THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA LOS ANGELES A TEXT-BOOK OF EUCLID'S ELEMENTS. A TEXT-BOOK OF EUCLID'S ELEMENTS FOR THE USE OF SCHOOLS BOOKS I. VI. AND XI. BY H. S. HALL, M.A. FORMERLY SCHOLAR OF CHRIST'S COLLEGE, CAMBRIDGE ; AND F. H. STEVENS, M.A. FORMERLY SCHOLAR OF QUEEN'S COLLEGE, OXFORD J MASTERS OF THE MILITARY SIDE, CLIFTON COLLEGE. Eon toon MACMILLAN AND CO. AND NEW YORK 1893 All Rights referred RICHARD CLAY AND SONS, LIMITED, LONDON AND BUNGAY. First Edition 1888; Second Edition (Boole XL added) 1889; Reprinted 190, ISftl, 1892, 1893. Engineering ft Mathe'naticrt A Sconces v Itomy Zj-^ \ H PREFACE TO THE FIRST EDITION. THIS volume contains the first Six Books of Euclid's Elements, together with Appendices giving the most im- portant elementary developments of Euclidean Geometry. The text has been carefully revised, and special atten- tion given to those points which experience has shewn to present difficulties to beginners. In the course of this revision the Enunciations have been altered as little as possible; and, except in Book V., very few departures have been made from Euclid's proofs: in each case changes have been adopted only where the old text has been generally found a cause of difficulty; and such changes are for the most part in favour of well-recog- nised alternatives. For example, the ambiguity has been removed from the Enunciations of Propositions 18 and 19 of Book I.: the fact that Propositions 8 and 20 establish the complete identical equality of the two triangles considered has been strongly urged; and thus the redundant step has been removed from Proposition 34. In Book II. Simson's ar- rangement of Proposition 13 has been abandoned for a well-known alternative proof. In Book III. Proposition 25 is not given at length, and its place is taken by a VI PREFACE. simple equivalent. Propositions 35 and 36 have been treated generally, and it has not been thought necessary to do more than call attention in a note to the special cases. Finally, in Book VI. we have adopted an alterna- tive proof of Proposition 7, a theorem which has been too much neglected, owing to the cumbrous form in which it has been usually given. These are the chief deviations from the ordinary text as regards method and arrangement of proof : they are points familiar as difficulties to most teachers, and to name them indicates sufficiently, without further enumeration, the general principles which have guided our revision. A few alternative proofs of difficult propositions are given for the convenience of those teachers who care to use them. With regard to Book V. we have established the princi- pal propositions, both from the algebraical and geometrical definitions of ratio and proportion, and we have endeavoured to bring out cleai'ly the distinction between these two modes of treatment. In compiling the geometrical section of Book V. we have followed the system first advocated by the late Pro- fessor De Morgan ; and here we derived very material assistance from the exposition of the subject given in the text-book of the Association for the Improvement of Geo- metrical Teaching. To this source we are indebted for the improved and more precise wording of definitions (as given on pages 286, 288 to 291), as well as for the order and substance of most of the propositions which appear between pages 297 and 306. But as AVC have not (except in the points above mentioned) adhered verbally to the text of the Association, we are anxious, while expressing in the fullest manner our obligation to their work, to exempt the PREFACE. vii Association fi-om all responsibility for our treatment of the subject. One purpose of the book is to gradually familiarise the student with the use of legitimate symbols and abbrevia- tions; for a geometrical argument may thus be thrown into a form which is not only more readily seized by an advanced reader, but is useful as a guide to the way in which Euclid's propositions may be handled in written work. On the other hand, we think it very desirable to defer the intro- duction of symbols until the beginner has learnt that they can only be properly used in Pure Geometry as abbrevia- tions for verbal argument: and we hope thus to prevent the slovenly and inaccurate habits which are very apt to arise from their employment before this principle is fully recognised. Accordingly in Book I. we have used no contractions or symbols of any kind, though we have introduced verbal alterations into the text wherever it appeared that con- ciseness or clearness would be gained. In Book II. abbreviated forms of constantly recurring words are used, and the phrases therefore and is equal to are replaced by the usual symbols. In the Third and following Books, and in additional matter throughout the whole, we have employed all such signs and abbreviations as we believe to add to the clear- ness of the reasoning, care being taken that the symbols chosen are compatible with a rigorous geometrical method, and are recognised by the majority of teachers. It must be understood that our use of symbols, and the removal of unnecessary verbiage and repetition, by no means implies a desire to secure brevity at all hazards. On the contrary, nothing appears to us more mischievous than an abridgement which is attained by omitting Viii PREFACE. steps, or condensing two or more steps into one. Such uses spring from the pressure of examinations ; but an examination is not, or ought not to be, a mere race; and while we wish to indicate generally in the later books how a geometrical argument may be abbreviated for the pur- poses of written work, we have not thought well to reduce the propositions to the bare skeleton so often presented to an Examiner. Indeed it does not follow that the form most suitable for the page of a text-book is also best adapted to examination purposes ; for the object to be attained in each case is entirely different. The text-book should present the argument in the clearest possible manner to the mind of a reader to whom it is new : the written proposition need only convey to the Examiner the assurance that the proposition has been thoroughly grasped and remembered by the pupil. From first to last \ve have kept in mind the undoubted fact that a very small proportion of those who study Ele- mentary Geometry, and study it with profit, are destined to become mathematicians in any real sense; and that to a large majority of students, Euclid is intended to serve not so much as a first lesson in mathematical reasoning, as the first, and sometimes the only, model of formal and rigid argument presented in an elementary education. This consideration has determined not only the full treatment of the earlier Books, but the retention of the formal, if some what, cumbrous, methods of Euclid in many places where proofs of greater brevity and mathematical elegance are available. We hope that the additional matter introduced into the book will provide sufficient exercise for pupils whose study of Euclid is preliminary to a mathematical edu- cation. PREFACE. IX The questions distributed through the text follow very easily from the propositions to which they are attached, and we think that teachers are likely to find in them all that is needed for an average pupil reading the subject for the first time. The Theorems and Examples at the end of each Book contain questions of a slightly more difficult type : they have been very carefully classified and arranged, and brought into close connection with typical examples worked out either partially or in full ; and it is hoped that this section of the book, on which much thought has been expended, will do something towards removing that extreme want of freedom in solving deductions that is so commonly found even among students who have a good knowledge of the text of Euclid, In the course of our work we have made ourselves acquainted with most modern English books on Euclidean Geometry : among these we have already expressed our special indebtedness to the text-book recently published by the Association for the Improvement of Geometrical Teach- ing; and we must also mention the Edition of Euclid's Ele- ments prepared by Dr. J. S. Mackay, whose historical notes and frequent references to original authorities have been of the utmost service to us. Our treatment of Maxima and Minima on page 239 is based upon suggestions derived from a discussion of the subject which took place at the annual meeting of the Geometrical Association in January 1887. Of the Riders and Deductions some are original; but the greater part have been drawn from that large store of floating material which has furnished Examination Papers for the last 30 years, and must necessarily form the basis of any elementary collection. Proofs which have been X PREFACE. found in two or more books without acknowledgement have been regarded as common property. As regards figures, in accordance with a usage not uncommon in recent editions of Euclid, we have made a distinction between given lines and lines of construction. Throughout the book we have italicised those deductions on which we desired to lay special stress as being in them- selves important geometrical results : this arrangement we think will be useful to teachers who have little time to devote to riders, or who wish to sketch out a suitable course for revision. We have in conclusion to tender our thanks to many of our friends for the valuable criticism and advice which we received from them as the book was passing through the press, and especially to the Rev. H. C. Watson, of Clifton College, who added to these services much kind assistance in the revision of proof-sheets. H. S. HALL, F. H. STEVENS. July, 1888. PREFACE TO THE SECOND EDITION. IN the Second Edition the text of Books I VI. has been revised ; and at the request of many teachers we have added the first twenty-one Propositions of Book XI. together with a collection of Theorems and Examples illustrating the elements of Solid Geometry. September, 1889. CONTENTS. BOOK I. DEFINITIONS, POSTULATES, AXIOMS SECTION I. PROPOSITIONS 1 26 SECTION II. PARALLELS AND PARALLELOGRAMS. PROPOSITIONS 27 34 50 SECTION III. THE AREAS OF PARALLELOGRAMS AND TRIANGLES. PROPOSITIONS 35 48 66 Theorems and Examples on Book I. ANALYSIS, SYNTHESIS 87 I. ON THE IDENTICAL EQUALITY OF TRIANGLES ... 90 II. ON INEQUALITIES 93 III. ON PARALLELS . 95 IV. ON PARALLELOGRAMS 96 V. MISCELLANEOUS THEOREMS AND EXAMPLES . . . 100 VI. ON THE CONCURRENCE OF STRAIGHT LINES IN A TRI- ANGLE 102 VII. ON THE CONSTRUCTION OF TRIANGLES WITH GIVEN PARTS 107 VIII. ON AREAS 109 IX. ON Loci 114 X. ON THE INTERSECTION OF Loci 117 Xll CONTENTS. BOOK II. PAGE DEFINITIONS, &c. ..... ... 120 PROPOSITIONS 1 14 122 THEOREMS AND EXAMPLES ON BOOK II. .... 144 BOOK III. DEFINITIONS, &c. . 149 PROPOSITIONS 1 37 153 NOTE ON THE METHOD OF LIMITS AS APPLIED TO TANGENCY . 213 Theorems and Examples on Book III. I. ON THE CENTRE AND CHORDS OF A CIRCLE . . . 215 II. ON THE TANGENT AND THE CONTACT OF CIRCLES. The Common Tangent to Two Circles, Problems on Tangency, Orthogonal Circles 217 III. ON ANGLES IN SEGMENTS, AND ANGLES AT THE CENTRES AND CIRCUMFERENCES OF CIRCLES. The Orthocentre of a Triangle, and properties of the Pedal Triangle, Loci, Simson's Line . . . 222 IV. ON THE CIRCLE IN CONNECTION WITH EECTANGLES. Further Problems on Tangency 233 V. ON MAXIMA AND MINIMA 239 VI. HARDER MISCELLANEOUS EXAMPLES 246 BOOK IV DEFINITIONS, &c. 250 PROPOSITIONS 1 1C 251 NOTE ON REGULAR POLYGONS 274 Theorems and Examples on Book IV I. ON THE TRIANGLE AND ITS CIRCLES. Circumscribed, Inscribed, and Escribed Circles, The Nine-points Circle 277 II. MISCELLANEOUS EXAMPLES . ... 283 CONTENTS. Xlll BOOK V. PAGE INTRODUCTORY 285 DEFINITIONS 286 SUMMARY, WITH ALGEBRAICAL PROOFS, OF THE PRINCIPAL THEOREMS OF BOOK V. 292 PROOFS OF THE PROPOSITIONS DERIVED FROM THE GEOMETRICAL DEFINITION OF PROPORTION . . . . . . 297 BOOK VI. DEFINITIONS 307 PROPOSITIONS 1 D 308 Theorems and Examples on Book VI. I. ON HARMONIC SECTION . 359 II. ON CENTRES OF SIMILARITY AND SIMILITUDE . . . 363 III. ON POLE AND POLAR 365 IV. ON THE EADICAL Axis OF Two OR MORE CIRCLES . . 371 V. ON TRANSVERSALS 374 VI. MISCELLANEOUS EXAMPLES ON BOOK VI 377 BOOK XI. DEFINITIONS 383 PROPOSITIONS 121 - . . .393 EXERCISES ON BOOK XI 418 THEOREMS AND EXAMPLES ON BOOK XI. . . 420 EUCLID'S ELEMENTS. BOOK I. DEFINITIONS. 1. A point is that which has position, but no mag- nitude. 2. A line is that which has length without breadth. The extremities of a line are points, and the intersection of two lines is a point. 3. A straight line is that which lies evenly between its extreme points. Any portion cut off from a straight line is called a segment of it. 4. A surface is that which has length and breadth, but no thickness. The boundaries of a surface are lines. 5. A plane surface is one in which any two points being taken, the straight line between them lies wholly in that surface. A plane surface is frequently referred to simply as a plane. NOTE. Euclid regards a point merely as a mark of position, and he therefore attaches to it no idea of size and shape. Similarly he considers that the properties of a line arise only from its length and position, without reference to that minute breadth which every line must really have if actually drawn, even though the most peri'ect instruments are used. The definition of a surface is to be understood in a similar way. H. E. 1 2 EUCLID'S ELEMENTS. 6. A plane angle is the inclination of two straight lines to one another, which meet together, but are not in the same straight line. The point at which the straight lines meet is called the vertex of the angle, and the straight lines themselves the arms of the angle. When several angles are at one point O, any one of them is expressed by three letters, of which the letter that refers to the vertex is put between the other two. Thus if the straight lines OA, OB, OC meet at the point O, the angle contained by the straight lines OA, OB is named the angle AOB or BOA ; and the angle contained by OA, OC is named the angle AOC or COA. Similarly the angle con- tained by OB, OC is referred to as the angle BOC or COB. But if there be only one angle at a point, it may be expressed by a single letter, as the angle at O. Of the two straight lines OB, OC shewn in the adjoining figure, we recognize that OC is more in- clined than OB to the straight line OA : this we express by saying that the angle AOC is greater than the angle AOB. Thus an angle must be regarded as having magnitude. It should be observed that the angle AOC is the sum of the angles AOB and BOC; and that AOB is the difference of the angles AOC and BOC. The beginner is cautioned against supposing that the size of an angle is altered either by increasing or diminishing the length of its [Another view of an angle is recognized in many branches of mathematics ; and though not employed by Euclid, it is here given because it furnishes .more clearly than any other a conception of what is meant by the magnitude of an angle. Suppose that the straight line OP in the figure is capable of revolution about the point O, like the hand of a watch, but in the opposite direction ; and suppose that in this way it has passed successively from the position OA to the positions occupied by OB and OC. . Such a line must have undergone more turning in passing from OA to OC, than in passing from OA to OB ; and consequently the angle AOC is said to be greater than the angle AOB.] DEFINITIONS. 7. When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the an- gles is called a right angle ; and the straight line which stands on the other is called a perpendicular to it. 8. An obtuse angle is an angle which is greater than one right angle, but less than two right angles. 9. An acute angle is an angle which is less than a right angle. (in [In the adjoining figure the straight line OB may be supposed to have arrived at its present position, from the position occu- pied by OA, by revolution about the point O in either of the two directions indicated by the arrows : thus two straight lines drawn from a point may be considered as forming two angles, (marked (i) and (ii) in the figure) of which the greater (ii) is said to be reflex. If the arms OA, OB are in the same i-^i straight line, the angle formed by them BOA on either side is called a straight angle.] 10. Any portion of a plane surface bounded by one or more lines, straight or curved, is called a plane figure- The sum of the bounding lines is called the perimeter of the figure. Two figures are said to be equal in area, when they enclose equal portions of a plane surface. 11. A circle is a plane figure contained by one line, which is called the circum- ference, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal to one another : this point is called the centre of the circle. A radius of a circle is a straight line drawn from the centre to the circumference. 4 KUCT.in's ELEMENTS. 12. A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference. 13. A semicircle is the figure bounded by a diameter of a circle and the part of the circumference cut off by the diameter. 14. A segment of a circle is the figure bounded by a straight line and the part of the circumference which it cuts off. 15. Rectilineal figures are those which are bounded by straight lines. 16. A triangle is a plane figure bounded by three straight lines. Any one of the angular points of a triangle may be regarded as its vertex; and the opposite side is then called the base. 17. A quadrilateral is a plane figure bounded by four straight lines. The straight line which joins opposite angular points in a quadri- lateral is called a diagonal. 18. A polygon is a plane figure bounded by more than four straight lines. 19. An equilateral triangle is a triangle whose three sides are equal. 20. An isosceles triangle is a triangle two of whose sides are equal. 21. A scalene triangle is a triangle which has three unequal sides. DF.FIXITIONS. 22. A right-angled triangle is a triangle which lias a right an. are appended to a theorem, and stand for Quod erat Demonstrandum, which was to be proved. A Corollary is a statement the truth of which follows readily from an established proposition ; it is therefore appended to the proposition as an inference or deduction, which usually requires no further proof. The following symbols and abbreviations may be employed in writing out the propositions of Book I., though their use is not recommended to beginners. " for therefore, par 1 (or ||) for parallel, = is, or are, equal to, par m parallelogram, angle, sq. square, rt. L right angle, rectil. rectilineal, A triangle, st. line straight line, perp. perpendicular, pt. point; and all obvious contractions of words, such as opp., adj., diag., &c., for opposite, adjacent, diagonal, &c. BOOK I. PROP. 1. 11 SECTION I. PROPOSITION 1. PROBLEM. To describe an equilateral triangle on a yiven Jirtite straight line. Let AB be the given straight line. It is required to describe an equilateral triangle on AB. Construction. From centre A, with radius AB, describe the circle BCD. Post. 3. From centre B, with radius BA, describe the circle ACE. Post. 3. From the point C at which the circles cut one another, draw the straight lines CA and CB to the points A and B. Post. 1. Then shall ABC be an equilateral triangle. Proof. Because A is the centre of the circle BCD, therefore AC is equal to AB. Def. 11. And because B is the centre of the circle ACE, therefore BC is equal to BA. Def. 11. But it has been shewn that AC is equal to AB ; therefore AC and BC are each equal to AB. But things which are equal to the same thing are equal to one another. Ax. 1. Therefore AC is equal to BC. Therefore CA, AB, BC are equal to one another. Therefore the triangle ABC is equilateral ; and it is described on the given straight line AB. Q.E. F. 12 EUCLID'S KLKMKNTS. PROPOSITION 2. PROBI.K.M. From a given point to draw a strai-jht line equal to a given straight Let A be the given point, and BC the given straight Hne. It is required to draw from the point A a straight line equal to BC. Construction. JoinAB; Post. 1. and on AB describe an equilateral triangle DAB. I. 1. From centre B, with radius BC, describe the circle CGH. Post. 3. Produce DB to meet the circle CGH at G. Post. 2. From centre D, with radius DG, describe the circle GKF. Produce DA to meet the circle GKF at F. Post. 2. Then AF shall be equal to BC. Proof. Because B is the centre of the circle CGH, therefore BC is equal to BG. Def. 11. And because D is the centre of the circle GKF, therefore DF is equal to DG ; Def. 11. and DA, DB, parts of them are equal; Def. 19. therefore the remainder AF is equal to the remainder BG. Ax. 3. And it has been shewn that BC is equal to BG : therefore AF and BC are each equal to BG. But things which are equal to the same thing are equal to one another. Ax. 1. Therefore AF is equal to BC ; and it has been drawn from the given point A. Q. E. F. [This Proposition is rendered necessary by the restriction, tacitly imposed l>y Euclid, that compasses shall not be used to transfei distances.] BOOK I. PROP. 3. PROPOSITION 3. PROBLEM. From the greater of two given straight lint's to cut off a part equal to the less. Let AB and C be the two given straight lines, of which AB is the greater. It is required to cut off from AB a part equal to C. Construction. From the point A draw the straight line AD equal to C ; I. 2. and from centre A, with radius AD, describe the circle DEF, meeting AB at E. Post. 3. Then AE shall be equal to C. Proof. Because A is the centre of the circle DEF, therefore AE is equal to AD. Def. 11. But C is equal to AD. Constr. Therefore AE and C are each equal to AD. Therefore AE is equal to C ; and it has been cut off from the given straight line AB. Q.E.F EXERCISES. 1. On a given straight line describe an isosceles triangle having each of the equal sides equal to a given straight line. 2. On a given base describe an isosceles triangle having each of the equal sides double of the base. 3. In the figure of i. 2, if AB is equal to BC, shew that D, the vertex of the equilateral triangle, will fall on the circumference of the circle CGH. 14 EUCLID'S ELEMENTS. Obs. Every triangle has six parts, namely its three sides and three angles. Two triangles are said to be equal in all respects, when they can be made to coincide with one another by superposition (see note on Axiom 8), and in this case each part of the one is equal to a corresponding part of the other. PROPOSITION 4. THEOREM. If two triangles have two sides of the one equal to two sides of t/ie other, each to each, and have also the angles contained by those sides equal; then shall their bases or third sides be equal, and the triangles shall be equal in area, and their remaining angles shall be equal, each to each, namely those to which the equal sides are opposite : that i* to say, the triangles sJiall be equal in all respects. D Let ABC, DEF be two triangles, which have the side AB equal to the side DE, the side AC equal to the side DF, and the contained angle BAC equal to the contained angle EOF. Then shall the base BC be equal to the base EF, and the triangle ABC shall be equal to the triangle DEF in area ; and the remaining angles shall be equal, each to each, to which the equal sides are opposite, namely the angle ABC to the angle DEF, and the angle ACB to the angle DFE. For if the triangle ABC be applied to the triangle DEF, so that the point A may be on the point D, and the straight line AB along the straight line DE, then because AB is equal to DE, therefore the point B must coincide with the point E. BOOK I. PROP. 4. 15 And because AB falls along DE, and the angle BAG is equal to the angle EOF, Hyp. therefore AC must fall along DF. And because AC is equal to DF, Hyp. therefore the point C must coincide with the point F. Then B coinciding with E, and C with F, the base BC must coincide with the base EF; for if not, two straight lines would enclose a space ; which is impossible. Ax. 10. Thus the base BC coincides with the base EF, and is therefore equal to it. Ax. 8. And the triangle ABC coincides with the triangle DEF, and is therefore equal to it in area. Ax. 8. And the remaining angles of the one coincide with the re- maining angles of the other, and are therefore equal to them, namely, the angle ABC to the angle DEF, and the angle ACB to the angle DFE. That is, the triangles are equal in all respects. Q. E.D. NOTE. It follows that two triangles which are equal in their several parts are equal also in area; but it should be observed that equality of area in two triangles does not necessarily imply equality in their several parts: that is to say, triangles may be equal in area, without being of the same sluipe. Two triangles which are equal in all respects have identity of form and magnitude, and are therefore said to be identically equal, or congruent. The following application of Proposition 4 anticipates the chief difficulty of Proposition 5. . In the equal sides AB, AC of an isosceles triangle ABC, the points X and Y are taken, so that AX is equal to AY ; and BY and CX are joined. Shew that BY is equal to CX. In the two triangles XAC, YAB, XA is equal to YA, and AC is equal to AB ; Hyp. that is, the two sides XA, AC are equal to the two sides YA, AB, each to each; and the angle at A, which is contained by these B C sides, is common to both triangles : therefore the triangles are equal in all respects; i.4. so that XC is equal to YB. Q. E.D. 16 EUCLID'S ELEMENTS. PROPOSITION 5. THEOREM. T/ie a/tyles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced, the angles on the other side of the base shall also be equal to one another. Let ABC be an isosceles triangle, having the side AB equal to the side AC, and let the straight lines AB, AC be produced to D and E : then shall the angle ABC be equal to the angle ACB, and the angle. CBD to the angle BCE. Construction. In BD take any point F ; and from AE the greater cut off AG equal to AF the less. I. 3. Join FC, GB. Proof. Then in the triangles FAC, GAB, {FA is equal to GA, Constr. and AC is equal to AB, Ihin. i,, j. j i . li also the contained angle at A is common to the two triangles ; therefore the triangle FAC is equal to the triangle GAB in all respects ; i. 4. that is, the base FC is equal to the base GB, and the angle ACF is equal to the angle ABG, also the angle AFC is equal to the angle AGB. Again, because the whole AF is equal to the whole AG, of which the parts AB, AC are equal, Hf/P- therefore the remainder BF is equal to the remainder CG. BOOK I. PROP. 5. 17 Then in the two triangles BFC, CGB, ( BF is equal to CG, Proved. -D and FC is equal to GB, Proved. j also the contained angle BFC is equal to the { contained angle CGB, Proved. therefore the triangles BFC, CGB are equal in all respects; so that the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. I. 4. Now it lias been shewn that the whole angle ABG is equal to the whole angle ACF, and that parts of these, namely the angles CBG, BCF, are also equal ; therefore the remaining angle ABC is equal to the remain- ing angle ACB ; and these are the angles at the base of the triangle ABC. Also it has been shewn that the angle FBC is equal to the angle GCB ; and these are the angles on the other side of the base. Q.E.D. COROLLARY. Jfence if a triangle is equilateral it is also equiangular. EXERCISES. 1. AB is a given straight line and C a given point outside it : shew how to find any points in AB such that their distance from C shall be equal to a given length L. Can such points always he found 1 2. If the vertex C and one extremity A of the base of an isosceles triangle are given, find the other extremity B, supposing it to lie on a given straight line PQ. 3. Describe a rhombus having given two opposite angular points A and C, and the length of each side. 4. AMNB is a straight line ; on AB describe a triangle ABC such that the side AC shall be equal to AN and the side BC to MB. 5. In Prop. 2 the point A may be joined to either extremity of BC. Draw the figure and prove the proposition in the case when A is joined to C. H. E. 2 1?A EUCLID'S ELEMENTS. The following proof is sometimes given as a substitute for the first part of Proposition 5 : PROPOSITION 5. ALTERNATIVE PROOF. A A B C qual to one another. If it be possible, on the same base AB, and on the same side of it, let there be two triangles ACB, ADB, having their sides AC, AD, which are terminated at A, equal to one another, and likewise their sides BC, BD, which are termi- nated at B, equal to one another. CASE I. When the vertex of each triangle is without the other triangle. Construction. Join CD. . I'ost. 1. Proof. Then in the triangle ACD, because AC is equal to AD, Hyp- therefore the angle ACD is equal to the angle ADC. I. 5. But the whole angle ACD is greater than its part, the angle BCD, therefore also the angle ADC is greater than the angle BCD ; still more then is the angle BDC greater than the angle BCD. Again, in the triangle BCD, because BC is equal to BD, ffyp- therefore the angle BDC is equal to the angle BCD: I. 5 but it was shewn to be greater ; which is impossible. 20 EUCLID'S ELEMENTS. CASE II. When one of the vertices, as D, is within the other triangle ACB. A B Construction. As before, join CD ; Post. 1. and produce AC, AD to E and F. Post. 2. Then in the triangle ACD, because AC is equal to AD, Hyp. therefore the angles ECD, FDC, on the other side of the base, are equal to one another. I. 5. But the angle ECD is greater than its part, the angle BCD; therefore the angle FDC is also greater than the angle BCD: still more then is the angle BDC greater than the angle BCD. Again, in the triangle BCD, because BC is equal to BD, Hyp. therefore the angle BDC is equal to the angle BCD : I. 5. but it has been shewn to be greater ; which is impossible. The case in which the vertex of one triangle is on a side of the other needs no demonstration. Therefore AC cannot be equal to AD, and at tlw same time, BC equal to BD. Q.E.D. NOTE. The sides AC, AD are called conterminous sides ; similarly the sides BC, BD are conterminous. PROPOSITION 8. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal, then the angle which is contained by the two sides of the one shall be equal to the angle which is contained by the two sides of the other. 21 Let ABC, DEF be two triangles, having the two sides BA, AC equal to the two sides ED, DF, each to each, namely BA to ED, and AC to DF, and also the base BC equal to the base EF: then shall the angle BAC be equal to the angle EOF. Proof. For if the triangle ABC be applied to the triangle DEF, so that the point B may be on E, and the straight line BC along EF ; then because BC is equal to EF, Hyp. therefore the point C must coincide with the point F. Then, BC coinciding with EF, it follows that BA and AC must coincide with ED and DF : for if not, they would have a different situation, as EG, GF: then, on the same base and on the same side of it there would be two triangles having their conterminous sides equal. But this is impossible. i. 7. Therefore the sides BA, AC coincide with the sides ED, DF. That is, the angle BAC coincides with the angle EOF, and is therefore equal to it. Ax. 8. Q. K. D. NOTE. In this Proposition the three sides of one triangle are given equal respectively to the three sides of the other; and from this it is shewn that the two triangles may be made to coincide with one another- Hence we are led to the following important Corollary. COROLLARY. If in two triangles the three sides of the one are equal to the three sides of the other, each to each, then the triangles are equal in all respects. 22 EUCLID'S ELKMENT.S. The following proof of Prop. 8 is worthy of attention as it irf inde- pendent of Prop. 7, which frequently presents difficulty to a beginner. PROPOSITION 8. ALTERNATIVE PROOF. A D Let ABC and DEF be two triangles, which have the sides BA, AC equal respectively to the sides ED, DF, and the base BC equal to the base EF : then shall the angle BAG be equal to the angle EOF. For apply the triangle ABC to the triangle DEF, so that B may fall on E, and BC along EF, and so that the point A may be on the side of EF remote from D, then C must fall on F, since BC is equal to EF. Let A'EF be the new position of the triangle ABC. If neither DF, FA' nor DE, EA' are in one straight line, join DA'. CASE I. When DA' intersects EF. Then because ED is equal to EA', therefore the angle EDA' is equal to the angle EA'D. Again because FD is equal to FA', therefore the angle FDA' is equal to the angle FA'D. Hence the whole angle EDF is equal to the whole anle EA'F; that is, the angle EDF is equal to the angle BAG. i. 5. i. 5. Two cases remain which may he dealt with in a similar manner: namely, CASE II. When DA' meets EF produced. CASE III. When one pair of sides, as DF, FA', are in one straight line. BOOK I. PROP. 9. 23 PROPOSITION 9. PROBLEM. To Insect a given angle, that is, to divide it into two equal parts. A Let BAC be the given angle: it is required to bisect it. Construction. In AB take any point D; and from AC cut off AE equal to AD. I. .'1 Join DE; and on DE, on the side remote from A, describe an equi- lateral triangle DEF. I. 1. Join AF. Then shall the straight line AF bisect the angle BAC. Proof. For in the two triangles DAF, EAF, {DA is equal to EA, Conatr. and AF is common to both; and the third side DF is equal to the third side EF; DC/. 19. therefore the angle DAF is equal to the angle EAF. I. 8. Therefore the given angle BAC is bisected by the straight line AF. Q.E.F. EXERCISES. 1. If in the above figure the equilateral triangle DFE were de- scribed on the same side of DE as A, what different cases would arise? And under what circumstances would the construction fail? 2. In the same figure, shew that AF also bisects the angle DFE. 3. Divide an angle into four equal parts. 24 EUCLID'S ELEMENTS. PROPOSITION 10. PROBLEM. To bisect a given finite straight line, that is, to divide it into two equal parts. A D B Let AB be the given straight line : it is required to divide it into two equal parts. Constr. On AB describe an equilateral triangle ABC, i. 1. and bisect the angle ACB by the straight line CD, meeting AB at D. I. 9. Then shall AB be bisected at the point D. Proof. For in the triangles ACD, BCD, e AC is equal to BC, Def. 19. -p. and CD is common to both; jalso the contained angle ACD is equal to the con- { tained angle BCD; Constr. Therefore the triangles are equal in all respects: so that the base AD is equal to the base BD. i. 4. Therefore the straight line AB is bisected at the point D. Q. E. F. EXERCISES. 1. Shew that the straight line which bisects the vertical angle of an isosceles triangle, also bisects the base. 2. On a given base describe an isosceles triangle such that the sum of its equal sides may be equal to a given straight line. BOOK I. PROP. 11. 25 PROPOSITION 11. PROBLEM. To draw a straight line at right angles to a given straight line, from a given point in the same. AD C E B Let AB be the given straight line, and C the given point in it. It is required to draw from the point C a straight line at right angles to AB. Construction. In AC take any point D, and from CB cut off CE equal to CD. i. 3. On DE describe the equilateral triangle DFE. I. 1. Join CF. Then shall the straight line CF be at right angles to AB. Proof. For in the triangles DCF, ECF, ( DC is equal to EC, Constr. p and CF is common to both ; land the third side DF is equal to the third side ( EF: Def. 19. Therefore the angle DCF is equal to the angle ECF: I. 8. and these are adjacent angles. But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of these angles is called a right angle; Def. 1. therefore each of the angles DCF, ECF is a right angle. Therefore CF is at right angles to AB, and has been drawn from a point C in it. Q.E.F. EXERCISE. In the figure of the above proposition, shew that any point in FC, or FC produced, is equidistant from D and E. 26 EUCLID'S ELEMENTS. PHOPOSITIOX 12. PROBLEM. To draw a straight line perpendicular to a given straight line of unlimited length, from a given point without it. C /G B Let AB be the given straight line, which may be pro- duced in either direction, and let C be the given point with- out it. It is required to draw from the point C a straight line perpendicular to AB. Construction. On the side of AB remote from C take any point D; and from centre C, with radius CD, describe the circle FDG, meeting AB at F and G. Post. 3. Bisect FG at H ; I. 10. and join CH. Then shall the straight line CH be perpendicular to AB. Join CF and CG. Proof. Then in the triangles FHC, GHC, ( FH is equal to GH, Constr. TV and HC is common to both; land the third side CF is equal to the third side I CG, being radii of the circle FDG ; Def. 11. therefore the angle CHF is equal to the angle CHG; i. b. and these are adjacent angles. But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of these angles is called a right angle, and the straight line which stands on the other is called a perpendicular to it. Therefore CH is a perpendicular drawn to the given straight line AB from the given point C without it. Q. E. F. NOTE. The given straight line AB must be of unlimited length, that is, it must be capable of production to an indefinite length in either direction, to ensure its being intersected in two points by the circle FDG. BOOK I. PROP. 12. 27 EXERCISES ON' PROPOSITIONS 1 TO 12. 1. Shew that the straight line which joins the vertex of an isosceles triangle to the middle point of the base is perpendicular to the base. 2. Shew that the straight lines which join the extremities of the base of an isosceles triangle to the middle points of the opposite sides, are equal to one another. 3. Two given points in the base of an isosceles triangle are equi- distant from the extremities of the base : shew that they are also equi- distant from the vertex. 4. If the opposite sides of a quadrilateral are equal, shew that the opposite angles are also equal. 5. Any two isosceles triangles XAB, YAB stand on the same base AB: shew that the angle XAY is equal to the angle XBY; and that the angle AXY is equal to the angle BXY. 6. Shew that the opposite angles of a rhombus are bisected by the diagonal which joins them. 7. Shew that the straight lines which bisect the base angles of an isosceles triangle form with the base a triangle which is also isosceles. 