9- ^• But let the solid parallelepipeds SE, QF be upon equal bases SB, CD, and be of the same altitude, and let their insisting straight lines be at right angles to the bases ; and place the bases SB, CD in the same plane, so that CL, LB be in a straight lincj and let the angles SLB, CLD be unequal ; the solid SE is also in this case equal to the solid CF : produce DL, TS until they meet in A, and from B draw BH parallel to DA ; and let HB, OD produced meet in Q, and complete the solids AE, LR : therefore the solid AE, of which the base is the parallelogram LE, and AK the one opposite to it, is equal*" to the solid SE, off 29. 11. which the base is LE, and to which SX is opposite ; for they are upon the same base LE, and of the same altitude, and their in- sisting straight lines, viz. LA, LS, BH, BT ; MG, MV, EK, EX are in the same straight lines AT, GX : and because the 2 G 234 THE ELEMENTS Book XI. parallelogram AB is equal k to SB, for they are upon the same <-. y — ^ base LB, and between the same parallels LB, AT ; and that the g 35. 1. base SB is equal to the base CD ; there- fore the base AB is equal to the base CD, and the angle ALB is equal to the angle O CLD: therefore, by the first case, the so- lid AE is equal to the solid CF ; but the so- lid AE is equal to the solid SE, as was demonstrated ; therefore the solid SE is equsl to the solid CF. But if the insisting straight lines AG, HK, BE, LM ; CN, RS, DF, op be not at right angles to the bases AB, CD ; in this case likewise, the solid AE is equal to the solid CF : from the points G, K, E, M ; N, S, F, P draw the straight lines GQ, hll. 11. KT, EV, MX; NY, SZ, FI, PU, perpendicular ^ to the plane in which are the bases AB, CD ; and let them meet it in the points Q, T, V, X ; Y, Z, I, U, and join QT, TV, VX, XQ ; YZ, ZI, lU, UY : then, because GQ, KT are at right angles to Q T C R Y Z i 6. 11. the same plane, they are parallel ' to one another : and MG, EK are parallels ; therefore the plane MQ, ET, of which one passes through MG, GQ, and the other through EK, KT, which are parallel to MG, GQ, and not in the same plane with them> kl5. 11. are parallel J' to one another: for the same reason, the planes MV, GT are parallel to one another : therefore the solid QE is a parallelepiped : in like manner, it may be proved, that the SO'- lid YF is a parallelepiped : but, from what has been domonstrat- ed, the solid EQ is equal to the solid FY, because they are upon equal bases MK, PS, and of the same altitude, and have their insisting straight lines at right angles lo the bases: and the so- OF EUCLID, 23-5 lid EQ is equal 1 to the solid AE ; and the solid FY, to the solid Book XI. CF : because they are upon the same bases and of the same alti- ^ — v ^^ tude : therefore the solid AE is equal to the solid CF. Where- 1 29. or fere, solid*t)arallelepipeds, &c. Q. E. D. 30. 11 PROP. XXXII. THEOR. SOLID parallelepipeds which have the same alti- See n. tude, are to one another as their bases. Let AB, CD be solid parallelepipeds of the same altitude : they are to one another as their bases ; that is, as the base AE to the base CF, so is the solid AB to the solid CD. To the straight line FG apply the parallelogram FH equal =i a Cor. 45. to AE, so that the angle FGH be equal to the angle LCG, and ^• complete the solid parallelepiped GK upon the base FH, one of whose insisting lines is FD, whereby the solids CD, GK must be of the same altitude: therefore the solid AB is equal'' to thebis. ll- solid GK, because they are upon e- B D K qual bases AE, FH, and are of the same altitude : and because the solid parallelepi- ped CK is cut by the plane DG, which is parallel to its opposite planes, the base HF is^ to the base FC, as the so- c 25. 11 lid HD to the solid DC : but the base HF is equal to the base AE, and the solid GK to the solid AB : therefore, as the base AE to the base CF, so is the solid AB to the solid CD. Wherefore, solid parallelepipeds, &c. Q. E. D. Cor. From this it is manifest that prisms upon triangular bases, of the same altitude, are to one another as their bases. Let the prisms, the bases of which are the triangles AEMj CFG, and NBO, PDQ the triangles opposite to them, have the same altitude ; and complete the parallelograms AE, CF, and the solid parallelepipeds AB, CD, in the first of which let MO, and in the other let GQ be one of the insisting lines. And be- cause the solid parallelepipeds AB, CD have the same altitude, they are to one another as the base AE is to the base CF j where- 236 THE ELEMENTS Book XI. fore the prisms, whicli are their halves^, are to one another as ^-"■v— ^ the base AE to the base CF ; that is, as the triangle AEM to the d 28. 11. triangle CFG. ^ PROP. XXXIII. THEOR. SIMILAR solid parallelepipeds are one to another in the triplicate ratio of their homologous sides. Let AB, CD be similar solid parallelepipeds, and the side AE homologous to the side CF : the solid AB has to the solid CD the triplicate ratio of that which AE has to CF. Produce AE, GE, HE, and in these produced take EK equal to CF, EL equal to FN, and EM equal to FR ; and complete the parallelogram KL, and the solid KG : because KE, EL are equal to CF, FN, and the angle KEL equal to the angle CFN, because it is equal to the ang;le AEG, which is equal to CFN, by reason that the solids AB, CD are similar; therefore the parallelogram KL is similar and equal to tlie parallelogram CN : for the same reason, the parallelogram MK is similar and equal to CR, and also OE to FD. -^ O Therefore three parallelograms of the solid KO are equal and similar to three parallelo- grams of the so- lid CD : and the three opposite ones in each solid are a24. 11. equal* and simi- lar to these: there- fore the solid KO bC. 11. is equally and si- milar to the solid CD \ \ N \ \ R h1\p A E M \^ ^ O cl.6. complete the parallelogram GK, and com- plete the solids EX, LP upon the bases GK, KL, so that EH be an insisting straight line in each of them, whereby they must be of the same altitude with the solid AB: and because the solids AB, CD are similar, and, by permutation, as AE is to CF, so is EG to FN, and so is EH to FR : and FC is equal to EK, and FN to EL, and FR to EM ; therefore as AE to EK, so is EG to EL, and so is HE to EM : but, as AE to EK, so^ is the paral- lelogram AG to the parallelogram GK i and as GE to EL, so is<= OF EUCLID. 237 GK to KL ; and as HE to EM, so « is PE to KM : therefore as Book XI. the parallelogram AG to the parallelogram GK, so is GK to KL, * — r— ' and PE t^vM : but as AG to GK, so^ is the solid AB to the c 1. 6. solid EX^nd as GK to KL, so* is the solid EX to the solid d 25. 11. PL ; and as PE to KM, so ^ is the solid PL to the solid KO ; and therefore as the solid AB to the solid EX, so is EX to PL, and PL to KO : but if four magnitudes be continual proportionals, the first is said to have to the fourth the triplicate ratio of that which it has to the second: therefore the solid AB has to the solid KO the triplicate ratio of that which AB has to EX: but as AB is to EX, so is the parallelogram AG to the parallelogram GK, and the straight line AE to the straight line EK. Where- fore the solid AB has to the solid KO, the triplicate ratio of that which AE has to EK. And the solid KO is equal to the solid CD, and the straight line EK is equal to the straight line CF. Therefore the solid AB has to the solid CD, the triplicate ratio of that which the side AE has to the homologous side CF, &c. Q. E. D. CoR. From this it is manifest, that, if four straight lines be continual proportionals, as the first is to the fourth, so is the so- lid parallelepiped described from the first to the similar solid similarly described from the second; because the first straight line has to the fourth the triplicate ratio of that which it has to the second. PROP. D. THEOR. SOLID parallelepipeds contained by parallelograms See n,. equiangular to one another, each to each, that is, of which the solid angles are equal, each to each, have . to one another the ratio which is the same with the ratio compounded of the ratios of their sides. Let AB, CD be solid parallelepipeds, of which AB is contain- ed by the parallelograms AE, AF, AG, equiangular, each to each, to the parallelograms CH, CK, CL, which contain the so- lid CD. The ratio which the solid AB has to the solid CD is the same with that which is compounded of the ratios of the aides AM to DL, AN to DK, and AO to DH. 238 THE ELEMENTS Book XI. Produce MA, NA, OA to P, Q, R, so that AP be equal to ^■^v^ - ^ DL, AQ to DK, and AR to DH ; and complete the solid paral- lelepiped AX contained by the parallelograms AS, AT, AV si- milar and equal to CH, CK, CL, each to each. TBw-efore the a C. 11. solid AX is equal * to the solid CD. Complete likewise the solid AY, the base of which is AS, and of which AO is one of its in- sisting straight lines. Take any straight line a, and as MA t» AP, so make a to b ; and as NA to AQ, so make b to c ; and as AO to AR, so c to d : then, because the parallelogram AE is equiangular to AS, AE is to AS, as the straight line a to c, as is demonstrated in the 23d prop, book 6, and the solids AB, AY, being betwixt the parallel planes BOY, EAS, are of the same b 32 11. altitude. Therefore the solid AB is to the solid AY, as ^ the base AE to the base AS ; that is, as the straight line a is to c. c 25. 11. And the solid AY is to the solid AX, as <= the base OQ is to the D L B li \ ^ K \ \ E c C P \" N \ \ P \ . \ M A Q R T \ \ V X base QR ; that is, as the straight line OA to AR ; that is, as the straight line c to the straight line d. And because the solid AB is to the solid AY, as a is to c, and the solid AY to the solid AX as c is to d ; ex ctquali^ the solid AB is to the solid AX, or CD which is equal to it, as the straight line a is to d. But 4 def. A. the ratio of a to d is said to be compounded "^ of the ratios of a to b, ^- b to c, and c to d, which are the same with the ratios of the sides MA to AP, NA to AQ, and OA to AR, each to each. And the sides AP, AQ, AR are equal to the sides DL, DK, DH, each to each. Therefore the solid AB has to the soHd CD the ratio which is the same with that which is compounded of the ratios of the sides AM to DL, AN to DK, and AO to DH. Q. E, D. OF EUCLID. PROP. XXXIV. THEOR. THE bases and altitudes of equal solid parallelepi- See n. peds are reciprocally proportional ; and, if the bases and altitudes be reciprocally proportional, the solid parallelepipeds are equal. Let AB, CD be equal solid parallelepipeds ; their bases are reciprocally proportional to their altitudes; that is, as the base EH is to the base NP, so is the altitude of the solid CD to the altitude of the solid AB. First, Let the insisting straight lines AG, EF, LB, HK ; CM, NX, OD, PR be at right angles to the bases. As the base EH to the base NP, so is CM to K B R D FT H M ,0 X N AG. If the base EH be equal to the base NP, then because the solid AB is like- wise equal to the solid CD, CM shall be equal to AG. Because if the bases EH, NP be equal, but the alti- tudes AG, CM be not equal, neither shall the solid AB be equal to the solid CD. But the solids are equal, by the hypo- thesis. Therefore the altitude CM is not unequal to the altitude AG ; that is, they are equal. Wherefore, as the base EH to the base NP, so is CM to AG. Next, Let the bases EH, NP not be equal, but EH greater than the other : since then the solid AB is equal to the solid CD, CM is therefore greater than AG ; for, if it be not, neither also in this case would the solids AB, CD be equal, which, by the hypothesis, are equal. Make then CT equal toH AG, and complete the solid parallelepiped CV, K B E 240 THE ELEMENTS Book XI. of which the base is NP, and altitude CT. Because the solid AB is equal to the solid CD, therefore the solid AB is to the so- lid CV, as a the solid CD to the solid CV. But as the solid AB to the solid CV, so'' is the base EH to the base NP"; for the so- lids AB, C'V are of the same altitude; and as the solid CD to C V, so c is the base MP to the base PT, and so ^ is the straight line MC to CT ; and CT is equal to AG. Therefore, as the base EH to the base NP, so is MC to AG. Wherefore the bases of the solid parallelepipeds AB, CD are reciprocally proportional to their altitudes. Let now the bases of the solid parallelepipeds AB, CD be re- ciprocally proportional to their altitudes, viz. as the base EH to the base NP, so the alti- c 25. 11. dl. 6. tude of the solid CD to the K altitude of the solid AB ; the solid AB is equal to the solid CD. Let the in- sisting lines be, as before, at right angles to the bases. Then, if the base EH be H equal to the base NP, since EH is to NP, as the alti- tude of the solid CD is to the B R \( ur \ L \ \ M D K E N is to tne altitude of the solid AB, there- e A. 5. fore the altitude of CD is equal « to the altitude of AB. But so- lid parallelepipeds upon equal bases, and of the same altitude, f 31. 11. are equal f to one another : therefore the solid AB is equal to the solid CD. But let the bases EH, NP be unequal, and let EH be the greater of the two. Therefore, since as the base EH to the base NP, so is CM the alti- tude of the solid CD to AG the altitude of AB, CM is greater e than B AG. Again, take CT K equal to AG, and com- plete, as before, the so- lid CV. And because the base EH is to the H base NP, as CM to AG, and that AG is equal to j^ C T, therefore the base EH is to the base NP, as MC to CT. But as -the base EH is to NP, so •> is the solid AB to the solid CV ; for the solids AB, CV are of the same altitude ; and as MC to CT, so is the base MP to the base G I. OF EUCLID. 241 PT, and the solid CD to the solid ^CV: and therefore as the BookXf, solid AB to the solid C V, so is the solid CD to the solid CV ; v—y—^ that is, each of the solids AB, CD has the same ratio to the solid c 25. 11. CV ; and therefore the solid AB is equal to the solid CD. Second general case. Let the insisting straight lines FE, BL, GA, KH; XN, DO, MC, RP not be at right angles to the bases of the solids ; and from the points F, B, K, G ; X, D, R, M draw perpendiculars to the planes in which are the bases EH, NP meeting those planes in the points S, Y, V, T ; Q, I, U, Z ; and complete the solids FV, XU, which are pa- rallelepipeds, as was proved in the last part of prop. 31, of this book. In this case, likewise, if the solids AB, CD be equal, their bases are reciprocally proportional to their altitudes, viz. the base EH to the base NP, as the altitude of the solid CD to the altitude of the solid AB. Because the solid AB is equal to the solid CD, and that the solid BT is equal s to the g 29. qr solid BA, for they are upon the same base FK, and of the 30. 11. same . altitude ; and that the solid DC is equal s to the solid DZ, being upon the same base XR,. and of the same altitude ; therefore the solid BT is equal to the solid DZ : but the bases are reciprocally proportional to the altitudes of equal solid pa- rallelepipeds of which the insisting straight lines are at right angles to their bases, as before was proved : therefore as the base FK to the base XR, so is the altitude of the solid DZ to the altitude of the solid BT : and the base FK is equal to the base EH, and the base XR to the base NP : wherefore as the base EH to the base NP, so is the altitude of the solid DZ to the altitude of the solid BT : but the altitudes of the solids DZ, DC, as also of the sohds BT, BA, are the same. There- fore as the base EH to the base NP, so is the altitude of the 2H 242 THE ELEMENTS Book XI. solid CD to the altitude of the solid AB ; that is, the bases of the ^-^■-^ solid parallelepipeds AB, CD are reciprocally proportional to their altitudes. Next, Let the bases of the solids AB, CD be reciprocally pro- portional to their altitudes, viz. the base EH to the base NP, as the altitude of the solid CD to the altitude of the solid AB ; the solid AB is equal to the solid CD : the same construction being made ; because, as the base EH to the base NP, so is the altitude of the solid CD to the altitude of the solid AB ; and that the base EH is equal to the base FK ; and NP to XR ; therefore the base FK is to the base XR, as the altitude of the solid CD to the altitude of Pv D g 29. o* 30. 11. AB : but the altitudes of the solids AB, BT are the same, as also of CD and DZ ; therefore as the base FK to the base XR, so is the altitude of the solid DZ to the altitude of the solid BT t wherefore the bases of the solids BT, DZ are reciprocally propor- tional to their altitudes ; and their insisting straight lines are at right angles to the bases ; wherefore, as was before proved, the solid BT is equal to the solid DZ : but BT is equal g to the solid BA, and DZ to the solid DC, because they are upon the same bases, and of the same altitude. 1 herefore the solid AB is equal to the solid CD. Q. E. D. OF EUCLID. PROP. XXXV. THEOR. IF from the vertices of two equal plane angles there See N. be drawn two straight lines elevated above the planes in which the angles are, and containing equal angles with the sides of those angles, each to each ; and if in the lines above the planes there be taken any points, and from them perpendiculars be drawn to the planes in which the first-named angles are; and from the points in which they meet the planes, straight lines be drawn to the vertices of the angles first- named ; these straight lines shall contain equal angles with the straight lines which are above the planes of the angles. Let BAC, EDF be two equal plane angles; and from the points A, D let the straight lines AG, DM be elevated above the planes of the angles, making equal angles with their sides, e£^ch to each, viz. the angle GAB equal to the angle MDE, and G/VC to MDF; and in AG, DM let any points G, M be ta- ken, and from them let perpendiculars GL, MN be drawn to the planes BAC, EDF meeting these planes in the points L, Nj ^nd join LA, ND ; the an^le GAL is equal to the angle MDN? D Make AH equal to DM, and through H draw HK parallel to GL: but GL is perpendicular to the plane BAC; where- fore HK is perpendicular » to the same plane : from the points * 8. H K, N, to the straight lines AB, AC, DE, DF, draw perpen- Uicula/s KB, KC, NE, NF; and join HB, BC, ME, EF; 244 THE ELEMENTS 11. d 3. 11. Book XI. because HK is perpendicular to the plane BAG, the plane » — v—i^ HBK which passes through HK is at right angles ^ to the plane bl8. 11. BAG; and AB is drawn in the plane BAG at right angles to the common section BK of the two planes ; therefore AB is c 4. def. perpendicular = to the plane HBK, and makes right angles ^ with every straight line meeting it in that plane: but BH meets ^sf- it in that plane ; therefore ABH is a right angle : for the same reason, DEM is a right angle, and is therefore equal to the angle ABH: and the angle HAB is equal to the angle MDE. Therefore in the two triangles HAB, MDE there are two angles in one equal to two angles in the other, each to each, and one side equal to one side, opposite to one of the equal angles in each, viz. HA equal to DM ; therefore the remaining sides are equal *, each to each : wherefore AB is equal to DE. In the same manner, if HG and MF be joined, it may be demon- strated that AG is equal to DF : therefore, since AB is equal to DE) BA and AC are equal to ED and DF ; and the angle c 25. 1. BAG is equal to the angle EDF; wherefore the base BC is f 4. 1. equal f to the base EF, and the remaining angles to the remain- ing angles : the angle ABG is therefore equal to the angle DliF: and the right angle ABK is equal to the right angle DEN, whence the remaining angle GBK is equal to the re- maining angle FEN : for the same reason, the angle BGK is equal to the angle EFN : therefore in the two triangles BGK, EFN, there are two angles in one equal to two angles in the other, each to each, and one side equal to one side adjacent to the equal angles in each, viz. BG equal to EF ; the other sides, tiierefore, are equal to the other sides; BK then is equal to EN : and AB is equal to DE ; wherefore AB, BK are equal 'to DE, EN; and they contain right angles; wherefore the base AK is equal to the base DN : and since AH is equal to OF EUCLID. 245 DM, the square of AH is equal to the square of DM: but theBookXI. squares of AK, KH are equal to the square s of AH, because ^^- y mJ AKH is a right angle: and the squares of DN, NM are equal g 47. 1. to the square of DM, for DNM is a right angle : wherefore the squares of AK, KH are equal to the squares of DN, NM ; and of those the square of AK is equal to the square of DN : therefore the remaining square of KH is equal to the remaining square of NM ; and the straight line KH to the straight line NM : and be- cause HA, AK are equal to MD, DN each to each, and the base HK to the base MN as has been proved ; therefore the angle HAK is equal h to the angle MDN. Q. E. D. h 8. 1. Cor. From this it is manifest, that if, from the vertices of two equal plane angles, there be elevated two equal straight lines containing equal angles with the sides of the angles, each to each ; the perpendiculars drawn from the extremities of the equal straight lines to the planes of the first angles are equal to one another. Another Demonstration of the Corollary, Let the plane angles B AC, EDF be equal to one another, and let AH, DM be two equal straight lines above the planes of the angles, containing equal angles with BA, AC ; ED, DF, each to each, viz. the angle HAB equal to MDE, and HAC equal to the angle MDF ; a'nd from H, M let HK, MN be perpendiculars to the planes BAC, EDF : HK is equal to MN. Because the solid angle at A is contained by the three plane angles BAC, BAH, HAC, which are, each to each, equal to the three plane angles EDF, EDM, MDF containing the solid angle at D ; the solid angles at A and D are equal : and therefore coincide with one another ; to wit, if the plane angle BAC be ap- plied to the plane angle EDF, the straight line AH coincides with DM, as was shown in prop. B of this book : and because AH is equal to DM, the point H coincides with the point M : wherefore HK which is perpendicular to the plane BAC coincides with MN i which is perpendicular to the plane EDF, because these i 13. 11. planes coincide with one another : therefore HK is equal to MN. Q. E. D. 246 Book XI. THE ELEMENTS PROP. XXXVI. THEOR. See N. IF three straight lines be proportionals, the solid parallelepiped described from all three as its sides, is equal to the equilateral parallelepiped described from the mean proportional, one of the solid angles of which is contained by three plane angles equal, each to each, to the three plane angles containing one of the solid angles of the other figure. Let A, P, C be three proportionals, viz. A to B, as B to C. The solid described from A, B, C is equal to the equilateral so- lid described from B, equiangular to the other. Take a solid angle D contained by three plane angles EDF, FDG, GDE; and make each of the straight lines ED, DF, DG equal to B, and complete the solid parallelepiped DH ; make LK equal to A, and at the point K in the straight line a. 26. 11. LK make a a solid angle contained by the three plane angles LKM, MKN, NKL equal to the angles EDF, FDG, GDE, O H M ^ M^ ^ E D B b 14. 6. rach to each ; and make KN equal to B, and KM equal to C ; and complete the solid parallelepiped KO : and because, a^ A is to B, so is B to C, and that A is equal to LK, and B to each of the straight lines DE, DF, and C to KM; there- fore LK is to ED, as DF to KM ; that is, the sides about the equal angles are reciprocally proportional; therefore the pa- rallelogram LM is equal •> to EF : and because EDF, LKM are two equal plane angles, and the two equal straight lines DG, KN are drawn from their vertices above their planes, and con- tain equal angles with their sides; therefore the perpendicu- >ars from the points G, N, to the planes EDF, LKM, are OF EUCLID. 247 equal <= to one another: therefore the solids KO, DH are of the Book XI. same altitude ; and they are upon equal bases LM, EF ; and "^ — r— ' therefore they are equal ^ to one another: but the solid KO is c Cor. 35. described from the three straight lines A, B, C, and the solid DH H- from the straight line B. If, therefore, three straight lines, £cc. d 31. 11. Q. E. D. PROP. XXXVII. THEOR. IF four straight lines be proportionals, the similar See n. solid parallelepipeds similarly described from them shall also be proportionals. And if the similar paral- lelepipeds similarly described from four straight lines be proportionals, the straight lines shall be propor- tionals. Let the four straight lines AB, CD, EF, GH be proporti'^nals, viz. as AB to CD, so EF to GH ; and let the similar parallelepi- peds AK, CL, EM, GN be similarly described from them. AK is to CL, as EM to GN. Make » AB, CD, O, P continual proportionals, as also EF, GH, a 11. 6. Q, R : and because as AB is to CD, so EF to GH ; and that CD \ ^ \ \ fv^^ A \ \rAl B C I> M ■^^-,^^^ \ ^^ N ^ E T R F G H Q isbto O, as GH to Q, and O to P, as Q to R ; therefore, e^b 11.5. a(/uali c, AB is to P, as EF to R : but as AB to P, so d is the solid c 22. 5. AK to the solid CL ; and as EF to R, so ^ is the solid EM to the solid GN : therefore •> as the solid AK to the solid CL, so is the dCor.JcJ. solid EM to the solid GN. 248 THE ELEMENTS Book XI. But let the solid AK he to the solid CL, as the solid EM to the »— v^-' solid GN : the straight line AB is to CD, as EF to GH. e 27. 11. Take AB to CD, as EF to ST, and from ST describe « a solid parallelepiped SV similar and similarly situated to either of the solids EM, GN: and because AB is to CD, as EF to ST ; and that from AB, CD the solid parallelepipeds AK, CL are similarly described; and, in like manner, the solids EM, SV from the straight lines EF, S L ; therefore AK is to CL, as EM to SV : f9. 5. \ Nf V \1 \ A. 3 C o -V N fx H Q R but, by the hypothesis, AK is to CL, as EM to GN : therefore GN is equal f to SV : but it is likewise similar and similarly situ- ated to SV ; therefore the planes which contain the solids GN, SV are similar and equal, and their homologous sides GH, ST equal to one another : and because as AB to CD, so EF to ST, and that ST is equal to GH ; AB is to CD, as EF to GH. There- fore, if four straight linos. Sec. Q. E. D. PROP. XXXVHL THEOR. SeeN. " IF a plane be perpendicular to another plane, and " a straight line be drawn from a point in one of the " planes perpendicular to the other plane, this straight *' line shall fall on the common section of the planes." " Let the plane CD be perpendicular to the plane AB, and let « AD be their common section ; if any point E be taken in the " plane CD, the perpendicular drawn from E to the plane AB « shall fall on AD. - / OV EUCLID. 249 11. '•■ For, if it does not, let it, if possible, fall elsewhere, as EF ; Book XL ••• and let it meet the plane AB in the point F ; and from F draw », ^-— v— ' " in the plane AB, a perpendicular FG to DA, which is also per- a 12. 1. " pendicular'' to the plane CD ; and join EG: then because FG b 4. def. " is perpendicular to the plane " CD, and the straight line EG, ( " which is in that plane, meets it ; " therefore FGE is a right an- " gle«: but EF is also at right " angles to the plane AB ; and " therefore EFG is a right angle : u " wherefore two of the angles of " the triangle EFG are equal to- " gether to two right angles : " which is absurd : tnerefore the perpendicular from the point " E to the plane AB, does not fall elsewhere than upon the " straight line AD : it therefore falls upon it. If therefore a « plane," S^c. Q. E. D. PROP. XXXIX. THEOR. IN a solid parallelepiped, if the sides of two of the See N. opposite planes be divided each into two equal parts, the common section of the planes passing through the points of division, and the diameter of the solid paral- lelepiped cut each other into two equal parts. Let the sides of the opposite planes CF, AH of the solid parallelepiped AF, be divided each into two equal parts in the points K, L, M, N; X, O, P, R; and join KL, MN, XO, PR : and be- cause DK, CL are equal and parallel, KL is parallel * to DC : for the same reason, MN is pa- rallel to BA : and BA is parallel to » 33. 1, 2 I 250 Book XI b 9. 11. c 29. 1. d4. 1. e li 1. 9, 33. !• f 15 1. 6 26.1. THE ELEMENTS DC ; therefore, because KL, BA are each of them parallel to DC, and not in the same plane with it, KL is parallel *> to BA : and because KL, MN are each of them parallel to BA, and not in the same plane with it, KL is parallel^ to MN ; wherefore KL, MN are in one plane. In like manner, it may be proved, that XO, PR are in one plane. Let YS be the common section of the planes KN, XR ; and DG the diameter of the solid paral- lelepiped AF : YS and DG do meet, and cut one another into two equal parts. Join DY, YE, BS, SG. Because DX is parallel to OE, the alternate ano;!es DXY, YOE are equal*: to one another: and be- cause DX is equal to OE, and XY to D K F YO, and contain equal angles, the base DY is equal ^ to the base YE, and the other angles are equal ; therefore the angle XYD is equal to the angle OYE, and DYE is a straight f^ line : foi* the same reason BSG is a straight B line, and BS equal to SG : and because CAis equal and pa- rallel to DB, and al- so ecjual and paral- lel to EG ; therefore DB is equal and parallel'' to EG : and DE, BG join their extremities; therefore DE is equal and parallel » to BG : and DG, YS are drawn from points in the one to points in the other ; and are therefore in one plane : whence it is mani- fest, that DG, YS must meet one another ; let them meet in T ; and because DE is parallel to BG, the alternate angles EDT, BGT are equal «; and the angle DTY is equal ^ to the angle GTS: therefore in the triangles DTY, GTS there are two an- ples in the one equal to two angles in the other, and one side equal to one side, opposite to two of the equal angles, viz. DY to GS; for they are the halves of DE, BG : therefore the re- maining sides are equals, each to each. Wherefore DT is equal to TG, and YT equal to TS. Vv^herefore, if in a solid, 8cc. ^' E- I^' OF EUCLID. PROP. XL. THEOR. IF there be two triangular prisms of the same al- titude, the base of one of which is a parallelogram, and the base of the other a triangle ; if the parallelo- gram be double of the triangle, the prisms shall be equal to one another. Let the prisms ABCDEF, GHKLMN be of the same altitude, the first whereof is contnined by the two triangles ABE, CDF and the three parallelograms AD, DE, EC; and the other by the two triangles GHK, LMN and the three parallelograms LHj HN, NG ; and let one of them have a parallelogram AF, and the other a triangle GHK for its base ; if the parallelogram AF be double of the triangle GHK, the prism ABCDEF is equal to the prism GHKLMN. Complete the solids AX, GO ; and because the parallelogram AF is double of the triangle GHK ; and the parallelogram HK double a of the same triangle ; therefore the parallelogram AFa 34. 1. is equal to HK. But solid parallelepipeds upon equal bases, and of the same altitude, are equal ^ to one another. Therefore theb31.il. solid AX is equal to the solid GO ; and the prism ABCDEF is half c of the solid AX; and the prism GHKLMN half <= of the c 28. 11. solid GO. Therefore the prism ABCDEF is equal to the prism GHKLMN. Wherefore, if there be two, &c. Q. E. D. THE ELEMENTS OF EUCLID. BOOK XII. LEMMA I. B. XII. Which is the first proposition of the tenth book, and is necessary *— -v"*^ to some of the propositions of this book. See N. IF from the greater of two unequal magnitudes there be taken more than its half, and from the re- mainder more than its half, and so on, there shall at length remain a magnitude less than the least of the proposed magnitudes. D K— H— Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken more than its half, and from the remainder more than its half, and so on, there shall at length remain a magnitude less than C. For C may be multiplied so as at length to become greater than AB. Let it be so multi- plied, and let DE its multiple be greater than AB, and let DE be divided into DF, FG, GE, each equal to C. From AB take BH greater than its half, and from the remainder AH take HK greater than its half, and so on, until there be as many divisions in AB as there are in DE: and let the divisions in AB be AK, KII, HB ; and the divisions in DE be DF, FG, GE. And because DE is greater than AB, and B C K F' G— THE ELEMENTS, See. 252 that EG taken from DE is not greater than its half, but BH ta- B. XII. ken from AB is greater than its half; therefore the remainder ^— -y^*^ GD is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than the half of GD, but HK is greater than the half of HA ; therefore the re- mainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK ; that is, AK is less than C. Q. E. D. And if only the halves be taken away, the same thing may in the same way be demonstrated. PROP. I. THEOR. SIMILAR polygons inscribed in circles are JLo one another as the squares of their diameters. Let ABCDE, FGHKL be two circles, and in them the si- milar polygons ABCDE, FGHKL ; and let BM, GN be the diameters of the circles ; as the square of BM is to the square of GN, so is the polygon ABCDE to the polygon FGHKL. Join BE, AM, GL, FN : and because the polygon ABCDE is similar to the polygon FGHKL, and similar polygons are divided into similar triangles; the triangles ABE, FGL are similar and A F D equiangular'' ; and therefore the angle AEB is equal to the angle b 6. 6, FLG : but AEB is equal <= to AMB, because they stand up- c 21. 3, on the same circumference ; and the angle FLG is, for the same reason, equal to the angle FNG ; therefore also the angle AMB is equal to FNG ; and the right angle BAM is equal to the right d angle GFN ; wherefore the remaining angles in the tri-d 51- 3, angles ABM, FGN are equal, and they are equiangulr.r to one 254 THE ELEMENTS B. XII. another : therefore as BM to GN, so e is BA to GF ; and there- ^- " ->r— -^ fore the duplicate ratio of BM to GN is the same ^ with the du- e 4 6. plicate ratio of B A to GF : but the ratio of the square of BM to f 10. def. the square of GN, is the duplicate e ratio of that which BM has 5. & 22.5. to GN ; and the ratio of the polys^on ABODE to the polygon g20. 6. FGHKL is the duplicate sof that which BA has to GF : there- fore as the square of BM to the square of GN, so is the polygon ABCDE to the polygon FGHKL. Wherefore similar polygons, &c. Q. E. D. PROP. n. THEOR. See N. CIRCLES are to one another as the squares of their dicimeters. Let A BCD, EFGH be two circles, and BD, FH their diame- ters : as the square of BD to the square of FH, so is the circle ABCD to the circle EFGH. For, if it be not so, the square of BD shall be to the square of FH, as the circle ABCD is to some space either less than the circle EFGH, or greater than it*. First, let it be to a space S less than the circle EFGH ; and in the circle EFGH describe the square EFGH : this square is greater than half of the circle EFGH ; because if, through the points E, F, G, H, there be drawn tangents to the circle, the • For there is some square equal to the circle ABCD : let P be the side of it, and to three straight lines BD, FH,- and P, there can be a fourth propor- tional ; let this be Qj therefore the squares of these four straight lines are proportionals : that is, to the squares of BD, FH and the circle ABCD, it is possible there may be a fourth proportional. Let this be S. And in like man- ner are to be understood .some things in some of the following propositions. OF EUCLID. 255 square EFGH is half* of the square described about the circle ; B. XII. and the circle is less than the square described about it; there- ^-^-v-i-^ fore the square EFGH is greater than half of the circle. Divide a 41. 1. the circumferences EF, FG, GH, HE each into two equal parts in the points K, L, M, N, and join EK, KF, FL, LG, GM, WH, HN, NE: therefore each of the triangles EKF, FLG, GMH, HNE is greater than half of the segment of the circle it stands in ; because, if straight lines touching the circle be drawn through the points K, L, M, N, and parallelograms upon the straight lines EF, FG, GH, HE be completed ; each of the triangles EKF, FLG, GMH, HNE shall be the half a of the parallelogram in which it is: but every segment is less than the parallelogram in which it is: wherefore each of the triangles EKF, FLG, GMH, HNE is greater than half the segment of the circle which con- tains it: and if these circumferences before named be divided each into two equal parts, and their extremities be joined by straight lines, by continuing to do this, there will at length re- main segments of the circle which, together, shall be less than the excess of the circle EFGH above the space S: because, by the preceding lemma, if from the greater of two unequal raag^ nitudes there be taken more than its half, and from the rem in- der more than its half, and so on, there shall at length remair. a magnitude less than the least of the proposed magnitudes. Let then the segments EK, KF, FL, LG, GM, MH, HN, NF be those that remain and are together less tl)an the excess of the circle EFGH above S: therefore the rest of the circle, viz. the polygon EKFLGMHN, is greater than the space S. Describe likewise in the circle AliCD the polygon AXBOCPDR similar to the polygon EKFLGMHN : as, therefore, the square of BD is to the square of FH, so ^ is the polygon AXBOCPDR to the b 1. 12: polygon EKFLGMHN: but the square of BD is also to the square of FII, a-, the circle ABCD i§ to the space S: thersfove *56 THE ELEMENTS B. Xll. as the circle ABCD is to the space S, so is *= the polygon AXBOCPDR to the polygon EKFLGMHN: but the circle ABCD is greater than the polygon contained in it : wherefore, the space S is greater ^ than the polygon EKFLGMHN: but it is likewise less, as has been demonstrated ; which is impossible. Therefore the square of BD is not to the square of FH as the circle ABCD is to any space less than the circle EFGH. In the same manner jt may be demonstrated, that neither is the square of FH to the square of BD, as the circle EFGH is to any space less than the circle ABCD. Nor is the square of BD to the square of FH, as the circle ABCD is to any space greater than the circle EFGH : for, if possible, let it be so to T,-a space great^ er than the circle EFGH : therefore, inversely, as the square of FH to the square of BD, so is the space T to the circle ABCD. But as the space f T is to the circle ABCD, so is the circle EFGH to some space, which must be less «* than the circle ABCD, because the space T is greater, by hypothesis, than the circle EFGH. Therefore, as the square of FH is to the square t tot, as in the foregoing- note, at *, it was explained how it was possible there could be a fourth proportional to the squares of BD, FH, and the circle ABCD, which was named S. So, in like manner, there can be a fourth pro- portional to this other space, named T, and the circles ABCD, EFGH. And the like is t^ be understood in some of the following propositions. OF EUCLID. 257 of BD, so is the circle EFGH to a space less than the circle B. XII. ABCD, which has been demonstrated to be impossible : there- ^— y-^J fore the square of BD is not to the square of FH as the circle ABCD is to any space greater than the cii'cle EFGH: and it has been demonstrated, that neither is the square of BD to the square of FH, as the circle ABCD to any space less than the circle EFGH : wherefore, as the square of BD to the square of FH, so is the circle ABCD to the circle EFGH^. Circles, therefore, are, kc. Q. E. D. PROP. III. THEOR. EVERY pyramid having a triangular base may be See n. divided into two equal and similar pyramids having triangular bases, and which are similar to the whole pyramid, and into two equal prisms which together are greater than half of the whole pyramid. Let there be a pyramid of which the base is the triangle ABC, and its vertex the point D : the pyramid ABCD may be divided into two equal and similar pyramids having triangular bases, and similar to the whole ; and into two equal prisms which together af-e greater than half of the whole pyramid. Divide AB, BC, CA, AD, DB, DC, each into two equal parts in the points E, F, G, H, K, L, and join FH, EG, GH, HK, KL, LH, EK, KF, FG. Because AE is equal to EB, and AH to HD, HE is parallel » to DB: for the same reason, HK is parallel to AB : therefore HEBK is a parallelogram, and HK equal ^ to EB : but EB is equal to AE ; there- fore also AE is equal to HK : and AH is equal to HD ; wherefore EA, AH are equal to KH, HD, each to each ; and the angle EAH is equal c to the angle KHD ; therefore the base EH is equal to the base KD, and the triangle B a 2. 6. b 34. 1. c29.1. :j: Because as a fourth proportional to the squares of BD, FH and the circle ABCD is possible, and that it can neither be less nor greater than the circle F'FGH, it must be equal to it. if 2 K 25S THE ELEMENTS B. XII. AEH equal <* and similar to the triangle HKD : for the same reason, the triangle AGH is equal and similar to the triangle HLD : and because the two straight lines EH, HG which meet one another are parallel to KD, DL that meet one another, and are not in the same plane with them, they contain equals angles; therefore the angle EHG is equal to the angle KDL. Again, because EH, HG are equal to KD, DL, each to each, and the angle EHG equal to the angle KDL; therefore the base EG is equal to the base KL : and the triangle EHG equal ^ and similar to the triangle KDL : for the same reason, the triangle AEG is also equal and similar to the triangle HKL. Therefore the py- ramid of which the base is the triangle AEG, and of which the f C. II. vertex is the point H, is equal f and similar to the pyramid the base of which is the triangle HKL, and ver- tex the point D : and because HK is parallel to AB a side of the triangle ADB, the tri- angle ADB is equiangular to the triangle g 4. 6. HDK, and their sides are proportionals e : therefore the triangle ADB is similar to the triangle HDK : and for the same rea- son, the triangle DBG is similar to the tri- angle DKL ; and the triangle ADC to the triangle HDL ; and also the triangle ABC to the triangle AEG : but the triangle AEG is similar to the triangle HKL, as before was proved ; therefore the triangle ABC is b 21. 6. similar ^ to the triangle HKL. And the pyramid of which the base is the triangle ABC, and vertex the point D, is therefore i B. 11. & similar » to the pyramid of which the base Jl. def. is the triangle HKL, and vertex the same point D : but the py- 11- ramid of which the base is the triangle HKL, and vertex the point D, is similar, as has been proved, to the pyramid the base of which is the triaui^le AEG, and vertex the point H: where- fore the pyramid the base of which is the triangle ABC, and ver- tex the point D, is similar to the pyramid of which the base is the triangle AEG, and vertex H : therefore each of the pyramids AEGH, HKLD is similar to the whole pyramid ABCD : and be- k 41. 1 cause BE is equal to EC, the parallelogram EBEG is double ^ of the triangle GEC: but when there are two prisms of the same alti- tude, of which one has a parallelogram for its base, and the other a triangle that is half of the parallelogram, these prisms are equal a 40. 11. * to one another; thei'efore the prism having the parallelogram EBIG for its base, and the straight line KH opposite to it, is equal to the prism having the triangle GFC for its base, and tVse triangle HKL opposite to it ; for they are of the same aiti- OF EUCLID. 559 tude, because they are between the parallel l' planes ABC, HKL: B. Xlt. and it is manifest that each of these prisms is greater than either ^'-r-i«i of the pyramids of which the triangles AEG, HKL are the b 15. 11. bases, and the vertices the points H, D ; because, if EF be join- ed, the prism having the parallelogram EBFG for its base, and KH the straight line opposite to it, is greater than the pyramid of which the base is the triangle EBF, and vertex the point K ; but this pyramid is equal ^ to the pyramid the base of which is C C. It the triangle AEG, and vertex the point H ; because they are contained by equal and similar planes : wherefore the prism having the parallelogram EBi^^G for its base, and opposite side KH, is greater than the pyramid of which the base is the trian- gle AEG, and vertex the point H : and the prism of which the base is the parallelogram EBFG, and opposite side KH, is equal to the prism having the triangle GFC for its base, and HKL the triangle opposite to it: and the pyramid of which the base is the triangle AEG, and vertex H, is equal to the pyramid of which the base is the triangle HKL, and vertex D : therefore the two prisms before mentioned are greater than the two pyramids of which the bases are the triangles AEG, HKL, and vertices the points H, D. Therefore the whole pyramid of which the base is the triangle ABC, and vertex the point D, is divided into two equal pyramids similar to one another, and to the whole pyra- mid ; and into two equal prisms; and the two prisms are toge» ther greater than half of the whole pyramid. Q. E. D. THE ELEMENTS PROP. IV. THEOR. See Jf. IF there be two pyramids of the same altitude, up- on triangular bases, and each of them be divided in- to two equal pyramids similar to the whole pyramid, and also into two equal prisms ; and if each of these pyramids be divided in the same manner as the first two, and so on ; as the base of one of the first two pyramids is to the base of the other, so shall all the prisms in one of them be to all the prisms in-the other, that are produced by the same number of divisions. Let there be two pyramids of the same altitude upon the tri- angular bases ABC, DEF, and having their vertices in the points G, H ; and let each of them be divided into two equal pyramids similar to the whole, and into two equal prisms ; and let each of the pyramids thus made be conceived to be divided in the like manner, and so on ; as the base ABC is to the base DEF, so are all the prisms in the pyramid ABCG to all the prisms in the pyramid DEFH made by the same number of di- visions. Make the same construction as in the foi'egoing proposition : and because BX is equal to XC, and AL to LC, therefore XL a 2. 6. is parallel » to AB, and the triangle ABC similar to the tri- angle LXC : for the same reason, the triangle DEF is similar to RVF : and because BC is double of CX, and EF double of FV, therefore BC is to CX, as EF to FV : and upon BC, CX are described the similar and similarly situated rectilineal figures ABC, LXC ; and upon EF, FV, in like manner, are described the similar figures DEF, RVF : therefore, as the tri- b 22. 6. angle ABC is to the triangle LXC, so ^ is the triangle DEF to the triangle RVF, and, by permutation, as the triangle ABC to the triangle DEF, so is the triangle LXC to the triangle RV^F : and because the planes ABC, OMN, as also the planes cl5. 11. DEF, STY, are parallel <=, the perpendiculars drawn from the points G, H to the bases ABC, DEF, which, by the hypothe- sis, are equal to one another, shall be cut each into two equal d 17. 11. ^ parts by the planes OMN, STY, because the straight lines GC, HF are cut into two equal parts in the points N, Y by the same planes: therefore the prisms LXCOMN, RVFSTY are of the same altitude ; and, therefore, as the base LXC to OF EUCLID. 261 the base RVF; that is, as the triangle ABC to the triangle B. XII. DEF, so a is the prism having the triangle LXC for its base, ^ — v— -^ and OMN the triangle opposite to it, to the prism of which the a Cor. 32- base is the triangle RVF, and the opposite triangle STY: and ^^• because the two prisms in the pyramid ABCG are equal to one another, and also the two prisms in the pyramid DEFH equal to one another, as the prism of which the base is the pa- rallelogram KBXL and opposite side MO, to the prism having the triangle LXC for its base, and OMN the triangle opposite to it, so is the prism of which the base •> is the parallelogram b 7. 5. PEVR, and opposite side TS, to the prism of which the base is the triangle RVF, and opposite triangle STY. Therefore, componendo, as the prisms KBXLMO, LXCOMN together G H are unto the prism LXCOMN, so are the prisms PEVRTS, RVFSTY to the prism RVFSTY; and, permutando, as the prisms KBXLMO, LXCOMN are to the prisms PEVRTS, RVFSTY, so is the prism LXCOMN to the prism RVFSTY: but as the prism LXCOMN to the prism RVFSTY. so is, as has been proved, the base ABC to the base DEF: therefore, as the base ABC to the base DEF, so are the two prisms in the pyramid ABCG to the two prisms in the pyramid DEFH : and likeAvise if the pyramids now made, for example, the two OMNG, STYH, be divided in the same manner ; as the base OMN is to the base STY, so shallthe two prisms in the pyramid OMNG be to the two prisms in the pyramid STYH : but the base OMN is to the base STY, as the base ABC to the base DEF ; there- fore: as the base ABC to the base DEF, so are the two prisms 263 THE ELEMENTS B'. XII. in the pyramid ABCG to the two prisms in the pyramid DF to b 34. II. one another. Therefore the solid parallelepiped BGML is equal to the solid parallelepiped EHPO. And the pyramid ABCG is the sixth part of the solid BGML, and the pyramid DEFH is the sixth part of the solid EHPO. Therefore the pyramid ABCG is equal to the pyramid DEFH. Therefore the bases, £cc. Q. E. D. PROP. X. THEOR. EVERY cone is the third part of a cylinder which has the same base, and is of an equal altitude with it. Let a cone have the same base with a cylinder, viz. the circle ABCD, and the same altitude. The cone is the third part of the cylinder ; that is, the cylinder is triple of the cone. If the cylinder be not triple of the cone, it mnst either be greater than the triple, or less than it. First, Let it be greater than the triple ; and describe the square ABCD in the circle ; this square is greater than the half of the circle ABCD*. * As was shown in prop. 2- of this book. sro THE ELEMENTS B. XIl. Upon the square ABCD erect a prism of the same dltitude with ^•^y-*"^ the cylinder; this prism is p^reater than half of the cylinder; because if a square be described about the circle, and a prism erected upon the square, of the same altitude with the cylinder, the inscribed square is half of that circumscribed; and upon these square bases are erected solid parallelepipeds, viz, the prisms of the same altitude ; therefore the prism upon the square ABCD is the half of the prism upon the square descri- bed about the circle : because they are to one another as their a 32. 11. bases ^ .- and the cylinder is less than the prism upon the square described about the circle ABCD : therefore the prism upon the square ABCD of the same altitude with the cylinder, is greater than half of the cylinder. Bisect the circumferences AB, BC, CD, DA, in the points E, F, G, H ; and join AE, EB, BF, FC, CG, GD, DH, HA : then, each of the triangles AEB, BFC, CGD, DHA is greater than the half of the segment of the circle in which it stands, as was ^ shown in prop 2. of this book. Erect prisms upon each of these triangles of the same altitude with the cylinder ; each of these prisms is greater than half of the segment of the cylinder in which it is : be- cause if, through the points E, F, G, H, parallels be drawn to AB, BC, CD, DA, and parallelograms be completed upon the same AB, BC, CD, DA, and solid parallelepipeds be erected upon the parallelograms ; the prisms upon the triangles AEB, BFC, CGD, DIL'V arc the halves of the solid b 2 Cor. parallelepipeds ^. And the segments of the cylinder Avhich are 7- 12. upon the segments of the circle cut off by AB, BC, CD, DA, are less than the solid parallek])ipeds which contain them. Therefore the prisms upon the triangles AEB, BFC, CGD, DHA, are greater than half of the segments of the cylinder in which they are; therefore, if each of the circumferences be di- vided into two equal parts, and straight lines be drawn from the points of division to the extremities of the circumferences, and upon the triangles thus made, prisms be erected of the same c Lem- altitude with the cylinder, and so on, there must at length re- mu.. main some segments of the cylinder which together are less <=■ than the excess of the cylinder above the triple of the cone. Let ihem be those upon the segments of tlic circle AE, EB, ]J1', OF EUCLID. 271 FC, CG, GD, DH, HA. Therefore the rest of the cylinder, that B. XII. is, the prism of which the base is the polys^on AEBFCGDH, and ^ —^.mJ of which the altitude is the same with that of the cylinder, is greater than the triple of the cone : but this prism is triple ^ of d 1. Cor. the pyramid upon the same base, of which the vertex is the same 7- 12- with the vertex of the cone ; therefore the pyramid upon the base AEBFCGDH, having the same vertex with the cone, is greater than the cone, of which the base is the circle ABCD : but it is also less, for the pyramid is contained within the cone ; which is impossible. Nor can the cylinder be less than the triple of the cone. Let it be less, if possible : therefore, inversely, the cone is greater than the third part of the cylinder. In the circle ABCD describe a square ; this square is greater than the half of the circle : and upon the square ABCD erect a pyramid having the same vertex with the cone ; this pyramid is greater than the half of the cone ; because, as was before demonstrated, if a square be described about the circle, the square ABCD is the half of it; and if, upon these squares, there be erect- ed solid parallelepipeds of the same altitude with the cone, which are also prisms, the prism upon the square ABCD shall be the half of that which is upon the square described about the circle ; ior they are to one ano- ther as their bases* ; as are also the H- ! / a 32. 11. third parts of them : therefore the pyramid, the base of which is the square ABCD, is half of the pyramid upon the square described about the' circle : but this last pyramid is greater than the cone which it contains ; therefore the pyramid upon the square ABCD, having the same vertex with the cone, is greater than the half of the cone. Bisect the circumferences AB, BC, CD, DA in the points E, F, G, H, and join AE, EB, BF, FC, CG, GD, DH, HA: therefore each of the triangles AEB, BFC, CGD, DHA is greater than half of the segment of the cir- cle in which it is: upon each of these triangles erect pyramids having the same vertex with the cone. Therefore each of these pyramids is greater than the half of the segment of the cone in Avhich it is, as before was demonstrated of the prisms and seg- ments of the cylinder ; and thus dividing each of the circumfe- rences into two equal parts, and joining the points of division and their extremities by straight lines, and upon the triangles erect- ing pyramids having their vertices the same with that of the cone, ?ind so on, there must at length remain some segments of the fone, which together shall be less than the excess of the cone 272 THE ELEMENTS H B. XII. above the third part of the cyHnder. Let these be the segments V— V— ' upon AE, EB, BF, FC, CG, GD, DH, HA. Therefore the rest of the cone, that is, the pyramid, of which the base is the polygon AEBFCGDH, and of which the ver- tex is the same with that of the cone, is greater than the third part of the cylinder. But this pyramid is the third part of the prism upon the same base AEBFCGDH, and of the same altitude with the cylinder. Therefore this prism is greater than the cylinder of which the base is the circle ABCD. But it is also less, for it is contained within the cylinder ; which is impos- sible. Therefore the cylinder is not less than the triple of the cone. And it has been demonstrated that neither is it greater than the triple. Therefore the cylinder is triple of the cone, or, the cone is the third part of the cylinder. Wherefore, eveiy ' one, Sec. Q. E. D. PROP. XL THEOR. See N. CONES and cylinders of the same altitude are to one another as their bases. Let the cones and cylinders, of which the bases are the circles ABCD, EFGH, and the axes KL, MN, and AC, EG the diame- ters of their bases, be of the same altitude. As the circle ABCD to the circle EFGH, so is the cone AL to the cone EN. If it be not so, let the circle ABCD be to the circle EFGH, as the cone AL to some solid either less than the cone EN, or greater than it. First, let it be to a solid less than EN, viz. to the solid X ; and let Z be the solid which is equal to the ex- cess of the cone EN above the solid X; therefore the cone EN is equal to the solids X, Z together. In the circle EFGH de- scribe tlie square EFGFI, therefore this square is greater than tlie half of the circle : upon the square EFGH erect a pyra- mid of ihe same altitude with tlie cone ; this pyramid is greater than half of the cone. For, if a square be described about th^ circle, and a pyramid be erected upon it, having the same ver- OF EUCLID. 27a. tex with the cone*, the pyramid inscribed in the cone is half B. XII. of the pyramid circums,cribed about it, because they are to one ^ —^mj another as their bases » : but the cone is less than the circum- a 6. 12. scribed pyramid ; therefore the pyramid of which the base is the square EFGH, and its vertex the same with that of the cone, is greater than half of the cone : divide the circumferences EF, FG, GH, HE, each into two equal parts in the points O, P, R, S, and join EO, OF, FP, PG, GR, RH, HS, SE : therefore each of the triangles EOF, FPG, GRH, HSE is greater than half of the segment of the circle in which it is: upon each of these triangles erect a pyramid having the same vertex with the cone ; each of these pyramids is greater than the half of the segment of the cone in which it is : and thus, dividing each of these cir- cumferences into two equal parts, and from the points of division drawing straight lines to the extremities of the circumferences, and upon each of the triangles thus made erecting pyran)ids having the same vertex with the cone, and so on, there must at length remain some segments of the cone which are togt;ther less'' than the solid Z : let these be the segments upon EO, OF, bLcm.l. * Vertex is put in place of ahitude which is in the Greek, because the pyra- mid, in what follows, is supposed to be circumscribed about the cone, and so must have the same vertex. And the same change is made m some places fol- lowing. 3 M 274 THE ELEMENTS B. XII. FP, PG, GR, RH, HS, SE : therefore the remainder of the cone,^ ^■*-»-^-^ viz. the pyramid of which the base is the polygon EOFPGRHS, and its vertex the same with that of the cone, is greater than the solid X: in the circle ABCD describe the polygon ATBYCVDQ similar to the polygon EOFPGRHS, and upon it erect a pyramid having the same vertex with the cone AL : a 1. 12. and because as the square of AC is to the square of EG, so ^ is the polygon ATBYCVDQ to the polygon EOFPGRHS; and b 2. 12. as the square of AC to tlie square of EG, so is ^ the circle c 11. $. ABCD, to the circle EFGH ; therefore the circle ABCD is <= to the circle EFGH, as the polygon ATBYCVDQ to the poly- \ N I \ \l gon EOFPGRHS : but as the circle ABCD to the circle EFGH, so is the cone AL to the the selid X ; and as the polygon d 6. 12. ATBYCVDQ to the polygon EOFPGRHS, so is <^ the pyra- mid of which the base is the first of these polygons, and ver- tex L, to the pyramid of which the base is the other polygon, and its vertex N : therefore, as the cone AL to the solid X, so is the pyramid of which the base is the polygon ATBYCVDQ, and vertex L, to the pyramid the base of which is the polygon EOFPGRHS, and vertex N: but the cone AL is greater than eU. 5. the pyramid contained in it; therefore the solid X is greater* than the pyramid in the cone EN ; but it is less, as was shown, OF EUCLID. S75 Vhich is absurd : therefore the circle ABCD is not to the circle B. XII. EFGH as the cone AL to any solid which is less than the cone ^^^-^m ^ EN. In the same manner it may be demonstrated that the circle EFGH is not to the circle ABCD as the cone EN to any solid less than the cone AL. Nor can the circle ABCD be to the cir- cle EFGH as the cone AL to any solid greater than the cone EN: for, if it be possible, let it be so to the solid I, which is greater than the cone EN : therefore, by inversion, as the circle EFGH to the circle ABCD, so is the solid I to the cone AL : but as the solid I to the cone AL, so is the cone EN to some solid, which must be less * than the cone AL, because the solid I is a 14. 5- greater than the cone EN : therefore, as the circle EFGH is to the circle ABCD, so is the cone EN to a solid less than the cone AL, which was shown to be impossible : therefore the circle ABCD IS not to the circle EFGH as the cone AL is to any solid greater than the cone EN : and it has been demonstrated that neither is the circle ABCD to the circle EFGH, as the cone AL to any solid less than the cone EN : therefore the circle ABCD is to the circle EFGH, as the cone AL to the cone EN: but as the cone is to the cone, so ^ is the cylinder to the cylinder, because b 15. 5. the cylinders are triple <= of the cone, each to each. Therefore, c 10. 1% as the circle ABCD to the circle EFGH, so are the cylinders up- on them of the same altitude. Wherefore, cones and cylinders of the same altitude are to one another as their bases. Q. E. D. PROP. Xn. THEOR. SIMILAR cones and cylinders have to one ano- see n. ther the triplicate ratio of that which the diameters of their bases have. Let the cones and cylinders of which the bases are the circles ABCD, EFGH, and the diameters of the bases AC, EG, and KL, MN the axis of the cones or cylinders, be similar : the cone of which the base is the circle ABCD, and vertex the point L, has to the cone of which the base is the circle EFGH, and ver- tex N, the triplicate ratio of that which AC has to EG. For, if the cone ABCDL has not to the cone EFGHN the tri- plicate ratio of that which AC has to EG, the cone ABCDL shall have the triplicate of that ratio to some «olid which is less or 276 THE ELEMENTS B. XII. greater than the cone EFGHN. First, let it have it to a less, ^ >— v-^ viz. to the solid X. Make the same construction as in the pre- ceding proposition, and it may be demonstrated the very same way as in that proposition, that the pyramid of which the base is the polyi^on EOFPGRHS, and vertex N, is greater than the soHd X. Describe also in the circle ABCD the polygon ATBYCv'DQ simi ar to the polygon EOFPGRHS, upon which erect a pyramid having the same vertex with the cone ; and let LAQ be one of the triangles containing the pyramid upon the polygon ATBYCVDQ, the vertex of which is L ; and let NES be one of the triangles containing the pyramid upon the polygon EOFPGRHS, of which the vertex is N ; and join KQ, MS: because then the cone ABCDL is similar to the cone a 24. def. EFGHN, AC is » to EG, as the axis KL to the axis MN ; and 11. as AC to EG, so b is AK to EM ; therefore as AK to EM, b 15. 5. so is KL to MN ; and, alternately, AK to KL, as EM to MN: and the right angles AKL, EMN are equal; therefore the sides about these equal angles being proportionals, the c 6. 