8. ABC is an isosceles triangle having AB equal to AC; and the angles at B and C are bisected by straight lines which meet at O : shew that OA bisects the angle BAG. 9. Shew that the triangle formed by joining the middle points of the sides of an equilateral triangle is also equilateral. 10. The equal sides BA, CA of an isosceles triangle BAG are pro- duced beyond the vertex A to the points E and F, so that AE is equal to AF; and FB, EC are joined: shew that FB is equal to EC. 11. Shew that the diagonals of a rhombus bisect one another at right angles. 12. In the equal sides AB, AC of an isosceles triangle ABC two points X and Y are taken, so that AX is equal to AY; and CX and BY are drawn intersecting in O : shew that (i) the triangle BOG is isosceles; (ii) AO bisects the vertical angle BAG ; (iii) AO, if produced, bisects BC at right angles. 13. Describe an isosceles triangle, having given the base and the length of the perpendicular drawn from the vertex to the base. 14. In a given straight line find a point that is equidistant from two given points. In what case is this impossible ? 28 EUCLID'S ELEMENTS. PROPOSITION 13. THEOREM. If one straight line stand upon anotlier straight line, then the adjacent angles shall be either two right angles, or together equal to two right angles. B B I. 11. the two Let the straight line AB stand upon the straight line DC : then the adjacent angles DBA, ABC shall be either two right angles, or together equal to two right angles. ' CASE I. For if the angle DBA is equal to the angle ABC, each of them is a right angle. Def. 1 . CASE II. But if the angle DBA is not equal to the angle ABC, from B draw BE at right angles to CD. Proof. Now the angle DBA is made up of angles DBE, EBA; to each of these equals add the angle ABC; then the two angles DBA, ABC are together equal to the three angles DBE, EBA, ABC. Ax. 2. Again, the angle EBC is made up of the two angles EBA, ABC; to each of these equals add the angle DBE. Then the two angles DBE, EBC are together equal to the three angles DBE, EBA, ABC. Ax. '2. But the two angles DBA, ABC have been shewn to be equal to the same three angles ; therefore the angles DBA, ABC are together equal to the angles DBE, EBC. Ax. 1. But the angles DBE, EBC are two right angles; Constr. therefore the angles DBA, ABC are together equal to two right angles. Q. E. D. BOOK I. PROP. 13. 29 DEFINITIONS. (i) The complement of an acute angle is its defect from a right angle, that is, the angle by which it falls short of a right angle. Thus two angles are complementary, when their sum is a right angle. (ii) The supplement of an angle is its defect from two right angles, that is, the angle by which it falls short of two right angles. Thus two angles are supplementary, when their sum is two right angles. COROLLARY. Angles which are complementary or supple- mentary to the same angle are equal to one another. EXERCISES. 1. If the two exterior angles formed by producing a side of a tri- angle both ways are equal, shew that the triangle is isosceles. 2. The bisectors of the adjacent angles which one. straight line makes with another contain a right angle. NOTE. In the adjoining figure AOB y is a given angle ; and one of its arms AO \ is produced to C : the adjacent angles \ AOB, BOG are bisected by OX, OY. Then OX and OY are called respect- ively the internal and external bisectors of the angle AOB. C O A, Hence Exercise 2 may be thus enunciated : The internal and external bisectors of an angle are at right angles to one another. 3. Shew that the angles AOX and COY are complementary. 4. Shew that the angles BOX and COX are supplementary; and also that the angles AOY and BOY are supplementary. 30 EUCLID'S ELEMENTS. PROPOSITION 14. THKOREM. If , at a point in a straight line, two other straight lines, on opposite sides of it, make the adjacent angles together equal to two right angles, then these two straight linen it/tall be in one and the same straight liv. C B D . At the point B in the straight line AB, let the two straight lines BC, BD, on the opposite sides of AB, make the adjacent angles ABC, ABD together equal to two right angles : then BD shall be in the same straight line with BC. Proof. For if BD be not in the same straight line with BC, if possible, let BE be in the same straight line with BC. Then because AB meets the straight line CBE, therefore the adjacent angles CBA, ABE are together equal to two right angles. I. 13. But the angles CBA, ABD are also together equal to two right angles. H'JP- Therefore the .angles CBA, ABE are together equal to the angles CBA, ABD. Ax. 11. From each of these equals take the common angle CBA; then the remaining angle ABE is equal to the remaining angle ABD; the part equal to the whole; which is impossible. Therefore BE is not in the same straight line with BC. And in the same way it may be shewn that no other line but BD can be in the same straight line with BC. Therefore BD is in the same straight line with BC. Q.E.I). EXERCISE. ABCD is a rhombus; and the diagonal AC is bisected at O. If O is joined to the angular points B and D; shew that OB and OD are in one straight line. BOOK I. PROP. 15. 31 Obs. When two straight lines intersect at a point, four angles are formed; and any two of these angles which are not adjacent, are said to be vertically opposite to one another. PROPOSITION 15. THEOREM. Jftu:o straight lines intersect one anotlier, then the vertically opposite angles shall be equal. A Let the two straight lines AB, CD cut one another at the point E: then shall the angle AEC be equal to the angle DEB, and the angle CEB to the angle AED Proof. Because AE makes with CD the adjacent angles CEA, AED, therefore these angles are together equal to two right angles. I. 13. Again, because DE makes with AB the adjacent angles AED, DEB, therefore these also are together equal to two right angles. Therefore the angles CEA, AED are together equal to the angles AED, DEB. From each of these equals take the common angle AED; then the remaining angle CEA is equal to the remaining angle DEB. Ax. 3. In a similar way it may be shewn that the angle CEB is equal to the angle AED. Q. E. r>. COROLLARY 1. From this it is manifest t/tat, if two straight lines cut one another, the angles which they make at the point where they cut, are together equal to four right angles. COROLLARY 2. Consequently, when any number of straight lines meet at a point, the sum of the angles made by con- secutive lines is equal to four right angles. 11. E. 3 32 EUCLID'S ELEMENTS. PROPOSITION 16. THEOREM. If one side of a, triangle be produced, then the exterior angle shall be greater than either of the interior opposite angles. Because Let ABC be a triangle, and let one side BC be produced to D : then shall the exterior angle ACD be greater than either of the interior opposite angles CBA, BAG. Construction. Bisect AC at E : I. 10. Join BE ; and produce it to F, making EF equal to BE. i. 3. Join FC. Proof. Then in the triangles AEB, CEF, AE is equal to CE, Constr. and EB to EF ; Constr. also the angle AEB is equal to the vertically opposite angle CEF; I. 15. therefore the triangle AEB is equal to the triangle CEF in all respects : I. 4 . so that the angle BAE is equal to the angle ECF. But the angle ECD is greater than its part, the angle ECF: therefore the angle ECD is greater than the angle BAE; that is, the angle ACD is greater than the angle BAC. In a similar way, if BC be bisected, and the side AC produced to G, it may be shewn that the angle BCG is greater than the angle ABC. But the angle BCG is equal to the angle ACD: I. If), therefore also the angle ACD is greater than the angle ABC. Q. E. D. BOOK I. PROP. 17. 33 PROPOSITION 17. THEOREM. Any two angles of a triangle are together less than two right angles. A Let ABC be a triangle: then shall any two of its angles, as ABC, ACB, be together less than two right angles. Construction. Produce the side BC to D. Proof. Then because ACD is an exterior angle of the triangle ABC, therefore it is greater than the interior opposite angle ABC. i. 16. To each of these add the angle ACB : then the angles ACD, ACB are together greater than the angles ABC, ACB. Ax. 4. But the adjacent angles ACD, ACB are together equal to two right angles. 1.13. Therefore the angles ABC, ACB are together less than two right angles. Similarly it may be shewn that the angles BAC, ACB, as also the angles CAB, ABC, are together less than two right angles. Q. E. D. NOTE. It follows from this Proposition that every triangle must have at least tico acute angles: for if one angle is obtuse, or a right angle, each of the other angles must be less than a right angle. EXERCISES. 1. Enunciate this Proposition so as to shew that it is the converse of Axiom 12. 2. If any side of a triangle is produced both ways, the exterior angles so formed are together greater than two right angles. 3. Shew how a proof of Proposition 17 may be obtained by joining each vertex, in turn to any point in the opposite side. 32 34 EUCLID'S ELEMENTS. PROPOSITION 18. THEOREM. If one side of a triangle be greater titan another, then the angle opposite to the greater side sltall be greater than the angle opposite to the less. B . C Let ABC be a triangle, in which the side AC is greater than the side AB : then shall the angle ABC be greater than the angle ACB. Construction. From AC, the greater, cut off a part AD equal to AB. I. 3. Join BD. Proof. Then in the triangle ABD, because AB is equal to AD, therefore the angle ABD is equal to the angle ADB. I. 5. But the exterior angle ADB of the triangle BDC is greater than the interior opposite angle DCB, that is, greater than the angle ACB. I. 16. Therefore also the angle ABD is greater than the angle ACB; still more then is the angle ABC greater than the angle ACB. Q.E.D. Euclid enunciated Proposition 18 as follows: The greater side of every triangle has the greater angle opposite to it. [This form of enunciation is found to be a common source of diffi- culty with beginners, who fail to distinguish what is assumed in it and what is to be proved.] [For Exercises see page 38.] BOOK. I. PROF. 19. 3ft PROPOSITION 19. THEOREM, If one angle of a triangle be greater than another, tJien the side opposite to the greater angle shall be greater than the side opposite to the less. Let ABC be a triangle in which the angle ABC is greater than the angle ACB : then shall the side AC be greater than the side AB. Proof. For if AC be not greater than AB, it must be either equal to, or less than AB. But AC is not equal to AB, for then the angle ABC would be equal to the angle ACB ; i. 5. but it is not. Hyp. Neither is AC less than AB ; for then the angle ABC would be less than the angle ACB ; 1.18. but it is not : Hyp. Therefore AC is neither equal to, nor less than AB. That is, AC is greater than AB. Q.E.D. NOTE. The mode of demonstration used in this Proposition is known as the Proof by Exhaustion. It is applicable to cases in which one of certain mutually exclusive suppositions must necessarily be true; and it consists in shewing the falsity of each of these supposi- tions in turn with one exception: hence the truth of the remaining supposition is inferred. Euclid enunciated Proposition 19 as follows : The greater angle of every triangle is subtended by the greater side, or, has the greater side opposite to it. [For Exercises see page 38.] 36 EUCLID S ELEMENTS. PROPOSITION 20. THEOREM. Any two sides of a triangle are together greater tJmn the third side. B C Let ABC be a triangle: then shall any two of its sides be together greater than the third side : namely, BA, AC, shall be greater than CB ; AC, CB greater than BA; and CB, BA greater than AC. Construction. Produce BA to the point D, making AD equal to AC. I. 3. Join DC. Proof. Then in the triangle ADC, because AD is equal to AC, Constr. therefore the angle ACD is equal to the angle ADC. i. 5. But the angle BCD is greater than the angle ACD ; Ax. 9. therefore also the angle BCD is greater than the angle ADC, that is, than the angle BDC. And in the triangle BCD, because the angle BCD is greater than the angle BDC, Pr. therefore the side BD is greater than the side CB. i. 19. But BA and AC are together equal to BD ; therefore BA and AC are together greater than CB. Similarly it may be shewn that AC, CB are together greater than BA ; and CB, BA are together greater than AC. Q. E. D. [For Exercises see page 38.J BOOK I. PROP. 21. 37 PROPOSITION 21. THEOREM. If from the ends of a side of a triangle, there be drawn two straight lines to a point within the triangle, then these straight lines shall be less than the other two sides of the triangle, but shall contain a greater angle. A B C Let ABC be a triangle, and from B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to a point D within the triangle : then (i) BD and DC shall be together less than BA and AC ; (ii) the angle BDC shall be greater than the angle BAC. Construction. Produce BD to meet AC in E. Proof, (i) In the triangle BAE, the two sides BA, AE are together greater than the third side BE : I. 20. to each of these add EC : then BA, AC are together greater than BE, EC. Ax. 4. Again, in the triangle DEC, the two sides DE, EC are to- gether greater than DC : I. 20. to each of these add BD ; then BE, EC are together greater than BD, DC. But it has been shewn that BA, AC are together greater than BE, EC : still more then are BA, AC greater than BD, DC. (ii) Again, the exterior angle BDC of the triangle DEC is greater than the interior opposite angle DEC ; I. 16. and the exterior angle DEC of the triangle BAE is greater than the interior opposite angle BAE, that is, than the angle BAC ; I. 16. still more then is the angle BDC greater than the angle BAC. Q.E.D. 38 EUCLID'S ELEMENTS. EXERCISES OK PROPOSITIONS 18 AND 19. 1. The hypotenuse is the greatest side of a right-angled triangle. 2. If two angles of a triangle are equal to one another, the sides also, which subtend the equal angles, are equal to one another. Prop. 6. Prove this indirectly by using the result of Prop. 18. 3. BC, the base of an isosceles triangle ABC, is produced to any point D ; shew that AD is greater than either of the equal sides. 4. If in a quadrilateral the greatest and least sides are opposite to one another, then each of the angles adjacent to the least side is greater than its opposite angle. 5. In a triangle ABC, if AC is not greater than AB, shew that any straight line drawn through the vertex A and terminated by the base BC, is less than AB. 6. ABC is a triangle, in which OB, OC bisect the angles ABC, ACB respectively: shew that, if A B is greater than AC, then OB is greater than OC. ON PROPOSITION 20. 7. The difference of any two sides of a triangle is less than the third side. 8. In a quadrilateral, if two opposite sides which are not parallel are produced to meet one another; shew that the perimeter of the greater of the two triangles so formed is greater than the perimeter of the quadrilateral. 9. The sum of the distances of any point from the three angular points of a triangle is greater than half its perimeter. 10. The perimeter of a quadrilateral is greater than the sum of its diagonals. 11. Obtain a proof of Proposition 20 by bisecting an angle by a straight line which meets the opposite side. ON PROPOSITION 21. 12. In Proposition 21 shew that the angle BDC is greater than the angle BAG by joining AD, and producing it towards the base. 13. The sum of the distances of any point within a triangle from its angular points is less than the perimeter of the triangle. BOOK. I. PROP. 22. 39 PROPOSITION' 22. PROBLEM. To describe a triangle having its sides equal to three given straight lines, any two of which are together greater than the third. Let A, B, C be the three given straight lines, of which any two are together greater than the third. It is required to describe a triangle of which the sides shall be equal to A, B, C. Construction. Take a straight line DE terminated at the point D, but unlimited towards E. Make DF equal to A, FG equal to B, and GH equal to C. I. 3. From centre F, with radius FD, describe the circle DLK. From centre G with radius GH, describe the circle MHK, cutting the former circle at K. Join FK, GK. Then shall the triangle KFG have its sides equal to the three straight lines A, B, C. Proof. Because F is the centre of the circle DLK, therefore FK is equal to FD : Def. 11. but FD is equal to A ; Constr. therefore also FK is equal to A. Ax. 1. Again, because G is the centre of the circle MHK, therefore G K is equal to G H : J)ef. 1 1 . but G H is equal to C ; Constr. therefore also GK is equal to C. Ax. 1. And FG is equal to B. Constr. Therefore the triangle KFG has its sides KF, FG, GK equal respectively to the three given lines A, B, C. Q.E.F. 40 EUCLID'S ELEMENTS. EXERCISE. On a given base describe a triangle, whose remaining sides shall be equal to two given straight lines. Point out how the construction fails, if any one of the three given lines is greater than the sum of the other two. PROPOSITION 23. PROBLEM. At a given point in a given straight line, to make an angle equal to a given angle. Let AB be the given straight line, and A the given point in it ; and let DCE be the given angle. It is required to draw from A a straight line making with AB an angle equal to the given angle DCE. Construction. In CD, CE take any points D and E ; and join DE. From AB cut off AF equal to CD. I. 3. On AF describe the triangle FAG, having the remaining sides AG, GF equal respectively to CE, ED. I. 22. Then shall the angle FAG be equal to the angle DCE. Proof. For in the triangles FAG, DCE, ( FA is equal to DC, Constr. Because . Obs. The following properties of a square, though not formally enunciated by Euclid, are employed in subsequent proofs. [See i. 48.] (i) The squares on equal straight lines are equal. (ii) Equal squares stand upon equal strait/Id lines. 86 EUCLID'S ELEMENTS. PROPOSITION 48. THEOREM. If the square described on one side of a triangle be equal to the sum of the squares described on the other two sides, then the angle contained by these two sides shall be a ri(/ht angle. B C Let ABC be a triangle ; and let the square described on BC be equal to the sum of the squares described on BA, AC : then shall the angle BAG be a right angle. Construction. From A draw AD at right angles to AC; 1. 11. and make AD equal to AB. I. 3. Join DC. Proof. Then, because AD is equal to AB, Constr. therefore the square on AD is equal to the square on AB. To each of these add the square on CA; then the sum of the squares on CA, AD is equal to the sum of the squares on CA, AB. But, because the angle DAC is a right angle, Constr. therefore the square on DC is equal to the sum of the squares on CA, AD, I. 47. And, by hypothesis, the square on BC is equal to the sum of the squares on CA, AB; therefore the square on DC is equal to the square on BC : therefore also the side DC is equal to the side BC. Then in the triangles DAC, BAC, {DA is equal to BA, Constr. and AC is common to both; also the third side DC is equal to the third side v BC; Proved. therefore the angle DAC is equal to the angle BAC. I. 8. But DAC is a right angle ; Coustr. therefore also BAC is a riht anle. q. E. D. THEOREMS AND EXAMPLES ON BOOK I. INTRODUCTORY. HINTS TOWARDS THE SOLUTION OF GEOMETRICAL EXERCISES. ANALYSIS. SYNTHESIS. It is commonly found that exercises in Pure Geometry present to a beginner far more difficulty than examples in any other branch of Elementary Mathematics. This seems to be due to the following causes. (i) The main Propositions in the text of Euclid must be not merely understood, but thoroughly digested, before the exercises depending upon them can be successfully attempted. (ii) The variety of such exercises is practically unlimited ; and it is impossible to lay down for their treatment any definite methods, such as the student has been accustomed to find in the rules of Elementary Arithmetic and Algebra. (iii) The arrangement of Euclid's Propositions, though per- haps the most convincing of all forms of argument, affords in most cases little clue as to the way in which the proof or con- struction was discovered. Euclid's propositions are arranged synthetically : that is to say, they start from the hypothesis or data ; they next pro- ceed to a construction in accordance with postulates, and pro- blems already solved ; then by successive steps based on known theorems, they finally establish the result indicated by the enun- ciation. Thus Geometrical Synthesis is a building up of known results, in order to obtain a new result. But as this is not the way in which constructions or proofs are usually discovered, we draw the attention of the student to the following hints. Begin by assuming the result it is desired to establish ; then by working backwards, trace the consequences of the assumption, and try to ascertain its dependence on some simpler J eheorem which is already known to be true, or on some condition which suggests the necessary construction. If this attempt is suc- cessful, the steps of the argument may in general be re-arranged in reverse order, and the construction and proof presented in a synthetic form. 88 EUCLID'S ELEMENTS. This unravelling of the conditions of a proposition in order to trace it back to some earlier principle on which it depends, is called geometrical analysis : it is the natural way of attack- ing most exercises of a more difficult type, and it is especially adapted to the solution of problems. These directions are so 'general that they cannot be said to amount to a method: all that can be claimed for Geometrical Analysis is that it furnishes a mode of searching for a suggestion, and its success will necessarily depend on the skill and ingenuity with which it is employed : these may be expected to come with experience, but a thorough grasp of the chief Pro- positions of Euclid is essential tc attaining them. The practical application of these hints is illustrated by the following examples. 1. Construct an isosceles trianyie having (jiven the base, and tlie sum of one of the equal sides and the perpendicular drawn from the vertex to the bane. Let AB be the given base, and K the sum of one side and the perpendicular drawn from the vertex to the base. ANALYSIS. Suppose ABC to be the required triangle. From C draw CX perpendicular to AB : then AB is bisected at X. Now if we produce XC to H, making XH equal to K, it follows that CH = CA ; and if AH is joined, we notice that the angle CAH = the angle CHA. i. 5. Now the straight lines XH and AH can be drawn before the 2>osition of C is knoivn ; Hence we have the following construction, which we arrange synthetically. THEOKEMS AND EXAMPLES ON BOOK I. 89 SYNTHESIS. Bisect AB at X : from X draw XH perpendicular to AB, making XH equal to K. Join AH. At the point A in HA, make the angle HAC equal to the angle AHX ; and join CB. Then ACB shall be the triangle required. First the triangle is isosceles, for AC = BC. i. 4. Again, since the angle HAC = the angle AHC, Conxtr. .-. HC = AC. i.G. To each add CX ; then the sum of AC, CX = the sum of HC, CX = HX. That is, the sum of AC, CX = K. g. E. F. 2. To divide a given straight line so that the square on one part may be double of the square on the other. Jc Let AB be the given straight line. ANALYSIS. Suppose AB to be divided as required at X : that is, suppose the square on AX to be double of the square on XB. Now we remember that in an isosceles right-angled triangle, the square on the hypotenuse is double of the square on either of the equal sides. This suggests to us to draw BC perpendicular to AB, and to make BC equal to BX. Join XC. Then the square on XC is double of the square on XB, i. 47. .-. XC = AX. And when we join AC, we notice that the angle X AC = the angle XC A. i. 5. Hence the exterior angle CXB is double of the angle XAC. i. 32. But the angle CXB is half of a right angle : i. 32. .'. the utujle XAC /' one-fourth of a right angle. This supplies the clue to the following construction ; 90 EUCLID'S ELEMENTS. SYNTHESIS. From B draw BD perpendicular to AB ; and from A draw AC, makiny BAG one-fourth of a right angle. From C, the intersection of AC and BD, draw CX, making the angle ACX equal to the angle BAC. i. 23. Then AB shall be divided as required at X. For since the angle XCA = the angle XAC, .-. XA = XC. 1.6. And because the angle BXC = the sum of the angles BAC, ACX, i. 32. .-. the angle BXC is half a right angle; and the angle at B is a right angle ; therefore the angle BCX is half a right angle ; i. 32. therefore the angle BXC = the angle BCX ; .-. BX = BC. Hence the square on XC is double of the square on XB : i. 47. that is, the square on AX is double of the square on XB. Q.E.F. I. OX THE IDENTICAL EQUALITY OF TRIANGLES. See Propositions 4, 8, 26. 1. If in a triangle the perpendicular from the vertex on the base bisects the base, then the triangle is isosceles. 2. If the bisector of the vertical angle of a triangle is also per- pendicular to the base, the triangle is isosceles. 3. If the bisector of the vertical angle of a triangle also bisects the base, the triangle is isosceles. [Produce the bisector, and complete the construction after the manner of i. 16.] 4. If in a triangle a pair of straight lines drawn from the ex- tremities of the base, making equal angles with the sides, are equal, the triangle is isosceles. 5. If in a triangle the perpendiculars drawn from the extremities of the base to the opposite sides are equal, the triangle is isosceles. 6. Two triangles ABC, ABD on the same base AB, and on opposite sides of it, are such that AC is equal to AD, and BC is equal to BD : shew that the line joining the points C and D is perpendicular to AB. 7. If from the extremities of the base of an isosceles triangle per- pendiculars are drawn to the opposite sides, shew that the straight line joining the vertex to the intersection of these perpendiculars bisects the vertical angle. THEOREMS AND EXAMPLES ON BOOK I. 91 8. ABC is a triangle in which the vertical angle BAG is bisected by the straight line AX : from B draw BD perpendicular to AX, and produce it to meet AC, or AC produced, in E; then shew that BD is equal to DE. 9. In a quadrilateral ABCD. AB is equal to AD, and BC is equal to DC : shew that the diagonal AC bisects each of the angles which it joins. 10. In a quadrilateral ABCD the opposite sides AD, BC are equal, and also the diagonals AC, BD are equal: if AC and BD intersect at K, shew that each of the triangles A KB, DKC is isosceles. 11. If one angle of a triangle be equal to the sum of the other two, the greatest side is double of the distance of its middle point from the opposite angle. 12. Two riyht-antjled triangles which have their hypotenuses equal, and one side of one equal to one side of the other, are identically equal. B Let ABC, DEF be two A s right-angled at B and E, having AC equal to DF, and AB equal to DE : then shall the A s be identically equal. For apply the A ABC to the A DEF, so that A may fall on D, and AB along DE; and so that C may fall on the side of DE remote from F. Let C' be the point on which C falls. Then since AB=DE, .. B must fall on E ; so that DEC' represents the A ABC in its new position. Now each of the / s DEF, DEC' is a rt. i_ ; Hyp. .-. EF and EC' are in one st. line. i. 14. Then in the A C'DF, because DF=DC', .-. the L DFC'=the / DC'F. i. 5. Hence in the two A s DEF, DEC', ( the / DEF = the / DEC', being rt. i_ s ; Because \ and the / DFE = the / DC'E; Proved. ( also the side D E is common to both ; .. the A S DEF, DEC' are equal in all respects; i. 26. that is, the A S DEF, ABC are equal in all respects. Q.E.D. 92 EUCLID'S ELEMENTS. 13. If two triawjles have two sides of the one equal to tvo xldes <n ; i. 34. ted at Y. y.E.D. See Propositions 33, 34, and the deductions from these Props. given on page 64. 1. The straight line drawn through the middle point of a tide of a triangle parallel to the base, bisects the remaining side. Let ABC be a A , and Z the middle point of the side AB. Through Z, Z Y is drawn par 1 A to BC ; then shall Y be the middle point of AC. Through Z draw ZX par 1 to AC. i. 31. Then in the A s AZY, ZBX, because ZY and BC are par 1 , .-.the Z AZY = the / ZBX; i. 29, and because Z X and AC are par 1 .-. the L ZAY = the / BZX; also AZ = ZB: .-. AY = ZX. But ZXCY is a par 1 " by construction ; .-. ZX = YC. Hence AY = YC; 2. The straight line which joins the middle points of tiro sides of a triangle, is parallel to the third side. Let ABC be a A , and Z, Y the middle A points of the sides AB, AC: then shall ZY be par 1 to BC. Produce ZY to V, making YV equal to ZY. Join CV. Then in the A s AYZ, CYV, ( AY = CY, Hyp. Because? andYZ=YV, Coitittr. ( and the / AYZ = the vert. opp. L CYV ; .-. AZ = CV, and the / ZAY = the / VCY; hence CV is par 1 to AZ. i. 1'7. But CV is equal to AZ, that is, to BZ : Hyp. .-. CV is equal and par 1 to BZ : .. ZV is equal and par 1 to BC : i. 33. that is, ZY is par 1 to BC. Q.E.D. [A second proof of this proposition may be derived from i. 38, 39.] THEOREMS AND EXAMPLES ON BOOK I. 97 3. The straight line which joins the middle points of two sides of a triangle is equal to half the third side. 4. Shew that the three straight lines which join the middle points of the fiCics of a triangle, divide it into four triangles which are identi- cally equal. 5. Any straight line drawn from the vertex of a triangle to the bane is bisected by the straight line ichick joins the middle points of the other sides of the triangle. 6. Given the three middle points of the sides of a triangle, con- struct the triangle. 7. AB, AC are two given straight lines, and P is a given point between them ; required to draw through P a straight line termi- nated by AB, AC, and bisected by P. 8. A BCD is a parallelogram, and X, Y are the middle points of the opposite sides AD, BC: shew that BX and DY trisect the dia- gonal AC. 9. If the middle points of adjacent sides of any quadrilateral be joined, the figure thus formed is a parallelogram. 10. Shew that the straight lines which join the middle points of opposite sides of a quadrilateral, bisect one another. 11. The straight line which joins the middle points of the oblique sides of a trapezium, is parallel to the two parallel sides, and passes through the middle points of the diagonals. 12. The straight line which joins the middle points of the oblique sides of a trapezium is equal to half the sum of the parallel sides ; and the portion intercepted between the diagonals is equal to half the difference of the parallel sides. Definition. If from the extremities of one straight line per- pendiculars are drawn to another, the portion of the latter intercepted between the perpendiculars is said to be the Ortho- gonal Projection of the tirst line upon the second. B Y Q A \^^ Y Q Thus in the adjoining figures, if from the extremities of the straight line AB the perpendiculars AX, BY are drawn to PQ, then XY is the orthogonal projection of AB on PQ. 72 98 EUCLID'S ELEMENTS. 13. A niven straight line AB is bisected at C; shew that the pro- jections of AC, CB on any other straight line are equal. Q Let XZ, ZY be the projections of AC, CB on any straight line PQ: then XZ and ZY shall be equal. Through A di-aw a straight line parallel to PQ, meeting CZ, BY or these lines produced, in H, K. i. 31. Now AX, CZ, BY are parallel, for they are perp. to PQ; i. 28. .-. the figures XH, HY are par" 1 '; .-. AH = XZ, and HK = ZY. i. 34. But through C, the middle point of AB, a side of the A ABK, CH has been drawn parallel to the side BK ; .-. CH bisects AK: Ex. 1, p. 90. that is, AH = HK; .-. XZ = ZY. Q.E.D. H. If three parallel straight lines make equal intercepts on a fourth straight line which meets them, they will also make equal inter- cepts on any other straight line ivhich meets them. 15. Equal and parallel straight lines have equal projections on any other straight line. 10. AB is a given straight line bisected at O ; and AX, BY are perpendiculars drawn from A and B on any other straight line : shew that OX is equal to OY. 17. AB is a given straight line bisected at O : and AX, BY and OZ are perpendiculars drawn to any straight line PQ, ichich does not pass between A and B: shew that OZ is equal to half the sum of AX, BY. [OZ is said to be the Arithmetic Mean between AX and BY.] 18. AB is a given straight line bisected at O; and through A, B and O parallel straight lines are drawn to meet a given straight line PQ in X, Y, Z : shew that OZ is equal to half the sum, or half the difference of AX and BY, according as A and B lie on the same side or on opposite sides of PQ. THEOREMS AND EXAMPLES ON BOOK I. 99 19. To divide a given finite straight line into any number of equal parts. [For example, required to divide the straight line AB into Jive equal parts. From A draw AC, a straight line of un- limited length, making any angle with AB. In AC take any point P, and mark off successive parts PQ, QR, RS, ST each equal to A P. Join BT ; and through P, Q, R, S draw parallels to BT. It may be shewn by Ex. 14, p. 98, that these parallels divide AB into five equal parts.] 20. If through an angle of a parallelogram any straight line is drawn, the perpendicular drawn to it from the opposite angle is equal to the sum or difference of the perpendiculars drawn to it from the two remaining angles, according as the given straight line falls without tlie parallelogram, or intersects it. [Through the opposite angle draw a straight line parallel to the given straight line, so as to meet the perpendicular from one of the remaining angles, produced if necessary: then apply i. 34, i. 26. Or proceed as in the following example.] 21. From the angular points of a parallelogram perpendiculars are drawn to any straight line which is without the parallelogram: shew that the sum of the perpendiculars drawn from one pair of opposite angles is equal to the sum of those drawn from the other pair. [Draw the diagonals, and from their point of intersection let fall a perpendicular upon the given straight line. See Ex. 17, p. 98.] 22. The sum of the perpendiculars drawn from any point in the base of an isosceles triangle to the equal sides is equal to the perpendi- cular drawn from either extremity of the base to the opposite side. [It follows that the sum of the distances of any point in the base of an isosceles triangle from the equal sides is constant, that is, the same whatever point in the base is taken.] 23. In the base produced of an isosceles triangle any point is taken : shew that the difference of its distances from the equal sides is constant. 24. The sum of the perpendiculars drawn from any point within an equilateral triangle to the three sides is equal to the perpendicular drawn from any one of the angular points to the opposite side, and is therefore constant. 100 EUCLID'S ELEMKNTS. PEOBLEMS. [Problems marked (*) admit of more than one solution.] *25. Draw a straight line through a given point, so that the part of it intercepted between two given parallel straight lines may be of gi%-en length. 26. Draw a straight line parallel to a given straight line, so that the part intercepted between two other given straight lines may be of given length. 27. Draw a straight line equally inclined to two given straight lines that meet, so that the part intercepted between them may be of given length. 28. AB, AC are two given straight lines, and P is a given point without the angle contained by them. It is required to draw through P a straight line to meet the given lines, so that the part intercepted between them may be equal to the part between P and the nearer line. V. MISCELLANEOUS THEOREMS AND KXAMPI.KS. Chiefly 011 i. 32. 1. A is the vertex of an isosceles triangle ABC, and BA is produced to D, so that AD is equal to BA ; if DC is drawn, shew that BCD /* . 2. The bisectors of the angles of a triangle are concurrent. Let ABC be a A. Bisect the / 8 ABC, A BCA, by straight lines which must meet at some point O. Ax. 12. Join AO. It is required to prove that AO bisects the L BAG. From O draw OP, OQ, OR perp. to the sides of the A . Then in the A S OBP, OBR, t the / OBP = the / OBR, Conxtr. Because \ and the / OPB = the / ORB, being rt. i_ s , | and OB is common ; .-. OP^OR. i. 26. 104 EUCLID'S ELEMENTS. Similarly from the A" OCR, OCQ, it may be shewn that OP=OQ, .-. OP, OQ, OR are all equal. Again in the A S OR A, OQA, the / ORA, OQA are rt. L. ', land the hypotenuse CA is Because' common also O R = O Q ; Proved. .-. the / RAO = the/ QAO. B P Ex. 12, p. 91. That is, AO is the bisector of the / BAC. Hence the bisectors of the three / s meet at the point O. o.. K. 3. The bisectors of two exterior angles of a triangle ami tli,'. bisector of the third angle are concurrent. Let ABC be a A, of which the sides AB, AC are produced to any points D and E. Bisect the / s DBC, ECB by straight lines which must meet at some point O. Ax. 12. Join AO. It is required to prove tliat AO bisects the angle BAC. From O draw OP, OQ, OR perp. to the sides of the A . Then in the A 3 OBP, OBR, Cthe L OBP = the / OBR, Constr. also the L OPB = the z ORB, being rt. L 8 , and OB is common ; .-. OP^OR. Similarly in the A" OCP, OCQ, it may be shewn that OP = OQ : .-. OP, OQ, OR are all equal. Again in the A 8 ORA, OQA, f the L 8 ORA, OQA are rt. L", Because - and the hypotenuse OA is common, alsoOR = OQ; Prm-ed. .-. the / RAO = the / QAO. Ex. 12, p. 5)1. That is, AO is the bisector of the / BAC. .-. the bisectors of the two exterior / s DBC, ECB. and of the interior / BAC meet at the point O. Because THEOREMS AND EXAMPLES OX BOOK I. 105 4. The medians of a triangle are concurrent. Let ABC be a A . Let BY and CZ be two of its medians, and let them intersect at O. Join AO, and produce it to meet BC in X. It i# required to shew tJuit AX ?'* the remaining median of the A . Through C draw CK parallel to BY: produce AX to meet CK at K. Join BK. In the A AKC, because Y is the middle point of AC, and YO is parallel to CK, . . O is the middle point of AK. Ex. 1, p. 96. Again in the A ABK, since Z and O are the middle points of AB, AK, .-. ZO is parallel to BK, Ex. 2, p. 9G. that is, OC is paiullel to BK : .. the figure BKCO is a par m . But the diagonals of a par bisect one another, Ex. o, p. 04. .-. X is the middle point of BC. That is, AX is a median of the A . Hence the three medians meet at the point O. Q.E.I>. COROLLARY. The three medians of a triangle cut one another at a point of trisection, the greater segment in each being towards the angular point. For in the above figure it has been proved that AO = OK, also that OX is half of OK; .-. OX is half of OA : that is, OX is one third of AX. Similarly OY is one third of BY, and OZ is one third of CZ. Q.E.D. By means of this Corollary it may be shewn that in any triangle the shorter median bisects the greater side. [The point of intersection of the three medians of a triangle is called the centroid. It is shewn in mechanics that a thin triangular plate will balance in any position about this point : therefore the centroid of a triangle is also its centre of gravity.] 106 EUCLID'S ELEMENTS. 5. The perpendiculars drawn from the vertices of a triangle to the opposite sides are concurrent. Let ABC be a A, and AD, BE, CF the three perp' drawn from the vertices to the opposite sides : then shall these perp 8 be concurrent. Through A, B, and C draw straight lines MN, NL, LM parallel to the opposite sides of the A . Then the figure BAMC is a par" 1 . J),-f. 26. .-. AB=MC. i.34. Also the figure BACL is a par" 1 . .-. AB = LC, .-. LC = CM : that is, C is the middle point of LM. So also A and B are the middle points of M N and N L. Hence AD, BE, CF are the perp a to the sides of the A LMN from their middle points. Ex. 3, p. 54. But these perp 8 meet in a point: Ex. 1, p. 103. that is, the perp 8 drawn from the vertices of the A ABC to the opposite sides meet in a point. Q.E.D. [For another proof see Theorems and Examples on Book m.] DEFINITIONS. (i) The intersection of the perpendiculars drawn from the vertices of a triangle to the opposite sides is called its ortho- centre. (ii) The triangle formed by joining the feet of the perpen- diculars is called the pedal triangle. THEOEEMS AND EXAMPLES ON BOOK I. 107 K VII. ON THE CONSTRUCTION OF TRIANGLES WITH GIVEN PARTS. No general rules can be laid down for the solution of problems in this section ; but in a few typical cases we give constructions, which the student will find little difficulty in adapting to other questions of the same class. 1. Construct a right-angled triangle, having given the hypotenuse and the sum of the remaining sides. [It is required to construct a rt. angled A , having its hypotenuse equal to the given straight line K, and the sum of its remaining sides equal to AB. From A draw AE making with BA an / equal to half a rt. i_ . From centre B, with radius equal to K, de- scribe a circle cutting AE in the points C,C'. From C and C' draw perp 3 CD, C'D' to AB ; and join CB, C'B. Then either of the A 3 CDB, C'D'B will satisfy the given conditions. NOTE. If the given hypotenuse K be greater than the perpendicu- lar drawn from B to AE, there will be two solutions. If the line K be equal to this perpendicular, there will be one solution ; but if less, the problem is impossible.'} 2. Construct a right-angled triangle, having given the hypotenuse and the difference of the remaining sides. 3. Construct an isosceles right-angled triangle, having given the sum of the hypotenuse and one side. 4. Construct a triangle, having given the perimeter and the angles at the base. P Q R B [Let AB be the perimeter of the required A , and X and Y the L ' the base. From A draw AP, making the L BAP equal to half the L X. From B draw BP, making the L ABP equal to half the / Y. From P draw PQ, making the / APQ equal to the / BAP. From P draw PR, making the / BPR equal to the /ABP. Then shall PQR be the required A .] at 108 EUCLID'S ELEMENTS. 5. Construct a right-angled triangle, having given the perimeter and one acute angle. 6. Construct an isosceles triangle of given altitude, so that its base may be in a given straight line, and its two equal sides may pass through two fixed points. [See Ex. 7, p. 4i).] 7. Construct an equilateral triangle, having given the length of the perpendicular drawn from one of the vertices to the opposite side. 8. Construct an isosceles triangle, having given the base, and the difference of one of the remaining sides and the perpendicular drawn from the vertex to the base. [See Ex. 1, p. 88.] 9. Construct a triangle, having given the base, one of the angles at the base, and the sum of the remaining sides. 10. Construct a triangle, having given the base, one of the angles at the base, and the difference of the remaining sides. 11. Construct a triangle, having given the base, the difference of the angles at the base, and the difference of the remaining sides. A B [Let AB be the given base, X the difference of the / s at the base, and K the difference of the remaining sides. Draw BE, making the / ABE equal to half the / X. From centre A, with radius equal to K, describe a circle' cutting BE in D and D'. Let D be the point of intersection nearer to B. Join AD and produce it to C. Draw BC, making the L DBC equal to the / BDC. Then shall CAB be the A required. Ex. 7, p. 101. NOTE. This problem is possible only when the given difference K is greater than the perpendicular drawn from A to BE.] 12. Construct a triangle, having given the base, the difference of the angles at the base, and the sum of the remaining sides. 13. Construct a triangle, having given the perpendicular from the vertex on the base, and the difference between each side and the adjacent segment of the base. THEOREMS AM) EXAMPLES ON BOOK I. 109 14. Construct a triangle, having given two sides and the median which bisects the remaining side. [See Ex. 18, p. 102.] 15. Construct a triangle, having given one side, and the medians which bisect the two remaining sides. [See Fig. to Ex. 4, p. 105. Let BC be the given side. Take two-thirds of each of the given medians ; hence construct the triangle BOC. The rest of the con- struction follows easily.] 16. Construct a triangle, having given its three medians. [See Fig. to Ex. 4, p. 105. Take two-thirds of each of the given medians, and construct the triangle OKC. The rest of the construction follows easily.] VIII. ON AREAS. See Propositions 35 48. It must be understood that throughout this section the word equal as applied to rectilineal figures will be used as denoting equality of area unless otherwise stated. 1. Shew that a parallelogram is bisected by any straight line which passes through the middle point of one of its diagonals, [i. 29, 26.] 2. Bisect a parallelogram by a straight line drawn through a given point. 3. Bisect a parallelogram by a straight line drawn perpendicular to one of its sides. 4. Bisect a parallelogram by a straight line drawn parallel to a given straight line. 5. ABCD is a trapezium in which the side AB is parallel to DC. ' Shew that its area is equal to the area of a parallelogram formed by drawing through X, the middle point of BC, a straight line parallel to AD. [i. 29, 26.] 6. A trapezium is equal to a parallelogram whose base is half the sum of the parallel sides of the given figure, and whose altitude is equal to the perpendicular distance between them. 7. ABCD is a trapezium in which the side AB is parallel to DC; shew that it is double of the triangle formed by joining the extremities of AD to X, the middle point of BC. 8. Shew that a trapezium is bisected by the straight line which joins the middle points of its parallel sides. [i. 38.] 110 EUCLID'S ELEMENTS. In the following group of Exercises the proofs depend chiefly on Propositions 37 and 38, and the two converse theorems. 9. If two straight lines AB, CD intersect at X, and if the straight lines AC and BD, which join their extremities are parallel, shew that the triangle AXD is equal to the triangle BXC. 10. If two straight lines AB, CD intersect at X, so that the triangle AXD is equal to the triangle XCB, then AC and BD are parallel. 11. A BCD is a parallelogram, and X any point in the diagonal AC produced; shew that the triangles XBC, XDC are equal. [See Ex. 13, p. 64.] 12. ABC is a triangle, and R, Gl the middle points of the sides AB, AC; shew that if BQ and CR intersect in X, the triangle BXC is equal to the quadrilateral AQXR. [See Ex. 5, p. 73.] 13. If the middle points of the sides of a quadrilateral be joined in order, the parallelogram so formed [see Ex. 9, p. 97] is equal to half the given figure. 1-1. Two triangles of equal area stand on the same base but on opposite sides of it : shew that the straight line joining their vertices is bisected by the base, or by the base produced. 15. The straight line which joins the middle points of the dia- gonals of a trapezium is parallel to each of the two parallel sides. 16. (i) A triangle is equal to the sum or difference of two triangles on the same base (or on equal bases), if the altitude of the former is equal to the sum or difference of the altitudes of the latter. (ii) A triangle is equal to the sum or difference of two triangles of the same altitude if the base of the former is equal to the sum or differ- ence of the bases of the latter. Similar statements hold good of parallelograms. 17. ABCD is a parallelogram, and O is any point outside it; shew that the sum or difference of the triangles CAB, OCD is equal to half the parallelogram. Distinguish between the two cases. On the following proposition depends an important theorem in Mechanics : we give a proof of the first case, leaving the second case to be deduced by a similar method. THEOREMS AND EXAMPLES OJT BOOK i. Ill 18. (i) A BC D is a parallelogram, and O is any point without the angle BAD and its opposite vertical angle ; shew that the triangle OAC is equal to the sum of the triangles OAD, OAB. (ii) If O is within the angle BAD or its opposite vertical angle, the triangle OAC is equal to the difference of the triangles OAD, OAB. CASE I. If O is without the /DAB and its opp. vert. / , then OA is with- out the par m A BCD : therefore the perp. drawn from C to OA is equal to the sum of the perp 8 drawn from B and D to OA. [See Ex. 20, p. 99.] Now the A 9 OAC, OAD, OAB are upon the same base OA ; and the altitude of the A OAC with respect to this base has been shewn to be equal to the sum of the altitudes of the A S OAD, OAB. Therefore the A OAC is equal to the sum of the A s OAD, OAB. [See Ex. 1C, p. 110.] Q.E.D. 19. A BCD is a parallelogram, and through O, any point within it, straight lines are drawn parallel to the sides of the parallelogram; shew that the difference of the parallelograms DO, BO is double of the triangle AOC. [See preceding theorem (ii).] 20. The area of a quadrilateral is equal to the area of a triangle having two of its sides equal to the diagonals of the given figure, and the included angle equal to either of the angles between the dia- gonals. 21. ABC is a triangle, and D is any point in AB: it is required to draw through D a straight line DE to meet BC produced in E, no that the triangle DBE may be equal to the triangle ABC. B C E [Join DC. Through A draw AE parallel to DC. i. 31. Join DE. The A EBD shall be equal to the A ABC. i. 37.] H. E, 112 EUCLID'S ELEMENTS. 22. On a base of given length describe a triangle equal to a given triangle and having an angle equal to an angle of the given triangle. 23. Construct a triangle equal in area to a given triangle, and having a given altitude. 24. On a base of given length construct a triangle equal to a given triangle, and having its vertex on a given straight line. 25. On a base of given length describe (i) an isosceles triangle ; (ii) a right-angled triangle, equal to a given triangle. 26. Construct a triangle equal to the sum or difference of two given triangles. [See Ex. 16, p. 110.] 27. ABC is a given triangle, and X a given point: describe a triangle equal to ABC, having its vertex at X, and its base in the same straight line as BC. 28. ABCD is a quadrilateral: on the base AB construct a triangle equal in area to ABCD, and having the angle at A common with the quadrilateral. [Join BD. Through C draw CX parallel to BD, meeting AD pro- duced in X ; join BX.] 29. Construct a rectilineal figure equal to a given rectilineal figure, and having fewer sides by one than the given figure. Hence sliew how to construct a triangle equal to a given rectilineal figure. 30. ABCD is a quadrilateral : it is required to construct a triangle equal in area to ABCD, having its vertex at a given point X in DC, and its base in the same straight line as AB. 31. Construct a rhombus equal to a given parallelogram. 32. Construct a parallelogram which shall have the same area and perimeter as a given triangle. 33. Bisect a triangle by a straight line drawn through one of its angular points. 34. Trisect a triangle by straight lines drawn through one of its angular points. [See Ex. 19, p. 102, and i. 38.] 35. Divide a triangle into any number of equal parts by straight lines drawn through one of its angular points. [See Ex. 19, p. 99, and i. 38.] THEOREMS AND EXAMPLES OX BOOK I. 113 3C. Bisect a triangle by a straight line drawn through a given point in one of its sides. [Let ABC be the given A, and P the given point in the side AB. Bisect AB at Z ; and join CZ, CP. Through Z draw ZQ parallel to CP. Join PQ. Then shall PQ bisect the A . See Ex. 21, p. 111.] B Q 37. Trisect a triangle by straight lines drawn from a given point in one of its sides. [Let ABC be the given A , and X the given point in the side BC. Trisect BC at the points P, Q. Ex. 19, p. 99. Join AX, and through P and Q draw PH and QK parallel to AX. Join XH, XK. These straight lines shall trisect the A ; as may be shewn by joining AP, AQ. See Ex. 21, p. 111.] P X Q 38. Cut off from a given triangle a fourth, fifth, sixth, or any part required by a straight line drawn from a given point in one of its sides. [See Ex. 19, p. 99, and Ex. 21, p. 111.] 39. Bisect a quadrilateral by a straight line drawn through an angular point. [Two constructions may be given for this problem: the first will be suggested by Exercises 28 and 33, p. 112. The second method proceeds thus. Let ABCD be the given quadrilateral, and A the given angular point. Join AC, BD, and bisect BD in X. Through X draw PXQ parallel t:> AC, meeting BC in P ; join AP. Then shall AP bisect the quadrilateral. Join AX, CX, and use i. 37, 38.] 40. Cut off from a given quadrilateral a third, a fourth, a fifth, or any part required, by a straight line drawn through a given angular point. [See Exercises 28 and 35, p. 112.] 114 EUCLID'S ELEMENTS. [The following Theorems depend on i. 47.] 41. In the figure of i. 47, shew that (i) the sum of the squares on AB and AE is equal to the sum of the squares on AC and AD. (ii) the square on EK is equal to the square on AB with four times the square on AC. (iii) the sum of the squares on EK and FD is equal to five times the square on BC. 42. If a straight line be divided into any two parts the square on the straight line is greater than the squares on the two parts. 43. If the square on one side of a triangle is less than the squares on the remaining sides, the angle contained by these sides is acute ; if greater, obtuse. 44. ABC is a triangle, right-angled at A; the sides AB, AC are intersected by a straight line PQ, and BQ, PC are joined : shew that the sum of the squares on BQ, PC is equal to the sum of the squares on BC, PQ. 45. In a right-angled triangle four times the sum of the squares on the medians which bisect the sides containing the right angle is equal to five times the square on the hypotenuse. 46. Describe a square whose area shall be three times that of a given square. 47. Divide a straight line into two parts such that the sum of their squares shall be equal to a given square. IX. ON LOCI. It is frequently required in the course of Plane Geometry to find the position of a point which satisfies given conditions. Now all problems of this type hitherto considered have been found to be capable of definite determination, though some admit of more than one solution : this however will not be the case if only one condition is given. For example, if we are asked to find a point which shall be at a given distance from a given point, we observe at once that the problem is indetenninate, that is, that it admits of an indefinite number of solutions ; for the condition stated is satisfied by any point on the circumference of the circle described from the given point as centre, with a radius equal to the given distance : moreover this condition is satisfied by no other point within or without the circle. Again, suppose that it is required to find a point at a given distance from a given straight line. THEOREMS AND EXAMPLES ON BOOK. I. 115 Here, too, it is obvious that there are an infinite number of such points, and that they lie on the two parallel straight lines which may be drawn on either side of the given straight line at the given distance from it : further, no point that is not on one or other of these parallels satisfies the given condition. Hence we see that when one condition is assigned it is not sufficient to determine the position of a point absolutely, but it may have the effect of restricting it to some definite line or lines, straight or curved. This leads us to the following definition. DEFINITION. The Locus of a point satisfying an assigned condition consists of the line, lines, or part of a line, to which the point is thereby restricted; provided that the condition is satisfied by every point on such line or lines, and by no other. A locus is sometimes defined as the path traced out by a point which moves in accordance with an assigned law. Thus the locus of a point, which is always at a given distance from a given point, is a circle of which the given point is the centre : and the locus of a point, which is always at a given distance from a given straight line, is a pair of parallel straight lines. "\Ve now see that in order to infer that a certain line, or system of lines, is the locus of a point under a given condition, it is necessary to prove (i) that any point which fulfils the given condition is on the supposed locus ; (ii) that every point on the supposed locus satisfies the given condition. 1. Find the locus of a point which is ahvaijs equidistant from two given points. Let A, B be the two given points, (a) Let P be any point equidistant from A and B, so that AP = BP. Bisect AB at X, and join PX. Then in the A S AXP, BXP, ( AX = BX, Coimtr. Because -|and PX is common to both, also AP=BP, .-. the L PXA = the L PXB; and they are adjacent / s ; .-. PX is perp. to AB. .. Any point which is equidistant from A and B is on the straight line which bisects AB at right angles. 116 EUCLID'S ELEMENTS. (/3) Also every point in this line is equidistant from A and B. For let Q be any point in this line. Join AQ, BQ. Then in the A S AXQ, BXQ, ( AX=BX, Because -I and XQ is common to both ; (also the / AXQ = the L BXQ, being rt. L s ; .-. AQ=BQ. i. 4. That is, Q is equidistant from A and B. Hence we conclude that the locus of the point equidistant from two given points A, B is the straight line which bisects AB at right 2. To find the locus of the middle point of a straight line drnicn from a given iioint to meet a given straight line of unlimited lemjth. Let A be the given point, and BC the given straight line of un- limited length. (a) Let AX be any straight line drawn through A to meet BC, and let P be its middle point. Draw AF perp. to BC, and bisect AF at E. Join EP, and produce it indefinitely. Since AFX is a A , and E, P the middle points of the two sides AF, AX, .-. EP is parallel to the remaining side FX. Ex. 2, p. 96. .-. P is on the straight line which passes through the fixed point E, and is parallel to BC. ()3) Again, every point in EP, or EP produced, fulfils the required condition. For, in this straight line take any point Q. Join AQ. and produce it to meet BC in Y. Then FAY is a A , and through E, the middle point of the side AF, EQ is drawn parallel to the side FY, . . Q is the middle point of AY. Ex. 1, p. %. Hence the required locus is the straight line drawn parallel to BC, and passing through E, the middle point of the perp. from A to BC. THEOREMS AND EXAMPLES ON BOOK I. 117 3. Find the locus of a point equidistant from two yiven inter- secting straight lines. [See Ex. 3, p. 49.] 4. Find the locus of a point at a given radial distance from the circumference of a given circle. 5. Find the locus of a point which moves so that the sum of its distances from two given intersecting straight lines of unlimited length is constant. 6. Find the locus of a point when the differences of its distances from two given intersecting straight lines of unlimited length is constant. 7. A straight rod of given length slides between two straight rulers placed at right angles to one another: find the locus of its middle point. [See Ex. 2, p. 100.] 8. On a given base as hypotenuse right-angled triangles are described : find the locus of their vertices. 9. AB is a given straight line, and AX is the perpendicular drawn from A to any straight line passing through B: find the locus of the middle point of AX. 10. Find the locus of the vertex of a triangle, when the base and area are given. 11. Find the locus of the intersection of the diagonals of a paral- lelogram, of which the base and area are given. 12. Find the locus of the intersection of the medians of a triangle described on a given base and of given area. X. ON THE INTERSECTION OF LOCI. It appears from various problems which have already been considered, that we are often required to find a point, the position of which is subject to two given conditions. The method of loci is very useful in the solution of problems of this kiivi : for corresponding to each condition there will be a locus on which the required point must lie ; hence all points which are common to these two loci, that is, all the points of intersection of the loci, will satisfy both the given conditions. 118 EUCLID'S ELEMENTS. EXAMPLK 1. To construct a triangle, having given the base, the altitude, and the length of the median wliich bisects the base. Let AB be the given base, and P and Q the lengths of the altitude and median respectively : then the triangle is known if its vertex is known. (i) Draw a straight line CD parallel to AB, and at a distance from it equal to P : then the required vertex must lie on CD. (ii) Again, from the middle point of AB as centre, with radius equal to Q, describe a circle : then the required vertex must lie on this circle. Hence any points which are common to CD and the circle, satisfy both the given conditions: that is to say, if CD intersect the circle in E, F each of the points of intersection might be the vertex of the required triangle. This supposes the length of the median Q to be greater than the altitude. EXAMPLE 2. To find a point equidistant from three gici'ii point* A, B, C, which are not in the same straigltt line. (i) The locus of points equidistant from A and B is the straight line PQ, which bisects AB at right angles. Ex. 1, p. 115. (ii) Similarly the locus of points equidistant from B and C is the straight line RS which bisects BC at right angles. Hence the point common to PQ and RS must satisfy both con- ditions: that is to say, the point of intersection of PQ and RS will be equidistant from A, B, and C. These principles may also be used to prove the theorems relating to concurrency already given on page 103. EXAMPLE. To prove that the bisectors of the angles of a triangle are concurrent. Let ABC be a triangle. Bisect the / s ABC, BCA by straight lines BO, CO : these must meet at some point O. Ax. 12. Join OA. Then shall OA bisect the / BAC. Now BO is the locus of points equi- distant from BC, BA; Ex. 3, p. 49. .. . Similarly CO is the locus of points " equidistant from BC, CA. .-. OP = OQ; hence OR = OQ. .. O is on the locus of points equidistant from AB and AC : that is OA is the bisector of the / BAC. Hence the bisectors of the three i. a meet at the point O. THEOREMS AND EXAMPLES ON BOOK I. 119 It may happen that the data of the problem are so related to one another that the resulting loci do not intersect : in this case the problem is impossible. For example, if in Ex. 1, page 118, the length of the given median is less than the given altitude, the straight line CD will not be intersected by the circle, and no triangle can fulfil the conditions of the problem. If the length of the median is equal to the given altitude, one point is common to the two loci ; and consequently only one solution of the problem exists : and we have seen that there are two solutions, if the median is greater than the altitude. In examples of this kind the student should make a point of investigating the relations which must exist among the data, in order that the problem may be possible ; and he must observe that if under certain relations two solutions are possible, and under other relations no solution exists, there will always be some intermediate, relation under which one and only one solution is possible. EXAMPLES. 1. Find a point in a given straight line which is equidistant from two given points. 2. Find a point which is at given distances from each of two given straight lines. How many solutions are possible? 3. On a given base construct a triangle, having given one angle at the base and the length of the opposite side. Examine the relations which must exist among the data in order that there may be two solu- tions, one solution, or that the problem may be impossible. 4. On the base of a given triangle construct a second triangle equal in area to the first, and having its vertex in a given straight line. 5. Construct an isosceles triangle equal in area to a given triangle, and standing on the same base. 6. Find a point which is at a given distance from a given point, and is equidistant from two given parallel straight lines. BOOK II. BOOK II. deals with the areas of rectangles and squares. DEFINITIONS. 1. A Rectangle is a parallelogram which has one of its angles a right angle. It should be remembered that if a parallelogram has one right angle, all its angles are right angles. [Ex. 1, p. 64.] 2. A rectangle is said to be contained by any two of its sides which form a right angle : for it is clear that both the form and magnitude of a rectangle are fully determined when the lengths of two such sides are jjiven. Thus the rectangle ACDB is said to be contained by AB, AC; or by CD, DB : and if X and Y are two straight lines equal respectively to AB and AC, then the rectangle contained by X and Y is equal to the rectangle contained by AB, AC. [See Ex. 12, p. 64.] B C X- Y- After Proposition 3, we shall use the abbreviation reel. AB, AC to denote the rectangle contained by AB and AC. 3. In any parallelogram the figure formed by either of the parallelograms about a diagonal together with the two complements is called $ gnomon. Thus the shaded portion of the annexed figure, consisting of the parallelogram EH together with the complements AK, KG is the gnomon AHF. The other gnomon in the figure is that which is made up of AK, GF and FH, namely the gnomon AFH. INTRODUCTORY. 121 INTRODUCTORY. Pure Geometry makes no \ise of number to estimate the magnitude of the lines, angles, and figures with which it deals : hence it requires no units of magnitude such as the student is familiar with in Arithmetic. For example, though Geometry is concerned with the relative lengths of straight lines, it does not seek to express those lengths in terms of yards, feet, or inches: similarly it does not ask how many square yards or square feet a given figure contains, nor how many degrees there are in a given angle. This constitutes an essential difference between the method of Pure Geometry and that of Arithmetic and Algebra ; at the same time a close connection exists between the results of these two methods. In the case of Euclid's Book II., this connection rests upon the fact that the number of units of .area in a rectangular figure is found by multiplying together the numbers of units of length in two adjacent sides. For example, if the two sides AB, AD of the rectangle A BCD are respectively four and three inches long, and if through the points of division parallels are drawn as in the annexed figure, it is seen that the rectangle is divided into three rows, each containing four- square inches, or into four columns, each containing three square inches. " Hence the whole rectangle contains 3x4, or 12, square inches. Similarly if AB and AD contain in and n units of length respectively, it follows that the rectangle A BCD will contain mn units of area: further, if AB and AD are equal, each containing m units of length, the rectangle becomes a square, and contains m* units of area. [It must be understood that this explanation implies that the lengths of the straight lines AB, AD are commensurable, that is, that they can be expressed exactly in terms of some common unit. This however is not always the case : for example, it may be proved that the side and diagonal of a square are so related, that it is impossible to divide either of them into equal parts, of which the other contains an exact number. Such lines are said to be incommen- 122 EUCLID'S ELEMENTS. surable. Hence if the adjacent sides of a rectangle are incommen- surable, we cannot choose any linear unit in terms of which these sides maybe exactly expressed; and thus it will be impossible to sub- divide the rectangle into squares of unit area, as illustrated in the figure of the preceding page. We do not here propose to enter further into the subject of incommensurable quantities : it is suffi- cient to point out that further knowledge of them will convince the student that the area of a rectangle may be expressed to any required degree of accuracy by the product of the lengths of two adjacent sides, whether those lengths are commensurable or not. ] From the foregoing explanation we conclude that t/ie rectangle contained by two straight lines in Geometry corresponds to the product of two numbers in Arithmetic or Algebra ; and that the square described on a straight line corresponds to the square of a number. Accordingly it will be found in the course of Book II. that several theorems relating to the areas of rectangles and squares are analogous to well-known algebraical formulae. In view of these principles the rectangle contained by two straight lines AB, BC is sometimes expressed in the form of a product, as AB.BC, and the square described on AB as AB 2 . This notation, together with the signs + and , will be employed in the additional matter appended to this book; but it is not admitted into Euclid's text because it is desirable in the first instance to emphasixe the distinction between geometrical mag- nitudes themselves and the numerical equivalents by which they may be expressed arithmetically. PROPOSITION 1. THEOREM. If there are two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the sum of the rectangles con- tained by the undivided straigJit line and the several parts of the divided line. Let P and AB be two straight lines, and let AB be divided into any number of parts AC, CD, DB : then shall the rectangle contained by P, AB be equal to the sum of the rectangles contained by P, AC, by P, CD, and by P, DB. BOOK II. PROP. 1. A C D B 123 K L F P From A draw AF perp. to AB; I. 11. and make AG equal to P. i. 3. Through G draw GH par 1 to AB ; I. 31. and through C, D, B draw CK, DL, BH par 1 to AG. Now the fig. AH is made up of the figs. AK, CL, DH : and of these, the fig. AH is the rectangle contained by P, AB; for the fig. AH is contained by AG, AB ; and AG = P : and the fig. AK is the rectangle contained by P, AC ; for the fig. AK is contained by AG, AC; and AG P : also the fig. CL is the rectangle contained by P, CD ; for the fig. CL is contained by CK, CD ; and CK = the opp. side AG, and AG = P : i. 34. similarly the fig. DH is the rectangle contained by P, DB. .. the rectangle contained by P, AB is equal to the sum of the rectangles contained by P, AC, by P, CD, and by P, DB. Q.E.D. CORRESPONDING ALGEBRAICAL FORMULA. In accordance with the principles explained on page 122, the result of this proposition may be written thus : P.AB=P.AC+P.CD + P.DB. Now if the line P contains p units of length, and if AC, CD, DB contain a, b, c units respectively, then AB = a + b + c, arid we have p (a + b + c) pa +pb +pc. 124 KUCLIDS ELEMENTS. PROPOSITION 2. THEOREM. If a straight line is divided into any two parts, the square on the whole line is equ. i. 6. i. 34. BOOK II. PROP. 4. 127 Again, the complement AG = the complement GE. I. 43. But the fig. AG = the rect. AC, CB ; for CG = CB. .'. the two figs. AG, GE- twice the rect. AC, CB. the sq. on AB = the fig. AE = the figs. HF, CK, AG, GE = the sqq. on AC, CB together with twice the rect. AC, CB. .'. the sq. on AB = the sum of the sqq. on AC, CB with twice the rect. AC, CB. Q.E.D. * For the purpose of oral work, this step of the proof may conveniently be arranged as follows : Now the sq. on A B is equal to the fig. AE, that is, to the figs. HF, CK, AG, GE; that is, to the sqq. on AC, CB together with twice the rect. AC, CB. COROLLARY. Parallelograms about the diagonals of a square are themselves squares. CORRESPONDING ALGEBRAICAL FORMULA. The result of this important Proposition may be written thus : AB 2 =r AC 2 + CB 2 + 2AC . CB. Let AC = a, and CB = 6; then &B and we have a + b 2 = a II. K 128 EUCLID'S ELEMENTS. PROPOSITION 5. THEOREM. If a straight line is divided equally and also unequally, the rectangle contained by the unequal parts, and the square on the line between the points of section, are together equal to the square on half the line. A p Q B Let the straight line AB be divided equally at P, and unequally at Q : then the rect. AQ, QB and the sq. on PQ shall be to- gether equal to the sq. on PB. On PB describe the square PCDB. Join BC. Through Q draw QE par 1 to BD, cutting BC in F. Through F draw LFHG par 1 to AB. Through A draw AG par 1 to BD. Now the complement PF = the complement FD : to each add the fig. QL; then the fig. PL = the fig. QD. But the fig. PL = the fig. AH, for they are par equal bases and between the same par 18 . .-. the fig. AH = the fig. QD. To each add the fig. PF; then the fig. AF = the gnomon PLE. Now the fig. AF = the rect. AQ, QB, for QB = QF ; .'. the rect. AQ, QB = the gnomon PLE. To each add the sq. on PQ, that is, the fig. HE ; n. 4. then the rect. AQ, QB with the sq. on PQ the gnomon PLE with the fig. HE = the whole fig. PD, which is the sq. on PB. i. 46. i. 31. i. 43. 18 on i. 36. BOOK II. PROP. 5. 129 That is, the rect. AQ, QB and the sq. on PQ are together equal to the sq. on PB. Q.E.D. COROLLARY. From this Proposition it follows that the difference of the squares on two straight lines is equal to the rectangle contained by their sum and difference. For let X and Y be the given A P Q Q st. lines, of which X is the greater. Draw AP equal to X, and pro- X duce it to B, making PB equal to Y AP, that is to X. From PB cut off PQ equal to Y. Then AQ is equal to the sum of X and Y, and QB is equal to the difference of X and Y. Now because AB is divided equally at P and unequally at Q, .-. the rect. AQ, QB with sq. on PQ = the sq. on PB; n. 5. that is, the difference of the sqq. on PB, PQ=the rect. AQ, QB, or, the difference of the sqq. on X and Y = the rect. contained by the sum and the difference of X and Y. CORRESPONDING ALGEBRAICAL FORMULA. This result may be written AQ.QB + PQ 2 =PB 3 . Let AB = 2a; and let PQ = 6; then AP and PB each = a. Also AQ = a + 6; andQB = a-6. Hence we have (a or EXERCISE. In the above figure shew that AP is half the sum of AQ and QB and that PQ is half their difference. 92 130 EUCLID'S ELEMENTS. PROPOSITION 6. THEOREM. Jf a straight line is 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 on half the line bisected, is equal to the square on the straight line made up of the half and the part produced. A P B Q i. 46. 1.31. i.43. Let the straight line AB be bisected at P, and pro- duced to Q : then the rect. AQ, QB and the sq. on PB shall be to- gether equal to the sq. on PQ. On PQ describe the square PCDQ. Join QC. Through B draw BE par 1 to QD, meeting QC in F. Through F draw LFHG par 1 to AQ. Through A draw AG par 1 to QD. Now the complement PF = the complement FD. But the fig. PF = the fig. AH ; for they are par" 18 on equal bases and between the same par 18 . I, 36, .-. the fig. AH = the fig. FD. To each add the fig. PL; then the fig. AL = the gnomon PLE. Now the fig. AL = the rect. AQ, QB, for QB = QL ; .'. the rect. AQ, QB = the gnomon PLE. To each add the sq. on PB, that is, the fig. HE; then the rect. AQ, QB with the sq. on PB = the gnomon PLE with the fig. HE = the whole fig. PD, which is the square on PQ. That is, the rect. AQ, QB and the sq. on PB are together equal to the sq. on PQ. Q.E.D. BOOK II. PROP. 6. CORRESPONDING ALGEBRAICAL FORMULA. This result may be written AQ.QB + PB 2 =PQ 2 . Let AB=2a; and let PQ=6; then AP and PB each = a. AlsoAQ = a + &; and QB = &- + PD 2 is constant, being equal to 36. The squares on the diagonals of a trapezium are together equal to the sum of the squares on its two oblique sides, with twice the rectangle contained by its parallel sides. PROBLEMS. 37. Construct a rectangle equal to the difference of two squares. 38. Divide a given straight line into two parts so that the rect- angle contained by thein may be equal to the square described on a given straight line which is less than half the straight line to be divided. 39. Given a square and one side of a rectangle which is equal to the square, find the other side. 40. Produce a given straight line so that the rectangle contained by the whole line thus produced and the part produced, may be equal to the square on another given line. 41. Produce a given straight line so that the rectangle contained by the whole line thus produced and the given line shall be equal to the square on the part produced. 42. Divide a straight line AB into two parts at C, such that the rectangle contained by BC and another line X may be equal to tho square on AC. PAET II. BOOK III. Book II F. deals with the properties of Circles. DEFINITIONS. 1. A circle is a plane figure bounded by one line, which is called the circum- ference, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal to one another : this point is called the centre of the circle. 2. A radius of a circle is a straight line drawn from the centre to the circumference. 3. A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference. 