6. triangle AKL is similar <= to the triangle EMN. Again, be- cause AK is to KQ, as EM to MS, and that these sides are OF EUCLID. 2rr about equal angles AKQ, EMS, because these angles are, each B. XII, of them, the same part of four right angles at the centres K, M ; ^ -y— ^ therefore the triangle AKQ is similar * to the triangle EMS : a 6. §. and because it has been shown that as AK to KL, so is EM to MN, and that AK is equal to KQ, and EM to MS, as QK to KL, so is SM to MN ; and therefore the sides about the right angles QKL, SMN being proportionals, the triangle LKQ is si- milar to the triangle NMS : and because of the similarity of the triangles AKL, EMN, as LA is to AK, so is NE to EM; and by the similarity of the triangles AKQ, EMS, as KA to AQ, so ME to ES ; ex aquali^, LA is to AQ, as NE to ES. Again, be- b 22- 5. cause of the similarity of the triangles LQK, NSM, as LQ to QK, so NS to SM : and from the similarity of the triangles KAQ, MES, as KQ to QA, so MS to SE ; ex xqvali b, LQ is to QA, as NS to SE : and it was proved that QA is to AL, as SE to EN ; therefore, again, ex aquali, as QL to LA, so is SN to NE : wherefore the triangles LQA, NSE, having the sides about all their angles proportionals, are equiangular = and similar to c 5. 6. one another : and therefore the pyramid of which the base is the triangle AKQ, and vertex L, is similar to the pyramid the base of which is the triangle EMS, and vertex N, because their solid angles are equal ^ to one another, and they are contained d B. 11. by the same number of similar planes : but similar pyramids which have triangular bases have to one another the triplicate « ratio of that which their homologous sides have ; therefore e 8. 12. the pyramid AKQL has to the pyramid EMSN the triplicate ratio of that which AK has to EM. In the same manner, if straight lines be drawn from the points D, V, C, Y, B, T to K, and from the points H, R, G, P, F, O to M, and py- ramids be erected upon the triangles having the same vertices v/ith the cones, it may be demonstrated that each pyramid in the first cone has to each in the other, taking them in the same order, the triplicate ratio of that Avhich the side AK has to the side EM; that is, which AC has to EG: but as one antece- dent to its consequent, so are all the antecedents to all the consequents f ; therefore as the pyramid AKQL to the pyra-fl2-5. mid EMSN, so is the whole pyramid the base of which is the polygon DQATBYCV, and vertex L, to the whole pyramid of which the base is the polygon HSEOFPGR, and vertex N. Wherefore also the first of these two last named pyramids has to the other the triplicate ratio of that which AC has to EG. But, by the hypothesis, the cone of which the base is the cir- cle ABCD, and vertex L, has to the solid X, the triplicate j ratio of that which AC has to EG ; therefore, as the cone of | which the base is the circle ABCD and vertex L, is to the 278 THE ELEMENT^ B. XII. solid X, s'o is tire pyramid the base of which Is the polygon ^'- y -^ DQATBYCV, and vertex L, to the pyramid the base of which is the polygon HSEOFPGR and vertex N: but the said cone is greater than the pyramid contained in it, therefore the solid a H. 5. X is greater ^ than the pyramid, the base of which is the poly- gon HSEOFPGR, and vertex N ; but it is also less ; which is impossible : therefore the cone of which the base is the circle ABCD, and vertex L, has not to any solid which is less than the cone of which the base is the circle EFGH and vertex N, the tri- plicate ratio of that which AC has to EG. In the same manner it may be demonstrated that neither has the cone EFGHN to any solid which is less than the cone ABCDL, the triplicate ratio of that which E(i has to AC. Nor can the cone ABCDL have to any solid which is greater than the cone EFGHN, the triplicate ratio of that which AC has to EG : for, if it be possi- ble, let it have it to a greater, viz. to the solid Z : therefore, inversely, the solid Z has to the cone ABCDL, the triplicate ratio of that which EG has to AC : but as the solid Z is to OF EUCLID. 279 the cone ABCDL, so is the cone EFGHN to some solid, B. XII. which must be less ^ than the cone ABCDL, because the solid *— v-— ' Z is greater than the cone EFGHN : therefore the cone EFGHN a 14. 5. has to a solid which is less than the cone ABCDL, the tripli- cate ratio of that which EG has to AC, which was demonstrat- ed to be impossible : therefore the cone ABCDL has not to any solid greater than the cone EFGHN the triplicate ratio of that which AC has to EG : and it was demonstrated that it could not have that ratio to any solid less than the cone EFGHN: there- fore the cone ABCDL has to the cone EFGHN the triplicate ratio of that which AC has to EG : but as the cone is to the cone, so ^ is the cylinder to the cylinder ; for every cone is the b 15. 5 third part of the cylinder upon the same base, and of the same altitude : therefore also the cylinder has to the cylinder the tri- plicate ratio of that which AC has to EG. Wherefore, similar cones, £cc. Q. E. D. PROP. Xni, THEOR. IF a cylinder be cut by a plane parallel to its op- See n posite planes, or bases ; it divides the cylinder into two cylinders, one of which is to the other as the axis of the first to the axis of the other. Let the cylinder AD be cut by the plane GH, parallel to the opposite planes AB, CD, meeting the axis EF in the point K, and let the line GH be the common section of the plane GH and the surface of the cy- linder AD : let AEFC be the paral- lelogram, in any position of it, by the revolution of which about the straieht line EF the cyHnder AD is described; and let GK be the common section of the plane GH, and the plane AEFC : and because the parallel planes AB, GH are cut by the plane AEKG, AE, KG, their common sec- tions with it, are parallel » ; where- fore AK is a parallelogram, and GK equal to EA the straight line from the centre of the circle AB : for the s^rae reason, eaph of ^he straight lines R B H D a 16. l:i. Y Q 280 THE ELEMENTS R B. XII. drawn from the point K to the line GH may be proved to be *— V— ^ equal to those which are drawn from the centre of the circle AB to its circumference, and are therefr-re all equal to one uno- al5. def, ther. Therefore the line GH is the circumference of a circle *, ^' of which the centre is the point K: therefore the piane GH di- vides the cylinder AD into the cylinders AH, GD ; for they are the same which would be described by the revolution of the parallelograms AK, GF, about the straight lines EK, KF : and it is to be shown, that the cylinder AH is to the cylinder HC, as the axis EK to the axis KF. Produce the axis EF both ways ; and take any number oC straight lines EN, NL, each equal to EK ; and any number r X, XM, each equal to FK ; and let planes parallel to AB,CD pass through the points L, N, X, M : therefore the common sec- tions of these planes with the cylinder produced are circles, the centres of which are the points L, N, X, M, as was proved of the plane GH ; and these planes cut off the cylinders PR, RB, DT, TQ : and because the axes LN, NE, EK are all equal, therefore the cylinders PR, RB, h 11. 12. BG are ^ to one another as their bases ; but their bases are equal, and therefore the cylinders PR, RB, BG are equal : and because the axes LN, NE, EK are equal to one another, as also the cylinders PR, RB, BG, and that there are as many axes as cylinders ; therefore, whatever multi- ple the axis KL is of the axis KE, the same multiple is the cylinder PG of the cylinder GB : for the same reason, whatever multiple the axis MK is of the axis KF, the same multiple is the cylin- der QG of the cylinder GD : and if the axis KL be equal to the axis KM, the cylinder PG is equal to the cylinder GQ ; and if the axis KL be greater than the axis KM, the cylinder PG is greater than the cylinder QG ; and if less, less: since there- fore there are four magnitudes, viz. the axes EK, KF, and the cylinders BG, GD, and that of tlie axis EK and cylindei' yG, there has been taken any equimultiples whatever, viz, th^ G V B H D Q OF EUCLID. 281 axis KL and cylinder PG ; and of the axis KF and cylinder GD, B. XII. any equimultiples whatever, viz. the axis KM and cylinder GQ : *-— v*^ and it has been demonsti'ated, if the axis KL be greater than the axis KM, the cylinder PG is greater than the cylinder GQ ; and if equal, equal ; and if less, less : therefore ^ the axis EK is to d 5. def. the axis KF, as the cylinder BG to the cylinder GD. Where- ^■ fore, if a cylinder, 8cc. Q. E. D. PROP. XIV. THEOR. CONES and cylinders upon equal bases are to one another as their altitudes. Let the cylinders EB, FD be upon the equal bases AB, CD : as the cylinder EB to the cylinder FD, so is the axis GH to the axis KL. Produce the axis, KL to the point N, and make LN equal to the axis GH, and let CM be a cylinder of which the base is CD, and axis LN, and because the cylinders EB, CM have the same altitude, they are to one another as their bases 3^: but a IJ. 12. their bases are equal, therefore also the cylinders EB, CM are equal. And because the cylin- der FM is cut by the plane F/ ^ CD parallel to its opposite planes, as the cylinder CM to the cylinder FD, so is ** the axis LN to the axis KL. But the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH : therefore as the cylinder EB to the cylin- der FD, so is the axis GH to the axis KL : and as the cylin- der EB to the cylinder FD, so is c the cone ABG to the cone CDK, because the cylinders are c 15. 5. triple ^ of the cones : therefore also the axis GH is to the axis d 10. 12 KL, as the cone ABG to the cone CDK, and the cylinder EB to the cylinder FD. Wherefore, cones, &c. Q. E. D. b 13- 12, 2N 282 B. XII. THE ELEMENTS PROP. XV. THEOR. See N. THE bases and altitudes of equal cones and cy- linders are reciprocally proportional, and, if the bases and altitudes be reciprocally proportional, the cones^ and cylinders are equal to one another. Let the circles ABCD, EFGH, the diameters of which are Ac, EG, be the bases, and KL, MN the axis, as also the alti- tudes, of equal cones and cylinders ; and let ALC, ENG be the cones, and AX, EO the cylinders : the bases and altitudes of the cylinders AX, EO are reciprocally proportional ; that is, as the base ABCD to the base EFGH, so is the altitude MN to the altitude KL. Either the altitude MN is equal to the altitude KL, or these altitudes are not equal. First, let them be equal ; and the cylinders AX, EO being also equal, and cones and cylinders a 11.12. of the same altitude being to one another as their basest, there- b A. 5. fore the base ABCD is equal ^ to the base EFGH ; and as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL. But let the alti- R tudes KL, MN be unequal, and MN the greater of the two, and from MN take MP equal to KL, and, through the point P, cut the cylinder EO by the plane TVS parallel to the op- posite planes of the circles EFGH, RO ; therefore the com- mon section of the plane TYS and the cylinder EO is a cir- cle, and consequently ES is a cylinder, the base of which is the circle EFGH, and altitude MP : and because the cylinder AX is equal to the cylinder EO, as AX is to the cyHnder ES, so c7. 5. 'is the cylinder EO to the same ES. But as the cylinder A^ to the cylinder ES, so ^ is the base ABCD to the base EFGH ; for the cylinders AX, ES are of the same altitude ; and as the d 13. 12. cylinder EO to the cylinder ES, so ^ is the altitude MN to the altitude MP, because the cylinder EO is cut by the plane OF EUCLID. 2S2 TYS parallel to its opposite planes. Therefore, as the base B. XII. ABCD to the base EFGH, so is the altitude MN to the altitude ^— y^-* MP : but MP is equal to the altitude KL ; wherefore as the base ABCD to the base EFGH, so is the altitude MN to the al- titude KL ; that is, the bases and altitudes of the equal cylinders AX, EO are reciprocally proportional. But let the bases and altitudes of the cylinders AX, EO be reciprocally proportional, viz. the base ABCD to the base EFGH, as the altitude MN to the altitude KL : the cylinder AX is equal to the cylinder EO. First, let the base ABCD be equal to the base EFGH ; then because as the base ABCD is to the base EFGH, so is the alti- tude MN to the altitude KL ; MN is equally to KL, and therefore b. A. 5. the cylinder AX is equal ^ to the cylinder EO. a 11. 12. But let the bases ABCD, EFGH be unequal, and let ABCD be the greater ; and because as ABCD is to the base EFGH, so is the altitude MN to the altitude KL : therefore MN is great- er b than KL. Then, the same construction being made as be- fore, because as the base ABCD to the base EFGH, so is the al- titude MN to the altitude KL : and because the altitude KL is equal to the altitude MP ; therefore the base ABCD is » to the base EFGH, as the cylinder AX to the cylinder ES ; and a^^ the altitude MN to the altitude MP or KL, so is the cylinder EO to the cylinder ES : therefore the cylinder AX is to the cylinder ES, as the cylinder EO to the same ES ; whence the cylinder AX is equal to the cylinder EO : and the same reasoning holds in cones. Q. E. D. PROP. XVI. PROB. TO describe in the greater of two circles that have the same centre, a polygon of an even number of equal sides, that shall not meet the lesser circle. Let ABCD, EFGH be two given circles having the same cen- tre K : it is required to inscribe in the greater circle ABCD a polygon of an even number of equal sides, that shall not meet the lesser circle. Through the centre K draw the straight line BD, and from the point G, where it meets the circumference of the lesser 2S4 THE ELEMENTS B. XII. circle, draw GA at right angles to BD, and produce it to C; ^^"•v--^ therefore AC touches ^ the circle EFGH : then, if the circum- a 16. 3. ference BAD be bisected, and the half of it be again bisected, b Lem- and so on, there must at length remain a circumference less ^ "^a. than AD; let this be LD ; and from the point L draw LM per- pendicular to BD, and produce it to N ; and join LD, DN. There- fore LD is equal to DN : and be- cause LN is parallel to AC, and that AC touches the circle EFGH ; therefore LN does not meet the circle EF'GH : and much less shall the straight lines LD, DN meet the circle EFGH: so that if straight lines equal to LD be applied in the circle ABCD from the point L around to N, there shall be described in the circle a polygon of an even number of equal sides not meeting the lesser circle.. Which was to be done. LEMMA n. IF two trapeziums ABCD, EFGH be inscribed in the circles, the centres of which are the points K, L ; and if the sides AB, DC be parallel, as also EF, HG ; and the other four sides AD, BC, EH, FG be all equal to one another; but the side AB greater than EF, and DC greater than HG : the straight line KA from the centre of the circle in which the greater sides are, is greater than the straight line LE drawn from the centre to the circumference of the other circle. If it be possible, let KA be not greater than LE ; then KA must be cither equal to it, or less. F'irst, let KA be equal to LE : therefore, because in two equal circles, AD, BC in the one are equal to EH, FG in the other, the circumferences a2B. 3. AD, BC arc equal » to the circumferences EH, FG ; but be- cause the straight lines AB, DC are respectively greater than EF, GH, the circumferences AB, DC are greater than EF, HG : therefore the whole circumference ABCD is greater than the v.hole EFGH; but it is also equctl to it, which is OF EUCLID. 28-5 impossible : therefore the straight line KA is not equal to B. Xll. But let KA be less than LE, and make LM equal to KA, and from the centre L, and distance LM, describe the circle MNOP, meeting the straight lines LE, LF, LG, LH, in M, N, O, P ; and join MN, NO, OP, PM, which are respectively parallel » to, and less than EF, FG, GH, HE : then because EH a 2. 6. is greater than MP, AD is greater than MP j and the circles >ABCD, MNOP are equal; therefore the circumference AD is greater than MP ; for the same reason, the circumference BC is greater than NO ; and because the straight line AB is greater than EF, which is greater than MN, much more is AB greater than MN : therefore the circumference AB is greater than MN ; and, for the same reason, the circumference DC is greater than PO : therefore the whole circumference ABCD is greater than the whole MNOP ; but it is likewise equal to it, which is impossi- ble : therefore K A is not less than LE ; nor is it equal to it ; the straight line KA must therefore be greater than LE. Q. E. D. CoR. And if there be an isosceles triangle, the sides of Avhich are equal to AD, BC, but its base less than AB the greater of the two sides AB, DC ; the straight line KA may, in the same manner, be demonstrated to be greater than the straight line drawn from the centre to the circumference of the circle de- scribed about the triangle. THE ELEMENTS PROP. XVII. PROB. See N. TO describe in the greater of two spheres whiclx. have the same centre, a solid polyhedron, the super- ficies of which shall not meet the lesser sphere. Let there be two spheres about the same centre A ; it is re-: quired to describe in the greater a solid polyhedron, the superfi- cies of which shall not meet the lesser sphere. Let the spheres be cut by a plane passing through the centre ; the common sections of it with the spheres shall be circles ; because the sphere is described by the revolution of a semicir- cle about the diameter remaining unmoveable ; so that in what- ever position the semicircle be conceived, the common section of the plane in which it is with the superficies of the sphere is the circumference of a circle ; and this is a great circle of the sphere, because the diameter of the sphere, which is likewise a IS. 3. the diameter of the circle, is greater * than any straight line in the circle or sphere : let then the circle made by the section of the plane with the greater sphere be BCDE, and with the lesser sphere be FGH ; and draw the two diameters BD, CE at right angles to one another; and in BCDE, the greater of b 16. 12. the two circles, describe ^ a polygon of an even number of e- qual sides not meeting the lesser circle FGH : and let its sides, in BE, the fourth part of the circle, be BK, KL, LM, ME ; join KA and produce it to N ; and from A draw AX at right angles to the plane of the circle BCDE meeting the superficies of the sphere in the point X; and let planes pass through AX, and each of the straight lines BD, KN, which, from what has been said, shall produce great circles on the superficies of the sphere, and let BXD, KXN be the semicircles thus made upon the diameters BD, KN : therefore because X A is at right angles to the plane of the circle BCDE, every plane which c 18. 11. passes through XA is at right <=■ angles to the plane pf the circle BCDE ; wherefore the semicircles BXD, KXN are at right angles to that plane: and because the semicircles BED, BXD, KXN, upon the equal diameters BD, KN, are equal to one another, their halves BE, BX, KX, arc equal to one an- other: therefore, as many sides of the polygon as are in BE, so many there are in BX, KX equal to the sides BK, KL, LM, ME: let these polygons be described, and their sides be BO, OP, PR, RX ; KS, ST, TY, YX, and join OF EUCLID. 287 OS, PT, RY ; and from the points O, S, draw OV, SQ perpen- b. Xii. diculars to AB, AK : and because the plane BOXD is at right ^-^-ymJ angles to the plane BCDE, and in one of them BOXD, OV is drawn perpendicular to AB the common section of the planes, therefore O V is perpendicular a to the plane BCDE : for the a 4. def. same reason SQ is perpendicular to the same plane, because li- the plane KSXN is at right angles to the plane BCDE. Join VQ; and because in the equal semicircles BXD, KXN the circumferences BO, KS are equal, and OV, SQ are perpen- dicular to their diameters, therefore d OV is equal to SQ, d26.1. and BV equal to KQ: but the whole BA is equal to the whole KA, therefore the remainder VA is equal to the remainder QA: as therefore BV is to VA, so is KQ to QA, wherefore VQ is parallel « to BK: and because OV, SQ are each ofe2. 6. them at right angles to the plane of the circle BCDE, OV is parallel*" to SQ; and it has been proved that it is also equal f 6. It. to it ; therefore QV, SO are equal and parallel s : and because g 33. 1- QV is parallel to SO, and also to KB ; OS is parallel i^ to BK ; b 9. 11. and therefore BO, KS which join them are in the same plane 2S8 THE ELEMENTS B. XII. in which these parallels are, and the quadrilateral figure KBOS V->vJ is in one plane : and if PB, TK be joined, and perpendiculars be drawn from the points P, T to the straight lines AB, AK, it may be demonstrated, that TP is parallel to KB in the very same way that SO was shown to be parallel to the same KB ; a 9. 11. wherefore * TP is parallel to SO, and the quadrilateral figure SOPT is in one plane : for the same reason, the quadrilateral b 2. 11. TPRY is in one plane : and thp figure YRX is also in one plane''. Therefore, if from the points O, S, P, T, R, Y there be drawn straight lines to the point A, there shall be formed a solid po- lyhedron between the circumferences BX, KX composed o£ pyramids, the bases of which are the quadrilaterals KBOS, SOPT, TPRY, and the triangle YRX, and of which the com- mon vertex is the point A : and if the same construction be made upon each of the sides KL, LM, ME, as has been done upon BK, and the like be done also in the other three qua- drants, and in the other hemisphere ; there shall be for- med a solid polyhedron described in the sphere, compo- OF EUCLIt). 289 sed of pyramids, the bases of which are the aforesaid quadri- B. XII. lateral figures, and the triangle YRX, and those formed in ^-> v --^ the like manner in the rest of the sphere, the common vertex of them all being the point A : and the superficies of this so- lid polyhedron does not meet the lesser sphere in which is the circle FGH: for, from the point A draw * AZ perpendicular a 11. 11. to the plane of the quadrilateral KBOS meeting it in Z, and join BZ, ZK : and because AZ is perpendicular to the plane KBOS, it makes right angles with every straight line meeting it in that plane ; therefore AZ is perpendicular to BZ and ZK : and because AB is equal to AK, and that the squares of AZ, ZB, are equal to the square of AB; and the squares of AZ, ZK to the square of AK ^ ; therefore the squares of AZ, ZB b 4,7- 1- are equal to the squares of AZ, ZK : take from these equals the square of AZ, the remaining square of BZ is equal to the I'emaining square of ZK ; and therefore the straight line BZ is equal to ZK ; in the Hke manner it may be demonstrated, that the straight lines drawn from the point Z to the points O, S are equal to BZ or ZK : therefore the circle described from the centre Z, and distance ZB, shall pass through the points K, O, S, and KBOS shall be a quadrilateral figure in the circle: and because KB is greater than QV, and QV equal to SO, there- fore KB is greater than SO : but KB is equal to each of the straight lines BO, KS ; wherefore each of the circumferences cut oft" by KB, BO, KS is greater than that cut oft" by OS ; and these three circumferences, together with a fourth equal to one of them, are greater than the sUme three together with that cut off" by OS ; that is, than the whole circumference of the cir- cle ; therefore the circumference subtended by KB is greater than the fourth part of the whole circumference of the circle KBOS, and consequently the angle BZK at the centre is great- er than a right angle : and because the angle BZK is obtuse, the square of BK is greater '= than the squares of BZ, ZK;cl2. 2. that is, greater than twice the square of BZ. Join KV, and because in the triangles KBVj OBV, KB, BV are equal to OB, BV, and that they contain equal angles; the angle KVB is equal ^ to the angle OVB : and OVB is a right angle ; there- d 4. 1. fore also KVB is a right angle: and because BD is less than twice DV, the rectangle contained by DB, BV is less than twice the rectangle DVB ; that is e, the square of KB is less e 8. 6. than twice the square of KV : but the square of KB is greater than twice the square of BZ : therefore the square of KV is greater than the square of BZ : and because BA is equal to AK, and that the squares of BZ, ZA are equal together to the square of BA, and the squares of KV, VA to the square of AK; therefore the squares of BZ, ZA are equal to the squares of KV, VA ; and of these the square of KV is greater than the 2 O 290 THE ELEMENTS B. XII. square of BZ ; therefore the square of VA is less than the ^— •v-^-^ square of ZA, and the straight line AZ greater than VA : much more then is AZ greater than AG ; because, in the pre- ceding proposition, it was shown that KV falls without the circle FGH : and AZ is perpendicular to the plain KBOS, and is therefore the shortest of all the straight lines that can be drawn from A, the centre of the sphere to that plane. Therefore the plane KBOS does not meet the lesser sphere. And that the other planes bet^veen the quadrants BX, KX fall without the lesser sphere, is thus demonstrated : from the point A draw AI perpendicular to the plane of the quadrila- teral SOPT, and join lO ; and, as was demonstrated of the plane KBOS and the point Z, in the same way it may be shown that the point I is the centre of a circle described about SOPT ; and that OS is greater than PT ; and PT was shown to be pa- rallel to OS: therefore, because the two trapeziums KBOS, SOPT inscribed in circles have their sides BK, OS parallel, as also OS, PT ; and their other sides BO, KS, OP, ST all equal to one another, and that BK is greater than OS, and OS a 2. km. gi-eater than PT, therefore the straight line ZB is greater"* ^2- than lO. Join AO which will be equal to AB ; and because AIO, AZB are right angles, the squares of AI, lO are equal to the square of AO or of AB ; that is, to the squares of AZ, ZB ; and the square of ZB is greater than the square of lO, therefore the square of AZ is less than the square of AI ; and the straight line AZ less than the straight line AI : and it was proved that AZ is greater than AG ; much more then is AI greater than AG ; therefore the plane SOPT falls wholly with- out the lesser sphere : in the same manner it may be demon- strated that the plane TPRY falls without the same sphere, as also the triangle YRX, viz. by the cor. of 2d lemma. And after the same way it may be demonstrated that all the planes which contain the solid polyhedron, fall without the lesser sphere. Therefore in the greater of two spheres which have the same centre, a solid polyhedron is described, the superficies of which does not meet the lesser sphere. Which was to be done. But the straight line AZ may be demonstrated to be greater than AG otherwise, and in a shorter manner, without the help of prop. 16, as follows. From the point G draw GU at right angles to AG, and join AU. If then the circumference BE be bisected, and its half again bisected, and so on, there will at length be left a circumference less than the circumference which is subtended by a straight line equal to GU inscribed in the circle BCDE : let this be the circumference KB : therefore the straight line KB is less than GU : and because the angle BZK is obtuse, as was proved in the preceding, therefore BK is greater than BZ : but GU is greater than BK ; much more OF EUCLID. 291 then is GU greater than BZ, and the square of GU than the B. XII. square of BZ ; and AU is equal to AB ; therefore the square ^^»v— J of AU, that is, the squares of AG, GU, are equal to the square of AB, that is, to the squares of AZ, ZB ; but the square of BZ is less than the square of GU ; therefore the square of AZ is greater than the square of AG, and the straight line AZ con- sequently greater than the straight line AG. Cor. And if in the lesser sphere there be described a solid polyhedron by drawing straight lines betwixt the points in which the straight lines from the centre of the sphere drawn to all the angles of the solid polyhedron in the greater sphere meet the superficies of the lesser ; in the same order in which are joined the points in which the same lines from the centre meet the superficies of the greater sphere ; the solid polyhe- dron in the sphere BCDE has to this other solid polyhedron the triplicate ratio of that which the diameter of the sphere BCDE has to the diameter of the other sphere : for if these two solids be divided into the same number of pyramids, and in the same order, the pyramids shall be similar to one ano- ther, each to each ; because they have the solid angles at their common vertex, the centre of the sphere, the same in each py- ramid, and their other solid angle at the bases equal to one another, each to each ^, because they are contained by three a B. 11. plane angles equal each to each : and the pyramids are contained by the same number of similar planes ; and are therefore similar ^ ^ H- to one another, each to each : but similar pyramids have to ^^" one another the triplicate '^ ratio of their homologous sides, c Cor. 8. Therefore the pyramid of which the base is the quadrilateral 12. KBOS, and vertex A, has to the pyramid in the other sphere of the same order, the triplicate ratio of their homologous sides ; that is, of that ratio which AB from the centre of the greater sphere has to the straight line from the same centre to the superficies of the lesser sphere. And in like manner, each pyramid in the greater sphere has to each of the same order in the lesser, the triplicate ratio of that which AB has to the se- midiameter of the lesser sphere. And as one antecedent is to its consequent, so are all the antecedents to all the consequents. Wherefore the whole solid polyhedron in the greater sphere has to the whole solid polyhedron in the other, the triplicate ratio of that which AB the semidiameter of the first has to the semi- diameter of the other ; that is, which the diameter BD of the greater has to the diameter of the other sphere^ a ir. 12. THE ELEMENTS PROP. XVIII. THEOR. SPHERES have to one another the triplicate ra- tio of that which their diameters have. Let ABC, DEF be two spheres, of which the diameters are BC, EF. The sphere ABC has to the spliere DEF the triplicate ratio of that which BC has to EF. For, if it has not, the sphere ABC shall have to a sphere ei- ther less or greater than UEF, the triplicate ratio of that which BC has to EF. First, let it have that ratio to a less, viz. to the sphere GHK ; and let the sphere DEF have the same centre with GHK ; and in the greater sphere DEF describe ^ bCor 17. 12. c 14. 5. a solid polyhedron, the superficies of which does not meet the lesser sphere GHK ; and in the sphere ABC describe another similar to that in the sphere DEF: therefore the solid polyhe- dron in the sphere ABC has to the solid polyhedron in the sphere DEF, the triplicate ratio ^ of that which BC has to EF. Bqt the sphere ABC has to the sphere GHK the triplicate ra- tio of that which BC has to EF ; therefore, as the sphere ABC to the sphere GHK, so is the solid polyhedron in the sphere ABC to the solid polyhedron in the sphere DEF : but the sphere ABC is greater than the solid polyhedron in it; therefore ^ al- so the sphere GHK is greater than the solid polyhedron in the sphere DEF: but it is also less, because it is contained within it, which is impossible : therefore the sphere ABC has not to any sphere less than DEF the triplicate ratio of that which BC has to EF. In the same manner it may be demonstrated, that the sphere DEF has not to any sphere less than ABC the triplicate ratio of that which EF has to BC. Nor can the sphere ABC' have to any sphere greater than DEF, the tripli- cate ratio of that which BC has to EF: for, if it can, let it OF EUCLID. 293 have that ratio to a greater sphere LMN: therefore, by inver- B. XII. sion, the sphere LMN has to the sphere ABC, the tripHcate ^^\^-J ratio of that which the diameter EF has to the diameter BC. But, as the sphere LMN to ABC, so is the sphere DEF to some sphere, which must be less <= than the sphere ABC, because the c 14. S, sphere LMN is greater than the sphere DEF : therefore the sphere DEF has to a sphere less than ABC the triplicate ratio of that which EF has to BC ; which was shown to be impos- sible : therefore the sphere ABC has not to any sphere greater than DEF the triplicate ratio of that which BC has to EF : and it was demonstrated, that neither has it that ratio to any sphere less than DEF. Therefore the sphere ABC has to the sphere DEF, the triplicate ratio of that which BC has to EF. Q. E. Do FINIS. NOTES, CRITICAL AND GEOMETRICAL : coirrAiNiNC AN ACCOUNT OF THOSE THINGS IN WHICH THIS EDITION DIFFERS FROM THE GREEK TEXT, AND THE REASONS OF THE ALTERATIONS WHICH HAVE BEEN MADE, AS ALSO OBSERVATIONS ON SOME OF THE PROPOSITIONS. BY ROBERT SIMSOJV, M. D. EMEKITUS PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF GLASGOW. PHILADELPHIA, PRINTED BY THOMAS AND GEORGE PALMER, 116, HIGH STREET. 1806. NOTES, &c. DEFINITION I. BOOK I. It is necessary to consider a solid, that is, a magnitude which has length, breadth, and thickness, in order to understand aright the definitions of a point, line, and superficies ; for these all arise from a solid, and exist in it : the boundary, or bounda- lies, which contain a solid are called superficies, or the boun- dary which is common to two solids which are contiguous, or which divides one solid into two contiguous parts, is called a superficies : thus, if BCGF be one of the boundaries which contain the solid ABCDEFGH, or which is the common boun- dary of this solid, and the solid BKLCFNMG, and is there- fore in the one as well as in the other solid, called a superficies, and has no thickness : for if it have any, this thickness must cither be a part of the thickness of the solid AG, or the solid BM, or a part of the thickness of E each of them. It cannot be a part of the thickness of the solid BM ; because if this solid be re- moved from the solid AG, the superficies BCGF, the boundary of the solid AG remains still the same as it was. Nor can it be a part of the thickness of the solid AG ; because, if this be re- moved from the solid BM, the superficies BCGF, the boundary of the solid BM, does nevertheless remain: therefore the super- ficies BCGF has no thickness, but only length and breadth. The boundary of a superficies is called a line, or a line is the common boundary of tAvo superficies that are contiguous, or which divides one superficies into two contiguous parts : thus, if R ' be one of the boundaries which contain the superficies ABCD, or which is the common boundary of this superficies, and of the superficies KBCL which is contiguous to it, this boundary BC is called a line, and has no breadth : for if it have any, this must be part either of the breadth of the super^ ficies ABCD, or of the superficies KBCL, or part of each of 3 P 298 NOTES. H M D Book I. them. It is not the part of the breadth of the superficies KBCL; -•'~^^'>»-' for, if this superficies be removed from the superficies ABCD, the line BC, which is the boundary of the superficies ABCD, remains the same as it was : nor can the breadth that BC is sup- posed to have be a part of the breadth of the superficies ABCD; because, if this be removed from the superficies KBCL, the line BC, which is the boundary of the superficies KBCL, does never- theless remain : therefore the line BC has no breadth : and be- cause the line BC is a superficies, and that a superficies has no thickness, as was shown ; therefore a line has neither breadth nor thickness, but only length. The boundary of a line is called a point, or a point is the common boundary or extremity of two lines that are contiguous : thus, if B be the extremity of the line AB, or the common extremi- ty of the two lines AB, KB, this extremity is called a point, and has no length : for, if it have any, this length must either be part of the length of the line AB, or of the line KB. It is not part A B K of the length of KB ; for if the line KB be removed from AB, the point B, which is the extremity of the line AB, remains the same as it was : nor is it part of the length of the line AB ; for, if AB be removed from the line KB, the point B, which is the extremity of the line KB, does nevertheless remain : there- fore the point B has no length : and because a point is in a line, and a line has neither breadth nor thickness, therefore a point has no length, breadth, nor thickness. And in this manner the definitions of a point, line, and superficies are to be understood. DEF. VII. B. I. Instead of this definition as it is in the Greek copies, a more distinct one is given from a property of a plane superficies, which is manifestly supposed in the Elements, viz. that a straight line drawn from any point in a plane to any other in it. is wholly in that plane. DEF. VIII. B. I. It seems that he who made this definition designed that it should comprehend not only a plane angle contained by two straight lines, but likewise the angle which some conceive to be made by a straight line and a curve, or by two curve lines, w hich meet one another in a plane : but though the meaning of NOTES. 299 the words i-x tuB-etetiy that is, in a straight line, or in the same Book I. direction, be plain, when two straight lines are said to be in a ^*^"^^">w' straight line, it does not appear what ought to be vuiderstood by these words, when a straight line and a curve, or two curve lines, are said to be in the same direction ; at least it cannot be explained in this place ; which makes it probable that this defi- nition, and that of the angle of a segment, and what is said of the angle of a semicircle, and the angles of segments, in the 16th and 31st propositions of book 3, are the additions of some less skilful editor : on which account, especially since they are quite useless, these definitions are distinguished from the rest by inverted double commas. DEF. XVII. B. I. The words, " which also divides the circle into two equal " parts," are added at the end of this definition in all the copies, but are now left out as not belonging to the definition, being only a corollary from it. Proclus demonstrates it by conceiving one of the parts into Avhich the diameter diAides the circle, to be applied to the other ; for it is plain they must coincide, else the straight lines from the centre to the circumference would not be all equal : the same thing is easily deduced from the 3 1st prop, of book 3, and the 24th of the same; from the first of which it follows that semicircles are similar segments of a circle : and from the other, that they are equal to one another. DEF. XXXIII. B. I. This definition has one condition more than is necessary ; be- cause every quadrilateral figure which has its opposite sides equal to one another, has likewise its opposite angles equal*; and on the contrary. Let ABCD be a quadrilateral figure of which the opposite sides AB, CD are equal to one another ; as also AD and BC : join BD ; the two sides AD, DB are equal to the two CB, BD, and the base AB is equal to the base CD ; therefore by prop. 8, of book 1, the angle ADB is equal to the angle B C CBD ; and by prop. 4, B. 1, the angle BAD is equal to the angle DCB, and ABD to BDC ; and therefore also the angle ADC is equal to the angle ABC. aOQ NOTES. Book I. And if the angle BAD be equal to the opposite angle BClD, ^-^f"-^-^*'^ and the angle ABC to ADC ; the opposite sides are equal ; be- cause, by prop. 32, book 1, all the angles of the quadrilateral figure AHCD ai^e together equal to a D four right angles, and the two angles BAD, ADC are together equal to the two angles BCD, ABC : where- fore BAD, ADC are the half of all the four angles ; that is, BAD and B C ADC are equal to two right angles: and therefore AB, CD are parallels by prop. 28, B. 1. In the same manner AD, BC are parallels : therefore ABCD is a parallelogram, and its opposite sides are equal by prop. 34, book I, PROP. VII. B. I. There are two cases of this proposition, one of which is not in the Greek text, but is as necessary as the other : and that the case left out has been formerly in the text, appears plainly from this, that the second part of prop. 5, which is necessary to the demonstration of this case, can be of no use at all in the Elements, or anywhere else, but in this demonstration ; because the second part of prop. 5 clearly follows from the first part, and prop. 1 3, book 1 . This part must therefore have been added to prop. 5, upon account of some proposition betwixt the 5th and 13th, but none of these stand in need of it except the 7th proposition, on account of which it has been added : besides, the translation from the Arabic has this case explicitly demon- strated. And Proclus acknowledges, that the second part of prop. 5 was added upon account of prop. 7, but gives a ridicu- lous reason for it, " that it might afford an answer to objections '• made against the 7th," as if the case of the 7th, Avhich is left out, were, as he expressly makes it, an objection against the proposition itself. Whoever is curious, may read what Proclus says ,of this in his conmientary on the 5th and 7th propositions ; for it is not Morth while to relate his trifles at full length. It was thought proper to change the enunciation of this 7th prop, so as to preserve the very same meaning ; the literal translation from the Greek being extremely harsh, and difficult to be understood by beginners. NOTES. PROP. XI. B. I. A corollary is added to this proposition, which is necessary to Prop. 1, B. 11, and otherwise. PROP. XX. and XXI. B. I. Proclus, in his commentary, relates, that the Epicureans de- rided this proposition, as being manifest even to asses, and need- ing no demonstration ; and his answer is, that though the truth of it be manifest to our senses, yet it is science which must give the reason why two sides of a triangle are greater than the third : but the right answer to this objection against this and the 2 1st, and some other plain propositions, is, that the number of axioms ought not to be increased without necessity, as it must be if these propositions be not demonstrated, Mons. Clairault, in the pre- face to his Elements of Geometry, published in French at Paris, ann. 1741, says, That Euclid has been at the pains to prove, that the two sides of a triangle which is included within another are together less than the two sides of the triangle which in- cludes it ; but he has forgot to add this condition, viz. that the triangles must be upon the saine base ; because, unless this be added, the sides of the included triangle may be greater than the sides of the triangle which includes it, in any ratio which is less than that of two to one, as Pappus Alexandrinus has demon- strated in prop. 3. B. 3. of his mathematical collections. PROP. XXII. B. I. Some authors blame Euclid, because he does not demonstrate that the two circles made use of in the construction of this problem must cut one another : but this is very plain from the determination he has given, viz. that any two of the straight lines DF, FG, GH must be great- er than the third: for who is so dull, though only beginning to learn the Elements, as not to per- ceive that the circle described from the centre F, at the distance FD, must meet FH betwixt F and H, because FD is less than FH ;^ ^^ F G H and that, for the like reason, the circle described from the centre G, at the distance GH, p^' GM, must meet DG betwixt D 502 NOTES. Book I. and G ; and that these circles must meet one another, because '-«^^^'''^*«*^ FD and GH are together greater than FG ? And this determina- tion is easier to be understood than that which Mr. Thomas Simpson derives from it, and puts instead of Euclid's, in the 49th page of his Elements of Geome- DM F G H try, that he may supply the omission he blames Euclid for ; which determination is, that any of the three straight lines must be less than the sum, but greater than the difference of the other two : from this he shows the circles must meet one another, in one case ; and says, that it may be proved after the same man- ner in any other case : but the straight line GM, which he bids take from GF, may be greater than it, as in the figure here an- nexed ; in which case his demonstration must be changed into another. PROP. XXIV. B. I. To this is added, " of the two sides DE, DF, let DE be " that which is not greater than the other ;" that is, take that side of the two DE, DF which is not greater than the other, in order to make with it the angle EDG D equal to BAG ; because, without this restriction, there might be three differ- ent cases of the proposition, as Campa- nus and others make. Mr. Thomas Simpson, in p. 262 of the second edition of his Elements of Geometry, printed anno 1760, observes in his notes, that it ought to have been shown that the point F falls below the line EG ; this probably Euclid omitted as it is very easy to perceive, that DG being equal to DF, the point G is in the circumference of a cir- cle described from the centre D, at the distance DF, and must be in that part of it which is above the straight line EV, because DG falls above DF, the angle EDG being greater than the an- -Ic EDF. PROP. XXIX. B. I. The proposition which is usually called the 5t1i postulate, or 1 Ith axiom, bv some the 12th, on which this 29th depends, has NOTES. 303 given a great deal to do, both to ancient and modern geome- Book I. ters : it seems not to be properly placed among the axioms, as, '-"^"^^"'^ indeed, it is not self-evident ; but it may be demonstrated : thus DEFINITION 1. The distance of a point from a straight line, is the perpendi- cular drawn to it from the point. DEF. 2. One straight line is said to go nearer to, or further from, another straight line, when the distance of the points of the first from the other straight line become less or greater than they were ; and two straight lines are said to keep the same distance from one another, when the distance of the points of one of them from the other is always the same. AXIOM. A straight line cannot first come nearer to another straight line, and then go further from it, before A t) it cuts it ; and in like manner, a straight "~^ line cannot go further from anotherD- E straight line, and then come nearer to F -~-..,.__— ■ — ^ H it ; nor can a straight line keep the G same distance from another straight line, and then come nearer to it, or go further from it ; for a straight line keeps always the same direction. For example, the straight line ABC cannot first come nearer to the straight line DE, as from the B point A to the point B, and then, A -~^ p q. . * from the point B to the point C, go D P figure further from the same DE : and in F tT H above. like manner, the straight line FGH cannot go further from DE, as from F to G, and then, from G to H, come nearer to the same DE : and so in the last case, as in fig. 2. PROP. 1. If two equal straight lines AC, BD, be each at right angles to the same straight line AB ; if the points C, D be joined by the straight line CD, the straight line EF drawn from any point E in AB unto CD, at right angles to AB, shall be equal to AC, or BD. If EF be not equal to AC, one of them must be greater than tlie other ; let AC be the greater ; then, because FE is J04 NOTES. Book I. less than CA, the straight line CFD is nearer loathe straight s^'V"^ line AB at the point F than at the point C, that is, CF comes nearer to AB from the point C to F : but because DB is greater than FE, the straight line CFD is further from AB at the point D than at F, that is, FD goes further from AB from F to D : therefore the straight line CFD first comes nearer to the straight line AB, and then goes further from it, before it cuts it ; which is impossible. If FEbe said to be greater than C A, or DB, the straight line CFD first goes further from the straight line AB, and then comes nearer to it ; which is also impossible. Therefore FE is not unequal to AC, that is, it is equal to it. PROP. C ^ If two equal straight lines AC, BD be each at right angles to the same straight line AB ; the straight line CD which joins their extremities makes right angles with AC and BD. Join AD, BC ; and because, in the triangles CAB, DBA, CA, AB are equal to DB, BA, and the angle CAB equal to a 4 1 the angle DBA ; the base BC is equal a to the base AD : and in the triangles ACD, BDC, AC, CD are equal to BD, DC, and the base AD is equal to the base BC : therefore the angle ACD is b 8. 1. equalhto the angle BDC: from any point E in AB draw EF unto CD, at right angles to AB : therefore, by prop. 1, EF is equal to AC, or BD ; wherefore, as has been just now shown, the angle ACF is equal to the angle EEC : in the same manner, the angle BDF is equal to tlie angle EFD ; but the an- gles ACD, BDC are equal ; therefore the angles EEC and EFD c 10. clef. 1. '^J'e equal, and right angles^; wherefore also the angles ACD, BDC are right angles. CoR. Hence, if two straight lines AB, CD be at right angles to the same straight line AC, and if betwixt them a straight line BD be drawn at right angles to either of them, as to AB ; then BD is equal to AC, and BDC is a right angle. if AC be not equal to BD, take BG equal to AC, and join CO : therefore by this proposition, the angle ACG is a right angle ; but ACD is also a right angle ; wherefore the an- NOTES. ^6i gles ACD, ACG are equal to one another, which is impossible. Book I. Therefore BD is equal to AC ; and by this proposition BDC is v^"v^s» a right angle. PROP. 3. If two straight lines which contain ah angle be produced, there may be found in either of them a point from which the perpendicular drawn to the other shall be greater than any given straight line. Let AB, AC be two straight lines which make an angle with one another, and let AD be the given straight line ; a point may be found either in AB or AC, as in AC, from which the per- pendicular drawn to the other AB shall be greater than AD. In AC take any point E, and draw ET perpendicular to AB; produce AE to G, so that EG be equal to AE ; and produce FE to H, and make EH equal to FE, and join HG. Because, in the triangles AEF, GEH, AE, EF are equal to GE, EH, each to each, and contain equal a angles, the angle GHE is a 15. 1, therefore equal b to the angle AFE which is a right angle : b 4. 1. draw GK perpendicular to AB ; and because the straight lines FK, HG are at right angles to FH, and A , F K B M KG at right angles to FK ; KG is equal N- - to FH, by cor. pr. 2, that is, to the double of FE. In the same manner, if AG be produced to L, so that GL be equal to AG, and LM be drawn perpendicular to AB, then LM is double of GK, and so on. In AD take AN equal to FE, and AO equal to KG, that is, to the double of FE, or AN ; also, take AP equal to LM, that is, to the double of KG, or AO ; and let this be done till the straight line taken be greater than AD : let this straight line so taken be AP, and because AP is equal to LM, therefore LM is greater than AD, Which was to be done. PROP. 4. If two straight lines AB, CD make equal angles EAB, ECD with another straight line EAC towards the same parts of it 5 AB and CD are at right angles to some straight line, 3 O 306 NOTES. a 15. 1. b 4. 1. Book I. Bisect AC in F, and draw FG perpendicular to AB ; take ^-*•■'^''^"*^ CH in the straight line CD equal to AG, and on the contrary- side of AC to that on which AG is, and join FH : therefore, in the triangles AFG, CFH, the sides FA, AG are equal to FC, CH, each to each, and the angle FAG, that a is, EAB is equal to the angle FCH ; wherefore b the angle AGF is equal to CHF, and AFG to the angle CFH : to these last add the common angle AFH ; therefore the two angles AFG, AFH are equal to the two angles CFH, HFA, which two last are equal together to two right angles c, therefore also AFG, C H D AFH are equal to two right angles, and consequently d GF and FH are in one straight line. And because AGF is a right an- gle, CHF which is equal to it is also a right angle ; therefore the straight lines AB, CD are at right angles to GH. c 13. d 14. PROP. 5, a 23. 1. b 13. 1. If two straight lines AB, CD be cut by a third ACE so as to make the interior angles BAC, ACD, on the same side of it, together less than two right angles ; AB and CD being pro- duced shall meet one another towards the parts on which are the two angles which are less than two right angles. At the point C in the sti'aight line CE make a the angle ECF equal to the angle EAB, and draw to AB the straight line CG at right angles to CF : then, because the angles ECF, EAB are equal to one an- p other, and that the angles ECF, FCA are together equal b to two right an- gles, the angles EAB, FCA are equal to two right angles. But, by the hypothesis, the angles EAB, ACD are together less than two right an- gles ; therefore the angle FCA is greater than ACD and CD falls between CF and AB: and because CF and CD make an angle with one another, by prop. 3 a point may be found in either of them CD, from which the perpendicular drawn to CF shall be greater than the straight line CG, Let M K / / X \D N ^ A C ) ( J B H • NOTES. 307 this point be H,' and draw HK perpendicular to CF, meeting Book I. AB in L : and because AB, CF contain equal angles with AC v-.*'~v'^»w' on the same side of it by prop. 4^ AB and CF are at right angles to the straight line MNO which bisects AC in N and is perpen- dicular to CF : therefore, by cor. prop. 2, CG and KL which are at right angles to CF are equal to one another : and HK is greater than CG, and therefore is greater than KL, and conse- quently the point H is in KL produced. Wherefore the straight line CDH drawn betwixt the points C, H, which are on contrary sides of AL, must necessarily cut the straight line AL. PROP. XXXV. B. L The demonstration of this proposition is changed, because, if the method which is used in it was followed, there would be three cases to be separately demonstrated, as is done in the translation from the Arabic ; for, in the Elements, no case of a proposition that requires a different demonstration, ought to be omitted. On this account, we have chosen the method which Mons/ Clairault has given, the first of any, as far as I know, in his Elements, page 21, and which afterward Mr. Simpson gives in his page 32. But whereas Mr. Simpson makes use of prop. 26, b. 1, from which the equality of the two ti-iangles does not immediately follow, because, to prove that, the 4 of b. 1, must likewise be made use of, as may be seen in the very same case in the 34 prop. b. I, it was thought better to make use only of the 4 ofb. 1. PROP. XLV. B. I. The straight line KM is proved to be parallel to FL from the 33 prop.; whereas KH is parallel to FG by construction, and KHM, FGL have been demonstrated to be straight lines. A corollary is added from Commandine, as being often used. PROP. XIIL B. IL IN this proposition only acute angled triangles are mentioned, Book II. whereas it holds true of every triangle : and the demonstrations s^^n'^^^ of the cases omitted are added ; Commandine and Clavius have likewise given their demonstrations of these cases. PROP. XIV. B. II. In the demonstration of this, some Greek editor has ignorant- ly inserted the words " but if not, one of the two BE, ED is the S''^>*'' ticians and logicians, inveigh too severely against indirect or Apagogic demonstrations, and sometimes ignorantly enough, not being aware that there are some things that cannot be de- monstrated any other way : of this the present proposition is a very clear instance, as no direct demonstration can be given of it : because, besides the definition of a circle, there is no princi- ple or property relating to a circle antecedent to this problem, from which either a direct or indirect demonstration can be de- duced : wherefore it is necessary that the point found by the construction of the problem be proved to be the centre of the circle, by the help of this definition, and some of the preceding propositions : and because, in the demonstration, this proposi- tion must be brought in, viz. straight lines from the centre of a circle to the circumference are equal, and that the point found by the construction cannot be assumed as the centre, for this is the thing to be demonstrated ; it is manifest some other point must be assumed as the centre ; and if froni this assumption an absurdity follows, as Euclid demonstrates there must, then it is not true that the point assumed is the centre ; and as any point whatever was assumed, it follows that no point, except that found by the construction, can be the centre, from which the necessity pf an indirect demonstration in this case is evident. PROP. XIII. B. III. . As it is much easier to imagine that two circles may touch one another within in more points than one, upon the same side, than upon opposite sides ; the figure of that case ought not to have been omitted; but the construction in the Greek text would not have suited with this figure so well, because the centres of the circles must have been placed near to the cir- cumferences : on which account another construction and de- monstration is given, which is the same with the second part pf that which Campanus h^s translated from the Arabic-, NOTES. 309 where without any reason the demonstration is divided into two Book III. PROP. XV. B. III. The co; J verse of the second part of this proposition is want- ing, though in the preceding, the converse is added, in a like case, both in the enunciation and demonstration ; and it is now added in this. Besides, in the demonstration of the first part of this 1 5th, the diameter AD (see Commandine's figure) is proved to be greater than the straight line BC by means of another straight line MN ; whereas it may be better done without it : on which accounts we have given a different demonstration, like to that which Euclid gives m the preceding 14th, and to that which Theodosius gives in prop. 6, b. l,of his Spherics, in this very affair. PROP. XVI. B. III. In this we have not followed the Greek nor the Latin trans- lation literally, but have given what is plainly the meaning of the proposition, without mentioning the angle of the semicircle, or that which some call the cornicular angle which they con- ceive to be made by the circumference and the straight line which is at right angles to the diameter, at its extremity ; which angles have furnished matter of great debate between some of the modern geometers, and given occasion of deducing strange consequences from them, whicn are quite avoided by the man- ner in which we have expressed the proposition. And in like manner, we have given the true meaning of prop. 3 1, b. 3, with- out mentioning the angles of the greater or lesser segments : these passages, Vieta, with good reason, suspects to be adulte- rated, in the 386th page of his Oper. Math. PROP. XX. B. III. The first words of the second part of this demonstration, " x.ix.XxcrB-u 3)) TtuXiv" are Avrong translated by Mr. Briggs and Dr. Gregory " Rursus inclinetur ;" for the translation ought to be " Rursus inflectatur," as Commandine has it : a straight line is said to be inflected either to a straight or curve line, when a straight line is drawn to this line from a point, and from the point in which it meets it, a straight line making an angle with the former is drawn to another point, as is evi- dent from the 90Lh prop, of Euclid's Data : for this the whole line betwixt the first and last points, is inflected or broken at 310 NOTES._ Book III. the point of inflection, where the two straight lines meet. And ^■^'■^''''"'W in the like sense two straight lines are said to be inflected from two points to a third point, when they make an angle at this point ; as may be seen in the description given by Pappus Alex- andrinus of Appollonius's boohs de Locis planis, in the preface to his 7th book : we have made the expression fuller from the 90th prop, of the Data. PROP. XXI. B. III. There are two cases of this proposition, the second of which viz. when the angles are in a segment not greater than a semi- circle, is wanting in the Greek : and of this a more simple demonstration is given than that which is in Commandine, as being derived only from the first case, without the help of tri- angles. PROP. XXIII. and XXIV. B. III. In proposition 24 it is demonstrated, that the segment AFB must coincide with the segment CFD, (see Commandine's figure), and that it cannot fall otherwise, as CGD, so as to cut the other circle in a third point G, from this, that, if it did, a circle could cut another in more points than two : but this ought to have been proved to be impossible in the 23d prop, as well as that one of the segments cannot fall within the other : this part then is left out in the 24th, and put in its proper place, the 23d proposition. PROP. XXV. B. III. This proposition is divided into three cases, of which two have the same construction and demonstration ; therefore it is now divided only into two cases. PROP. XXXIII. B. III. This also in the Greek is divided into three cases, of which tv/o, viz. one in which the given angle is acute, and the other in which it is obtuse, have exactly the same construction and de- 7Bonstration ; on which account, the demonstration of the last case is left out as quite superfluous, and the addition of some unskilful editor ; besides the demonstration of the case when the angle given is a right angle, is done a roundabout way, and is therefore changed to a more simple one, as was done by Clavius. NOTES. PROP. XXXV. B. III. As the 25th and 3. Id propositions are divided into more cases, so this 35th is di\ided into fewer cases than are necessary. Nor can it be supposed that EucUd omitted them because they are easy ; as he has given the case, w^hich by far, is the easiest of them all, viz. that in w^hich both the straight lines pass through the centre : and in the following proposition he separately de- monstrates the case in which the straight line passes through the centre, and that in which it does not pass through the cen- tre : so that it seems Theon, or some other, has thought them too long to insert : but cases that require different demonstra- tions, should not be left out in the Elements, as was before taken notice of: these cases are in the translation from the Arabic, and are now put into the text. PROP. XXXVII. B. III. At the end of this, the words, " in the same manner it may " be demonstrated, if the centre be in AC," are left out as the addition of some ignorant editor. DEFINITIONS OF BOOK IV. WHEN a point is in a straight line, or any other line, this Book IV. point is by the Greek geometers said xzs-TiirB-xiy to be upon, or in ^^'~^'">*^ that line, and when a straight line or circle meets a circle any way, the one is said uTTTio-B-en to meet the other : but when a strciight line or circle meets a circle so as not to cut it, it is said {(pccTFTta-^-ui, to touch the circle ; and these two terms are never promiscuously used by them : therefore, in the 5th definition of book 4, the compound upcKrTr.rxi must be read, instead of the simple etTTTtiTcit: and in the 1st, 2d, 3d, and 6th definitions in Commandine's translation, " tangit," must be read instead of " contingit :" and m the 2d and 3d definitions of book 3, the same change must be made : but in the Greek text of proposi- tions nth, 12th, 13th, 18th, 19th, book 3, the compound verb is to be put for the simple. PROP. IV. B. IV. In this, as also in the 8th and 1 3th proposition of this book, it is demonstrated indirectly, that the circle touches a straight line : whereas in the 17th, 33d, and 37th propositions of book 3, the same thing is directly demonstrated :. and this way wc 312 NOTES. Book IV. have chosen to use in the propositions of this book, as it is ^•-^"^^^^w shorter, PROP. V. B. IV. The demonstration of this has been spoiled by some unskil- ful hand : for he does not demonstrate, as is necessary, that the two straight lines which bisect the sides of the triangle at right angles must meet one another ; and, without any reason, he di- vides the proposition into three cases ; whereas, one and the same construction and demonstration serves for them all, as Campanus has observed ; which useless repetitions are now left out : the Greek text also in the corollary is manifestly \'i- tiated, where mention is made of a given angle, though there neither is, nor can be any thing in the proposition relating to a given angle. PROP. XV. and XVI. B. IV. In the corollary of the first of these, the words equilateral and equiangular are wanting in the Greek : and in prop. 16, instead of the circle of ABCD, ought to be read the circumfer- ence ABCD : where mention is made of its containing fifteen equal parts. DEF. III. B. V. _ , y. MANY of the modern mathematicians reject this definition : ^^.