4. A semicircle is the figure bounded by a diameter of a circle and the part of the circumference cut off' by the diameter. From these definitions we draw the following inferences 1 . (i) The distance of a point from the centre of a circle is less than the radius, if the point is within the circumference : and the distance of a point from the centre is greater than the radius, if the point is without the circumference. (ii) A point is within a circle if its distance from the centre is less than the radius : and a point is without a circle if its distance from the centre is greater than the radius. (iii) Circles of equal radius are equal in all respects; that is to say, their areas and circumferences are equal. (iv) A circle is divided by any diameter into two parts which are equal in all respects. 150 EUCLID'S ELEMENTS. 5. Circles which have the same centre are said to be concentric. 6. An arc of a circle is any part of the circumference. 7. A chord of a circle is the straight line which joins any two points on the circumference. From these definitions it may be seen that a chord of a circle, which does not pass through the centre, divides the circumference into two unequal arcs ; of these, the greater is called the major arc, and the less the minor arc. Thus the major arc is greater, and the minor arc less than the semicircumference. The major and minor arcs, into which a cir- cumference is divided by a chord, are said to be conjugate to one another. 8. Chords of a circle are said to be equidistant from the centre, when the perpendiculars drawn to them from the centre are equal : and one chord is said to be further from the centre than another, when the per- pendicular drawn to it from the centre is greater than the perpendicular drawn to the other. 9. A secant of a circle is a straight line of indefinite length, which cuts the circumference in two points. 10. A tangent to a circle is a straight line which meets the circumference, but being produced, does not cut it. Such a line is said to touch the circle at a point; and the point is called the point of contact. DEFINITIONS. Jf a secant, which cuts a circle at the points P and Q, gradually changes its position in such a way that P remains fixed, the point Q will ultimately approach the fixed point P, until at length these points may be made to coincide. When the straight line PQ reaches this limiting position, it becomes the tangent to the circle at the point P. Hence a tangent may be defined as a straight line which passes through two coinci- dent points on the circumference. 151 11. Circles are said to touch one another when they meet, but do not cut one another. _ When each of the circles which meet is outside the other, they are said to touch one another externally, or to have external contact" when one of the circles is within the other, they are said to touch one another internally, or to have internal contact. 12. A segment of a circle is the figure bounded by a chord and one of the two arcs into which the chord divides the circumference. The chord of a segment is sometimes called its base. 152 EUCLID'S ELEMENTS. 13. An angle in a segment is one formed by two straight lines drawn from any point in the arc of the segment to the 'extremities of its chord. [It will be shewn in Proposition 21, that all angles in the same segment of a circle are equal.] 14. An angle at the circumference of a circle is one formed by straight lines drawn from a point on the circumference to the extremities of an arc : such an angle is said to stand upon the arc, which it subtends. 15. Similar segments of circles are those which contain equal angles. 16. A sector of a circle is a figure bounded by two radii and the arc inter- cepted between them. SYMBOLS AND ABBREVIATIONS. In addition to the symbols and abbreviations page 10, we shall use the following. for circle, Q for circumference. BOOK in. PROP. 1. 153 PROPOSITION 1. PROBLEM. To find the centre of a given circle. E Let ABC be a given circle: it is required to find its centre. In the given circle draw any chord AB, and bisect AB at D. i. 10. From D draw DC at right angles to AB; I. 11. and produce DC to meet the O ce at E and C. Bisect EC at F. i. 10. Then shall F be the centre of the ABC. First, the centre of the circle must be in EC : for if not, let the centre be at a point G without EC. Join AG, DG, BG. Then in the A s GDA, GDB, ( DA^DB, Conxtr. Because - and GD is common ; [ and GA GB, for by supposition they are radii: .'. the ^.GDA^the _GDB; I. 8. .'. these angles, being adjacent, are rt. angles. But the L. CDB is a rt. angle ; Constr. .'. the /.GDB = the /.CDB, Ax. 11. the part equal to the whole, which is impossible. .'. G is not the centre. So it may be shewn that no point outside EC is the centre ; .'. the centre lies in EC. .'. F, the middle point of the diameter EC, must be the centre of the ABC. Q.E.F. COROLLARY. The straight line which bisects a chord of a circle at right angles pauses through the centre. [For Exercises, see page 156.] 154 EUCLID'S ELEMENTS. PROPOSITION 2. THEOREM. If any two points are taken in the circumference of a circle, the chord which joins them falls within the circle. Let ABC be a circle, and A and B any two points on its O ce : then shall the chord AB fall within the circle. Find D, the centre of the 0ABC; in. 1. and in AB take any point E. Join DA, DE, DB. In the A DAB, because DA- DB, in. Def. 1. .*. the L DAB = the L DBA. I. 5. But the ext. L DEB is greater than the int. opp. /.DAE; i. 16. .'. also the L DEB is greater than the L DBE; .'. in the A DEB, the side DB, which is opposite the greater angle, is greater than DE which is opposite the less: I. 19. that is to say, DE is less than a radius of the circle ; .'. E falls within the circle. So also any other point between A and B may be shewn to fall within the circle. .'. AB falls within the circle. Q. E. D. DEFINITION. A part of a curved line is said to be concave to a point when, any chord being taken in it, all straight lines drawn from the given point to the intercepted arc are cut by the chord : if, when any chord is taken, no straight line drawn from the given point to the intercepted arc is cut by the chord, the curve is said to be convex to that point. Proposition 2 ^proves that the whole circumference of a circle is concave to its centre. BOOK III. PROP. 3. 155 PROPOSITION 3. THEOREM. If a straight line drawn through the centre of a circle bisects a chord which does not pass through the centre, it shall cut it at right angles : and, conversely, if it cut it at right angles, it shall bisect it. B Let ABC be a circle ; and let CD be a st. line drawn through the centre, and AB a chord which does not pass through the centre. First. Let CD bisect AB at F : then shall CD cut AB at rt. angles. Find E, the centre of the circle; in. 1. and join EA, EB. Then in the A s AFE, BFE, ( AF = BF, Hyp. Because -| and FE is common ; ( and AE = BE, being radii of the circle ; .'. the L AFE = the L BFE; i. 8. .'. these angles, being adjacent, are rt. angles, that is, DC cuts AB at rt. angles. Q.E. D. Conversely. Let CD cut AB at rt. angles : then shall CD bisect AB at F. As before, find E the centre ; and join EA, EB. In the AEAB, because EA= EB, in. Def. 1. .'.the L EAB = the L EBA. I. 5. Then in the A s EFA, EFB, ( the L. EAF = the L EBF, Proved. Because < and the L. EFA = the L. EFB, being rt. angles; Hyp. ( and EF is common; .'. AF=BF. l. 26. Q. E.D. [For Exercises, see page 156.] 156 EUCLID'S ELEMENTS. EXERCISES. ON PROPOSITION 1. 1. If two circles intersect at the points A, B, shew that the line which joins their centres bisects their common chord AB at right angles. 2. AB, AC are two equal chords of a circle; shew that the straight line which bisects the angle BAG passes through the centre. 3. Two chords of a circle are given in position and magnitude: find the centre of the circle. 4. Describe a circle that shall pass through three given points, which are not in the same straight line. 5. Find the locus of the centres of circles which pass thromjli tico given points. 6. Describe a circle that shall pass through two given points, and have a given radius. ON PROPOSITION 2. 7. A straight line cannot cut a circle in more titan two jwints. ON PROPOSITION 3. 8. Through a given point within a circle draw a chord which shall be bisected at that point. 9. The parts of a straight line intercepted between the circum- i'erences of two concentric circles are equal. 10. The line joining the middle points of two parallel chords of a circle passes through the centre. 11. Find the locus of the middle points of a system of parallel chords drawn in a circle. 12. If two circles cut one another, any two parallel straight lines drawn through the points of intersection to cut the circles, are equal. 13. PQ and XY are two parallel chords in a circle : shew that the points of intersection of PX. QY, and of PY, QX, lie on the straight line which passes through the middle points of the given chords. BOOK III. PROP. 4. 157 PROPOSITION 4. THEOREM. If in a circle two chords cut one another, which do not both pass through the centre, they cannot both be bisected at their point of intersection. Let ABCD be a circle, and AC, BD two chords which intersect at E, but do not both pass through the centre: then AC and BD shall not be both bisected at E. CASE I. If one chord passes through the centre, it is a diameter, and the centre is its middle point; .'. it cannot be bisected by the other chord, which by hypo- thesis does not pass through the centre. CASE II. If neither chord passes through the centre; then, if possible, let E be the middle point of both; that is, let AE = EC; and BE= ED. Find F, the centre of the circle: in. 1. Join EF. Then, because FE, which passes through the centre, bisects the chord AC, Hyp- .'. the L. FEC is a rt. angle. in. 3. And because FE, which passes through the centre, bi- sects the chord BD, Hyp. .'. the L FED is a rt. angle. /. the L. FEC = the L FED, the whole equal to its part, which is impossible. .'. AC and BD are not both bisected at E. Q. E. D. [For Exercises, see page 158.] 158 EUCLID'S ELEMENTS. PROPOSITION 5. THEOREM. If two circles cut one another, they cannot have the same centre. Let the two s AGO, BFC cut one another at C: then they shall not have the same centre. For, if possible, let the two circles have the same centre; and let it be called E. Join EC; and from E draw any st. line to meet the O ces at F and G. Then, because E is the centre of the 0AGC, Hyp- :. EG = EC. And because E is also the centre of the BFC, Hyp. :. EF= EC. /. EG = EF, the whole equal to its part, which is impossible. .'. the two circles have not the same centre. Q. E. i>. EXERCISES. ON PROPOSITION 4. 1. If a parallelogram can be inscribed in a circle, the point of intersection of its diagonals must be at the centre of the circle. 2. Kectangles are the only parallelograms that can be inscribed in a circle. ox PROPOSITION o. 3. Two circles, which intersect at one point, must also intersect at another. BOOK III. PROP. 6. 159 PROPOSITION 6. THEOREM. If two circles touch one another internally, they cannot have the same centre. B Let the two O s ABC, DEC touch one another internally at C: then they shall not have the same centre. For, if possible, let the two circles have the same centre; and let it be called F. Join FC; and from F draw any st. line to meet the O " 3 at E and B. Then, because F is the centre of the ABC, Hyp. .'. FB=^FC. And because F is the centre of the DEC, Hyp. :. FE = FC. .'. FB = FE; the whole equal to its part, which is impossible. .". the two circles have not the same centre. Q.E.I). NOTE. From Propositions 5 and 6 it is seen that circles, whose circumferences have any point in common, cannot be concentric, unless they coincide entirely. Conversely, the circumferences of concentric circles can have no point in common. H. E. 11 160 EUCLID'S ELEMENTS. PROPOSITION 7. THEOREM. If from any point within a circle which in not the centre, straight lines are drawn to the circumference, the greatest is that which passes through the centre ; and the least is that which, when produced backwards, passes through the centre : and of all other such lines, that which is nearer to the greatest is always greater than one more remote : also two equal straight lines, and only two, can be drawn from the given point to the circumference, one on each side of the diameter. Let ABCD be a circle, within which any point F is taken, which is not the centre: let FA, FB, FC, FG be drawn to the O ee , of which FA passes through E the centre, and FB is nearer than FC to FA, and FC nearer than FG : and let FD be the line which, when produced backwards, passes through the centre : then of all these st. lines (i) FA shall be the greatest; (ii) FD shall be the least; (iii) FB shall be greater than FC, and FC greater than FG; (iv) also two, and only two, equal st. lines can be drawn from F to the O ce - Join EB, EC, EG. (i) Then in the A FEB, the two sides FE, EB are together greater than the third side FB. i. 20. But E'B = EA, being radii of the circle; .'. FE, EA are together greater than FB; that is, FA is greater than FB. BOOK III. PROP. 7. 161 Similarly FA may be shewn to be greater than any other st. line drawn from F to the O ce ; .'. FA is the greatest of all such lines. (ii) In the A EFG, the two sides EF, FG are together greater than EG ; I. 20. and EG = ED, being radii of the circle; .'. EF, FG are together greater than ED. Take away the common part EF; then FG is greater than FD. Similarly any other st. line drawn from F to the O may be shewn to be greater than FD. .'. FD is the least of all such lines. (iii) In the A 3 BEF, CEF, ( BE = CE, IU. Def. 1. Because < and EF is common; ( but the L. BEF is greater than the L. CEF; .'. FB is greater than FC. i. 24. Similarly it may be shewn that FC is greater than FG. (iv) At E in FE make the L. FEH equal to the _ FEG. i. 23. Join FH. Then in the A 8 GEF, HEF, ( GE = HE, in. Def. 1. Because \ and EF is common; (also the L. GEF = the L HEF; Constr. .'. FG = FH. I. 4. And besides FH no other straight line can be drawn from F to the O ce equal to FG. For, if possible, let FK = FG. Then, because FH FG, Proved. .'. FK= FH, that is, a line nearer to FA, the greatest, is equal to a line which is more remote; which is impossible. Proved. .'. two, and only two, equal st. lines can be drawn from F to the O ce . Q.E.D. 11-2 162 EUCLID'S ELEMENTS. PROPOSITION 8. THEOREM. If from any point without a circle straight lines are drawn to the circumference, of those which fall on tJie concave cir- cumference, the greatest is that which passes through the centre ; and of others, that which is nearer to the greatest is always greater than one more remote : f^< / rther, of those which fall on the convex circumference, the least is that which, when produced, passes through the centre; and of others that which is nearer to the least is always less than one more remote: lastly, from the given point there can be drawn to the circumference two, and only two, equal straight lines, one on each side of the shortest line. Let BG D be a circle of which C is the centre ; and let A be any point outside the circle : let ABD, AEH, AFG, be st. lines drawn from A, of which AD passes through C, the centre, and AH is nearer than AG to AD : then of st. lines drawn from A to the concave O ce , (i) AD shall be the greatest, and (ii) AH greater than AG : and of st. lines drawn from A to the convex O ce } (iii) AB shall be the least, and (iv) AE less than AF. (v) Also two, and only two, equal st. lines can be drawn from A to the O ce - Join CH, CG, CF, CE. (i) Then in the A ACH, the two sides AC, CH are together greater than AH : i. 20. but CH = CD, being radii of the circle: .'. AC, CD are together greater than AH: that is, AD is greater than AH. Similarly AD may be shewn to be greater than any other st. line drawn from A to the concave O ce ; .'. AD is the greatest of all such lines. BOOK III. PROP. 8. 163 Cii) In the A s HCA, GCA, ( HC = GC, ni. Def, 1. Because -I and CA is common; ( but the L. HCA is greater than the L. GCA; .'. AH is greater than AG. i. 24. (iii) In the A A EC, the two sides AE, EC are together greater than AC : I. 20. but EC=BC; m. Def. 1. .'. the remainder AE is greater than the remainder AB. Similarly any other st. line drawn from A to the convex O ce niay be shewn to be greater than AB; .'. AB is the least of all such lines. (iv) In the A AFC, because AE, EC are drawn from the extremities of the base to a point E within the triangle, .'. AF, FC are together greater than AE, EC. I. 21. But FC = EC, in. Def, 1. .'. the remainder AF is greater than the remainder AE. (v) At C, in AC, make the L ACM equal to the /_ ACE. Join AM. Then in the two A 8 EC A, MCA, ( EC = MC, in. Def. 1. Because < and CA is common ; ( also the L EC A = the /_ MCA; Constr. .'.AE^AM; 1.4. and besides AM, no st. line can be drawn from A to the O ce , equal to AE. For, if possible, let AK = AE : then because AM = AE, Proved. AM = AK; that is, a line nearer to the shortest line is equal to a line which is more remote ; which is impossible. Proved. .'. two, and only two, equal st. lines can be drawn from A to the O ce . Q.E.D. Where are the limits of that part of the circumference which is concave to the point A? 164 EUCLID'S ELEMENTS. Obs. Of the following proposition Euclid gave two distinct proofs, the first of which has the advantage of being direct. PROPOSITION 9. THEOREM. [FIRST PROOF.] If from a point within a circle more tlvan two equal straight lines can be drawn to the circumference, that point is the centre of the circle. Let ABC be a circle, and D a point within it, from which more than two equal st. lines are drawn to the O ce , namely DA, DB, DC : then D shall be the centre of the circle. Join AB, BC : and bisect AB, BC at E and F respectively. i. 10. Join DE, DF. Then in the A s DEA, DEB, f EA = EB, Conslr. Because < and DE is common; ( and DA= DB; .'. the L DEA = the i_ DEB; .'. these angles, being adjacent, are rt. angles. Hence ED, which bisects the chord AB at rt. angles, must pass through the centre. in. 1. Cor. Similarly it may be shewn that FD passes through the centre. .'. D, which is the only point common to ED and FD, must be the centre. Q.E.D. BOOK III. PROP. 9. 165 PROPOSITION 9. THEOREM. [SECOND PROOF.] If from a point within a circle more than two equal straight lines can be drawn to the circumference, that point is the centre of the circle. Let ABC be a circle, and D a point within it, from which more than two equal st. lines are drawn to the O ce , namely DA, DB, DC : then D shall be the centre of the circle. For, if not, suppose E to be the centre. Join DE, and produce it to meet the O ce at F, G. Then because D is a point within the circle, not the centre, and because DF passes through the centre E; .'. DA, which is nearer to DF, is greater than DB, which is more remote : in. 7. but this is impossible, since by hypothesis, DA, DB, are equal. .'. E is not the centre of the circle. *Ancl wherever we suppose the centre E to be, other- wise than at D, two at least of the st. lines DA, DB, DC may be shewn to be unequal, which is contrary to hypo- thesis. .'. D is the centre of the O ABC. Q.E.D. * NOTE. For example, if the centre E were supposed to be within the angle BDC, then DB would be greater than DA; if within the angle ADB, then DB would be greater than DC ; if on one of the three straight lines, as DB, theii DB would be greater than both DA and DC, 166 EUCLID'S ELEMENTS. 06s. Two proofs of Proposition 10, both indirect, were given by Euclid. PROPOSITION 10. THEOREM. [FIRST PROOF.] One circle cannot cut another at more t/ian two points. If possible, let DABC, EABC be two circles, cutting one another at more than two points, namely at A, B, C. Join AB, BC. Draw FH, bisecting AB at rt. angles; I. 10, 11. and draw GH bisecting BC at rt. angles. Then because AB is a chord of both circles, and FH bisects it at rt. angles, .'. the centre of both circles lies in FH. in. I. Cor. Again, because BC is a chord of both circles, and GH bisects it at right angles, .'. the centre of both circles lies in GH. in. I. Cor. Hence H, the only point common to FH and GH, is the centre of both circles ; which is impossible, for circles which cut one another cannot have a common centre. in. 5. .'. one circle cannot cut another at more than two points. COROLLARIES, (i) Two circles cannot meet in three points without coinciding entirely. (ii) Two circles cannot have a common arc without coinciding entirely. (iii) Only one circle can be described through three points, which are not in the same straight line. BOOK III. PROP. 10. 167 PROPOSITION 10. THEOREM. [SECOND PROOF.] One circle cannot cut another at more than two points. D If possible, let DABC, EABC be two circles, cutting one another at more than two points, namely at A, B, C. Find H, the centre of the DABC, in. 1. and join HA, HB, HC. Then since H is the centre of the DABC, .'. HA, HB, HC are all equal. in. Def. 1. And because H is a point within the EABC, from which more than two equal st. lines, namely HA, HB, HC are drawn to the O ce , .'. H is the centre of the EABC : in. 9. that is to say, the two circles have a common centre H ; but this is impossible, since they cut one another, in. 5. Therefore one circle cannot cut another in more than two points. Q.E.D. NOTE. This proof is imperfect, because it assumes that the centre of the circle DABC must fall within the circle EABC; whereas it may be conceived to fall either without the circle EABC, or on its circumference. Hence to make the proof complete, two additional cases are required. 168 EUCLID'S ELEMENTS. PROPOSITION 11. THEORKM. If two circles touch one another internally, the straight line which joins their centres, being produced, shall pass through the point of contact. Let ABC and ADE be two circles which touch one another internally at A ; let F be the centre of the ABC, and G the centre of the ADE: then shall FG produced pass through A. If not, let it pass otherwise, as FGEH. Join FA, GA. Then in the A FGA, the two sides FG, GA are together greater than FA : I. 20. but FA = FH, being radii of the ABC : Hyp. .'. FG, GA are together greater than FH. Take away the common part FG ; then GA is greater than GH. But GA = GE, being radii of the ADE : Hyp. .'. GE is greater than GH, the part greater than the whole ; which is impossible. .'. FG, when produced, must pass through A. Q.E.D. EXERCISES. 1. If the distance between the centres of two circles is equal to the difference of their radii, then the circles must meet in one point, but in no other; that is, they must touch one another. 2. If two circles whose centres are A and B touch one another internally, and a straight line be drawn through their point of contact, cutting the circumferences at P and Q; shew that the radii AP and BQ. are parallel. BOOK III. PROP. 12. 169 PROPOSITION 12. THEOREM. Jf two circles touch one another externally, the straight line which joins their centres shall pass through the point of contact. Let ABC and ADE be two circles which touch one another externally at A; let F be the centre of the ABC, and G the centre of the ADE : then shall FG pass through A. If not, let FG pass otherwise, as FHKG. Join FA, GA. Then in the A FAG, the two sides FA, GA are together greater than FG : I. 20. but FA FH, being radii of the ABC ; Hyp. and GA = GK, being radii of the ADE ; Hyp. .'. FH and GK are together greater than FG : which is impossible. .'. FG must pass through A. EXERCISES. 1. Find the locus of the centres of all circles which touch a given circle at a given point. 2. Find the locus of the centres of all circles of given radius, which touch a given circle. 3. If the distance between the centres of two circles is equal to the sum of their radii, then the circles meet in one point, but in no other ; that is, they touch one another. 4. If two circles ivhose centres are A and B touch one another externally, and a straight line lie drawn through their point of contact cutting the circumferences at P and Q; shew that the radii AP and BQ. are parallel. 170 EUCLID'S ELEMENTS. PROPOSITION 13. THEOREM. Two circles cannot touch one another at more than one point, whether internally or externally. Fig. 1 Fig. 2 G D G If possible, let ABC, EOF be two circles which touch one another at more than one point, namely at B and D. Join BD; and draw GF, bisecting BD at rt. angles, i. 10, 11. Then, whether the circles touch one another internally, as in Fig. 1, or externally as in Fig. 2, because B and D are on the O ces of both circles, .'. BD is a chord of both circles : .'. the centres of both circles lie in GF, which bisects BD at rt. angles. in. 1. Cor. Hence GF which joins the centres must pass through a point of contact; III. 11, and 12. which is impossible, since B and D are without GF. .". two circles cannot touch one another at more than one point. Q.E.D. NOTE. It must be observed that the proof here given applies, by virtue of Propositions 11 and 12, to both the above figures : we have therefore omitted the separate discussion of Fig. 2, which finds a place in most editions based on Simson's text. BOOK III. PROP. 13. 171 EXERCISES ON PROPOSITIONS 1 13. 1. Describe a circle to pass through two given points and have its centre on a given straight line. When is this impossible ? 2. All circles which pass through a fixed point, and have their centres on a given straight line, pass also through a second fixed point. 3. Describe a circle of given radius to touch a given circle at a given point. How many solutions will there be? When will there be only one solution? 4. From a given point as centre describe a circle to touch a given circle. How many solutions will there be? 5. Describe a circle to pass through a given point, and touch a given circle at a given point. [See Ex. 1, p. 169 and Ex. 5, p. 156.] When is this impossible? 6. Describe a circle of given radius to touch two given circles. [See Ex. 2, p. 169.] How many solutions will there be ? 7. Two parallel chords of a circle are six inches and eight inches in length respectively, and the perpendicular distance between them is one inch : find the radius. 8. If two circles touch one another externally, the straight lines, which join the extremities of parallel diameters towards opposite parts, must pass through the point of contact. 9. Find the greatest and least straight lines which have one extremity on each of two given circles, which do not intersect. 10. In any segment of a circle, of all straight lines drawn at right angles to the chord and intercepted between the chord and the arc, the greatest is that which passes through the middle point of the chord ; and of others that which is nearer the greatest is greater than one more remote. 11. If from any point on the circumference of a circle straight lines be drawn to the circumference, the greatest is that which passes through the centre; and of others, that which is nearer to the greatest is greater than one more remote ; and from this point there can be drawn to the circumference two, and only two, equal straight lines. 172 EUCLIDS ELEMENTS. PROPOSITION 14. THEOREM. Equal chords in a circle are equidistant from the centre and, conversely, chords which are equidistant from ti centre are equal. Let ABC be a circle, and let AB and CD be chords, of which the perp. distances from the centre are EF and EG. First, Let AB = CD : then shall AB and CD be equidistant from the centre E. Join EA, EC. Then, because EF, which passes through the centre, is perp. to the chord AB; Hyp. .'. EF bisects AB ; in. 3. that is, AB is double of FA. For a similar reason, CD is double of GC. But AB = CD, .'. FA = GC. Now EA = EC, being radii of the circle; .'. the sq. on EA = the sq. on EC. But the sq. on EA = the sqq. on EF, FA ; for the L. EFA is a rt. angle. And the sq. on EC = the sqq. on EG, GC ; for the L. EGC is a rt. angle. .'. the sqq. on EF, FA = the sqq. on EG, GC. Now of these, the sq. on FA = the sq. on GC ; for FA = GC. .'. the sq. on EF = the sq. on EG, .'. EF=EG ; that is, the chords AB, CD are equidistant from the centre. Q.E.D. Hyp. i. 47. BOOK III. PROP. 14. 173 Conversely. Let AB and CD be equidistant from the centre E ; that is, let EF = EG : then shall AB^CD. For, the same construction being made, it may be shewn as before that AB is double of FA, and CD double of GC; and that the sqq. on EF, FA = the sqq. on EG, GC. Now of these, the sq. on EF the sq. on EG, for EF = EG : Hyp. .'. the sq. on FA = the square on GC ; .'. FA = GC; and doubles of these equals are equal ; that is, AB-CD. Q.E.D: EXERCISES. 1. Find the locus of the middle points of equal chords of a, circle. 2. If two chords of a circle cut one another, and make equal angles with the straight line which joins their point of intersection to the centre, they are equal. 3. If two equal chords of a circle intersect, shew that the segments of the one are equal respectively to the segments of the other. 4. In a given circle draw a chord which shall be equal to one given straight line (not greater than the diameter) and parallel to another. 5. PQ is a fixed chord in a circle, and AB is any diameter : shew that the sum or difference of the perpendiculars let fall from A and B on PQ is constant, that is, the same tor all positions of AB. 174 EUCLID'S ELEMENTS. PROPOSITION 15. THEOREM. Tlie diameter is the greatest chord in a circle ; and of others, that which is nearer to the centre is greater than one more remote: conversely, the greater chord is nearer to the centre tJian the less. Let ABCD be a circle, of which AD is a diameter, and E the centre ; and let BC and FG be any two chords, whose perp. distances from the centre are EH and EK : then (i) AD shall be greater than BC : (ii) if EH is less than EK, BC shall be greater than FG : (iii) if BC is greater than FG, EH shall be less than EK. (i) Jbin EB, EC. Then in the A BEC, the two sides BE, EC are together greater than BC ; I. 20. but BE = AE, ill. Def. 1. and EC ED ; .'. AE and ED together are greater than BC ; that is, AD is greater than BC. Similarly AD may be shewn to be greater than any other chord, not a diameter. (ii) Let EH be less than EK; then BC shall be greater than FG. Join EF. Since EH, passing through the centre, is perp. to the chord BC, .'. EH bisects BC ; ill. 3. BOOK In. PROP. 15. 175 that is, BC is double of HB. For a similar reason FG is double of KF. Now EB = EF, in. Def. 1. .'. the sq. on EB = the sq. on EF. But the sq. on EB = the sqq. on EH, HB; for the L. EHB is a rt. angle ; I. 47. also the sq. on EF = the sqq. on EK, KF; for the L EKF is a rt. angle. .'. the sqq. on EH, HB = the sqq. on EK, KF. But the sq. on EH is less than the sq. on EK, for EH is less than EK ; .'. the sq. on HB is greater than the sq. on KF .'. HB is greater than KF : hence BC is greater than FG. (iii) Let BC be greater than FG ; then EH shall be less than EK. For since BC is greater than FG, .', HB is greater than KF ; ,'. the sq. on HB is greater than the sq, on KF, But the sqq. on EH, HB = the sqq. on EK, KF : Proved, ,'. the sq. on EH is less than the sq. on EK ; .'. EH is less than EK. EXERCISES. 1. Throuyh a given point within a circle draiv the least possible chord. 2. AB is a fixed chord of a circle, and XY any other chord having its middle point Z on AB: what is the greatest, and what the least length that XY may have? Shew that XY increases, as Z approaches the middle point of AB. 3. In a given circle draw a chord of given length, having its middle point on a given chord. When is this problem impossible? H. E. 12 176 EUCLID'S ELEMENTS. 06s. Of the following proofs of Proposition 16, the second (by reductio ad absurdum) is that given by Euclid ; but the first is to be preferred, as it is direct, and not loss simple than the other. PROPOSITION 16. THEOREM. [ALTERNATIVE PROOF.] The straight line drawn at right angles to a diameter of a circle at one of its extremities is a tangent to the circle: and every other straight line drawn through this point cuts the circle. Let AKB be a circle, of which E is the centre, and AB a diameter; and through B let the st. line CBD be drawn at rt. angles to AB : then (i) CBD shall be a tangent to the circle; (ii) any other st. line through B, as BF, shall cut the circle. (i) In CD take any point G, and join EG. Then, in the A EBG, the L EBG is a rt. angle; Hyp. .'. the L EGB is less than a rt. angle; i. 17. .'. the L EBG is greater than the L. EGB; .'. EG is greater than EB: I. 19. that is, EG is greater than a radius of the circle; .'. the point G is without the circle. Similarly any other point in CD, except B, may be shewn to be outside the circle : hence CD meets the circle at B, but being produced, does not cut it; that is, CD is a tangent to the circle. ui.Def. 10, BOOK III. PROP. 16. 177 (ii) Draw EH perp. to BF. I. 12. Then in the A EH B, because the L. EHB is a rt. angle, .'. the L EBH is less than a rt. angle; I. 17. .'. EB is greater than EH; I. 19. that is, EH is less than a radius of the circle : .'. H, a point in BF, is within the circle; .'. BF must cut the circle. Q. K. D. PROPOSITION 16. THEOREM. [EUCLID'S PROOF.] The straight line drawn at right angles to a diameter of a circle at one of its extremities, is a tangent to the circle: and no other straight line can be drawn through this point so as not to cut the circle. E Let ABC be a circle, of which D is the centre, and AB a diameter; let AE be drawn at rt. angles to BA, at its extremity A : (i) then shall AE be a tangent to the circle. For, if not, let AE cut the circle at C. Join DC. Then in the A DAC, because DA = DC, in. Def. 1. .'. the L. DAC = the L DCA. But the L. DAC is a rt. angle; H_!. .'. the ^ s ADC, ABC together = two rt. angles. Similarly it may be shewn that the _ s BAD, BCD together = two rt, angles. Q. E. I). EXERCISES. 1. If a circle can be described about a parallelogram, the parallelogram must be rectangular. 2. ABC is an isosceles triangle, and XY is drawn parallel to the base BC: shew that the four points B, C, X, Y lie on a circle. 3. If one side of a quadrilateral inscribed in a circle is produced, thf exterior angle is equal to the opposite interior angle of the quadri- lateral. BOOK III. PROP. 22. 189 PROPOSITION 22. [Alternative Proof.] Let A BCD be a quadrilateral inscribed in the ABC : then shall the / " ADC, ABC together = two rt. angles. Join FA, FC. Then the zAFC at the centre = twice the / ADC at the o ce , standing on the same arc ABC. in. 20. Also the reflex angle AFC at the centre = twice the / ABC at the O ce , standing on the same arc ADC. in. 20. Hence the / s ADC, ABC are together half the sum of the L AFC and the reflex angle AFC ; but these make up four rt. angles: i. lo. Cor. 2. .-. the L s ADC, ABC together=two rt. angles. Q.E.I). DEFINITION. Four or more points through which a circle may be described are said to be concyclic. CONVERSE OF PROPOSITION 22. If a pair of opposite angles of a quadrilateral are together equal to tico right angles, its vertices are concyclic. Let ABCD be a quadrilateral, in which the opposite angles at B and D together = two rt. angles ; then shall the four points A, B, C, D be concyclic. Through the three points A, B, C describe a circle : iv. 5. then this circle must pass through D. For, if not, it will cut AD, or AD produced, at some other point E. Join EC. Then since the quadrilateral ABCE is inscribed in a circle, .-. the z s ABC, AEC together = two rt. angles. m. 22. But the L s ABC, ADC together= two rt. angles ; Hyp. hence the / s ABC, AEC = the L s ABC, ADC. Take from these equals the / ABC: then the z A EC = the / ADC ; that is, an ext. angle of a triangle = an int. opp. angle; which is impossible. 1. 16. .'. the circle which passes through A, B, C cannot pass otherwise than through D : that is the four vertices A, B, C, D are concyclic. Q.E.D. 190 EUCLID'S ELEMENTS. DEFINITION. Similar segments of circles are those which contain equal angles. PROPOSITION 23. THEOREM. On the same chord and on the same side of it, there cannot be two similar segments of circles, not coinciding with one another. If possible, on the same chord AB, and on the same side of it, let there be two similar segments of circles ACS, ADB, not coinciding with one another. Then since the arcs ADB, ACB intersect at A and B, .'. they cannot cut one another again; in. 10, .'. one segment falls within the other. In the outer arc take any point D ; join AD, cutting the inner arc at C: join CB, DB. Then because the segments are similar, /. the /.ACB = the /.ADB; in. Def. that is, an ext. angle of a triangle = an int. opp. angle ; which is impossible. I. 16. Hence the two similar segments ACB, ADB, on the same chord AB and 011 the same side of it, must coincide. Q.E.D. EXERCISES ON PROPOSITION 22. 1. The straight lines which bisect any angle of a quadrilateral figure inscribed in a circle and the opposite exterior angle, meet on the circumference. 2. A triangle is inscribed in a circle : shew that the sum of the angles in the three segments exterior to the triangle is equal to four right angles. 3. Divide a circle into two segments, so that the angle contained by the one shall be double of the angle contained by the other. HOOK IIT. 1'KOI'. '24. 191 PROPOSITION' 24. THKORKM. Similar segments of circles OH equal chords are equal to one anotJier. B Let AEB and CFD be similar segments on equal chords AB, CD: then shall the segment ABE = the segment CDF. For if the segment ABE be applied to the segment CDF, so that A falls on C, and AB falls along CD; then since AB = CD, .'. B must coincide with D. .'. the segment AEB must coincide with the segment CFD ; for if not, on the same chord and on the same side of it there would be two similar segments of circles, not co- inciding with one another : which is impossible. in. 23. ,', the segment AEB = the segment CFD. Q. E. D. EXERCISES. 1. Of two segments standing on the same chord, the greater segment contains the smaller angle. 2. A segment of a circle stands on a chord AB, and P is any point on the same side of A B as the segment: shew that the angle APB is greater or less than the angle in the segment, according as P is within or without the segment. 3. P, Q, R are the middle points of the sides of a triangle, and X is the foot of the perpendicular let fall from one vertex on the opposite side : shew that the four points P, Q, R, X are coney die. [See page 96, Ex. 2: also page 100, Ex. 2.] 4. Use the preceding exercise to shew that the middle points of the sides of a triangle and the feet of the perpendiculars let fall from the vertices on the opposite sides, are concyclic. H. E, 13 192 EUCLID S KI.KMKXTS. PROPOSITION 25. PROBLEM*. An arc of a circle being given, to describe the whole cir- cumfei-ence of which the given arc is a part. Let ABC be an arc of a circle: it is required to describe the whole O ce of which the arc ABC is a part. In the given arc take any three points A, B, C. Join AB, BC. Draw DF bisecting AB at rt. angles, I. 10. 11. and draw EF bisecting BC at rt. angles. Then because DF bisects the chord AB at rt. angles, .'. the centre of the circle lies in DF. in. 1. Cor Again, because EF bisects the chord BC at rt. angles, .'. the centre of the circle lies in EF. in. 1. Cor. .'. the centre of the circle is F, the only point common to DF, EF. Hence the O ce of a circle described from centre F, with radius FA, is that of Avhich the given arc is a part. Q. K. F. * NOTE. Euclid gave this proposition a somewhat different form, as follows: A segment of a circle being given, to describe the circle of u-Tticli it is a segment. Let ABC be the given segment standing on the chord AC. Draw DB, bisecting AC at rt. angles. 1. 10. Join AB. At A. in BA, make the L BAE equal to the / ABD. i. 23. Let AE meet BD, or BD produced, at E. Then E shall be the centre of the required circle. [Join EC ; and prove (i) EA = EC ; i. 4. (ii) EA=EB. i. 6.] BOOK TTT. 1'KOP. 2(1 193 PROPOSITION 26. THEOREM. In equal circles the arcs which subtend equal angles, ellie.r at the centres or at fhe circumferences, shall be equal. K Let ABC, DEF be equal circles and let the /_ s CGC, EHF, at tlie centres be equal, and consequently the _ 8 BAG, EOF at the O ces equal: in. 20. then shall the arc BKC = the arc ELF. Join BC, EF. Then because the s ABC, DEF are equal, .'. their radii are equal. Hence in the A 8 BGC, EHF, I BG = EH, Because < .and GC = HF, (and the /_ BGC = the L EHF : ]fi/p. .'. BC = EF. l! 4. Again, because the L. BAG = the L EOF, .'. the segment BAC is similar to the segment EOF; in. Def. 15. and they are on equal chords BC, EF; .'. the segment BAC = the segment EOF. in. 24. But the whole ABC = the whole DEF; .'. the remaining segment BKC = the remaining segment ELF, .'. the arc BKC = the arc ELF. Q.E. D. [For an Alternative Proof ami Exercises see pp. 197, 198.] 13-3 194 ETt 'I. ID'S KI.KMKNTS. PROPOSITION 27. THEOREM. In equal circles the angles, whether at the centres or thft circumferences, which stand on equal arcs, shall be equal. Let ABC, DEF be equal circles, and let the arc BC = the arc EF : then shall the L. BGC = the _ EHF, at the centres ; and also the L BAG = the L EOF, at the O ces . If the _ s BGC, EHF are not equal, one must be the greater. If possible, let the L. BGC be the greater. At G, in BG, make the L. BGK equal to the L. EHF. I. 23. Then because in the equal s ABC, DEF, the L. BGK = the L EHF, at the centres : Constr. .'. the arc BK="the arc EF. in. 26. But the arc BC = the arc EF, Hyp. .'. the arc BK = the arc BC, a part equal to the whole, which is impossible. .'. the L. BGC is not unequal to the L. EHF that is, the L. BGC = the <_ EHF. And since the _ BAC at the O ce is half the _ BGC at the centre, in. 20. and likewise the L. EOF is half the _ EHF, .'. the L. BAC = the L. EOF. Q. E. D. [For Exercises see pp. 197, 198.] BOOK in. 1'Kor. 28. 195 PROPOSITION 28. THEOREM. lit, equal circles the arcs, which are cut off by equal chorda, shall be equal, the major arc equal to the major arc, and the minor to the minor. Let ABC, DEF be two equal circles, and let the chord BC = the chord EF : then shall the major arc BAG = the major arc EOF: and the minor arc BGC = the minor arc EHF. Find K and L the centres of the s ABC, DEF : and join BK, KC, EL, LF. Then because the s ABC, DEF are equal, .'. their radii are equal. Hence in the A* 1 BKC, ELF, ( BK = EL, Because -< KC = LF, (and BC = EFj .'. the L. BKC = the t_ ELF; .'. the arc BGC = the arc EHF; and these are the minor arcs. in. 1. Hyp. I. 8. in. 26. Hyp. But the whole O ee ABGC = the whole O cu DEHF; .'. the remaining arc BAG = the remaining arc EOF and these are the major arcs. Q.E.D. [1'or Exc-rci.se.s see pp. lit?, IDS.] 196 EUCLID'S KI.KMKNTS. PROPOSITION 29. THEOREM. In equal circles tlte chords, ivhich cut off equal arcs, shall be equal. A D Let ABC, DEF be equal circles, and let the arc BGC the arc EHF : then shall the chord BC = the chord EF. Find K, L the centres of the circles. Join BK, KC, EL, LF. Then in the equal s ABC, DEF, because the arc BGC = the arc EHF, /. the _ BKC = the _ ELF. Hence in the A s BKC, ELF, | BK = EL, being radii of equal circles; Because - KC = LF, for the same reason, [and the _ BKC -^ the _ ELF; Pnn-- of tJie arcs cut off by it, then the bisector of the angle APB cuts the conjugate arc in the same point, whatever be the position of P. 8. Two circles intersect at A and B ; anil through these points straight lines are drawn from any point P on the circumference of one of the circles: shew that when produced they intercept on the other circumference an arc which is constant for all positions of P. 9. A triangle ABC is inscribed in a circle, and the bisectors of the angles meet the circumference at X, Y, Z. Find each angle of the triangle XYZ in terms of those of the original triangle. ON PROPOSITIONS 28, 29. 10. The straight lines which join the extremities of parallel chords in a circle (i) towards the same parts, (ii) towards opposite parts, are equal. 11. Through A, a point of intersection of two equal circles two straight lines PAQ, XAY are drawn : shew that the chord PX is equal to the chord QY. 12. Through the points of intersection of two circles two parallel straight lines are drawn terminated by the circumferences : shew that the straight lines which join their extremities towards the same parts are equal. 13. Two equal circles intersect at A and B ; .and through A any straight line PAQ. is drawn terminated by the circumferences: shew that BP^BQ. 14. ABC is an isosceles triangle inscribed in a circle, and the bisectors of the base angles meet the circumference at X and Y. Shew that the figure BXAYC must have four of its sides equal. What relation must subsist among the angles of the triangle ABC, in order that the figure BXAYC may be equilateral? 1U8 EUCLID'S ELEMENTS. NOTE. We have given Euclid's demonstrations of Propositions n6, 27, 28, 29 ; but. it should be noticed that all these propositions ulso admit of direct proof by the method of superposition. To illustrate this method we will apply it to Proposition 26. PROPOSITION 26. [Alternative Proof.] In equal circles, the arcs which subtend equal aiujlca, IK hut her at ike centres or circumferences, shall be equal. D K Let ABC, DEF be equal circles, and let the / s BGC, EHF at the centres be equal, and consequently the / B BAG, EOF at the O"** equal : m. '20. then shall the arc BKC = the arc ELF. For if the ABC be applied to the DEF, so that the centre G may fall on the centre H, then because the circles are equal, Hyp. :. their O 068 must coincide ; hence by revolving the upper circle about its centre, the lower circle remaining fixed, B may be made to coincide with E, and consequently G B with H E. And because the / BGC = the / EHF, /. G C must coincide with H F : and since GC= H F, Hyp. .: C must fall on F. Now B coinciding with E, and C with F, and the o ce of the ABC with the o ce of the DEF, .. the arc BKC must coincide with the arc ELF. /. the arc BKC = the arc ELF. BOOK III. PROP. 30. 199 PROPOSITION 30. PROBLEM. To bisect a given arc. D Let ADB be the given arc: it is required to bisect it. Join AB; and bisect it at C. i. 10. At C draw CD at rt. angles to AB, meeting the given arc at D. I. 11. Then shall the arc ADB be bisected at D. Join AD, BD. Then in the A s ACD, BCD, {AC = BC, Constr. and CD is common; and the L ACD the L BCD, being rt. angles: .'. AD = BD. I. 4. And since in the 0ADB, the chords AD, BD are equal, .'. the arcs cut off by them are equal, the minor arc equal to the minor, and the major arc to the major: in. 28. and the arcs AD, BD are both minor arcs, for each is less than a semi-circumference, since DC, bisecting the chord AB at rt. angles, must pass through the centre of the circle. HI. 1. Cor. .'. the arc AD = the arc BD : that is, the arc ADB is bisected at D. Q. E. F. EXERCISES. 1. If a tangent to a circle is parallel to a chord, the point of contact will bisect the arc cut off by the chord. 2. Trisect a quadrant, or the fourth part of the circumference, of a circle. K LX'Lllr.S ELEMENTS. PROPOSITION 31. THEOREM. The angle in a semicircle is a right angle : tlie 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 BC is a diameter, and E the centre; and let AC be a chord dividing the circle into the segments ABC, ADC, of which the segment ABC is greater, and the segment is ADC less than a semicircle: then (i) the angle in the semicircle BAC shall be a rt. angle ; (ii) the angle in the segment ABC shall be less than a rt. angle ; (iii) the angle in the segment ADC shall be greater than a rt. angle. In the arc ADC take any point D; Join BA, AD, DC, AE; and produce BA to F. (i) Then because EA = EB, in. !) BEC ; /. the L BAG is half of the straight angle BEC: and a straight angle two rt. angles; .-. the / BAG ^ one half of two rt. angles, one rt. angle. NOTE 2. From Proposition 31 we may derive a simple practical solution of Proposition 17, namely, To draw a tangent to a circle from a given external point. Let BCD be the given circle, and A the given exter- nal point: it is required to draw from A a tangent to the BCD. Find E, the centre of the circle, and join AE. On AE describe the semi- circle ABE, to cut the given circle at B. Join AB. Then AB shall he a tangent to the BCD. For the / ABE, being in a semicircle, is a rt. angle. m. 31. .. AB is drawn at rt. angles to the radius EB, from its ex- tremity B; /. AB is a tangent to the circle. in. 16. Q.E.F. Since the semicircle might be described on either side of AE, it is clear that there will be a second solution of the problem, as shewn by the dotted lines of the figure. 203 PROPOSITION' 32. THEOREM. If a straight line touch a circle, and from the point of contact a chord be drawn, the angles which this chord makes with the tanyent shall be equal to the angles in the alternate segments of the circle. Let EF touch the given QABC at B, and let BD be a chord drawn from B, the point of contact: then shall (i) the L. DBF = the angle in the alternate segment BAD: (ii) the L DBE = the angle in the alternate segment BCD. From B draw BA perp. to EF. Take any point C in the arc BD; and join AD, DC, CB. I. 11. (i) Then because BA is drawn perp. to the tangent EF, at its point of contact B, .'. BA passes through the centre of the circle: ill. 19. .'. the L ADB, being in a semicircle, is a rt. angle: in. 31. .'. in the AABD, the other ;L s ABD, BAD together = a rt. ingle ; I. 32. that is, the L S ABD, BAD together = the L ABF. From these equals take the common L ABD; .'. the L DBF - the L. BAD, which is in the alternate seg- ment. 204 B (ii) Because ABCD is a quadrilateral inscribed in a circle, .'. the L* BCD, BAD together = two rt. angles: in. 22. but the _ s DBE, DBF together = two rt. angles; I. 13. /. the <_* DBE, DBF together = the ^ s BCD, BAD: and of these the L DBF = the L. BAD; Proved. .'. (he L. DBE = the L DCB, which is in the alternate seg- ment. Q. E. D. EXERCISES. 1. State and prove tlie converse of this proposition. 2. Use this Proposition to shew that the tangents drawn to a circle from an external point are equal. 3. If two circles touch one another, any straight line drawn through the point of contact cuts otf similar segments. Prove this for (i) internal, (ii) external contact. 4. If two circles touch one another, and from A, the point of con- tact, two chords APQ, AXY are drawn: then PX and QY are parallel. Prove this for (i) internal, (ii) external contact. ft. Two circles intersect at the points A, B: and one of them passes through O, the centre of the other : prove that OA bisects the angle between the common chord and the tangent to the first circle at A. G. Two circles intersect at A and B; and through P, any point on the circumference of one of them, straight lines PAC, PBD are drawn to cut the other circle at C and D: shew that CD is parallel to the tangent at P. 7. If from the point of contact of a tangent to a circle, a chord be drawn, the perpendiculars dropped on the tangent and chord from the middle point of either arc cut off by the chord are equal. BOOK III. 1'ROP. 33. PROPOSITION- 33. PROBLEM. 205 On a given straight lirte to describe a segment of a circle ftf/fill contain an angle equal to a, given angle. H Let AB be the given st. line, and C the giv r en angle: it is required to describe on AB a segment of a circle which shall contain an angle equal to C. At A in BA, make the _ BAD equal to the L. C. i. 23. Prom A draw AE at rt. angles to AD. I. 11. Bisect AB at F; I. 10. and from F draw FG at rt. angles to AB, cutting AE at G. Join GB. Then in the A s AFG, BFG. I AF = BF, Constr. Because - and FG is common, (and the L AFG = the L. BFG, being rt. angles; .'. GA = GB : i. 4. .'. the circle described from centre G, with radius GA, will pass through B. Describe this circle, and call it ABH: then the segment AHB shall contain an angle equal to C. Because AD is drawn at rt. angles to the radius GA from its extremity A, .'. AD is a tangent to the circle: in. 16. and from A, its point of contact, a chord AB is drawn; ,'. the L. BAD = the angle in the alt. segment AHB. in. 32. But the L. BAD = the L. C : Constr. ,'. the angle in the segment AHB = the L. C. .'. AHB is the Segment required. Q. K.F. 206 EUCLID'S ELEMENTS. NOTK. In the particular case when the given angle C is a rt. angle, the segment required will be the semicircle described on the given |-| st. line AB; for the angle in a [Q semicircle is a rt. angle. in. 31. EXERCISES. [The following exercises depend on the corollary to Proposition 21 given on page 187, namely The locus of the vertices of triangles which stand on the same bam' and have a given vertical angle, is the arc of the segment standing on this base, and containing an angle equal to the given angle. Exercises 1 and 2 afford good illustrations of the method of find- ing required points by the Intersection of Loci. See page 117.] 1. Describe a triangle on a given base, having a given vertical angle, and having its vertex on a given straight line. 2. Construct a triangle, having given the base, the vertical angle and (i) one other side. (ii) the altitude. (iii) the length of the median which bisects the base. (iv) the point at which the perpendicular from the vertex meets the base. 3. Construct a triangle having given the base, the vertical angle, and the point at which the base is cut by the bisector of the vertical angle. [Let AB be the base, X the given point in it, and K the given angle. On AB describe a segment of a circle containing an angle equal to K; complete the O ce by drawing the arc APB. Bisect the arc APB at P: join PX, and produce it to meet the O ee at C. Then ABC shall be the required triangle.] 4. Construct a triangle having given the base, the vertical angle, and the sum of the remaining sides. [Let AB be the given base, K the given angle, and H the given line equal to the sum of the sides. On AB describe a segment containing an angle equal to K, also another segment containing an angle equal to half the / K. From centre A, with radius H, describe a circle cutting the last drawn segment at X and Y. Join AX (or AY) cutting the first segment at C. Then ABC shall be the required triangle.] 5. Construct a triangle having given the base, the vertical angle, and the difference of the remaining sides. BOOK in. PROP. 34. 207 PROPOSITION 34. PROBLEM. From a given circle to cut off a segment which shall contain an angle equal to a given angle. Let ABC be the given circle, and D the given angle: it is required to cut off from the ABC a segment which shall contain an angle equal to D. Take any point B on the O ce , and at B draw tile tangent EBF. in. 17. At B, in FB, make the L FBC equal to the z. D. I. 23. Then the. segment BAC shall contain an angle equal to D. Because EF is a tangent to the circle, and from B, its point of contact, a chord BC is drawn, .'. the _ FBC the angle in the alternate segment BAC. in. 32. But the L. FBC = the L D ; Constr. .'. the angle in the segment BAC = the L D. Hence from the given 0ABC a segment BAC has been cut off, containing an angle equal to D. Q. E. F. EXERCISES. 1. The chord of a given segment of a circle is produced to a fixed point : on this straight line so produced draw a segment of a circle similar to the given segment. 2. Through a given point without a circle draw a straight line that will cut off a segment capable of containing an angle equal to a given angle. H. E. H 208 EUCLID'S ELEMENTS. PKOPOSITION 35. THEOREM. If two chords of a circle cut one another, t/te rectangle contained by the segments of one shall be equal to the rect- an'jle contained by the segments of the other. Let AB, CD, two chords of the QACBD, cut one another at E: then shall the rect. AE, EB = the rect. CE, ED. Find F the centre of the 0ACB: in. 1. From F draw FG, FH perp. respectively to AB, CD. i. 12. Join FA, FE, FD. Then because FG is drawn from the centre F perp. to AB, .'. AB is bisected at G. in. 3. For a similar reason CD is bisected at H. Again, because AB is divided equally at G, and unequally at E, .'. the rect. AE, EB with the sq. on EG - the sq. on AG. II. 5. To each of these equals add the sq. on G F ; then the rect. AE, EB with the sqq. on EG, GF = the sum of the sqq. on AG, GF. But the sqq. on EG, GF - the sq. on FE; I. 47. and the sqq. on AG, GF the sq. on AF; for the angles at G are rt. angles. .'. the rect. AE, EB with the sq. on FE = the sq. on AF. Similarly it may be shewn that the rect. CE, ED with the sq. on FE = the sq. on FD. But the sq. on AF the sq. on FD; for AF = FD. .'. the rect. AE, EB with the sq. on FE the rect. CE, ED with the sq. on FE. From these equals take the sq. on FE: then the rect. AE, EB^the rect. CE, ED. ci.KD. in. PROP. 35. 209 COROLLARY. If through a fixed point within a circle ant/ number of chords are drawn, the rectangles contained by their segments are all equal. NOTE. The following special cases of this proposition deserve notice. (i) when the given chords both pass through the centre : (ii) when one chord passes through the centre, and cuts the other at right angles : (iii) when one chord passes through the centre, and cuts the other obliquely. Iii each of these cases the general proof requires some modifica- tion, which may be left as an exercise to the student. EXERCISES. 1. Two straight lines AB, CD intersect at E, so that the rectangle AE, EB is equal to the rectangle CE, ED; shew ttiat the four points A, B, C, D are concyclic. 2. The rectangle contained by the segments of any chord drawn through a given point within a circle is equal to the square on half the shortest chord which may be drawn through that point. 3. ABC is a triangle right-angled at C ; and from C a perpen^ dicular CD is drawn to the hypotenuse : shew that the square on CD is equal to the rectangle AD, DB. 4. ABC is a triangle; and AP, BQ the perpendiculars dropped from A and B on the opposite sides, intersect at O: shew that the rectangle AC, OP is equal to the rectangle BO, OQ. 5. Two circles intersect at A and B, and through any point in AB their common chord two chords are drawn, one in each circle; shew that their four extremities are concyclic. 6. A and B are two points within a circle such that the rectangle contained by the segments of any chord drawn through A is equal to the rectangle contained by the segments of any chord through B : shew that A and B are equidistant from the centre. 7. If througlt, E, a point without a circle, two secants EAB, ECD are drawn; shew that the rectangle EA, EB is equal to the rectangle EC, ED. [Proceed as in in. 35, using n. 6.] 8. Through A, a point of intersection of two circles, two straight lines CAE, DAF are drawn, each passing through a centre and termi- nated by the circumferences : shew that the rectangle CA, AE is equal to the rectangle DA, AF. 14-2 210 EUCLID'S ELEMENTS. PROPOSITION 36. THEOREM. If from any point without a circle a tangent and a secant be drawn, then the rectangle contained by the whole secant and the part of it without the circle shall be equal to the square on the tangent. Let ABC be a circle; and from D a point without it, let there be drawn the secant DC A, and the tangent DB: then the rect. DA, DC shall be equal to the sq. on DB. Find E, the centre of the ABC: in. 1. and from E, draw EF perp. to AD. I. 12. Join EB, EC, ED. Then because EF, passing through the centre, is perp. to the chord AC, .'. AC is bisected at F. in. 3. And since AC is bisected at F and produced to D, .'.the rect. DA, DC with the sq. on FC = the sq. on FD. u. G. To each of these equals add the sq. on EF : then the rect. DA, DC with the sqq. on EF, FC = the sqq. on EF, FD. But the sqq. on EF, FC = the sq. on EC ; for EFC is a rt. angle; the sq. on EB. And the sqq. on EF, FD = the sq. on ED ; for EFD is a rt. angle; -the sqq. on EB, BD; for EBD is a rt. angle. in. 18. *. the rect. DA, DC with the sq. on EB = the sqq. on EB, BD. From these equals take the sq. on EB : then the rect. DA, DC = the sq. on DB. q. E.D. NOTE. This proof may easily be adapted to the case when the secant passes through the centre of the circle. BOOK III. PROP. 36. 211 COROLLARY. If from a given point without a circle any number of secants are drawn, tJie rectangles contained by the whole secants and the parts of them without the circle are all equal ; for each of these rectangles is equal to the square on the tangent drawn from the given point to the circle. For instance, in the adjoining figure, each of the rectangles PB, PA and PD, PC and PF, PE is equal to the square on the tangent PQ: .'. the rect. PB, PA = the rect. PD, PC - the rect. PF, PE. NOTE. Remembering that the segments into which the chord AB is divided at P, are the lines PA, PB, (see Part I. page 131) we are enabled to include the corollaries of Propositions 35 and 36 in a single enunciation. If any number of chords of a circle are drawn through a given point within or without a circle, tlie rectangles contained by the segments of the chords are equal. EXERCISES. 1. Use this proposition to shew that tangents drawn to a circle from an external point are equal. 2. If two circles intersect, tangents drawn to them from any point in their common chord produced are equal. 3. If two circles intersect at A and B, and PQ. is a tangent to both circles; shew that AB produced bisects PQ. 4. If P is any point on the straight line AB produced, shew that the tangents drawn from P to all circles which pass through A and B are equal. 5. ABC is a triangle right-angled at C, and from any point P in AC, a perpendicular PQ is drawn to the hypotenuse: shew that the rectangle AC, AP is equal to the rectangle AB, AQ. 6. ABC is a triangle right-angled at C, and from C a perpen- dicular CD is drawn to the hypotenuse: shew that the rect. AB, AC is equal to the square on AC, 212 EUCLID'S ELEMENT*. PROPOSITION 37. THEOREM. If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other ineets it, and if t/te rectangle contained by t/te whole line ivhich cuts the circle and the part of it without t/te circle be equal to the square on the line which meets the circle, then the line ^vhich meets the circle shall be a tangent to if. Let ABC be a circle; and from D, a point without it, let there be drawn two st. lines DC A and DB, of which DC A cuts the circle at C and A, and DB meets it; and let the rect. DA, DC = the sq. on DB: then shall DB be a tangent to the circle. From D draw DE to touch the ABC: m.17. let E be the point of contact. Find the centre F, and join FB, FD, FE. in. 1. Then since DCA is a secant, and DE a tangent to the circle, .'. the rect. DA, DC = the sq. on DE, in. 36. But, by hypothesis, the rect. DA, DC = the sq. on DB: .'. the sq. on DE = the sq. en DB, .'. DE - DB. Hence in the A s DBF, DEF. [ DB = DE, Proved. Because - and BF = EF; in. Def. 1. | and DF is common; .'. the L DBF = the L DEF. i. 8. But DEF is a rt. angle ; in. 18. .'. DBF is also a rt. angle; and since BF is a radius. .'. DB touches the 0ABC at the point B. Q. E. D. NOTE OX THE METHOD OF LIMITS AS APPLIED TO TANGENCY. \Q Euclid defines a tangent to a circle as a straight line which meets the circumference, but being produced, does not cut it : and from this definition lie deduces the fundamental theorem that a tangent is per- pendicular to the radius drawn to the point of contact. Prop. 1C. But this result may also be established by the Method of Limits, which regards the tangent as the ultimate position of a secant when its lico points of intersection with the circumference are brought into coin- cidence [See Note on page 151] : and it may be shewn that every theorem relating to the tangent may be derived from some more general proposition relating to the secant, by considering the ultimate case when the two points of intersection coincide. 1. To prove by the Method of Limits that a tangent 1o a circle is at right angles to the radius drawn to the point of contact. Let ABD be a circle, whose centre is C; and PABQ a secant cutting the C M in A and B; and let P'AQ' be the limiting position of PQ when the point B is brought into coincidence with A : then shall CA be perp. to P'Q'. Bisect AB at E and join CE: then CE is perp. to PQ. in. B. Now let the secant PABQ change its position in such a way that while the point A remains fixed, the point B con- tinually approaches A, and ultimate! y coincides with it ; then, however near B approaches to A, the st. line CE is always perp. to PQ, since it joins the centre to the middle point of the chord AB. But in the limiting position, when B coincides with A, and the secant PQ becomes the tangent P'Q', it is clear that the point E will also coincide with A ; and the perpendicular C E becomes the radius CA. Hence CA is perp. to the tangent P'Q' at its point of contact A. Q. E.D. NOTE. It follows from Proposition 2 that a straight line cannot cut the circumference of a circle at more than two points. Now when the two points in which a secant cuts a circle move towards coinci- dence, the secant ultimately _ becomes a tangent to the circle: we infer therefore that a tangent cannot meet a circle otherwise than at its point of contact. Thus Euclid's definition of a tangent may be deduced from that given by the Method of Limits, 214 EUCLID'S ELEMENTS. 2. By thin Method Proposition 32 may be derived ax <>cial case from Proposition 21. For let A and B be two points on the Q ce of the ABC; and let BCA, BPA be any two angles in the segment BCPA : then the / BPA = the / BCA. m. 21. Produce PA to Q. Now let the point P continually approach the fixed point A, and ultimately coincide with it ; then, however near P may approach to A, the / BPQ = the / BCA. in. 21. But in the limiting position when P coincides with A, and the secant PAQ becomes the tangent AQ', it is clear that BP will coincide with BA, and the / BPQ becomes the z BAQ'. Hence the z BAQ' = the L BCA, in the alternate segment. Q. E. r>. The contact of circles may be treated in a similar manner by adopting the following definition. DEFINITION. If one or other of two intersecting circles alters its position in such a way that the two points of intersection continually approach one another, and ultimately coincide ; in the limiting posi- tion they are said to touch one another, and the point in which the two points of intersection ultimately coincide is called the point of contact. EXAMPLES ON LIMITS. 1. Deduce Proposition 19 from the Corollary of Proposition 1 and Proposition 3. 2. Deduce Propositions 11 and 12 from Ex. 1, pag3 156. 3. Deduce Proposition 6 from Proposition u. 4. Deduce Proposition 13 from Proposition 10. 5. Shew that a straight line cuts a circle in two different points, two coincident points, or not at all, according as its distance from the centre is less than, equal to, or greater than a radius. 6. Deduce Proposition 32 from Ex. 3, page 188. 7. Deduce Proposition 36 from Ex. 7, page 209. 8. The angle in a semi-circle is a right angle. To what Theorem is this statement reduced, when the vertex of the right angle is brought into coincidence with an extremity of the diameter? 9. From Ex. 1, page 190, deduce the corresponding property of a triangle inscribed in a circle. THEOREMS AND EXAMPLES ON BOOK III. 215 THEOREMS AND EXAMPLES ON BOOK III. I. OK THE CENTRE AND CHORDS OF A CIRCLE. See Propositions 1, 3, 14, 15, 25. 1. All circles which pass through a fixed point, and have their centres on a given straight line, pass also through a second fixed point. Let AB be the given st. line, and P the given point. P' From P draw PR perp. to AB ; and produce PR to P', making RP' equal to PR. Then all circles which pass through P, and have their centres on AB, shall pass also through P'. For let C be the centre of an;/ one of these circles. Join CP, CP'. Then in the A' CRP, CRP' CR is common, Because J andRP=RP', Comtr. I and the / CRP = the / CRP', being rt. angles; /. CP = CP'; i.4. .-. the circle whose centre is C, and which passes through P, must pass also through P'. But C is the centre of any circle of the system ; .. all circles, which pass through P, and have their centres in AE, pass also through P'. Q. E. r>. 2. Describe a circle that shall pass through three given points not in the same straight lim>. 216 EUCLID'S KI.KMKNTS. 3. Describe a circle that shall pass through two given points and have its centre in a given straight line. When is this impossible? 4. Describe a circle of given radius to pass through two given points. When is this impossible? 5. ABC is an isosceles triangle ; and from tbe vertex A, as centre, a circle is described cutting the base, or the base produced, at X and Y. Shew that BX = CY. 6. If two circles which intersect are cut by a straight line parallel to the common chord, shew that the parts of it intercepted between the circumferences are equal. 7. If two circles cut one another, any two sti-aight lines drawn through a point of section, making equal angles with the common chord, and terminated by the circumferences, are equal. [Ex. 12. p. 156.] 8. If two circles cut one another, of all straight lines drawn through a point of section and terminated by the circumferences, the greatest is that which is parallel to the line joining the centres. 9. Two circles, whose centres are C and D, intersect at A, B; and through A a straight line PAQ is drawn terminated by the circumferences: if PC, QD intersect at X, shew that the angle PXQ is equal to the angle CAD. 10. Through a point of section of two circles which cut one another draw a straight line terminated by the circumferences and bisected at the point of section. 11. AB is a fixed diameter of a circle, whose centre is C; and from P, any point on the circumference, PQ is drawn perpendicular to AB; shew that the bisector of the angle CPQ always intersects the circle in one or other of two fixed points. 12. Circles are described on the sides of a quadrilateral as diameters: shew that the common chord of any two consecutive circles is parallel to the common chord of the other two. [Ex. 1), p. 97.] 13. Two equal circles touch one another externally, and through the point of contact two chords are drawn, one in each circle, at right angles to each other: shew that the straight line joining their other extremities is equal to the diameter of either circle. 14. Straight lines are drawn from a given external point to the circumference of a circle : find the locus of their middle points. [Ex. 3, p. 97.] 15. Two equal segments of circles are described on opposite sides of the same chord AB; and through O, the middle point of AB, any straight line POQ is drawn, intersecting the arcs of the segments at P and Q: shew that OP = OQ. THEOREMS AXP EXAMPLES OX BOOK ITT. 217 II. OX THE TAXGEXT AXD THE CONTACT OF CIRCLES. See Propositions 11, 12, 16, IT, 18, 19. 1. All equal chords placed in a given circle touch a fixed concen- tric circle. 2. If from an external point two tangents are drawn to a circle, the angle contained by them is double the angle contained by the chord of contact and the diameter drawn through one of the points of contact. 3. Two circles touch one another externally, and through the point of contact a straight line is drawn terminated by the circum- ferences : shew that the tangents at its extremities are parallel. 4. Two circles intersect, and through one point of section any straight line is drawn terminated by the circumferences : shew that the angle between the tangents at its extremities is equal to the angle between the tangents at the point of section. 5. Shew that two parallel tangents to a circle intercept on any third tangent a segment which subtends a right angle at the centre. 6. Two tangents are drawn to a given circle from a fixed external point A, and any third tangent cuts them produced at P and Q: shew that PQ subtends a constant angle at the centre of the circle. 7. In any quadrilateral circumscribed about a circle, the sum of one pair of opposite sides is equal to the sum of the other pair. 8. If the sum of one pair of opposite sides of a quadrilateral is equal to the sum of the other pair, sheiv that a circle may be inscribed in the figure, [Bisect two adjacent angles of the figure, and so describe a circle to touch three of its sides. Then prove indirectly by means of the last exercise that this circle must also touch the fourth side.] 9. Two circles touch one another internally: shew that of all chords of the outer circle which touch the inner, the greatest is that which is perpendicular to the straight line joining the centres. 10. In any triangle, if a circle is described from the middle point of one side as centre and with a radius equal to half the sum of the other two sides, it will touch the circies described on these sides as diameters. 11. Through a given point, draw a straight line to cut a circle, .so that the part intercepted by the circumference may be equal to a given straight line. In order that the problem may be possible, between what limits must the given line lie, when the given point is (i) without the circle, (ii) within it ? 218 EUCLID'S ELEMENTS. 12. A series of circles touch a given straight line at a given point', shew that the tangents to them at the points where they cut a given parallel straight line all touch a fixed circle, whose centre is the given point. 13. If two circles touch one another internally, and any third circle be described touching both ; then the sum of the distances of the centre of this third circle from the centres of the two given circles is constant. 14. Find the locus of points such that the pairs of tangents drawn from them to a given circle contain a constant angle. 15. Find a point such that the tangents drawn from it to two given circles may be equal to two given straight lines. When is this impossible ? 16. If three circles touch one another two and two ; prove that the tangents drawn to them at the three points of contact are con- current and equal. THE COMMON TANGENTS TO Two CIRCLES. 17. To draw a common tangent to two circles. First, if the given circles are external to one another, or if they intersect. Let A be the centre of the greater circle, and B the centre of the less. / C! \ ~~~~~~ -^E From A, with radius equal to the diff ce of the radii of the given circles, describe a circle: and from B draw BC to touch the last drawn circle. Join AC, and produce it to meet the greater of the given circles at D. Through B draw the radius BE par 1 to AD, and in the same direction. Join DE: then DE shall be a common tangent to the two given circles. For since AC = the diff ce between AD and BE, Comtr. :. CD = BE: and CD is par 1 to BE; Constr. .'. DE is equal and par 1 to CB. i. 33. But since BC is a tangent to the circle at C, /. the L ACB is a rt. angle; in. 18. hence each of the angles at D and E is a rt. angle: i. 29. .'. DE is a tangent to both circles. O.E.F, THEOREMS AND EXAMPLES ON BOOK III. 219 It follows from hypothesis that the point B is outside the circle used in the construction : .-. two tangents such as BC may always be drawn to it from B ; hence tico common tangents may always be drawn to the given circles by the above method. These are called the direct common tangents. When the given circles are external to one another and do not intersect, two more common tangents may be drawn. For, from centre A, with a radius equal to the sum of the radii of the given circles, describe a circle. From B draw a tangent to this circle; and proceed as before, but draw BE in the direction opposite to AD. It follows from hypothesis that B is external to the circle used in the construction ; .. two tangents may be drawn to it from B. Hence two more common tangents may be drawn to the -given circles : these will be found to pass between the given circles, and are called the transverse common tangents. Thus, in general, four common tangents may be drawn to two given circles. The student should investigate for himself the number of common tangents which may be drawn in the following special cases, noting in each casa where the general construction fails, or is modified : (i) When the given circles intersect : (ii) When the given circles have external contact : (iii) When the given circles have internal contact : (iv) When one of the given circles is wholly within the other. 18. Draw the direct common tangents to two equal circles. 19. If the two direct, or the two transverse, common tangents are drawn to two circles, the parts of the tangents intercepted be- tween the points of contact are equal. 20. If four common tangents are drawn to two circles external to one another; shew that the two direct, and also the two transverse, tangents intersect on the straight line which joins the centres of the circles. 21. Two given circles have external contact at A, and a direct common tangent is drawn to touch them at P and Q : shew that PQ subtends a right angle at the point A. 22. Two circles have external contact at A, and a direct common tangent is drawn to touch them at P and Q : shew that a circle described on PQ as diameter is touched at A by the straight line which joins the centres of the circles. 220 EUCLID'S ELEMENTS. 