^.,' the very learned Dr. Barrow has explained it at large at the end of his third lecture of the year 1666, in which also he answers the objections made against it as well as the subject would al- low : and at the end gives his opinion upon the whole, as fol- lows : " I shall only add, that the author had, perhaps, no other " design in making this definition, than (that he might more " fully explain and embellish his subject) to give a general " and summary idea of ratio to beginners, by premising •' this metaphysical definition, to the more accurate defini- " tions of ratios that are the same to one another, or one of " which is greater, or less than the other : I call it a meta- " physical, for it is not properly a mathematical definition, " since nothing in mathematics depends on it, or is deduced, " nor, as I judge, can be deduced from it : and the defini- " tion of analogy, which follows, viz. Analogy is the simi- NOTES. 313 ** litude of ratios, is of the same kind, and can serve for no Book V. " purpose in mathematics, but only to give beginners some ^-^"^^'^^i' " general, though gross and confused notion of analogy : but " the whole of the doctrine of ratios, and the whole of mathe- " matics, depend upon the accurate mathematical definitions " which follow this : to these we ought principally to attend, as " the doctrine of ratios is more perfectly explained by them ; " this third, and others like it, may be entirely spared without " any loss to geometry ; as Ave see in the 7th book of the Ele- " ments, where the proportion of numbers to one another is " defined, and treated of, yet without giving any definition of " the ratio of numbers ; though such a definition was as neces- " sary and useful to be given in that book, as in this : but in- " deed shere is scarce any need of it in either of them: though " I think that a thing of so general and abstracted a nature, and " thereby the more difficult to be conceived and explained, can- " not be more commodiously defined than as the author has " done : upon which account I thought fit to explain it at large, " and defend it against the captious objections of those Avho " attack it." To this citation from Dr. Barrow I have nothing to add, except that I fully believe the Sd and 8th definitions are not Euclid's, but added by some unskilful editor^ DEF. XI. B. V. It was necessary to add the word " continual" before " pro- " portionals" in this definition ; and thus it is cited in the 33d prop, of book 1 1 . After this definition ought to have followed the definition of compound ratio, as this was the proper place for it ; duplicate and triplicate ratio being species of compound ratio. But Theon has made it the 5th def. of book 6, where he gives an absurd and entirely useless definition of compound ratio : for this rea- son we have placed another definition of it betwixt the 1 1th and 12th of this book, which, no doubt, Euclid gave ; for he cites it expressly in prop. 23, book 6, and which Clavius, Herigon, and Barrow have likewise given, but they retain also Theon's, which they ought to have left out of the Elements. DEF. Xm. B. V. This, and the rest of the definitions following, contain the ex- plication of some terms which are used in the 5th and following books; which, except a few, are easilv enougli understood from. 2 R ■ iU NOTES. Book V. the propositions of this book where they are first mentioned : ^.^'-^^'^•^.^ they seem to have been added by Theon, or some other. How- ever it be, they are explamed something more distinctly for the sake of learners. PROP. IV. B. V. In the construction preceding the demonstration of this, the words d £ry;^£, any whatever, are twice wanting in the Greek, as also in the Latin translations ; and are now added, as being wholly necessary. Ibid, in the demonstration ; in the Greek, and in the Latin translation of Commandine, and in that of Mr. Henry Briggs, Avhich was published at London in 1620, together with the Greek text of the first six books, which translation in this place is followed by Dr. Gregory in his edition of Euclid, there is this sentence following, viz. " and of A and C have been taken " equimultiples K, L ; and of B and D, any equimultiples " whatever («.' irvx^) M, N ;" which is not true, the words " any whatever," ought to be left out : and it is strange that neither Mr. Briggs, who did right to leave out these words in one place of prop. 1 3 of this book, nor Dr. Gregory who chang- ed them into the word " some" in three places, and left them out in a fourth of that same prop. 13, did not also leave them out in this place of prop. 4, and in the second of the two places where they occur in prop. 17, of this book, in neither of which they can stand consistent with truth : and in none of all these ])laces, even in those which they corrected in their Latin trans- lation, have they cancelled the words » £Tt;;^sin the Greek text, as they ought to have done. The same words d irvx? are found in four places of prop. 1 1, of this book, in the first and last of which they are necessary, but in the second and third, though they are true, they are quite superfluous ; as they likewise are in the second of the two pla- ces in which til cy are found in the 12th prop, and in the like places of prop. 22, 23, of this book ; but are wanting in the last place of prop. 23, as also in prop. 25, book 1 1. COR. IV. PROP. B. V. This corollary has been unskilfully annexed to this propo- sition, and has been made instead of the legitimate dem.on- stration, which, without doubt, Theon, or some other editor, has taken away, not from this, but from its proper place in NOTES. 315 this book: the author of it designed to demonstrate, that if four Book V. magnitudes E, G, F, H be proportionals, they are also proper- ^.-'■^'■>*-' tionals inversely ; that is G is to E, as H to F ; which is true ; but the demonstration of it does not in the least depend upon this 4th prop, or its demonstration: for, when he says, " be- " cause it is demonstrated that if K be greater than M, L is " greater than N," £cc. This indeed is shown in the demon- stration of the 4th prop, but not from this, that E, G, F, H are proportionals; for this last is the conclusion of the proposition. Wherefore these words, " because it is demonstrated," Sec. are wholly foreign to his design : and he should have proved, that if K be greater than M, L is greater than N, from this, that E, G, F, H are proportionals, and from the 5th def. of this book which he has not ; but is done in proposition B, which we have given in its proper place, instead of this corollary ; and another corollary is placed after the 4th prop, which is often of use ; and is necessary to the demonstration of prop. 18 of tliis book. PROP. V. B. V. In the construction which precedes the demonstration of this proposition, it is required that EB may be the same multiple of CG, that AE is of CF -., that is, that EB be divided into as many equal parts, as there are parts in AE equal to CF : from which it is evident, that this construction is not Euclid's ; for he does not show the way of di\ading straight lines, and far less other magnitudes, into any number of equal parts, until the 9th pi'oposition of book 6 ; and he never requires any thing to be done in the construction, of which he had not before given the method of doing. For this reason, we have changed the construction to one, which, with- out doubt, is Euclid's, in which nothing is re- quired but to add a magnitude to itself a certain E— ' number of times ; and this is to be found in the translation from the Arabic, though the enuncia- tion of the proposition and the demonstration are there very much spoiled. Jacobus Peletarius, who was the first, as far as I know, who took no- B D tice of this error, gives also the right construc- tion in his edition of Euclid, after he had given the other Avhich he blames. He says, he Avould not leave it out, because it Avas fine, and might sharpen one's genius to invent others like it ; whereas there is not the least difference between the two demon- C 316 NOTES. Book V. strations, except a single word in the construction, which very ^^.^'^r^-m^ probably has been owing to an unskilful librarian. Clavius like- wise gives both the ways ; but neither he nor Peletarius takes notice of the reason why the one is preferable to the other. PROP. VI. B. V. There are tAvo cases of this proposition, of which only the first and simplest is demonstrated in the Greek ; and it is pro- bable Theon thought it was sufficient to give this one, since he was to make use of neither of them in his mutilated edition of the 5th book ; and he might as well have left out the other, as also the 5th proposition, for the same reason. The demonstra- tion of the other case is now added, because both of them, as also the 5th proposition are necessary to the demonstration of the 18th proposition of this book. The translation from the Arabic gives both cases briefly. PROP. A. B. V. This proposition is fi-equently used by geometers, and it is necessary in the 25th prop, of this book, 31st of the 6th, and 34th of the 1 1th, and 15th of the 12th book. It seems to have been taken out of the Elements by Theon, because it appeared evident enough to him, and others, who substitute the confused and indistinct idea the vulgar have of proportionals, in place of that accurate idea which is to be got from the 5th definition of this book. Nor can there be any doubt that Eudoxus or Eu- clid gave it a place in the Elements, when we see the 7th and 9th of the same book demonstrated, though they are quite as easy and evident as this. Alphonsus Borellus takes occasion from this proposition to censure the 5th definition of this book very severely, but most unjustly. In p. 126 of his Euclid restor- ed, printed at Pisa in 1658, he says, " Nor can even this least t* degree of knowledge be obtained from the foresaid property," viz. that which is contdned in 5th def. 5. " That, if four " magnitudes be proportionals, the third must necessarily be " greater than the fourth, when the first is greater than the " second ; as Clavius acknowledges in the 1 6th prop, of the " 5th book of the Elements." But though Clavius makes no such acknowledgment expressly, he has given Eorellus a han- dle to say this of him ; because when Clavius, in the above cited place, censures Commandine, and that very justly, for de- monstrating this proposition by help of the 16th of the fifth ; yet he himself gives no demonstration of it, but thinks it plain NOTES. 317 from the nature of proportionals, as he writes in the end of the Book V. 14th and 1 6th prop, book 5 of his edition, and is followed by He- ^--'"^•''^^m/ rigon in Schol. 1, prop. 14th, book 3, as if there was any nature of proportionals antecedent to that which is to be derived and understood from the definition of them. And, indeed, though it is very easy to give a right demonstration of it, nobody, as far as I know, has given one, except the learned Dr. Barrow, who, in answer to Borellus's objection, demonstrates it indirectly, but very briefly and clearly, from the 5th definition, in the 322d page of his Lect. Mathem. from which definition it may also be easily demonstrated directly. On which account we have placed it next to the propositions concerning equimultiples. PROP. B. B. V. This also is easily deduced from the 5th def. b, 5, and therc- fore is placed next to the other ; for it was very ignorantly made a corollary from the 4th prop, of this book. See the note on that corollary. PROP. C. B. V. This is frequently made use of by geometers, and is necessary to the 5th and 6th propositions of the 10th book. Clavius, in his notes subjoined to the 8th def. of book 5, demonstrates it only in numbers, by help of some of the propositions of the 7th book ; in order to demonstrate the property contained in the 5th definition of the 5th book, when applied to numbers, from the property of proportionals contained in the 20th def. of the 7th book. And most of the commentators judge it difficult to prove that four magnitudes which are proportionals according to the 20th def. of 7th book, are also proportionals according to the 5tli def. of 5th book. But this is easily made out, as follows. First, If A, B, C, D be four magnitudes, such that A is the same multiple, or the same part of B, which C is of D ; A, B, C, D are proportionals. This is demonstrated in proposition C. Secondly, If AB contain the same parts of CD, that EF does of GH : in this case likewise AB is to CD, as EF to GH. D K-V H Lf \ C E G M 318 NOTES. B H D Kt A C E G M Book V. Let CK be a part of CD, and GL the same part of GH ; v^'v"'^^ and let AB be the same multiple of CK, that EF is of GL : therefore, by prop. C, of 5th book, AB is to CK, as EF to GL: and CD, GH are equimultiples of CK, GL the second and fourth ; wherefore, by cor. prop. 4, book 5, AB is to CD, as EF to GFL And if four magnitudes be pro- portionals according to the 5th def. of book 5, they are also proportionals according to the 20th def. of book 7. First, If A be to B, as C to D ; then if A be any multiple or part of B, C is the same multiple or part of D, by prop. D, of book 5. Next, If AB be to CD, as EF to GH ; then if AB contains any parts of CD, EF contains the same parts of GH : for let CK be a part of CD, and GL the same part of GH, and let AB be a multiple of CK ; EF is the same multiple of GL : take M the same multiple of GL that AB is of CK ; there- fore by prop. C, of book 5, AB is to CK, as M to GL ; and CD, GH are equimultiples of CK, GL : wherefore by cor. prop. 4, b. 5, AB is to CD, as M to GH. And, by the hypothesis, AB is to CD, as EF to GH ; therefore M is equal to EF, by prop. 9, book 5, and consequently EF is the same multiple of GL that ABisofCK. PROP. D. B. V. This is not unfrequently used in the demonstration of other propositions, and is necessary in that of prop. 9,b. 6. It seems Theon has left it out for the reason mentioned in the notes at prop. A. PROP. VIII. B. V. In the demonstration of this, as it is now in the Greek, there are two cases (see the demonstration in Hervagius, or Dr. Gregory's edition), of which the first is that in which AE is less than EB ; and in this it necessarily follows, that H© the multiple of EB is greater than ZH, the same multiple of AE, which last multiple, by the construction is greater than A : whence also H© must be greater than A. But in the second case, viz. that in which EB is less than AE, though ZH be NOTES. ;i9 greater than A, yet H0 may be less than the same A ; so that Book V. there cannot be taken a multiple of A which is the first that is v.^~v<^^*. greater than K or H0, because A itself is greater than it : up- on this account, the author of this demonstration found it ne- cessary to change one part of the construction that was made use of in the first case : but he has, without any necessity, changed also another part of it, viz. when he orders to take N that multiple of A which is the y '. first that is greater than ZH ; , for he might have taken that multiple of A which is the first . that is gi-eater than Ho, or K, H— as was done in the first case : he likewise brings in this K E — . H-^ into the demonstration of both cases, without any reason ; for E 4- it serves to no purpose but to lengthen the demonstration. There is also a third case, © B A B A which is not mentioned in this demonstration, viz. that in which AE in the first, or EB in the second of the two other cases, is greater than D ; and in this any equimultiples, as the doubles, of AE, KB are to be taken, as is done in this edition, where all the cases are at once demonstrated : and from this it is plain that Theon, or some other unskilful editor, has vitiated this proposition. PROP. IX. B. V. Of this there is given a more explicit demonstration than that which is now in the Elements. PROP. X. B. V. It was necessary to give another demonstration of this pro- position, because that which is in the Greek and Latin, or other editions, is not legitimate : for the words greater, the same, or equal, lesser, have a quite different meaning when applied to magnitudes and ratios, as is plain from the 5 th and 7 th defini- tions of book 5. By the help of these let us examine the de- monstration of the 10th prop, which proceeds thus : " Let A *' have to C a greater ratio than B to C ; I say that A is greater '< than B. For if it is not greater, it is either equal, or less. " But A cannot be equal to B, because then each of them " would have the same ratio to C ; but they have not. There- " fore A is not equal to B." The force of v/hich reasoning is 32Q NOTES. Book V. this, if A had to C, the same ratio that B has to C, then if *«^''^"'''^*^ any equimultiples whatever of A and B be taken, and any- multiple whatever of C ; if the multiple of A be greater than the multiple of C, then, by the 5th def. of book 5, the multiple of B is also greater than that of C ; but, from the hypothesis that A has a greater ratio to C, than B has to C, there must, by the 7th def. of book 5, be certain equimultiples of A and B, and some multiple of C such, that the multiple of A is greater than the multiple of C, but the multiple of B is not greater than the same multiple of C ; and this proposition directly contradicts the preceding ; wherefore A is not equal to B. The demonstration of the 10th prop, goes on thus : " But nei- " ther is A less than B ; because then A would have a less ra- " tio to C, than B has to it : but it has not a less ratio, there- " fore A is not less than B," &c. Here it is said, that " A " would have a less ratio to C, than B has to C," or, which is the same thing, that B would have a greater ratio to C, than A to C; that is, by 7th def. book 5, there mvist be some equimultiples of B and A, and some multiple of C such, that the multiple of B is greater than the multiple of C, but the multiple of A is not greater than it : and it ought to have been proved that this can never happen if the ratio of A to C be greater than the ratio of B to C ; that is, it should have been proved, that, in this case, the multiple of A is always greater than the multiple of C, whenever the multiple of B is greater than the multiple of C ; for, when this is demonstrated, it will be evident that B cannot have a greater ratio to C, than A has to C, or, which is the same thing, that A cannot have a less ratio to C, than B has to C : but this is not all proved in the 10th proposition ; but if the 10th were once demonstrated, it would immediately follow from it, but cannot without it be easily demonstrated, as he that tries to do it will find. Where- fore the 10th proposition is not sufliciently demonstrated. And it seems that he who has given the demonstration of the 10th proposition as we now have it, instead of that which Eudoxus or Euclid had given, has been deceived in applying what is manifest, when understood of magnitudes, unto ratios, viz. that a magnitude cannot be both greater and less than another. That those things which are equal to the same arc equal to one another, is a most evident axiom when vmderstood of magnitudes ; yet Euclid does not make use of it to infer that those ratios which are the same to the same i-atio, are the same to one another ; but explicitly demonstrates this in prop. 11, of hook 5 . The demonstration we have given of the 1 0th prop, is NOTES. 321 lio doubt the same with that of Eudoxus or Euclid, as it is im- Book V. mediately and directly derived from the definition of a greater v-^'v''^>«-' Tatio, viz. the 7. of the 5. The abovementioned proposition, viz. If A have to C a greater ratio than B to C ; and if of A and B there be taken certain equimultiples, and some multiple of C ; then if the multiple of B be greater than the multiple of C, the multiple of A is also greater than the same, is thus demonstrated. Let D, E be equimultiples of A, B, and F a multiple of C, such, that E the multiple of B is greater than F ; D the multiple of A is also greater than F. Because A has a greater ratio to C, than B to C, A is greater than B, by the 10th prop. B. 5; therefore D the multiple of A is greater than E the same multiple of B : and E is greater than F ; much more therefoi'e D is greater than F. A C B C D PROP. XIII. B. V. In Commandine's,Briggs's, and Gregory's translations, at the beginning of this deinonstration, it is said, " And the multi- " pie of C is greater than the multiple of D ; but the multiple " of E is not greater than the multiple of F ;" which words are a literal translation from the Greek : but the sense evidently requires that it be read, " so that the multiple of C be greater " than the multiple of D ; but the multiple of E be not greater " than the multiple of F." And thus this place v, as restored to the true i^eading in the first editions of Commandine's Euclid, printed in 8 vo. at Oxford; but in the later editions, at least in that of 1747, the error of the Greek text Avas kept in. There is a corollary added to prop. 13, as it is necessary to the 20th and 2 1st prop, of this book, and is as useful as the proposition. PROP. XIV. B. V The two cases of this, which are not in the Greek, are add- ed; the demonstration of them not being exactly the same with that of the first case. 2 S NOTES. PROP. XVII. B. V. The order of the words in a clause of this is changed to one more natural: as was also done in prop. 1. PROP. XVIII. B. V. The demonstration of this is none of Euclid's, nor is it legi- timate ; for it depends upon this hypothesis, that to any three magnitudes, two of which, at least, are of the same kind, there may be a fourth proportional ; which, if not proved, the demonstration now in the text is of no force : but this is as- sumed without any proof; nor can it, as far as I am able to discern, be demonstrated by the propositions preceding this ; so far is it from deserving to be reckoned an axiom, as Cla- vius, after other commentators, would have it, at the end of the definitions of tlie 5th book. Euclid does not demonstrate it, nor does he show how to find the fourth propoitional, be- fore the 12th prop, of the 6th book: and he never assumes any thing in the demonstration of a proposition, which he had not before demonstrated.; at least, he assumes nothing the existence of which is not evidently possible ; for a certain conclusion can never be deduced by the means of an uncertain proposition : vipon this account, we have given a legitimate demonstration of this proposition instead of that in the Greek and other edi- tions, which very probably Theon, at least some other, has put in the place of Euclid's, because he thought it too prolix : and as the 17th prop, of v/hich this 18th is the converse, is de- monstrated by help of the 1st and 2d propositions of this book; so, in the demonstration now given of the 18th, the 5th prop, and both cases of the 6th are necessary, and these two propo- sitions are the converses of the 1st and 2d. Now the 5th and 6th do not enter into the demonstration of any proposition in this book as we now have it ; nor can they be of use in any proposition of the Elements, except in this 18th, and this is a manifest proof, that Euclid made use of them in his demon- stration of it, and that the demonstration now given, which is exactly the converse of that of the 1 7th, as it ought to be, dif- fers nothing from that of Eudoxus or Euclid : for the 5th and 6th have undoubtedly been put into the 5th book for the sake of some propositions in it, as all the other propositions about eq\iimultiples have been. Hieronymus Saccherius, in his book named Euclides ab om- ni nsivo viiidicutus, printed at. Milan, anno 1733, in 4to, ac- NOTES. 32: knowledges this blemish in the demonstration of the 18th, and Book V. that he may remove it, and render the demonstration we now ^•^'■^''^>*. have of it legitimate, he endeavours to demonstrate the follow- ing proposition, which is in page 1 1 5 of his book, viz. " Let A, B, C, D be four magnitudes, of which the two " first are of one kind, and also the two others either of the " same kind with the two first, or of some other the same " kind with one another. I say the ratio of the third C to the " fourth D, is either equal to, or greater, or less than the I'atio « of the first A to the second B." And after two propositions premised as lemmas, he proceeds thus. " Either among all the possible equimultiples of the first " A, and of the third C, and, at the same time, among all " the possible equimultiples of the second B, and of the " fourth D, there can be found some one multiple EF of the " first A, and one IK of the second B, that are equal to one " another; and also, in the same case, some one multiple " GH of the third C equal to LM the multiple of the fourth " D, or such equality is no where to be found. If the first " case happen, " [i. e. if such A E F " equality is to " be found] it is B I K " manifest from « what is be- C G H " fore demon- " strated, that D L M " A is to B as " C to D; but if such simultaneous equality be not to be " found upon both sides, it will be found either upon one " side, as upon the side of A [and B ;] or it will be found " upon neither side ; if the first happen ; therefore, from " Euclid's definition of greater and lesser ratio foregoing, " A has to B, a greater or less ratio than C to D ; accord- " ing as GH the multiple of the third C is less, or greater " than LM the multiple of the fourth D : but if the second " case happen ; therefore upon the one side, as upon the side " of A the first and B the second, it may happen that the " multiple EF, [viz. of the first] may be less than IK the " multiple of the second, while, on the contrary, upon the other " side, [viz. of C and D] the multiple GH [of the third C] is " greater than the other multiple LM [of the foui-th D :] and " then, from the same definition of Euclid, the ratio of the first " A to the second B, is less than the ratio of the third C to the *' fourth D ; or on the contrary. J24 ^OTES. Book v. " Therefore the axiom [i. e. the proposition before set down], V*^^^^"""^-^ " remains demonstrated," 8cc. Not in the least ; but it remains still undemonstrated : for what he says may happen, may, in innumerable cases, never happen ; and therefore his demonstration does not hold : for example, if A be the side, and B the diameter of a square ; and C the side, and D the diameter of another square ; there • can in no case be any multiple of A equal to any of B ; nor any one of C equal to one of D, as is well known ; and yet it can never happen that when any multiple of A is greater than a multiple of B, the multiple of C can be less than the mul- tiple of D, nor when the multiple of A is less than that of B, the multiple of C can be greater than that of D, viz. taking equimultiples of A and C, and equimultiples of B and D : for A, B, C, D are proportionals ; and so if the multiple of A be greater. Sec. than that of B, so must that of C be greater. Sec. than that of D ; by 5th def. b. 5. The same objection holds good against the demonstration which some give of the 1st prop, of the 6th book, v.'hich we have made against this of the 18th prop, because it depends upon the banie insufficient foundation with the other, PROP. XIX. B. V. A corollary is added to this, which is as frequently used as the proposition itself. The covoUary which is subjoined to it in the Greek, plainly shows that the 5th book has been vitiated by editors who were not geometers : for the conversion of ratios does not depend upon this 19th, and the demonstration which several of the commentators on Euclid gave of conver- sion, is not legitimate, as Clavius has rightly observed, who has given a good demonstration of it which Ave have put in pro- position E ; but he makes it a corollary from the 19th, and be- fpns it Avith the words, " Hence it easily follows," though it does not at all follow from it. PROP. XX. XXI. XXII. XXIII. XXIV. B. V. The demonstrations of the 20th and 21st propositions, are shorter than those Euclid gives of easier propositions, either in the preceding, or following books : wherefore it Avas pro- per to make them more explicit, and the 22d and 23d propo-> hitions are, as they ought to be, extended to any number of NOTES. 325 Hiagnitudes : and, in like manner, may the 24th be, as is taken Book V. notice of in a corollary ; and another corollary is added, as use- v^^n^^*^ fill as the proposition, and the words " any whatever" are sup- plied near the end of prop. 23, which are wanting in the Greek text, and the translations from it. In a paper writ by Phillippus Naudxus, and published after his death, in the History of the Royal Academy of Sciences of Berlin, anno 1745, page 50, the 23d prop, of the 5th book is censured as being obscurely enunciated, and, because of this, prolixly demonstrated: the enunciation there given is not Eu- clid's, but Tacquet's, as he acknowledges, which, though not so well expressed, is, upon the matter, the same with that which is now in the Elements. Nor is there any thing obscure in it, though the author of the paper has set clown the proportionals in a disadvantageous order, by which it appears to be obscure: but, no doubt, Euclid enunciated this 23d, as well as the 2 2d, so as to extend it to any number of magnitudes, which taken two and two are proportionals, and not of six only ; and to this general case the enunciation which Naudxus gives, cannot be well applied. The demon;.tration which is given of this 23d, in that paper, is quite wrong; because, if the proportional magnitudes be plane or solid figures, there can no rectangle (which he impro- perly calls difiroduct) be conceived to be made by any two of them : and if it should be said, that in this case straight lines are to be taken which are proportional to the figures, the demon- stration would this way become much longer than Euclid's : but, even though his demonstration had been right, who does not see that it could not be made use of in the 5th book. PROP. F, G, H, K. B. V. These propositions are annexed to the 5th book, because they are frequently made use of by both ancient and modern geome- ters : and in many cases compound ratios cannot be brought into demonstration, without making use of them. Whoever desires to see the doctrine of ratios delivered in this 5th book solidly defended, and the arguments brought against it by And. Tacquet, Alph. Borellus, and others, fully refuted, may read Dr. Barrow's Mathematical Lectures, viz. the 7th and 8th of the year 1666. The 5th book being thus corrected, I most readily agree to what the learned Dr. Barrow says,* " That there is nothing " in the whole body of the Elements of a more subtile invention, * Pag-e 336. 326 NOTES. Book V, « nothing more solidly established, and more accurately hand- ^-^'^'"^ii' " led, than the doctrine of proportionals." And there is some ground to hope, that geometers will think that this could not have been said with as good reason, since Theon's time till the present. DEF. II. AND V. OF B. VI. Book VI. THE 2d definition does not seem to be Euclid's, but some \in ^•^~'^'^>»-^ skilful editor's : for there is no mention made by Euclid, nor, as far as I know, by any other geometer, of reciprocal figures: it is obscurely expressed, which made it proper to render it more distinct : it would be better to put the following definition in place of it, viz. DEF. II. Two magnitudes are said to be reciprocally proportional to two others, when one of the first is to one of the other magni- tudes, as the remaining one of the last two is to the remaining one of the first. But the 5th definition, which since Theon's time, has been kept in the Elements, to the great detriment of learners, is now justly thrown out of them, for the reason given in the notes on the 23d prop, of this book. PROP. I. and II. B. VI. To the first of these a corollary is added, which is often used : and the enunciation of the second is made more general. PROP. III. B. VI. A second case of this, as useful as the first, is given in prop. A: viz. the case in which the exterior angle of a triangle is bi- sected by a straight line : the demonstration of it is very like to that of the first case, and upon this account may, probably, have been left out, as also the enunciation, by some unskilful editor : at least, it is certain, that Pappus makes use of this case, as an elementary proposition, without a demonstration of it, m prop. 39 of his 7th book of Mathematical Collection's. NOTES. PROP. VII. B. VI. To this a case is added which occurs not unfrequently in de- monstrations. PROP. VIII. B. VI. It seems plain that some editor has changed the demonstra- tion that Euclid gave of this proposition : for, after he has de- monstrated, that the triangles are equitingular to one another, he particularly shows that their sides about the equal angles are proportionals, as if this had not been done in the demonstration of the 4th prop, of this book : this superfluous part is not found in the translation from the Arabic, and is now left out. PROP. IX. B. VI. This is demonstrated in a particular case, viz. that in which the thii'd part of a straight line is required to be cut off ; which is not at all like Euclid's manner : besides, the author of the demonstration, from four magnitudes being proportionals, con- cludes that the 'bird of them is the same multiple of the fourth, which the first is of the second ; now, this is no where demon- strated in the 5th book, as we now have it : but the editor as- sumes it from the confused notion which the vulgar have of pro- portionals : on this account, it was necessary to give a general and legitimate demonstration of this proposition. PROP. XVIII. B. VI. The demonstration of this seems to be vitiated: for tht- proposition is demonstrated only in the case of quadrilateral figures, without mentioning how it may be extended to figures of five or more sides : besides, from two triangles being equi- angular, it is inferred, that a side of the one is to the homolo- gous side of the other, as another side of the first is to the side homologous to it of the other, without permutation of the proportionals ; which is contrary to Euclid's manner, as is clear from the next proposition : and the sam^e fault occurs again in the conclusion, where the sides about the equal angles are not shown to be proportionals, by reason of again neglect- ing permutation. On these accounts, a demonstration is given in Euclid's manner, like to that he makes use of in the "Otl> 328 NOTES. Book VI. prop, of this book ; and it is extended to five-sided figures, by ^-^'^•^'^'^m^ whicij it may be seen how to extend it to figures of any number of sides. PROP. XXIII. B. VI. Nothing is usually reckoned more difficult in the elements of geometry by learners, than the doctrine of compound ra- tio, which Thcon has rendered absurd and ungcometrical, by substituting the 5th definition of the 6th book in place of the right definition, which without doubt Eudoxus or Euclid gave, in its proper place, after the definition of triplicate ratio &c. in the 5th book. Theon's definition is this ; a ratio is said to be compounded of ratios 'orctv ai tuv Xoyuv TtyiXiKor/jne i^' ^ixvtag 7reXA«57Aa(7/«5-.^«5-a!< 7rotai^'>«-' explain it by long commentaries, when they ought rather to have taken it quite away from the Elements. For, by comparing def. 5, book 6, with prop. 5, book 8, it will clearly appear that this definition has been put into the Elements in place of the right one which has been taken out of them : because, in prop. 5, book 8, it is demonstrated that the plane number of which the sides are C, D has to the plane number of which the sides are E, Z, (see Hergavius' or Gregory's edition,) the ratio which is compounded of the ra- tios of their sides ; that is, of the ratios of C to E, and D to Z-: and by def. 5. book 6. and the explication given of it by all the commentutoi's, the ratio which is compounded of the ra- tios of C to E, and D to Z, is the ratio of the product made by tiie multiplication of the antecedents C, D to the product of the consequents E, Z, that is, the ratio of the plane number of which the sides are C, D to the plane number of which the sides are E, Z. Wherefore the proposition which is the 5th def. of book 6, is the very same with the 5th prop, of book 8, and therefore it ought necessarily to be cancelled in one of these places ; because it is absurd that the same proposition should stand as a definition in one place of the Elements, and be de- monstrated in another place of them. Now, there is no doubt that prop. 5, book 8, should have a place in the Elements, as the same thing is denaonstrated in it concerning plane num- bers, which is demonstrated in prop. 23d, book 6, of equiangu- lar parallelograms ; wherefore def. 5, book 6, ought not to be in the Elements. And from this it is evident that this definition is not Euclid's, but Theon's, or some other unskilful geometer's. But nobody, as far as I know, has hitherto shown the true use of compound ratio, or for what purpose it has been in- troduced into geometry: for every proposition in wnich . compound ratio is made use of, may v/ithout it be both enun- ciated and demonstrated. Now the use of compound ratio consists wholly in this, that by means of it, circumlocutions may be avoided, and thereby propositions may be more brief- ly either enunciated or demonstrated, or both may be done, for instance if this 23d proposition of the sxith book were to be enunciated, without mentioning compound ratio, it might be done as follows. If two parallelograrns be equiangular, and if as a side of the first to a side of the second, so any assumed straight line be made to a second straight line ; and as the other side of the first to the other side of the second, so the se- cond sti'aight line be made to a third. The first parallelogram is to the second, as the first straight line to the third. And the NOTES. 331 demonstration would be exactly the same as we now have it. Baok Vi. But the ancient geometers, when they observed tnis enunciu.- '^~i'^^'^^>^ tion could be made shorter, by giving a name to the ratio which the first straight line has to the last, by which name the intermediate ratios might likewise be signified, of the first to the second, and of the second to the thiid, and so on, if there Avere more of them, they called this ratio of the first to the last, the ratio compovmded of the ratios of the first to the se- cond, and of the second to the third straight line : that is, in the present example, of the ratios which are the same with the ratios of the sides, and by this they expressed the proposi- tion more briefly thus : if there be two equiangular parallel- ograms, they have to one another the ratio which is the same with that which is compounded of ratios that are the same with the ratios of the sides. Which is shorter than the preceding enunciation, but has precisely the same meaning. Or yet shorter thus : equiangular parallelograms have to one another the ratio which is the same with that which is com- pounded of the ratios of their sides. And these two enuncia- tions, the first especially, agree to the demonstration which is now in the Greek. The proposition may be more briefly de- monstrated, as Candulla does, thus : let ABCD, CEJ: G be two equiangular parallelograms, and complete the parallelo- gram CDHG ; then, because there are three parallelograms AC, CH, CF, the first AC (by the definilion of compound ratio) has to the third CF, the ratio A D H which is compounded of the ratio of the first AC to the second CH, and of the ratio of CH, to the third CF ; but B the paralleogram AC is to the pa- rallelogram CH, as the straight line BC to CG ; and the parallelogram CH is to CF, as the straight line E . F CD is to CE ; therefore the parallelogram AC has to CF the ratio which is compounded of ratios that are the same v.ith the ratios of the sides. And to this demonstration agrees the enun- ciation which is at present in the text, viz. Equiangular parallel- ograms have to one another the ratio which is compovmded of the ratios of the sides : for the vulgar reading, " wiiich is com- " pounded of their sides," is absurd. But, in this edition, vie have kept the demonstration which is in the Greek text, though not so short as Candalla's ; because the way of finding the ratio which is compounded of the ratios of the sides, that is, of find- ing the ratio of the parallelograms, is shown in that, but not in Candalla's demonstration ; whereby beginners may learn, in likg 1 c 332 NOTES. Book VI. cases, how to find the ratio which is compounded of two or more v^'"'<'">*^ given ratios. From what has been said, it may be observed, that in any magnitudes whatever of the same kind A, B, C, D, Sec the ratio compounded of the ratios of the first to the second, of the second to the third, and so on to the last, is only a name or expression by which the ratio which the first A has to the last D is signified, and by wliich at the same time the ratios of all the magnitudes A to B, B to C, C to D from the first to the last, to one another, whether they be the same, or be not the same, are indicated ; as in magnitudes which are continual proportionals A, B, C, D, &.c. the duplicate ratio of the first to the second is only a name, or expression by which the ratio of the first A to the third C is signified, and by which, at the same time, is shown that there are two ratios of the magni- tudes, from the first to the last, viz. of the first A to the se- cond B, and of the second B to the third or last C, which are the same with one another ; and the triplicate ratio of the first to the second is a nam^ or expression by which the ratio of the first A to the fourth D is signified, and by which, at the same time is snown that there are three ratios of the magni- tudes fiom the first to the last, viz. of the first A to the se- cond B, and of B to the third C, and of C to the fourth or last D, which are all the same with one another ; and so in the case of any other multipiicate ratios. And that this is the right explication of the meaning of these ratios is plain from tne definitions of duplicate and triplicate ratio in which Euclid makes use of the Avord .-viici, is said to be, or is called; which V/ord, he, no doubt, made use of also in the definition of compound ratio, which Theon, or some other, has expung- ed from the Elements ; for the very same word is still retained in the wrong definition of compound ratio, which is now the 5th of the 6th book : but in the citation of these definitions it is some times retained, as in the demonstration of prop. 19, book 6, " the first is sajd to have, '^z-hv asy6t«(, to the third the " duplicate ratio," &c, wi:ich is wrong translated by Comman- dine and others, " has" instead of " is said to have :" and sometemes it is left out, as in the demonstration of prop. 33. of the 1 1th book, in which Ave find " the first has, '3;%j«, to the " third the triplicate ratio ;'* but without doubt 's^ <, " has," in this place signifies the same as 's;v^^v Xiyirai^ is said to have: so likewise in prop. 23, B. 6, we find this citation, " but the " ratio of K to M is compounded, c-vyxitrxi, of the ratio of " K to L, and the ratio of L to M," which is a shorter way of expressing the sa.nie thing, which, according to the d-efinition. NOTES. 333 ought to have been expressed by o-!;v>6««-.^«< AsysTai, is said so be Book VI. compounded. Si^'N'^^/ From these remarks, together with the propositions subjoined to tne 5th book, aii tnat is found concerning compound ratio, either in the ancient or modern geometers, may be understood and explained. PROP. XXIV. B. VI. It seems that some unskilful editor has made up this demon- stration as we now have it, out of two others ; one of whici- may be made from the 2d prop, and the otner from the 4th or tuis book : for after he has, from the 2d of this book, and compo- sition and permutation, demonstrated that the sides about the angle common to the two parallelograms are propoi'tionals, he might nave immeciia.tely concluded that the sides about the other equal angles were proportionals, viz. from prop 34, B. 1, and prop. 7, book 5. This he does not, but proceeds to show that the triangles and parallelograms are equiangular ; and in a te- dious way, by help of prop. 4, of this book, and the 22d of book 5, deduces the same conclusion : from which it is plain that this ill composed demonstration is not Euclid's : these su- perfluous things are now left out, and a more simple demonstra- tion is given from the 4th prop, of this book, the same which is in tne translation from the Arabic, by the help of the 2d prop, and composition ; but in this the author neglects permutation, and does not show the parallelograms to be equiangular, as is proper to do for the sake of beginners. PROP. XXV. B. VI. It is very evident that the demon sti^ation which Euclid had given of this proposition has been vitiated by some unskilful hand : for, after this editor had demonstrated that " as the " rectilineal figure ABC is to the rectilineal KGH, so is the " parallelogram BE to the parallelogram EF ;" nothing more should have been" added but this, " and the rectilineal figure " ABC is equal to the parallelogram BE: therefore" the recti- " lineal KGH is equal to the parallelogram EF," viz. from prop. 14, book 5. But betwixt these two sentences he has in- serted this ; '' wherefore by permutation, as the rectilineal fi- " gure ABC to the parallelogram BE, so is the rectilineal KGH 354 NOTES. ^Book VI. " to the parallelogram EF ;" by which, it is plain, he thought ^i^'V''^^ it was not so evident to conclude tliat the second of four pro- portionals is equal to the fourth from the equality of the first and third, which is a thing demonstrated in the l4th prop, of B. 5, as to conclude that the third is equal to the fourth, from the equality of the first and stcond, wiiich is no where demon- strated in the Elements as Ave now have them : but though this proposition, viz. the third of four proportionals is equal to the fourth, if the third be equal to the second, had been given in the Elements by Euclid, as very probably it was, yet he would not have made use of it in this place ; because, as was said the conclusion could have been immediately deduced without this superfluous step by permutation : this v/e have shown at the greater length both because it affords a certain proof of the Adtiation of the text of Euclid ; for the very same blunder is found twice in the Greek text of prop. 23, book 1 1, and twice in prop. 2, B. 12, and in the 5, 11, 12, and 18th of that book ; in which places of book 12, except the last of them, it is rightly left out in the Oxford edition of Commandine's translation ; and also that geometers may beware of making use of permu- tation in the like cases ; for the moderns not unfrequently com- mit this mistake, and among others Commandine himself in his commentary on prop. 5, book 3, p. 6, b. of Pappus Alexandri- nus, and in other places : the vulgar notion of proportionals has, it seems, pre-occupied many so much, that they do not suf- ficiently understand the true nature of them. Besides, though the rectilineal figure ABC, to which another is to be made similar, may be of any kind whatever ; yet in the demonstration the Greek text has " triangle" instead of " recti- " lineal figure," which error is corrected in the above named Oxford edition. PROP. XXVII. B. VI. The second case of this has '«aa»?, otherwise, prefixed to it, as if it was a different demonstration, which probably has been done by some unskilful librarian. Dr. Gregory has rightly left it out : the scheme of this second case ougiit to be marked with the same letters of the alphabet which are in the scheme of the first, as is now done. PROP. XXVIII and XXIX. B. VI. These two problems, to the first of which the 27th prop, iiv iiecessai-y, are the most general and useful in all the Elements, NOTES. 3*35 and are most frequently made use of by the ancient gometers Book VI. in the solution of other problems ; and therefore are very igno- *<^^"v-Xi,^ rantly left out by Tacquet and Dechales in their editions of the Elements, who pretend that they are scarce of any use. The cases of these pi'oblems, wherein it is required to apply a rect- angle which shall be equal to a given square, to a given straight line, either deficient or exceeding by a square ; as also to apply a rectangle which shall be equal to another given, to a given straight line, deficient or exceeding by a square, are very often made use of by geometers. And, on this account, it is thought proper, for the sake of beginners, to give their constructions as follows : 1 . To apply a rectangle which shall be equal to a given square, to a given straight line, deficient by a square : but the given square must not be greater than that upon the half of the given line. I n^ D /G / c K B Let AB be the given straight line, and let the square upon the given straight line C be that to which the rectangle to be ap- plied must be equal, and this square by the determination, is not greater than that upon half of the straight line AB. Bisect AB in D, and if the square upon AD be equal to the square upon C, the thing required is done : but if it be not equal to it, AD must be greater than C, according L H to the determination : draw DE at right angles to AB and make it equal to C : A produce ED to F, so that EF be equal to AD or DB and from the centre E, at the distance EF, describe a circle meeting AB in G, E and upon GB describe the square GBKH, and complete the rectangle AGHE ; also join EG. And because AB is bisected in D, the rectangle AG GB together with the square of DG is equal a to (the square of DB, that is, of EF or EG, that is, a 5. 2. to) the squares of ED, DG : take away the square of DG from each of these equals ; therefore the remaining rectangle AG, GB, is equal to the square of ED, that is, of C : but the rectangle AG, GB is the rectangle AH, because GH is equal to GB ; therefore the rectangle AH is equal to the given square upon the straight line C. Wherefore the rectangle AH, equal to the given square upon C, has been applied to the given 336 NOTES. Book VI. straight line AB, deficient by the squai^e GK. Which was to ^^-^"^^"^^fm^ be done. 2. To apply a rectangle Avhich shall be equal to a given squtU'e, to a given straight line, exceeding by a square. Let AB be the given straight line, and let the square upon the given straight line C be that to M^hich the rectangle to be applied must be equal. Bisect AB in D, and dravi^ BE at right angles to it, so that BE be equal to C ; and having joined DE, from the centre D at the distance DE describe a circle meeting AB pi'oduced in G ; upon BG describe the square BGHK, and complete the rc;ct- angle AGHL. And because AB is bisected in D, and produced to G, the rectangle AG, GB together with the square of DB a 6. 2. is equal a to (the square of DG or DE, that is, to) the squares F of EB, BD. From each of these equals take the square of DB ; C therefore the remaining rectangle AG, GB is equal to the square of BE, that is, to the square upon C. But the rectan- gle AG, GB is the rectangle AH, because GH is equal to GB : therefore the rectangle AH is equal to the square upon C. Wherefore the rectangle AH, equal to the given square upon C, has been applied to the given straight line AB, exceeding by the square GK. Which was to be done. 3. To apply a rectangle to a given straight line which shall be equal to a given rectangle, and be deficient by a square. But the given rectangle must not be greater than the square upon the half of the given straight line. Let AB be the given straight line, and let the given rectan- gle be that which is contained by the straight lines C, D Av.iich is not greater than the square upon tlie half of AB ; it is re- quired to apply to AB a rectangle equal to the rectangle C, D, deficient by a square. Draw AE, BF at right angles to AB, upon the same side of it, and make AE equal to C, and BF to D : join EF, and bisect it in G ; and from the centre G, at tiie cUatance GE, describe a circle mes^ting AE again in H ; join HF, and draw GK parallel to it, and GL parallel to AE, meeting AB in L. NOTES. 337 Because the angle EHF in a semicircle is equal to the right Book VI. angle EAB, AB and HF are parallels, and AH and BF are ^-^'■^'"^i^ parallels ; wherefore AH is equal to BF, and the rectangle EA, AH equal to the rectangle EA, BF, that is, to the rectangle C, D : and because EG, GF are equal to one another, and AE, LG, BF parallels : therefore AL and LB are equal ; also EK is equal to KH a, and the rectangle C, D from the a 3. 3. determination, is not greater than the square of AL the half of AB ; wherefore the rectangle EA, AH is not gi'eater than the square of AL, that is of KG : add to each the square of KE ; therefore the square b of AK is not greater than the b 6. 2. squares of EK, KG, that is, ^ C than the squre of EG ; and consequently the straight line AK or GL is not greater than GE. Now, if GE be equal to GL, the circle EHF touches AB in L, and there. fore the square of AL is c equal to tiie rectangle EA, AH, that is, to the given rect- angle C, D ; and that wiiich was required is done : but if EG, GL be unequal, EG must be the greater : and c 36. therefore the circle EHF cuts the straight line AB ; let it cut it in the points M, N, and upon NB describe the square NBOP, and complete the rectangle ANPQ : because ML is equal to dj. 3. 3. LN, and it has been proved that AL is equal to LB ; therefore AM is equal to NB, and the rectangle AN, NB equal to the rectangle N A, AM, that is, to the rectangle e E A, AH, or the g q^^ gg. rectangle C, D : but the rectangle AN, NB is the rectangle 3. AP, because PN is equal to NB : therefore the rectangle AP is equal to the rectangle C, D ; and the rectangle AP equal to the given rectangle C, D has been applied to the given straight line AB, deficient by the square BP. Wiiich Vv^as to be done. 4. To apply a rectangle to a given straight line that shall be equal to a given rectangle, exceeding by a square Let AB be the given straight line, and the rectangle C, D the given rectangle, it is required to apply a rectangle to AB equal to CD, exceeding by a square. Draw AE, BF at right angles to AB, on the contrary sides of it, and make AE equal to C, and BF equal to D : join EF, and bisect it in G ; and from the centre G, at the distance ^ U 338 NOTES. Book VI GE, describe a circle meeting AE again in H ; join HF, and v.^^>^'^^ draw GL parallel to AE ; let the circle meet AB pro- duced in M, N, and upon BN describe the square NBOP, and complete the rectangle ANFQ : because the angle EHl- in a semi- circk is equal to the right angle EAB, AB cUid HF are parallels, and therefore AH and BF are eqUcil, and the rectangle E A, AH equal to the rectangle E A,BF,that is, to the rectangle C, D : and because ML is equal to LN, and AL to LB, therefoi'e MA is equal to BN, and the rectangle AN, NB to MA, AN, that is, a 35. 3. a to the rectangle EA, AH, or the rectangle C, D : therefore the rectangle AN, NB, that is, AP, is equal to the rectangle C, D ; and to the given straight line AB the rectangle AP has been applied equal to the given rectangle C, D, exceeding by the square BP. Which was to be done. Willebrordus Snellius was the first, as far as I know, who gave these constructions of the 3d and 4th problems in his Ap- pollonius Batavus : and afterwards the learned Dr. Halley gave them in the Scholium of the 18th prop, of the 8th book of Ap- polonius's conies restored by him. The 3d problem is otherwise enunciated thus : To cut a given straight line AB in the point N, so as to make the rect- angle AN, NB equal to a given space : oi", which is the same thing, having given AB the sum of the sides of a rect- angle, and the magnitude of it being likewise given, to find its sides. And the fourth problem is the same with this. To find a point N in the given straight line AB produced, so as to make the rectangle AN, NB equal to a given space : or, wiiich is the same thing, having given AB the difterence of the sides of a rectangle, and the magnitude of it, to find the sides. NOTES. PROP. XXXI. B. VI. In the demonstration of this, the inversion of proportionals is twice neglected,' and is now added, that the conclusion m^y be legitimately made by help of the 24th prop, of B. 5. as Clavius had done. PROP. XXXII. B. VI. The enunciation of the preceding 26th prop, is not general enough ; because not only two simii^ir paralleiogranis tiuit have an angle common to both, are about the same cliumeter ; but likewise two similar parallelograms that have vertically opposite angles, have their diameters in the same straight line : but ti.ere seems to have been another, and that a direct demonstration of these cases, to which this 32d proposition was needful : and the 32d may be otherwise and something more briefly demonstrated as follows : PROP. XXXII. B. VI. If two triangles which have two sides of the one, &c. Let Gx\F, HFC be two triangles which have two sides AG, GF, proportional to the two sides fH, HC, viz. AG to GF, as FH to HC ; and let AG be paral- lel to FH, and GF to HC ; AF and FC are in a straight line. Draw CK parallel a to FH, and let it meet GF produced in K ; because AG, KC are each of them parallel to FH, they are parallel b to one another, and therefore the alternate angles AGF, FKC are G F D ;H a 31.1. b 30. 1. B K 14. 1. equal : and AG is to GF, as (FH to HC, that is c) CK to KF ; c 34. 1 wherefore the triangles AGF, CKF are equiangular d, and the d 6. 6. angle AFG equal to the angle CFK : but GFK is a straight line, therefore AF and FC are in a straight linee. The 26th prop, is demonstrated from the 32d, as follows : If two similar and similarly placed parallelograms have an angle common to both, or vertically opposite angles ; their di- ameters are in the same straight line. First, Let the parallelogram-j ABCD, AEFG have the angle BAD common to both, and be similar, and similarly placed ; ABCD, AEFG are about the same diameter. 340 NOTES. a Cor. 5, Book VI. Produce EF, GF, to H, K, and join FA, FG : then be- y-^^^^^"*-^ caust tiie parallelograms ABCD, AE,FG are similar, DA is to AB, as GA to AE : where- AG D 19. fore the remainder DG is a to the remainder EB, as G A to AE : but DG is equal to FH, EB to HC, E [ ^i^ j h and AE to GF : therefore as FH to HC, so is AG to GF ; and FH, HC are parallel to AG, GF ; and the triangles AGF, FHC are joined at one angle, in the point 6. F ; wherefore Ai' , FC are in the same straight line b. Next, Let the parallelograms KFHC,GFEA, which are simi- lar and similarly placed, have their angles KFH, GFE vertically opposite ; their diameters AF, FC are in the same straight line. Because AG, GF are parallel to FH, HC ; and that AG is to GF, as FH to HC: therefore AF, FC ai'e in the same straight line b. A G E \ r \ B K PROP. XXXHI. B. VL The words " because they are at the centre," are left out, as the addition of some unskilful hand. In the Greek, as also in the Latin translation, the words « irv)(,h " any whatever," are left out in the demonstration of both parts of the proposition, and are now added as quite neces- sary ; and in the demonstration of the second part, where the triangle BGC is proved to be equal to CGK, the illative par- ticle ci£» in the Greek text ought to be omitted. The second part of tl;e proposition is an addition of Thecn's, as he tells us in his commentary on Ptolemy's Miyd?,-,; I^wTa^ic^ p. 50. PROP. B.C. D. B.VL These three propositions are added, because they are fre- quently made use ol by geometers. NOTES. DEF. IX. and XL B. XI. THE similitude of plane figures is defined from the equality of their angles, and tiie proportionality of tiie siaes about the equal angles ; for from the proportionality of the sidv^s oniy, or ouiy from tne equality of the angles, the similitude of the figures does not follow, except in the case wnen the figures are trian- gles : the similar position of the sides whicli contain the figures, to one another, depenuing partly upon eacr. of these : and, for the same reason, those are similar soiid figures which have all their solid angles equal, each to each, and are contained by the same number of similar plane figures : for there are some solid figures contained by similar plane figures, of the same number, and even of the same magnitude, that are neither similar nor equal, as shall be demonstrated after the notes on the 10th defi- nition : upon this account it was necessary to amend the defini- tion of similar solid figures, and to place the definition of a solid angle before it: and from this and the 10th definition, it is suf- ficiently plain how much the Elements have been spoiled by unskilful editors. DEF. X. B. XI. Since the meaning of the word " equal" is known and established before it comes to be used in this definition ; therefore the proposition which is the 10th definition of this book, is a theorem, the truth or falsehood of which ought to be demonstrated, not assumed ; so that Theon, or some other editor, has ignorantly turned a theorem which ought to be demonstrated into this 10th definition: that figures are similar, ought to be proved from the definition of similar figures; that they are equal ought to be demonstrated from the axiom, " Magnitudes that wholly coincide, are equal "to one another;" or from prop. A. of book 5, or the 9th prop, or the 14th of the same book, from one of which the equality of all kind of figures must ultimately be deduced. In the preceding books, Euclid has given no definition of equal figures, and it is certain he did not give this: for v/hat is 342 NOTES. Book XI. called the 1st def. of the 5d book, is really a theorem in v^''^'^^*^ Wiiich these circles are said to be equal, that have the straight lines from their centres to the circumferences equal, which is plain, from the defiration of a circle; and thertfoie has by- some editor been improperly placed among the dennidoris. The equality of figures ought not to be defined, but demonstrated : therefore, though it were true, that solid figures contained by the same nunaber of similar and equal plane figures are equal to one another, yet he would justly deserve to be blamed wi^,o would make a definition of this proposition w!,ich ought to be demonstrated. But if this pi'oposidon be not true, must it not be confessed, that geometers have, for these thirteen hundred years, been mistaken in tiiis elementary matter ? And tl is should teach us modesty, and to acknowledge how little, through the weakness of our minds, we are able to prevent mistakes even m the principles of sciences which are justly reckoned amongst the most certain ; for that the proposition is not universally true, can be shewn by many examples; the following is sufficient. Let there be any plane rectilineal figure, as the triangle ■A 12. 