23. Two circles whose centres are C and C' have external contact at A, and a direct common tangent is drawn to touch them at P and Q : shew that the bisectors of the angles PCA, QC'A meet at right angles in PQ. And if R is the point of intersection of the bisectors, shew that RA is also a common tangent to the circles. 24. Two circles have external contact at A, and a direct common tangent is drawn to touch them at P and Q : shew that the square on PQ is equal to the rectangle contained by the diameters of the circles. 25. Draw a tangent to a given circle, so that the part of it intercepted by another given circle may be equal to a given straight line. When is this impossible? 26. Draw a secant to two given circles, so that the parts of it intercepted by the circumferences may be equal to two given straight lines. PEOBLEHS ON TAXGENCY. The following exercises are solved by the Method of Inter- section of Loci, explained on page 117. The student should begin by making himself familiar with the following loci. (i) The locus of the centres of circles which pass through two given points. (ii) The locus of the centres of circles which touch a given straight line at a given point. (iii) The locus of the centres of circles which touch a given circle at a given point. (iv) The locus of the centres of circles which touch a given straight line, and have a given radius. (v) The locus of the centres of circles which touch a given circle, and have a given radius. (vi) The locus of the centres of circles which touch tico given straight lines. In each exercise the student should investigate the limits and relations among the data, in order that the problem may be possible. 27. Describe a circle to touch three given straight lines. 28. Describe a circle to pass through a given point and touch a given straight line at a given point. 29. Describe a circle to pass through a given point, and touch a given circle at a given point. THEOREMS AND fcXAMPLKSi ON" bOOK HI. 221 30. Describe a circle of given radius to pass through a given point, and touch a given straight line. 31. Describe a circle of given radius to touch two given circles. 32. Describe a circle of given radius to touch two given straight lines. 33. Describe a circle of given radius to touch a given circle and a given straight line. 34. Describe two circles of given radii to touch one another and a given straight line, on the same side of it. 35. If a circle touches a given circle and a given straight line, shew that the points of contact and an extremity of the diameter of the given circle at right angles to the given line are collinear. 36. To describe a circle to touch a given circle, and alao to touch a (jii'en straight line at a given point. Let DEB be the given circle, PQ the given st. line, and A the given point in it : it is required to describe a circle to touch the DEB, and also to touch PQ at A. At A draw AFperp. to PQ: 1. 11. then the centre of the required circle must lie in AF. HI. 19. Find C, the centre of the DEB, in. 1. and draw a diameter BD perp. to PQ: P A Q join A to one extremity D, cutting the o co at E. Join CE, and produce it to cut AF at F. Then F is the centre, and FA the radius of the required circle. [Supply the proof : and shew that a second solution is obtained by joining AB, and producing it to meet the o ce : also distinguish between the nature of the contact of the circles, when PQ cuts, touches, or is without the given circle.] 37. Describe a circle to touch a given straight line, and to touch a given circle at a given point. 38. Describe a circle to touch a given circle, have its centre in a given straight line, and pass through a given point in that straight line. [For other problems of the same class see page 235.] 222 EUCLID'S ELEMENTS. OKTHOGOXAL CIRCLES. DEFINITION". Circles which intersect at a- point, so that the two tangents at that point are at right angles to one another, are said to be orthogonal, or to cut one another ortho- gonally. 39. In two intersecting circles the angle between the tangents at one point of intersection is equal to the angle between the tangents v,t the other. 40. If two circles cut one another ortliogonally , the tangent to each circle at a point of intersection ivill pass through the centre of the other circle. 41. If tico circles cut one another orthogonally, the square on the distance between their centres is equal to the sum of the squares on their radii. 42. Find the locus of the centres of all circles which cut a given circle orthogonally at a given point. 43. Describe a circle to pass through a given point and cut a given circle orthogonally at a given point. III. ON ANGLES IN SEGMENTS, AND ANGLES AT THE CENTRES AND CIRCUMFERENCES OF CIRCLES. See Propositions 20, 21, 22; 26, 27, 28, 29; 31, 32, 33, 34. 1. If two chords intersect within a circle, they form an angle equal to that at the centre, subtended by half the sum of the arcs they cut off. Let AB and CD be two chords, intersecting at E within the given ADBC: then shall the / A EC be equal to the angle at the centre, subtended by half the sum of the arcs AC, BD. Join AD. Then the ext. z AEC = the sum of the int. opp. L B EDA, EAD ; that is, the sum of the L s CD A, BAD. But the Z"CDA, BAD are the angles at the O subtended by the arcs AC, BD ; .-. their sum = half the sum of the angles at the centre subtended by the same arcs; or, the L A EC = the angle at the centre subtended by half the sum of the arcs AC, BD. Q. E. i>. THEOREMS AND EXAMPLES ON BOOK III 223 2. If two chords ichen produced intersect outside a circle, they form an, aiKjle equal to that at the centre subtended by half the difference of the area they cut off. 3. The sum of the arcs cut off by two chords of a circle at right angles to one another is equal to the semi-circumference. 4. AB, AC are any two chords of a circle; and P, Glare the middle points of the minor arcs cut off by them: if PQ is joined, cutting AB and AC at X, Y, shew that AX = AY. 5. If one side of a quadrilateral inscribed in a circle is produced, the exterior angle is equal to the opposite interior angle. 6. If two circles intersect, and any straight lines are drawn, one through each point of section, terminated by the circumferences; shew that the chords which join their extremities towards the same parts are parallel. 7. A BCD is a quadrilateral inscribed in a circle ; and the opposite sides AB, DC are produced to meet at P, and CB, DA to meet at Q: if the circles circumscribed about the triangles PBC, QAB intersect at R, shew that the points P, R, Q are collinear. 8. If a circle is described on one of the sides of a right-angled triangle, then the tangent drawn to it at the point where it cuts the hypotenuse bisects the other side. 9. Given three points not in the same straight line : shew how to find any number of points on the circle which passes through them, without finding the centre. 10. Through any one of three given points not in the same straight line, draw a tangent to the circle which passes through them, without finding the centre. 11. Of two circles which intersect at A and B, the circumference of one passes through the centre of the other : from A any straight line is drawn to cut the first at C, the second at D ; shew that CB = CD. 12. Two tangents AP, AQ are drawn to a circle, and B is the middle poiut of the arc PQ, convex to A. Shew that PB bisects the angle APQ. 13. Two circles intersect at A and B ; and at A tangents are drawn, one to each circle, to meet the circumferences at C and D : if CB, BD are joined, shew that the triangles ABC, DBA are equiangular to one another. 14. T\vo segments of circles are described on the same chord and 011 the same side of it ; the extremities of the common chord are joined to any point on the arc of the exterior segment : shew that the a-rc intercepted on the interior segment is constant. H.E. 15 224 EUCLID'S ELEMENTS; 15. I( & series of triangles are drawn standing on a fixed base, and having a given vertical angle, shew that the bisectors of the verti- cal angles all pass through a fixed point. 1C. ABC is a triangle inscribed in a circle, and E the middle point of the arc subtended by BC on the side remote from A: it' through E a diameter ED is drawn, shew that the angle DEA is half the difference of the angles at B and C. [See Ex. 7, p. 101.] 17. If two circles touch each other internally ut a point A. any chord of the exterior circle which touches the interior is divided at its point of contact into segments which subtend equal angles at A. 18. If two circles touch one another internally, and a straight line is drawn to cut them, the segments of it intercepted between the circumferences subtend equal angles at the point of contact. THE ORTHOCENTBE OF A TRIANGLE. 19. The perpendiculars drawn from the vertices* of . .-. the remaining / AFO = art. angle: i. :>'J. that is, CF is perp. to AB. Hence the three perp' AD, BE, CF meet at the point O. y. E. i>. [For an Alternative Proof see page 106.] THEOREMS AND EXAMPLES ON BOOK III. 225 DEFINITIONS. (i) The intersection of the perpendiculars drawn from the vertices of a triangle to the opposite sides is called its ortho- centre. (ii) The triangle formed by joining the feet of the perpendi- culars is called the pedal or orthocentric triangle. '20. In an acute-angled triangle the perpendiculars drawn front the vertices to the opposite sides bisect tJie angles of the pedal trianyle tJirougli which they pass. In the acute-angled A ABC, let AD, BE, CF be the perp" drawn from the vertices to the opposite sides, meeting at the orthocentre O; and let DEF be the pedal triangle : then shall AD, BE, CF bisect respect- ively the z 8 FDE, DEF, EFD. For, as in the last theorem, it may Q~ 5 C be shewn that the points O, D, C, E are concyclic ; .'. the z ODE the z OCE, in the same segment. Similarly the points O, D, B, F are concyclic; .. the / ODF = the z OBF, in the same segment. But the / OCE = the / OBF, each being the comp' of the L BAC. .-. the / ODE = the z ODF. Similarly it may be shewn that the / " DEF, EFD are bisected by BE and CF. Q. E.D. COROLLARY, (i) Every two sides of the pedal triangle are equally inclined to that side of the original triangle in which they meet. For the z E DC = the comp' of the /ODE = the comp' of the z OCE = the L BAC. Similarly it may be shewn that the L FOB = the / BAC, .'. the L EDC = the L FDB = the z A. In like manner it may be proved that the L DEC = the z FEA = the z B, and the z DFB = the z EFA = ihe z C. COROLLARY, (ii) The triangles DEC, AEF, DBF are equiangular to one another and to the triangle ABC. NOTE. If the angle BAC is obtuse, then the perpendiculars BE, CF bisect externally the corresponding angles of the pedal triangle. 15-2 226 EUCLID'S ELEMENTS. 21. In any triangle, if the perpendiculars drawn from the on the opposite sides are produced to meet the circumscribed circle, then each side bisects that portion of the line perpendicular to it which lies between the orthocentre and the circumference. Let ABC be a triangle in which the perpen- diculars AD, BE are drawn, intersecting at O the orthocentre; and let AD be produced to meet the o ce of the circumscribing circle at G : then shall DO=DG. Join BG. Then in the two A s OEA, ODB, the / OEA = the /ODB, being rt. angles; and the / EOA the vert. opp. / DOB; .-. the remaining / EAO = the remaining / DBO. But the L CAG=the / CBG, in the same segment; .-. the / DBO = the / DBG. Then in the A S DBO, DBG, (the / DBO = the /DBG, P raced. Because 'the / BDO = the / BDG, f and BD is common; .-. DO=DG. i. 26. Q. E. D. 22. In an acute-angled triangle the three sides are the external bisectors of the angles of the pedal triangle: and in an obtuse-angled triangle the sides containing the obtuse angle are the internal bisectors of the corresponding angles of the pedal triangle. 23. If O 'is the orthocentre of the triangle ABC, nhew that the angles BOC, BAG are supplementary. 24. If O is the orthocentre of the triangle ABC, then any one of the four points O, A, B, C is the orthocentre of the triangle whose vertices are the other three. 25. The three circles which pass through two vertices of a triangle and its orthocentre are each equal to the circle circumscribed about the triangle. 26. D, E are taken on the circumference of a semicircle described on a given straight line AB : the chords AD. BE and AE, BD intersect (produced if necessary) at F and G : shew that FG is per- pendicular to AB. 27. A BCD is a parallelogram; AE and CE are drawn at right angles to AB, and CB respectively: shew that ED, if produced, will be perpendicular to AC. THEOREMS AXD EXAMPLES ON BOOK III. 227 28. ABC is a triangle, O is its orthocentre, and AK a diameter of the circumscribed circle: shew that BOCK is a parallelogram. 29. The orthocentre of a triangle is joined to the middle point of the base, and the joining line is produced to meet the circumscribed circle : prove that it will meet it at the same point as the diameter which passes through the vertex. 30. The perpendicular from the vertex of a triangle on the base, and the straight line joining the orthocentre to the middle point of the base, are produced to meet the circumscribed circle at P and Q : shew that PQ is parallel to the base. 31. The distance of each vertex of a triangle from the orthocentre is double of the perpendicular draicn from the centre of the circum- scribed circle on the opposite side. 32. Three circles are described each passing through the ortho- centre of a triangle and two of its vertices: shew that the triangle formed by joining their centres is equal in all respects to the original triangle. 33. ABC is a triangle inscribed in a circle, and the bisectors of its angles which intersect at O are produced to meet the circumference in PQR : shew that O is the orthocentre of the triangle PQR. 34. Construct a triangle, having given a vertex, the orthocentre, and the centre of the circumscribed circle. Loci. 35. Given the base and vertical angle of a triangle, find the locus of its orthocentre. Let BC be the given base, and X the given angle ; and let BAG be any triangle on the base BC, having its vertical L A * equal to the / X. /\ Draw the perp s BE, CF, intersecting f \ at the orthocentre O. It is required to find the locus of O. Since the z s OFA, OEA are rt. angles, .. the points O, F, A, E are concyclic ; B C ..the / FOE is the supplement of the / A: m. 22. .-. the vert. opp. / BOC is the supplement of the / A. But the / A is constant, being always equal to the / X ; .'. its supplement is constant; that is, the A BOC has a fixed base, and constant vertical angle; hence the locus of its vertex O is the arc of a segment of which BC is the chord. [See p. 187.] 228 EUCLID R ELEMENTS. 3G. Given the bane and vertical angle of a triangle, find tlte locus of the intersection of the bisectors of its angles. Let BAG be any triangle on the given base BC, having its vertical angle equal to the given L X; and let Al, Bl, Cl be the bisectors of its angles: [see Ex. 2, p. 103.] it is required to find the locus of the point I. Denote the angles of the A ABC by A, B. C; and let the L BIG be denoted by 1. Then from the ABIC, i. 32. (i) I + \ B + iC = two rt. angles, and from the A ABC, A + B -f C = two rt. angles ; i. 32. (ii) so that iA + ^B + iC = one rt. angle, .". , taking the differences of the equals in (i) and (ii), I - \ A = one rt. angle : or, ' I = one rt. angle + \ A. But A is constant, being always equal to the L X ; .. I is constant : .. , since the base BC is fixed, the locus of 1 is the arc of a segment of which BC is the chord. 37. Given the base and vertical angle of a triangle, find tli? of the centroid, that is, the intersection of the median*. Let BAG be any triangle on the given base BC, having its vertical angle equal to the given angle S; let the medians AX, BY, CZ intersect at the centroid G [see Ex. 4, p. 105] : it is required to find the locus of the point G. Through G draw GP, GQ par 1 to AB and AC respectively. Then ZG is a third part of ZC; Ex. 4, p. 105. and since GP is par 1 to ZB, .-. BP is a third part of BC. Similarly QC is a third part of BC ; .-. P and Q are fixed points. P X Q E.r. 19. p. <)9. Now since PG, GQ are par 1 respectively to BA, AC, Const r. .: the L PGQ=the L BAC, i. 29. = the / S, that is, the / PGQ is constant; and since the base PQ is fixed, the locus of G is the arc of a segment of which PQ is the chord, THEOREMS AND EXAMPLES OX BOOK III. 229 06s. In this problem the points A and G move on the area of siviilar segments. 38. Given the base and the vertical angle of a triangle ; find the locus of the intersection of the bisectors of the exterior base angles, 39. Through the extremities of a given straight line AB any two parallel straight lines AP, BQ are drawn ; find the locus of the inter- section of the bisectors of the angles PAB, QBA. 40. Find the locus of the middle points of chords of a circle drawn through a fixed point. Distinguish between the cases when the given point is within, on, or without the circumference. 41. Find the locus of the points of contact of tangents drawn from a fixed point to a system of concentric circles. 42. Find the locus of the intersection of straight lines which pass through two fixed points on a circle and intercept on its circumference an arc of constant length. 43. A and B are two fixed points on the circumference of a circle, and PQ is any diameter : find the locus of the intersection of PA and QB. 44. BAG is any triangle described on the fixed base BC and having a constant vertical angle ; and BA is produced to P, so that BP is equal to the sum of the sides containing the vertical angle: find the locus of P. 45. AB is a fixed chord of a circle, and AC is a moveable chord passing through A : if the parallelogram CB is completed, find the locus of the intersection of its diagonals. 46. A straight rod PQ slides between two rulers placed at right angles to one another, and from its extremities PX, QX are drawn perpendicular to the rulers: find the locus of X. 47. Two circles whose centres are C and D, intersect at A and B : through A, any straight line PAQ is drawn terminated by the circum- ferences ; and PC, QD intersect at X : find the locus of X, and shew that it passes through B. [Ex. 9, p. 216.] 48. Two circles intersect at A and B, and through P, any point on the circumference of one of them, two straight lines PA. PB are drawn, and produced if necessary, to cut the other circle at X and Y: find the locus of the intersection of AY and BX. 49. Two circles intersect at A and B; HAK is a fixed straight line drawn through A and terminated by the circumferences, and PAQ is any other straight line similarly drawn: find the locua of the intersection of HP and QK. 230 EUCLID'S ELEMENTS. 50. Two segments of circles are on the same chord AB and on the same side of it ; and P and Q are any points one on each arc : find the locus of the intersection of the bisectors of the angles PAQ, PBQ. 51. Two circles intersect at A and B ; and through A any straight line PAQ is drawn terminated by the circumferences: find the locus of the middle point of PQ. MISCELLANEOUS EXAMPLES ON ANGLES IN A CIBCLE. 52. ABC is a triangle, and circles are drawn through B, C, cutting the sides in P, Q, P', Q', ... : shew that PQ, P'Q' ... are parallel to one another and to the tangent drawn at A to the circle circumscribed about the triangle. 53. Two circles intersect at B and C, and from any point A, on the circumference of one of them, AB, AC are drawn, and produced if necessary, to meet the other at D and E : shew that D E is parallel to the tangent at A. 54. A secant PAB and a tangent PT are drawn to a circle from an external point P; and the bisector of the angle ATB meets AB at C : shew that PC is equal to PT. 55. From a point A on the circumference of a circle two chords AB, AC are drawn, and also the diameter AF: if AB, AC are produced to meet the tangent at F in D and E, shew that the triangles ABC, AED are equiangular to one another. 56. O is any point within a triangle ABC, and CD, OE, OF are drawn perpendicular to BC, CA, AB respectively : shew that the angle BOC is equal to the sum of the angles BAG, EOF. 57. If two tangents are drawn to a circle from an external point, shew that they contain an angle equal to the difference of the angles in the segments cut off by the chord of contact. 58. Two circles intersect, and through a point of section a straight line is drawn bisecting the angle between the diameters through that point : shew that this straight line cuts off similar segments from the two circles. 59. Two equal circles intersect at A and B ; and from centre A, with any radius less than AB a third circle is described cutting the given circles on the same side of AB at C and D : shew that the points B, C, D are collinear. 60. ABC and A'B'C' are two triangles inscribed in a circle, so that AB, AC are respectively parallel to A'B', A'C' : shew that BC' is parallel to B'C. THEOREMS AND EXAMPLES ON BOOK III, 231 61. Two circles intersect at A and B, and through A two straight lines HAK, PAQ are drawn terminated by the circumferences : if HP and KQ intersect at X, shew that the points H, B, K, X are concyclic. 62. Describe a circle touching a given straight line at a given point, so that tangents drawn to it from two fixed points in the given line may be parallel. [See Ex. 10, p. 183.] 63. C is the centre of a circle, and CA, CB two fixed radii: if from any point P on the arc AB perpendiculars PX, PY are drawn to CA and CB, shew that the distance XY is constant. 64. AB is a chord of a circle, and P any point in its circum- ference ; PM is drawn perpendicular to AB, and AN is drawn perpen- dicular to the tengent at P : shew that MN is parallel to PB. 65. P is any point on the circumference of a circle of which AB is a fixed diameter, and PN is drawn perpendicular to AB ; on AN and BN as diameters circles are described, which are cut by AP, BP at X and Y : shew that XY is a common tangent to these circles. 06. Upon the same chord and on the same side of it three seg- ments of circles are described containing respectively a given angle, its supplement and a right angle : shew that the intercept made by the two former segments upon any straight line drawn through an ex- tremity of the given chord is bisected by the latter segment. 67. Two straight lines of indefinite length touch a given circle, and any chord is drawn so as to be bisected by the chord of contact : if the former chord is produced, shew that the intercepts between the circumference and the tangents are equal. 68. Two circles intersect one another: through one of the points of contact draw a straight line of given length terminated by the cir- cumferences. 69. On the three sides of any triangle equilateral triangles are described remote from the given triangle : shew that the circles de- scribed about them intersect at a point. 70. On BC, CA, AB the sides of a triangle ABC, any points P, Q, R are taken; shew that the circles described about the triangles AQR, BRP, CPQ meet in a point. 71. Find a point within a triangle at which the sides subtend equal angles. 72. Describe an equilateral triangle so that its sides may pass through three given points. 73. Describe a triangle equal in all respects to a given triangle, and having its sides passing through three given points. EUCLID'S ELEMENTS. SIMSON'S LINE. 74. If from any point on the circumference of the circle circum- scribed about a triangle, perpendiculars are drawn to the three sides, the feet of these perpendiculars are collinear. Let P be any point on the o ce of the circle circumscribed about the A ABC ; and let PD, PE, PF be the perp" drawn from P to the three sides. It is required to prove that the points D, E, F are collinear. Join FD and DE: then FD and DE shall be in the same st. line. Join PB, PC. Because the / s PDB, PFB are rt. angles, :. the points P, D, B, F are concylic: /. the L PDF = the L PBF, in the same segment. in. 21. But since BACP is a quad 1 inscribed in a circle, having one of its sides AB produced to F, .-. ihec::t. / PBF = the opp. int. L ACP. Ex.?,, p. 188. .-. the L PDF = the L ACP. To each add the / PDE : then the Z"PDF, PDE = the / " ECP, PDE. But since the / " PDC. PEC are rt. angles, .'. the points P, D, E, C are concylic ; .-. the / 9 ECP, PDE together = two rt. angles: .-. the L PDF, PDE together = two rt. angles; .-. FD and DE are in the same st. line; i. 14. that is, the points D, E, F are collinear. O..E.D. [The line FDE is called the Pedal or Simson's Line of the triangle ABC for the point P ; though the tradition attributing the theorem to Robert Simson has been recently shaken by the researches of Dr. J. S. Mackay. ] 75. ABC is a triangle inscribed in a circle ; and from any point P on the circumference PD. PF are drawn perpendicular to BC and AB : if FD, or FD produced, cuts AC at E, shew that PE is perpendicular to AC. 76. Find the locus of a point which moves so that if perpendicu- lars are drawn from it to the sides of a given triangle, their feet are collinear. 77. ABC and AB'C' are two triangles having a common vertical angle, and the circles circumscribed about them meet again at P : shew that the feet of perpendiculars drawn from P to the four lines AB, AC, BC, B'C' are collinear. THEOREMS AND EXAMPLES ON BOOK III. 233 78. A triangle is inscribed in a circle, and any point P on the cir- cumference is joined to the ortlwccntre of the triangle ; shew that this joining line in />/r/3.] 15. A semicircle is described on AB as diameter, and any two chords AC, BD are drawn intersecting at P : shew that AB- = AC. AP + BD . BP. 16. Two circles intersect at B and C, and the two direct common tangents AE and DF are drawn : if the common chord is produced to meet the tangents at G and H, shew that GH J =AE 5 +BC*, 17. If from a point P, without a circle, PM is drawn perpendicular to a diameter AB, and also a secant PCD, shew that PM 2 = PC . PD + AM . MB. THEOREMS AND EXAMPLES ON BOOK III. 235 18. Three circles intersect at D, and their other points of inter- section are A, B, C ; AD cuts the circle BDC at E, and EB, EC cut the circles ADB, ADC respectively at F and G : shew that the points F, A, G are collinear, and F, B, C, G concyclic. 19. A semicircle is described on a given diameter BC, and from B and C any two chords BE, CF are drawn intersecting within the semicircle at O ; B F and C E are produced to meet at A : shew that the sum of the squares on AB, AG is equal to twice the square on the tangent from A together with the square on BC. 20. X a:id Y are two fixed points in the diameter of a circle equidistant from the centre C : through X any chord PXQ is drawn, and its extremities are joined to Y; shew that the sum of the squares on the sides of the triangle PYQ is constant. [See p. 147, Ex. 24.] PKOBLEJIS ON TAXGEXCY. 21. To describe a circle to pass throiujh two given points and to touch a given straight line. Let A and B be the given points, and CD the given st. line: it is required to describe a circle to pass through A and B and to touch CD. Join BA, and produce it to meet CD at P. ,-' Describe a square equal to the C P Q D rect. PA, PB; n. 14. and from PD (or PC) cut off PQ equal to a side of this square. Through A, B and Q describe a circle. Ex. 4, p. 156. Then since the rect. PA, PB = the sq. on PQ, . '. the ABQ touches CD at Q. in. 37. Q. E. F. NOTE, (i) Since PQ may be taken on either side of P, it is clear that there are in general two solutions of the problem. (ii) When AB is parallel to the given line CD, the above method is not applicable. In this case a simple construction follows from in. 1, Cor. and in 10' and it will bo found that only one solution exists. 236 EUCLID'S ELEMENTS. 22. To describe a circle to pass through tini yireti points <> " given triangle. Let ABC be the given circle, and DEF the given triangle: it is required to inscribe in the ABC a triangle equiangular to the A DEF. At any point A, on the O ce of the 0ABC, draw the tangent GAH. in. 17. At A make the L GAB equal to the L. DFE ; I. 23. and make the i_ MAC equal to the /_ DEF. I. 23. Join BC. Then ABC shall be the triangle required. Because GH is a tangent to the 0ABC, and from A its point of contact the chord AB is drawn, .'. the _ GAB = the _ ACB in the alt. segment: in. 32, .'. the _ ACB = the _ DFE. ' Con*/,: Similarly the _ MAC = the L. ABC, in the alt. segment: .'. the L. ABC = the L. DEF. Hence the third L. BAG = the third _ EOF, for the three angles in each triangle are together equal to two rt. angles. I. 32. .'. the A ABC is equiangular to the A DEF, and it is inscribed in the 0ABC, BOOK IV. PROP, 3 253 PROPOSITION 3. PROBLEM. About a given circle to circumscribe a triangle equi to a given triangle. Let ABC be the given circle, and DEF the given A : it is required to circumscribe about the 0ABC a triangle equiangular to the A DEF. Produce EF both ways to G and H. Find K the centre of the 0ABC, m. 1, and draw any radius KB. At K make the L BKA equal to the L DEG ; i. 23. and make the L. BKC equal to the L. DFH. Through A, B, C draw LM, MN, NL perp. to KA, KB, KC. Then LMN shall be the triangle required. Because LM, MN, NL are drawn perp. to radii at their extremities, .'. LM, MN, NL are tangents to the circle, in. 1C. And because the four angles of the quadrilateral AKBM together four rt. angles ; I. 32. Cor. and of these, the _ s KAM, KBM, are rt. angles; Conxtr. .'. the .L 8 AKB, AMB, together two rt. angles. But the L s DEG, DEF together = two rt. angles ; i. 13. .'. the L s AKB, AMB = the L. * DEG, DEF : and of these, the /_ AKB = the L DEG ; Constr. :. the .AMB = the L. DEF. Similarly it may be shewn that the L LNM the L DFE. /. the third L MLN =the third L. EOF. I. 32. .'. the A LMN is equiangular to the A DEF, and it is circumscribed about the ABC, Q.E. F. 254 EUCLID S ELEMENTS. PROPOSITION 4. PROBLEM. To inscribe a circle in a given triangle. Let ABC be the given triangle : it is required to inscribe a circle in the A ABC. Bisect the L " ABC, ACB by the st. lines Bl, Cl, which intersect at I. i > 9 t From I draw IE, IF, IG perp. to AB, BC, CA. i. 12, Then in the A s EIB, FIB, ( the L EBI = the L FBI ; Constr, Because -jand the L BEI = the L BFI, being rt. angles ; ( and Bl is common; '- IE= IF. i. 26.' Similarly it may be shewn that IF = IG. .'. IE, IF, IG are all equal. From centre I, with radius IE, describe a circle: this circle must pass through the points E, F, G ; and it will be inscribed in the A ABC. For since IE, IF, IG are radii of the EFG ; and since the L s at E, F, G are rt. angles ; .'. the 0EFG is touched at these points by AB, BC, CA: in. 16. .'. the EFG is inscribed in the A ABC. Q. E. F. NOTE. From page 103 it is seen that if Al be joined, then Al bisects tbe angle BAG. BOOK IV. PROP. 4. 255 Hence it folloics that the bisectors of the angles of a triangle are concurrent, the point of intersection being the centre of the inscribed circle. The centre of the circle inscribed in a triangle is sometimes called its in-centre. DEFINITION. A circle which touches one side of a triangle and the other two sides produced is said to be an escribed circle of the triangle. To draiv an escribed circle of a given triangle. Let ABC be the given triangle, of which the two sides AB, AC are produced to E and F : it is required to describe a circle touching BC, and AB, AC produced. Bisect the / 8 CBE, BCF by the st. lines Blj, CI 1; which intersect at \ 1 . i. 9. From l : draw ^G, ^H, IjK perp. to AE, BC, AF. i. 12. Then in the A 8 IjBG, IjBH, the / I 1 BG = the / IjBH, Constr. and the / l lG B = the L I.HB, being rt. angles ; also IB is common; Similarly it may be shewn that IjH = IjK; .'. IjG, IjH, IjK are all equal. From centre lj with radius ^G, dcsciibe a circle: this circle must pass through the points G, H, K : and it will be an escribed circle of the A ABC. For since IjH, IjG, I : K are radii of the HGK, and since the angles at H , G , K are rt. angles, .. the GHK is touched at these points by BC, and by AB, AC produced : .'. the GHK is an escribed circle of the A ABC. Q.E.F. It is clear that every triangle has three escribed circles. NOTE. From page 104 it is seen that if Al x be joined, then Al x bisects the angle BAC : hence it follows that The bisectors of tico exterior angles of a triangle and the bisector of the third angle are concurrent, the point of intersection being the centre of an escribed circle. H. E. 17 EUCLID'S ELEMENTS. PROPOSITION 5. PROBLEM. To circumscribe a circle about a given triangle. .A A, B/ Let ABC be the given triangle : it is required to circumscribe a circle about the A ABC. Draw DS bisecting AB at rt. angles; i. 11. and draw ES bisecting AC at rt. angles; then since AB, AC are neither par 1 , nor in the same st. line, .'. DS and ES must meet at some point S. Join SA ; and if S be not in BC, join SB, SC. Then in the A 8 ADS, BDS, ( AD=BD Because y joining the points of contact of the inscribed circle. (v) Of the four points I, Ij, I 2 , I 3 each is the orthocentre of the triangle whose vertices are the other three. (vi) Tlie four circles, each of which passes through three of the points I, IL I 2 , l a , are all equal. THEOREMS AND EXAMPLES OX BOOK. IV. 279 3. With the notation of page 277, shew that in a triangle ABC, if the angle at C is a right angle, 4. With the figure given on page 278, shew that if the circles whose centres are I, \ 1} I 2 , I 3 touch BC at D, D lf D 2 , D 3 , then (i) DD a =D 1 D 8 = 6. (ii) DD,= D 1 D a = c. (iii) D 2 D 3 = 6 + c. (iv) DD 1 = b~c. 5. Shew that the orthocentre and vertices of a triangle are the centres of the inscribed and escribed circles of the pedal triangle. [See Ex. 20, p. 225.] 6. Given the base and vertical angle of a triangle, find the locus of the centre of the inscribed circle. [See Ex. 36, p. 228.] 7. Given the base and vertical angle of a triangle, find the locus of the centre of the escribed circle which touches tlie base. 8. Given the base and vertical angle of a triangle, shew that the centre of the circumscribed circle is fixed. 9. Given the base BC, and the vertical angle A of a triangle, find the locus of the centre of the escribed circle which touches AC. 10. Given the base, the vertical angle, and the radius of the inscribed circle ; construct the triangle. 11. Given the base, the vertical angle, and the radius of the escribed circle, (i) which touches the base, (ii) which touches one of the sides containing the given angle ; construct the triangle. 12. Given the base, the vertical angle, and the point of contact with the base of the inscribed circle ; construct the triangle. 13. Given the base, the vertical angle, and the point of contact with the base, or base produced, of an escribed circle ; construct the triangle. 14. From an external point A two tangents AB, AC are drawn to a given circle ; and the angle BAG is bisected by a straight line which meets the circumference in I and \ 1 : shew that I is the centre of the circle inscribed in the triangle ABC, and Ij the centre of one of the escribed circles. 15. I is the centre of the circle inscribed in a triangle, and I 1( I 2 , I 3 the centres of the escribed circles; shew that \\ lt II 2 , II 3 are bisected by the circumference of the circumscribed circle. 16. ABC is a triangle, and I 2 , I 3 the centres of the escribed circles which touch AC, and AB respectively: shew that the points B, C, I 2 , I 3 lie upon a circle whose centre is on the circumference of the circle circumscribed about ABC. 280 EUCLID'S ELEMENTS. 17. With three given points as centres describe three circles touching one another two by two. How many solutions will there be? 18. Two tangents AB, AC are drawn to a given circle from an external point A; and in AB, AC two points D and E are taken so that DE is equal to the sum of DB and EC : shew that DE touches the circle. 19. Given the perimeter of a triangle, and one angle in magnitude and position : shew that the opposite side always touches a fixed circle. 20. Given the centres of the three escribed circles ; construct the triangle. 21. Given the centre of the inscribed circle, and the centres of two escribed circles ; construct the triangle. 22. Given the vertical angle, perimeter, and the length of the bisector of the vertical angle ; construct the triangle. 23. Given the vertical angle, perimeter, and altitude ; construct the triangle. 24. Given the vertical angle, perimeter, and radius of the in- scribed circle ; construct the triangle. 25. Given the vertical angle, the radius of the inscribed circle, and the length of the perpendicular from the vertex to the base ; construct the triangle. 26. Given the base, the difference of the sides containing the vertical angle, and the radius of the inscribed circle ; construct the triangle. [See Ex. 10, p. 258.] 27. Given a vertex, the centre of the circumscribed circle, and the centre of the inscribed circle, construct the triangle. 28. In a triangle ABC, I is the centre of the inscribed circle ; shew that the centres of the circles circumscribed about the triangles BIG, CIA, AIB lie on the circumference of the circle circumscribed about the given triangle. 29. In a triangle ABC, the inscribed circle touches the base BC at D ; and r, r^ are the radii of the inscribed circle and of the escribed circle which touches BC : shew that r . 7^=80 . DC. 30. ABC is a triangle, D, E, Fthe points of contact of its inscribed circle ; and D'E'F' is the pedal triangle of the triangle DEF : shew that the sides of the triangle D'E'F' are parallel to those of ABC. 31. In a triangle ABC the inscribed circle touches BC at D. Shew that the circles inscribed in the triangles ABD, ACD touch one another. THEOREMS AND EXAMPLES ON BOOK IV. 281 THE NINE-POINTS ClRCLE. 32. In any triangle the middle points of the sides, the feet of the perpendiculars drawn from the vertices to the opposite sides, and the middle points of the lines joining the arthocentre to the vertices are concyclic. In the A ABC, let X, Y, Z be the middle points of the sides BC, CA, AB ; let D, E, F be the feet of the perp s drawn to these sides from A, B, C ; let O be the orthocentre, and a, /3, 7 the middle points of OA, OB, OC: then shall the nine points X, Y, Z, D, E, F, a, /3, 7 be concyclic. Join XY, XZ, Xa, Ya, Za. Now from the A ABO, since AZ = Z B, and Aa = aO, Hyp. .: Za is par 1 to BO. Ex. 2, p. 96. And from the A ABC, since BZ = Z A, andBX = XC, Hyp. :. Z X is par 1 to AC. But BO makes a rt. angle with AC ; Hyp. /. the / XZa is a rt. angle. Similarly, the / XYa is a rt. angle. i. 29. /. the points X, Z, a, Y are concyclic: that is, a lies on the o co of the circle, which passes through X, Y, Z ; and Xa is a diameter of this circle. Similarly it may be shewn that /3 and 7 lie on the O ce of the circle which passes through X, Y, Z. Again, since aDX is a rt. angle, Hyp. .'. the circle on Xa as diameter passes through D. Similarly it may be shewn that E and F lie on the circumference of the same circle. /. the points X, Y, Z, D, E, F, a, /3, 7 are concyclic. Q.E.D. From this property the circle which passes through the middle po'nts of the sides of a triangle is called the Nine-Points Circle ; many of its properties may be derived from the fact of its being the circle circumscribed about the pedal triangle. 282 EUCLID'S ELEMENTS. 33. To prove that (i) the centre of the nine-points circle is the middle j>oint of the straight line which joins the orthocentre to the circumscribed centre: (ii) the radius of the nine-points circle is half the radius of the circumscribed circle : (iii) the centroid is collinear with the circumscribed centre, the nine-points centre, and the orthocentre. In the A ABC, let X, Y, Z be the middle points of the sides; D, E, F the feet of the perp s ; O the ortho- centre; S and N the centres of the circumscribed and nine-points circles respectively. (i) To prove that N is the middle point of SO. It may be shewn that the perp. to XD from its middle point bisects SO ; Ex. 14, p. 98. Similarly the perp. to EY at its middle point bisects SO : that is, these perp 3 intersect at the middle point of SO : And since X D and E Y are chords of the nine-points circle, /. the intersection of the lines which bisect XD and EY at rt. angles is its centre: in. 1. /. the centre N is the middle point of SO. (ii) To prove that the radius of the nine-points circle is half the radius of the circumscribed circle. By the last Proposition, Xa is a diameter of the nine-points circle. .. the middle point of Xa is its centre: but the middle point of SO is also the centre of the nine-points circle. (Proved.) Hence Xa and SO bisect one another at N. Then from the A 8 SNX, ONa ( SN = ON, Because < and NX = No, (and the / SNX = the / ONa; i. 15. .-. SX = Oa 1.4. = Aa. And SX is also par 1 to Act, /. SA = Xa. r. 33. But SA is a radius of the circumscribed circle; and Xa is a diameter of the nine-points circle; .'. the radius of the nine-points circle is half the radius of the eireuoi- scribed circle. THEOREMS AND EXAMPLES OX BOOK IV. 283 (ill) To prove that the centroid is collinear with points S, N , O. Join AX and draw ay par 1 to SO. Let AX meet SO at G. Then from the A AGO, since Aa = aO and ag is par 1 to OG, .-. A<7 = <7G. Ex. 13, p. 98. And from the A Xa#, since aN = NX, and NG is par 1 to ag, .: gG=GX. Ex. 13, p. 98. .-. AG=f of AX; .'. G is the centroid of the triangle ABC. That is, the centroid is collinear with the points S, N, O. Q.E.D. 34. Given the base and vertical angle of a triangle, find the locus of the centre of the nine-points circle. 35. The nine-points circle of any triangle ABC, whose orthocentre is O, is also the nine-points circle of each of the triangles AOB, BOC, COA. 36. If I, I 1( I 2 . (3 are the centres of the inscribed and escribed circles of a triangle ABC, then the circle circumscribed about ABC is t:ie nine-points circle of each of the four triangles formed by joining three of the points I, l lf I 2 , I 3 . 37. All triangles which have the same orthocentre and the same circumscribed circle, have also the same nine points circle. 38. Given the base and vertical angle of a triangle, shew that one angle and one side of the pedal triangle are constant. 39. Given the base and vertical angle of a triangle, find the locus of the centre of the circle which passes through the three escribed centres. NOTE. For another important property of the Nine-points Circle sec Ex. 60, p. 382. II. MISCELLANEOUS EXAMPLES. 1. If four circles are described to touch every three sides of a quadrilateral, shew that their centres are concyclic. 2. If the straight lines which bisect the angles of a rectilineal figure are concurrent, a circle may be inscribed in the figure. 3. Within a given circle describe three equal circles touching one another and the given circle. 4. The perpendiculars drawn from the centres of the three escribed circles of a triangle to the. sides which they touch, are con- current. 284 EUCLID'S ELEMENTS. 5. Given an angle and the radii of the inscribed and circumscribed circles; construct the triangle. 6. Given the base, an angle at the base, and the distance between the centre of the inscribed circle and the centre of the escribed circle which touches the base ; construct the triangle. 7. In a given circle inscribe a triangle such that two of its sides may pass through two given points, and the third side be of given length. 8. In any triangle ABC, I, \ lt I 2 , I 3 are the centres of the in- scribed and escribed circles, and Sj,, S 2 , S 3 are the centres of the circles circumscribed about the triangles BIG, CIA, AIB: shew that the triangle SjS-jSjhas its sides parallel to those of the triangle \j\. 2 \ 3 , and is one-fourth of it in area: also that the triangles ABC and SjSjjS-j have the same circumscribed circle. 9. O is the orthocentre of a triangle ABC : shew that where d is the diameter of the circumscribed circle. 10. If from any point within a regular polygon of n sides perpen- diculars are drawn to the sides the sum of the perpendiculars is equal to n times the radius of the inscribed circle. 11. The sum of the perpendiculars drawn from the vertices of a regular polygon of n sides on any straight line is equal to n times the perpendicular drawn from the centre of the inscribed circle. 12. The area of a cyclic quadrilateral is independent of the order in which the sides are placed in the circle. 13. Given the orthocentre, the centre of the nine-points circle, ai.d the middle point of the base ; construct the triangle. 14. Of all polygons of a given number of sides, which may be inscribed in a given circle, that which is regular has the maximum area and the maximum perimeter. 15. Of all polygons of a given number of sides circurnsciibod about a given circle, that which is regular has the minimum area and the minimum perimeter. 16. Given the vertical angle of a triangle in position and magni- tude, and the sum of the sides containing it : find the locus of the centre of the circumscribed circle. 17. P is any point on the circumference of a circle circumscribed about an equilateral triangle ABC: shew that PA- + PB- + PC 2 is constant. BOOK V. Book V. treats of Ratio and Proportion. INTRODUCTORY. The first four books of Euclid deal with the absolute equality or inequality of Geometrical magnitudes. In the Fifth Book magnitudes are compared by considering their ratio, or relative greatness. The meaning of the words ratio and proportion in their simplest arithmetical sense, as contained in the following defini- tions, is probably familiar to the student : The ratio of one number to another is the multiple, part, or parts that the first number is of the second; and it may therefore be measured by the fraction of which the first number is the numerator and the second the denominator. Four numbers are in proportion when the ratio of the first to the second is equal to that of the third to the fourth. But it will be seen that these definitions are inapplicable to Geometrical magnitudes for the following reasons : (1) Pure Geometry deals only with concrete magnitudes, re- presented by diagrams, but not referred to any common unit in terms of which they are measured : in other words, it makes no use of number for the purpose of comparison between different magnitudes. (2) It commonly happens that Geometrical magnitudes of the same kind are incommensurable, that is, they are such that it is impossible to express them exactly in terms of some common unit. For example, we can make comparison between the side and diagonal of a square, and we may form an idea of their relative great- ness, but it can be shewn that it is impossible to divide either of them into equal parts of which the other contains an exact number. And as the magnitudes we meet with in Geometry are more often incom- mensurable than not, it is clear that it would not always be possible to exactly represent such magnitudes by numbers, even if reference to a common unit were not foreign to the principles of Euclid. It is therefore necessary to establish the Geometrical Theory of Proportion on a basis quite independent of Arithmetical principles. This is the aim of Euclid's Fifth Book. 286 EUCLID'S ELEMENTS. We shall employ the following notation. Capital letters, A, B, C,... will be used to denote the magnitudes themselves, not any numerical or algebraical measures of them, and small letters, TO, n, p,... will be used to denote whole numbers. Also it w'.ll be assumed that multiplication, in the sense of repeated addition, can be applied to any magnitude, so that m . A or will be used for the words greater than, and < for less than. DEFINITIONS. 1. A greater magnitude is said to be a multiple of a less, when the greater contains the less an exact number of times. 2. A less magnitude is said to be a submultiple of a greater, when the less is contained an exact number of times in the greater. The following properties of multiples will be assumed as self-evident. (1) mA > = or < mB according as A > = or < B ; and conversely. (2) iA + mB+... = rn(A+B+...). (3) If A > B, then mA - mB = m (A - B). (4) mA+nA+... = (m + H + ...) A. (5) If m > n, then MI A nA = (w - ;i) A. (6) m . A = nj . A = nm . A = 7* . HI A. 3. The Ratio of one magnitude to another of the same kind is the relation which the first bears to the second in respect of quantuplicity. The ratio of A to B is denoted thus, A : B; and A is called the antecedent, B the consequent of the ratio. The term quantuplicity denotes the capacity of the first magnitude to contain the second with or without remainder. If the magnitudes are commensurable, their quantuplicity may be expressed numerically by observing what multiples of the two magnitudes are equal to one another. Thus if A ma, and B = na, it follows that A = wiB. In this case A = B, and the quantuplicity of A with respect to B is the arith- m metical fraction . DEFINITIONS. 287 But if the magnitudes are incommensurable, no multiple of the first can be equal to any multiple of the second, and therefore the quantuplicity of one with respect to the other cannot exactly be expressed numerically: in this case it is determined by examining how the multiples of one magnitude are distributed among the multiples of the other. Thus, let all the multiples of A be formed, the scale extending ad injinitum; also let all the multiples of B be formed and placed in their proper order of magnitude among the multiples of A. This forms the relative scale of the two magnitudes, and the quantuplicity of A with respect to B is estimated by examining how the multiples of A are distributed among those of B in their relative scale. In other words, the ratio of A to B is known, if for all integral values of TO we know the multiples ?;B and (>* + !) B between which mA lies. In the case of two given magnitudes A and B, the relative scale of multiples is definite, and is different from that of A to C, if C differs from B by any magnitude however small. For let D be the difference between B and C ; then however small D may be, it will be possible to find a number in such that mD > A. In this case, mB and ?wC would differ by a magnitude greater than A, and therefore could not lie between the same two multiples of A ; so that after a certain point the relative scale of A and B would differ from that of A and C. [It is worthy of notice that we can always estimate the arithmetical ratio of two incommensurable magnitudes within any required degree of accuracy. For suppose that A and B are incommensurable ; divide B into m equal parts each equal to /3, so that B = mft, where m is an integer. Also suppose /3 is contained in A more than n times and less than (n + 1) times; then A Ti/3 (n+l)8 s > , an d < B mp m/3 xi. j. A ,. ~ n , n + 1 that is. lies between and ; B mm A n 1 so that -^r differs from - by a quantity less than . And since we B TO ' TO can choose /3 (our unit of measurement) as small as we please, m can be made as great as we please. Hence can be made as small as we TO please, and two integers n and m can be found whose ratio will express that of a and b to any required degree of accuracy.] H. E. 19 288 EUCLID'S ELEMENTS. 4. The ratio of one magnitude to another is equal to that of a third magnitude to a fourth, when if any equi- multiples whatever of the antecedents of the ratios are taken, and also any equimultiples whatever of the con- sequents, the multiple of one antecedent is greater than, equal to, or less than that of its consequent, according as the multiple of the other antecedent is greater than, equal to, or less than that of its consequent. Thus the ratio A to B is equal to that of C to D when mC > = or < nD according as mA > = or < nB, whatever whole numbers m and n may be. Again, let m be any whole number whatever, and n another whole number determined in such a way that either ?A is equal to ?;B, or wiA lies between nB and (n + 1) B; then the definition asserts that the ratio of A to B is equal to that of C to D if mC = n D when mA = n B ; or if mC lies between nD and (n + 1) D when mA lies between nB and (n + l)B. In other words, the ratio of A to B is equal to that of C to D when the multiples of A are distributed among those of B in the same manner as the multiples of C are distributed among those of D. 5. When the ratio of A to B is equal to that of C to D the four magnitudes are called proportionals. This is ex- pressed by saying " A is to B as C is to D", and the proportion is written A : B : : C : D, or A : B = C : D. A and D are called the extremes, B and C the means; also D is said to be a fourth proportional to A, B, and C. Two terms in a proportion are said to be homologous when they are both antecedents, or both consequents of the ratios. [It will be useful here to compare the algebraical and geometrical definitions of proportion, and to sbew that each may be deduced from the other. According to the geometrical definition A, B, C, D are in propor- tion, when mC> = = = <;iD according as mA> = = = D Suppose = > ; then it will be possible to find some fraction B D m which lies between them, n and m being positive integers. Hence > ................................. ; and <.. ...(2). D m From (1), w?A>nB; from (2), mC . *_>. it is clear that the ratio compounded of two equal ratios is the dupli- cate ratio of either of them. 14. When four magnitudes are in continued proportion, the first is said to have to the fourth the triplicate ratio of that which it has to the second. It may be shewn as above that the ratio compounded of three equal ratios is the triplicate ratio of any one of them. 292 EUCLID'S ELEMENTS. Although an algebraical treatment of ratio and proportion when applied to geometrical magnitudes cannot be considered exact, it will perhaps be useful here to summarise in algebraical form the principal theorems of proportion contained in Book V. The student will then perceive that its leading propositions do not introduce new ideas, but merely supply rigorous proofs, based on the geometrical definition of proportion, of results already familiar in the study of Algebra. We shall only here give those propositions which are afterwards referred to in Book VI. It will be seen that in their algebraical form many of them are so simple that they hardly require proof. SUMMARY OF PRINCIPAL THEOREMS OP BOOK V. PROPOSITION 1. Ratios which are equal to the same ratio are equal to one another. That is, if A : B = X : Y and C : D = X : Y; then A : B = C : D. PROPOSITION 3. If four magnitudes are proportionals, they are also proportionals when taken inversely. That is, if A : B = C : D, then B:A=D:C. This inference is referred to as invertendo or inversely. PROPOSITION 4. (i) Equal magnitudes have the same ratio to the same magnitude. For if A=B, then A : C = B : C. (ii) The same magnitude has the same ratio to equal magnitudes. For if A=B, then C:A = C:B. SUMMARY OF PRINCIPAL THEOREMS OF BOOK V. 293 PROPOSITION 6. (i) Magnitudes which have the same ratio to the same magnitude are equal to one another. That is, if A : C = B : C, then A = B. (ii) Those magnitudes to which the same magnitude has the same ratio are equal to one another. That is, if C : A = C : B, then A=B. PROPOSITION 8. Magnitudes have the same ratio to one another which their equi- multiples have. That is, A : B = wiA : mB, where m is any whole number. PROPOSITION 11. If four magnitudes of the same kind are proportionals, they are also proportionals when taken alternately. If A : B = C : D, then shall A : C = B : D. A C For since B = D' , . . . . B A B C B .-. multiplying by - , we have ^ Q = D ' C ' that is, = - , or A : C = B : D. This inference is referred to as alternando or alternately. 294 EUCLID S ELEMENTS. PROPOSITION 12. If any number of magnitudes of the same kind are proportionals, then as one of the antecedents is to its consequent, so is the sum of the antecedents to the sum of the consequents. Let A : B = C : D = E : F=...; then shall A : B = A + C+E+... : B + D + F+.... For put each of the equal ratios ,5 , = , = , . . . equal to k ; D \j r then A = B/t, C=Dk, E = Fk,... =7 , = A = 9 _ I_ ; B D F This inference is sometimes referred to as addenda. PROPOSITION 13. (i) If four magnitudes are proportionals, the sum of the Jirst and second is to the second as the sum of the third and fourth is to the fourth. Let A : B = C : D, then shall A + B:B = C+D:D. For since R ~ D ' A . C .-.,+1-5+1! A+B C+D that is, -g- D , or A+B:B = This inference is referred to as componendo. (ii) If four magnitudes are proportionals, the difference of the first and second is to the second as the difference of the third and fourth is to the fourth. That is, if A : B = C : D, then A~B : B = C~D : D. The proof is similar to that of the former case. This inference is referred to as dividendo. SUMMARY OF PRINCIPAL THEOREMS OF BOOK. V. 295 PROPOSITION 14. If there are two sets of magnitudes, such that the first is to the second of the first set as the first to the second of the other set, and the second to the third of the first set as the second to the third of the other, and so on to the last magnitude: then the first is to the last of the first set as the first to the last of the other. First let there be three magnitudes, A, B, C, of one set, and three, P, Q, R, of another set, and let and then shall For since A : B=P: Q, B :C = Q : R; A : C=P : R. A P B Q that is, or Similarly if it can be proved that This inference is referred to as ex sequali. If A B ' B C~ P Q Q ' R ; A C~ P R A . c= P R. A B : Q Q P Q Q, R, L : M = Y Z; A : M = P Z. COROLLARY. and then shall For since A : B=P : Q, B : C=R : P; A : C=R : Q. A P , B R A B B' C A_ C Q ' A :C=R : Q. P R Q' P ; R 296 EUCLID'S ELEMENTS. PROPOSITION 15. If A : B = C : D, and E:B = F:D; then shall A + E:B = C + F:D. AC , E F For since B = D ' and B = D ; that is, A+.E : B = C + PROPOSITION 16. If two ratios are equal, their duplicate ratios are equal; and conversely. Let A : B = C : D; then shall the duplicate ratio of A : B be equal to the duplicate ratio of C : D. Let X be a third proportional to A, B; so that A : B=B : X; B _A " X ~ B ; B A_A A " X ' B ~ B ' B ; A A 2 that is, X = B*' But A : X is the duplicate ratio of A : B ; .-. the duplicate ratio of A : B = A 2 : B 2 . But since A : B = C : D ; A_C A B~D' A 2 _C_ 2 " B 2 ~ D 2 ' or A 2 : B 2 = C 2 : D 2 ; that is, the duplicate ratio of A : B = the duplicate ratio of C : D. Conversely, let the duplicate ratio of A : B be equal to the dupli- cate ratio of C : D ; then shall A : B = C : D, for since A 2 : B 2 = C 2 : D 2 , .-. A : B = C : D. PROOFS OF THE PROPOSITIONS OF BOOK V. 297 PROOFS or THE PROPOSITIONS OF BOOK V. DERIVED FROM THE GEOMETRICAL DEFINITION OF PROPORTION. Obs. The Propositions of Book V. are all theorems. PROPOSITION 1. Ratios which are equal to the same ratio are equal to one another. Let A : B :: P : Q, and also C : D :: P : Q; then shall A : B :: C : D. For it is evident that two scales or arrangements of multiples which agree in every respect with a third scale, will agree with one another. PROPOSITION 2. If two ratios are equal, the antecedent of the second is greater than, equal to, or less than its consequent according as the antecedent of the first is greater than, equal to, or less than its consequent. Let A : B :: C : D, then C > = or < D, according as A > = or < B. This follows at once from Def. 4, by taking m and n each equal to unity. 298 EUCLID'S ELEMENTS. PROPOSITION 3. If two ratios are equal, their reciprocal ratios are equal. Let A : B :: C : D, then shall B : A : : D : C. For, by hypothesis, the multiples of A are distributed among those of B in the same manner as the multiples of C are among those of D; therefore also, the multiples of B are distributed among those of A in the same manner as the multiples of D are among those of C. That is, B : A :: D : C. NOTE. This proposition is sometimes enunciated thus // four magnitudes are proportionals, they are also proportionals when taken inversely, and the inference is referred to as invertendo or inversely. PROPOSITION 4. Equal 'magnitudes have the same ratio to the same mag- nitude; and' the same magnitude has the same ratio to equal magnitudes. Let A, B, C be three magnitudes of the same kind, and let A be equal to B; then shall A : C : : B : C and C : A : : C : B. Since A B, their multiples are identical and therefore are distributed in the same way among the multiples of C. .'. A : C :: B : C, Def. 4. .'. also, invertendo, C:A::C:B. v. 3.' PROOFS OF THE PROPOSITIONS OF BOOK V. 299 PROPOSITION 5. Of two unequal magnitudes, the greater has a greater ratio to a third magnitude than the less has ; and the same magnitude has a greater ratio to the less of two magnitudes than it has to the greater. First, let A be > B; then shall A : C be > B : C. Since A > B, it will be possible to find m such that mA exceeds mB by a magnitude greater than C; hence if ?>*A lies between nC and (n + 1)C, mB B : C. Def. 6. Secondly, let B be < A ; then shall C : B be > C : A. For taking m and n as before, nC > mB, while nC is not > iA ; .'. C : B > C : A. Def. G. PROPOSITION 6. Magnitudes which have the same ratio to the same mag- nitude are equal to one another; and those to which the same magnitude has the same ratio are equal to one another. First, let A : C :: B : C; then shall A= B. For if A > B, then A : C > B : C, and if B > A, then B : C > A : C, v. 5. which contradict the hypothesis; .'. A=B. 300 EUCLID'S ELEMENTS. let C : A :: C : B; A= B. Because C : A :: C : B, .'., invertendo, A : C :: B : C, v. 3. A=B, by the first part of the proof. PROPOSITION 7. That magnitude which has a greater ratio than another has to the same magnitude is the greater of the two; and that magnitude to which the same has a greater ratio than it has to another magnitude is the less of the two. First, let A : C be > B : C ; then shall A be > B. For if A = B, then A : C :: B : C, v. 4. which is contrary to the hypothesis. And if A < B, then A : C < B : C ; v. 5. which is contrary to the hypothesis; .'. A> B. Secondly, let C : A be > C : B ; then shall A be < B. For if A = B, then C : A : : C : B, v. 4. which is contrary to the hypothesis. And if A > B, then C : A < C : B; v. 5. which is contrary to the hypothesis; /. A < B. 1'ROOFS OF THE PROPOSITIONS OF BOOK V. 301 PROPOSITION 8. Magnitudes have the same ratio to one another which their equimultiples have. Let A, B be two magnitudes; then shall A : B : : wiA : mB. If p, q be any two whole numbers, then m . p& > or < IH . qB according as jt>A > = or < ^B. But m.pkp.m^ and m.qB = q.mB; .'. p . ?>iA > = or < q . mB according as pA > - or < qB; .'. A : B : : mA : mB. Def. 4. COK. Let A : B :: C : D. Then since A : B : : mA : mB, and C : D :: nC : nD; .'. mA : mB :: nC : nD. v. 1. PROPOSITION 9. If two ratios are equal, and any equimultiples of the antecedents and also of the consequents are taken, the multiple of the first antecedent has to that of its consequent the same ratio as the multiple of the other antecedent has to that of its consequent. Let A : B :: C : D; then shall mA : nB :: mC : nD. Let p, q be any two whole numbers, then because A : B : : C : D, pm . C > = or < qn . D according as pm . A > or < qn . B, Def. 4. that is, p . mC > = or - or = or < C according as B > = or < D. If B> D, then A : B < A : D ; v. 5. but A : B :: C : D; .'. C : D < A : D; .'. A : D>C : D: .'. A>C. v. 7. Similarly it may be shewn that if B < D, then A < C, and if B = D, then A = C. PROPOSITION 11. If four magnitudes of tJie same kind are proportionals, they are also proportionals wJien taken alternately. Let A, B, C, D be four magnitudes of the same kind sucli that A : B :: C : D; then shall A : C : : B : D. Because A : B :: mA : mB, v. 8. and C : D :: mC : nD; .'. mA : mB :: nC : nD. v. 1. .'. mA > ~or or = or =or or < nF, .'. so is ??iA + raC + mE +...> = or < nB + nD + nF + ... and m and n are any whole numbers; /. A:B::A + C + E+... : B + D + F + . . . . Def. 4. NOTE. This inference is usually referred to as addendo. PROPOSITION 13. If four magnitudes are proportionals, the sum or differ- ence of the first and second is to the second as the sum or difference of the third and fourth is to the fourth. Let A : B : : C : D ; then shall A+B:B::C + D:D, and A~B:B::C~D:D. If m be any whole number, it is possible to find another number n such that mA = nB, or lies between nB and (n+l)B, .'. mA + mB = mB + nB, or lies between mB + nB and mB + (n + 1) B. H. E. 20 304 EUCLID'S ELEMENTS. But mA + mB m(A + B), and mB + nB (m + n) B ; .'. m(A + B) = (m + n) B, or lies between (m + n) B and (m 4- n + 1 ) B. Also because A : B : : C : D, .'. mC = nD, or lies between ?^D and (n+ 1)D; Def. 4. .'. m(C + D) = (m + w) D or lies between (m 4- n) D and (m + n + 1 ) D ; that is, the multiples of C + D are distributed among those of D in the same way as the multiples of A + B among those of B; .'. A+ B : B :: C + D : D. In the same way it may be proved that A-B:B::C-D:D, or B A:B::D C:D, according as A is > or < B ; NOTE. These inferences are referred to as componendo and divi- dendo respectively. PROPOSITION 14. If there are two sets of magnitudes, such that the first is to the second of the first set as the first to the second of the other set, and the second to the third of the first set as the, second to the third of the other, and so on to the last magni- tude : then the first is to the last of the first set as the first to the last of the other. First, let there be three magnitudes A, B, C, of one set and three, P, Q, R. of another set, v. 8, Cor. v. 9. v. 3. and let A : B P : Q, and B : C Q : R; then shall A : C P : R. Because A : B P:Q, .'. mA : mB mP : mQ; and because B : C Q : R, .'. mB : nC mQ. : nR, endo, nC : mB : : nR : mQ. PROOFS OF THE PROPOSITIONS OF BOOK V. 305 Now, if mA > nC, then mA : mB> nC : mB; .'. mP : mQ> nR : 7/iQ, and .'. mP> nR. Similarly it may be shewn that mP = or < nR, according as 7/tA = or < nC, .'. A : C :: P : R. Secondly, let there be any number of magnitudes, A, B, C, ... L, M, of one set, and the same number P, Q, R, ...Y, Z, of another set, such that v. 5. v. 7. 4. A : B P : Q, B : C Q : R, L : M Y Z; then shall A : M P Z. For A C P R, and C D R S; .'. by the first case A : D P s, and so on, until finally Z. A : M = NOTE. This inference is referred to as ex sequali. COROLLARY. If A : B :: P : Q, and B : C : : R : P ; then A : C :: R : Q. Proved. Hyp. PROPOSITION 15. If A B C : D, and E B F : D; then shall A + E B C + F : D. For since E B F : D, , invertendo, B E D : F. Also A B C : D, , ex cequali, A E C :F, Hyp. v. 3. v. 14. 20-2 306 EUCLID'S ELEMENTS. .'. , componendo, A + E' : E : : C + F : F. v. 13. Again, E : B :: F : D, Hyp- .'. , ex cequali, A + E : B : : C + F : D. v. 14. PROPOSITION 16. If two ratios are equal, their duplicate ratios are equal; and conversely, if the duplicate ratios of two ratios are equal, the ratios themselves are equal. Let A : B : : C : D ; then shall the duplicate ratio of A to B be equal to that of C to D. Let X be a third proportional to A and B, and Y a third proportional to C and D, so that A : B :: B : X, and C : D :: D : Y: then because A : B : : C : D, .*. B : X :: D : Y; .'. , ex cequali, A : X :: C : Y. But A : X and C : Y are respectively the duplicate ratios of A : B and C : D, Def. 13. .'. the duplicate ratio of A : B = that of C : D. Conversely, let the duplicate ratio of A : B = that of C : D; then shall A : B : : C : D. Let P be such that A : B : : C : P, .'. , invertendo, B : A :: P : C. Also, by hypothesis, A : X :: C : Y, .". , ex cequali, B : X : : P : Y; but A : B : : B : X, .'. A : B :: P : Y; V. 1. .'. C : P :: P : Y; v. 1. that is, P is the mean proportional between C and Y. /. P=D, .'. A : B :: C : D. BOOK VI. DEFINITIONS. 1. Two rectilineal figures are said to be equiangular when the angles of the first, taken in order, are equal respectively to those of the second, taken in order. Each angle of the first figure is said to correspond to the angle to which it is equal in the second figure, and sides adjacent to corresponding angles are called corresponding sides. 2. Rectilineal figures are said to be similar when they are equiangular and have the sides about the equal angles proportionals, the corresponding sides being homologous. [See Def. 5, page 288.] Thus the two quadrilaterals ABCD, EFGH are similar if the angles at A, B, C, D are respec- tively equal to those at E, F, G, H, and if the following proportions hold AB BC : EF FG, BC CD: FG GH, CD DA : GH HE, DA AB : HE EF. 3. Two figures are said to have their sides about two of their angles reciprocally proportional when a side of the first is to a side of the second as the remaining side of the second is to the remaining side of the first. 4. A straight line is said to be divided in extreme and mean ratio when the whole is to the greater segment as the greater segment is to the less. 5. Two similar rectilineal figures are said to be similarly situated with respect to two of their sides when these sides are homologous. 308 EUCLID'S ELEMENTS. PROPOSITION 1. THEOREM. T/w areas of triangles of the same altitude are to one another as their bases. H Q B CDKLM Let ABC, ACD be two triangles of the same altitude, namely the perpendicular from A to BD: then shall the A ABC : the A ACD :: BC : CD. Produce BD both ways, and from CB produced cut oft' any number of parts BG, GH, each equal to BC; and from CD produced cut off any number of parts DK, KL, LM each equal to CD. Join AH, AG, AK, AL, AM. Then the A s ABC, ABG, AGH are equal in area, for they are of the same altitude and stand on the equal bases CB, BG, GH, I. 38. .'. the A AHC is the same multiple of the A ABC that HC is of BC; Similarly the A ACM is the same multiple of ACD that CM is of CD. And if HC = CM, the A AHC = the A ACM; I. 38. and if HC is greater than CM, the A AHC is greater than the A ACM; I. 38, Cor. and if HC is less than CM, the A AHC is less than the A ACM. I. 38, Cor. Now since there are four magnitudes, namely, the A 8 ABC, ACD, and the bases BC, CD; and of the antecedents, any equimultiples have been taken, namely, the A AHG BOOK VI. PROP. 1. 309 and the base HC ; and of the consequents, any equi- multiples have been taken, namely the A ACM and the base CM; and since it has been shewn that the A AHC is greater than, equal to, or less than the A ACM, according as HC is greater than, equal to, or less than CM; .'. the four original magnitudes are proportionals, v. Def. 4. that is, the A ABC : the A ACD :: the base BC : the base CD. Q.E.D. COROLLARY. The areas of parallelograms of the same altitude are to one another as their bases. Let EC, CF be par m5 of the same altitude; then shall the par m EC : the par CF :: BC : CD. Join BA, AD. Then the A ABC : the A ACD :: BC : CD; Proved. but the par m EC is double of the A ABC, and the par m CF is double of the AACD; .'. the par m EC : the par m CF :: BC : CD. v. 8. NOTE. Two straight lines are cut proportionally when the seg- ments of one line are in the same ratio as the corresponding segments of the other. [See definition, page 131.] Fl 'g'1 Fig. 2 A X B A B X YD C D Thus AB and CD are cut proportionally at X and Y, if AX : XB ;: CY : YD. And the same definition applies equally whether X and Y divide AB, CD internally as in Fig. 1 or externally as in Fig. 2. 310 EUCLID'S ELEMENTS PROPOSITION 2. THEOREM. If a straight line be drawn parallel to one side of a triangle, it shall cut the other sides, or those sides produced, proportionally : Conversely, if the sides or the sides produced be cut pro- portionally, the straight line which joins t/te points of section, shall be parallel to the remaining side of the triangle. B Let XY be drawn par 1 to BC, one of the sides of the A ABC: then shall BX : XA : : CY : YA. Join BY, CX. Then the A BXY = the A CXY, being on the same base XY and between the same parallels XY, BC; i. 37. and AXY is another triangle; .'. the A BXY : the A AXY :: the A CXY : the A AXY. v. 4. But the A BXY : the A AXY :: BX : XA, VI. 1. and the A CXY : the A AXY : : CY : YA, .'. BX : XA :: CY : YA. v. 1. Conversely, let BX : XA :: CY : YA, and let XY be joined: then shall XY be par 1 to BC. As before, join BY, CX. By hypothesis BX : XA : : CY : YA ; but BX : XA :: the A BXY and CY : YA :: the A CXY .'. the A BXY : the A AXY : : the A CXY the ABXY^the A CXY; the A AXY, VI. 1. the A AXY; the A AXY. v. 1. v. 6. and they are triangles on the same base and on the same side of it. .'. XY is par 1 to BC. I. 39. Q. F. D. BOOK VI. PROP. 2. 311 EXERCISES. 1. Shew that every quadrilateral is divided by its diagonals into four triangles proportional to each other. 2. If uny two straight lines are cut by three parallel straight lines, they are cut proportionally. 3. From a point E in the common base of two triangles ACB, ADB, straight lines are drawn parallel to AC, AD, meeting BC, BD at F, G : shew that FG is parallel to CD. 4. In a triangle ABC the straight line DEF meets the sides BC, CA, AB at the points D, E, F respectively, and it makes equal angles with AB and AC: prove that BD : CD :: BF : CE. 5. If the bisector of the angle B of a triangle ABC meets AD at right angles, shew that a line through D parallel to BC will bisect AC. 6. From B and C, the extremities of the base of a triangle ABC, lines BE, CF are drawn to the opposite sides so as to intersect on the median from A: shew that EF is parallel to BC. 7. From P, a given point in the side AB of a triangle ABC, draw a straight line to AC produced, so that it will be bisected by BC. 8. Find a point within a triangle such that, if straight lines be drawn from it to the three angular points, the triangle will be divided into three equal triangles. 312 EUCLID'S ELEMENTS. PROPOSITION 3. THEOREM. If the vertical angle of a triangle be bisected by a straight line which cuts the base, the segments of the base shall have to one another the same ratio as the remaining sides of the triangle : Conversely, if the base be divided so that its segments have to one another the same ratio as the remaining sides of the triangle have, the straight line drawn from the vertex to the point of section shall bisect the, vertical angle. In the A ABC let the L. BAG be bisected by AX, which meets the base at X; then shall BX, : XC : : BA : AC. Through C draw CE par 1 to XA, to meet BA produced at E. i. 31. Then because XA and CE are par 1 , /. the L BAX = the int. opp. L. AEC, I. 29. and the L XAC = the alt. L. ACE. I. 29. But the /_BAX = the Z.XAC; Hyp. .'. the L. AEC = the ^.ACE; AC = AE. i. 6. Again, because XA is par' to CE, a side of the A BCE, .'. BX : XC :: BA : AE; VI. 2. that is, BX : XC : : BA : AC. BOOK VI. PROP. 3. 313 Conversely, let BX : XC :: BA : AC; and let AX be joined: then shall the L BAX = L XAC. For, with the same construction as before, because XA is par 1 to CE, a side of the A BCE, .'. BX : XC :: BA : AE. vi. 2. But by hypothesis BX : XC :: BA : AC; .'. BA : AE :: BA : AC; v. 1. .'. AE = AC; .'. the L ACE = the L AEC. i. 5. But because XA is par 1 to CE, /. the L XAC = the alt. L. ACE. i. 29. and the ext. L BAX = the int. opp. L AEC; I. 29. .*. the L BAX = the L XAC. Q.E.D. EXERCISES. 1. The side BC of a triangle ABC is bisected at D, and the angles ADB, ADC are bisected by the straight lines DE, DF, meeting AB, AC at E, F respectively: shew that EF is parallel to BC. 2. Apply Proposition 3 to trisect a given finite straight line. 3. If the line bisecting the vertical angle of a triangle be divided into parts which are to one another as the base to the sum of the sides, the point of division is the centre of the inscribed circle. 4. A BCD is a quadrilateral: shew that if the bisectors of the angles A and C meet in the diagonal BD, the bisectors of the angles B and D will meet on AC. 5. Construct a triangle having given the base, the vertical angle, and the ratio of the remaining sides. 6. Employ this proposition to shew that the bisectors of the angles of a triangle are concurrent. 7. AB is a diameter of a circle, CD is a chord at right angles to it, and E any point in CD: AE and BE are drawn and produced to cut the circle in F and G : shew that the quadrilateral CFDG has any two of its adjacent sides in the same ratio as the remaining two. 314 EUCLID'S ELEMENTS. PROPOSITION A. THEOREM. If one side of a triangle be produced, and the exterior angle so formed be bisected by a straight line which cuts the base produced, the segments between the bisector and the extremities of the base shall have to one another the same ratio as the remaining sides of the triangle have: Conversely, if the segments of the base produced have to one another the same ratio as the remaining sides of the tri- angle have, the straight line drawn from the vertex to the point of section shall bisect the exterior vertical angle. In the A ABC let BA be produced to F, and let the exterior L CAP be bisected by AX which meets the base produced at X : then shall BX : XC :: BA : AC. Through C draw CE par 1 to XA, i. 31. and let CE meet BA at E. Then because AX and CE are par', .'