11. ABC, and from a point D witiiin it draw a the straight Ijne DE at right angles to the plane ABC ; in DE take DE, DP equal to one another, upon the opposite sides of the plane, and let G be any point in EF ; join DA, DB, DC ; EA EB, EC ; FA, FB, FC ; G A, GB, GC : because the straight line EDF is at right angles to the plane ABC, it makes iight angles Avith DA, DB, DC which it meets in that plane ; and in the triangles EDB, FDB, ED and DB are equal to FD and DB, each to each, and they contain right angles ; therefore b 4. 1. ■ the base EB is equal b to the base FB; in the same manner EA is e- qual to FA, and EC to FC : and in the triangles EBA, FBA, EB, BA ai'c equal to FB, BA, and the base EA is e- qual to the base FA ; wherefore the angle c 8. 1. EBA is equalc to the ^ anele FBA, and the tri- „ . '^--"^ I ■ . "~ C angle EBA equal b to the triangle FBA, and the other angles equal to r 4. 6. the other angles ; there- p d< I. def. fore these triangles are ^ ^- similar d; in the same manner the triangle EBC is similar to NOTES. 343 the triangle FBC, and the triangle EAC to FAC ; therefore Book XI. there are two solid figures, each of which is contained by six v.^'"v->»/ triangles, one of them by three tri^^ngles, the common vertex of which is the pomt G, and their bases the straight lines AB, BC, CA, and by three other triangles the common vertex of which is the point E, and their bases the same lines AB, BC, C A ; the otner solid is contained by the same three tri- angles the common vertex of which is G, and their bases AB, BC, CA ; and by three other triangles of which the common vertex is the point F, and their bases the same straight lines AB, BC, CA: now the three triangles GAB, GBC, GCA are common to both solids, and the three othei-s EAB, EBC, EC A, of the first solid have been shown equal and similar to the three others FAB, FBC, FCA of the other solid, each to each ; therefore these two solids are contained by the same number of equal and similar planes : but that they are not equal is mani- fest, because the first of them is contained in the other : there- fore it is not univers Jly true tiiat solids are equal which are contained by the same number of equal and similar planes. Cor. From this it appears that two unequal solid angles may be contained by the same number of equal plane angles. For the solid angle at B, which is contained by the four plane angles EBA, EBC, GBA, GBC is not equal to the solid angle at the same point B which is contained by the four plane angles FBA, FBC, GBA, GBC ; for this last contains the other : and each of them is contained by four plane angles, which are equal to one another, each to each, or are the self same ; as has been proved : and indeed there may be innunierable solid angles all unequal to one another, wiiich are each of them contained by plane angles that are equal to one another, each to each : it is likewise manifest that the before-mentioned solids are not simi- lar, since their solid angles are not all equal. And that there may be innumerable solid angles all unequal to one another, which are each of them contained by the same plane angles disposed in the same order, will be plain from the three following propositions. PROP. I. PROBLEM. Three magnitudes. A, B, C being given, to find a fourth such, that every three shall be greater than the remaining one. Let D be the fourth : therefore D must be less than A, B, C 344 NOTES. Book XI. together : of the three A, B, C, let A be that which is not lesa "•— '''■■"^"'««' tiiaii eituer of the two B and C : and first let B and C together be not less than A : therefore B, C, D together are greater than A ; and because A is not less than B ; A, C, D together are gi'eater than B : in the like manner A, B, D together are greater than C : wherefore in the case in which B and C together are not less than A, any magnitude D which is less than A, B, C together will answer tiie problem. But if B and C together be less than A ; then, because it is reqviired tliat B, C, D together be greater than A, from each of these taking away B, C, the remadning one D must be greater than the excess of A above B and C : take therefore any mag- nitude D whici^. is less than A, B, C together, but greater than the excess of A above B and C : than B, C, D together are great- er than A ; and because A is greater than eitner B or C much more will A and D, together with either of the two t", C be greater than the other : and, by the construction, A, B, C are together greater than D. CoR. If besides it be required, that A and B together shall not be less than C and D together ; the excess of A and i; toge- ther above C must not be less than D, that is, D must not be greater than that excess. PROP. II. PROBLEM. Four magnitudes A, E, C, D being given, of which A and B together are not less than C and D together, and such that any three of them whatever are greater than the fourth ; it is required to find a fifth magnitude E such, that any two of the three A, B, E shall be greater than the third, and also that any two of the three C, D, E shall be greater than the third. Let A be not less than B: and C not less than D. First, Let the excess of C above D be not less than the excess of A above B : it is plain that a magnitude E can be taken which is less than the sum of C and D, but greater than the excess of C above D ; let it be taken : then E is greater like- wise than the excess of A above B ; wherefore E and B together are greater than A ; and A is not less than B : therefore A and E together are greater than P. : and, by the hypothesis, A and B together are not less than C and D together, and C and D together are greater than E ; therefore likewise A and B arc greater than E. NOTES. 345 But let the excess of A above B be greater than the excess of Book XI. C above D : and because, by the hypotnesis, tiie three B, C, ^— ''^''^^w D are together greater than the fourth A ; C and D together are greater than the excess of A above B : therefore a magni- tude may be taken whica is less than C and D together, but greater than the excess of A above B. Let this magnitude be E ; and because E is greater than the excess of A above B, B together with E is greater than A : and, as in the preceding case, it may be shown that A together with E is greater than B, and that A together with B is greater than E : thei-efore, in each of the cases, it has been shown that any two of the three A, B, E are greater than the third. And because in each of the cases E is greater than the excess of C above D, E together with D is greater than C ; and, by the hypothesis, C is not less than D ; therefore E together with C is greater than D ; and, by the construction, C and D toge- ther are greater than E : therefore any two of the three, C, D, E are e;reater than the third. PROP. III. THEOREM. There may be innumerable solid angles all unequal to one another, each of which is contdned by the same four plane an- gles, placed in the same order. Take three plane angles. A, B, C, of which A is not less than either of the other two, and such, that A and B toge- ther are less than two right angles : and by problem 1 and its corollary, find a fourth angle D such, that any three what- ever of the angles A, B, C, D be greater than the remaining angle, and such, that A and B together be not less than C and D together,: and by problem 2, find a fifth angle E such that any two of the angles A, B, E be greater than the third, -«nd also that any two of the angles C, D, E be greater than 2 X 546 NOTES. Book XI. the third : and because A and B together are less than two ^^'^f^*^ right angles, the double of A and B together is less than four right angles : but A and B together are greater than the angle E ; wherefore the double of A, B together is greater than the three angles A, B, E together, which three are conse- quently less than four right angles ; and every two of the same angles A, B, E are greater tliun the third ; therefore, by prop. 23, 1 ), a solid angle may be made contained by three plane angles equal to the angles A, B, E, each to each. Let this be the angle F contained by the three plane angles GFH, HFK, GFK wiiicn are equal to the angles A, B, E, each to each : and because the angles C, D together are not greater than the angles A, B together, therefore the angles C, D, E are not greater than the angles A, B, E : but these last three are less than four right angles, as has been demonstrated : where- fore also the angles C, D, E are together less than four right angles, and every two of them are greater than the third; there- fore a solid angle may be made which shall be contained by three a 23. 11. plane angles equal to the angles C, D, E, each to each a: and by prop. 26, 1 1, at the point F in the straight line FG a solid angle may be made equal to that which is contained by the three plane angles that are equal to the angles C, D, E : let this be made, and let the angle GFK, which is equal to E, be one of the three; and let KFL, GFL be the other two which are equal to the angles, C, D, each to each. Thus there is a solid angle constituted at the point F contained by the four plane angles GFH, HFK, KFL, GFL v/hich are equal to the angles A, B, C, D, each to each. Again, Find another angle M such, that every two of the three angles A, B, M l)e greater than the third, and also every tv/o of the three C, D, M be greater than the third : n. NOTES. 347 and, as in the preceding part, it may be demonsti-ated that Book XI. the three A, L-, M are less Vi^^v^**./ than four right angles, as also that the three C, D, M are less than four right angles. Make therefore a a solid angle at N contained by the three plane angles ONP, PNQ, ONQ, which are equal to A, B, M, each to eacli : and by prop. 26, 11, make at the point N in the straight line ON a solid angle contained by three plane angles of which one is the angle ONQ equal to M, and the other two are the angles QNR, ONR wiich are equal to the angles C, D, each to each. Thus, at the point N, there is a solid angle contained by the four plane angles ONP, PNQ, QNR, ONR which are equal to the angles A, B, C, D, each to each. And that the two solid angles at the points F, N, each of which is contained by the above-named fom- plane an- gles, are not equal to one another, or that they cannot coin- cide, will be plain by considering that the angles GFK, ONQ; that is, the angles E, M, are unequal by the construction ; and therefore the straight lines GF, i-K cannot coincide with ON, NQ, nor consequently can the solid angles, which therefore are unequal. And because from the four plane angles A, B, C, D, there can be found innumtrable other angles that will serve the same purpose with the angles E and M ; it is plain that innumerable other solid angles may be constituted which are each con.ained by the same four plane angles, and all of them unequal to one another. Q. E. D. And from this it appears that Clavius and other authors are mistaken, who assert that those solid angles are eqvial which are contained by the same number of plane angles that are equal to one another, each to each. Also it is plain that the 26th prop, of book 11, is by no means sufficiently demonstrated, because the equality of two solid angles, whereof each is contained by three plane angles which are equal to one another, each to each, 15 only assumed, and not demonstrated. NOTES. PROP. I. B. XI. The words at the end of this, " for a straight line cannot " meet a straight line in more than one point," are left out, as an addition by some unskilful hand ; for tnis is to be demon- strated, not assumed. Mr. Tnomas Simpson, in his notes at the end of the 2d edition of his Elements of Geometry, p. 252, after repeating the words of this note, adds, " Noav, can it possibly show any want of " skill in an editor (he means Euciid or Theon) to refer to an " axiom which Euclid himself hath laid down, book 1, No. 14, *' (he means Barrow's Euclid, for it is the 10th in the Greek), " and not to have demonstrated, what no man can demonstrate ?" But all that in tiiis case can follow from that axiom is, that, if two straight lines couid meet each other in two points, the parts of them betwixt these points must coincide, and so they would have a segment betwixt these points common to both. Now, as it has not been shown in Euciid, that they cannot have a com- mon segment, this does not prove that they cannot meet in two points from which their not having a common segment, is de- duced in the Greek edition : but, on the contrary, because they cannot have a common segment, as is shown in cor. of 1 Ith prop, book 1, of 4to edition, it follows plainly that they cannot meet in two points, which the remarker says no man can de- monstrate. Mr. Simpson, in the same notes, p. 265, justly observes, that in the corollary of prop. 1 l,book l,4to. edition, the straight lines AB, BD, BC, are supposed to be all in the same plane, which cannot be assumed in 1st prop, book 11. This, soon after the 4to. edition was published, I observed and corrected as it is now in this edition : he is mistaken in thinking the 10th axiom he mentions here to be Euclid's ; it is none of Euclid's but is the 10th in Dr. Barrow's edition, who had it from Herigon's Cur- sus, vol. 1, and in place of it the "corollary of 10th prop, book 1, xyas added. PROP. II. B. XI. This proposition seems to have been changed and vitiated by some editor : for all the figures defipcd in the first book of the Elements, arxl among them triangles, are, by the hypothesis, plane figures ; that is, such as are described in a plane ; Avhere- fore the second part of the enunciation needs no demonstration. Besides, a convex superficies may be terminated by three straight NOTES. 349 lines meeting one another; the thing that should have been de-Book XI. monstrated is, that two, or three straight iincs, that meet one ^•^'^^~^*'^ another are in one plane. And as this is not sufiiciently done, the enunciation and demonsti'ation are changed into those now put into the text. PROP. III. B. XI. In this proposition the following words near to the end of it are left out, viz. *' therefore DEB, DFB are not straight lines ; " in the like manner it may be demonstrated that there can be " no other straight line between the points D, B :" because from this, that two lines include a space, it only foilovv's that one of them is not a straigiit line : and the force of the argument lies in t^iis, viz. if the common section of the planes be not a straight line, then tvvo straight lines could not include a space, which is absurd ; therefore the common section is a straight line. PROP. IV. B. XI. The words " and the triangle AED to the triangle EEC" are omitted, because the Avhole conclusion of the 4th prop, book 1. has been so often repeated in the precedmg books, it was needless to repeat it here. PROP. V. B. XI. In this, near to the end, l5i-D," which are manifestly corruptea, or liave been added to the text ; for there was not the least neces- sity to go so far about to show that DC is in the s&me piai:ie in which are BD, DA because it immediately follows from prop. 7 preceding, that BD, DA, are in the plane in winch are the parallels Ab, CD: therefore, instead of these words, there ought only to be " because all three are in the plane in which are the parallels AB, CD." PROP. XV. B. XI. After the words " and because EA is parallel to GH," the following are added, " for each of them is parallel to DE, and " are not both in the same plane with it," as being manifestly forgotten to be put into the text. PROP. XVI. B. XI. In this, near to the end, instead of the Avords " but straight " lines which meet neither way" ought to be read, "but straight " lines in the same plane wiiich produced meet neither way :" because, though in citing this definition in prop. 27, book 1, it was not necessary to mention the words, " in the same plane," all the straight lines in the books preceding this being in the snme plane ; yet here it was quite necessary. PROP. XX. B. XI. In this, near the beginning, are the words, " But if not, " let f; AC be the greater :" but the angle BAG may happen to he equal to one of the other two : wherefoi'e this place should NOTES. 351 be read thus, " But if not, let the angle BAC be not less than Book XI. " either of the other two, but greater than DAB." v-o'^-^^v^ At the end of this proposition it is said, " in the same man- " ner it may be demonstrated," though there is no need of any demonstration; because the angle BAC being not less than either of the other two, it is evident that BAC together with one of them is greater than the other. PROP. XXII. B. XL And likewise in this, near the beginning, it is said, " But if " not, let the angles at B, E, H be unequal, and let the angle " at B be greater than either of those at E, H:" which Avords manifestly show this place to be vitiated, because the angle at B may be equal to one of the other two. They ought therefore to be read thus, " But if not, let the angles at B, E, H be une- " qual, and let the angle at B be not less than either of the " other two at E, H: therefore the straight line AC is not less " than either of the two DF, UK." PROP. XXIII. B. XI. The demonstration of this is made something shorter, by not repeating in the third case the things which were demon- strated in the first ; and by making use of the construction which Campanus has given; but he does not demonstrate the second and third cases; the construction and demonstration of the third ease are made a little more simple than in the Oreek text. PROP. XXIV. B. XI. The word " similar" is added to the enunciation of this pro- position, because the planes containing the solids wnich are to be demonstrated to be equal to one another, in the 25th propo- sition, ought to be similar and equal, that the equality of the solids may be inferred from prop. C, of this book : and in the Oxford edition of Commandine's translation, a corollary is add- ed to prop. 24, to show that the parallelograms mentioned in this proposition are similar, that the equality of the solids in prop. 25, may be deduced from the 10th def. of book 1 1. PROP. XXV. and XXVI. B. XI. In the 25th prop, solid figures, which are contained by the same number of similar and equal plane figures, are supposed 352 NOTES. Rook XI. to be equul to one another. And it seems that Theon, or some "W^v-^XBi-- other editor, that he might save himseif the trouble of demon- strating the soiid figures mentioned in this proposition to be equal to one another, has inserted the 10th clef, of tl^is book, to serve instead of a demonstration ; which was very ignorantly done. Likewise in the 26th prop, two solid angles are supposed to be equal : if each of them be contained by three plane angles Avhich are equal to one another, each to each. And it is strange enough, that none of the commentators on Euclid have, as far as I know, perceived that sometning is vx^ancing in the demon- strations of these two propositions. Clavius, indeed, in a note upon the 1 1th def. of tiiis book, affirms, that it is evident that those solid angles are even which are contained by the same number of plane angles, equal to one another, each to each, because they will coincide, if they be conceived to be placed within one another ; but this is sa.id without any proof, nor is it ahvays true, except when the solid angles are contained by three plane angles only, Avhich are equal to one another, each to each : and in this case the proposition is the same with this, that two spherical triangles that are equilateral to one another, are also equiangular to one another, and can coincide ; which ought not to be granted without a demonstration. Euclid does not assume this in the case of rectilineal triangles, but demon- strates, in prop. 8, book 1, that triangles which are equilateral to one another are also equiangvilar to one another ; and from this their total equality appears by prop. 4, book 1 . And Me- nelaus, in the 4th prop-, of his 1st book of spherics, explicitly demonstrates, that spherical triemgles which are mutually equi- lateral, are also equiangular to one another ; from which it is easy to show that they must coincide, pro\iding they have their sides disposed in the same order and situation. To supply these defects, it was necessary to add the thi'ee propositions marked A, B, C to this book. I'or the 25th, 26th, and 28th propositions of it, and consequently eight others, viz. the 27th, 31st, 32d, 33d, 34th, 36th, 37th, and 40th of the same, which depend upon them, h-ave hitherto stood upon an infirm foundation; as also, the 8th, 12th, cor. of 17th and '18th of j^ 12th book, v/hich depend upon the 9th definition. For it has been shov/n in the notes on def. 10, of this book, that solid figures which are contained by the same number of similar and equal plane figvu'cs, as also solid angles that are cont.dned by the same number of equal plane angles, are not always equal to one another. NOTES. 353 It is to be observed that Tacquet, in his Euclid, defines equal Book XI. solid angles to be such, " as being put within one another do ^-^"v^>-.' " coincide :" but this is an axiom, not a definition ; for it is true of all magnitudes whatever. He made this useless definition, that by it he might demonstrate the 36th prop, of tnis book, without the help of the 35th of the same : concerning wnich demonstration, see the note upon prop. 36. PROP. XXVIII. B. XL In this it ought to have been demonsti-ated, not assumed, that the diagonals are in one plane. Clavius had supplied l.Js defect. PROP. XXIX. B. XL There are three cases of this proposition ; the first is, when the two parallelograms opposite to the base AB have a side common to both ; the second is, when these parallelograms are separated from one another , and the thii'd, when there is a part of them common to both ; and to this last only, the demonstra- tion that has hitherto been in the Elements does agree. The first case is immediately deduced from the preceding 28th prop, which seems for this purpose to have been premised to this 29th, for it is necessary to none but to it, and to the 40th of this book, as we now have it, to which last it would, without doubt, have been premised, if Euclid had not made use of it in the 29th; but some unskilful editor has taken it away from the Elements, and has mutilated Euclid's demonstration of the other two cases, which is now restored, and serves for both at once. PROP. XXX. B. XL In the demonstration of this, the opposite planes of the solid CP, in the figure in this edition, that is, of the solid CO in Commandine's figure, are not proved to be parallel ; which it is proper to do for the sake of learners. PROP. XXXI. B. XL There are two cases of this proposition ; the first is, when the insisting straight lines are at right angles to the bases ; the other, when they are not : the first case is divided again into two others, one of which is, when the bases are equiangular parallelograms ; the other when thev are not equiangular : 2 Y ■ 354 NOTES. Book XI. the Greek editor makes no mention of the first of these two ^"^'■"''">>-^ last cases, but has inserted the demonstration of it as a part of that of the other : and tiierefore should have taken notice of it in a corollary ; but we thought it better to give these two cases separately : the demonstration also is made something shorter by following the way Euclid has made use of in prop. 14, book 6. Besides, in the demonstration of the case in which the insisting straight lines are not at right angles to the bases, the editor does not prove that the solids described in the construction are paral- lelopipeds, which it is not to be thought that Euclid neglected : also the words " of which the insisting straight lines are not in " the same straight lines," have been added by some unskilful hand ; for they may be in the same straight lines. PROP. XXXII. B. XI. The editor has forgot to order the parallelogram FH to be applied in tlie angle FGH equal to the angle LCG, which is necessary. Clavius has supplied this. Also, in the construction, it is required to complete the so- lid of which the base is FH, and altitude the same with that of the solid CD : but this does not determine the solid to be completed, since there may be innumerable solids upon the same base, and of the same altitude : it ought therefore to be said " complete the solid of which the base is FH, and one of " its insisting straight lines is FD ;" the same correction must be made in the following proposition 33. PROP. D. B. XI. It is very probable that Euclid gave this proposition a place in the Elements, since he gave the like proposition concerning equiangular parallelograms in the 23d B. 6. PROP. XXXIV. B. XI. In this the words, uv «/ itpis-raa-t ax. utrtv Izri tmv civruv ivhiav, " of which the insisting straight lines are not in the same " straight lines," arc thrice repeated ; but these words ought either to be left out, as they are by Clavius, or, in place of them, ought to be put, " whether the insisting straight lines be, or be " not, in the same straight lines:" for the other case is with- out any reason excluded ; also the words, *v roi -xln, of which NOTES. 35b " the altitudes," are twice put for uv ^ai 'iipic-Tucrcit, " of which Book XI. " the insisting straight lines;" which is a plain mistake: for '^■-^"^■^^^^ the altitude is always at right angles to the base. PROP. XXXV. B. XI. The angles ABH, DEM.are demonstrated to be right angles in a shorter way than in the Greek ; and in the same Avay ACH, DEM may be demonstrated to be right angles : also the i-epe- tition of the same demonstration, which begins with " in the " same manner," is left out, as it was probably added to the text by some editor ; for the words, " in like manner we may " demonstrate," are not inserted except when the demonstration is not given, or when it is something different froin the other if it be given, as in prop. 26, of this book. Companus has not this repetition. We have given another demonstration of the corollary, be- sides the one in the original, by help of which the following 36th prop, may be demonstrated without the 35th. PROP. XXXVI. B. XL Tacquet in his Euclid demonstrates this proposition without the help of the 35th ; but it is plain, that the solids mentioned in the Greek text in the enunciation of the proposition as equi- angular, are such that their solid angles are contained by thi-ee plane angles equal to one another, each to each ; as is evident from the construction. Now Tacquet does not demonstrate, but assumes these solid angles to be equal to one another ; for he supposes the solids to be already made, and does not give the construction by Avhich they are made : but, by the second demonstration of the preceding corollary, his demonstration is rendered legitimate likewise in the case where the solids are constructed as in the text. PROP. XXXVII. B. XI. In this it is assumed that the ratios which are triplicate of those ratios which are the same with one another, are likewise . the same with one another ; and that those ratios are the same with one another, of which the triplicate ratios are the same with one another ; but this ought not to be granted without a demonstration ; nor did Euclid assume the first and easiest of these two pi'opositions, but demonstrated it in the case of dupli- cate ratios, in the 22d prop, book 6. On this account, another demonstration is given of this proposition like to that which Euclid gives in prop. 22, book 6, as Clavius has done. 356 NOTES. Book XI. v^-v-^^ PROP. XXXVIII. B. XI. < When it is required to draw a perpendicular from a point in one plane wldch is at right angles to another plane, unto tliis last piane, it is done by drawing a perpcntlicular from the point * to tiie common section of the planes ; for this perpendicular will be perpendicular to the plane, by def. 4, of tlus book : and it would be foolish in this case to do it by the 1 1th prop, of •^y -ic, In the same: but Euclid a, ApoUonius, and other geometers, otl e (Tdi- when they have occasion for this problem, direct a perpendicu- tions. lar to be drawn from the point to the plane, and conclude that it will fail upon the common section of the planes, because this is the very same thing as if they had made use of the construc- tion above mentioned, and then concluded that the straight line must be perpendicular to the plane ; but is expressed in fewer words. Some editor, not perceiving tiiis, thought it was ne- cessary to add tiiis proposition, wiiich can never be of any use to the 1 1th book, and its being near to the end among proposi- tions with which it has no connexion, is a mark of its having been added to the text. PROP. XXXIX. B. XI. In this it is supposed, that the straight lines which bisect the sides of the opposite planes, are in one plane, which ought to have been demonstrated ; as is now done. BOOK XII. Book XII. THE learned Mr. Moore, professor of Greek in the Univer- ^^r-s^->^ sity of Glasgow, observed to me, that it plainly appears from Archimedes's epistle to Dositheus, prefixed to his books of the Sphere and Cylinder, which epistle he has restored from anci- ent manuscripts, that Eudoxus was the author of the chief pro- positions in this 12th book. PROP. II. B XII. At the beginning of this it is said, " if it be not so, the square " of BD shall be to the square of FH, as the circle AKCD is " to some space either less than the circle EFGH, or greater " than it." And the like is to be found near to the end of this proposition, as also in prop. 5, 11, 12, 18, of this book : con- NOTES. 357 cerning which, it is to be observed, that, in the demonstration Book XII. of theorems, it is sulhcient, in tiiis and the iik;_ cses, that a s^'v^^i' thing- made use of in the I'easoning can possibly exist, pro\i- ding tuis be evident, though it cannot be exhibited or fovmd by a geometrical coiistruction : so, in tiis place, it is assumed, that there may be a fouitn proportional to these three mui^nitudes, viz. the squares of D, t H, and tlie circle A CD ; because it is evident that tilers; is some square equa; to the circle ABCD though it cannot be found geometrically ; and to the three recti- lineal figures, viz. the squares of BD, FH, and the square which is equal to the circle ABCD, there is a fourth square proportional ; because to the three stridght lines which are their sides, there is a fourth straight line proportional a, and a 12. 6. this fourth square, or a space equal to it, is the space which in this proposition is denoted by the letter S : and tiie like is to be understood in tiie other places above cited : and it is proba- ble tnat this has been shown by Euclid, but left out by some editor ; for the lemma which some unskilful hand has added to this proposition explains nothing of it. PROP. III. B. XII. In the Greek text and the translations, it is said, " and " because the two straight lines BA, A- which meet one ano- " ther," &:c. here the angles BAC, KHL are demonstrated to be equal to one another by 10th prop. B. 11, which had been done before : because the triangle EAG was proved to be similar to the triangle KHL : this repetition is left out, and the triangles BAC, KHL, are proved to be similar in a shorter way by prop. 21, B. 6. PROP. IV. B. XII. A few things in this are moi'e fully explained than in the Greek text. PROP. V. B. XII. In this, near to the end, are' the words, »? s^;t§05-5sv ihx6>t, " as was before shown," and the same are found again m the end of prop. 18, of this book : but the demonstration referred to, except it be the useless lemma annexed to the 2d prop, is no where in these Elements, and has been perhaps left out by some editor who has forgot to cancel those words also. NOTES. PROP. VI. B. XII. A shorter demonstration is sriven of this ; and that which is in the Greek text may be made sliorter by a step than it is : for the author of it makes use of the 22d prop, of B. 5, twice : whereas once would have served his purpose ; because that proposition extends to any number of magnitudes which are proportionals taking two and two, as well as to three which are proportional to other three. COR. PROP. Vm. B. XII. a20. 6- bll.def. 11. c4. 6. The demonstration of this is imperfect, because it is not shown, that the triangular pyram.ids into which those upon multangular bases are divided, are similar to one another, as ought necessarily to have been done, and is done in the like case in prop. 12, of this book. The full demonstration of the corollary is as follows : Upon the polygonal bases ABCDE, FGHKL, let there be si- milar and similarly situated pyramids which have the points M, N for their vertices : the pyramid ABCDEM has to the pyramid FGHKLN the triplicate ratio of that which the side AB has to the homologous side FG. Let the polygons be divided into the triangles ABE, EBC, ECD ; FGL, LGH, LHK, which are similar a each to each ; and because the pyramids are similai', thereforeb the triangle EAM is similar to the triangles LFN, and the triangle ABM to FGN : wherefore c ME is to EA, as NL to LP ; and as AE C L A B F to EB, so is FL to LG, because the triangles EAB, LFG are similar ; therefore, ex jequali, as ME to EB, so is NL to LG : NOTES. 359 in like manner it may be shown that EB is to BM, as LG to Book XII. GN ; therefore again ex aquali^ as EM is to MB, so is LN to ^-^^^^s-/ GN : wherefore the triangles EMB, LNG having their sides proportionals are a equiangular, and similar to one another : a 5. 6. therefore the pyramids which have the triangles EAB LFG for their bases, and the points M, N for their veitices, are similar b to one another, for their solid angles are c equal, and theb 11. clef, solids themselves are contained by the same number of similar bi- planes : in the same manner, the pyramid EBCM may becb.ll. shown to be similar to the pyramia LGHN, and the pyramid ECDM to LHKN. And because the pyramids EABM, LFGN are similar, and have triangular bases, the pyramid EABM has d to LFGN the triplicate ratio of that which EB has to the ho-d 8. 12. mologous side LG. And in the same inanner, the pyramid EBCM has to the pyramid LGHN the triplicate ratio of that which EB has to LG. Therefore as the pyramid EABM is to the pyramid LFGN, so is the pyramid EBCM to the pyramid LGHN. In like manner as the pyramid EHCM is to LGHN, so is the pyramid ECDM to the pyramid LHKN. And as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents : therefore as the pyramid EABM to the pyramid LiGN so is the whole pyramid ABCDEM to the whole pyramid FGHKLN : and the pyramid EABM has to the pyramid LtGN the triplicate ratio of that which AB has to FG ; therefore the whole pyramid has to the whole pyramid the triplicate ratio of that which AB has to the homologous side FG. Q. E. D. PROP. XL and XH. B. XH. The order of the letters of the alphabet is not observed in these two propositions, according to Euclid's manner, and is now restored ; by which means the first part of prop. 12 may be de- monstrated in the same words with the first part of prop. II: on this account the demonstration of that first part is left out, and assumed from prop. 1 1 . PROP. XH. B. XH. In this proposition the common section of a plane parallel to the bases of a cylinder, with the cylinder itself, is supposed to be a circle, and it was thought proper briefly to demonstrate it ; from whence it is sufficiently manifest, that this plane divides the cylinder into two others ; and the same thing is understood to be supplied in prop. 14. NOTES. PROP. XV. B. XII. " And complete the cylinders AX, EO," both the enuncia- tion and exposition of the proposition represent the cylinders as well as the cones, as already described : wiierefore the reading ought rather to be, " and let the cones be ALC, ENG ; and the « cylinders AX, EO." The first case in the second part of the demonstration is want-- ing ; and something also in the second case of that part, before the repetition of the construction is mentioned ; which are now added. PROP. XVII. B. XII. In the enunciation of this proposition, the Greek words Ui Ts;v Ltii?ovx cpciipxv c-neiov TioXviO^ov iyTpor^xi jt4>iy«t/«v t*j5 iXutrtJOVc? (7(pxte^cii yMTccTYiv i-pripdvuiiv are thus translated by Ccmmandine and others, " in majore solidum polyhedrum describere quod " minoris sphxrs; superficiem non tangat ;" that is, " to de- " scribe in the greater sphere a solid polyhedron which shall " not meet the superficies of the lesser sphere :" whereby they refer the words Kara ryiv iTricpavuocy to these next to th.em T>)«- iKaaaovo? a-tpui^ui. I'ut they ought by no means to be tiius translated ; for the solid polyhedron doth not only meet the superficies of the lesser sphere, but pervades the whole of that sphere ; therefore the foresaid words are to be referred to TO e-Tsg5flv 7ro>.ii5?§«v, and ought thvis to be translated, viz to describe in the greater sphere a solid polyhedron Avhose superficies shall not meet the lesser sphere ; as the meaning of the proposition necessarily requires. The demonstration of the proposition is spoiled and mutilat- ed : for some easy things are very explicitly demonstrated, Avhile others not so obvious are not sufficiently explained : for example, when it is affirmed, that tl^iC square of Kb is greater than the double of the square FZ, in the first demonstra- tion, and that the angle f'ZK is obtuse, in the second ; both which ought to have been demonstrated. Besides, in the first demonstration it is said, "draw KC2 from the point K perpen- " dicular to BD ;" whereas it ought to have been said, " join " KV," and it should have deen demonstrated that KA^ is perpendicular to liD : for it is evident from the figure in Her- vagius's and Gregory's editions, and from the words of the NOTES. 561 demonstration, that the Greek editor did not perceive that Book XII. the perpendicular drawn from the point K to the straight line ^^^'v-^s-^ BD must necessarily fall upon the point V, for in the figure it is made to fall upon the point Q a different point from V, which is likewise supposed in the demonstration. Comman- dine seems to have been aware of this ; for in this figure he marks one and the same point with the two letters V, 12 ; and before Commandine, the learned John Dee, in the commen- tary he annexes to this proposition in Henry Billinsley's trans- lation of the Elements, printed at London, anno 1570, expressly takes notice of this error, and gives a demonstration suited to the construction in the Greek text, by Avhich he shows that the perpendicular drawn from the point K to BD, must necessarily fall upon the point V. Likewise it is not demonstrated that the quadrilatei'al figures SOFT, TPRY and the triangle YRX do not meet the lesser sphere, as was necessary to have been done : only Clavius, as far as I know, has observed this, and demonstrated it by a lem- ma, which is now premised to this proposition, something alter- ed and more briefly demonstrated. In the corollary of this proposition, it is supposed that a solid polyhedron is described in the other sphere similar to that which is described in the sphere BCDE ; but, as the construction by which this may be done is not given, it was thought proper to give it, and to demonstrate, that the pyramids in it are similar to those of the same order in the solid polyhedron described ii) the sphere BCDE. From the preceding notes, it is sufficiently eAident how much the Elements of Euclid, who was a most accurate geometer, have been vitiated and mutilated by ignorant editors. The opinion which the greatest part of learned men have entertained concerning the present Greek edition, viz. that it is very little or nothing different from the genuine work of Euclid, has without doubt deceived them, and made them less attentive and accurate in examining that edition ; whereby several errors, some of them ^ gross enough, have escaped their notice from the age in which Tlieon lived to this time. Upon which account there is some ground to hope, that the pains we have taken in correcting those errors, and freeing the Elements, as far as we could, from ble- mishes, will not be unacceptable to good judges, who can discern when demonstrations are legitimate, and when they are not. The objections which, since the first edition, have been made against some things in the notes, especially against the doctrine of proportionals, have either been fully answered in Dr. Bar- row's Lect. Mathemat. and in these notes, or are such, except 2 Z 362 NOTES. Book XII. one which has been taken notice of in the note on Prop. 1. Book ^-^'"-'''''"^^ 1 1 . as show that the person who made them has not sufficiently considered the things against which they are brought : so that it is not necessary to make any further answer to these objections and others like them against Euclid's definition of proportionals; of which definition Dr. Barrow justly says, in page 297 of the above named book, that " Nisi machinis impulsa validioribus " seternum persistet inconcussa." FINIS. «BttcIitfS ^aU. THIS EDITION SEVERAL ERRORS ARE CORRECTED, AND SOME PROPOSITIONS ADDED. BY ROBERT SIMPSOJ^T, M. D. EMERITUS PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF GLASGOW. PHILADELPHIA PRINTED BY WM. F. m'LAUGHLIN, NO. 28 NORTH SECOND STREET. 1806. PREFACE. EUCLID'S DATA is the first in order of tlie books WTitten by the ancient geometers to facilitate and pro- mote the method of resolution or analysis. In the general, a thing is said to be given which is either actually exhibited, or can be found out, that is, which is either knoA\'n by h}'pothesis, or that can be demon- strated to be known; and the propositions in the book of Euclid's Data show 'what things can be found out or known from those that by h}"pothesis are al- ready knoA\ n ; so that in the analysis or investigation of a problem, from the thiiigs that are laid down to be known or given, by the help of these propositions other things are demonstrated to be given, and from these, other things are again sho^^'n to be given, and so on, until that which was proposed to be fo-nid out in the problem is demonstrated to be given, and \\ hen this is done, the problem is solved, and its composi- tion is made and derived from the compositions of the Data v. hich were made use of in the analysis. And thiws the Data of Euclid are of the most general and necessary use in the solution of problems of ever}- kind. Euclid is reckoned to be the author of the Book of the Data, both by the ancient and modern geome- ters ; and there seems to be no doubt of his having wri'ttn a book on this subject, but which, in the course of so niany ages, has been much vitiated by 366 PREFACE. unskilful editors in several places, both in the order of the propositions, and in the definitions and demon- strations themselves. To correct the errors which are no^v found in it, and bring it nearer to the accu- racy with which it ^^ as, no doubt, at first written by Euclid, is the design of this edition, that so it may be rendered more useful to geometers, at least to be- ginners ^^ ho desire to learn the investigatory method of the ancients. And for their sakes, the composi- tions of most of the Data are subjoined to their de- monstrations, that the compositions of problems solv- ed by help of the Data may be the more easily made. Marinus the philosopher's preface, which, in the Greek edition, is prefixed to the Data, is here left out, as being of no use to understand them. At the end of it he says, that Euclid has not used the syn- thetical, but the analytical method in delivering them; in which he is quite mistaken; for, in the analysis of a theorem, the thing to be demonstrated is assumed in the analysis ; but in the demonstrations of the Data, the thing to be demonstrated, which is, that some- thing or other is given, is never once assumed in the demonstration, from which it is manifest, that every one of them is demonstrated synthetically; though, indeed if a proposition of the Data be turned into a problem, for example the 84th or 85th in the former editions, which here are the 85th and 86th, the de- monstration of the proposition becomes the anal}'sis of the problem. W^herein this edition differs from the Greek, and the reasons of the alterations from it, ^vill be shown in the notes at tlie end of the Data. (iBttcIiD^S 2Data, DEFINITIONS. I. SPACES, lines, and angles, are said to be given in magnitude, when equals to them can be found. II. A ratio is said to be given, when a ratio of a given magnitude to a given magnitude which is the same ratio with it can be found. III. Rectilineal figures are said to be given in species, which have each of their angles given, and the ratios of their sides given. IV. Points, lines, and spaces, are said to be given in position, which have always the same situation, and which are either actually exhibited, or can be found. A. An angle is said to be given in position Avhich is contained by straight lines given in position. V. A circle is said to be given in magnitude, when a straight line from its centre to the circumference is given in magnitude. VI. A circle is said to be given in position and magnitude, the cen- tre of which is given in position, and a straight line from it to the circumference is given in magnitude. VII. Segments of circles are said to be given in magnitude, when the angles in them, and their bases, are given in magnitude. VIII. Segments of circles are said to be given in position and magni- tude, when the angles in them are given in magnitude, and their bases are given both in position and magnitude. IX. A magnitude is said to be greater than another by a given mag- nitude, when this given magnitude being taken from it, the Remainder is eq\ial to the other magnitude. ;68 EUCLID'S 2. See N. X. A magnitude is said to be less than afiother by a given magni- tude, when this given magnitude being added to it, the whole is equal to the other magnitude. 1. • PROPOSITION I. SeeN. THE ratios of given magnitudes to one another is G'iven. Let A, B be two given magnitudes, the ratio of A to B is given. Because A is a given magnitude, there may a 1 def * ^^ found one equal to it ; let this be C : and dat. because B is given, one equal to it may be found ; let it be D ; and since A is equal to C, b 7. 5. and B to D ; therefore b A is to B, as C to D ; and consequently the ratio of A to B is given, because the ratio of the given magni- tudes C, D which is the same with it, has been A B C D found. PROP. IL IF a elvcn maQ:nitucIe has a ^Ivtn ratio to ano'ther magnitude, " and if unto the two magnitudes l^y " which the c-iven ratio is exhibited, and the civen " magnitude, a fou.rth proportional can be found ;" the otiicr maicnitude is G:i'»en. Let the given magnitude A have a given ratio to the mag- nitude B ; if a fourth proportional can be found to the three magnitudes above named, B is given in magnitude Because A is given, a magnitude may be a 1. def. found equal to it a; let this be C: and be- cause the ratio of A to B is given, a ratio which is the same •v^'ith it may be found, let this be the ratio of the given magnitude E A B C D to the given magnitude F : imto tiie magni- E F tudes E, F, C find a fourth proportional D, which, by the hypothesis, can be done. Wherefore, because A is to B, as E to F ; and b 11. 5. as E to F, so is C to D ; A is b to B, as C to * The figures in the marg'in sbow the number of the propositions iu the other euitloiis. DATA. 369 D. But A is equal to C ; therefore c B is equal to D. The c 14. 5. magnitude B is therefore given a because a magnitude D equal a 1. def, to it has been found. The limitation within the inverted commas is not in the Greek text, but is now necessarily added ; and the same must be understood in all the propositions of the book which depend Upon this second proposition, where it is not expressly mention- ed. See the note upon it. PROP. III. 3 IF any given magnitudes be added together, their sum shall be given. Let any given magnitudes AB, BC be added together, their sum AC is given. Because AB is given, a magnitude equal to it may be founda; a 1. def. let this be DE : and because BC is gi- A B C ven, one equal to it may be found ; let ■ j — this be EF : wherefore, because AB is equal to DE, and BC equal to EF ; the D E F whole AC is equal to the whole DF : ij — AC is therefore given, because DF has been found which is equal to it. PROP. IV. 4> IF a eiven magnitude be taken from a eiven mai>:. nitude ; the remainina; masmitude shall be eiven. From the given magnitude AB, let the given magnitude AC be taken ; the remaining magnitude CB is given. Because Aii is given, a magnitude equal to it may a be a 1. def. found ; let this be DE : and because . PR AC is given, one equal to it may be , found ; let tiiis be DF : wherefore be- ' cause AB is equal to DE, and AC to y~. F F Di' ; the remainder C 'S is equal to the remainder FE. CB is therefore ' ' given a, because FE which is equal to it has been found. 3 A 3ro , EUCLID'S 12. PROP. V. See N. lY qJ- three magnitudes, the first together with the second be given, and also the second together -s^ith the third; either the first is equal to the third, or one of tliem is gTeater than the other by a given mag- nitude. Let AB, BC, CD be three magnitudes, of which AT5 together with LC, that is AC, is given ; and also BC together with CD, that is ED, is given. Either A)' is equal to CD, or one of them is greater than the other by a given magnitude. I'ec-iuse AC, BD are each of them given, they are either equcil to one another, or not equal. First, let them be equal, and because A B CD AC is equal to BD, take away the com- [ ] mon part B C; therefore the remuin- der A" is equal to the remainder CD. Rut if they be unequal, let AC be greater than BD, and make CE equal to BD. Therefore CE is given, because BD is given. And the whole AC is given ; a 4. dut. therefore a AE the remainder is A E B C D given. And because EC is equal 1 1 ■ •! io BD, by taking liC from both, the remainder EP is equal to the remainder CD. And AE is given ; wherefore AB exceeds EB, that is CD by the given magnitude AE. 5. PROP. VL See N. IF a magnitude has a given ratio to a part of it, it shall also have a given ratio to the remaining part of it. Let the magnitude AB have a given ratio to AC a part of it ' it has also a given ratio to the remainder BC. Because the ratio of Ali to AC is given, a ratio maybe a 2. dcf found a which is the same to it : let this be the ratio of DE a given magnitude to the given magni- . PR tude DF. And because DE, DF are gi- ^ , b 4. dat. ven, the remainder FE is b given : and because AB is to AC, as DE to DF, by .^ -p p c E. 5. conversion c AB is to BC, as Di-. to EF. ^ Therefore the raiio of AB to BC is given, because theratioof the given magnitudes DE, EF, which is the same with it, has been found. DATA. wi Cor, Froin this it follows, that the parts AC, CB have a given ratio to one another: because as AB to BC, so is DE to EF ; by division d, AC is to CB, as DF to FE : and DF, FE ared 17. 5. given ; therefore a the ratio of AC to CB is given. a 2. def. PROP. VII. 6. IF two magnitudes \\'hich have a given ratio to one See n. another, be added together; the vrhoie magnitude skill have to each of them a gi\"en ratio. Let the magnitudes AB, BC which have a given ratio to one another, be added together ; the whole AC has to each of the magnitudes AB, BC a given ratio. Because the ratio of AB to BC is given, a ratio may be found a wnich is the same with it; let this be the ratio of the a 2. def. given magnitudes DE, EF : and be- cause DE, EF are given, the whole A B C DF is given b : and because as AB to [ b 3. dat. BC, so is DE to EF ; by composition D E F cAC is to CB, as DF to FE; and by 1 « 18. 5. conversiond, AC is to AB, as DF to d E. 5. DE : wherefore because AC is to each of the magnitudes AB, BC, as DF to each of the others DE, EF ; the ratio of AC tp each of the magnitudes AB, BC is givena. PROP. VIII. r. IF a given magnitude be divided into two paits See Note. which have a given ratio to one another, and if a fourth proportional can be foimd to the sum of the two magnitudes by which the given ratio is exhibited, one of them, and the given magnitude ; each of the parts is given. Let the given magnitude AB be divided into the parts AC, CB which have a given ratio to one another ; if a fourth pro- portional can be found to the above named magnitudes ; AC and CB are A C B each of them given. t Because the ratio of AC to CB is D F E given ; the ratio of AB to BC is — ; given a; therefore a ratio which is a 7. dat. 372 EUCLID'S b 2. clef, tlie same Avith it can be foundb, let this be the ratio of the given magnitudes DE, EF : and because the given magnitude AB has A C B to BC the given ratio of DE to EF, if ]- unto DE, EF, AB a fourth propor- D I' E tional can be found, this which is BC 1 c 2. dat. is given c ; and because AB is given d 4. dat. the other part AC is given d. In the same manner, and with the like limitation, if the dif- ference AC of two magnitudes AB, BC which have a given ra- tio, be given ; each of the magnitudes AB, BC is given. 8. PROP. IX. MAGNITUDES whch h^\e pjiven ratios to the same magnitude, have also a given ratio to one ano- ther. Let A, C have each of them a given ratio to B ; A has a given ratio to C. Because the ratio of A to B is given, a ratio which is the a 2, def. same to it may be found a ; let this be the ratio of the given magnitudes D, E : and because the ratio of B to C is given, a ratio which is the same Avith it may be founds ; let this be the ratio of the given magnitudes F, G : to F, G, E find a fourth propor- tional H, if it can be done; and because as A is to B, so is D to E ; and as V> to C, so is (F to G, and so is) E to H ; ex xcjuali^ as A to C, so is D to H ; therefore the ra- A li C D E tio of A to C is given a, because the ^ ratio of the given magnitudes D and H, which is the same with it, has been found : but if a fourth propor- tional to F, G, E cannot be found, then it can only be said that the ratio of A to C is compounded of the ratios of A to H, and B to C, that is, of the given ratios of D to i'^, and F to Ci . H G DATA. 373 PROP. X. IF two or more magnitudes have gi^'en ratios to orx another, and if they have given ratios, thougli they be not the same, to some other magnitudes ; these other magnitudes sh.Jl also have given ratios to one another. Let two or more magnitudes A, B, C have given ratios to one another ; and let them have given ratios, though they be not the same, to some other magnitudes D, E, F : the magni- tudes D, E, F have given ratios to one another. Because the ratio of A to B is given, and likewise the ratio of A to D ; therefore the ra- tio of D to B is givena; but A D the ratio of B to E is given, B E — therefore a the ratio of D to C — — F a 9. dat. E is given : and because the ratio of B to C is given : and also the ratio of B to E ; the ratio of E to C is givena ; and the ratio of C to F is given ; where- fore the ratio of E to F is given : D, E, F have therefore given ratios to one another. PROP. XI. 22. IF two magnitudes have each of them a given ratio to another magnitude, both of them together shall ha\'e a giA^en ratio to that other. Let the magnitudes AB, BC have a given ratio to the mag- nitude D ; AC has a given ratio to the same D. Because All, BC have each of them a given ratio to D, the ratio B of AB to BC is givena ; and by A 1 C a 9. dat. composition, the ratio of AC to D CB is given b : but the ratio of b 7. dat. BC to D is given ; therefore a the ratio of AC to D is given. S74 EUCLID'S PROP. XII. See N. IF the whole ha^-e to the whole a gi\en ratio, and the parts have to the parts given, but not the same, ratios, ever}' one of them, ■whole or part, shall liave to e^-ery one a given ratio. Let the whole AB have a given ratio to the whole CD, and the parts AE, EB have given, but not the same, ratios to the parts CF, FD, every one shall have to every one, whole or part, a given ratio. Because the ratio of AE to CF is given, as AE to CF, so make AB to CG ; the ratio therefore of AB to CG is given ; wherefore the ratio of the remainder EB to the remainder :i 19. 5. pQ j^g given, because it is the same a with the ratio of AB to CG ; and the ratio of EB to FD is given, wherefore the ratio of FD b 9 dat to FG is given b ; and by conver- sion, the ratio of FD to DG is c 6. dat. given c ; and because AB has to each of the magnitudes CD, CG a given ratio, the ratio of GD to CG is given b ; and therefore c the ratio of CD to DG is given : but the ratio of GD to DF is given, whereforeb the ratio of CD to DF is given, and conse- d cor. 6. quently (1 the ratio of CF to FD is given ; but the ratio of CF to e'Vo dat ^^ given, as also the ratio of FD to EB, whereforee the ra- tio of AE to EB is given; as also the ratio of AB to each of f 7. dat. thenif : the ratio therefore of every one to every one is given. 24 PROP. XIII. See N. jY ^^^^^ |-j,j,_^ Q.|> three proportional straight lines has a \'en ratio to the t ratio to the !::eeor.d A E B C F G D gi\'en ratio to the third, the lir.st sliali also have a f^iven Let A, B, C be three proportional straight Unes, that is, as A to B, so is B to C ; if A has to C a given ratio, A shall also have to B a given ratio. Because the ratio of A to C is given, a ratio which is the a 2. dcf. 5,ame with it may be foirnda ; let this be the ratio of the given fa 13. 6. DATA. 375 proportional F; therefore the rectangle contained by D and E is equal to the square of F, and the rect- angle D, E is given, because its sides D, E are given ; wherefore the square of F, and the straight line F is given : and because as A is to C, so is D to E ; but as A to C, so is c the square of A to the square of B ; and as D to E, so is c the square of D to the square of F : A B C therefore the square d of A is to the square of B, as the square of D to the square of F : D F E as therefore e the straight line A to the straight line B, so is the straight line D to the straight line F : therefore the ratio of A to B is given a, because the ratio of the given straight lines D, F which is the same with it has been found. c 2 cor. 20. 6. d 11. 5. e 22. 6. a 2. def PROP. XIV. A. IF a magnitude together with a given magnitude see n. has a given ratio to another magnitude ; the excess of tiiis other magnitude above a given magnitude has a s:i\'en ratio to the first magnitude : and if the ex- cess of a magnitude above a given magnitude has a given ratio to another magnitude : this other magni- tude together with a given magnitude has a .given ratio to the first magnitude. Let the magnitude AB together with the given magnitude BE, that is AE, have a given ratio to the magnitude CD ; the excess of CD above a given magnitude has a given ratio to AB. Because the ratio of AE to CD is given, as AE to CD, so make BE to FD ; therefore the rutio of BE to FD is given, and BE is given; wherefore FD is gi- ven a : and because as AEto CD, so is BE to FD, the remainder AB is b to the remainder CF, as AE to CD : but the ratio of AE to CD is given, therefore the ratio of AB to CF is given ; that is, CFthe excess of CD above the given magnitude FD has a given ratio to AB. Next, Let the excess of the magnitude AB above the given magnitude BE, that is, let AE have a given ratio to the mag-- A E a 2. dat b 19. 5. D 376 EUCLID'S nitude CD : CD together with a given magnitude has a given ratio to AB. Becuuse the ratio of AE to CD is given, as AE to CD, so make BE to FD ; therefore the ratio of BE to FD is given, and BE is given, A E B a2.dat. wherefore FD is given a. And because 1 as AE to CD, so is BE to FD, AB is c 12. 5. to CF, as c AE to CD : but the ratio of C D F AE to CD is given, therefore the ratio il of AB to CF is given : that is, CF which is equal to CD, toge- ther with the given magnitude DF, has a given ratio to AB. B PROP. XV. See Note. IF a magnitude, together with that to which ano- ther mag:iitude has a given ratio, be given ; the i:.um of tliirs other, and that to wliich the hrst magnitude has a given ratio, is given. Let AB, CD be two magnitudes, of Avhich AB together with BE, to which CD has a given ratio, is given ; CD is given, together with that magnitude to which AL'- has a given ratio. Because the ratio of CD to iiE is given, as BE to CD, so make AE to FD ; therefore the ratio of AE to FD is given, and a 2 dat. AE is given, wherefore a FD is given : . R F and because as BE to CD, so is AE to , b Cor. 19. FD: ABisb to FC, as BE to CD: and 5- the ratio of BE to CD is given, where- F C D fore the ratio of AB to FC is given : and !— FD is given, that is, CD together v/ith FC, to which AB has a given ratio, is given. 10. PROP. XVL See Note. IF the excess of a magnitude above a given magni- tude, has a given ratio to another magnitude ; the excess of both together above a given magnitude shall have to that other a given ratio : and if the excess of two magnitudes together above a giA'en magnitude, has to one of them a given ratio ; either the excess of the other above a gi^ en magnitude has to that one a given ration, or the other is gi\'en together with the magnitude to vdiich that one has a given ratio. DATA. 3?r Let the excess of the magnitude AB above a given magni- tude, have a given ratio to the magnitude BC ; the excess of AC, both of them together, above the given magnitude, has a given ratio to BC. Let AD be the given magnitude, the excess of AB above which, viz. DB has a given ratio . D R P to BC : and because DB has a giv- ^ en ratio to BC, the ratio of DC to ' CB is given a, and AD is given ; therefore DC, the excess of a 7. dat. AC above the given magnitude AD, has a given ratio to BC. Next, let the excess of tAVo magnitudes AB, BC together, above a given magnitude, have to . D R F r one of them BC a given ratio ; ei- ^^ i___ ther the excess of the other of 1 i I them AC above the given magnitude shall have to BC a given ratio ; or AB is given, together with the magnitude to which BC has a given ratio. Let AD be the given magnitude, and first let it be less than AB ; and because DC the excess AC above AD has a given ra- tio to BC, DB hash a given ratio to BC; that is, DB the excess b Cor. 6. of AB above the given magnitude AD, has a given ratio to BC.^** But let the given magnitude be greater than AB, and make A\r. equal to it ; and because EC, the excess of AC above A P., has to BC a given ratio, BC hasc a given ratio to B'vl ; and be-c 6. dat. cause AK is given, AB together with BE, to which BC has a given ratio, is given. PROP. XVIL 11. IF the excess of a magnitude above a given magni- See Note. tude has a given ratio to another magnitude ; the ex- cess of the same first magnitude above a given mag- nitude, shall have a given ratio to both the magnitudes together. And if the excess of either of two magni- tudes above a gi\'en magnitude lias a gi\'en ratio to JDOth magnitudes together: the excess of the same aboA-e a given magnitude shall have a gi\'en ratio to the other. Let the excess of the magnitude AB above a given magni- tude have a given ratio to the magnitude BC ; the excess of AB above a given magnitude has a given ratio to AC. 3 B 378 EUCLID'S Let AD be the given magnitude ; and because ll/'B, the ex- cess of AB above AD, has a given ratio to EC ; the ratio of DC a 7. dat, to DB is given a ; make the ratio of AD to DE the same with this ratio ; therefore the ratio of . F D B C AiJ to DE is given: and AD is __^^___r -b 2. dat. given, wherefore b D',, and the III remainder AE are given: and because as DC to DB, so is AD c 12. 5. to DE, AC isc to EB, as DC to DB; and the ratio of DC to DB is given; wherefore the ratio of AC to LB is given: and because the ratio of EB to AC is given, and that AE is given, therefore EB the excess of AB above the given magnitude AE, has a given ratio to AC. Next, Let the excess of AB above a given magnitude have a given ratio to AB and BC together, that is, to AC ; the excess of A3 above a given magnitude has a given ratio to BC. Let AE be the given magnitude; and because EB the excess of AB above AE has to AC a given ratio, as AC to EB, so make lven ratio to one another, a given magnitude be added; the ^vhole shall either have a given ratio to one another, or tb.e excess of one of them abo^e a .G:iven niae^nitude shall ha\^e a given ratio to the other. b' Let the two magnitudes AB, CD liave a given ratio to one another, and to AB let the given magnitude BE be added, and the given magnitude DF to CD ; the wholes AE, CF either have a given ratio to one another, or^the excess of one of them a 1. flat, above a given magnitude has a given* ratio to the othu"'^ Because li£, DF are each of them given, their ratio is given. DATA. 379 ind if this ratio be the same with A B the ratio of x\B to CD, the ratio of 1- AE to CF, which is the same b with CD F b 12. 5. the given ratio of AB to CD, shall 1 be given. But if the ratio of BE to DF be not the same with the ratio of AB to CD, either it is greater than the ratio of AB to CD, or, by inversion, the ratio of DF to BE is greater than the ratio of CD to AB : first, let . ^ ^ ^ the ratio of BE to DF be greater than the ratio of AB to CD ; and as "Z ^^T ^"7i AB to CD, so make BG to DF ; ___7 therefore the ratio of BG to DF is given ; uud DF is given, thereforec BG is given : and because c 2. dat. BE has a greater ratio to DF than (AB to CD, that is, than) BG to DF, BE is greaterd than BG ; and because as AB to CD,^ ^^- ^• so is BG to DF ; therefore AG isb to CF, as AB to CD : but the ratio of AB to CD is given, wherefore the ratio of AG to CF is given ; and because BE, BG are each of them given, GE is given : therefore AG the excess of AE above a given magni- tude GE, has a given ratio to CF. The other case is demon- sti'ated in the same manner. PROP. XIX. 15. IF li^om each of two magnitudes, which ha\e a Hven ratio to one another, a s-ivgw raa^cnitiide be taken, the remainders shall either have a given ratio to oii.e another, or the excess of one of them above a givea magnitude, sliall have a gi\en ratio to the other. Let the magnitudes AB, CD have a given ratio to one ano- ther, and from AB let the given magnitude AE be taken, and from CD, the given magnitude CF : the remainders EB, FD shall either have a given ratio to one another, or the excess of one of them above a given a p Tt magnitude shall have a given ratio to the other. „ ' p. Because AE, CF are each of , them given, their ratio is given a: a 1. dat. and if this ratio be the same with the ratio of AB to CD, the 5^ EUCLID'S ratio of the remainder EB to the remainder FD, which is the b 19. 5. sameb with the given ratio of AB to CD, shall be given. But if the ratio of AB to CD be not the same with the ratio of AE to CF, either it is greater than the ratio of AE to CF, or, by inversion, the ratio of CD to AB is greater than the ra- tio of CF to AE. First, let the ratio of AB to CD be greater than the ratio of AE to CF, and as AB to CD, so make AG to CF ; therefore the ratio of AG to . v c H CF is given, and Ct is given, c 2. dat. wherefore c AG is given : and p ' p '~T| because the ratio of AB to CD, that is, the ratio of AG to CF, ' d 10. 5. is greater than the ratio of AE to CF ; AG is greater d than AE : and AG, AE are given, therefore the remainder EG is given; and as AB to CD, so is AG to CF, and so isb the re- mainder GB to the remainder FD ; and the ratio of AB to CD is given : wherefore the ratio of GB to FD is given ; therefore GB, the excess of EB above a given magnitude EG, has a given ratio to FD. In the same manner the other case is de- monstrated. 16. PROP. XX. IF to one of t^vo magnitudes which have a given ratio to one another, a given magnitude be added, and from the other a given magnitude be taken ; the excess of tb.e sum above a sriven magnitude shall have o o a given ratio to the remainder. Let the two magnitudes AB, CD have a given ratio to one another, and to AB let the given magnitude EA be added, and from CD let the given magnitude CF be taken ; the excess 'of the sum EB above a given magnitude has a given ratio the re- mainder FD. Because the ratio of AB to CD is given, make as AB to CD, so AG to CF : therefore the ratio of AG to CF is given, a 2. d t. and CF is given, wherefore a AG p a C P is given ; and E A is given, there- . ^ fore the whole EG is given : and Z ' y t> because as AB to CD, so is AG t , h 19. 5. tQ (iy_^ ^yy^ gQ jgb the remainder GB to the i-emainder FD ; the ratio of GB to FD is given. And EG is given, therefore GB, the excess of the sum EB DATA'. f 381 above the given magnitude EG, has a given ratio to the remain- der FD. PROP. XXL C. IF two magnitudes have a given ratio to one ano- See n. ther, if a given magnitude be added to one of them, and the other be taken from a gi\en magnitude ; tlie sum, together with the magnitude to ^\ hich the re- mainder has a given ratio, is given : and the remain- der is given together with die magnitude to which the sum has a eiven ratio. o Let the two magnitudes AB, CD have a given ratio to one another ; and to AB let the given magnitude BE be added, and let CD be taken from the given magnitude FD: the sum AF. is given, together with the magnitude to which the remainder FC has a given ratio. Because the ratio of AB to CD is given, make as AB to CD, so i^B to FD : therefore the ratio of GB to FD is given, and Fi) is ariven, wherefore GB is ^^ * n t- J T.1- • • 4-1 ^ A BE a 2. dat, given a; and BE is given, the ^__ , whole G E is therefore given : and because as AB to CD, so is GB v r D toFD,andsoisbGAtoFC;the ____J J b 19. 5, ratio of GA to FC is given : and AEl together with G A is given, because GE is given ; therefore the sum AE together with GA, to which the remainder FC has a given ratio, is given. The second part is manifest from prop. 15. PROP. XXII. D. IF two magnitudes ha^v^e a eiven ratio to one anc- See n. ther, if fiom one of them a given magnitude be taken, and the other be taken from a given mxagniuide ; each of the remainders is given, together with the magni- tude to "v^ hich the other remainder has a given ratio. Let the two magnitudes AB, CD have a given ratio to one ^inother, and from AB let the given magnitude AE be taken, '3S2 EUCLID'S and let CD be taken from the given magnitude CF : the remain- der EB is given, together with the magnitude to which the other remainder DF has a given ratio. Because the ratio of AB to CD is given, make as AB to CD, so AG to CF : the ratio of AG to CF is therefore given, and a 2. dut. CF is given, wherefore a AG is . p- p ^. given ; and AE is given, and ' therefore the remainder EG is ' . given ; and because as AB to CD, p "D F b 19. 5. so is AG to CF : and so is b the remainder BG to the remainder ^ DF ; the ratio of BG to DF is given: and FB together with BG is given, because EG is given : therefore the remainder Y B together with BG, to which DF the other remainder has a given ratio, is given. The second part is plain from, this and prop. 15. 20. PROP. XXIII. See N. I^" from two given magnitudes there be taken maG-nitudcs which have a eiven ratio to one another, the remainders shall either have a e;iven ratio to cne another, or the excess of one of them' above a given magnitude shall have a given ratio to the other. Let AB, CD be two given magnitudes, and from them let the magnitudes AE, CF, wl.ich have a given ratio to one another, be taken ; the remainders EB, FD either have a given ratio to one another, or the excess of one of them above a given magnitude has a given rfitio to the other. Because AB, CD are each of A E B them given, the ratio of AB to ■ -1 CD is given : and if this ratio be the same with the ratio of AE C F D to CF) then the remahider EB 1 a 19. 5. ^^^s a the same given ratio to the remainder FD. But if the ratio of AB to CD be not the same with the ra- tio of AE to CF, it. is either greater than it, or, by inversion, the ratio of CD to AB is greater than the ratio of CF to AE : first, let the ratio of AB to CD be greater than the ratio of AE to CF ; and as AE to CF, so make AG to CD ; there- fore the ratio of AG to CD is given, because the ratio of b2.dat. AE to CF is G-iven ; and CD is civen. wherefore I> AG i^ DATA. 3i given ; and because the ratio of AB to CD is greater than the ratio of (AE to CF, that is, than . V C ^ the ratio of) AG to CD ; A P. is ^_ , ^ ^ greater c than AG : and AP,AG " ' c 10. 5. are given ; therefore tlie remain- p F n der ;"/G is given : and because as AE to CF, so is AG to CD, and ' so is a EG to DF ; the ratio of EG to FD is given : and GB is a 19- 5. given ; therefore EG, the excess of EB above a given magni- tude GB, has a given ratio to FD. The other case is shown in t^he same wav. PROP. XXIV. 13. IF there be three magnitudes, the first of which see n. has a given ratio to the second, and the excess of the second above a gi^^en magnitude has a given ratio to the third ; the excess of the first above a given mag- nitude shall also have a given ratio to the third. Let AB, CD, E, be the three magnitudes of v/hich AB has a given ratio to CD ; and the excess of CD above a given mag- nitude has a given ratio to E : the excess of AB above a given magnitude has a given ratio to E. Let CF be the given magnitude, the excess of CD above which, viz. FD has a given ratio to E : and because the ratio of AB to CD is given, as AB to CD, so make A AG to CF ; therefore the ratio of AG to CF is given ; and CF is given, wherefore a AG is p given : and because as AB to CD, so is AG "* to CF, and so is b QB to FD ; the ratio of GB to FD is given. And the ratio of FD to E is given, wherefore c the ratio of GB to E is given, and AG is given ; therefore GB the ex- cess of AB above a given magnitude AG has a given ratio to E. B Cor. 1. And if the first has a given ratio to the second, and the excess of the first above a given magnitude has a given ratio to the third ; the excess of the second above a given mag- nitude shall have a given ratio to the third. For, if the second be called the first, and the first the second, this corollary will hv the same with the proposition. F- D a 2. dat. b 19. 5. c 9. dat . 384 EUCLID'S Con. 2. Also, if the first has a given ratio to the second, and the excess of the third above a given magnitude has also u given ratio to the second, the same excess shall have a given ratio to the first; as is evident from the 9th dat. 17. PROP. XXV. IF there be three magnitudes, the excess of the first whereof above a gi\'en magnitude has a given ratio to the second ; and the excess of the third above a given magnitude has a given ratio to the same se- cond : the first shall either have a given ratio to the third, or the excess of one of them above a given magnitude shail have a given ratio to the other. Let AB, C, DE be three magnitudes, and let the excesses of each of the two AI?, DE above given magnitudes have given ratios to C ; AB, DE either have a given ratio to one another, or the excess of one of them above a given magnitude has a gi- ven ratio to the other. Let FB the excess of AB above the given magnitude AF have a given ratio to C ; and let GE the ex- A cess of DE above the given magnitude DG have a given ratio to C ; and because FB, GE have each of them a given ratio to C, they F-^ a 9. dat. have a given ratio a to one another. l?ut to FB GE the given magnitudes AF, DG are add- b 18. dat. ed ; therefore b the whole magnitudes AB, DE have either a given ratio to one another, or the excess of one of them above a given magnitude has a given ratio to the other. 18. PROP. XXVL IF there be three magnitudes, the excesses of one (jf which above given magnitudes have given ratios to the other two magnitudes ; tliese t\\o shall either have a given ratio to one anoti er, or the excess of one of them L:bove a given magnitude sliall httvc a given ratio to the other. D G- B C E t)ATA. 386 Lfet AB, CD, F.F be three magnitudes, andlet GD the ex- cess of one of them CO above the given magnitude CG have a given ratio to AB ; and also let KD the excess of the same CD above the given magnitude CK have a given ratio to EF, either AB has a given ratio to EF, or the excess of one of them above a given magnitude has a given ratio to the other. Because GD has a given ratio to AB, as GD to AB, so make CG to HA ; therefore the ratio of CG to HA is given ; and C(i is given, Avherefore a HA is given ; and because as Gi> * 2. dat. to AB, so is CG to HA, and so is b CD to HB ; the ratio of Cb b 12. 5. to HB is given : also because Kl) has a given ratio to i..F, as K J to P2F, so make CK to LE ; therefore rr the ratio of CK to LE is given : and CK is given, vi^herefore LE a is given : and be- cause as KD to EF, so is CK to L • , and so « bisCDtoLF; the ratio of 'D to LF is given: but the ratio of CD to HB is given : ^^ c9.dat, wherefore c the ratio of HB to LF is given : and from HB, LF the given magnitudes HA, B LE being taken, the remainders AB, EF shall either have a given ratio to one another, or the excess of one of them above a given magnitude has a given ratio to the j jg ^^^ other d. c L - G - K - E - D F Another Demonstration. Let AB, C, DE be three magnitudes, and let the excesses of one of them C above given magnitudes have given ratios to AB and DE ; either AB, D2 have a given ratio to one another, or the excess of one of them above a given magnitude has a given ratio to the other. Because the excess of C above a given magnitude has a given a 14. dat, ratio to AB ; therefore a AB together with a given magnitude has a given ratio to C : let this given magni- tude be AF, wherefore FB has a given ratio to C : also because the excess of C above a given magnitude has a given ratio to DE ; therefore a ' DE together with a given magnitude has a given ratio to C : let this given magnitude be , D(t, wherefore GE has a given ratio to C : and FB has a given ratio to C, therefore b the ratio of FB to GE is given : and from FB, (iE the given magnitudes AF, DG being taken, the remainders AB, DE either have a given ratio to one another, or the excess of one of them above a given magnitude has a given ratio to the other c. c 19. dat. ,3C b 9. dati ;86 EUCLID'S 19. PROP. XXVII. a 2. dat. b 19. 5. c 9. dat. IF there be three magnitudes, the excess of the first of which above a given magnitude has a given ratio to the second ; and the excess of the second above a given magnitude has also a given ratio to the third : the excess of the first above a given magnitude shall have a given ratio to the third. Let AB, CD, E be three magnitudes, the excess of the first of which AB above the given magnitude AG, viz. Ci B, has a given ratio to CD ; and FD the excess of CD above the given magnitude CF, has a given ratio to E : the excess of AB above a given magnitude has a given ratio to E. Because the ratio of GB to CD is given, as GB to CD, so make GH to CF ; therefore the ratio of GH . to CF is given ; and CF is given, w^herefore a GH is given; and AG is given, wherefore ^^ the whole AH is given : and because as GB to CD, so is GH to CF, and so is b the re- mainder HBto the remainder FD ; the ratio H-- of HB to FD is given : and the ratio of FD to E is given, wherefore c the ratio of HB to E is given : and AH is given ; therefore HB B the excess of AB above a given magnitude AH has a given ra tio to E. C F- D d 24. dat. Otherwise.^ Let AB, C, D be three magnitudes, the excess EB of the first of which AB above the given magnitude AE has a given ratio to C, and the excess of C cibove a given . magnitude has a given ratio to D : the excess of AB above a given magnitude has a given p ratio to D. Because EB has a given ratio to C, and the excess of C above a given magnitude has a giv- en ratio to D ; therefore d the excess of EB above a given magnitude has a given ratio to D : let this given magnitude be EF : therefore FB the excess of EB above EF has a given ra- ^ C D tio to D : and AF is given, because AE, I'.F F^ DATA. 387 are given : therefore FB the excess of AB above a given mag- nitude AF has a given ratio to D, PROP. XXVIII. 25. IF two lines given in position cut one another, the point or points in which they cut one another ai'e given. Let two lines AB, CD given in position cut one another in the point E ; the point E is gi- C ven. Because the lines AB, CD are given in position, they have always the same situations, and therefore the point, or points, in which they cut one another have always the same situation : and because the lines AB, CD can be found a, the point, or points, in which they cut one another, are likewise found ; and therefore are given in position a. See N. a 4. def. PROP. XXIX. 26. IF the extremities of a straight line be given in po- sition ; the straight line is given in position and mag- nitude. Because the extremities of the straight lin& are given, they can be found a ; let these be the points A, B, between which a 4. def. a straight line AB can be drawn b ; b 1 Post- this has an invariable position, be- A ■ ■ -B ulate cause between two given points there can be drawn but one straight line : and when the straight line AB is drawn, its magnitude is at the same time exhibited^ or given : therefore the straight line AB is given in positiop and magnitude. J8b EUCLID'S gr. PROP. XXX, IF one of the extremities of a straight Hne given in position and magnitude be given j the oiher extremity- shall also be given, Let the point A be given, to wit, one of the extremities of a straight line given in magnitude, and which lies in the straight line AC given in position ; the other extremity is also given. Because the straight line is given in magnitude, one equal a 1 def. to it can be found a ; let this be the straight line D : from the greater straight line AC cut oft' AB equal to the lesser D : therefore the A B C other extremity B of the straight -— 1 line A B is found: and the point B has always the same situation ; be- D cause any other point in AC, upon . the same side of A, cuts oft" between it and the point A a greater or less straight line than AB, that b 4 def. is, than D ; therefore the point B is given b : and it is plain another such point can be found in AC produced upon the other side of the point A. 28. PROP. XXXI. IF a straight line be drawn through a gi^en point parallel to a straight line given in position; that straight line is given in position. Let A be a given point, and BC a straight line given in posi- tion ; the straight line drawn through A parallel to BC is given in position. j^ 2 1 Through A draw a the straight line DAE parallel to BC ; the straight D A E line DAE has always the same posi- ', - tion, because no other straight line can be drawn tl. rough A parallel to BC; B C therefore the straight line DAE wliich ■ . . >> 4 def. has been found is givenb in position. DATA. 389 PROP. XXXII. 29. IF a straight line be drawn to a given point in a straight line given in position, and makes a given angle with it ; that straight line is given in position. Let AB be a straight line given in position, and C a given point in it, the straight line drawn to C, which makes a given angle with CB, is given in position. Because the angle is given, one equal to it can be found a : let this be the angle at D, at the given point C, in the given straight line AB, make bthe angle ECB equal to the angle at D : therefore the straight line EC has always the same situation, because any other straight line FC, drawn to the point C, makes with CB a greater or less angle than the angle ECB, or the angle at D : therefore the straight line EC, which has been found, is given in position. It is to be observed, that there are two straight lines EC, GC upon one side of AB that make equal angles with it, and which make equal angles with it when produced to the other side. PROP. XXXIII. IF a straight line be dra^^a"l from a given point to a straight" line given in position, and makes a given angle with it, that straight line is given in position. From the given point A, let the straight line AD be drawn to the straight line ; C given in position, and make with it a given angle ADC ; AD is given in po- E A F sition. ^- Through the point A, draw a the \ a 31. 1. straight line EAF parallel to BC ; and \ because through the given point A, the „ ^j;^ ~ straight line EAF is draAvn parallel to BC, which is given in position, EAF is therefore given in po- sitionb : and because the straight line AD meets the parallels b 31. dat. S90 EUCLID'S c 29. 1. B<", EF, the angle EADc is equal to the angle ADC ; and ADC is given, wherefore also the angle EAD is given : therefore, because the straight line DA is drawn to the given point A in tlie straight line EF given in position, and makes with it a given d 32. dat. angle EAD, AD is givend in position. 31. PROP. XXXIV. See N. IF from a given point to a straight line given in po- sition, a straight line be drawn which is given in mag- nitude ; the same is also given in position. B D- C Let A be a given point, and BC a straight line given in posi- tion, a straight line given in magnitude drawn from the point A to BC is given in position. Because the straight line is given in magnitude, one equal to a 1. def. it can be founds ; let this be the straight line D : from the point A draw AE perpendicular to BC ; and /^ because AE is the shortest of all the straight lines Avhich can be draAvn from the point A to BC, the straight line D, to which one equal is to be drawn from the point A to BC, cannot be less than AE. If therefore D be equal to AE, AE is the straight line given in magnitude drawn from the given point A to BC : and b 33. dat. it is evident that AE is given in position b, because it is drawn from the given point A to BC, which is given in position, and makes with BC the given angle AEC. But if the stredght line D be not equal to AE, it must be greater than it : produce AE, and make AF equal to D ; and from the centre A, at the distance AF, describe the circle GFH, and join Ao, AH : because the circle GFH is given in posi- c 6. dcf. tionc, and the straight line BC is also given in position ; there- fore their intersection G is gi- A d 28. dat yen d ; and the point A is gi- ven ; wherefore AG is given in e 29. dat. position e, that is, the straight line AG given in magnitude, (for it is equal to D) and drawn from the given point A to the straight line BC given in posi- tion, is also given in position: and in like manner AH is given, in po.sition : therefore in this case there are two straight lines DATA. 391 AG, AH of the same given magnitude which can be drawn from a given point A to a straight line BC given in position. PROP. XXXV. 32, IF a 'Straight line be drawn between two paia'ilel straight lines given in position, and makes given an- gles with them, the sti'aight hne is given in magni- tude. Let the straight line EF be drawn between the parallels AB, CD, which are given in position, and make the given angles BEF, EFD : EF is given in magnitude. In CD take the given point (i, and through G draw a GHa 31. 1. parallel to EF: and because CD meets the parallels GH, EF, the angle EFD is equal b to the angle . t? tj tj b 29. 1. HGD : and EFD is a given angle ;_ ^ ^ ^ wherefore the angle HGD is given ; and because HG is drawn to the given point G,inthe straight line CD, given in posi- tion, and makes a given angle HGD : 7; ^1 ~^. ^ the straight line HG is given in posi- tion c : and AB is given in position : therefore the point H is c 32. dat. given d ; and the point G is also given, wherefore GH is given d 28. the point A to the straight line EC given in position, in the given angle ADC : the F AH F straight line EAF drawn through A pa-""^ rallel to BC is given in position. In BC take a given point G, and draw Gil parallel to AD : and because HG is-; ' — -'; jr, drav,n to a given point G in the straight " D G DATA. 39S line BC given in position, in a given angle HGC, for it is equal a 29. 1. a to the given angle ADC ; HGis given in position b ; but it is b 32. dat. given also in magnitude, because it is equal to c AD which is c 34. 1. given in magnitude ; therefore because G one of the extremi- ties of the straight line GH given in position and magnitude is given, the other extremity H is given d ; and the straight line d 30. dat. EAF, which is drawn through the given point H parallel to bC given in position, is therefore given e in position. e 31 dat. PROP. XXXVIII. IF a straight line be drawn from a given point to two parallel straight lines given in position, the ratio of the segments between the given point and the pa- rallels shall be given. Let the straight line EFG be drawn from the given point E to the parallels AB, CD, the ratio of EF to EG is given. From the point E draw EHK perpendicular to CD ; and because from a given point E the straight line EK is drawn to CD which is given in position, in a given angle EKC ; EK is .34- given in position a ; and AB, CD are given in position ; there»a 33. dat. fore b the points H, K are given : and the point E is given ; b 28. dat. wherefore c EH, EK are given in magnitude, and the ratio d ofc 29. dat. them is therefore given. But as EH to EK, so is EF to EG, ,j j j^^_ because AB, CD are parallels ; therefore the ratio of EF to EG is given. PROP. XXXIX. 35,36. IF the ratio of the segments of a straight line be- See N. t\veen a given point in it and two parallel sti-aight lines, be given, if one of the parallels be given in po- sition, the other is also given in position. 3 D 894 EUCLID'S From the given point A, let the straight line AED be drawn to the two parallel straight lines FG, BC, and let the ratio of the segments AE, AD be given ; if one of the parallels BC be given in position, the other FG is also given in position. From the point A, draw AH perpendicular to BC, and let it meet FG in K : and because AH is drawn from the given point A to the straight line BC given in position, and makes a a 33. dat. given angle AHD ; AH is given * in position ; and BC is likewise given in position, therefore the point H is giv-_B en b : the point A is also given ; where- fore AH is given in magnitude c, and, because FG, BC are parallels, as AE to AD, so is AK to AH ; and the b 28. dat c 29. dat E K H D G F ratio of AE to AD is given, where- fore the ratio of AK to AH is given ; but AH is given in mag- nitude, therefore d AK is given in magnitude ; and it is also given in position, and the point A is given ; wherefore e the point K is given. And because the straight line FG is drawn through the given point K parallel to EC which is given in position, f 31. d.at. therefore f ¥G is given in position. d 2. dat. e 30. dat 37. 38. PROP. XL. See N. IF the ratio of the segments of a straight hne into which it is cut by three p:millel straight lines, be giv- en ; iftwoof the parallels are given in position, the position. third also is given ii Let AB, CD, HK be three parallel straight lines, of which AB, CD are given in position ; and let the ratio of the seg- DATA. 895 ments GE, GF into which the straight line GEF is cut by the three parallels, be given ; the third parallel HK is given in po- sition. In AB take a given point L, and draw LM perpendicular to CD, meeting HK in N ; because LM is drawn from the given point L to CD which is given in position and makes a given angle LMD ; LM is given in positional, and CD is given in ^ 33. dat, position, wherefore the point M is given b ; and the point L isb 28. dat. given, LM is therefore given in magnitude c ; and because thee 29. dat. ratio of GE to GF is given, and as GE to GF, so is NL to K A A E ] L B P /" V K / 1 M D NM ; the ratio of NL to NM is given ; and therefore d the ratio of ML to LN is given ; but LM is given in magnitude d, where- d force LN is given in magnitude ; and it is also given in position, and the point L is given; wherefore f the point N is given, f and because the straight line HK is drawn through the given point N parallel to CD which is given in position, therefore HK is given in position g, S' r Cor. <. 6. or ( 7.dat. 2. dat. 30. dat. 31. dat. PROP. XLL P. IF a straight line meets three parallel straight lines see Note, which are gi\'en in position, the segments into which they cut it have a given ratio. Let the parallel straight lines AB, CD, EF given in position, be cut by the straight line GHK ; the ratio of GH to HK is given. In AB take a given point L, and A G L B draw LM perpendicular to CD, meet- 7 ing EF in N ; therefore a LM is given , , / ^ in position; and ED, CF are given ^ '- / in position, wherefore the points M, / Nare given; and the point L is given; / therefore b the straight lines LM,MN ; r^re gFven in magnitude ; and the ratio i^ "• -^ D a 33. dat, b 29. dat. 396 EUCLID'S c 1. dat. of LM to MN is therefore givenc : but as LM to MN, so is GH to HK ; wherefore the ratio of GH to HK is given. 39. PROP. XLII. See ifote. IF each of the sides of a triangle be given in mag- nitude, tlie triangle is given in species. a 22. 1. b8. 1. c 1. def. d 1. def. e 3. def. Let each of the sides of the triangle ABC be given in mag- nitude, the triangle ABC is given in species. Make a triangle a DEF, the sides of which are equal, each to each, to the given straight lines AB, BC, CA, Avhich can be done ; because any two of them must be greater than the third ; and let DE be e- qual to AB, EF to BC, and FD to CA ; and be- cause the two sides ED, DF are equal to the two BA, AC, each to each, and the base EF equal to B the base BC ; the angle EDF, is equal b to the angle BAC ; therefore, because the an- gle EDF, which is equal to the angle BAC, has been found, the angle BAC is givenc, in like manner the angles at B, C are given. And because the sides AB, BC, CA are given, their ratios to one another are given d, therefore the triangle ABC is given e in species. 40. a 23. 1. PROP. XLIIL IF each of the angles of a triangle be given in mag- nitudc, the triangle is given in species. Let each of the angles of the triangle ABC be given in mag- nitude, the triangle ABC is given in species. Take a straight line DE given in >1 D position and magnitude, and at the points D, E make a the angle EDF equal to the angle BAC and tlie angle DEF equal to ABC ; there- t> C E I fore the other angles EFD, BCA are equal, and each of the angles at the points A, B, C, is given DATA. 39/ \('herefore each of those at the pomts D, E, F is given : and because the straight line FD is drawn to the given point D in DE which is given in position, making the given angle EDF ; therefore DF is given in positionb. In like manner EF also is ^ 32. dat, given in position ; wherefore the point F is given : and the points D, E are given; therefore each of the straight lines DE, EF, FD is given c in magnitude ; wherefore the triangle DEF is' 29. dat. ivhich is therefore given in species. C 4. 6. - 1 def. given in species d : and it is similar e to the triangle ABC : '^ '*-• ^^*^^' 4. 1 6. PROP. XLIV. 41. IF one of the angles of a triangle be given, and if tlie sides about it have a given ratio to one another; the triangle is given in species. Let the triangle ABC have one of its angles BAG given, and let the sides BA, AC about it have a given ratio to one another; the triangle ABC is given in species. Take a straight line DE given in position and magnitude, and at the point D in the given straight line DE, make the an- gle EDF equal to the given angle BAC ; wherefore the angle EDF is given ; and because the straight line FD is drawn to the given point D in ED which is given in position, making the given angle EDF; therefore FD A is given in position a. And because the ratio of BA to AC is given, make the ratio of ED to DF the same with it, and join EF ; and be- a 32. dat. D cause the ratio of ED to DF is given, ^ C E F and ED is given, therefore b DF is given in magnitude: and it^ 2. dat, is given also in position, and the point D is given, wherefore the point F is given c ; and the points D, E are given, where- ^ 30. dat^ fore DE, EF, FD are givend in magnitude : and the triangle 'I 29. dat. DEF is therefore given in species ; and because the triangles e 42. dat. ABC DEF have one angle BAC equal to one angle EDF, and the sides about these angles proportionals ; the triangles ai"e 1" similar ; but the triangle DEF is given in species, and there- ^ ^- ^'• fore also the triangle ABC. 398 EUCLID'S 42. PROP. XLV. See Note. JF the sicles of a trlimerle have to one another sriven is given in species. ratios; the triangi a 2. dat. b 22. 5. c 20 1. d A. 5. e 22. 1. f 42. dat, S 5. 6. D F F Let the sides of the triangle ABC have given ratios to one another, the triangle ABC is given in species. Take a straight line D given in magnitude ; and because the ratio of AB to BC is given, make the ratio of D to E the same with it ; and D is giv«jn, therefore a E is given. And because the ratio of EC to C A is given, to this make the ratio of E to F the same ; and E is given, and therefore a F ; and because as AB to BC, so is D to E ; by composition AB and BC together are to BC, as D and E to F ; but as BC to CA, so is E to F ; therefore, f.r aqualih^ as AB and BC are to CA, so are D and E to F, and AB and EC are greaterc than CA ; there- fore D and E are greaterd than F. In the same manner any two of the three D, E, F are greater than the third. Makee the triangle GHK whose sides are equal to D, E, F, so that GH be equal to D, HK to E, and KG to F ; and because D, E, F are each of them given, there- fore GH, HK, KG are each of them given in magnitude ; there- fore the triangle GHK is givenf in species : but as AB to BC, so is (D to E, that is) GH to HK ; and as BC to CA, so is (E to F, that is,) HK to KG ; therefore, ex cecjuali, as AB to AC, so is GH to GK. Wherefore K^ the triangle ABC is equiangular and similar to the triangle GHK ; and the triangle GHK is given in species ; therefore also the triangle ABC is given in species. Cor If a triangle is required to be made, the sides of which shall have the same ratios which three given straight lines D, E, F have to one another ; it is necessary that every two of them be trreater than the third. DATA. 39^ PROP. XLVI. IV the Sides of a right-angled triangle iibout one of the acute angles have a given ratio to one anotlier; the triangle is given in species. Let the sides AB, BC about the acute angle ABC of the tri- angle ABC, Avhich has a right angle at A, have a given ratio to one another; the triangle ABC is given in species. Take a straight line DE given in position and magnitude ; and because the ratio of AB to BC is given, make as AB to BC, so DE to EF ; and because DE has a given ratio to EF, and DE is given, therefore a EF is given ; and because as AB to BC, so is DE to EF ; and AB is lessb than BC, therefore DE is lessc than EF. From the point D draw DO at right an- gles to DE, and from the centre E at the distance EF, describe a circle which shall meet DG in two points ; let G be either of them, and join EG ; therefore the circumference of the circle is given d in position ; and the straight line DG is given e in position, because it is drawn to the given point D in DE given in position, in a given angle ; therefore f the point G is given ; and the points D, E are given, wherefore DE, EG, GD are given ff in magnitude, and the tri- angle DEG in speciesh. And because the triangles ABC, DEG have the angle BAC equal to the angle EDG, and the sides about the angles ABC, DEG proportionals, and each of the other angles BCA, EGD less than a right angle ; the triangle ABC is equiangular! and similar to the triangle DEG : but DEG is given in species ; therefore the triangle ABC is given in species : and, in the same manner, the triangle made by drawing a straight line from E to the other point in which the circle meets DG is given in species. a 2 dat. b 19. 1. c A. 5. d 6. def. e 32. dat, f 28. dat, g 29. dat, h 42. dat. i 7. 6, 400 EUCLID'S 44, PROP. XLVII. See Note. IF a triancie has one of its an ales which is not a right angle given, and if the sides about another angle have a given ratio to one another; the triangle is given in species. Let the triangle ABC have one of its angles ABC a given, but not a right angle, and let the sides BA, AC about another angle BAC have a given ratio to one another ; the triangle ABC is given in species. First, let the given ratio be the ratio of equality, that is, let the sides B A, AC and consequently the angles ABC, ACB be e- qual ; and because the angle ABC is given, a 32. 1. the angle ACB, and also the remaining a angle BAC is giveil ; therefore the trian- b 43. dat. gle ABC is given b in species : and it is B C evident that in this case the given angle ABC must be acute. Next, let the given ratio be the ratio of a less to a greater, that is, let the side AB adjacent to the given angle be less than the side AC : take a straight line DE given in position and magnitude, and make the angle DEF equal to the given angle c 32. dat. ABC ; therefore EF is given c in position ; and because the ratio of B A to AC is given, as B A to AC, so make ED to DO ; and because the ratio of ED to DG is and ED is civen, the given, D d 2. dat straight line DG is given d, and BA is less than AC, therefore ED e A. 5. is lessc than DG. From the centre D at the distance DG de- scribe the circle GF meeting EF in F, and join DP' ; and because the circle is given f in position, as also the straight line El'\ the point dat. F is given ff ; and the points D, E are given ; wherefore the straigiit '^^^- lines DE, EF, FD are givenh in ^ (].^t magnitude, and the triangle DhF 1. in species i, and because BA is less than AC, the angle ACB is 1 1. 7. 1. Icssk than the angle ABC, and therefore ACB is less I than f 6. def. g28. h 29. i 42 k 18. DATA. 401 c 32. dat. a right angle. In the same mannei-, because ED is less than DG or DF, the angle DFE is less than a right angle : and be- cause the triangles ABC, DEF have the angle ABC equal to the angle DEF, and the sides about the angles BAC EDF proportionals, and each of the other angles ACB, DFE less than a right angle ; the triangles ABC, DEF are m similar, and"^ ?"• ^• DEF is given in species, wherefore the triangle ABC is also gi- ven in species. Thirdly, Let the given ratio be the ratio of a greater to a less, that is, let the side AB adjacent to the given angle be ^ greater than AC ; and as in the last A case, take a straight line DE given in position atid magnitude, and make the angle DEF equal to the given angle ABC ; therefore FE is given c in posi- tion : also draw DG perpendicular to EF ; therefore if the ratio of BA to AC be the same with the ratio of ED to the perpendicular DG, the triangles ABC, DEG are similarm, because the angles ABC, DEG are equal, and DGE is a right angle : therefore the angle ACB is a right angle, and the triangle ABC is given inb species. But if, in this last case, the given ratio of BA to AC be not the same with the ratio of ED to DG, that is, with the ratio of BA to the perpendicular AM drawn from A to BC ; the ratio of BA to AC must be less thano the ratio of BA to AM, because AC is greater than AM. Make as B A to AC o 8. 5 so ED to DH ; therefore the ratio of A ED to DH is less than the ratio of (BA to AM, that is, than the ratio of) ED to DG ; and consequently, DH is great- erp than DG ; and because B A is great- er than AC, ED is greatere than DH. From the centre D, at the distance DH, describe the circle KHF which necessa- rily meets the straight line EF in two points, because DH is greater than DG, and less than DE. Let the circle meet EF in the points F, K which are given, as was shown in the preceding case ; and DF,DKbeingjoined,thetrianglesDEF, DEK are given in species, as was there shown. From the cen- tre A, at the distance AC, describe a circle meeting BC again in L : and if the angle ACB be less than a right angle, ALM must 3E b 43. dat, p 10. 5. e A. 5. 40S EUCLID'S m 7. 6. be greater than a right angle ; and on the contrary. In the same manner, if the angle DFE be less than a right angle, DKE must be greater than one ; and on the contrary. Let each of the an- gles ACB, DEF be either less or great- A er than a right angle : and because in the triangles ABC, DEF the angles ABC, DEF are equal, and the sides BA, AC, and ED, DF about two of the other angles proportionals, the triangle ABC is similar m to the triangle DEF. In the same manner, the triangle ABL is simi- lar to DEK. And the triangles DEF, DEK are given in species ; therefore al- so the triangles ABC, ABL are given in species. And from this it is evident, that in this third case, there are always two triangles of a different species, to which the things mentioned as given in the proposition can agree. 45. PROP. XLVIII. IF a triangle has one angle given, and if both the bides together about that angle have a given ratio to the remaining side; the triangle is given in species. Let the triangle ABC have the angle BAC given, and let the sides BA, AC together about that angle have a given ratio to BC ; the triangle ABC is given in species. Bisect a the angle BAC by the straight line AD ; therefore the angle BAD is given. And because as BA to AC, so is b BD to CD, by permutation, as AB to BD, so is AC to CD ; and as BA and AC to- gether toBC, soisc AB to BD. But the ratio of BA and AC together to BC is given, wherefore the I'atio of AB to BD is given, and the angle BAD is given ; d 47. dat. therefore d the triangle ABD is given in species, and the angle ABD is therefore given ; the angle BAC a 43. dat. is also given, wherefore the triangle ABC is given in species e. A triangle which shall have the things that are mentioned in the proposition to be given, can be found in the folio-wing a 9. b3, cl2. •a9. 1. DATA. m manner. Let EFG be the given angle, and let the ratio of H to K be the given ratio which the two sides about the angle EFG must have to the third side of the triangle ; therefore because two sides of a triangle are greater than the third side, the ratio of H to K must be the ratio of a greater to a less. Bisect a the angle EFG by the straight line FL, and by the ' 47th proposition find a triangle of which EFL is one of the angles, and in which the ratio of the sides about the angle op- posite to FL is the same with the ratio of H to K : to do which, take FE given in position and magnitude, and draw EL perpendicular to FL : then if the ratio of H to K be the same with the ratio of FE to EL, produce EL, and let it meet FG in P ; the triangle FEP is that which was to be found : for it has the given angle EFG ; and because this angle is bisected by FL, the sides EF, FP together are to EP, as b FE to EL, that is, as H to K. But if the ratio of H to K be not the same with the ratio of FE to EL, it must be less than N it, as was sho'wn in prop. 47, and in this case their are two tri- angles, each of which has the given angle EFL, and the ratio of the sides about the angle opposite to FL the same with the ratio of H to K. By prop. 47, find these triangles EFM, EFN, each of which has the angle EFL for one of its angles, and the ratio of the side FE to EM or EN the same with the ratioof H to K ; and let the angle EMF be greater, and ENF less than a right angle. Artd because H is greater than K, EF is greater than EL, and therefore the angle EFN, that is, the angle NFG, is less f than the angle ENF. To each of these add the angles f 18. 1. NEF, EFN : therefore the angles NEF, EFG are less than the angles NEF, EFN, FNE, that is, than two right angles ; there- fore the straight lines EN, FG must meet together when produ- ced ; let them meet in O, and produce EM to G. Each of the triangles, EFG, EFO has the things mentioned to be given in the proposition : for each of them has the given angle EFG ; and because this angle is bisected by the straight line FMN, the sides FF, FG together have to EG the third side the ratio of FE to EM, that is, of H to K. In like manner, the sides EF, FO together have to EO the ratio which H has to K. 404 EUCLID'S 46. PROP. XLIX. IF a triangle has one angle gi^"e^, and if the sides about another angle, both together, have a given ra- tio to the third side ; the triangle is given in species. Let the triangle ABC have one angle ABC given, and let the two sides B A, AC about another angle BAC have a given ratio to BC ; the triangle ABC is given in species. Suppose the angle BAC to be bisected by the straight line AD ; BA and AC together are to BC, as AB to BD, as was shown in the preceding proposition. But the ratio of BA and AC together to BC is given, therefore also the ratio of AB to a 44. dit. BD is given. And the angle ABD is given, wherefore a the triangle ABD is given in species : and consequently the angle B.AD, and its double the angle BAC A are given ; and the angle ABC is gi- ven. Therefore the triangle ABC is b 4". dat. given in species b. A triangle which shall have the things mentioned in the proposition to be given, may be thus found. Let EFG be the given angle, and the ra- tio of H to K the given ratio ; and by prop. 44, find the triangle EFL, which has the angle EFG for one of its angles, and the ratio of the sides EF, FL about this angle the same with the ratio of H to K ; and make the angle LEM equal to the angle FEL. And because the ratio of H toK is the ratio which two sides of a triangle have to the third, H must be greater than K ; and because EF is to FL, as H to K, therefore EF is great- er than FL, and the angle FEL, that is, LEM, is therefore less than the angle ELF. Wherefore the angles LFE, FEM are less than two right angles, as was sho^vn in the foregoing propo- sition, and the straight lines FL, EM must meet if produced ; let them meet in G, EFG is the triangle which was to be found ; for EFG is one of its angles, and because the angle FFG is bi- sected by EL, the two sides FE EG together have to the third side FG the ratio of EF to FL that is, the given ratio of H to K. DATA. 40S PROP. L. 76. IF from the vertex of a triangle gi\en in species, a sti-aigbt line be dra%\Ti to the base in a gi^ en angle ; it shall liave a gi\"en ratio to the base. From the vertex A of the triangle ABC which is given in species, let AD be drawn to the base BC in a given angle ADB : the ratio of AD to BC is given. Because the triangle ABC is given in spe- cies, tlie angle ABD is given, and the angle ADB is given, therefore the triangle ABD is given a in species ; wherefore the ratio of AD ^^^ I \ i 43. dat. to AB is given. And the ratio of AB to BC is given ; and therefore b the ratio of AD to BC is given. PROP. LI. 47. RECTILINEAL figures given in species, are di- ^ ided into triangles whicli are given in species. Let the rectiligeal figure .ABCDE be given in species ' ABCDE may be dixided into triangles given in species. Join BE, BD ; and because ABCDE is given in species, the angle BAE is given a, and the ratio of BA j^^ a 3. de£, to AE is given a ; wherefore the triangle BAE is given in species b, and the angle AEB is therefore given a. But the whole angle AED is given, and therefore the re- B maining angle BED is given, and the ratio of AEto EB is given, as also the ratio of AE to ED ; therefore the ratio of BE to C D ED is given c. And the angle BED is given, wherefore the tri- c 9. dat angle BED is given b in spjecies. In the same manner, the tri- angle BDC is given in species: therefore rectilineal figures which aix^ given in species are divided into triangles given in species. 406 EUCLID'S 48. PROP. LII. IF two triangles given in species be described upon the same straight line ; they shall have a given ratio to one another. Let the ti'iangle ABC, ABD given in species be described upon the same straight line AB ; the ratio of the triangle ABC to the triangle ABD is given. Through the point C, draw CE parallel to AB, and let it meet DA produced in E, and join BE. Because the triangle ABC is given in species, the angle BAC, that is, the angle ACE, is given ; and because the triangle ABD is given in spe- cies, the angle DAB, that is, the angle AEC E C is given. Therefore the ^--^ ~ ~]7] L H triangle ACE is given ill species ; wherefore the ratio of E A to AC a 3, def. is given a, and the ra- tio of CA to AB is given, as also the ratio of B A to AD ; there- b 9. dat. fore the ratio of b EA to AD is given, and the triangle ACB is c 37. 1. equal c to the triangle AEB, and as the triangle AEB, or ACB, ^ J g ' is to the triangle ADB, so is d the straight line EA to AD. But the ratio of EA to AD is given, therefore the ratio of the trian- gle ACB to the triangle ADB is given. PROBLEM. To find the ratio of two triangles ABC, ABD given in species, and which are described upon the same straight line AB. Take a straight line FG given in position and magnitude, and because the angles of the triangles ABC, ABD are given, at the points F, G of the straight line FG, make the angles e 23. 1. GFH, GFK e equal to the angles BAC, BAD ; and the angles FGH, FGK equal to the angles ABC, ABD, each to each. Therefore the triangles ABC, ABD are equiangular to the tri- angles FGH, FGK, each to each. Through the point H draw HL parallel to FG meeting KF produced in L. And because the angles BAC, BAD are equal to the angles GFH, GFK, each to each ; therefore the angles ACE, AEC are equal to FHL FLH, each to each, and the triangle AEC equiangular to the triangle FLH. Therefore as EA to AC, so is LF to FH ; and. ^ DATA. 40r ti as CA to AB, so HF to FG ; and as BA to AD, so is GF to FK ; whei-efore, ex aguali, as EA to AD, so is LF to FK. But, as was shown, the triangle ABC is to the triangle ABD, as the straight line EA to AD, that is, as LF to FK. The ratio there- fore of LF to FK has been found, which is the same with the ratio of t}ie ti'iangle ABC to the triangle ABD. PROP. LIIL 49. IF two rectilineal figures given in species be de- ^^^ Note. scribed upon the same straight hne ; they shall have a given j'atio to one another. Let any two rectilineal figures ABCDE, ABFG Avhich are given in species, be described upon the same straight line AB; the ratio of them to one another is given. Join AC, AD, AF; each of the triangles AED, ADC, ACB, AGF, ABF is given a in species. And becausethetrianglesa51.dat. ADE, ADC given in species are de- scribed upon the same straight line AD, the ratio of EAD to DAC is E . , , ^ b 2 1 given b ; and, by composition, the \ / ^^^"'y' b52. ca. ratio of EACD to DAC is given c \ X^^-^ /„ c7.dat. And the ratio of DAC to CAB is given b, because they are described upon the same straight line AC ; therefore the ratio of EACD to ACB ^^ ^ is given d ; and by composition, the jj ^y .y j_j__o ^ ^' *^**' ratio of ABCDE to ABC is given. In the same manner, the ratio of ABFG to ABF is given. But the ratio of the triangle ABC to the triangle ABF is given ; wherefore b, because the ratio of ABCDE to ABC is given, as also the ratio of ABC to ABF, and the ratio ABF to ABFG ; the ratio of the rectilineal ABCDE to the rectilineal ABFG is given d. PROBLEM. To find the ratio of two rectilineal figures given in species, and described upon the same straight line. Let ABCDE, ABFG be two rectilineal figures given in species, and described upon the same straight line AB, and join AC, AD, AF. Take a straight line HK given in positioi\ and magnitude, and by the 52d dat. find the ratio of the tri- angle ADE to the triangle ADC, and make 'che ratio of HK 408 EUCLID'S to KL the same with it. Find also the ratio of the triangle ACD to the triangle ACB. And make the ratio of KL to LM the same. Also, find the ratio of the triangle ABC to the triangle ABF, and make the ratio of LM to MN the same. And lastly, find the ratio of the triangle AFB to the triangle AFG, and make the ratio of MN to NO the D same. Then the ratio of ABCDE to ABFG is the same with the ra- tio of HM to MO. Because the triangle EAD is to the triangle DAC, as the straight line HK to KL ; and as the triangle DAC to CAB, so is the straight line KL to LM ; therefore, by using composition as often as the number of triangles requires, the rectilineal ABCDE is to the triangle ABC, as the straight line HM to ML. In like manner, because the triangle G AF is to FAB, as ON to NM, by composition, the rectilineal ABFG is to the triangle ABF, as MO to NM; and, by inversion, as ABF to ABFG, so is NM to MO. And the triangle ABC is to ABF, as LM to MN. Wherefore, because as ABCDE to ABC, so is HM to ML ; and as ABC to ABF, so is LM to MN ; and as ABF to ABFG, so is MN to MG ; ex xquali^ as the rectilineal ABCDE to ABFG, so is the straight line HM to MO; 50. PROP. LIV. IF two straight lines have a gi\ en ratio to one ano- ther; the similar rectilineal figin-es described upon them« similarly, shall have a given ratio to one another. Let the straight lines AB, CD have a given ratio to one ano- ther, and let the similar and similarly placed rectilineal figures E, F be described upon them ; the ratio of E to F is given. To AB, CD, let G be a third pro' portional : therefore as AB to CD, so is CD to G. And the ratio of AB to CD is given, wherefore the ratio of CD to G is given ; and conse- quently the ratio of AB to G is also a 9. dat. given a . But as AB to G, so is the b 2. Cor. figure E to the figure ^F. There- 20. 6. fore the ratio of E to F is given. DATA. 409 PROBLEM. To find the ratio of two similar rectilineal figures, E, F, simi- larly described upon straight lines AB, CD which have a given ratio to one another : let G be a third proportional to AB, CD. Take a straight line H given in magnitude ; and because the ratio of AB to CD is given, make the ratio of H to K the same with it ; and because H is given, K is given. As H is to K, so make K to L ; then the ratio of E to F is the same with the ratio of H to L : for AB is to CD, as H to K, wherefore CD is to G, as K to L ; and, ex xquali, as AB to G, so is H to L : but the figure E is tob the figure F, as AB to G, that is, as H to L. ^^ ^°^' PROP. LV. 51 IF two straight lines have a given ratio to one ano- ther ; the rectihneal figures given in species described upon them, shall have to one anodier a given ratio. Let AB, CD be two straight lines which have a given ratio to one another ; the rectilineal figures E, F given in species and described upon them, have a given ratio to one another. Upon the straight line AB, describe the figure AG similar and similarly placed to the figure F ; and because F is given in species, AG is also given in spe- ^V cies : therefore, since the figures / xtK E, AG which are given in spe- a /-^ -S B C D cies, are described upon the same I 1 straight line AB, the , ratio of E [ ^ j to AG is given a, and because the a 53. dat. ratio of AB to CD is given, and jj K L— — upon them are described the simi- lar and similarly placed rectilineal figures AG, F, the ratio of AG to F is given b : and the ratio of AG to E is given ; there- b 54. dat. fore the ratio of E to F is givenc. ^ g j^j.^ PROBLEM. To find the ratio of two rectilineal figures E, F given in spe- cies and described upon the straight lines AB, CD which have a given ratio to one another. Take a straight line H given in magnitude ; and because the rectilineal figures E, AG given in species are described up- on the same straight line AB, find their I'atio by the 53d dat, and make the ratio of H to K the same ; K is thex'efore given : and because the similar rectilineal figures AG, F are described 3 F ' 410 EUCLID'S 62. a 53. b 2. dat 14. 5. 53. upon the straight lines AB, CD, which have a given ratio, find their ratio by the 54th dat. and make the ratio of K to L the same : the figure E has to F the same ratio which H has to L : for, by the construction, as E is to AG, so is H to K ; and as AG to F, so is K to L ; therefore, ejc ayuali, as E to F, so is H to L. PROP. LVI. IF a rectilineal figure given in species be described upon a straight line given in magnitude ; the figure is given in magnitude. Let the rectilineal figure ABCDE given in species be de- scribed upon the straight line AB given in magnitude ; the figure ABCDE is given in magnitude. Upon AB let the square AF be described ; therefore AF is given in species and magnitude, and because the rectilineal figures ABCDE, AF given in species are described upon the same straight line AB, dat- the ratio of ABCDE to AF is given a: but the square AF is given in magnitude, therefore '' also the figure ABCDE is given in magnitude. PROB. To find the magnitude of a rectilineal ^ figure given in species described upon a straight line given in magnitude. Take the straight line GH equal to the given straight line AB, and by the 53d dat. find the ratio which the square AF upon AB has to the figure ABCDE ; and make the ratio of GH to HK the same ; and upon GH de- scribe the square GL, and complete the parallelogram LHKM ; the figure ABCDE is equal to LHKM : because AF is to ABCDE, as the straight line GH to HK, that is, as the figure GL to HM ; arid AF is equal to GL ; therefore ABCDE is equal to HMc. PROP. LVH. IF two rectilineal figu'-es are given in species, and if a side of one of tliem has a given ratio to a side of the other; the ratios of the remaining sides to tlie re- m.aining sides shall be gi^-en. DATA. 411 Let AC, DF be two rectilineal figures given in species, and let the ratio of the side AB to the side DE be given, the ratios of the remaining sides to the remaining sides are also given. a 3. def. Because the ratio of AB to DE is given, as also a the ratios of ^ jq ^^^^ AB to BC, and of DE to EF, the ratio of BC to EF is givenb. In the same manner, the I'atios of the other sides to the other sides are given. A^ The ratio Avhich BC has to EF may be found thus : take a straight line G given in magnitude, and be- cause the ratio of BC to BA is given, make the ratio of G to H the same ; and because the ratio of AB to DE is given, make the ratio of H to K the same ; and make the ratio of K to L the same with the given ratio of DE to EF. Since therefore as BC to BA, so is G to H ; and as BA to DE, so is H to K : and as DE to EF, so is K to L : ejc aquali, BC is to EF, as G to L ; therefore the ratio of G to L has been found, which is the same with the ratio of BC to EF. PROP. LVIII. IF t^vo similar rectilineal figures ha'\'e a given ratio See N. to one another, their homologous sides have also a G:iven ratio to one another. Let the two similar rectilineal figures A, B have a given ratio ro one another, their homologous sides have also a given ratio. Let the side CD be homologous to EF, and to CD, EF let^ g. Cor. the straight line Gbe a third proportional. As therefore a CD 20. 6, to G, so is the figure A to B ; and the ratio of A to B is given, there- fore the ratio of CD to G is given ; and CD, EF, G are proportionals; wherefore b the ratio of CD to EF is given. The ratio of CD to EF may be found thus : take a sti'aight line H given in magnitude ; and because the ratio of the figure A to B is given, make the ratio of H to K the same with it : and as the 13th dat. directs to be done, find a mean proportional L between H and K ; the ratio of CD to EF l±2 - E F G b 13. dat H K 412 EUCLID'S is the same with that of H to L. Let G be a third proportional to CD, EF ; therefore as CD to G, so is (A to B, and so is) H to K ; and as CD to EF, so is H to L, as is shown in the 13th dat. 54. PROP. LIX. ^eeN. IF two rectilineal figures given in species have a given ratio to one another, their sides shall likewise have given ratios to one another. Let the two rectilineal figures A, B given In species, have a given ratio to one another, their sides shall also have given ratios to one another. If the figure A be similar to B, their homologous sides shall have a given ratio to one another, by the preceding proposition ; and because the figures are given in species, the sides of each of a 3. def. them have given ratios a to one another ; therefore each side of b 9. dat. oiie of them has b to each side of the other a given ratio. But if the figure A be not similar to B, let CD, EF be any two of their sides ; and upon EF' conceive the figure EG to be described similar and similarly placed to the figure A, so that CD, EF be homologous sides ; therefore EG is given in spe- cies ; and the figure B is given c S3, dat. iri species ; wherefore c the ra- tio of B to EG is given ; and the ratio of A to B is given, H therefore b the ratio of the figure K A to EG is given ; and A is si- M d 58. dat. rnilar to EG ; therefore d the ra- L ■ tin of the side CD to EF is given ; and consequently b the ratios of the remaining sides to the remaining sides are given. The ratio of CD to EF may be found thus : take a straight line H given in magnitude, and because the ratio of the figure A to B is given, make the ratio of H to K the same with it. And by the 53d dat. find the ratio of the figure B to EG, and nnake the ratio of K to L the same : between H and L find a mean proportional M, the ratio of CD to EF is the same with the ratio of II to M ; because the figure A is to B as II to K ; and as B to EG, so is K to L ; ex ie(^ua/i, as A DATA. 413 to EG, so is H to L : and the figures A, EG are similar, and M is a mean proportional between H and L ; therefore, as wsls shown in the preceding proposition, CD is to EF as H to M. PROP. LX. 55. IF a rectilineal figure be given in species and mag- nitude, the sides of it sliall be given in magnitude. Let the rectilineal figure A be given in species and magnitude, its sides are given in magnitude. Take a straight line BC given in position and magnitude, and upon BC describe a the figure D similar, and similarly a IS. 6. placed, to the figure A, and let EF be the side of the figure A homologous to BC the side of D ; there- fore the figure D is given in species. And because upon the given straight line BC the figure D gi- ven in species is described, D is given b in magnitude, H | *lCf Z' * b 36. dat. and the figure A is gi- ven in magnitude, therefore the ratio of A to D is given : and the figure A is similar to D ; therefore the ratio of the side EF to the homologous side ^n BC is given c ; and BC is given, wherefore d EF is given: and the ratio of EF to EG is given e, therefore EG is given. '^ 2- '^^*^- And, in the same manner, each of the other sides of the figure ^ 3* '^'cf A can be shown to be given. PROBLEM. To describe a rectilineal figure A similar to a given figure D and equal to another given figure H. It is prop. 25, b, 6, Elem. Because each of the figures D, H is given, their ratio is gi- ven, which may be found by making f upon the given straight f Ccr. 45. line BC the parallelogram BK equal to D, and upon its side.. CK making f the parallelogram KL equal to H in the angle KCL equal to the angle MBC ; therefore the ratio of D to H, that is, of BK to KL, is the same with the ratio of BC to CL : and because the figux'es D, A are similar, and that the ratio of D to A, or H, is the same with the ratio of BC to CL ; by the 58th dat. the ratio of the homologous sides BC, EF is the same with the ratio of BC to the mean proportional between BC and CL. Find EF the mean proportional ; then EF is the '414 EUCLID'S side of the figure to be described, homologous to BC the side of D, and the --gure itself can be described by the 18th prop. g 2. Cor. book 6, which, by the construction, is similar to D ; and because 20. 6. D is to A, as g BC to CL, that is as the figure BK to KL; and h 14. 5. ^^^"^ ^ ^^ equal to BK, therefore A h is equal to KL, that is, to H. 5^ PROP, LXL See N. ^^ ^ parallelogram given in magnitude lias one of its sides and one of its angles given in magnitude, the other side also is given. Let the parallelogram ABDC given in magnitude, have the side AB and the angle B AC given in magnitude, the other side AC is given. Take a straight line EF given in position and magnitude ; and because the parallelogram AD A B is given in magnitude, a rectilineal J 7 ^ , r. figure equal to it can be found ^ • / / And a parallelogram equal to this /_ / h Cor 45. ^8'"''^ ^^^ t»e applied b to the given c D straight line EF in an angle equal to E F the given angle BAC. Let this be f the parallelogram EFHG having / the angle I'EG equal to the angle / BAC. And because the parallelo- [ grams AD, EH are equal, and have G the angles at A and E equal ; the c 14 6. sides about them are reciprocally proportional c ; therefore as AB to EF, so is EG to AC ; and AB, EF, EG are given, there- d 12. 6. fore also AC is given d. Whence the way of finding AC is ma- nifest. "• PROP. LXIL See N. IF a p?.rallelograni has a given angle, the rectangle contained by the sides about that angle has a i^ivQu ratio to the pcU'alielogram. A Let the parallelogram ABCD have the gi- ven angle ABC, the rectangle AB, BC has a given ratio to the parallelogram AC. From the point A draw AE perpendi- B~E cular to BC ; because the angle ABC is gi- ven, as also the angle AEB, the triangle a 43. dat. ABE is given a in species ; therefore the ratio of BA to AE is given. But as BA to b 1. 6. AE, so is b t!ie rectangle AB, BC to the rectangle AE, BC; therefore the ratio of &' DATA. 415 the l-ectangle AB, BC to AE, BC that is c, to the parallelogram c 35. 1. AC is given. And it is evident how the ratio of the rectangle to the pa- rallelogram may be found by making the angle FGH equal to the given angle ABC, and draAving, from any point F in one of its sides, FK perpendicular to the other GH ; for GF is to FK, as B A to AE, that is, as the rectangle AB, BC, to the parallelo- gram AC. Cor. And if a triangle ABC has a given angle ABC, the rect- angle AB, BC contained by the sides about that angle, shall have a given ratio to the triangle ABC. Complete the parallelogram ABCD ; therefore, by this pro- position, the rectangle AB, BC has a given ratio to the paral- lelogram AC ; and AC has a given ratLo to its half the triangle d d 41. 1. ABC ; therefore the rectangle AB, BC has a given e ratio to thee 9. dat. triangle ABC. And the ratio of the rectangle to the triangle is found thus : make the triangle FGK, as was shown in the proposition ; the ratio of GF to the half of the perpendicular FK is the same with the ratio of the rectangle AB, BC to the triangle ABC. Because, as was shown, GF is to FK, as AB, BC to the parallelogram AC ; and FK is to its half, as AC is to its half, which is the tri- angle ABC ; therefore, ex equali^ GF is to the half of FK, as AB, BC rectangle is to the triangle ABC. PROP.LXIIL 5^. IF two parallelograms be equiangular, as a side of the first to a side of the second, so is the other side of the second to the straight line to which tlie other side of the first has the same ratio ^vhich the first pa- rallelogram has to the second. And consequently, if t'lc ratio of the first parallelogram to the second be given, the ratio of the other side of the first to that straight line is given; and if the ratio of tlie other side of tlie first to that straight line be given, the ratio of t!-ic fii-st parallelogram to the second is given. Let iVC, DF be two equiangular parallelograms, as BC, a side of the first, is to EF, a side of the second, so is DE, the o-tlier side of the second, to the straiglit line to which AB, the 416 EUCLID'S other side of the first has the same ratio wliich AC has to DF. Produce the straight line AB, and make as BC to EF, so DE to BG, and complete the parallelo- gram BGHC ; therefore, because BC or GH, is to EF, as DE to BG, the sides about the equal angles BGH, DEF are & 14. 6. reciprocally proportional ; wherefore a the parallelogram BH is equal to DF ; and AB is to BG, as the parallelogram AC is to BH , that is, to DF ; as there- fore BC is to EF, so is DE to BG, Avhich is the straight line to which AB has the same ratio that AC has to DF. And if the ratio of the parallelogram AC to DF be given, then the I'atio of the straight line AB to BGis given ; and if the ratio of AB to the straight line BG be given, the ratio of the paral- lelogram AC to DF is given. 74. 73. PROP. LXIV. See Note. IF two parallelograms have unequal but given an- gles, and if as a side of the first to a side of the se- cond, so the other side of the second be made to a certain straight line ; if the ratio of the first parallelo- gram to the second be given, the ratio of the other side of the first to that straight line shall be given. And if the ratio of the other side of the first to that straight line be given, the ratio of the first parallelo- gram to the second shall be given. Let ABCD, EFGH be two parallelograms which have the unequal, but given, angles ABC, EFG ; and as BC to FG, so make EF to the straight line M. If the ratio the parallelo- gram AC to EG be given, the ratio of AB to M is given. At the point B of the straiglit line BC make the angle CBK equal to the angle EFG, and complete the parallelogram KBCL. And because the ratio of AC to EG is given, and that a 35. 1 AC is equal a to the parallelogram KC, therefore the ratio of KC to EG is given; and KC, EG ai'-a equiangular; there- fa 63. dat. fore as BC to FG, so is ^ EF to the straight line to which KB has a given ratio, viz. the same which the parallelogram KC has to EG ; but as BC to FG, so is EF to the straight line M ; therefore KB has a given ratio to M ; and the ratio DATA. •♦ir 34" y C b 63. dat. .K G ©f AB to BK is given, because the triangle ABIC is given in species <= ; therefore the ratio of AB to M is given d. c 43. dav. And if the ratio of AB to M be ^iven, the ratio of the pa-^^^'^i*** ralielogram AC to EG is given : for since the ratio of KB to BA is given, as also the ratio of AB to M, K A L D the ratio of KB to M is given ^ ; and be- cause the parallelograms KC, EG are equiangular, as BC to FG, so is ^ EF to the straight line to which KB has the same ratio which the parallelogram KC has to EG ; but as BC to FG, so is EF to M ; therefore KB is to M, as the parallel- ogram KC is to EG ; and the ratio of KB to M is given, therefore the ratio of the parallelogram KC, that is, of AC to EG, is given. Cor. And if two triangles ABC, EFG, have two equal angles, or two unequal, but given angles ABC, EFG, and if as BC a side of the first to FG a side of the second, so the other side of the se- cond EF be made to a straight line M ; if the ratio of the trian- gles be given, the ratio of the other side of the first to the straight line M is given. Complete the parallelograms ABCD, EFGH ; and because the ratio of the triangle ABC to the triangle EFG is given, the ratio of tile parallelogram AC to EG is given e, because the parallelo-e 15. 3. grams are doub4e f of the triangles ; and because BC istoFG, as £41. 1. EF to M, the ratio of AB to M is given by the 63d dat. if the an- gles ABC, EFG are equal ; but if they be unequal, but given an- gles^ the ratio of AB to M is given by this proposition. And if the ratio of AB to M be given, the ratio of the paral- lelogram AC to EG is given by the same proposition ; and there- fore the ratio of the triangle ABC to EFG is given. 73. PROP. LXV. ^Sf If two equiangular parallelograms have a given ratio to one another, and ii one side has to one side a given ratio ; the other side shall also have to the other side a gl^•en ratio. Let the two equiangular parallelograms AB, CD have a given ratio to one another, and let the side EB have a given ratio to the side FD ; the other side AE has also a given ratio to the other side CF, 3G o. 418 EUCLID'S Because the two equiangular parallelograms AB, CD have a given ratio to one another ; as EB, a side of the first, is to FD, a 63.dat. a side of the second, so is ^ FC, the other side of the second, to the straight line to which AL, the other side of the first, has the same given ratio \yhich the first parallelogram AB has to the other CD. Let this straight line be EG; therefore the ratio of AE to EG is given ; and EB is to FD, as FC to EG, therefore the ratio of FC -^j to EG is given, because the ratio of EB to FD is given ; E^ and because the ratio of AE to EG, as also the ratio of FC G' to EG is given ; the ratio of b9, dat. AE to CF is given i^. The ratio of AE to CF may be found thus : take a straight line H given in magnitude ; and because the ratio of the paral- lelogram AB to CD is given, make the ratio of H to K the same with it. And because the ratio of FD to EB is given, make the ratio of K to L the same : the ratio of AE to CF is the same with the ratio of H to L. Make as EB to FD, so FC to EG, therefore, by inversion, as FD to EB, so is EG toFC ; and as AE to EG, so is a (ilie parallelogram AB to CD, and so is) H to K ; but as LG to FC, so is (FD to EB, and so is) K to L ; therefore, ex aequali, as AE to FC, so is H to L. 69. PROP. LXVI. IF two parallelogiaiiis have unequal, but given an- p'les, and a criven ratio to one another ; if one side has to one side a giA en ratio, the other side has also a given ratio to the other side. Let the two parallelograms ABCD, EFGH which have the given unequal angles ABC, EFG, have a given ratio to one an- other, and let the ratio of BC to EG be given ; the ratio also of AB to EF is given. At the point B of the straight line BG make the angle CBK equal to the given angle EFG, and complete the parallelo- gram BKLC ; and because each of the angles BAK, AKB is -43 d S'^'^"' ^'^^ triangle ABK. is given ^ in species ; therefore the 'ratio of AB to Bl^ is given ; ynd because, by the hypothesis; DATA. A19 the ratio of the parallelogram AC to EG is given, and that AC is equal ^ to BL ; therefore the ratio of BL to EG is given : and b 35. 1. because BL is equiangular to EG, and, by the hypothesis, the ratio of BC to FG is given ; therefore *= the ratio of kb to EFisc65.clat. given, and the ratio of KB to BA is given; the ratio therefore 'i of AB to \ , " , d 9. dau EF is given. The ratio of AB to EF may be found thus : take the straight line MN given E in position and magnitude ; and make the angle NMO equal to the given F angle BAK, and the angle MNO equal to the given angle EFG or AKB : and because the parallelogram BL is equiangular to EG, and has a given ratio to it, and that the ratio of BC to FG is given ; find by the 65th dat. the ratio of KB to LF ; and make the ratio of NO to OP the same with it: then the ratio of AB to EF is the same with the ratio of MO to OP : for since the triangle ABK is equiangular to MON, as AB to BK, so is MO to ON : and as KB to EF, so is NO to OP ; therefore, ex squali, as AB to EF, so is MO to OP. PROP. LXVII. IF the sides of two equiangular parallelograms have See x. given ratios to one another ; the pai-allelograms sliall have a given ratio to one another. Let ABCD, EFGH be two equiangular parallelograms, and let the ratio of AB to EF, as also the ratio of BC to F(j, be given ; the ratio of the parallelogram AC to EG is given. Take a straight line K given in magnitude, and because the ratio of AB to EF is given, make A D E II the ratio of K to L the same with . it; therefore L is given*: and \ because the ratio of BC to FG BV. is given, make the ratio of L to K P.I the same : therefore M is L F given a : and K is given, where- M — . lore^ the ratio of K to Mis given : but the parallelogram AC isbl. dat. to the parallelogram EG, as the straight line K to the straigh.t a 2. dat. 426 EUCLID'S line M, as is demonstrated in the 2^d prop, of B. 6, Elein. there- fore the ratio of AC to EG is given. From this it is plain how the ratio of two equiangular paral- lelograms may be found when the ratios of their sides are given. 70. PROP. LXVIir, SceN. IF the sides of two parallelograms \^•llich have un- equal, but given angles, have given ratios to one another ; the parallelograms shall have a given ratio to one another. Let tvl'o parallelograms ABCD, EFGH which have the given unequal angles ABC, EFG have ihe ratios of their sides, viz. of AB to EF, and of BC to FG, given ; the ratio of the parallelo- gram AC to EG is given. At the point B of the straight line BC make the angle CBK equal to the given angle EFG, and complete the parallelogram KBCL : and because each of the angles B.VK, BK A, is given, a 43. dat. the triangle ABK is given ^ in species: therefore the ratio of AB to BK is given ; and the ratio of AB to EF is given, where- h 9. dat. foi-eb the ratio of BK to EF is j^ ^ L given : and the ratio of BC to FG is given ; andtlie angle KBC is equal to the angle El'G ; c 67. dat. therefore c the ratio of the pa- rallelogram KC to EG is given; d. 35. 1. hut KC is equal "^ to AC; there- fore the ratio of AC to EG is given. The ratio of the parallelogram AC to EG may be found tiius : take the straight line MN given in position and magni- tude, and make the angle MNO equal to the given angle KAB» and the angle NMO equal to the given angle AKB or FEli : and because the ratio of AB to EF is given, make the ratio of NO to P the same : also make the ratio of P to Q the same with the given ratio of BC to FG, the parallelogram AC is to EG, as rvIO to Q. Because the angle KAB is equal to the angle MNO, and the angle AKB rquul to the angle N.MO; the triangle AKB is equiangular to MiVIO: therefore as KB to BA, so is MO to ON; and as BA to KV, so is NO to P; wherefore, ex a)- G C lograms KC, LF. And because the angles AGC, DHF, or their equals, the angles KBC, LEF are either equal, or unequal, but given ; and that the ratio of AG to DH, that is, of KB to LE, is given, as also the ratio of BC to EF ; therefore ^ the I'atio of the partvlleiogram KC to LF isgi-a 67. or ven ; wherefore also the ratio of the triangle ABC to DEF is^^*^^^ given' C41.1. i.15,5. PROP. LXIX. 61. IF a parallelogram which has a given angle be ap- plied to one side of a rectilineal figure given in species ; if the figure have a given ratio to the parallelogram, the parallelogram is given in species. Let ABCD be a rectilineal figure given in species, and to one side of it AB, let the parallelogram ABEF, having the given angle ABE, be applied ; if the figure ABCD has a given ratio •to the parallelogram BF, the p;vraileIogram BF is given in species. 423 EUCLID'S Throuf^h the point A draw AG parallel to BC, and through the point C draw CG parallel to AB, and produce GA, CB to aS.def. the points H, K: because the angle ABC is given =>, and the ratio of AB to BC is given, the figure ABCD being given in species; therefore, the parallelogram BG is given ^ in species. And because upon the same straight line AB the two rectilineal figures BD, BG given in species are described, the ratio of BD .b53.dat. to BG is given b; and, by hypothesis, the ratio of BD to the e 9.dat. parallelogram BF is given ; wherefore '^ the ratio of BF, that is^, d35.1. of the parallelogram BH, to BG is given, and therefore e the ra- -e 1. 6. tio of the straight line KB to BC is given ; and the ratio of BC to B A is given, wherefore the ratio of KB to BA is given "^ : and because the angle ABC is given, the adjacent angle ABK is gi- ven ; and the angle ABE is given, therefore the remaining angle KBE is given. The angle EKB is also given, because it is equal to the angle ABK ; therefore the triangle BKE is given in spe- cies, and consequently the ratio of EB to BK is given ; and the ratio of KB to B \ is given, wherefore "= the ratio of EB to B A is given ; and the an- gle ABE is given, there- fore the parallelogram BF is given in species. A parallelogram similar to BF may be found thusipi" take a straight line LM gi- ven in position and magnitude ; and because the angles ABK» ABE are given, make the angle NLM equal to ABK, and the angle NLO equal to ABF.. And because the ratio of BF' to BD is given, make the ratio of LM to P the same with it ; and be- cause the ratio of the figure BD to BG is given, find this ratio by the 53d dat. and make the ratio of P to Q the same. Also, be- cause the ratio of CB to BA is given, make the ratio of Q to R the same ; and take LN equal to R ; through the pomi M draw OM parallel to LN, and complete the parallelogram NLOS; then / this is similar to the parallelogram BF. Because the angle ABK is equal to NLM, and the angle ABE to NLO, the angle KBE is equal to MLO ; and the angles EKE, LMO are equal, because the angle ABK is equal to NLlNI ; therefore the triangles BKE, LMO are equiangular to one another ; wherefore as BE to BK, so is LO to LM ; and because as the figure BF to BD, so is the straight line LM to P ; and as BD to BG, so is P to Q ; ex xquali, as BF, that is 'Miirl to BG, so is LM to Q: but BB is to^ BG, as KB to BC ; as therefore KB to BC, so is LM to Q ; and be- cause BE hi to BK, as LO to LM; and as BK to BC, so is LM to Q : and as BC to BA, so Q was made to R; there- DATA. 43» fore, ex aquali, as BE to BA, so is LO to R, that is to LN ; and the angles ABE, NLO arc equal ; therefore the parallelo- gram BF is similar to LS. PROP. LXX. 62, ra IF two straight lines have a given ratio to one ano-SeeNote, thcr, and upon one of them be described a rectihiieal figure given in species, and upon the other a paraiklo- gram having a given angle ; if the figure have a given ratio to the paralielogram, tlie paralleiogrLm is given in species. Let the two straight lines AB, CD have a given ratio to one another, and upon AB let the figure AEB given in species be described, and upon CD the parallelogram DF having the given angle FCD ; if the ratio of AEB to DF be given, the parallelo- gram DF is given in species. Upon the straight line AB, conceive the parallelogram AG to be described similar, and similarly placed to FD ; and because the ratio of AB to CD is given, and upon them are described the similar rectiline.il figures AG, FD ; the ratio of AG to FD is gi- ven »; and the ratio of FD to AEB is given ; therefore ^ the ratio of AEB to AG is given; and the angle ABG is given, because it is equul to the angle FCD ; because tht^re- fore the parallelogram AG which has a given angle ABG is applied to a side AB of the figure AEB t:,i- ven in species, and the ratio of AEB to AG is given, the paral-G69.dat* lelogram AG is given <= in species ; but FD is similar to AG ; therefore FD is given in species. A parallelogram similar to FD may be found thus: take a straight line H given in magnitude ; and because the ratio of the figure AEB to FD is given, make the ratio of H to K the same with it : also, because the ratio of the straight line CD to AB is given, find by the 54th dat. the ratio which the figure FD de- scribed upon CD has to the figure AG described upon AB si- milar to FD ; and make the ratio of K to L the same with this ratio : and because the ratios of H to K, and of K to L are 4a4r EUCLID'^S b 9. dat. given, the ratio of H to L is given ^ ; because, therefore, ns- AEB to FD, so is H to K ; and as FD to AG, so is K lo L ; ex xquali, a« AEB to AG, so is H to L ; therefore the ratio of AEB to AG, is given ; and the figure AEB is given in spe- cies, and to its side AB the parallelogram AG is applied in the given angle AEG ; therefore by the 69th dat. a paralklogrann may be found similar to AG : let this be the puraMelogram MN ; MN also is similar to FD ; for, by the construction, MN is similar to AG, and AG is similar to FD j therefore the paral lelogram FD is similar to MN. «^ PROP. Lxxr. IF the extremes of three proportional straight lines- have given ratios to the extremes of other three propor- tional straight lines ; the means shall also have a given ratio to one another : and if one extreme has a given ratio to one extreme, and the mean to the mean ; like- wise the other extreme shall have to the other a given ratio. Let A, B, C be three proportional straight lines, and D, E, F three other ; and let the ratios of A to D, and of C to F be given ; then the ratio of B to E is also given. Because the ratio of A to D, as also of C to F is given, the a 6r. dat. ratio of the rectangle A, C to the rectangle D, F is given ^ ; b 17. 6. but the square of B is equal ^ to the rectangle A, C ; and the square of E to the rectangle ^^ D, F ; therefore the ratio of the c58.dat. square of B lo the square of E is given j wherefore ^ also the ratio of the straight line B to E is given. Next, let the ratio of A to D, and of B to E be gi- ven ; then the ratio of C to F is also given. Because the ratio of B to E is given, the ratio of » J. X d 54; dat. the square of B to the square of E is given ^ ; there- n k f fore ^ the ratio of the rectangle A, C to the rectangle D, F is given ; and the ratio of the side A to the side P is given ; therefore the ratio of the other side C to eG5. dat. the other F is given e. Cou. And if the extremes of four proportionals have to the extremes of four other proportionals given ratios, and one of the means a given ratio to one of the means ; the other mean shall have u given ratio to the other mean, as may be shown in the same manner as in the foregoing proposition. DATA, 42S PROP. LXXII. 82, IF four straight lines be proportionals ; as the first is to the straight line to Avhich the second has a given ratio, so is the third to a straight line to \\ hich the fourth has a given ratio. Let A, B, C, D be four proportional straight lines, viz. as A to B, so C to D ; as A is to the straight line to which B has a given ratio, so is C to a straight line to which D has a given ratio. Let E be the straight line to which B has a given ratio, and as B to E, so make D to F ; the ratio of B to E is given ^ , and therefore the ratio of D to F ; and because as A to B, so is C to D ; and as B to E so D to F ; therefore, ex jcquali, as A to E, so is A li C to F ; and E is the straight line to which B has a C U given ratio, and F that to v.'hich D has a given ratio ; therefore as A is to the straight line to which B has a given I'atio, so is C to a line to which D has a gi- ven ratio. a Hyg. E F PROP. LXXIIL 8S. IF four straight lines be proportionals ; as the first ^^^ ^• is to the straight line to which the second has a given ratio, so is a straight line to which the third has a given ratio to the foiirdi. Let the straight line A be to B, as C to D ; as A to tlve straight line to which B has a given ratio, so is a straight line to which C has a given ratio to D. Let E be the straight line to which B has a given ratio, and as B to E, so make F to C; because the ratio of B to E is given, the ratio of C to F is gi- ABE ven : and because A is to B, as "^^ to D ; and as B F C D to E, so F to C ; therefore, ex xquali in proportione perturbata », A is to E, as F to D ; that is, A is to E a 31 Si to which B has a given ratio, as F, to which C has a given ratio, is to D. 3 H 426 EUCLID'S ^- PROP. LXXIV. IF a triangle has a given obtuse angle ; the excess of the square of the side which subtends the obtuse angle, aboN'c the squares of the sides which contain it, shall have a inven ratio to the triane-le. o o Let the triangle ABC have a giren obtuse angle ABC ; and produce the straight line ('B, and from the point A draw AD perpendicular to BC : the excess of the square of AC abcv.. the a 12. 2. squares of AB, BC, thatis-"*, the double of the rectangle con- tained by DB, BC, has a given ratio to the triangle ABC. Because the angle ABC is given, the angle ABD is also gi- ven ; and the angle ADB is given ; wherefore the triangle b43.dat. A13D is given ^ in species; and therefore the ratio of AD to c 1. 6. DB is given : and as AD to DB, so is ^ the rectangle AD, BC to the rectangle DB, BC ; wherefore the ratio of the rect- angle AD, BC to the rectangle DB, BC is given, as also the ratio of twice the rectangle DB, BC to the A E rectangle AD, BC: but the ratio of the rect- angle AD, BC to the triangle ABC is gi- d 41. 1. Yen, because it is double ^ of the triangle ; therefore the ratio of twice the rectangle e 9. dat. DB, BC to the triangle ABC is given « ; and twice the rectangle DB, BC is the ex- D B C cess ^of the square of AC above the squares of AB, BC ; there- fore this excess has a given ratio to the triangle ABC. And the ratio of this excess to the triangle AEC may be found thus : take a straiglit line EF given in posititn and magnitude ; and because the angle ABC is given, at the point F of the straight line EF, make the angle EFG equal to the angle ABC ; pro- duce GF, and draw EH perpendicular to FG ; then the ratio of the excess of the square of AC above the squares of AB, BC to the triangle ABC, is the same with the ratio of quadruple the straight line HF to HE. Because the angle ABD is equal to the angle EFH, and the angle ADB to LHF, eacli being a right angle ; the triangle £4.6. ADB is equiangular to EHF ; therefore *" as BD to DA, so 2Cor.4 5. FH to HE ; and as quadruple of liD to DA, so is s quadru- ple of FH to HE : but as twice BD is to DA, so is = twice the rectangle DB, BC to the rectangle AD, BC ; and as DA h C»r. 5. to the ha)f of it, so is '^ the rectunjile AD, BC to its half the DATA. 427 triangle ABC; therefore, ex aquali, as twice BD is to the half of DA, that is, as quadruple of BD is to DA, that is, as qua- di'plt^ of FH to HE, so is twice the rectangle DB, BC to the triangle ABC. PROP. LXXV. 65. IF a triangle has a given acute angle, the space l^y which the square of the side subtending the acute angle, is h.iis than the squares of the sides which contain i^, sli;.'i have a given ratio to the triangle. Let the triangle ABC have a given acute angle ABC, and draw AD perpendicular to BC, the space by which the square of AC is i-ss than the squares of ^B, BC, that is S the double o^^the^ 13. 2i rectangle contained by CB, BD, has a given ratio to the triangle ABC. Because tlse angles ABD, ADB are each of them given, the tridngle ABD is given in species; and therefore the ratio of BD to DA is given : and as BD to DA, A so is the rectangle CB, BD to the rectangle CB, AD ; therefore the ratio of these rect- angles is given, as also the ratio of twice the rectangle CB, BD to the rectangle CB, AD ; but the rectangle CB, AD has a given ratio to its half the triangle ABC ; therefore ^ the B D C 1,9 j^^ ratio of twice the rectangle CB, BD to the triangle ABC is given ; and twice the rectangle CB, BD is ^ the space by which the square of AC is less than the squartsof VB, BC ; therefore the ratio of this space to the triangle ABC is given : and the ratio may be found as in the preceding proposition. LEMMA. IF from the vertex A of an iscosceles triangle ABC, any straight line AD be drawn to the base BC, the square of the side AB is equal to the rectangle BD, DC of the segments of the base toge- ther with the square of AD ; but if AD be drawn to the base produced, the square of AD is equal to the rectangle BD, DC together with the square of AB. Case 1. Bisect the base BC in E, and A join AE which will be perpendicular ^ to BC ; wherefore the square of AB is equal ••to the squares of i\E, EB ; but the square of EB is equal =^ to the rectangle BD, DC together with the square of DE ; there- fore the square of AB is equal to the 423 EUCLID'S b 47. 1. squares of AE, ED, that is, to ^ the square of AD, together with the rectangle BD, DC ; the other case is shown in the same way by 6. 2. Elem. 67. PROP. LXXVI. a5.&.32 1. b 45. dat c 50. dat d 1. 6. e41 1. fr,7, 1. jr9, dat. IF a triangle have a ghen angle, the excess of the square of the straight Une which is equal to the t\^^o sides that contain the gi^^en angle, above the square of the third side, shall hdve a given ratio to the triangle. Let the triangle ABC have the given angte BAC, the excessof the square of the straightline which is equal to BA, AC togeiher abovu the square of BC, shall have a given ratio to the triangle ABC. Pioduce BA, and take AD equal to AC, join DC and pro- duce it to E, and through the point B draw BE parallel to AC ; join AE, and draw AF perpendicular to DC ; and be- cause AD is equal to AC, BD is equal to BE : and BC is drawn from the vertex B of the isosceks triangle DBE ; there- fore, by the lemma, the square of BD, that is, of BA and AC together, is equal to the rectangle DC, CE together with the squai'e of BC : and, therefore, the square of BA, AC to- gether, that is, of BD, is greater than D the square of BC by the rectangk DC, CE ; and this rectangle has a given ratio to the ti'iangle ABC : because the angle BAC is given, the adjacent angle CAD is given ; and each of the angles ADC, DC A is given, for each -^ of them is the half ^ of the given angle q y^jL ^ BAC ; therefore the triangle aDC is "\1 given ^ in species; and AI' is drawn ^^K from its vertex to the base in a given angle ; wherefore the ratio of AF to the base CD is given = ; and as CD to At", so is'^ the rectangle DC, CE to the rectangle AF, CE ; and the ratio of the rectangle AF, CE to its Iialf^ the tri- angle ACE is givtn ; therefor'.; the ratio of the rectangle DC, CE to the triangle ACE, tliat is*", to the triangle ABC, is given sr and the rectangle DC, CE is the excess of the square of BA, AC together above the square of BC : therefore the ratio of this excess to the triangle ABC is given. Tne ratio wliich the rectangle DC, CE has to the triangle ABC is found thusc take the straight line GH given in posi- DATA. 42!9 tion and magnitude, and at the point G in GH make the angle HGK equal to the given angle CAD, and take GK equal to GH, join iiH, and draw GL perpendicular to it; then the ratio of HK to the half of GL is the same with the ratio of the rect- angle DC, CE to the triangle ABC : because the angles HGK, DAC at the vertices of th^ isosceles triangles GHK, ADC are equal to one another, these triangles are sin\ilar ; and because GL, AF are perpendicular to the bases HK, DC, as HK to GL, so is*^ (DC to AF, and so is) the rectangle DC, CE to the rect- angle AF, CE ; but as GL to its half, so is the rectangle, AF, CE to its half, which is the triangle ACE, or the triangle ABC ; therefore, ex sequali, HK is to the half of the straight line GL, as the rectangle DC, CE is to the triangle ABC. Cor. And if a triangle have a given angle, the space by whicli the square of the straight line which is the difference of the sides which contain the given angle is less than the square of the third side, shall have a given ratio to the triangle. This is demonstrated the same way as tlie preceding proposition, by help of the second case of the lemma. C4. 6 PROP. LXXVII. L. IF the perpendicular drawn from a given angle of a SeeNote. triangle to the opposite side, or base, has a gi^'en ratio to the base, the triangle is given in species. Let the triangle ABC have the given angle BAC, and let the perpendicular AD drawn to the base BC have a given ratio to it, the triangle ABC is given in species. If ABC be an isosceles triangle, it is evident ^ that if anya5.&32'. 1. G one of its angles be given, the rest are also given ; and there- fore the triangle is given in species, without the consideration of the ratio of the perpendicular to the base, whicli in this case is given by prop. 30. But when ABC is not an isosceles triangle, take any straight line EF given in position and magnitude; and upon it describe 4S0 EUCLID'S the segment of a circle EGF containing; an angle equal to the given angle BAG, draw GH bisecting EF at right angles, and join EG, GF : then, since the angle EGF is equal to the angle BAG, and that EGF is an isosceles triangle, and ABC is not, the angle FEG is not equal to the angle CBA : draw EL ma- king the angle FEL equal to the angle CBA ; join FL, and draw LM perpendicular to EF ; then, because the triangles ELF, BAG are equiangular, as also are the triangles MLE, DAB, as ML to LE, so is DA to AB ; and as LE to EF, so is AB to BC ; wherefore, ex squali, as LM to ILF, so is AD to BC ; and because the ratio of AD to BC is given, therefore the ratio b2.dat. of LM to EF is given ; and EF is given, wherefore ^ LM also is given. Complete the parallelogram LMFK ; and because LM is given, FK is given in magnitude ; it is also given in position, c30. dat. and the point F is given, and consequently "^ the point K ; and be- cause through K the straight line KL is drawn parallel to EF d 51.dat. which is given in position, thereibreti KL is given in position r B Pv D C E O H M F and the circumference ELF is given in position ; therefore tha e ^8. dat. point L is given «". And because the points L, E, F, are given, f 29.dat.tl^^ straight lines LE, EF, FL, are given ' in magnitude ; there- "42. dat. ^^^^ the triangle LEF is given in species ?; and the triangle ABC is similar to LEF, wherefore also ABC is given in species. Because LM is less than GH, the ratio of LM to EF, that is, the given ratio of AD to BC, n\ust be less than the ratio of GH to EF, which the straight line, in a segment of a circle contain- ing an angle equal to the given angle, that bisects the base of the segment at right angles, has unto the base. CoR. 1. If two triangles, ABC, LEF have one angle BAG equal to one angle ELF, and if the perpendicular AD be to the base BC, as the perpendicular LM to the b?,se EF, the triangles ABC, LEF are sinular. Describe the circle EGF about the triangle ELF, and draw LN pardlcl to EF, join EN, NF, and draw NO perpeneicu- lar to EF ; because the angles ENF, ELF are equal, and that DATA. 431 the angle EFN is equal to the alternate angle FNL, that is, t© the angle FEL in the sume segment ; therefore the triangle NEF is similar to LEF ; and in the segment EGF there can be no other triangle upon the base EF, which has the ratio of its perpendicular to that base the same with the ratio of LM or NO to EF, because the perpendicular must be greater or less than LM or NO ; but, as has been shown in the preceding demon- stration, a triangle similar to ABC can be described in the seg- ment EGF upon the base EF, and the ratio of its perpendicular to tiie base is the same, as was there shown, with the ratio of AD to BC, that is, of LM to EF ; therefore that triangle must be either LEF, or NEF, which therefore are similar to the tri- angle ABC. Cor. 2. If a triangle ABC has a given angle BAC, and if the sitriiight line AR drawn from the given angle to the opposite sidt EC, ill a given angle ARC, has a given ratio to BC, the tri- angle ABC is given ir» species. Draw AD perpendicular to BC ; therefore the triangle ARD is gi\en in species ; wherefore the ratio of AD to AR is given : and the ratio of AR to BC is given, and consequently ^ the ratiohS-dat. of AD to BC is given ; and the triangle ABC is therefore given in specic^s'. 177. dat. CoR. 3. If two triangles ABC, LEF have one angle BAC equal to one angle ELF, and if straight lines drawn from these angles to the bases, making with them given and equal angles, have the same ratio to the bases, each to each ; then the trian- gles are similar \ for having drawn perpendiculars to the bases from the equal angles, as one perpendicular is to its base, so is . ^ the other to its buse i^ ; wherefore, by Cor. 1, the triangles are J^ ^ 32 5. similar. A triangle similar to ABC may be found thus : having de- scribed the segment EGF, and drawn the straight line GH, as was oircctcd in the proposition, find IK, which has to EF the given ratio of AD to BC ; and place FK at right angles to EF from ihc point F ; then because, as has been shown, the ratio of AD to BC, that is of FK to El', must be less than the ratio of GH to EF ; therefore FK is less than GH; and consequently the parallel to EF, drawn through the point K, must meet the circumference of the segment in two points : let L be either of them, and join EL, LF, and draw LM perpendicular to EF : then, because the angle BAC is eqiial to the angle ELF, and that aD is to Be, .as KF, that is LM, to EF, the triangle ABC is similar to the triangle LEF, by Cor. l. 439 EUCLID'S 80. PROP. LXXVIII. IF a triangle ha^e one angle given, and if the ratio of the rectangle of the sides which contain the given an- gle to the square of the third side be given, the triangle is given in species. Let the triangle ABC have the given angle BAC, and let the ratio of the rectangle BA, AC to the square of BC be gi* ven ; the triangle ABC is given in species. From the point A, draw AD perpendicular to BC, the rect- a 41. 1. angle AD, BC has a given ratio to its half^ the triangle ABC; and because the angle BAC is given, the I'atio of the triangle bCor62. ABC to the rectangle BA, AC is given "^ ; and by the hypo- dat. thesis, the ratio of the rectangle BA, AC to the square of BC is c9.dat. given; therefore "= the ratio of the rectangle AL>, BC lo the dl. 6. square of BC, that is d, thei ratio of the straight line AD to BC e77.dat. is given ; wherefore the triangle ABC is given in species e. A triangle similar to ABC may be found thus : take a straight line EF given in position and magnitude, and make the angle FEG equal to the given angle BAC, and draw FH perpendicular to EG, and BK perpendicular to AC ; therefore the triangles ABK, EFH are similar, and the rect- angle AD, BC, or the rectangle BK, AC which is equal to it, is to the rectangle BA, AC, as the straight line BK to BA, that is, as FH to FE. Let the given ratio of the rect- angle BA, AC to the square of BC be the same with the ratio of the straight line EF to FL ; therefore, ex xquali, the ratio of the rectangle AD, BC to the square of BC, that is, the ratio of the straight line AD to BC, is the same uith the ratio of HF to FL ; and because AD is not greater than the straight line MN in the segment of the circle described about the triangle ABC, which liisects BC at right angles ; the ratio of AD to BC, that is, of HF to FL, must not be greater than the ratio of MN to BC : let it be so, and, by the 77th dat. find a triangle OPQ which has one of its angles POQ equal to the given an- gle BAC, and the ratio of the pL-rpendicular OR, drawn from that angle to the base PQ the same with the ratio of HF to FL ; then the triangle ABC is similar to OPQ: because, as has bee» B D J> c 10. dat. d 7. dat. DATA. 433 shown, the ratio of AD to BC is the same with the ratio of (HF to FL, that is, by the construction, with the ratio of) OR to PQ : and the angle BAC is equal to tlie angle POQ ; there- fore the triangle ABC is similar f to the triangle POQ. f 1- Cnr. 77. dat. Ot/termise, Let the triangle ABC have the given angle BA.C, and let the ratio of the rectangle BA, AC to the square of BC be given ; the triangle \^C is given in species. Because the angle BAC is given, the excess^ of the square of both the sides BA, AC together above the square of the third side BC has a given ^ ratio to the triangle ABC. Let the a 76, dat. figure D be equal to this excess ; thevefoie the ratio of D to the triangle ABC is given : and the ratio of the triangle ABC to the rectangle BA, AC is given'', because BAC is a given I) Cor. 62. angle ; and the rectangle BA, AC has ^ ^ dat a given ratio to the square of BC : wherefore ^ the ratio of D to the square of BC is given ; and, by composition '^ the ratio of the space D together with the square of BC to the square of BC B C is given ; but D together with the square of BC is equal to the square of both BA and AC together; therefore the ratio of the square of BA, AC together to the square of BC is given ; and the ratio of BA, AC together to BC is therefore given e; and thee59.dat angle BAC is given, wherefoi'e *" the triangle ABC is given inf48.dat. species. The composition of this, which depends upon those of the 76th and 48th propositions, is more complex than the preced- ing composition, which depends upon that of prop. 77, which is easy. PROP, LXXIX. K. IF a triangle have a given angle, and if the straight SeeNotc. line drawn from that angle to the base, making a given angle with it, divides the base into segments m hich have a given ratio to one another ; the triangle is given in species. Let the triangle ABC have the given angle BAC, and let the straight line AD drawn to the base BC making the given angle ADB, divide BC into the segments BD, DC which have 31 434 EUCLID'S a given ratio to one anotl:ier ; the triangle ABC is given In species. a 5. 4. Describe ^ the circle BAC about the triangle, and from its cc;Ure E, draw EA, EB, EC, ED ; becuuse the angle BAC is b 20. 3. given, the angle BEC at the centre, which is the double ^ of it, is given. And the ratio of BE to EC is given, because they c 44. dat. ^[-(j equal to one another; therefore'^ the triangle BEC is given in species, and the ratio of EB to BC is given ; also the d7.dat. i-^tio of CB to BD is given '^ because the ratio of BD to DC e9.dat. ig oiven ; therefore the ratio of EB to BD is given <=, and the angle EBC is given, wherefore the triangle EBD is given < in species, and the ratio of LB, that is, of EA to ED, is there- fore given ; and the angle EDA is given, because each of the angles BDE, BDA is given; therefore the triangle AED is f 4?". dat.given'*" in species, and the angle AED gi- ven ; also the angle DEC is given, because each of the angles BED, BEC is given ; therefore the angle AEC is given, and the ratio of EA to EC, which are equal, is gi- ven ; and the triangle AEC is therefore given ^' in species, and the angle ECA gi- ven ; and the angle ECB is given, wherefore the angle ACB is 543. dat. given, and the angle BAC is also given ; therefore s the triangle ABC is given in species. A triangle similar to ABC may be found, by taking a straight line given in position and magnitude, and dividing it in th^ given ratio which the segments BD, DC are required to have to one another; then, if upon that straight line a segment of a circle be described containing an angle equal to the given, angle BAC, and a straight line be drawn from the point of division in un angle equal to the given angle ADB, and from the point wliere it meets the circumference, straight lines be drawn to the extremity of the first line, these, together with the first line, shall contain a triangle similar to ABC, as may easily be shown. The demonstration may be also made in the manner of that of the 77th prop, and that of the 77th may be made in the man- ner of this. I A a 26. 1, /N :^ • 43. dat B D C bU. cUt, DATA. 433 PROP. LXXX. IF the sides about an angle of a triangle have a given ratio to one another, and if the perpendicuiar drawn iiom that angle to the base has a given ratio to the I'^asc ; tlie triangle is given in species. Let the sides BA, AC, about the angle BAG of the triangk- ABC have a given ratio to one another, and let the perpendicu- lar AD have a given ratio to the base BC ; the triangle ABC is given in species. First, let the sides AB, AC be equal to one another, there- fore the perpendicular AD bisecls -^ the base BC ; and the ratio of AD to BC, and there- fore to its half DB, is given ; and the angle ADB is given ; wherefore the triangle * ABD, and consequently the triangle ABC, is given '^ in species. But let the sides be unequal, and BA be greater than AC ; and make the angle Ci\ E equal to the angle ABC; becaiuie the angle AEB is common to the triangles AEB, CEA, they .-.re similar; therefore as AB to BE, so is CA lo AE, imd. by permutation, as BA to AC, so i» BE to EA, and'bo is EA to EC ; and the ratio of BA to AC is given, therefore the ratio of BE to EA, and tiie ratio of EA to EC, as also the ratio of BE to EC is given = ; wherefore the rat-.o of EB to c 9. dat. BC is given '^ ; and the mtio of AD lo BC Ad6.dat. is given by the hypothesis, therefore '^ the ratio of AD to BE is given ; and the ratio BE to EA was shown to be given ; where- fore the ratio of AD to A.]L is given, and ADE is a right angle, therefore the triangle ADE is given e in species, and the angle AEB given ; thera-e46.dat. tio of BE to EA is likewise given, therefore ^ the lii.Uigle ABE is given in species, and consequently the angle E -Mi, as also the angle ABE, that is, the angle CAE, is given; iheriibre the angle BAC is given, and the angle ABC beirig also given, the triangle ABC is given ^ in species. f 43. dar. How to find a triarigle v.hich shall have the things which are mentioned to be given in the proposition, is evident in the first case ; and to find it tlie more easily in tlie other case, it is to be observed that, if the strai^lu line EF equal to 436 lUCLID'S 6.2. EA be placed in EB towards B, the point F divides the base BC into the segments BF, FC which have to one another the ratio of the sides BA,' AC, because BE, EA or EF, and 19- 5. EC were shown to be proportionals, therefore * BF is to FC as BE to EF, or EA, that is, as BA to AC ; and AE cannot be less than the altitude of the triangle ABC, but it may be equal to it, which, if it be, the triangle, in this case, as also the ratio of the sides, may be thus found ; having given the ratio of the perpendicular to the base, take the straight line GH given in position and magnitude, for the base of the tri- angle to be found ; and let the given ratio of the perpendicu- lar to the base be that of the straight line K to GH, that is, let K be equal to the perpendicular ; and suppose Gi^H to be the triangle which is to be found, therefore having made the angle HLM equal to LGH, it is required that LM be perpendicular to GM, and equal to K ; and because GM, ML, MH are propor- tionals, as was shown of BE, EA, EC, the rectangle GMH is equal to the square of ML. Add the common square of NH, (having bisected GH in N), and the square of SM is equal e to the squares of the given straight lines NH and ML or K ; there- fore the square of NM and its side NM, is given, as also the point M, viz. by taking the straight line NM, the square ■ of ■which is equal to the squares of NH, ML. Draw ML equal to K, at right angles to GM ; and because ML is given in position and magnitude, therefore the point L is given, join LG, LH ; then the triangle LGH is that which was to be found, for the sq':'.ire of NM is equal to the squares of NH and ML, and taking awav the common square of ^ L K S 3 NH, the rectangle GMH is equal s to the square of ML : therefore as GM to ML, so is ML to MH, and h 6. 6. ^'^^ triangle LGM is ^^ there- fore equicingular to HLM, and the angle HLM equal to G NQ H M P the angle LGM, and the straight line LM, drawn from the vertex of the triangle making the angle HLM equal to LGK, is perpendicular to the base and equal to the given straight line K, as was required ; and tiie ratio of the sides GL, LH is the same with the ratio of CM to ML, that is, with the ratio of the straight line which is made up of GN the half of the given base and of NM, the square of which is equal to the squares of GN and K, to the straight line K. And whether this ratio of GM to ML is greater or less than the ratio of the sides of any other triangle upon the base GH, and of which the altitude is equal to the straight line K, DATA. ' 437 fcliat is, the vertex of which is in the parallel to GH drawn through the point L, may be thus found. Let OGH be any such triangle, and draw OP, making the angle HOP equal to the angle OGH ; therefore, as before, GP, PO, PH are pro- portionals, and PO cannot be equal to LM, because the rect- angle GPH would be equal to the rectangle GMH, which is impossible ; for the point P cannot fall upon M, because O would then fall on L ; nor can PO be less than LM, therefore it is greater ; and consequently the rectangle GPH is greater than the rectangle GMH, and the straight line GP greater than GM: therefore the ratio of GM to MH is greater than the ratio of GP to PH, and the ratio of the square of GM to the square of ML is therefore' greater than the ratio of the > 2 Cor. square of GP to the square of PO, and the ratio of the straight 2^* *• line GM to ML, greater than the ratio of GP to PO. But as GM to ML, so is GL to LH ; and as GP to PO,-so is GO to OH ; therefore the ratio of GL to LH is greater than the ratio of GO to OH ; wherefore the ratio of GL-to LH is the greatest of all others ; and consequently the given ratio of the greater side to the less, must not be greater than this ratio. But if the ratio of the sides be not the same with this great- est ratio of GM to ML, it must necessarily be less than it : let any less ratio be given, and the same things being suppo- sed, viz. that GH is the base, and K equal to the altitude of the triangle, it may be found as follows. Divide GH in the point Q, so .that the ratio of GQ to QH may be the same with the given ratio of the sides ; and as GQ to QH, so make GP to PQ, and so will f PQ be to PH ; wherefore the square f 19.5.;; of GP is to the square of PQ, as' the straight line GP to PH : and because GM, ML, MH are proportionals, the square of GM is to the square of ML, as' the straight line GM to MH : but the ratio of GQ to QH, that is, the ratio GP to PQ, is less than the ratio of GM to ML ; and therefore the ratio of the square of GP to the square of PQ is less than the ratio of the square of GM to that of ML ; and consequently the ratio of the straight line GP to PH is less than the ratio of GM to MH ; and, by division, the ratio of GH to HP is less than that of GH to HM ; wherefore k the straight line HP is k 10. 5. greater than HM, and the rectangle GPH, that is, the square of PQ, greater than the rectangle GiMH, that is, than the square of ML, and the straight line PQ is therefore greater than ML. Draw LR parallel to GP : and from P draw PR at right angles to GP : because PQ is greater than ML or PR, the circle described from the centre P, at the distance PQ, must ntcessarily cut LR in two points j let these be O, S, and 43S EUCLID'S join OG, OK; SG, SIT: each of the triangles OGH, SGH have the things mehi'ioned lo be given in the proposition : join OP, SP; and beciiuse as GV % P'Q, or PO, so is PO to PH, the triangle OGP is ecpiiingular to HOP ; as, there- fore OG lo GP, so is no to OP; and, bv permutation, as bo to OH so is GP to PO, or 'PQ : and Jo is (iQ lo QH : therefore the triangle OCiH has the ratio of its sicits GO, OH tlVe same with the given ratio of GQ to QH : and the perpendi- cular has lo the base the given ratio .of K to GH, because the perpendicular is equal to LMor K : the like may be shown in tiie, same way of the triangle SGPI. This construction by which the triangle OGH is found, is shorter than that which would be deduced from the demonstra- tion of the datum, by reason that the base GH is given in position and magnitude, v/hich w.as not, supposed in the de- monstraticn : the same thing is to be observed in the next pro- position. M. PROP. LXXXI. IF the sides about an angle of a triangle be unequal and ha^'C a given ratio to one another, and if the perpen- dicular from tiiat angle to ,tbe base, di\ides it into seg- ments that ha've a gi^en i-atio to one another, the triaii- gle is given in species. Let ABC be a triangle, th.e sides of which about the angle BAG are unequal, and have a given ratio to one another, and let the perpendicular AD to the, base BC -livide it into the seg- ments BD, DC, which have a given ratio to one another, the triangle ABC is given in species. Let AB be greater than AC, and make the angle CAE equal to the angle ABC ; imd because the angle ALB is com- a4 6' mon to the trianglis ABE, CAE, they are =^ equiangular to one another : thercfcre as AB lo BE, so is CA to AE, and, DATA, 4sa by permutaiiOi), as AB lo AC, so BE lo EA, aiul so is EA to EC : but the rutio of BA. lo 7\C is ;4iven, therefore the ratio of BE to EA, as also the ratio of EA to EC is given ; wherefore'' the ratio, of BE to i.C, as also*^ the ratio of EC to CB is givtn : and the ratio of BC to CD: is given'!, because the ratio of ^ BD to DC is given ; therefore ^ the ratio of EC to CD is given, and consequenlly ''. tlic ratio of DE to EC : and the ratio of EC Q. K, L II N to EA was shown to be given, therefore ^ the. ratio, of DE to. EA is given ; and ADE is a right angle, wherefore « the triajigle ADE is given in species, and the angle AED given : aod the ratio of CE to EA is given, therefore f the triungle AEG is gj^ ven in species, and consequently the angle ACE is given, as also the adjacent angle ACB. In the same manner, because the ratio of BE to EA is given, the triangle BEA is given in spe- cies, and the angle ABE is therefore given: a^^d the angle ACB is given ; wherefore the triangle ABC is given s in spe- cies. But the ratio of the greater side BA to the other AC must be less than the ratio of the greater segment BD to DC : be- cause the square of BA is lo the square of AC, as the squares of BD, DA to the squares of DC, DA; and the squares of BD, DA have lo the squares of DC, DA a less ratio than the square of BD has to the square of DC t) because the square of BD is greater than the square of DC ; therefore tlie square of BA has lo the squ;\re pf AC a less ratio than the square of BD has lo that of DC : and consequently the ratio of BA to AC ;?i less than the ratio of BD to DC. This being premised, a triangle which shall have the thing-; mentioned to be given in the proposition, and to which the triangle ABC is similar, may be found thus: take a straight line GH given in position and magnitude, and divide it in K, so that the ratio of CK to KM may be the same with the givcrt ratio of BA to AC; divide also GH irv L, so that the ratio of GL to Lli may be the same with the given ratio of BD to e 4&; da,t, f 44. dat. sr 43. dar. t If A be greater then B, and C any third magnitude; then A and C toge- ther have to B and C together a less ratio than A has to B. Let A be to B as C to D, and because A is greater than B, C is greater than D : but as A is to B, so A and C to B and D ; and A and C have to B and C a less fzfio than A and C have to B and D, because C is greater thui>. P, thejefore A ai-d C have to B and C a less ra-io 'ban A to 3. 440 EUCLID'S DC, and draw LM at right angles to GH : and because the ratio of the sides of a triangle is less than the ratio of the seg- ments of the base, as has been shown, the ratio GK to KH is less than the ratio of GL to LH : wherefore the point L must fall betwixt K and H : also make as GK to KH, so GN h 19. 5 .to NK, and so shall ^ NK be to NH. And from the centre N, at the distance NK, describe a circle, and let its circumference meet LM in O, and join OG, OH ; then OGH is the triangle- v/hich was to be described : because GN is to NK, or NO, as NO to NH, the triangle OGN is equiangular to HON; there- fore as OG to GN, so is HO to ON, and, by permutation, as GO to OH, so is GN to NO, or NK, that is, as GK to KH, that is, in the given ratio of the sides, and by the construction, GL, LH have to one another the given ratio of the segments of the base. 60. PROP. LXXXII. IF a parallelogram given in species and magnitude be increased or diminished by a gnomon given in magnitude, the sides of the gnomon are given in mag- nitude. First, Let the parallelogram AB given in species and magni- tude be increased by the given gnomon ECBDFG, each'of the straight lines CE, DP is given. Because AB is given in species and magnitude, and that the gnomon ECBDFG is given, therefore the whole space AG is given in magnitude : but AG is also given in spieces, be- "S.dcf. cause it is similar^ to AB : therefore the sides of AG are gi- |2.andven'' ; each of the straight lines AE, AF 24. o. jg therefore given ; and each of the straight b 60. dat. lines CA, AD is given S therefore each of c 4. dat. the remainders EC, DF is given <^, Next, let the parallelogram AG given in species and magnitude, be diminished by the given gnomon ECBDFG, each of the straight lines CE, DF is given. Because the parallelogram AG is given. H as also its gnomon ECBDFG, the remaining space AB is given DATA. AU in magnitude : but it is also given in species : because it is simi- lar ^ to AG ; thereibrt: ^ its sides CA, AD are given, and each oi^ the straJLjht lines EA, AF is given ; therefore EC, DF are each of them '^i^en. The gnomon and its sides CE, DF may be found thus in the firsi case. Let H be the given sp.ict to which the gnomon must be made equid. and find ^ a parallelogram simil a- to AB and d equal to the figures AB and H togetlier, and place its sides AE, AF from the point A, upon the straight lines AC, AD, and com- plete the parallelogram AG which is about the same diunieter«'e with AB ; because therefore AG is equal to both AB and H, take away the common p.\rt AB, the remaining gjiomon ECBDFCr is equal to the remaining figure H ; therefore a gnomon equal to H, and iis sides CE, DF are found : and in like manner they may be found in the other case, in wiiich the given figure H must be less than the fi:^ure FE from which it is to be taken. 25.0. 26. 6. PROP. LXXXIII. 58. IF a parallelogram equal to a given space be applied to a given straight line, deficient by a parallelogram given in species, the sides of the defect are gi\en. Let the parallelogram AC equal to a given space be applied to the given straight line AB, deficient by the parallelogram BDCL given in species', each of the straight lines CD, DB are given. Bisect AB in E ; therefore EB is given in magnitude : upon EB describe ^ the parallelogram EF similar to DL and similarly a 18. Cr. placed ; therefore EF is given in species, and G H F is about the same diameter ^ with DL; let f;:^^ | ] b 2fi. 6. BCG be the diameter, and construct the figure ; therefore, because the figure EF gi- ven in sj erii s is described upon the cTIVl. straight line EB, EF is given ^ in magnitude, ^^ and tiie gnomon ELti is equal t^ to me gi- ven figure I' : tUerclbrc ^ since EF is diminished by the givei gnomoi' ELH, the sides EK, FH of the gnomon are given but EK is equal to DC, and FH to DB ; vdierefore CD, DBe82.dat. j^rc each of them given. 3 K c 5(^. dat. d 36. and 43. 1. 442 EUCLID'S This demonstration is the analysis of the problem in the J8tli prop, of book 6, the construction and demonstration of which proposition is the composition of the analysis ; and because the given space AC or its equal the gnomon ELH is to be taken from the figure EF described upon the half of AB similar to BC, therefore AC must not be greater than EF, as is shovn in the 27th prop. B. 6. 59. PROP. LXXXIV. IF a parallelogram equal to a given space be applied to a given straight line, exceeding by a parallciogram given in species ; the sides of the excess are given. G ] F H A E \ Let the parallelogram AC equal to a given space be applied to the given straight line AB, exceeding by the parallelogram BDGL given in species ; each of the straight lines CD, DB are given. Bisect AB in E ; therefore EB is given in magnitude ; upon a 18. 6. EB d(:scribe=i the parallelogram EF similar to LD, and similar- ly placed ; therefore EF is given in species, and is about the b 26. 6. same diameter *» with LD. Let CBG be the diameter and construct the figure : therefore, because the figure EF given in species is described upon the given A. ^_, IbJii_jD straight line EB, EF is given in magni- c 56.clat. tude c, and the gnomon ELH is equal d 36. and to the given figure '^ AC ; wherefore, 43. 1. siiice EF is encreased by the given gnomon ELH, its sides EK e 82. dat. I'^i ^''*^ given « ; but EK is equal to CD, and FH to BD ; there- fore CD, DB are each of them given. Tbis demonstration is the analysis of the problem in the 29th prop, book 6, the construction and demonstration of which is the composition of the analysis. CoK. If a parallelogram given in species be applied to a given stnj:;ht line, exceeding by a parallelogram equal to a given space ; the sides of the parallelogram are given. Let the pardlle!o;.^ram ADCE given in species be applied to the given straight line AB exceeding by tHe parallelogram BDCG equjil to a given space ; the sides AD, DC of the parallelogram are given. K L C \ r ^ K A B D DATA. U3 Draw the diameter DE of the parallelogram AC, and con- struct the figure. Because the parallelogram AK is equal=* to*^'^- ■^• BC which is given, therefore AK is E G C b24 6 given ; and BK is similar^ to AC, there- fore BK is given in species. And since the parallelogram AK given in magni- „ tude is applied to the given straight line AB, exceeding by the parallelogi-am BK given in species, therefore by this pro- position, BD, DK the sides of the excess are given, and the straight line AB is given ; therefore the whole AD, as also DC, to which it has a given ratio, is given. PROB. To apply a parallelogram similar to a given one, to a given straight line AB, exceeding by a parallelogram equal to a given' space. To the given straight line AB apply <= the parallelogram AK = 29. £ equal to the given space, exceeding by the parallelogram BK similar to the one given. Draw DF the diameter of BK, and through the point A draw AE parallel to BF meeting DF pro- duced in E, and complete the parallelogram AC. The parallelogram BC is equal » to AK, that is, to the given space ; and the parallelogram AC is similar ^ to BK ; therefore the parallelogram AC is applied to the straight line AB similar to the one given and exceeding by the parallelogram BC which is equal to the given space. PROP. LXXXV. 84. IF two straight lines contain a parallelogram given in magnitiide, in a given angle ; if the difference of the straight lines be given, they shall each of them be given. Let AB, BC contain the parallelogram AC given in magni- tude, in the given angle ABC, and let the excess of BC above AB be given ; each of the straight lines AB. BC is given. Let DC be the given excess of BC above A E BA, therefore the remainder BD is equal to BA. Complete the parallelogram AD ; and because AB is equal to BD, the ratio of ABto BD is given ; and the angle ABD ' :^ — ^ is given, therefore the parallelogram AD is given in species; and because the given parallelogram AC is 4'U - EUCLID'S applied to the given straight line DC, exceeding by the paral- lelogram AD given in specitis, the sides of the excess are gi- j,84. dat. ven *: therefore BD is given ; and DC is given, wherefore me whole BC is given : and AB is given, therefore AB, BC arc each of them given. 85. PROP. LXXXVI. IF tv;0 straight lines contain a parallelogram gi\cn in magnitude, in a given angle ; if both of them together be gi\'en, they sliall each of them be given. Let the two straight lines AB, BC contain the parallelogi'am AC given in magnitude, in the given angle ABC, and kt \B, BC together be given ; each of tlie straight lines AB, BC is given. Produce CB, and make BD equal to AB, and complete the parallelogram ABDE. Because DB is equal to BA, ana the angle ABD given, because the adjacent an- E A gle ABC is given, the parallelogram AD is given in species : and because AB, BC to- gether are given, and AB is equal to BD ; therefore DC is given : and because the gi- ven parallelogram AC is applied to the gi- ven straight line DC, deficient by the parallelogram AD given a83. dat,!"^ species, the sides AB, BD of the defect are given =* ; and DC is given, wherefore the remainder BC is given ; and each of the straight lines AB, BC is therefore given. 87. ( PROP. LXXXVI I. IF t^^•o straight lines contain a parallelogram gi\-en in magnitude, in a given angle ; if the excess of the square of the greater above the square of the lesser be gnen, each of the straight lines shall be gi\en.' Let the two straight lines AB, BC contain the given parallelo- gram AC in the given angle ABC ; if the excess of the squaie of BC above the square of BA be given ; AB and BC arc each of them given. Let tlie given excess of the square of BC above the square of BA be the rectangle CB, BD : take this from the square DATA. 445 of BC, the remainder, which is * the rectangle BC, CD is e»a2. 2. qu,.l to tlie square of AB ; and because the angle ABC of Iht purullelogram AC is given, the ratio of the rectangle of the sides AB, BC to the parallelogram AC is given ^ ; and AC b 62. dar. is ii,ivcn, therefore the rectangle AB, BC is given ; and the rectangle CB, BD is given ; therefore the ratio of the rectangle CB, BD to the rectangle AB, BC, that is':, the ratio of thee 1.6. straight line DB to BA is given: therefore 'I the ratio of the d 54. tht." square of DB to the square of BA is given, A and the square of BA is equal to the rect- angle BC'. CD: wherefore the ratio of the rectangle BC, CD to the square of BD is given, as also the ratio of four times the B PD C rectangle BC- CD to the square of BU ; and, by composition e, the ratio of four times the i-^ctangle BC, CDeT. dat, together with the square of BC to the square of BD is given : but four times the rectangle BC, CD together with the square of BD is equal f to the square of the straight lines BC, CD^S. 2. taken together: therefore the ratio of the square of BC, CD together to tlie square of liD is given ; wherefores the ratio ofS^^.dat, the straiglu line BC together with CD to BD is given : and, by composition, the riitio of BC together with CD and DB, that is, the ratio of twice BC to BD, is given ; therefore the ratio of BC to BD is given, as also<= the ratio af the square of BC to the rectangle CB, BD : but the rect^ingle CB, BD is given, being the given excess of the squares of BC, BA ; therefore the square of BC, and the straight line BC is given : and the ratio of BC to BD, as also of BD lo BA has been shown to be given ; therefore ii the ratio of BC to BA is given ; and BC is given, h 9. da', wherefore Bi^ is given. Tlie preceding demonstration is the analysis of this problem, viz. A parallelogram AC which has a given angle ABC being gi- ven in magnitude, and the excess of the squt.re of hC one of its sides above the square of the olher BA being given ; to fmd the sides: and the composition is as follows. Let EFG be the given angle lo which the angle ABC is re- quired to be equal, and from aiiy point E in i'"E, araw ECi perpendicular to FG ; let the rect- angle EG, GH be the given space to which the paralielogran^ AC is to be made equal ; auU the rectan- gle HG, GL, be tlie givtn excess of the squares of BC, 3A. Takv, in the straight line GE, GK equal to FE, and make GM F G L O H N double of GK : join AIL, and in GL produced, take LN equal to LM : bisect CN in O.. and between CfL GO {ind a mean pro- 446 EUCLID'S portional BC : as OG to GL, so make CB to BD ; and malie the angle CBA equal to Gr E, and as LG to GK so make DB to B A ; and complete the parallelogram AC : AC is equal to the rect.. angle EG, GH, and the excess of the squares of CB, BA is equal to the rectangle HG, GL. Because as CB to BD, so is OG to GL, the square of CB J 1. 6. is to the rectangle CB, BD as * the rectangle HG, GO to the rectangle HG, GL : and the square of CB is equal to the rectangle HG, GO, because GO, BC, GH are proportionals ; b 14. 5. therefore the rectangle CB, BD is equal ^ to HG, GL. And because as CB to BD, so is OG to GL ; twice CB is to BD, as twice OG, that is, GN to GL ; and, by division, as BC together with CD is to BD, so is NL, that is, LM, to LG ; c22. 6. therefore <= the square of BC together with CD is to the square of BD, as the square of ML to the square of LG : but the d 8. 2. square of BC and CD together is equal <1 to four times the rectangle BC, CD together with the square of BD : therefore four times the rectangle BC, CD together with the square of BD is to the square of BD, as the square of ML to the square of LG : and, by division, four times the rectangle BC, CD is to the square of BD, as the square of MG to the square of GL ; wherefore the rectangle BC, CD is to the square of BD as (the square of KG the half of MG to the square of GL» that is, as) the square of AB to the square of BD, because as LG to GK, so DB was made to BA : therefore i' the rectan- gle BC, CD is equal to the square of AB. To each of these add the rectangle CB, BD, and the square of BC becomes equal to the square of AB together with the rectangle CB, BD ; therefore this rectangle, that is, the given rectangle HG, GL, is the excess of the squares of BC, AB. From the point A, draw AP perpendicular to BC, and hrcause the angle ABP is equal to the angle EFG the triangle ABP is equiangular to EFG: and DB was made to BA, as LG to GK ; therefore as the rectangle CB, BD to CB, BA, so is the rectangle HG, M PD F G O HN GL to HG, GK; and as the rectangle CB, BA to AP, BC, so is (the straight' line BA to AP, and so is FE or GK ta DATA. 447 EG, and so is) the rectangle HG, GK to HG, GE ; therefore ex xqualii as the rectangle CB, BD to AP, BC, so is the rect- angle HG, GL to EG, GH: and the rectangle CB, BD is equal to HG, GL ; therefore the rectangle AP, BC, that is, the paral- lelogram AC, is equal to the given rectangle EG, GH. PROP. LXXXVIII. n;. IF two straight lines contain a parallelogram given in magnitude, in a given angle ; if the sum of the squiires of its sides be given, the sides shall each of them be given. Let the two straight lines AB, BC contain the parallelogram ABCD given in magnitude in the given angle ABC, and let the sum of the squares of AB, BC be given ; AB, BC are each of them given. First, let ABC be a right angle ; and because twice the rect- angle contained by two equal straight lines is equal to both their squares ; but if two straight lines are un- A D equal, twice the rectangle contained by them is l J less than the sum of their squares, as is evident I from the 7th prop, book 2, Elem ; therefore twice B C the given space, to which space the rectangle of which the sides are to be found is equal, must not be greater than the given sum of the squares of the sides : and if twice that space bt; equal to the given sum of the squares, the sides of the rectangle must ne- cessarily be equi*l to one another ; therefore in this case describe a square ABCD equal to the given rectangle, and its sides AB, BC are those which were to be found : for the rectangle AC is equal to the given space, and the sum of the squares of its sides AB, BC is equal to twice the rectangle AC, that is, by the hypo- thesis, to the given space to which the sum of the squares was required to be equal. But if twice the given rectangle be not equal to the given sum of the squares of the sides, it must be less than it, as has been shown. Let ABCD be the rectangle, joi'n AC and draw BE *i8 EUCLID'S perpendicular lo it. and complete the rectangle AEEi', and de-- scribe the circle ABC about the triai gle i\EC; AC is its duimc- aCor.5.4. tgj, a ; and because the triangle ABC is similar ^ to AEB, as AC to ^^•^- CB so is AB to BE ; thercibre the rectangle AC, BE is <.qt;ul to AB, BC ; and the rectangle AB, BC is given, \vherLIbr^J .iC, BE is given: and because t!^e sum of the squares of aB, B-^ is c 4r. 1. given, the square of AC which is equal ^ to that sum is given ; and AC itself is therefore given in magnitudi^ : let AC be likewise given d32. dat. in position, and the jjoint A; therefore AF is given '^ in position : and the rectangle AC, BE is given, as lias e 61.dat. been shown, and 7\.C is given, vvhei'cfore « BE is given in magnitude, as also AF which is equal to it ; and AF is also giv- ,^ ^ Ij en in position, and the point A is given ; ^^V f 30. dat. wherefore f the point F is given, and the g31. dat. straight line FB in position a : and the cir- cumference ABC is given in position, li28. dat. wherefore ^ the point B is given : and the G K points A, C are given ; therefore the straight lines AB, BC are. i29.dat. given > in position and magnitude. The sides AB, BC of the rectangle may be found thus ; let the rectangle GH, GK be the given space to which the rect- angle AB, BC is equal ; and let GH, GL be the given rect- angle to which the sum of the squares of A3, BC is equal : k 14. 2. ^-^^^ ^ ^ square equal to the rectangle GH, GL : and let its side AC be given in position ; upon AC as a diameter describe the semicircle ABC, and as AC to GH, so make GK to AF, and from the point A place AF at right angles to AC : therefore the rect- ri6. 6. angle CA, AF is equaP to GH, GK; and, by the hypothesis, twice the rectangle GH, GK is less than GH, GL, that is, than the square of AC; wherefore twice the rectangle C.\, AF is less than the square of AC, and the rectangle CA, AF itself less than half the square of AC, that is, than the rectangle contain- ed by the diameter AC and its half ; wherefore AF is less than the semidiameter of the circle, and consequently the straight line drawn through the point F parallel to AC must meet the circum- ference in two points: let B be cither of them, and joint AB, BC, and complete the rectangle ABCD, ABCD is the rectangle which was to be found: draw BE perpendicular to AC; there- in 34. 1. fo'"'-' 3E is equal '" to AF, and because the angle ABC in a semi- circle is a right angle, the rectangle AB, BC is equal ^ to AC, BE, that is, to the rectangle CA, AF, which is equal to the given rectangle GH, GK ; and the squares of AB, BC are together equal <^ to the square of AC, that is, to the given rectangle GH, .GL. DATA. 449 B L But if the given angle ABC of the parallelogram AC be not a right angle, in this case, because ABC is a given angle, the ra- tio of the rectangle contained by the sides AB, BC to the paral- lelogram AC is given"; and AC is given, therefore the rcctan-n62.dat. gle AB, BC is given ; and the sum of the squares of AB, BC is given ; therefore the sides AB, BC are given by the preceding case. The sides AB, BC and the parallelogram AC may be found thus : let EFG be the given angle of the parallelogram, and from any point E in FE draw EG perpendicular to FG ; and let the rectangle EG, FH be the given space to which the pa- rallelogram is to be made equal, and let EF, A D FK be the given rectangle to which the sum of the squares of the sides is to be equal. And, by the preceding case, find the sides of a rectangle which is equal to the given rectangle EF, FH, and the squares of the sides of which are together equal to the gi- ven rectangle EF, FK; therefore, as was shown in that case, twice the rectangle EF, FH must not be greater than the rectangle EF, FK ; let it be so, and let AB, BC be the sides of the rectangle joined in the an- gle ABC equal to the given angle EFG, and complete the parallelogram ABCD, which will be that which was to be found; draw Ai> perpendicular to BC, and because the angle ABL is equal to EFG, the triangle ABL is equiangular to EFG ; and the parallelogram AC, that is, the rectangle AL, BC is to the rectangle AB, BC as (the straight line AL to AB, that is, as EG to EF, that is, as) the rectangle EG, FH to EF, FH -, and, by the construction, the rectangle AB, BC is equal to EF, FH, therefore the rectangle AL, BC, or, its equal, the paralle- logram AC, is equal to the given rectangle EG, FH ; and the squares of AB, BC are together equal, by construction, to the given rectangle EF, FK. L 450 EUCLID'S 86, PROP. LXXXIX. IF two straight lines contain a given parallelogram in a given angle, and if the excess of the square of one of them above a given space, has a given ratio to the square of the other ; each of the straight lines shall be given. Let the two straight lines AB, BC contain the given parallelo- gram AC in the p;iven angle ABC, and let the excess of the square of BC above a given space have a given ratio to the square ©f AB, each of the straight lines AB, BC is given. Because the excess of the square of BC above a given space lias a given ratio to the square of BA, let the rectangle CB, BD be the given space ; take this from the square of BC, the re- a 2. 2. mainder, to wit, the rectangle =^ BC, CD has a given ratio to the square of BA : draw AE perpendicular to BC, and let the square of BF be equal to the rectangle BC, CD, then, because the angle ABC, as also BEA, is given, the b 43 dat. triangle ABE is given ^^ in species, and the ratio of AE to AB given : and because the ratio of the rectangle BC, CD, that is, of the square of BF to the square of BA, is gi- ven, the ratio of the straight line BF to BA c 58. dat. is given « ; and the ratio of AE to AB is given, wherefore ^Lthe d9, dat. ratio of AE to BF is given ; as also the ratio of the rectangle e 35. 1. AE, BC, that is, c of the parallelogram AC to the rectangle KB, BC ; and AC is given, wherefore the rectangle FB, BC is given. The excess of thg square of BC above the square of BF, that is, above the rectangle BC, CD, is given, for it is equal ^ to the given rectangle CB, BD ; therefore, because the rectangle con- tained by the straight lines FB, BC is given, and also the excess of the square of BC above the square of BF ; FB, BC are each (87-. clat of them given ^ ; and the ratio of FB to BA is given ; therefore, AB, BC are given. The cornpos'Uion is as follows ; Let CilK be the given siigle to which the angle of the paral- lelogram is to be made equal, and from any point G in IIG, draw CK. perpendicular to HK j let GK, IIL be the rectangle to which DATA. 4:5 i H KM A 7 / c the parallelogram is to be made equal, and N let LH, HM be the rectangle equal to the giten space which is to be taken from the square of one of the sides; and let the ratio of the remainder to the square of the, other side be the same with the ratio of the square of the given straight line NH to the square of the given straight line HG. By help of the 87th dat. find two straight lines BC, BF, which contain a rectangle equal to the given rectangle NH, HL, and such that the excess of the square of BC F above the square of BF be equal to the gi- ven rectangle LH, HM ; and join CB, BF in the angle FBC equal to the given angle GHK: and as NH to HG, so make FB to BA, and complete the parallelogram AC, BED and draw AE perpendicular to BC ; then AC is equal to the rectangle f K, IIL ; and if from the square of BC, the given rectangle LH, HM be taken, the remainder shall have to the square of BA the same ratio which the square of NH has to the square of HG. Because, by the construction, the square of BC is equal to the square of BF, together with the rectangle LH, HM ; if from the square of BC there be taken the rectangle LH, HM, there re- mains the square of BF which has s to the square of BA the same „ 22. 6. ratio vvhich the squai'e of NH has to the square of HG, because, as NH to HG, so FB was made to BA ; but as HG to GK, so is BA. to AE, because the triangle GHK is equiangular to ABE; therefore, ex aguali, as NH to GK, so is FB to AF ; wherefore ^^h 1. Gf the rectangle NH, HL is to the rectangle GK, HL, as the rect- angle FB, BC to AE, BC ; but by the construction, the rectangle NH, HL is equal to FB, BC ; therefore ' the rectangle GK, HL i 14. 5: is equal to the rectangle AE, BC, that is, to the parallelogram AC. The analysis of this problem might have been made as In the 86th prop, in the Greek, and the composition of it may be made as that which is in prop. 87th of this edition. 45C EUCLID'S 0, PROP. XC. IF hvo straight lines contain a given pai'alielogram in a given angle, and if the square of one of them together with the space Vvhich has a gi\'cn ratio to the squai'e of the other be given, each of the straight lines shall be given. Let the two straight lines AB, BC contain the given parallelo- gram AC in the given angle ABC, and let tlie square of BC to- gether with the space which has a given ratio to the square of AB be given, AB, BC are each of them given. Let the square of BD be the space which has the given ratio to the square of AB ; therefore, by the hypothesis, the square of BC together with the square of BD is given. From the point A, draw AL perpendicular to BC ; and because the angles ABE, a43.dat. BE A are given, the triangle ABE is given » in species; there- fore the ratio of BA to AE is given : and because the ratio of the square of BD to the square of BA is given, the ratio of the b 58 dat. straight line BD to B A is given ^ ; and the ratio of BA to AE c 9. dat is given ; therefore ^ the ratio of AE to BD is given, as also the ratio of the reciangle AE, BC, that is, of the parallelogram AC to the rectangle DB, BC ; and AC is given, therefore the rect- angle DB, BC is given ; and the square of BC together with the D M A "7 y BE C GH K d88.dat square of BD is given ; therefore ^ because the rectangle contain- ed by the two straight lines DB, BC is given, and the sum of their squares is given : the sti-aight lines DB, BC are each of them given ; and the ratio of DB to BA is given ; therefore AB, BC are given. The comflosiiion is as fellows : Let FGH be the given angle to vvhich the angle of the paral- lelogram is to be made equal, and from any point F in OF draw FH perpendicular to GH ; and let the rectangle FH, GK be tliat to which the parallelogram is to be made equal; and let the rectangle KG, GL be the space to which the square DATA. 453 of one of the sides of the parallelogram together with the space which has a given ratio to the square of the other side, is to be made equal ; and let this given ratio be the same which the square of the given straight line MG has to the square of GF. By the 88th dat. find two straight lines DB, BG which con- tain a rectangle equal to the given rectangle MG, GK, and such that the sum of their squares is equal to the given rectan- gle KG, GL : therefore, by the determination of the pro- blem in that proposition, twice the rectangle MG, GK must not be greater than the rectangle KG, GL. Let it be so, and join the straight lines DB, BC in the angle DEC equal to the given angle FGH ; and, as MG to GF, so make DB to BA, and complete the parallelogram AC : AC is equal to the rect- D M BE 7 a K angle FH, GK; and the square of BC together with the square of BD, which, by the construction, has to the square of BA the given ratio which the square of MG has to the square of GF, is equal, by the construction, to the given rectangle KG, GL. Draw AE perpendicular to BC. Because, as DB to BA, so is MG to GF ; and as BA to AE, so GF to FH ; ex aguali, as DB to AE, so is MG to FH ; there- fore, as the rectangle DB, BC to AE, BC, so is the rectangle MG, GK to FH, GK ; and the rectangle DB, BC is equal to the rectangle MG, GK ; therefore the rectangle AE, BC,^ that is, the parallelogram AC, is equal to the rectangle FH, GK. PROP. XCI. 88. IF a straight line drawn vrithin a circle given in magnitude cuts off a segment Vvbich contains a given angle ; the straight line is given in magnitude. In the circle ABC given in magnitude, let the straight line AC be drawn, cutting" off the segment AEC which contains the given angle AEC ; the straight line AC is given in magnitude. " Take D the centre of the circle ^ join AD and produce Hal. 3. ■454 EUCLID'S to E, and join EC : the angle ACE being b ol. 3. a right ^ angle is given ; and the angle >c4j. dat. AEC is given ; therefore = the triangle ACE is given in species, and the ratio of EA to AC is therefore given ; and EA is given in d 5. def. magnitude, because the circle is given ^ in e a. dat. magnitude; AC is therefore given ^ inmag« nitude. 89. PROP. XCII. IF a straight line giA^en in magnitude be drawn within a circle given in magnitude, it shall cut off a segment containing a given angle. Let the straight line AC given in magnitude be drawn within the circle ABC given in magnitude ; it shall cut off a segment contaning a given angle. Take D the centre of the circle, join AD and produce it to E, and join EC : and be- cause each of the straight lines EA and AC a 1. dat. is given, their ratio is given ^ ; and the an- gle ACE is a right angle, therefore the tri- b46.dat. angle ACE is given '' in species, and conse- quently the angle AEC is given. 90. PROP. XCIIL IF from any point in the circumference of a circle given in position two straight lines be drawn meet- ing the circumference and containing a given angle ; if the point in w^hich one of them meets the circum- ference again be given, the point in which the other meets it is also given. From any point A in the circumference of a circle ABC gi- ven in position, let AB, AC be drawn to the circumference, ma- king the given angle BAC ; if the point B be given, the point C is also given. Take D the centre of the circle, and join BD, DC ; and because each of the a 29. dat. points B, D is given, BD is given ^^ in po- sition ; and because the angle BAC is gi- B b 20. 3. ven, the angle BDC is given ^, therefore DATA* 455 because tlie straight line DC is drawn to the given point D in the straight hne BD given in position in the given angle BDC, DC. is given <= in position : and the circumference ABC is given c 32. dat. in position, therefore^ the point C is given. d28.dat. PROP. XCIV. ^* If from a given point a straight line be drawn touch- ing a circle given in position ; the straight line is given in position and magnitude. Let the straight line AB be drawn from the given point A touching the circle BC given in position ; AB is given in posi- tion and magnitude. Take D the centre of the circle, and join DA, each of the points D, A is given, the straight line AD is given * in position and magnitude : and DBA is a right'' angle, wherefore DA is a diameter* of the circle DBA, described about the tri- angle '1 DBA; and that circle is there- fore given ^ in position : and the circle BC is given in position, therefore the point 13 is given^ ; the point A is also given; therefore the ^28. dat. straight line AB is given * in position and magnitude. DB: because d6. def. PROP. XCV. 92. IF a straight line be drawn from a given point without a circle given in position ; the rectangle contained by the segments betwixt the point and the circumference of the circle is given. Let the straight line ABC be drawn from the given point A without the circle BCD given in posi- tion, cutting it in B, C ; the rectangle BA, AC is given. From the point A draw ^ AD touch- ing the circle; therefore AD is givcn^ in position and magnitude ; and because AD is given, the square of AD is gi- ven <=, which is equal 'i to the rectangls rectangle BA, AG is gis'en- a 17. BA. b94.dat- B A, AC ; therefore the ^ 56. dat. 456 EUCLID'S 93. PROP. xcvr. IF a straight line be drawn througli a given point within a circle given in position, the rectangle contained by the segments betwixt the point and the circumference of the circle is given. Let the straight line BAG be drawn through the given point A within the circle BCE given in position ; the rectangle BA, AC is given. Take D the centre of the circle, join AD, and produce it to the points E, F ; because the points A, D are given, the straight line a29.dat. AD is given ^ in position ; and the circle BEC is given in position; therefoi'e the points B' b 28. dat. £, p are given ^ ; and the point A is given, therefore EA, AF are each of them given ^ ; ^ c 35. 3. ^"d ^he rectangle EA, AF is therefore given ; and it is equal '^ to the rectangle BA, AC, which consequently is given. M. PROP. XCVIL IF a straight line be drawn within a circle given in magnitude cutting off a segment containing a given angle ; if the angle in the segment be bisected by a straight line produced till it meets the circumference, the straight lines which contain the given angle shall both of them together have a given ratio to the straight line which bisects the angle : and the rectangle contained by botli these lines together which contain the given angle, and the part of the bisecting line cut off below the base of the segment, shall be given. Let the straight line BC be drawn within the circle ABC gi- ven in magnitude, cutting off a segment containing the given angle BAC, and let the angle BAC be bisected by the straight line AD ; BA together with AC has a given ratio to AD; and the rectangle contained by B A and AC together, and the straight line ED cut off from AD below BC the base of the segment, is given. .Toiii BD ; and because BC is drawn within the circle ABC DATA. 457 given in mag;nitude ciitting off the segment BAG, containing the given angle BAG ; BC is given ^ in magnitude ; by the same a 91.dat. reason BD is given ; therefore '» the ratio ofBC to BD is given : b 1. dat. and because the angle BAG is bisected by AD, as BA to AC, so is^BE to EG ; and, by permutation, as AB to BE, so is AGc3. 6. to GE: wherefore oint E where it meets the circumference again let EF be drawn parallel to DA meeting BC in F; the point F is given, as also the rectangle AD, EF. Produce FF to the circumference in G, and join AG : be- cause GEA is a right angle, the straight line AG is ^ the dia- a Cor, 5. meter of the circle ABC ; and BC is also a diameter of U ;*• therefore the point H where they meet is the centre of the circle, and consequently H is given : and the point D is given, wherefore DH is given in magnitude : and because AD is pa- *^^ EUCLID'S b4 6. rallel to FG, and GH equal to HA ; DH is equal *> to HF, and AD equal to GF : and DH is given, therefore HF is given ia A A inagnitude ; and it is also given in position, and the point H is c 30. dat. given, therefore ^ the point F is givt-n. And because the straight line KFG is drawn from a given point F without or v^iihin the circle ABC given in position, d 'JS. or therefore *i the rectangle EF, FG is given : and GF is equal to 96. dat. ^Q^ wherefore the x-ectangle AD, EF is given. 4 PROP. c. IF from a given point in a straight line given in posi- tion, a straight line be drawn to any point in the circum- ference of a circle given in position ; and from this point a straight line be drawn making Avith the first an angle equal to the difference of a right angle and the angle contained by the straight line given in position, and the straight line which joins the given point and the centre of the circle ; and from the point in which the second line meets the circumference again, a third straight line be drawn making with the second an angle equal to that which the first makes with the second : the point in which this third line meets the straight line given in po- sition is given ; as also the rectangle contained by the first straight line and the segment of the third betwixt the circumference and the straight line given in position, is given. Let the straight line CD be drawn from the given point C in the straight line AB given in position, to the circumference of the circle DEF given in position, of which G is the centre ; join CG, and from the point D let DF be drawn making the angle CDF equal to the difference of a right angle and the angle BCG, and from the point F let FE be drawn making DATA. 46J a 26. h 8. i: the angle DFE equal to CDF, meeting AB in H : the point H given ; as aho the rectangle CD, FH. Let CD, FH meet one another in the point K, from which draw KL perpendicular to DF; and let DC meet the circumference again in M, and let FH meet the same in E, and join JVIG, GF, GH. Because the angles MDF, DFE are equal to one another, the circumfer- ences MF, DE are equal ^ ; and add- ing or taking away the common part ME, the circumference DM is equal to EF ; therefore the straight line DM is equal to the straight line EF, and the angle GMD to the angle b GFE ; and the angles GMC, GFH are equal to one another, because they are ei- ther the same with the angles GMD, GFE, or adjacent to thein : and be- cause the angles KDL, LKD are toge- ther equal <^ to a right angle, that is, by the hypothesis, to the angles KDL, GCB: the angle GCB, or GCH is equal to the angle (LKD, that is, to the angle) LKF or GKH : therefore the points C, K, H, G are in the circumference of a circle ; and the angle GCK is there- fore equal to the angle GHF ; and the angle GMC is equal to GFH, and the straight line GM to GF ; therefore ^ CG is equal d 26. 1- to GH, and CM to klF : and because CG is equal to GH, the angle GCH is equal to GHC ; but the angle GCH is given : therefore GHC is given, and consequently the angle CGH is given ; and CG is given in position, and the point G ; there- fore e GH is given in position ; and CB is also given in position, e 52. dat. whereof the point H is given. And because HF is equal to CM, the rectangle DC, FH is equal to DC, CM : but DC, CM is given f, because the point C f 95. or is given, therefore the rectangle DC, FH is given. 5^- ^^.t. c 32. 1. AC H B FINIS. NOTES ON EUCLID'S DATA. DEFINITION II. This is made more explicit than in the Greek text, to pre- vent a mistake which the author of the second demonstration of the 24th proposition in the Greek edition has fallen into, of think- ing that a ratio is given to which another ratio is shown to be equal, though this other be not exhibited in given magnitudes. See the Notes on that proposition, which is the 13th in this edi- tion. Besides, by this definition, as it is now given, some pro- positions are demonstrated, which in the Greek are not so well done by help of prop. 2. DEF. IV. In the Greek text, def. 4. is thus : " Points, lines, spaces", " and angles are said to be given in position which have always " the same situation ;" but this is imperfect and useless, because there are innumerable cases in which things may be given ac- cording to this definition, and yet their position cannot be found ; for instance, let the triangle ABC be given in position, and let it be proposed to draw a straight line BD from the angle at B to the opposite side AC, which shall cut A off the angle DBC, which shall be the seventh part of the angle ABC ; suppose this is done, therefore the straight line BD is invariable in its position, that is, has always the same situation; for any other straight line drawn from the point B on either side of BD cuts off an angle greater or lesser than the 'seventh part of the angle ABC ; therefore, according to this definition, tlie straight line BD is given in position, as also =» the point D inj^28. dat' which it meets the straight line AC which is given in position. But from the things here given, neither the straight line BD nor the point D can be found by the help of Euclid's Elements only, by which every thing in his Data is supposed may be found. 464 NOTES ON This definition is therefore of no use. We have amended it by adding-, "and which are either actually exhibited or can be found ;" for nothing is to be reckoned given, v/hich cannot be found, or is not actually exhibited. The definition of an angle given by position is taken out of th« 4th, and given more distinctly by itself in the definition marked A. DEF. XI, Xil, XIII, XIV, XV. The 11th and I2th are omitted, because they cannot be given in English so as to have any tolerable sense ; and, therefore, wherever the terms defined occur, the words which express their meaning are made use of in their place. The 13th, 14th, 15th are omitted, as being of no use. It is to be observed in general of the Data in this book, that they are to be understood to be given geometrically, not always arithmetically, that is, they cannot always be exhibited in num- bers; for instance, if the side of a square be given, the ratio of b.44.dat. it to its diameter i- given'' geometrically, but not in numbers; c 2. dat. dud the diameter is given ^ ; but though the number of any equal parts in the side be given, for example 10, the number of them in the diameter cannot be given : and the like holds in many other cases. PROPOSITION I. In this it is shown that A is to B, as C to D, from this, that A is to C, as B to D, and then by permutation ; but it follows directly, vicithout these two steps, from 7. 5. PROP. II. The limitation added at the end of this proposition between the inverted commas is quite necessary, because without it the proposition cannot always be demonstrated : for the author hav- ing said*, " because A is given, a magnitude equal to it can a 1. def. " ^^ found ^ ; let this be C ; and because the ratio of A to B b 2. def. " '* given, a ratio which is the same to it can be found ^," adds, "let it be found, and let it be the ratio of C to A." Now, from the second definition nothing more follows, than that some ratio, suppose the ratio of E to Z, can be founds which 15 the same with the ratio of A to B ; and when the author supposes that the ratio of C to A, which is also the • See Dr. Gregory's edition of tbe Data. EUCLID'S DATA. 4^ same with the ratio of A to B, can be found, he necessarily sup- poses tliat to the three magnitudes E, Z, C, a fourth propor- tional A may be found ; but this cannot always be done by the Elements of Euclid; from which it is plain Euclid must have understood the proposition under the limitation which is now added to his text. An example will make this clear: let A be a given angle, and B another an- gle to which A has a given ratio, for instance, the ratio of the given straight line E to the given one Z ; then, hav- ing found an angle C equal to A, how can the angle a be found to which C has the same ratio that E has to Z ? certainly no way, until it be shown how to find an angle to which a given angle has a given ratio which cannot be done by Euclid's Elements, nor probably by any Geometry known in his time. Therefore, in all the propositions of this book which depend up- on this second, the abovementioned limitation must be under- stood, though it be not explicitly mentioned. PROP. V. The order of the propositions in the Greek text, between prop. 4 and prop. 25, is now changed into another which is more natural, by placing those which are more simple before those which are more complex ; and by placing together those which are of the same kind, some of which were mixed among others of a different kind. Thus, prop. 12, in the Greek, is now- made the 5th, and those which were the 22d and 23d are made the Uth and 12th, as they are more simple than the proposi- tions concerning magnitudes, the excess of one of which above a given magnitude has a given ratio to the other, after which these two were placed ; and the 24th in the Greek text is, for the same reason, made the 1 3th. PROP. VI, VII. • These are universally true, though, in the Greek text, they are demonstrated by prop. 2, Which has a limitation : they ar^ therefore now shown without it. 3N m> NOTES ON PROP. XIL In the 23d prop, in the Greek text, which here is the ISth, the words, " ftu ry; eivmi h," are wrong translaXed by Claud. Hardy, in his edition of Euclid's Data, printed at Paris, anno 1625, which was the first edition of the Greek text ; and Dr. Gregory follows him in transhuing them by the words, "etsi non easdem," as if the Greek had been $i x«< jw-ij ?»$ xvrtif, as in prop* 9, of the Greek text. Euclid's mtaning is, that the ratios mentioned in the proposition must not be the same ; for, if they were, the proposition would not be true» Whatever ratio the whole has to the whole, if the ratios of the parts of the first to the parts of the other be the same with this ratio, one part of the first may be double, triple, &c. of the other part of it, or have any other ratio to it, and consequently cannot have a given ratio to it ; wherefore, these words must be rendered by " non autem easdem," but not the same ratios, as Zambertu« has translated them in bis edition. PROP. XIII. Some very ignorant editor has given a second demonstration of this proposition in the Greek text, which has been as igno- rantly kept in by Claud. Hardv and Dr. Gregory, and has been retained in the translations of Zambertus and others ; Carolus Renaldinus gives it only : the author of it has thought that a ratio was given if another ratio could be shown to be the same to it, though this last ratio be not found : but this is altogether ab- surd, because from it would be deduced, that the ratio of the sides of any two squares is given, and the ratio of the diameters of any two circles, &c. And it is to be observed, that the mo- derns frequently take given ratios, and ratios that are always the same, for one and the same thing : and Sir Isaac Newton has fallen into this mistake in the 17th lemma of his Principia, edit, 1713, and in other places ; but this should be carefully avoided, as it may lead into other errors. PROP. XIV, XV- Euclid, in this book, has several propositions concerning magnitudes, the excess of one of which above a given magni' EUCLID'S DATA. 467 tude has a given ratio to the other ; but he has given none con- cerning magnitudes whereof one together with a given magni- * nitude has a given ratio to the other ; though these last occur as frequently in the solution of problems as the first : the rea- son of which is, that the last may be all demonstrated by help of the first ; for, if a magnitude, together with a given magni- tude has a given ratio to another magnitude, the excess of this other above a given magnitude shall have a given ratio to the first, and on the contrary; as we have dijmoiistrated in prop. 14. And for a like reason prop. 15 has been added to the Datu. One example will make the thing clear: suppose it were to be de- monstrated, that if a magnitude A together with a given mag- nitude has a given ratio to another magnitude B, that the two magnitudes A and B, together with a given magnitude, have a given ratio to that other magnitude B ; which is the same pro- position with respect to the last kind of magnitudes above-men- tioned, that the first part of prop. 16, in this edition, is in I'e- spect of the first kind: this is shown thus; from the hypothesis, and by the first part of prop. 14, the excess of B above a given magnitude has unto A a given ratio ; and, therefore, by the first part of prop. 17, the excess of B above a given magnitude has unto B and A together a given ratio ; and by the second part of prop. 14, A and B together with a given magnitude has unto B a given ratio ; which is the thing that was to be demonstrated. In like manner, the other propositions concerning the last kind of magnitudes may be shown. PROP. XVI, XVII. In the third part of prop. 10, in the Greek text, which is the 1 6th in this edition, after the ratio of EC to CB has been shown to be given : from this, by inversion and conversion the ratio of BC to BE is demonstrated to be given ; but without these two steps, the conclusion should have been made only by citing tlie 6th proposition. And in like manner, in the first part of prop. 11, in the Greek, which in this edition is the 17th from the ratio of DB to Be being given, the ratio of DC to DB is shown to be given by inversion and composition, instead of citing prop. 7, and the same fault occurs in the second part of the same prop. 11. PROP. XXI, XXII. These now are added, as being wanting to complete the sub- ject treated of in the four preceding propositions. 4&e NOTES ON PROP. XXIII. This, which is prop. 20, in the Greek text, was separated from prop. 14, 15, 16, in that text, after which it should have been immediately placed, as being, of the same kind ; it is now put into its proper place ; but prop. 21 in the Greek is left out, as being the same with prop. 14, in that text, which is here prop. 18. PPvOP. XXIV. This, which is prop. 13, in the Greek, is now put into its pro- per place, having been disjoined from the three following it in this edition, which are of the same kind. PROP. XXVIII. This, which in the Greek text is prop. 25, and several of the following propositions are there deduced from def. 4, which is not sufficient, as has been mentioned in the note on that defini- tion : they are therefore now shown more explicitly. PROP. XXXIV, XXXVI. Each of these has a determination, which is now added, which occasions a change in their demonstrations. PROP. XXXVII, XXXIX, XL, XLI. The 35th and 36th propositions in the Greek text are joined into one, which makes the 39th in this edition, because the same enunciation and demonstration serves both : and for the same reason prop. 37, 38, in the Greek, are joined into one, which here is the 40th. Prop. 37 is added to the Data, as it frequently occurs in the solution of problems ; and prop. 41 is added to complete the rest. PROP. XLII. This is prop. 39, in the Greek text, where the whole construc- tion of prop. 22, of book I. of the Elements is put, without need, into the demonstration, but is now only cited. PROP. XLV. This is prop. 42, in the Greek, where the three straight lines made use of in the construction are said, but not shown, to be such that any two of them is greater thaa the third, which, is BOW done. EUCLID'S DATA. ;4d9 PROP. XLVII. This IS prop. 44, in the Greek text ; but the demonstration of it is changed into another, wherein the several cases of it are shown, which, though .necessary, is not done in the Greek. PROP. XLVIII. There are two cases in this proposition, arising from the two cases of the third part of prop. 47, on which the 48th depends ; and in the composition these two cases are explicitly given. PROP. LII. The construction and demonstration of this, which is prop. 48, in the Greek, are made something shorter than in that text. PROP. LIII. Prop. 63, in the Greek text is omitted, being only a case of prop. 49, in that text, which is prop. 53, in this edition. PROP. LVIII. This is not in the Greek text, but its demonstration is con- tained in that of the first part of prop. 5 4, in that text; which proposition is concerning figures that are given in species: this 58th is true of similar figures, though they be not given in spe*' cies, and as it frequently occurs, it was necessary to add it. PROP. LIX, LXI. This is the 54th in the Greek ; and the 77th in the Greek, being the very same with it, is left out, and a shorter demon- stration is given of prop. 61. PROP. LXII. This, which is most frequently useful, is not in the Greek, and is necessary to prop. 87, 88, in this edition, as also, though not mentioned, to prop. 86, 87, in the former editions. Prop, 66, in the Greek text, is made a corollary to it. PROP. LXIV. This contains both prop. 74, and 73, in the Greek text ; the first case of the 74th is a repetition of prop. 55, from v/hicli it is separated in that text by many propositions ; and as there is no order in these propositions, as they stand in the Greek, they are now put into the order which seemed most convenient and natui'al. 4ro NOTES ON The demonstration of the first part of prop. 73, in the Greek, is grossly vitiated. Dr. Gregory says, that the sentences he has inclosed betwixt two stars are superfluous, and ought to be can- celled ; but he has not observed, that what follows them is ab- surd, being to prove that the ratio [see his figure] of Ai to 1 K is given, which by the hypothesis at the beginning of the proposition is expressly given ; so that the whole of this part was to be altered, which is done in this prop. 64. PROP. LXVII, LXVIII. Prop. TO, in the Greek text, is divided into these two, for the sak.e of distinctness ; and the demonstration of the 67th is rendered shorter than that of the first part of prop. 70, in the Greek, by means of prop. 23, of book 6, of the Elements. PROP. LXX. This is prop. 62, in the Greek text ; prop. 78, in that text, is only a particular case of it, and is therefore omitted. Dr. Gregory, in the demonstration of prop. 62, cites the 49th prop. dat. to prove that the ratio of the figure AEB to the pa- rallelogram AH is given ; whereas this was shown a few lines before : and besides, the 49th prop, is not applicable to these two figures ; because AH is not given in sptcies, but is by the step for which the citation is brought, proved to be given in species. PROP. LXXIII. Prop. S3, in the Greek text, is neither welj enunciated nor demonstrated. The 73d, which in this edition is put in place of it, is really the same, as will appear by considering [see Dr. Gregory's edition] that A, B, r, E in the Greek text are four proportionals ; and that the proposition is to show that a, wliich has a given ratio to E, is to r, as B is to a straight line to which A has a given ratio ; or, by inversibJi, that r is to ^, as a straight line to which A has a given ratio is to B ; that is, if the proportionals be placed in this order, viz. r, E, A, B, that the first r is to A lo which the second E has a given ratio, as a straight line to which the third A has a given ratio is to the fourth B ; which is the enunciation of this 73d, and was thus changed that it might be made like to that of prop. 72, in this 'jcUtion, which is the~82d in the Greek text ; and the demon- EUCLID'S DATA. 471 Stration of prop. 73 is the same with that of prop. 72, only mak- ing use of prop. 23j instead of prop. 22, of book 5, of the Elements. PROP. LXXVII. This is put in place of prop. 79, in the Greek text, which is not a datum, but a theorem premised as a lemma to prop. 80 in that text : and prop. 79 is made cor. I to prop. 77, in this edition. CI. Hardy, in his edition of the Data, takes notice, that in prop. 80, of the Greek text, the parallel KL in the fiij;ure of prop. 77, in this edition, must meet the circumference, but does not demonstrate it, which is done here at the end of cor. 3, prop. 77, in the construction for finding a triangle similar to ABC. PROP. LXXVIII. The demonstration of this, which is prop. 80, in the Greek, is rendered a good deal shorter by help of prop. 77. PROP. LXXIX, LXXX, LXXXI. These are added to Euclid's Data, as propositions which are often useful in the solution of problems. PROP. LXXXII. This, which is prop. 60, in the Greek text, is placed before the 83d and 84th, which, in the Greek, are the 58th and 59th, because the demonstration of these two m this edition ar-e dedu- ced from that of prop. 82, from which they naturally follow. PROP. LXXXVIII, XC. Dr. Gregory, in his preface to Euclid's Works, which he published at Oxford in 1703, after having told that he had sup- plied the defects of the Greek text of the Data in innumerable places from several manuscripts, and corrected CI. Hardy's translation by Mr. Bernard's, adds, that the 86th theorem, " or proposition," seemed to be remarkably vitiated, but which could not be restored by help of the manuscripts ; then he gives three different translations of it in Latin, according to which he thinks, it may be read ; the two first have no distinct meanings and the third, which he says is the best, though it contains a 4r7'^ NOTES ON tfue proposition, which is the 90th in this edition, has no con- nection in the least with the Greek text. And it is strange that Dr. Gregory did not observe, that, if prop. 86 was changed into this, the demonstration of the 86th must be cancelled, and ano- ther, put in its place: but the truth is, both the enunciation and the demonstration of prop. 86 are quite entire and right, only- prop. 87, which is more simple, ought to have been placed be- fore it ; and the deficiency which the doctor justly observes to be in this part of Euclid's Data, and which, no doubt, is owing to the carelessness and ignorance of the Greek editors, should have been supplied, not by changing prop. 86, v/hich is both entire and necessary, but by adding the two propositionsj which are the 88th and 90th in this edition. PROP, xcvni, c. These were communicated to me by tvpo excellent geometers, the first of them by the Right Honourable the Earl of Stanhope, and the other by Dr. Matthew Stewart ; to which I have added the demonstrations. Though the order of the propositions has been in many places changed from that in former editions, yet this will be of little disadvantage, as the ancient geometers never cite the Data, and the moderns very rarely. As that part of the composition of ^ problem which is its con- struction may not be so readily deduced from the analysis by be- ginners: for their sake the following example is given, in which the derivation of the several parts of the construction from the analysis is particularly shown, that they may be assisted to do the like in other problems. PROBLEM. Having given the magnitude of a parallelogram, the angle of which ABC is given, and also the excess of the square of its side BC above the square of tlie side AB ; to find its sides, and de- ■scribe it. The analysis of this is the same with the demonstration of the 87th prop, of the Data, and the construction lhr.t is given of the problem at the end of that proposition is thus derived from t|^e analysis. EUCLID'S DATA. iti Let Et^G be equal to the given angle ABC, and because in the analysis it is said that the ratio of the rectangle AH, BC to the parallelogram AC is given by the 62d prop. dut. therefore, from a point in FE, the perpendicular EG is drawn to FG, as the ratio of FE to EG is the ratio of the rectangle F G L O H N a6, BC to the parallelogram AC by what is shown at the end of prop. 62. Next, the magnitude of AC is exhibited by mak- ing the rectangle EG, GH equal to it ; and the given excess of the square of BC above the square of BA, to v^hich excess the rectangle CB, BD is equal, is exhibited by the rectangle HG» GL : thi,n in the analysis, the rectangle AB, BC is said to be given, and this is equal to the rectangle FE, GH, because the rectangle AB. BC is to the parallelogram AC, as (FE to EG, that is, as the rectangle) FE, GH to EG, GH ; and the paral» lelogram AC is equal to the rectangle EG, GH, therefore thft rectangle AB, BC, is equal to FE, CiH: and consequently the ratio of the I'ectangle CB, BD, that is, of the rectangle HG, GL, to AB, BC, that is, of the straight line DB to BA, is the same with the ratio (of the rectangle GL, GH to FE, GH, that is) of the straight line GL to FE, which ratio of DB to BA is the next thing said to be given in the analysis ; from this it is plain that the square of FE is to the square of GL, as the square of BA, which is equal to the rectangle BC, CD, is to the square of BD : the ratio of which spaces is the next thing said to be given : and from this it follows that four times the square of FE is to the square of GL, as four times the rect- angle BC, CD is to the square of BD ; and, by composition, four times the square of FE together with the square of GL, is to the square of GL, as four times the i'ectangle BC, CD, together with the square of BD, is to the square of BD, that is (8. 6,) as the square of the straight lines BC, CD taken to- gether is to the square of BD, which ratio is the next thing said to be given in the analysis: and because four times the square of FE and the square of GL are to be added together J therefore in the perpendicular EG ther& is taken KG equal to 30 NOTES ON FE, and iMG equal to the double of it, because thereby the squares of MG, GL, that is, joining ML, the square of ML is equal to four limes the square of FE and to the square of GL : and because the square of ML is to the square of GL, as the square of the straight line made up of BC and CD is to the square of BD, therefore (22. 6.) ML is to LG, as BC together •with CD is to BD ; and, by composition, ML and LG together, tliat is, producing GL to N, so that ML be equal to LN, the straight line NG is to GL, as twice BC is to BD ; and by taking GO equal to the half of NG, GO is to GL, as BC to BD, the ratio of which is said to be given in the analysis : and from this it follows, that the rectangle HG, GO is to HG, GL, as the square of BC is to the rectangle CB, BD, which is equal to the lectangle HG, GL ; and therefore the square of BC is equal to the rectangle HG, GO ; and BC is consequently found by taking a mean proportional betwixt HG and GO, as is said in the con- struction : and because it was shown that GO is to GL, as BC to BD, and that now the three first are found, the fourth BD is found by 12. 6. It was likewise shown that LG is to FE, or GK, as DB to BA, and the three first are now found, arid there- by the fourth BA. Make the afigle ABC equal to EFG, and complete the parallelogram of which the sides are AB, BC, and the construction is finished ; the rest of the composition con- lains the demonstration. AS the propositions from the 13th to the 28th may be thought by beginners to be less useful ihan the rest, because they can- not so readily see how they are to be made use of in the solution of problems ; on this account the two following problems are added, to show that they are equally useful with the other pro- ])ositions, and from which it may be easily judged that many other problems depend upon these propositions. PROBLEM I. To find three straight lines such, that the ratio of the first to the second is given ; and if a given straight line be taken from the second, the ratio of the remainder to the third is given ; also the rectangle contained by the first and third is given. EUCLID'S DMA. 475 Let AB be the first straight line, CD the second, and EF the third ; and because the ratio of AB to CD is given, and that if a given straight line be taken from CD, the ratio of the re- mainder to EF is given ; therefore ^ the excess of flie first ABa24. dar. above a given straight line has a given ratio to the third EF : let BH be that given straight line ; therefore AH, the excess of AB above it, has a given ratio to EF : and consequently^ the rectangle BA, AH, has a A H B bl.6. given ratio to the rectangle AB, EF, which last rectangle is given by the hypothesis ; therefore ^ the rectangle BA, AH is given, ^ ^ U £. 2. dat. and BH the excess of its sides is given ; where- A H 1 B C 1 G i D E 1 F K NM L -1-1-1- O fore the sides AB, AH are given d; and be- -p y, d85.dat. cause the ratios of AB to CD, and of AH to EF are given, CD and EF are <= given. The Composition* Let the given ratio of KL to KM be that which AB is requir- ed to have to CD ; and let DG be tlie given straight line wliich is to be taken from CD, and let the given ratio of KM to KN be that which the remainder must have to EF ; also let the given rectangle NK, KO be that to which the rectangle AB, EF is re- quired to be equal: find the given straight line BH which is to be taken from AB, which is done, as plainly appears from prop. 24, dat. by making as KM to KL, so GD to HB. To the given straight line BH apply e a rectangle equal to LK, KO exceeding e 29. C>, by a square, and let BA, AH be its sides : then is AB the first of the straight lines required to be found, and by making as LK to KM, so AB to DC, DC will be the second : and lastly, make as KM to KN, so CG to EF, and EF is the third. For as AB to CD. so is HB to GD, each of these ratios being the same with the ratio of LK to KM ; therefore ^ AH is to CG, f 19. 5, as (AB to CD, that is, as) LK to KM ; and as CG to EF, so is KM to KN ; wherefore, ex squali, as AH to EF, so is LK to KN : and as the rectangle BA, AH to the rectangle BA, EF, so. is g the rectangle LK, KO to the rectangle KN, KO : and by the g i. 6. construction, the rectangle BA, AH is equal to LK, KO : there- fore ^ the rectangle AB, EF is equal to the given rectangle NK, hl4. ^. KO : and AB has to CD the given ratio of KL to KM ; and iVoin CD the given straight line GD being taken, the remainder CG has to EF the given ratio of KM to KN. Q. E. D. 476. NOTES ON a 2*.dat b 44. dat c 32. dat, d47. i e 34 dat £28. dat g33 dat h 29 dat i «. dat. PROB. II. TO find three straight lines such, that the ratio of the first to the second is given ; and if a given straight Hne be taken from the second, the ratio of the remainder to the third is given ; also tlie sum of the squares of the first and third is given. Let AB be the first straight Hne, BC the second, and BD the third : and because the ratio of AB to BC is given, and that if a given straight line be taken from BC, the ratio of the remain- , der to BD is given ; therefore ^ the excess of the first AB above a given straight line, has a given ratio to the third BD : let AE be that given straight line, therefore the i*emainder EB has a given ratio to BD : let BD be placed at right angles to EB and join DE ; then the triangle EBD is ^ given in species; where- fore the angle BE^D is given : let AE, which is given in magni- tude, be given also in position, as also the point E, and the , straight line ED will be given ^ in position : join AD, and be- cause the sum of the squares of AB, BD, that h^, the square of AD is givtn, therefore the straight line AD is given in mag- . nitude ; and it is also i^iven e in position, because from the given point A it is drawn to the straight line ED given in position : therefore the point D, in which the two straight lines AD, ED • given in position cut one another, is given f; and the straight , line DB which is at right angles to AB is given s in position, and AB is given in position, therefore f the point B is given : and the . points A, D are given, wherefore ^ the straight lines AB, BD are given: and the ratio of AB to BC is given, and therefore* BC is criven. The Corr.fiosUion, Let the given ratio of FG to GB be that which AB is requir- ed to have to BC, and let HK be the given straight line which is to be taken from BC, -dud let the ratio which the remainder is L D B N ISl req'.iircd to have to BD be the given ratio of HG to GL, and plaqe GL at right angles to FH, and join LF, LH : next, as HG EUCLID'S DATA. 477 * is to GF, so make HK to AE ; produce AE to N, so that AN be the straight line to the square of which the sum of the squares of AB, BD is required to be equal ; and make the angle NED equal to the angle GFL ; and from the centre A at the distance AN describe a circle, and let its circumference meet ED in D, and draw DB perpendicular to AN, and DM, making the angle BDM equal to the angle GLH. Lastly, produce BM to C, so that MC be equal to HK. then is AB the first, BC the second, and BD the third of the straight lines that were to be found. For the triangles EBD, FGL, as also DBM, LGH being eqi- angular, as EB to BD, so is FG to GL ; and as DB to BM, so is LG to GH ; therefore, ex aequali, as EB to BM, so is (FG to GH, and so is) AE to HK or MC ; wherefore \ AB is to BC, k 12. 5> as AE to HK. that is, as FG to GH, that is, in the given ratio ; and from the straight line BC taking MC, which is equal to the given straight line HK, the remainder BM has to BD the given ratio of HG to GL ; and the sum of the squares of AB, BD is equal d to the square of AD or AN, which is the given space, d 47. 1. Q. E. D. I believe it would be in vain to try to deduce the preceding construction from an algebraical solution of the problem. FINIS. I TBS ELEMENTS OF PLANE AND SPHERICAL TRIGONOMETRY. J ■ ■■!■' PHILADELPHIA J PRINTED FOR, AND PUBLISHED BT, MATHEW CARET, NO. 122, MARKET-STREET, 18Q6, PLANE TRIGONOMETRY. LEMMA I. FIG. 1. IjET ABC be a rectilineal angle, if about the point li as a cen- tre, and with any distance BA, a circle be described, meeting BA, BC, the straight lines including the angle ABC in A, C; the angle ABC will be to four right angles, as the arch AC fi the whole circumference. Produce AB till it meet the circle again in F, and through B "draw DE perpendicular to AB, meeting the circle in D, E. By 33. 6. Elem. the angle ABC is to a right angle ABD, as the arch AC to the arch AD; and quadrupling the consequents, the angle ABC will be to four right angles, as the arch AC td four time3 the arch AD, or to the whole circumference. LEMMA II. FIG. 2. LET ABC be a plane rectilineal angle as before ; alxiut B jls a centre with any two distances BD, BA, let two circles be describ- ed meeting BA, BC, in D, E, A, C ; the arch AC will be to thr. whole circumference of which it is an arch, as the arch DE is lo the whole circumference of which it is an arch. By Lemma 1. the arch AC is to the whole circumference of which it is an arch, as the angle ABC is to four right angles ; and by the same Lemma 1. the arch DE is to the whole circum- ference of which it is an arch, as the angle ABC is to four right angles ; therefore the arch AC is to the whole circumference of which it is an arch, as the arch DE to the whole circumference of which it is an arch. DEFINITIONS. FIG. 5> I. LET ABC be a plane rectiHrteal angle ; if about B as a centre^, with BA any distance, a circle ACF be described meeting BA< BC in A, C ; the arch AC is called the measure of the angle ABC. II. The rircumlerrnce of a circle is snpp'o'^cd to be divided into PLANE TRIGONOMETRY. 360 equal parts called degrees, and each degree into 60 equal parts called minutes, and each minute into 60 equal parts cal- led seconds, &c. And as many degrees, minutes, seconds, &c. as are contained in any arch, of so many degrees, minutes, se- conds, &c. is the angle, of which that arch is the measure, said to be. Cor. Whatever be the radius of the circle of which the measure of a given angle is an arch, that arch will contain the same mmiber of degrees, minutes, seconds, 8cc. as is manifest from Lemma 2. in. Let AB be produced till it meet the circle again in E, the angl^ CBF, which, together with ABC, is equal to two right angles, is culled the Suplilement of the angle ABC. IV. A straight line CD drawn through C, one of the extremities of the arch AC perpendicular upon the diameter passing through the other extremity A, is called the Sine of the arch AC, oj- of the angle ABC, of which it is the measure. CoK. The Sine of a quadrant, or of a right angle, is equal to the radius. V. The segment DA of the diameter passing through A, one ex- tremity of the arch AC between the sine CD, and that ex- tremity, is called the Versed Si?ie of the arch AC, or angle ABC. VI. A straight line AE touching the circle at A, one extremity of the arch AC, and meeting the diameter BC passing through the other extremity C in E, is called the Tangent of the arch AC, or of the angle ABC. vn. The straight line BE between the centre and the extremity of the tangent AE, is called the Secant oS. the arch AC, or an- gle ABC. Cor. to def. 4, 6, 7. the sine, tangent, and secant of any angle ABC, are likewise the sine, tangent, and secant of its supple- ment CBF. It is manifest from def. 4. that CD is the sine of the angle CBF. Let CB be produced till it meet the circle again in G ; and it is manifest that AE is the tangent, and BE the secant, of the. angle ABG or EBF, from def. 6, 7. PLANE TRIGONOMETRY. 48; C©R. to def. 4, 5, 6, 7. The sine, versed sine, tangent, and se- cant, of any arch which is the measure of any given angle ABC, is to the sine, versed sine, tangent, and secant, of any other arch which is the meysure of the same angle, as the radius of the first is to the radius of the second. Let AC, MN be measures of the angle ABC, according to def. 1. CD the sine, DA the versed sine, AE the tangent and BE the secant of the arch AC, according to def. 4, 5, 6, 7. and NO the sine, OM the versed sine, MP the tangv ut, and BP the secant of the arch MN, according to the same definitions. Since CD, NO, AE, MP are parallel, CD is to NO as the radius CB to the radius NB, and AE to MP as AB to BM, and BC or BA to BD as BN or BM to BO ; and, by conversion, DA to MO as AB to MB. Hence the corol- lary is manifest ; therefore, if the radius be supposed to be divided into any given numbvrof equal parts, the sine, versed sine, tangent, and secant of any given angle, will each contain a given number of these parts; and, by trigonometrical tables, the length of the sine, versed sine, tangent, and secant of any angle may be found in parts of which the radius contains a given number; and, vice versa, a number exprcasing the length of the sine, versed sine, tangent, and secant being given, the angle of which it is the sine^ versed sine, tangent, and secant may be found. VilL Fig. S The difference of an angle from a right angle is called the coml}lement of that angle. Thus, if BH be drawn perpendi- cular to AB, the angle CBH will be the complement of the . angle ABC, or of CBF. IX. Let HK be the tangent, CL or DB, which is equal to it, the sine, and BK the secant of CBH, the complement of ABC, accord- ing to def. 4, 6, 7. HK is called the co-tangent, BD the co-sine, and BK the co-secant of the angle ABC. CoR. 1. The radius is a mean proportional between the tangent, and co-tangent. For, since HK, BA are parallel, the angles HKB, ABC will be equal, and the angles KHB, BAE are right ; therefore the tri- angles BAE, KHB are similar, and therefore AE is to AB, as BH or BA to HK. CoR. 2. The radius is a mean proportional between the co-sine and secant of any angle ABC. S^ncf. CD, AE are parallel, BD is to BC or BA, as BA to BE. *ti-i PLANE TRIGONOMETRY. PRO?. I. FIG. 5. IN a riglit anf^lcd plane triangle, if the liypotheniise be made radius, the sides become the sines oi' the angles opposite to them ; and if either side be made radius, the remaining side is the tangent of the angle opposite to it^ and the hypothenuse the secant of the same angle. Let ABC be a right angled triangle ; if the hypothenuse BC be made radius, either of the sides AC will be the sine of the angle ABC opposite to it; and if either side BA be made ra- dius, the other side AC will be the tangent of the angle ABC opposite to it, and the hypothenuse BC the secant of the same angle. About B as a centre, with BC, BA for distances, let two circles CD, EA be described, meeting BA, BC in D, E : since CAB is a right angle, BC being radius, AC is the sine of the angle ABC by def. 4. and BA being radius, AC is the tangent, and BC the secant of the angle ABC, by def. 6, 7. Cor. J. Of the hypothenuse u side and an angle of a right angled triangle, any two being given, the third is alsp given. Cor. Of the two sides and an angle of a right angled triangle, ^ny two being given, the third is aho given. PROP. n. FIG. 6, 7. THE sides of a plane triangle arc to one anotiier as the- bines of the angles opposite to them. In right angled triangles, thjs prop, is manifest from prop. 1. for if the hypothenuse be made radius, the sides are the sines of the angles opposite to them, and the radius is the sine of a right angle (cor. to def. 4.)) which is opposite to the hypothenuse. In any oblique angled triangle ABC, any two sides AB, AC will be to one another as the sines of the angles ACB, ABC Avhich are opposite to thenri. From C, B draw CE, BD perpendicular upon the opposite sides AB, AC produced, if need be. Since CEB, CDB are right angles, BC being radius, CE is the sine of the angle CBA, and BD the sine of the angle ACB; but the two triangles CAE, DAB have each a right angle at D and E ; and likewise the cqmmou angle CAB; therefore they are liiaiilar, i.nd tynsc- PLANE TRIGONOMETRY,. 48^ .^uently, CA is to AB, as CE to DB ; that is, the side^ are as tt»e sines of the angles opposite to them. Cor. Hence of two sides, and two angles opposite to them, in a plane triangle, any three being given, the fourth is also §;iven. PROP. III. FIG. 8. In a plane triangle, the sum of any t^\o sides is to tlieir difference, as the tangent of half the sum of tlu; angles at the base, to the tangent of half their differ.-, ence. Let ABC be a plane triangle, the sum of any two sides, AB, AC will be to their diflercnce as the tangent of half the sum of the angles at the base ABC, ACB to the tangent of half their difference. About A as a centre, with AB the greater side for a distance, let a circle be described, meeting AC produced in E, F, and BC in D ; join DA, EB, FB ; and draw FG parallel to BC, meeting EB in G. The angle EAB (32. 1.) is equal to the sum of the angles at the base, and the angle EFB at ihe circumference is equal to the half of EAB at the centre (20. 3.); therefore EFB is half the sum of the angles at the base; but the angle ACB (32. 1.) is equal to the angles CAD and ADC, or ABC together ; therefore Fad is the difference of the angles at the base, and FBD at the circumference, or BFG, on account of the parallels FG, BD, is the half of that difference ; but since the angle EBF in a semi- circle is a right angle (I. of this) FB being radius, BE, BG, are the tangents of the angles EFB, BFG ; but it is manifest that EC is the sum of the sides BA AC, and CF their difference ; and since BC FG are parallel (2*. 6.) EC is to CF, as EB to BG ; that is, the sum of the sides is to their difference, as t!ie tangent of half the sum of the angles at the base to the tangent of hal! their difference. PROP. IV. FIG. 18. In any plane triangle BAG, ^\hose Uvo sides aie BA, AC, and base BC, the less of the two sides which let be BA, is to the greater AC as the radius is to the tangent pf an r.n^Ic, and the radius is to the tangeixt of tlie ex- 486 PLANE TRIGONOMETRY. cess of this angle above half a right angle as the tangent of half the supi of the angles B and C at the base, is to the tansrent of half their difference. At the point A, draw the straight line EAD perpendicular to BA; make AE, AF, each equal to AB, and AD to AC; join BE, BF, BD, and from D draw DG perpendicular upon BF. And because BA is at right angles to EF, and EA, AB, AF are equal, each of the angles EBA ABF is half a right angle, and the whole EBF is a right angle; also (4. 1. El.) EB is equal to BF. And since EBF, FGD are right angles, EB is parallel to CD, and the triangles EBF, FGD are similar; therefore EB is to BF as DG to GF, and EB being equal to BF, FG must be equal to GD. And because BAD is a right angle, BA the less side is to AD or AC the greater, as the radius is to the tangent of the angle ABD ; and because BGD is a right a'ngle, EG is to GD or GF as the radius is to the tangent of GBD, which is the excess of the angle ABD above ABF half a right angle. But because. EB is parallel to GD, BG is to C^F as ED is to DF, that is, since ED is the sum of the sides BA, AC and FD *heir dif- ference (3. of this), as the tangent of half the sum ot the angles B, C, at the base to the tangent of half their difference. There- fore, in any plane triangle? Stc Q. E. D. PROP. V. FIG. 9, 10. IN any plane triangle, twice the rectangle contained by any two sides is to the difference of the surn of the squares of these tw^o sides, and the sqnarc of the base as the ra- dius is to the co-sine of the angle included by the two sides. Let ABC be a plane triangle, twice the rectangle ABC con- to.ined by any two sides BA, BC is to the differencs of the sum of the squares of Bx\, BC, and the square of the base AC, as the radius to the co-sine of the angle ABC. From A, draw AD perpendicular upon the opposite side BC, then (by 12. and 13. 2. El.) the difference of the sun\ of the squares of AB, BC, and the square of the base AC, is equal to twice the rectangle CBD; but twice the rectangle CBA is to twice the rectangle CBD; that is, to the difference of the sura PLANE TRIGONOMETRY. t%7 of the squares of AB, BC, and the square of AC, (1. 6.) as AB to BD ; that is, by prop. I. as radius to the sine of BAD, which is the complement of the avigle ABC, that is, as radius to the co-sine of ABC. FROP. VI. FIG. 11. IN any plane triangle ABC, whose t^^o sides arc AB, AC, and base BC, the rectangle contained by half the peri- meter, and the excess of it abo^e the base BC, is to the rectangle contained by the sti-aight lines by AA'hich the hall' of the perimeter exceeds the other two sides AB, AC, as the square of the radius is to the square of the tangent of half the angle BAC opposite to the base. Let the angles BAC, ABC be bisected by the straight lines AG, BG ; and producing the side AB. let the exterior angle CBH be bisected by the straight line BK, meeting AG in K; and from the points G, K, let there be drawn perpendicular upon the sides the straight lines GD, GE, GF, KH, KL, KM. Since therefore (4. 4.) G is the centre of the circle inscribed in the triangle ABC, GD, GF, GE will be equal, and AD will be equal to AE, BD to BF, and CE to CF. In like manner KH, KL, KM will be equal, and BH will be equal to BM, and AH to AL, because the angles HBM, HAL are bisected by the straight lines BK, KA : and because in the triangles KCL, KCM, the sides LK, KM are equal, KC is common and KLC, KMC are right angles, CL will be equal lo CM : since there- fore BM is equal to BH, and CM to CL ; BC will be equal to BH and CL together; and, adding AB and AC together, AB, AC, and BC will together be equal to AH and AL together : but AH, AL are equal : wherefore each of them is equal to half the perimeter of the triangle ABC : but since AD, AE arc equal, and BD, BF, and also CE, CF, AB together with FC, will be equal to half the perimeter of the triangle to which AH or AL was shewn to be equal ; taking away therefore the com- mon AB, the I'emainder FC will be equal to the remainder BH : in the same manner it is demonstrated, that BF is equal to CL : and since the points B, D, G, F, are in a circle, the angle DGF will be equal to the exterior and opposite angle FBH, (22. 3.) ; ■wherefore their halves BGD, HBK will be equal to one another: the right angled triangles BCiD, HBK, will therefore be equi- angular, and GD will be to BD, as BH to HK, and the rectan- -488 PLANE rRIGONOMETRY. gle contained by GD, HK will be equal to the rectangle DBti or BFC : but since AH is to HK, as AD to DG, the rectangle HAD (22. 6.) will be to the rectangle contained by HK, DG, or the rectangle BFC, (as the square of AD is to the square of DG, that is) as the square of the radius to the square of the tangent of the angle DAG, that is, the half of BAG : but HA is half the perimeter of the triangle ABC, and AD is the excess of the same above HD, that is, above the base BC : but BV or CL is the excess of HA or AL above the side AC, and FC, or Hl3, is the excess of the same HA above the side AB ; therefore the ; rectangle contained by half the perimeter, and the excess of the same above the base, viz. the rectangle HAD, is to the rectangle contained by the straight lines by which the half of the peri- meter exceeds the other two sides, that is, the rectangle BFC, as the square of the radius is to the square of the tangent of half the angle BAC opposite to the base. Q. E. D. PROP. VII. FIG. 12, 15. IN a plane triangle, the base is to the sum of the sides, as the difference of the sides is to the sum or difference of tlie segments of the base made by the perpendicular upon it from the ^'ertex, according as the squai"C of the greater side is greater or less than the sum of the squares of the lesser side and the base. Let y\BC be a plane triangle ; if from A the vertex be drawn a straight line AD perpendicular upon the base BC, the base BC uill be to the sum of the sides BA, AC, as the difference of tire same sides is to the sum or difference of the segments CD, BD, according as the square of AC the greater side is greater or less than the sum of the squares of the lesser side AB, and the base BC. About A as a centre, -vvith AC the greater side for a distance^ let a circle be described meeting AB produced in E, F, and CB in G : it is manifest that FB is the sum, and BE the difference of the sides; and since AD is perpendicular to GC, GD CD will be equal ; consequently GB will be equal to the sum or dif- ference of the segments CD BD, according as the perpendicular AD meets the base, or the base produced ; that is, (by conv. 12. and 13. 2.) according as the square of AC is greater or less than the sum of the squares of AB, BG : but (by 35. 3.) the rectan-* PLANE TRIGONOMETRY. 48g ^le CBG is equal to the rectangle EBF ; that is, (16. 6.) BC is to BF, as BE is to BG ; that is, the base is to the sum of the sides, as the difference of the sides is to the sum or differ- ence of the segments of the base made by the perpendicular from the vertex, according as the square of the greater side is greater or less than the sum of the squares of the lesser side and the base. Q. E. D* tROP. VIII. PROB. FIG. 14. ^ The sum and difference of two magnitudes being given, to find them. Half the given sum added to half the given difference, Will be the greater, and half the difference subtracted from half the sum^ will be the less. For, let AB be the given sura, AC the greater, and BC the less. Let AD be half the given snm ; and to AD, DB, which are equal, let DC be added, then AC will be equal to BD, and DC together ; that is, to BC, and twice DC ; consequently twice DC is the difference, and DC half that difference : but AC the greater is equal to AD, DC ; that is, to half the sum added to half the difference, and BC the less is equal to the excess of BD, half the sum above DC half the difference. Q. E. F. SCHOLIUM. Of the six parts of a plane triangle (the three sides and three angles) any three being given^ to find the other three is the bu- siness of plane trigonometry ; and the several cases of that pro- blem may be resolved by means of the preceding ^propositions, as ill the two following, with the tables annexed. In these, the solution is expressed by a fourth proportional to three given lines ; but if the given parts be expressed by numbers from trigonome- trical tables, it may be obtained arithmetically by the common Rule of Three. 2o, AC. Aii, AC, unci ii wo sides and anar.- ^le opposite to one if them. ihe un- rles A anci Ali, AC, ak.d A, two sides and tht included angle. 'ilic all- eles B and C. *S Bc, J; S, C ; 8, A : : AB and also, S, C : S, B, : : AB S AC. (2.) \ AC : Ab : : b, B : b, C ^J (2.) This case admits of S Lwo solutions ; tor C may be Ij reuter or icss than a quad- S int. (Cor. to def. 4.) ^^ -vB + AC : aB— AC C-t-B : T, C— B : Fig. 16, 17. (3.):j 2 2 <) the sum and difference of ^ the angles C, B being given, S each of them is given. (7.) «, Othervjine. Vig. 18. S BA : AC : : R : T, ABC, $ and also R : T, ABC— 45" S ! T , B+C ; r. B— C : (4,) v 2 Z ^ therefore B and C are given ds before. (7.) ^- /-/" r^ f /^j~y'^y. \ s s s s PLANE TRIGONOMETRY. S s :: s s s S GIVEN. SOUGHT. s th AB, BC, CA the ree sides. A, B, C. the three ingles. -2 ACx^b : ACrH-vB.y — vBy: :R:CoS,C. If AB? + CB5r be greater than ABq. ig. 16. 2 ACxCB : ABy— ACy — CBy : : R : Co S, C If ABy be greater than ACq-\- Ciiq. Fig. 17. (4.) Otherwise, Let AB+BC+AC=2P. !'Xl — /^B : P — AC X P— liC : : Rg : Ty, -| C, and iience C is known. (5.) Otherivise. Let AD be perpendicular io BC. ).. If AB^ be less than ACy.+CBy. Fig. 16. BC : BA+AC : : BA— AC : BD^DC, and BC the sum of BD, DC is given ; there- fore each of them is given. •(7.) 2. If ABy be greater than VCy+CBfy. Fig. 17. BC : BA + AC : : BA— AC : BD +DC;andBCthedifference of BD, DC is given, there- fore each of them is given. (7.) And CA : CD : : R : Co S, C. (1.) and C being found, A and B are found Dv case 2. or 3. S ^ s s s s s s s I s. s ^ s s s. s s s s <» s s s s s s s s s s s s s s s SPHERICAL TRIGONOMETRY, DEFINITIONS. I. X HE pole of a circle of the sphere is a point in the superficies of the sphere, from which all straight lines drawn to the cir- cumference of the circle are equal. ir. A great circle of the sphere is any whose plane passes through the centre of the sphere, and whose centre therefore is the same with that of the sphere. lU. A spherical triangle is a figure upon the superficies of a sphere comprehended by three arches of three great circles, each of which is less than a semicircle. IV. A spherical angle is that which on the superficies of a sphere is contained by two arches of great circles, and is the same with the inclination of the planes of these great circles. PROP. I. GREAT circles bisect one another. As they have a common centre their common section will be A diameter of each which will bisect them. PROP. 11. FIG. I. THE arch of a great circle bet\^ixt the pole and the circumference of another is a quadrant. Let ABC be a great circle, and D its pole: if a great circle DC pass through D, and meet ABC in C, the arch DC will be a quadrant. Let the great circle CD meet ABC again in A, and let AC be the coroaion section of the great circles, which will pass through 49'4. SPHERICAL TRIGONOMETRY. E the cQhtre of the sphere: join DE, DA, DC : by def. 1. DA, DC are equal, and AE, EC are also equal, and DE is common ; therefore (8. 1.) the angles DEA, DEC are equal ; wherefore the arches DA, DC are equal, and consequently each of them is ^ quadrant. Q. E. D. PROP. III. FIG. 2. IF a sfreat circle be described meetinf^r two ^reat cir- cics AB, AC passing through its pole A in B, C, the angle at the centre of the sphere upon the circumierence BC, is the same with the sphericd angle BAC, and the arch BC is called the measure of the spherical ansrle BAC. Let the planes of the great circles AB, AC intersect one an- other in the straight line AD passing through D their common centre ; join DB, DC. Since A is the pole of BC, AB, AC will be quadrants, and the angles ADB, ADC right angles ; therefore (6. def. 1 1.) the ai)gle CBD is the inclination of the planes of the circles AB, AC; that is, (def. 4.) the spherical angle BAC. Q. E. D. CoR. If through the point A, two quadrants AB, AC be drawn, the point A will be the pole of the great circle BC, passing through their extremities B, C. Join AC, and draw AE a straight line to any other point E in BC ; join DE : since AC, AB are quadrants, the angles ADB, ADC arc right angles, and AD will be perpendicular to the plane of BC : therefore the angle ADE is a right angle, and AD, DC are equal to AD, DE, each to each ; therefore AE, AC are equal, and A is the pole of BC, by def. 1. Q. E. D. PROP. IV. FIG. 3. IN isosceles spherical triangles, the angles at the base are equal. Let ABC be an isosceles triangle, and AC, CB the equal sides ; the angles BAC, ABC, at the bai;e AB, are equal. Let D be the centre of the sphere, and join DA, DB, DC; in DA Vake any point- E, from which draw, in the plane \DC3 the strujght line EF at right angles to ED needing CD SPHERICAL TRIGONOMETRY* -fRS ill F, and draw, in the plane ADB, EG at right angles to the same ED; therefore the rectilineal angle FEG is (6. clef. 11.) the inclination of the planes ADC, ADB, and therefore is the same with the spherical angle BAC ; from F draw FH perpendicu- lar to DB. and from H draw, in the plane ADB, the straight tine HG at right angles to HD meeting EG in G, and join GF. Be- cause DE is at right angles to EF and EG, it is perpendicular to the plane FEG, (4. 1 1.) and therefore the plane FEG, is perpen- dicular to the plane ADB, in which DE is : (18. 1 1.) in the same manner the plane FHG is perpendicular to the plane ADB; and therefore GF the common section of the planes FEG, FHG is perpendicular to the plane ADB ; (19. 11.) and because the an- gle FHG is the inclination of the planes BDC, BDA, it is the same with the spherical angle ABC; and the sides AC,CBofthe spherical triangle being equal, the angles EDF, HDF, which stand upon them at the centre of the sphere, are equal ; and in the triangles EDF, HDFj the side DF is common, and the angles DEF, DHF are right angles; therefore EF, FH are equal; and in the triangles FEG, FHG the side GF is common, and the sides EG, GH will be equal by the 47. Land therefore the angle FEG is equal to FHG ; (8. 1.) that is, the spherical angle BAC is equal to the spherical angle ABC. PROP. V. FIG. 3. IF, in a spherical triangle ABC, two of the angles BAC, ABC be equal, the sides BC, AC opposite to them, are equal. Read the constructloii and demonstration of the preceding- proposition, unto the words, " and the sides AC, CB," &cc. and the rest of the demonstration will be as follows, viz. And the spherical angles BAC, ABC, being equal, the recti- lineal angles FEG, FHG, which are the same "with them, are equal ; and in the triangles FGE, FGHthe angles at G are right; angles, and the side FG opposite to two of the equal angles is coti.mon ; therefore (26. 1.) EF is equal to FH ; and in the right angled triangles DEF, DHF the side DF is common ; wherelore (47. 1.) ED is equal to DH, and the angles EDF, HDF, are therefore equal. (4. 1.) and consequently the sides AC, RCof tht; spherical triangle are equal. 49« SPHERICAL TRIGONOMETRY. PROP. VI. FIG. 4. ANY two sides of a spherical triangle are greater than the tliird. Let ABC be a spherical triangle, any two sides AB, BC wili be greater than the other side AC. Let D be the centre of the sphere ; join DA, DB, DC. The solid angle at D, is contained by three plane angles, ADB, ADC, BDC ; and by 20. 11. any two of them ADB, BDC are greater than the third ADC ; that is, any two sides AB, BC of the spherical triangle ABC, are greater than the third AC. PROP. VII. FIG. 4. THE three sides of a spherical triangle are less than a circle. Let ABC be a spherical triangle as before, the three sides A&j BC, AC are less than a circle. Let D be the centre of the sphere : the solid angle at D is contained by three plane angles BDA, BDC, ADC, which to- gether are less than four right angles, (^l. 11.) therefore the sides AB, BC, AC together, will be less than four quadrants; that rs, less than a circle. PROP. VIIL FIG. 5. IN a spherical triangle the greater angle is opposite to the greater side; and conversely. Let ABC be a spherical triangle, the greater angle A is op- posed to the greater side BC. Let the angle Bx\D be made equal to the angle B, and then BD, DA will be equal, (5. of this) and therefore AD, DC are equal to BC ; but AD, DC are greater than AC, (6. of this), therefore BC is greater than AC, that is, the greater angle A is opposite to the greater side BC. The converse is demonstrated as prop. 19. I. El. Q. li.D. SPHERICAL TRIGONOMETRY. 497 PROP. IX. FIG. 6. IN any spherical triangle ABC, if the sum of the sides AB, BC, be greater, equal, or less than a semicircle, the internal angle at the base AC will be greater, equal, or less than the external and opposite BCD ; and there- fore the sum of the angles A and ACB will be greater, equal, or less than two right angles. Let AC, AB produced meet in D. 1. If AB, BC be equal to a semicircle, that is, to AD, JBC, BD will be equal, that is, (4. of this) the angle D, or the angle A will be equal to the angle BCD. 2. If AB, BC together be greater than a semicircle, that is, greater than ABD, BC will be greater than BD ; and therefore (8. of this) the angle D, that is, the angle A, is greater than the angle BCD. 3. In the same manner it is shown, that if AB, BC together be less than a semicircle, the angle A is less than the angle BCD. And since the angles BCD, BCA are equal to two right angles, if the angle A be greater than BCD, A and ACB together will be greater than two right angles. If A be equal to BCD, A and ACB together will be equal to two right angles ; and if A be less than BCD, A and ACB will be less than two right angles. Q. E. D. PROP. X. FIG. r. • IF the angular points A, B, C of the spherical trian- gle ABC be the poles of three great circles, these great circles by their intersections will form another triangle FDE, which is called supplemental to the former ; that is, the sides FD, DE, EF are the supplements of the measures of the opposite angles C, B, A, of the trian- gle ABC, and the measures of the angles F, D, E of the triangle FDE, will be the supplements of the sides AC. BC, BA, in the triangle ABC. 3 R SPHERICAL TRIGONOMETRY. Let AB produced meet DE, EF in G, M, and AC meet FD, FE in K, L, and BC meet FD, DE in N, H. Since A is the pole of FE, and the circle AC passes through A, EF will pass through the pole of AC, (13. Ifi. 1. rh.) and since AC passes through C, the pole of FD, FD will pass through the pole of AC ; therefore the pole of AC is in the point F, in which the arches DF, EF intersect each other. In the same manner D is the pole of BC, and E the pole of AB. And since F, E are the poles of AL, AM, FL and EM are quadrants, and FL, EM together, that is, FE and ML together, :ire equal to a semicircle. But since A is the pole of ML, ML is the measure of the angle BAC, consequently FE is the sup- plement of the measure of the angle BAC. In the same man- ner, ED, DF are the supplements of the measures of the angles ABC, BCA. Since likewise CN, BH are quadrants, CN, BH together, that is, NH, BC together are equal to a semicircle; and since D is the pole of NM, NH is the measure of the angle FDh, there- fore the measure of the angle FDE is the supplement of the side BC. In the same manner, it -is shown that the measures of the angles DEF, EFD are the supplements of the sides AB, AC, ii> the triangle ABC. Q. E. D. PROP. XL FIG. r. THE three angles of a spherical triangle are greater than two right angles, and less than six right angles. The measures of the angles A, B, C, in the triangle ABC, together with the three sides of the supplemental triangle DEF, are (10. of this) equal to three semicuiles; but the three sides •of the triangle FDE, are (7. of this) less than two semicircles; therefore the measures of the angles A, B, C are greater than a semicircle ; and hence the angles A, B, C are greater than two right angles. All the external and internal angles of any triangle are equal to six right angles ; therefore all the internal angles are less than six right angles. PROP. XIL FIG. 8. IF from any point C, which is not the pole of the great cbxie AjjD, there be drawn arches of great circles CA, SPHERICAL TRIGONOMETRY. 499 CD, CE, CF, &c. the greatest of these is CA, which passes througli H the pole o'l ABD, and CB the remain- der of ACB is the least, and of any others CD, CE, CF, &c. CD, which is nearer to CA, is greater than CE, which is more remote. Let the common section of the planes of the great circles ACB, ADB be AB: and from C, draw CG perpendicular to AB, which will also be perpendicular to the plane ADB (4.dcf. 11.); join GD, GE, GF, CD, CE, CF, CA, CB. Of all the straight lines (Iruwn from G to the circumference ADU, GA is the greatest, and GB the least (7. 3.); and GD vil/u h is tuartr lo GA is greater than GE, which is more re- mole. Tiu- triangles CGA, CGD are right angled at G, and they have the comniou siue CG ; therefore the squares of CG, GA togithi.-r, tiiat is. the square of CA, is greater than the squares of t'G. GD together that is, the square of CD ; and CA is great- er tiidii CU, and thevi-fore the arch CA is greater than CD. In tlie same manner, since GD is greater than GE, and GE than GF, ix-c. it is shown tiiat CD is grealer than CE, and CE than CF, 8cc. and coiiseqv.enlly, the arch CD greater than the arch CE, and the arch CE greater than the arch CF, &c. And since GA is the greatest, and GB the least of all the straight lines drawn from G to the circumference ADB, it is manifest that CA is the greatest, and Cii the least of all the straight lines drawn fro ■. I.' to the circumference ; and therefore the arch CA is the gretttesi, and CB tiie least of all the circles drawn through C, meeting ADB. Q. E. D. PROP. Xin. FIG. 9. IN a right angled spherical triangle the sides arc of tli^ same affection v* ith the opposite angles ; that is, if the sides be greater or less than quadrants, the opposite an- gles w^iil be greater or less than right angles. Let ABC be a spherical triangle right angled at A, any side AB, will be of the same affection with the opposite angle ACB. Case 1. Let AB be less than a quadrant, kl AE be a quadrant, and let EC be a great circle passing through E, C. Since A is a right angle, and AE a quadrant, li is the pole of the great cir-. 5Q® SPHERICAL TRIGONOMETRY. cle AC, and ECA a right angle : but EC A is greater than BCAj therefore BCA is less than a right angle. Q. E. D. •fig. 10. Case 3' Let AB be greater than a quadrant, make AE a quad- rant, and let a great circle pass through C, E, ECA is a right angle as before, and BCA is greater than ECA, that is, greater than a right angle. Q. E. D. PROP. XIV. IF the two sides of a right angled spherical triangle be of the same affection, the hypothenuse will be less than a quadrant ; and if they be of different affection, the hy- pothenuse will be greater than a quadrant. Let ABC be a right angled spherical triangle, if the two sides AB, AC be of the same or of different affection, the hypothe- nuse BC will be less or greater than a quadrant. Fig. 9. Case 1 . Let AB, AC be each less than a quadrant. Let AE, AG be quadrants ; G will be the pole of AB, and E the pole of AC, and EC a quadrant; but, by prop. 12. CE is greiter than CB, since CB is farther off from CGD than CE. In the same manner, it is shown that CB, in the triangle CBD, where the two sides CD, BD are each greater than a quadrant, is less than CE, that is, less than a (quadrant. Q E. D. Fig. 10, Case 2. Let AC be less, and AB greater than a quadrant; then the hypothenuse BC will be greater than a quadrant ; for let AE be a quadrant, then E is the pole of AC, and EC will be a quadrant. But CB is greater than CE by prop. 12. since AC passes through the pole of ABD. Q. E. D. PROP. XV. IF the hypothenuse of a right angled triangle be great- er or less than a quadrant, the sides will be of different or the same affection. This is the converse of the preceding, and demonstrated in the same manner. SPHERICAL TRIGONOMETRY. 504 PROP. XVI. In^ any spherical triangle ABC, if the perpendicu- lar AD from A upon the base BC, fall within the tri- angle, , the angles B and C at the base will be of the same affection ; and if the perpendiculai' fall without the triangle, the angles B and C will be of diil'erent affection. 1. Let AD fall within the triangle; then (13. of this) since Fig. 11. ADB, ADC are right angled spherical triangles, the angles B, C must each be of the same aflection as AD. 2. Let AD fall without the triangle, then (13. of this) the Fig. 12. angle B is of the same affection as AD ; and by the same the angle ACD is of the same affection as AD ; therefore the angle ACB and AD are of different affection, and the angles B and ACB of different affection. CoR. Hence if the angles B and C be of the same affection, the perpendicular will ftill within the base ; for, if it did not (16. of this), B and C would be of different affection. And if the angles B and C be of opposite affection, the perpendicular ■will fall without the triangle; for, if it did not (16. of this), the angles B and C would be of the same affection, contrary to the supposition. PROP. XVIL FIG. 13. IN right angled spherical triangles, the sine of ei- ther of the sides about the right angle, is to the radius of the sphere, as the tangent of the remaining side is to the tangent of the angle opposite to that side. Let ABC be a triangle, having the right angle at A ; and let AB be either of the sides, the sine of the side AB will be to the radius, as the tangent of the other side AC to the tan- gent of the angle ABC, opposite to AC Let D be the cen- tre of the sphere ; join AD, BD, CD, and let AE be drawn perpendicular to BD, which therefore will be the sine of the arch AB, and from the point E, let there be drawn in the plane BDC the straight line EF at right angles to BD, meeting DC in E, and let AF be joined. Since therefore the straiglit line BE is at ri^ht angles to both EA and EF, it will also be |02 SPHERICAL TRIGONOMETRY. at right angles to the plane AEF (4, 11.) wherefore the plane ABD, which passes through DE, is perpendicular to the plane AEF (18. 11.), and the plane AEF pt rpendicular to x\BD : the plane ACD or AFD is also perpendicular to the same ABD: therefore the common section, viz. the straight line AF, is at right angles to the plane ABD (19. II.): and FAE, FAD are right angles (3. def. 11.); therefore AF is the tangent ol the arch AC; and in the rectilineal triangle AEF, having aright' angle at A, AE will be to the radius as AF to the tangent of the angle AEF (1. PI. Tr.) ; but AE is the sine of the arch AB, and AF the tangent of the arch AC, and the angle AEF is the inclination of the planes CBD, ABD (6. def. 11.), or the sphe- rical angle ABC : therefore the sine of the arch AB is to the radius as the tangen*, of the arch AC, to the tangent of the opposite angle ABC. CoR. 1. If therefore of the two sides, and an angle oppo- site to one of them, any two be given, the third will also be given. CoR. 2. And since by this proposition the sine of the side AB is to the radius, as the tangent of the other side AC to the tangent of the angle ABC opposite to that side ; and as the radius is to the co-tangent of the angle ABC, so is the tangent of the same angle ABC to the radius (Cor. 2. def. PI. Tr.), by equality, the sine of the side AB is to the co-tangent of the angle ABC adjacent to it, as the tangent of the other side AC to the radius. PROP. XVIII. FIG. 13. IN right angled spherical triangles the sine of the hypothenusc is to the radius, as the sine of either side is to the sine of the angle opposite to that side. Let the triangle ABC be right angled at A, and let AC be either of the sides; the sine of the hypnthenuse BC will be to the radius as the sine of the arch AC is to the sine of the angle ABC. . Let D be the centre of the sphere, and let CG I)e drawn per- pendicular to DB, which will therefore be the sine of the hy- pothenusc BC ; aad from the point G let there be drawn in the plane ABD the straight line GM perpendicular to DB, and let CH be joined : CH will be at right angles to the plane ABD, as was shown in the preceding proposition of the straight line FA: wherefore CHD, CHG are right angles, and CH is the sine of the arch AC ; and in the triangle CHG, having the right SPHERICAL TRIGONOMETRY. 50! angle CHG, CG is to the radius as CH to the sine of the angle CGH (1 PI Tr.): but since CG, HG are at right angles to DGB, wliich is the common section of the planes CBD, ABD, the ani>!e CGH will be equal to the inclination of these planes (6. clef. 11.); that is, to the spherical angle ABC. The sine, therefore, of the hypothenuse CB is to the radius as the sine of the side AC is to the sine of the opposite angle ABC. Q. E. D. Cor. Of theSf three, viz. the hypothenuse, a side, and the an- gle opposite to that side, any two being given, the third is also given by prop. 2. PROP. XIX. FIG. 14. IN right angled spherical triangles, the co-sine of the hypothenuse is to the radius as the co-tangent of either of the angles is to the tangent of the remaining angle. Let ABC be a spherical triangle, having a right angle at A, the co-sine of the hypothenuse BC will be to the radius as the co-tangent of the angle ABC to the tangent of the angle ACB. Describe the circle DE, of which B is the pole, and let it meet AC in F, and the circle BC in E ; and since the circle BD pas- ses through the pole B of the circle DF, DF will also pass through the pole of BO (13. 18. 1. Theod. Sph.). And since AC is per- pendicular to BD, AC will also pass through the pole of BD ; wherefore the pole of the circle BD will be found in the point where the circles AC, DE meet, that is, in the point F : the arches FA, FD are therefore quadrants, and likewise the arches BD, BE: in the triangle CEF, rigiu angled at the point E, CE is the csmplement of the hypothenuse BC of the tri;ingle ABC, EF is the complement of the arch ED, which is the mtasure of the angle ABC, and FC the hypothenuse of tl^.e triangle CEF, is the complement of AC, and the arch AD, vvhicii is the nuusure of the anglt CFE, is the complement of AB. But (17. of this) in the triangle CEF, the sine of the side CE is to the radius, as the tangent of the other side i-- to the tan^'ent of the angle ECF opposite to it, that is, in the triangle ABC, the co-sine of the h> pothenuse BC is to the radius, ai the co- tangent of the angle ABC is to the tangent of t';e angle ACB. Q. E. D. CoR. 1. Of these three, viz. the hypothenuse ai. i the two au- ?jles, any two being given, the third will also be given. ^04 SPHERICAL TRIGONOMETRY. Cor. 2. And since by this proposition the co-sine of the hy- pothenuse BC is to the radius as the co-tangent of the angle ABC to the tangent of the angle ACB. But as the radius is to the co-tangent of the angle ACB, so is the tangent of the same to the radius (Cor. 2. def. PI. Tr.) ; and, ex xquo, the co-sine of i the hypothenuse BC is to the co-tangent of the angle ACB, as the co-tangent of the angle ABC to the radius. PROP. XX. FIG. 14. IN right angled spherical triangles, the co-sine of an angle is to the radius, as the tangent of the side adjacent to that angle is to the tangent of the hypothenuse. The same construction remaining; in the triangle CEF (17. of this), the sine of the side EF is to the radius, as the tangent of the other side CE is to the tangent of the angle CFE opposite to it ; that is, in the triangle ABC, the co-sine of the angle ABC is to the radius as (the co-tangent of the hypothenuse BC to the co-tangent of the side AB, adjacent to ABC, or as) the tangent of the side AB to the tangent of the hypothenuse, since the tan- gents of two arches are reciprocally pi-oportional to their co-tan- gents. (Cor. 1. def. PI. Tr.) CoR. And since by this proposition the co-sine of the angle ABC is to the radius, as the tangent of the side AB is to the tan- gent of the hypothenuse BC ; and as the radius is to the co-tan- gent of BC, so is the tangent of BC to the radius ; by equality, the co-sine of the angle ABC will be to the co-tangent of the hy- pothenuse BC, as the tangent of the side AB, adjacent to the an^ gle ABC to the radius. PROP. XXI. FIG. 14. IN right angled spherical triangles, the co- sine of either of the sides is to the radius, as the co-sine of the hypo- thenuse is to the co-sine of the other side. The same construction remaining ; in the triangle CEF, the sine of the hypothenuse CF is to the radius, as the sine of the side CE to the sine of the opposite angle CFE (18. of this) ; that is, in the triangle ABC the co-sine of the side CA is to the SPHERICAL TRIGONOMETRY. . 505 radius as the co-sine of the hypothenuse BC to the co-sine of the other side BA. Q. E. D. PROP. XXII. FIG. 14. IN right angled spherical triangles, the co-sine of either of the sides is to the radius, as the co-sine of the angle opposite to that side is to the sine of the other angle. The same construction remaining ; in the triangle CEF, the sine of the hypothenuse CF is to the radius as the sine of the side EF is to the sine of the angle ECF opposite to it ; that is, in the triangle ABC, the co-sine of the side CA is to the radius, as the co-sine of the angle ABC opposite to it, is to the sin.e of the other angle. Q. E. D. S S i,06 SPHERICAL TRIGONOMETRY. OF THE CIRCULAR PARTS. Fig. 15. IN any right angled spherical triangle ABC, the complement of the hypothenuse, the complements of the angles, and the two sides are called The circular parts of the triangle, as if it ■were following; each other in a circular order, from whatever part we begin: thus, if we begin at the complement of the hypothenuse, and proceed towards the side BA, the parts fol- lowing in order will be the complement of the hypothenuse, the complement of the angle B, the side BA, the side AC, (for the right angle at A is not reckoned among the parts), and, lastly, the complement of the angle C. And thus at whatever part we begin, if any three of these five be taken, they either will be all contiguous or adjacent, or one of them will not be conti- guous to either of the other two: in the first case, the part which is between the other two is called the Middle part^ and the other two are called Adjacent extremes. In llie second case, the part which is not contiguous to either of the other two i» called the Middle part^ and the other two Opposite extremes, I'or example, if the three parts be the complement of the hy- pothenuse BC, the complement of the angle B, and the sida BA ; since these three are contiguops to each other, the com- plement of the angle B will be the middle part, and the com- plement of the hypothenuse BC and the side BA will be adjacent extremes : but if the complement of the hypothenuse BC, and the sides BA, AC be taken ; since the complement of the hypo- thenuse is not adjacent to either of the sides, viz. on account of the complements of the two angles B and C intervening be- tween it and the sides, the complement of the hypothenuse BC will be the middle part, and the sides, BA, AC opposite ex- tremes. The most acute and ingenious Baron Napier, the inven- tor of Logarithms, contrived the two following rules concerning these parts, by means of which all the cases of right angled spherical triangles are resolved with the greatest ease. RULE L The rectangle contained by the radius and the sine of the mid- dle part, is equal to the rectangle contained by the tangents of the adjacent parts. SPHERICAL TRIGONOMETRY. 507 RULE II. The rectangle contained by the radius, and the sine of the mid- die part is equal to the rectangle contained by the co-sinea of the opposite parts. These rules are demonstrated in the following manner: First, Let either of the sides, as BA, be the middle part, and Fij;. ir- thereforc the complement of the angle B, and the side AC will be adjacent extremes. And by Cor. 2. prop. 17. of this, S, BA is to the Co-T, B as T, AC is to the radius, and therefore RxS, BA=Co-T, BxT, AC. The same side BA being the middle part, the complement of the hypothenuse, and the complement of the angle C, are oppo- site ouremes ; and by prop. 18. S, BC is to the radius, as S, BA to S, C ; therefore RxS, BA=S, BCxS, C. Secondly, Let the complement of one of the angles, as B, be the middle part, and the complement of the hypothenuse, and the side BA will be adjacent extremes: and by Cor. prop. 20. Co-S, B is to Co-T, BC, as T, BA is to the I'adius, and therefore RxCo-S, B=:Co-T, BCxT, BA. Again, Let the complement of the angle B be the middle part, and the complement of the angle C, and the side AC will be op- posite extremes : and by prop. 22. Co-S, AC is to the radius, as Co-S, B is to S, C: and therefore RxCo-S, B=Co-S ACxS, C Thirdly, Let the complement of the hypothenuse be the mid- dle part, and the complements of the angles B, C, will be adja- cent extremes: but by Cor. .2. prop. 19. Co-S, BC is to Co-T, B as Co-T, B to the radius: therefore RxCo-S, BC=Co-T CxCo-T, C. Again, Let the complement of the hypothenuse be the mid- dle part, and the sides AB, AC will be opposite extremes : but by prop. 21. Co-S, AC is to the radius, Co-S, BC to Co-S, BA ; therefore RxCo-S, BC=Co.S, BAxCo-S, AC. Q. E. D. 5«8 SPHERICAL TRIGONOMETRY. SOLUTION OF THE SIXTEEN CASES OF RIGHT ANGLE© SPHE- RICAL TRIANGLES. GENERAL PROPOSITION. IN a right angled spherical triangle, of the three sides and three angles, any two being given, besides the right angle, the other three may be found. In the following Table the solutions are derived from the pre- ceding propositions. It is obvious that the same solutions may be derived from Baron Napier's two rules above demonstrated, which, as they are easily remembered, are commonly used in practice. Case Given So't 1 AC,C B R : Co-S, AC : : b, C : Co-S, B : and B is ofthe same species with CA, by 22. and 13. 2 AC, B c Co-S, AC : R : : Co-b, B : S, C : by 22. 3 B, C AC S, C ; Co-S, B : : R : Co-S, AC : by 22. and AC is of the same species with B. 13. 4 BA, AC BC R : Co-S, BA : : Co-S, AC : Co-S, BC. 21. and if both BA, AC be greater or less than a quadrant, BC will be less than a quadrant. But if they be of different affec- tion, BC will be greater than a quadrant. 1 4. 5 BA, BC AC Co-S, B A : R : : Co-S, BC : Co-S, AC. 2 1 . and if BC be greater or less than a quad- rant, BA, AC will be of different or the same affection: by 15. 6 BA, AC B S, BA : R : : T, CA : T, B. 17. and B is of the same affection with AC 13. #^ s s s s s \ \ SPHERICAL TRIGONOMETRY. 509 ^ Case S S S I s s s s s Given So't 10 BA, B AC, B AC "Ba BC, C AC AC, C BC R : b, BA : : T, B : T, Av^, 17. And AG is of the same affection with B. 13. T, B : R : : T, CA : S, BA. 17. R : Co-S, C : : T, BC : T, CA. 20. If BC be less or greater than a quadrant, C and B will be of the same or different affec- tion. 15. 13. S s s 1 , BC : R : : T, CA : Co-is, G. 20. If BC \ be less or greater than a quadrant, CA and S AB, and therefore CA and C, are of the s same or different affection. 15. S Co-S, C : R : : T, AG : 1 , BC. 20. And BC is less or greater than a quadrant, ac- cording as C and AC or C and B are of the same or different affection. 14. 1. 11 12 BC,CA C BC, B AC K : b, BC : : S, B : b, AC. 18. And AG ^ is of the same affection with B. c S I \ \- S S' s s AC, B bG b, B : S, AC : : R : S, BC : 18. g S, BC : R : : S, AG : b, B : 18. And B is of the same affection with AC U BC, AC 15 B, C BC r, C : R : : Co-T, B : Co-S, BC. 19. And according as the angles B and C are of dif- ferent or the same affection, BC will be greater or less than a quadrant. 14. 16 BC, C B R : Go-b, BC : : 1 ,C : Go-T ,B. 19. If BG be less or greater than a quadrant, C and B will be of the same ordifferentaffection. 15. 519 SPHERICAL TRIGONOMETRY. The second, eighth, and thirteenth cases, which are cominon- ly called ambiguous, adnnit of two solutions : for in these it is not determined whether the side or measure of the angle sought be greater or less than a quadrant. PROP. XXIII. FIG. 16. IN spherical triangles, whetlier right angled or oblique angled, the sines of the sides are proportional to the sines of the angles opposite to them. First, let ABC be a right angled triangle, having a right an- gle at A; therefore by 'prop. 18. the sine of the hypothenuse BC is to the radius (or the sine of the right angle at A) as the sine of the side AC to the sine of the angle B. And in like manner, the sine of BC is to the si;ie of the angle A, as the sine of AB to the sine of the angle C ; wherefore (11. 5.) the sine of the side AC is to the sine of the angle B, as the sine of AB to the sine of the angle C. Secondly, let BCD be an oblique angled triangle, the sine of either of the sides BC, will be to the sine of either of the other two CD, as the sine of the angle D opposite to BC is to the sine of the angle B opposite to the side CD. Through the point C, let there be drawn an arch of a great circle CA perpendicular upon BD ; and in the right angled triangle ABC (18. of this) the sine of BC is to the i-adius, as the sine of AC to the sine of the angle B; and in the triangle ADC (by 18. of this): and, by in- version, the radius is to the sine of DC as the sine of the angle D to the sine of AC : therefore ex xquo perturbate, the sine of BC is to the sine of DC, as the sine of the angle D to the sine of the angle B. Q. E. D. PROP. XXIV. FIG. ir, 18. IN oblique angled spherical triangles having drawn a perpendicular arch from any of the angles upon the oppo- site side, the co-sines of the angles at the base are pro- portional to the sines of the ^^ertical angles. Let BCD be a triangle, and the arch CA perpendicular to the base BD ; the co-sine of the angle B will be to the co-sine of the angle D, as the sine of the angle BC A to the sine of the an- trle DCA. SPHERICAL TRIGONOMETRY. SIf For by 22. the co-sine of the angle B is to the sine of the angle BCA as (the co-sine of the side AC is to the radius ; that is, by prop. 22. as) the co-sine of the angle D to the sine of the angle DCA ; and, by permutation, the co-sine of the angle R is to the co-sine of the angle D, as the sine of the angle BCA to the sine of the angle DCA. Q. E. D. PROP. XXV. FIG. 17, 18. THE same things remaining, the co-sines of the sides BC, CD, are proportional to tlie co-sines of the bases BA, AD. For by 21. the co-sine of BC is to the co-sine of BA, as (th* co-sine of AC to the radius ; that is, by 21. as) the co-sine of CD is to the co-sine of AD : wherefore, by permutation* the co-sines of the sides BC, CD are proportional to the co-sines of the bases BA, AD. Q. E. D. PROP. XXVI. FIG. ir, 18. THE same construction remaining, the sines of the bases BA, AD are reciprocally proportional to the tan- gents of the angles B and D at the base. For by 17. the sine of BA is to the radius, as the tangent of AC to the tangent of the angle B ; and by 17. and inversion the radius is to the sine of AD, as the tangent of D to the tan- gent of AC : therefore, ex aquo perturbate, the sine of BA is to the sine of AD, as the tangent of D to the tangent of B. PROP. XXVII. FIG. 17, IS. THE CO- sines of the vertical angles are reciprocally proportional to the tangents of the sides. For by prop. 20. the co-sine of the angle BCA is to the ra- dius as the tangent of CA is to the tangent of BC ; and by the same prop. 20^ and by inversion, the radius is to the co-sine of the angle DCA, as the tangent of DC to the tangent of CA: therefore, ex aequo perturbate, the co-sine of the angle BCA is 512 SPHERICAL TRIGONOMETRY. to the ce^sine of the angle DCA, as the tangent of DC is to the tangent of BC. Q. E. D. LEMMA. FIG. 19, 20. IN right angled plain triangles, the hypotheniise is to the radius, as the excess of the hypothenuse above either of the sides to the versed sine of the acute angle adja- cent to that side, or as the sum of the hypothenuse, and either of the sides to the versed sine of the exterior angle of the triangle. Let the triangle ABC have a right angle at B ; AC will be to the radius as the excess of AC above AB, to the versed sine of the angle A adjacent to AB ; or as the sum of AC, AB to the versed sine of the exterior angle CAK. Vv'ith any radius DE, let a circle be described, and from D the centre let DF be drawn to the circumference, making the angle EDF equal to the angle BAC, and from the point F, let FG be drawn perpendicular to DE; let AH, AK be made equal to AC, and DL to DE : DG therefore is the co-sine of the an- gle EDF or BAC, and GE its versed sine : and because of the equiangular triangles ACB, DFG, AC or AH is to DF or DE, as AB to DG : therefore (19. 5.) AC is to the radius DE as BH toGE, the versed sine of the angle EDF or BAC: and since AH is to DE, as AB to DG (12. 5.), AH or AC will be to the radius DE as KB to LG, the versed sine of the angle LDF or KAC. Q. E. D. PROP. XXVIII. FIG. 21, 22. IN any spherical triangle, the rectangle contained by the sines of two sides, is to the square of the radius, as the excess of the versed sines of the third side or base, and the arch, which is the excess of the sides is to the versed sine of the angle opposite to the base. Let ABC be a spherical triangle, the rectangle contained by the sines of AB, BC will be to the square of the radius, as the excess of the versed sines of the base AC, and of the arch, which is the excess of AB, BC to the versed sine of the angle ABC opposite to the base. SPHERICAL TRIGONOMETRY. us Let D be the centre of the sphere, and let AD, BD, CD be joined, and let the sines AE, CF, CG of the arches AB, BC, AC be drawn ; let the side BC be greater than BA, and let BH be made equal to BC ; AH will therefore be the excess of the sides BC, BA ; let HK be drawn perpendicular to AD, and since AG is the versed sine of the base AC, and AK the versed sine of the arch AH, KG is the excess of the versed sines of the base AC, and of the arch AH, which is the excess of the sides BC, B A : let GL likewise be drawn parallel to KH, and let it meet FH in L, let CL, DH be joined, and let AD, FHmeet each other in M. Since therefore in the triangles CDF, HDF, DC, DH are equal, DF is common, and the angle FDC equal to the angle FDH, because of the equal arches BC, BH, the base HF will be equal to the base FC, and the angle HFD equal to the right angle CFD : the straight line DF therefore (4. 11.) is at right angles to the plane CFH : wherefore the plane CFH is at right angles to the plane BDH, which passes through DF (18. 11.). In like man- ner, since DG is at right angles to both GC and GL, DG will be perpendicular to the plane CGL ; therefore the plane CGL is at right angles to the plane BDH, which passes through DG : and it was shown, that the plane CFH or CFL was perpendicular to the same plane BDH ; therefore the common section of the planes CFL, CGL, viz. the straight line GL is perpendicular to the plane BDA (19. 11.), and therefore CLF is a right angle: in the trian- gle CFL having the right angle CLF^ by the lemma CF, is to the radius as LH, the excesSf'vifT'df CF or FH above EL, is to the versed sine of the angle CFL ; but the angle CFL is the in- clination of the planes BCD, BAD, since FC, FL are drawn in them at right angles to the common section BF : the spherical angle ABC is therefore the same with the angle CFL ; and there- fore CF is to the radius as LH to the versed sine of the spherical angle ABC ; and since the triangle AED is equiangular (to the triangle MFD, and therefore) to the triangle MGL, AE will be to the radius of the sphere AD, (as MG to ML ; that is, becaaSe of the parallels as) GK to LH : the ratio therefore which is com- pounded of the ratios of AE to the radius, and of CF to the same radius ; that is, (23. 6.) the ratio of the rectangle contained by AE, CF to the square of the radius, is the same with the ratio compounded of the ratio of GK to LH, and the ratio of LH to the versed sine of the angle ABC ; that is, the same with the ratio of GK to the versed sine of the angle ABC ; therefore, the rectangle contained by AE, CF, the sines of the sides AB, BC, is to the square of the radius as GK, the excess of the versed sines AG,. AK, of the base AC, and the arch AH, which is the 3T SPHERICAL TRIGONOMETRY. excess of the sides to the versed sine of the angle ABC opposite to the base AC. Q. E. D. PROP. XXIX. FIG. 23. THE rectangle contained by half of the radius, and the excess of the versed sines of two arches, is equal to the rectangle contained by the sines of half the sum, and half the difference of the same arches. Let AB, AC be any two arches, and let AD be made equal to AC the less ; the arch DB therefore is the sum, and the arch CB the difference of AC, AB : through E the centre of the circle^ let there be drawn a diameter DEF, and AE joined, and CD likewise perpendicular to it in G ; and let BH be perpendicular to AE, and AH will be the versed sine of the arch AB, and AG the versed sine of AC, and HG the excess of these versed sines: let BD, BC, BF be joined, and FC also meeting BH in K. Since therefore BH, CG are parallel, the alternate angles BKC, KCG will be equal ; but KCG is in a semicircle, and therefore a right angle ; therefore BKC is a right angle ; and in the trian gles DFB, CBK, the angles FDB, BCK, in the same segment are equal, and FBD, BKC are right angles ; the triangles DFB, CBK are therefore equiangular ; wherefore DF is to DB, as BC to CK, or HG ; and therefore the rectangle contained by the di- ameter DF, and HG is equal to that contained by DB, BC ; wherefore the I'ectangle contained by a fourth part of the diame- ter, and HG, is equal to that contained by the halves of DB, BC ; but half the chord DB is the sine of half the arch DAB, that is, half the sum of the arches AB, AC ; and half the chord of BC is the sine of half the arch BC, which is the difference of ABj AC, "Whence the proposition is manifest. PROP. XXX. FIG. 19, 24. THE rectangle contained by half of the radius, and the versed sine of any arch, is equal to the square of the sine of half the same arch. Let AB be an arch of a circle, C its centre, and AC, CB, BA being joined : let AB be bisectd*d in D, and let CD be SPHERICAL TRIGONOMETRY. 516 joined, which will be perpendicular to BA, and bisect it in E (4. !.). BE or AE therefore is the sine of the arch DB or AD, the half of AB : let BF be perpendicular to AC, and AF will be the versed sine of the arch BA : but, because of the similar triangles CAE, BAF, CA is to AE, as AB, that is, twice AE, to AF ; and by halving the antecedents, half of the radius CA is to AE, the sine of the arch AD, as the same AE to AF the vers- ed sine of the arch AB. Wherefore by 16. 6. the proposition is manifesJt. PROP. XXXI. FIG. 25. IN a spherical triangle, the rectangle contained by the sines of the two sides, is to tlie square of the radius, as the rectangle contained by the sine of the arch which is half the sum of the base, and the excess of the sides, and the sine of the arch, which is half the difference of the same to the square of die sine of half the angle opposite to the base. Let ABC be a spherical triangle, of which the two sides are AB, BC, and base AC, and let the less side BA be produced, so that BD shall be equal to BC : AD therefore is the excess of BC, BA; and it is to be shown, that the rectangle contained by the sines of BC, BA is to the square of the radius, as the rectangle contained by the sine of half the sum of AC, AD, and the sine of half the difference of the same AC, AD to the square of the sine of half the angle ABC, opposite to the base AC. Since by prop. 28. the rectangle contained by the sines of the sides BC, BA is to the square of the radius, as the excess of the versed sines of the basse AC and AD, to the versed sine of the angle B; that is (1. 6.), as the rectangle contained by half the radius, and that excess, to the rectangle contained by half the radius, and the versed sine of B; therefore (29. 30. of this), the recvangle contained by the sines of the sides BC, BA is to the square of the radius, as the rectangle contained by the sine of the arch, which is half the sum of AC, AD, and the sine of the arch which is half the difference of the same AC, AD is to the square of the sine of half the- angle ABC. Q. E. D. 516 SPHERICAL TRIGONOMETRY. SOLUTION OF THE TWELVE CASES OF OBLiqUE ANGLED SPHE- RICAL TRIANGLES. GENERAL PROPOSITION. IN an oblique angled spherical triangle, of the three sides and three angles, any three being given, the other three may be found. 27. ^..- Given B, D, and BC, two an- .^les and a side opposite one of them. B, C, and 3C, two an- gles and tht side betweer: ihem. BC, CD and B. BC, DB, and B. ^' 50 I C. D. aD CD Co-S, BC : R : : Co-T, B : T, BCA. 19. Likewise by 24. Co-S, B : S, BCA : : Co-S, D : S, DCA; wherefore BCD IS the sum or difference of the angles DCA, BCA according as the perpendi- cular CA falls within or without the tri- -ingle BCD ; that is ('6. of this), accord- ing as the angles B, D are of the same or different affection. Co-S, BC : R : : Co-T, B : T, BCA. 19. and also by 24. S, BCA : S, DCA : : Co-S, B : Co-S, D ; and according IS the angle BCA is less or greater than BCD, the perpendicular CA falls within or without the triangle BCD ; and there- fore (16. of this) the angles B, D will be of the same or different affection. R : Co-S, B : : T, BC « T, BA. 20. ind Co-S, BC : Co-S, BA : : Co-S, DC : Co-S, DA. 25. and BD is the sum or lifference of BA, DA. R : Co-S, B : : T, BC ; T, BA. 20. uid Co-S, BA : Co-S, BC : : Co-S, DA ; Co-S, DC. 25. and according as DA, .\C are of the same or different affec- ion DC will be less or greater than a juadrant. 14. "•^ SPHERICAL TRIGONOMETRY. S\7 S Given. B, D and BC. So't DB D. DC 0. DC 5 R : Co-S, B I : T, BC : T, BA. 20. and T, D : T, B : : S, BA : S, DA. 26. and BD is the sum or difference of BA DA. 6 BC, BD and B. BC, DC And B. R : Co-S, B : : T, BC : T, BA. 20. .nd S, DA : S, BA : : T, B : T, D; and according as BD is greater or less than B \, the angles B, D are of the same or ilifferent affection. 16. 7 Co-S, BC : R : : Co- T, B : T, BCA. 19. and T, DC : T, BC : : Co-S, BCA : Co-S, DCA. 27. the sum or difference of the angles BCA, DCA is equal to the ingle BCD. 8 B, C and BC. Co,S, BC : R : : Co-T, B : T, BCA. 19. also by 27. Co-S, DCA : Co-S, BCA : : T, BC : T, DC 27. if DCA and B ■ic of the same affection; that is (13.), .f AD and CA be similar, DC will be less than a quadrant. 14. and if AD, CA be not of the same affection, DC is great- er than a quadrant. 14. 9 BC, DC and B. S, CD : S, B : : S, BC : S, D. 10 B, D and BC. S, D : S, BC : : S, B : S, DC. 11 BC, BA AC. Fig. 25. B. S, AB X S, BC : R^^ : : S, AC-f AD 2 X S, AC— AD : Sg, ABC. See Fig. 23. 2 2 AD being the difference of the sides BC, BA. SPHERICAL TRIGONOMETRY. S 12 Given So't A, B, C. Th Fig. 7. sides. See Fig. 7. In the triangles DEF, DE, EF, FD ire respectively the supplements of the measures of the given angles B, A, C in the triangle BAG; the sides of the triangle DEF are therefore given, and by the preceding case the angles D, E, F may be found, and the sides BC, BA, AC are the supplements of the mea- sures of these angles. The 3d, 5th, 7th, 9th, 10th cases, which are commonly called ambiguous, admit of two solutions, either of which will answer the conditions required ; for, in these cases, the measure of the angle or side sought, may be either greater or less than a quad- rant, and the two solutions will be supplements to each other