. the ext. L. FAX = the int. opp. L A EC, and the /. XAC = the alt. /_ ACE. i. 29. But the L. FAX = the L. XAC; Hyp. .'.the ^AEC^the ^.ACE; .'. AC = AE. I. 6. Again, because XA is par 1 to CE, a side of the A BCE, Constr. .'. BX : XC :: BA : AE; VI. 2. that is, BX : XC : : BA : AC. BOOK VI. PROP. A. 315 Conversely, let BX : XC :: BA : AC, and let AX be joined: then shall the L. FAX = the L XAC. For, with the same construction as before, because AX is par 1 to CE, a side of the A BCE, /. BX : XC :: BA : AE. VI. 2. But by hypothesis BX : XC :: BA : AC; .'. BA : AE :: BA : AC; v. 1. .'. AE = AC, /. the L ACE -the L. AEC. I. 5. But because AX is par 1 to CE, .'. the z. XAC = the alt. L. ACE, and the ext. L FAX = the int. opp. L. AEC ; I. 29. /. the L. FAX = the L. XAC. Q.E.D. Propositions 3 and A may be both included in one enunciation as follows : // the interior or exterior vertical anyle of a triangle be bisected by a straight line which also cuts the base, the base shall be divided internally or externally into segments which have the same ratio as the sides of the triangle : Conversely, if the base be divided internally or externally into seg- ments which have the same ratio as the sides of the triangle, the straight line drawn from the point of division to the vertex will bisect the interior or exterior vertical angle. EXERCISES. 1. In the circumference of a circle of which AB is a diameter, a point P is taken ; straight lines PC, PD are drawn equally inclined to AP and on opposite sides of it, meeting AB in C and D ; shew that AC : CB :: AD : DB. 2. From a point A straight lines are drawn making the angles BAG, CAD, DAE, each equal to half a right angle, and they are cut by a straight line BCDE, which makes BAE an isosceles triangle: shew that BC or DE is a mean proportional between BE and CD. 3. By means of Propositions 3 and A, prove that the straight lines bisecting one angle of a triangle internally, and the other two externally, are concurrent. 316 EUCLID'S ELEMENTS. PROPOSITION 4. THEOREM. If two triangles be equiangular to one anotlier, the sides about the equal angles shall be proportionals, those sides which are opposite to equal angles being homologous. C E Let the A ABC be equiangular to the A DCE, having the L. ABC equal to the L. DCE, the L. BCA equal to the L. CED, and consequently the L CAB equal to the L. EDO: I. 32. then shall the sides about these equal angles be propor- tionals, namely AB : BC :: DC : CE, BC : CA :: CE : ED, and AB : AC :: DC : DE. Let the A DCE be placed so that its side CE may be contiguous to BC, and in the same straight line with it. Then because the .L 8 ABC, ACB are together less than two rt. angles, i. 17. and the L. ACB = the /.DEC; Hyp. .'. the L * ABC, DEC are together less than two rt. angles; .'. BA and ED will meet if produced. Ax. 12. Let them be produced and meet at F. Then because the L ABC = the L DCE, Hyp. .'. BF is par 1 to CD; I. 28. and because the /_ACB = the L DEC, Hyp. .'. AC is par 1 to FE, I. 28. .'. FACD is a par m ; ,'. AF = CD, and AC = FD. i. 34. BOOK VI. PROP. 4. 317 Again, because CD is par 1 to BF, a side of the A EBF, VI. 2. V. 11. a side of the A FEE, vi. 2. v. 11. Proved. v. 14. Q. E. D. .'. BC : CE FD DE; but FD AC; .*. BC : CE AC DE; and, alternately, BC : CA CE ED. Again, because AC is par 1 FE, a sid< .'. BA : AF BC CE; but AF CD; /. BA : CD BC CE; and, alternately, AB : BC DC CE. Also BC : CA CE ED; .'., ex cequali, AB : AC DC DE. [For Alternative Proof see Page 320.] EXERCISES. 1. If one of the parallel sides of a trapezium is double the other, shew that the diagonals intersect one another at a point of trisection. 2. In the side AC of a triangle ABC any point D is taken : shew that if AD, DC, AB, BC are bisected in E, F, G, H respectively, then EG is equal to HF. 3. AB and CD are two parallel straight lines; E is the middle point of CD ; AC and BE meet at F, and AE and BD meet at G : shew that FG is parallel to AB. 4. ABCDE is a regular pentagon, and AD and BE intersect in F : shew that AF : AE :: AE : AD. 5. In the figure of i. 43 shew that EH and GF are parallel, and that FH and GE will meet on CA produced. 6. Chords AB and CD of a circle are produced towards B and D respectively to meet in the point E, and through E, the line EF is drawn parallel to AD to meet CB produced in F. Prove that EF is a mean proportional between F B and FC. 318 EUCLID'S ELEMENTS. PROPOSITION o. THEOREM. If the sides of two triangles, talcen in order about each of their angles, be proportionals, the triangles shall be equi- angular to one another, having those angles equal which are opposite to the homologous sides. B C Let the A 8 ABC, DEF have their sides proportionals, so that AB : BC :: DE : EF, BC : CA :: EF : FD, and consequently, ex cequali, AB : CA :: DE : FD. Then shall the triangles be equiangular. At E in FE make the L FEG equal to the /L ABC; and at F in EF make the L EFG equal to the z_ BCA; i. 23. then the remaining L EG F = the remaining /.BAG. i. 32. .'. the A GEF is equiangular to the A ABC; .'. GE : EF :: AB : BC. vi. 4. But AB : BC :: DE : EF; Hyp. :. GE : EF :: DE : EF; v. 1. .'. GE= DE. Similarly GF = DF. Then in the triangles GEF, DEF ( GE=DE, Because < GF= DF, (and EF is common; .'. the i. GEF = the L DEF, i. 8. and the L GFE = the L DFE, and the L EGF = the L EOF. But the L GEF = the /. ABC; Constr. :. the L. DEF = the L. ABC. Similarly, the L. EFD = the L. BCA, BOOK VI. PROP. 6. 319 .'. the remaining L. FDE = the remaining ^CAB; I. 32. that is, the A DEF is equiangular to the A ABC. Q. E.D. PROPOSITION 6. THEOREM. If two triangles have one angle of the one eqttal to one angle of the other, and the sides about the equal angles pro- portionals, the triangles shall be similar. In the A s ABC, DEF let the L BAG = the L EOF, and let BA : AC :: ED : DF. Then shall the A s ABC, DEF be similar. At D in FD make the L FDG equal to one of the L. s EOF, BAC : at F in DF make the L DFG equal to the L ACB; I. 23. .'. the remaining /_ FGD = the remaining L_ ABC. I. 32. Then the A ABC is equiangular to the A DGF; .'. BA : AC :: GD : DF. VI. 4. But BA : AC :: ED : DF; Hyp. :. GD : DF :: ED : DF, .'. GD = ED. Then in the A s GDF, EOF, (GD = ED, and DF is common; and the L GDF the L EDF; Constr. .'. the A s GDF, EDF are equal in all respects, I. 4. so that the A EDF is equiangular to the A GDF; but the A GDF is equiangular to the ABAC; Constr. .'. the A EDF is equiangular to the ABAC; .'. their sides about the equal angles are proportionals, vi. 4. that is, the A s ABC, DEF are similar. Q. E. D. H. E. 21 320 EUCLID'S ELEMBSTS. NOTE 1. From Definition 2 it is seen that two conditions are necessary for similarity of rectilineal figures, namely (1) the figures must be equiangular, and (2) the sides about the equal angles must be proportionals. In the case of triangles we learn from Props. 4 and 5 that each of these conditions follows from the other : this how- ever is not necessarily the case with rectilineal figures of more than three sides. NOTE 2. We have given Euclid's demonstrations of Propositions 4, 5, 6 ; but these propositions also admit of easy proof by the method of superposition. As an illustration, we will apply this method to Proposition 4. PROPOSITION 4. [ALTERNATIVE PROOF.] If two triangles be equiangular to one another, the sides about the equal angles shall be proportionals, those sides which are opposite to equal angles being hojnologous. B C E H F Let the A ABC be equiangular to the A DEF, having the / ABC equal to the L DEF, the L BCA equal to the L EFD, and conse- quently the / CAB equal to the L FDE: 1.32. then shall the sides about these equal angles be proportionals. Apply the A ABC to the A DEF, so that B falls on E and BA along ED: then BC will fall along EF, since the / ABC = the / DEF. Hyp. Let G and H be the points in ED and EF, on which A and C fall. Join GH. Then because the / EGH = the L EDF, Hyp. :. GH is par 1 to DF: .-. DG : GE :: FH : HE; .-. , componendo, DE : GE :: FE : HE, v. 13. .-. , alternately, DE : FE :: GE : HE, v. 11. that is, DE : EF :: AB : BC. Similarly by applying the A ABC to the A DEF, so that the point C may fall on F, it may be proved that EF : FD :: BC : CA. .-. , ex aquali, DE : DF :: AB : AC. Q. E. D. BOOK VI. PROP. 7. 321 PROPOSITION 7. THEOREM. If two triangles have one angle of the one equal to one angle of the other and the sides about one other angle in each proportional, so that the sides opposite to the equal angles are homologous, then the third angles are either equal or sup- plementary ; and in the former case the triangles are similar. Let ABC, DEF be two triangles having the L ABC equal to the L. DEF, and the sides about the angles at A and D pro- portional, so that BA : AC :: ED : DF; then shall the L* ACB, DFE be either equal or supple- mentary, and in the former case the triangles shall be similar. If the L. BAG = the L. EOF, then the z_BCA = the L. EFD; 1.32. and the A a are equiangular arid therefore similar, vi. 4, But if the L. BAC is not equal to the L EOF, one of them must be the greater. Let the /_ EOF be greater than the L BAC. At D in ED make the /_ EOF' equal to the L BAC. i. 23. Then the A s BAC, EOF' are equiangular, Constr. .'. BA : AC :: ED : DF'; VI. 4. but BA : AC :: ED : DF; Hyp. .'. ED : DF :: ED : DF', V. 1. .'. DF=DF', .'. the L. OFF' = the L. DF'F. I. 5. But the ^. s DF'F, DF'E are supplementary, i. 13. .'. the L s OFF', DF'E are supplementary: that is, the /_ 3 DFE, ACB are supplementary. Q. E.D. 21-2 222 EUCLID'S ELEMENTS. COROLLARIES TO PROPOSITION 7. A Three cases of this theorem deserve special attention. It has been proved that if the angles ACB, DFE are not mentttry, they are equal: and we know that of angles which are supplementary and unequal, one must be acute and the other obtuse. Hence, in addition to the hypothesis of this theorem, (i) If the angles ACB, DFE, opposite to the two homologous sides AB, DE are both acute, both obtuse, or if one of them is a right angle, it follows that these angles are equal ; and therefore the triangles are similar. (ii) If the two given angles are right angles or obtuse angles, it follows that the angles ACB, DFE must be both acute, and therefore equal, by Ji) : so that the triangles are similar. (iii) If in each triangle the side opposite the given angle is not less than the other given side; that is, if AC and DF are not less than AB and DE respectively, then the angles ACB, DFE cannot be greater than the angles ABC, DEF, respectively; therefore the angles ACB, DFE, are both acute; hence, as above, they are equal ; and the triangles ABC, DEF similar. BOOK VI. PROP. 7. 323 EXERCISES. ON PROPOSITIONS 1 TO 7. 1. Shew that the diagonals of a trapezium cut one another in the same ratio. 2. If three straight lines drawn from a point cut two parallel straight lines in A, B, C and P, Q, R respectively, prove that AB : BC :: PQ : QR. 3. From a point O, a tangent OP is drawn to a given circle, and OQR is drawn cutting it in Q. ancr R ; shew that 00.: OP :: OP : OR. 4. If two triangles are on equal bases and between the name parallel*, any straight line parallel to their bases will cut off equal areas from the two triangles. 5. If two straight lines PQ, XY intersect in a point O, so that PO : OX :: YO : OQ, prove that P, X, Q, Y are concyclic. 6. On the same base and on the same side of it two equal triangles ACB, ADB are described; AC and BD intersect in O, and through O lines parallel to DA and CB are drawn meeting the base in E and F. Shew that AE= BF. 7. BD, CD are perpendicular to the sides AB, AC of a triangle ABC, and CE is drawn perpendicular to AD, meeting AB in E : shew that the triangles ABC, ACE are similar. 8. AC and BD are drawn perpendicular to a given straight line CD from two given points A and B ; AD and BC intersect in E, and EF is perpendicular to CD : shew that AF and BF make equal angles with CD. 9. A BCD is a parallelogram ; P and Q. are points in a straight line parallel to AB ; PA and QB meet at R, and PD and QC meet at S : shew that RS is parallel to AD. 10. In the sides AB, AC of a triangle ABC two points D, E are taken such that BD is equal to CE ; if DE, BC produced meet at F, shew that AB : AC :: EF : DF. 11. Find a point the perpendiculars from which on the sides of a given triangle shall be in a given ratio. 324 EUCLID'S ELEMENTS. PROPOSITION 8. THEOREM. In a right-angled triangle if a perpendicular be drawn from the right angle to the hypotenuse, the triangles on each side of it are similar to the whole triangle and to one anot/ier. A B DC Let ABC be a triangle right-angled at A, and let AD be perp. to BC: then shall the A s DBA, DAC be similar to the A ABC and to one another. In the A s DBA, ABC, the L BDA = the L. BAG, being rt. angles, and the L ABC is common to both; .'. the remaining L. BAD = the remaining L BCA, I. 32. that is, the A s DBA, ABC are equiangular; .'. they are similar. vi. 4. In the same way it may be proved that the A 8 DAC, ABC are similar. Hence the A 8 DBA, DAC, being equiangular to the same A ABC, are equiangular to one another; .'. they are similar. vi. 4. Q. E. D. COROLLARY. Because the A 8 BDA, ADC are similar, .*. BD : DA :: DA : DC; and because the A 8 CBA, ABD are similar, .\ CB : BA :: BA : BD; and because the A 8 BCA, ACD arc similar, .'. BC : CA :: CA : CD. EXERCISES. 1. Prove that the hypotenuse is to one side as the second side i.-? to the perpendicular. 2. Sheic that the radius of a circle is a mean proportional between the segments of any tangent between its point of contact and a pair of parallel tangents. BOOK VI. PROP. 9. 325 DEFINITION. A less magnitude is said to be a sub- multiple of a greater, when the less is contained an exact number of times in the greater. [Book v. Def. 2.] \ PROPOSITION 9. PROBLEM. From a given straight line to cut off" any required sub- r niultipZe. A F Let AB be the given straight line. It is required to cut off a certain submultiple from AB. From A draw a straight line AG of indefinite length making any angle with AB. In AG take any point D ; and, by cutting off successive parts each equal to AD, make AE to contain AD as many times as AB contains the required submultiple. Join EB. Through D draw DF par 1 to EB, meeting AB in F. Then shall AF be the required submultiple. Because DF is par 1 to EB, a side of the AAEB, /. BF : FA :: ED : DA; vi. 2. .'., componendo, BA : AF :: EA : AD. v. 13. But AE contains AD the required number of times; Constr. .'. AB contains AF the required number of times; that is, AF is the required submultiple. Q. E. F. EXERCISES. 1. Divide a straight line into five equal parts. 2. Give a geometrical construction for cutting off two-sevenths of a given straight line. 326 EUCLID'S ELEMENTS. PROPOSITION 10. PROBLEM. To divide a, straight line similarly to a given divided straight line. / B K C Let AB be the given straight line to be divided, and AC the given straight line divided at the points D and E. It is required to divide AB similarly to AC. Let AB, AC be placed so as to form any angle. Join CB. Through D draw DF par 1 to CB, I. 31. and through E draw EG par 1 to CB, and through D draw DHK par 1 to AB. Then AB shall be divided at F and G similarly to AC. For by construction each of the figs. FH, HB is a par m ; .'. DH = FG, and HK = GB. 1.34. Now since HE is par 1 to KC, a side of the A DKC, .'. KH : HD :: CE : ED. VI. 2. But KH = BG, and HD = GF; .'. BG : GF :: CE : ED. v. 1. Again, because FD is par 1 to GE, a side of the A AGE, .'. GF : FA :: ED : DA, vi. 2. and it has been shewn that BG : GF :: CE : ED, ,'., ex cequali, BG : FA :: CE : DA : v. 14. .'. AB is divided similarly to AC. Q. E. F. EXERCISE. Divide a straight line internally and externally in a given ratio. Is this always possible ? BOOK VI. PROP. 11. 327 PROPOSITION 11. PROBLEM. To find a third proportional to two given straight lines. K BA Let A, B be two given straight lines. It is required to find a third proportional to A and B. Take two st. lines DL, DK of indefinite length, containing any angle : from DL cut off DG equal to A, and GE equal to B; and from DK cut off DH equal to B. I. 3. Join GH. Through E draw EF par 1 to GH, meeting DK in F. I. 31. Then shall HF be a third proportional to A and B. Because GH is par 1 to EF, a side of the A DEF; .'. DG : GE :: DH : HF. vi. 2. But DG A; and GE, DH each = B; Constr. :. A : B :: B : HF; that is, HF is a third proportional to A and B. Q. E. F. EXERCISES. 1. AB is a diameter of a circle, and through A any straight line is drawn to cut the circumference in C and the tangent at B in D : shew that AC is a third proportional to AD and AB. 2. ABC is an isosceles triangle having each of the angles at the base double of the vertical angle BAC ; the bisector of the angle BCA meets AB at D. Shew that AB, BC, BD are three proportionals. 3. Two circles intersect at A and B ; and at A tangents are drawn, one to each circle, to meet the circumferences at C and D : shew that if CB, BD are joined, BD is a third proportional to CB, BA. 328 EUCLID'S ELEMENTS. PROPOSITION 12. PROBLEM. To find afowrtJi proportional to three given straight lilies. ABC Let A, B, C be the three given straight lines. It is required to find a fourth proportional to A, B, C. Take two straight lines DL, DK containing any angle: from DL cut off DG equal to A, GE equal to B; and from DK cut off DH equal to C. I. 3. Join GH. Through E draw EF par' to GH. I. 31. Then shall HF be a fourth proportional to A, B, C. Because GH is par 1 to EF, a side of the A DEF; /. DG : GE :: DH : HF. vi. 2. But DG = A, GE = B, and DH = C; Constr. .'. A : B :: C : HF; that is, HF is a fourth proportional to A, B, C. Q.E. F. EXERCISES. 1. If from D, one of the angular points of a parallelogram A BCD, a straight line is drawn meeting AB at E and CB at F ; shew that CF is a fourth proportional to EA, AD, and AB. 2. In a triangle ABC the bisector of the vertical angle BAG meets the base at D and the circumference of the circumscribed circle at E : shew that BA, AD, EA, AC are four proportionals. 3. From a point P tangents PQ, PR are drawn to a circle whose centre is C, and QT is drawn perpendicular to RC produced : shew that QT is a fourth proportional to PR, RC. and RT. BOOK VI. PROP. 13. 329 PROPOSITION 13. PROBLEM. k 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 on AC describe the semicii'cle ADC. From B draw BD at rt. angles to AC. I. 11. Then shall BD be a mean proportional between AB and BC. Join AD, DC. Now the _ ADC being in a semicircle is a rt. angle; in. 31. and because in the right-angled A ADC, DB is drawn from the rt. angle perp. to the hypotenuse, .'. the A s ABD, DBC are similar; vi. 8. .'. AB : BD :: BD : BC; that is, BD is a mean proportional between AB and BC. Q. E. F. EXERCISES. 1. If from one angle A of a parallelogram a straight line be drawn cutting the diagonal in E and the sides in P, Q, shew that AE is a mean proportional between PE and EQ. 2. A, B, C are three points in order in a straight line : find a point P in the straight line so that PB may be a mean proportional between PA and PC. 3. The diameter AB of a semicircle is divided at any point C, and CD is drawn at right angles to AB meeting the circumference in D ; DO is drawn to the centre, and CE is perpendicular to OD : shew that DE is a third proportional to AO and DC. 330 EUCLID'S ELEMENTS. 4. AC is the diameter of a semicircle on which a point B is taken so that BC is equal to the radius: shew that AB is a mean propor- tional between BC and the sum of BC, CA. 5. A is any point in a semicircle on BC as diameter; from D any point in BC a perpendicular is drawn meeting AB, AC, and the cir- cumference in E, G, F respectively; shew that DG is a third propor- tional to DE and DF. 6. Two circles touch externally, and a common tangent touches them at A and B : prove that AB is a mean proportional between the diameters of the circles. [See Ex. 21, p. 219.] 7. If a straight line be divided in two given points, determine a third point such that its distances from the extremities may be proportional to its distances from the given points. 8. AB is a straight line divided at C and D so that AB, AC, AD are in continued proportion; from A a line AE is drawn in any direc- tion and equal to AC ; shew that BC and CD subtend equal angles at E. 9. In a given triangle draw a straight line parallel to one of the sides, so that it may be a mean proportional between the segments of the base. 10. On the radius OA of a quadrant OAB, a semicircle ODA is described, and at A a tangent AE is drawn ; from O any line ODFE is drawn meeting the circumferences in D and F and the tangent in E : if DG is drawn perpendicular to OA, shew that OE, OF, OD, and OG are in continued proportion. 11. From any point A, in the circumference of the circle ABE, as centre, and with any radius, a circle BDC is described cutting the former circle in B and C ; from A any line AFE is drawn meeting the chord BC in F, and the circumferences BDC, ABE in D, E respec- tively: shew that AD is a mean proportional between AF and AE. DEFINITION. Two figures are said to have their sides about two of their angles reciprocally proportional, when a side of the first is to a side of the second as the remaining side of the second is to the remaining side of the first. [Book vi. Def. 3.] BOOK VI. PROP. 14. 331 PROPOSITION 14. THEOREM. Parallelograms which are equal in area, and which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional: Conversely, parallelograms which have one angle of the one equal to one angle of the other, and the sides about these angles reciprocally proportional, are equal in area. L G l 'C Let the par" 15 AB, BC be of equal area, and have the L DBF equal to the L. QBE: then shall the sides about these equal angles be reciprocally proportional, that is, DB : BE :: GB : BF. Place the par so that DB, BE may be in the same straight line; .'. FB, BG are also in one straight line. I. 14. Complete the par m FE. Then because the par 1 " AB = the par BC, Hyp and FE is another par", .'. the par AB : the par m FE :: the par BC : the par 1 " FE; but the par m AB : the par FE :: DB : BE, vi. 1. and the par m BC : the par m FE :: GB : BF, .*. DB : BE :: GB : BF. v. 1. Conversely, let the L DBF be equal to the L QBE, and let DB : BE :: GB : BF. Then shall the par 1 " AB be equal in area to the par 1 " BC. For, with the same construction as before, by hypothesis DB : BE :: GB : BF; but DB : BE :: the par"' AB : the par FE, vi. 1. and GB : BF :: the par BC : the par" 1 FE, .'. the par 1 " AB : the par ra FE : : the par" 1 BC : the par FE; v. 1 . .'. the par AB =the par BC. Q. E. D. 332 EUCLID'S ELEMENTS. PROPOSITION 15. THEOREM. Triangle* which are equal in area, and ivhich Jtave one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional: Conversely, triangles which Jiave one angle of the one equal to one angle of the other, and the sides about these angles reciprocally proportional, are equal in area. Let the A s ABC, ADE be of equal area, and have the L CAB equal to the L. EAD : then shall the sides of the triangles about these angles be reciprocally proportional, that is, CA : AD :: EA : AB. Place the A s so that CA and AD may be in the same st. line; .'. BA, AE are also in one st. line. i. 14. Join BD. Then because the A CAB = the A EAD, and ABD is another triangle; /. the A CAB : the A ABD :: the A EAD : the A ABD; but the ACAB : the A ABD : : CA : AD, vi. I. and the A EAD : the A ABD : : EA : AB, .'. CA : AD :: EA : AB. v. 1. Conversely, let the L. CAB be equal to the L. EAD, and let CA : AD :: EA : AB. Then shall the ACAB= A EAD. For, with the same construction as before, by hypothesis CA : AD :: EA : AB ; but CA : AD :: the ACAB : the A ABD, vi. 1. and EA : AB :: the A EAD : the A ABD, .'. the ACAB : the A ABD :: the A EAD : the A ABD; v. 1. .'. the A CAB = the A EAD. Q. E.D. BOOK VI. PROP. 15. 333 KXERCISES. ON PROPOSITIONS 14 AND 15. 1. Parallelograms which are equal in area and which have their sides reciprocally proportional, liave their angles respectively equal. 2. Triangles which are equal in area, and which have the side* about a pair of angles reciprocally proportional, have those angles equal or supplementary. 3. AC, BD are the diagonals of a trapezium which intersect in O ; if the side AB is parallel to CD, use Prop. 15 to prove that the triangle AOD is equal to the triangle BOG. 4. From the extremities A, B of the hypotenuse of a right- angled triangle ABC lines AE, BD are drawn perpendicular to AB, and meeting BC and AC produced in E and D respectively : employ Prop. 15 to shew that the triangles ABC, ECD are equal in area. 5. On AB, AC, two sides of any triangle, squares are described externally to the triangle. If the squares are ABDE, ACFG, shew that the triangles DAG, FAE are equal in area. 6. ABCD is a parallelogram; from A and C any two parallel straight lines are drawn meeting DC and AB in E and F respectively; EG, which is parallel to the diagonal AC, meets AD in G : shew that the triangles DAF, GAB are equal in area. 7. Describe an isosceles triangle equal in area to a given triangle and having its vertical angle equal to one of the angles of the given triangle. 8. Prove that the equilateral triangle described on the hypotenuse of a right-angled triangle is equal to the sum of the equilateral triangles described on the sides containing the right angle. [Let ABC be the triangle right-angled at C ; and let BXC, CYA, AZB be the equilateral triangles. Draw CD perpendicular to AB ; and join DZ. Then shew by Prop. 15 that the A AYC = the A DAZ ; and similarly that the A BXC = the A BDZ.] 334 EUCLID'S ELEMENTS. PROPOSITION 16. THEOREM. If four straight lines are proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means: Conversely, if the rectangle contained by the extremes is equal to the rectangle contained by the means, the four straight lines are proportional. D E G Let the st. lines AB, CD, EF, GH be proportional, so that AB : CD :: EF : GH. Then shall the rect. AB, GH =the rect. CD, EF. From A draw AK perp. to AB, and equal to GH. I. 11, .'i. From C draw CL perp. to CD, and equal to EF. Complete the par ms KB, LD. Then because AB : CD :: EF : GH ; Hyp. and EF = CL, and GH = AK; Constr. :. AB : CD : : CL : AK; that is, the sides about equal angles of par ms KB, LD are reciprocally proportional ; .'.KB = LD. vi. 14. But KB is the rect. AB, GH, for AK = GH, Constr. and LD is the rect. CD, EF, for CL= EF; .'. the rect. AB, GH=--the rect. CD, EF. Conversely, let the rect. AB, GH = the rect. CD, EF: then shall AB : CD :: EF : GH. For, with the same construction as before, because the rect. AB, GH =the rect. CD, EF; Hyp- and the rect. AB, GH = KB, for GH = AK, Constr. and the rect. CD, EF = LD, for EF = CL ; .'. KB = LD; BOOK VI. PROP. 17. 335 that is, the par" 13 KB, LD, which have the angle at A equal to the angle at C, are equal in area; .'. the sides about the equal angles are reciprocally proportional : that is, AB : CD :: CL : AK; .'. AB : CD :: EF : GH. Q. E. D. PROPOSITION 17. THEOREM. If three straight lines are proportional the rectangle con- tained by the extremes is equal to the square on the mean: Conversely, if the rectangle contained by the extremes is equal to the square on the mean, the three straight lines are proportional. Let the three st. lines A, B, C be proportional, so that A : B :: B : C. Then shall the rect. A, C be equal to the sq. on B. Take D equal to B. Then because A : B : : B : C, and D = B ; .'. A : B :: D : C; .'. the rect. A, C the rect. B, D; vi. 16. but the rect. B, D = the sq. on B, for D = B; .'. the rect. A, C - the sq. on B. Conversely, let the rect. A, C = the sq. on B: then shall A : B :: B : C. For, with the same construction as before, because the rect. A, C = the sq. on B, Hyp. and the sq. on B = the rect. B, D, for D = B; .'. the rect. A, C = the rect. B, D, .*. A : B :: D : C, vi. 1G. that is, A : B : : B : C. q. E. D. H. E. 22 336 EUCLID'S ELEMENTS. EXERCISES. ON PKOPOSITIOXS 16 AND 17. 1. Apply Proposition 16 to prove that if two chords of a circle intersect, the rectangle c< stained by the segments of the one is equal to the rectangle contained by the segments of the other. 2. Prove that the rectangle contained by the sides of a right- angled triangle is equal to the rectangle contained by the hypotenuse and the perpendicular on it from the right angle. 3. On a given straight line construct a rectangle equal to a given rectangle. 4. ABCD is a parallelogram ; from B any straight line is drawn cutting the diagonal AC at F, the side DC at G, and the side AD pro- duced at E: shew that the rectangle EF, FG is equal to the square oil BF. 5. On a given straight line as base describe an isosceles triangle equal to a given triangle. 6. AB is a diameter of a circle, and any line ACD cuts the circle in C and the tangent at B in D ; shew by Prop. 17 that the rectangle AC, AD is constant. 7. The exterior angle A of a triangle ABC is bisected by a straight line which meets the base in D and the circumscribed circle in E : shew that the rectangle BA, AC is equal to the rectangle EA, AD. 8. If two chords AB, AC drawn from any point A in the cir- cumference of the circle ABC be produced to meet the tangent at the other extremity of the diameter through A in D and E, shew that the triangle AED is similar to the triangle ABC. 9. At the extremities of a diameter of a circle tangents are drawn ; these meet the tangent at a point P in Q and R : shew that the rect- angle QP, PR is constant for all positions of P. 10. A is the vertex of an isosceles triangle ABC inscribed in a circle, and ADE is a straight line which cuts the base in D and the circle in E ; shew that the rectangle EA, AD is equal to the square on AB. 11. Two circles touch one another externally in A ; a straight line touches the circles at B and C, and is produced to meet the straight line joining their centres at S : shew that the rectangle SB, SC is equal to the square on SA. 12. Divide a triangle into two equal parts by a straight line at right angles to one of the sides. BOOK VI. PROP. 18. 337 DEFINITION. Two similar rectilineal figures are said to lie similarly situated with respect to two of their sides when these sides are homologous. [Book vi. Def. 5.] PROPOSITION 18. PKOBLKM. On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure. Let AB be the given st. line, and CDEF the given rectil. figure: first suppose CDEF to be a quadrilateral. It is required to describe on the st. line AB, a rectil. iigure similar and similarly situated to CDEF. Join DF. At A in BA make the L. BAG equal to the L DCF, I. 23. and at B in AB make the L. ABG equal to the L. CDF; .'. the remaining /_ AG B = the remaining z_CFD;i. 32. and the AAGB is equiangular to the ACFD. Again at B in GB make the L. GBH equal to the L. FDE, and at G in BG make the /. BGH equal to the L. DFE; I. 23. .'. the remaining L. BHG = the remaining L DEF ; i. 32. and the A BHG is equiangular to the A DEF. Then shall ABHG be the required figure. (i) To prove that the quadrilaterals are equiangular. Because the L. AGB =the L. CFD, and the L. BGH =the L. DFE; Constr. .'. the whole L. AGH = the whole L. CFE. Ax. 2. Similarly the L. ABH = the L CDE ; and the angles at A and H are respectively equal to the angles at C and E ; Constr. .'. the fig. ABHG is equiangular to the tig. CDEF. 22-2 338 EUCLID'S ELEMENTS. (ii) To prove that the quadrilaterals have the sides about their equal angles proportional. Because the A 8 BAG, DCF are equiangular; vi. 4. v. 14. vi. 4. .'. the figs. ABHG, CDEF have their sides about the equal angles proportional ; .*. ABHG is similar to CDEF. Def. 2. In like manner the process of construction may be extended to a figure of five or more sides. Q.E.F. .'. AG : GB CF : FD. Lnd because the A s BGH DFE are equiangular; /. BG : GH DF : FE, .'., ex cequali, AG : GH CF : FE. Similarly it may )e shewn that AB : BH CD : DE. Also BA : AG DC : CF, and GH : HB FE : ED; DEFINITION. When three magnitudes are proportionals the first is said to have to the third the duplicate ratio of that which it lias to the second. [Book v. Def. 13.] BOOK VI. PROP. 19. 339 PROPOSITION 19. THEOREM. Similar triangles are to one another in the duplicate ratio of their homologous sides. Let ABC, DEF be similar triangles, having the /.ABC equal to the L DEF, and let BC and EF be homologous sides : then shall the A ABC be to the A DEF in the duplicate ratio of BC to EF. To BC and EF take a third proportional BG, so that BC : EF :: EF : BG. VI. 11. Join AG. Then because the A s ABC, DEF are similar, Hyp. .'. AB : BC :: DE : EF; .'., alternately, AB : DE :: BC : EF; v. 11. but BC : EF :: EF : BG; Constr. :. AB : DE :: EF : BG; v. 1. that is, the sides of the A s A3G, DEF about the equal angles at B and E are reciprocally proportional; .'. the AABG=:the A DEF. vi. 15. Again, because BC : EF : : EF : BG, Constr. .'. BC : BG in the duplicate ratio of BC to EF. Def. But the A ABC : the AABG :: BC : BG, VI. 1. .'. the A ABC : the AABG in the duplicate ratio of BC to EF: v. 1. and the AABG -the A DEF; Proved. .'. the A ABC : the A DEF in the duplicate ratio of BC : EF. Q.E.D. 340 EUCLID'S ELEMENTS. PROPOSITION 20. THEOREM. Similar polygons may be divided into the same number of similar triangles, having the same ratio each to each that the polygons have; and the polygons are to one another in the duplicate ratio of their homologous sides. H Let ABODE, FGHKL be similar polygons, and let AB be the side homologous to FG ; then, (i) the polygons may be divided into the same number of similar triangles; (ii) these triangles shall have each to each the same ratio that the polygons have; (iii) the polygon ABODE shall be to the polygon FGHKL in the duplicate ratio of AB to FG. Join EB, EC, LG, LH. (i) Then because the polygon ABODE is similar to the polygon FGHKL, Hyp. :. the .LEAB = the /. LFG, and EA : AB : : LF : FG ; vi. Def. 2. .'. the A EAB is similar to the A LFG ; vi. 6. .'. the /_ ABE = the L. FGL. But, because the polygons are similar, Hyp. :. the .ABC = the L. FGH, vi. Def. 2. .'. the remaining L. EBC = the remaining L. LGH. And because the A 8 ' ABE, FGL are similar, Proved. :. EB : BA :: LG : GF; and because the polygons are similar, Hyp. :. AB : BC = FG : GH : vi. Def. 2. .'., ex cequali, EB : BC :: LG : GH, v. 14. that is, the sides about the equal L. s EBC, LGH are proportionals ; .'. the A EBC is similar to the A LGH. vi. 6. BOOK. VI. PROP. 20. 341 In the same way it may be proved thajt the A ECD is similar to the ALHK. .'. the polygons have been divided into the same number of similar triangles. (ii) Again, because the A ABE is similar to the AFGL, .'. the A ABE is to the A FGL in the duplicate ratio of EB : LG; VI. 19. and, in like manner, the A EBC is to the A LGH in the duplicate ratio of EB to LG; /.the A ABE : the AFGL :: the A EBC : the A LGH. v. L In like manner it can be shewn that the A EBC : the A LGH :: the A EDO : the ALKH, .'. the A ABE : the AFGL :: the A EBC : the A LGH :: the A EDC : the ALKH. But when any number of ratios are equal, as each ante- cedent is to its consequent so is the sum of all the ante- cedents to the sum of all the consequents; v. 12. .'. the A ABE : the A LFG :: the tig. ABODE : the fig. FGHKL. (iii) Now the A EAB : the A LFG in the duplicate ratio of AB : FG, and the A EAB : the A LFG :: the fig. ABODE : the fig. FGHKL; .'. the fig. ABODE : the fig. FGHKL in the duplicate ratio of AB : FG. Q. E.IX COROLLARY 1 . Let a third proportional X be taken to AB and FG, then AB is to X in the duplicate ratio of AB : FG; but the fig. ABODE : the fig. FGHKL in the duplicate ratio of AB : FG. Hence, if three straight lines are proportionals, as the first is to the third, so is any rectilineal figure described on the first to a similar and similarly described rectilineal figure on the second. COROLLARY 2. It follows that similar rectilineal figures we to one another as the squares on their homologous sides. For squares are similar figures and therefore are to one another in the duplicate ratio of their sides. 342 EUCLID'S ELEMENTS. PROPOSITION 21. THEOREM. Rectilineal figures ivhich are similar to the same recti- lineal figure, are also similar to each other. Let each of the rectilineal figures A and B be similar to C: then shall A be similar to B. For because A is similar to C, .'. A is equiangular to C, and the sides about their equal angles are proportionals vi. Def. 2. Again, because B is similar to C, H'JP- .'. B is equiangular to C, and the sides about their equal angles are proportionals. vi. Def. 2. .'. A and B are each of them equiangular to C, and have the sides about the equal angles proportional to the cor- responding sides of C ; .'. A is equiangular to B, and the sides about their equal angles are proportionals; v. 1. .'. A is similar to B. Q. E. D. BOOK VI. PROP. 22. 343 PROPOSITION 22. THEOREM. If four straight lines be proportional and a pair of similar rectilineal figures be similarly described on the first and second, and also a pair on the third and fourth, these figures shall be proportional: Conversely, if a rectilineal figure on the first of four straight lines be to the similar and similarly described figure ou the second as a rectilineal figure on the third is to the similar and similarly described figure on the fourth, the four straight lines shall be proportional. Let AB, CD, EF, GH be proportionals, so that AB : CD :: EF ; GH ; and let similar figures KAB, LCD be similarly described on AB, CD, and also let similar figs. MF, NH be similarly described on EF, GH: then shall the fig. KAB : the fig. LCD :: the fig. MF : the fig. NH. To AB and CD take a third proportional X, and to EF and GH take a third proportional O; vi. 11. then AB : CD : CD : X, Constr. and EF : GH : GH : O. But AB : CD : EF : GH; Hyp. /. CD : X : GH : O, V. 1. .'., ejc cequali, AB : X : EF : 0. v. 14. But AB : X :: the fig. KAB : the fig. LCD, vi . 20, 6V. and EF : O :: the fig. MF the fig. NH: the fig. KAB : the fig.' LCD : : the fig~ MF : the fig. NH. v. 1. 344 EUCLID'S ELEMENTS. Conversely, let the fig. KAB : the fig. LCD :: the fig. MF : the fig. NH; then shall AB : CD :: EF : GH. To AB, CD, and EF take a fourth proportional PR : vi. 12. and on PR describe the fig. SR similar and similarly situated to either of the figs. MF, NH. vi. 18. Then because AB : CD :: EF : PR, Constr. .'. , by the former part of the proposition, the fig. KAB : the fig. LCD :: the fig. MF : the fig. SR. But the fig. KAB : the fig. LCD :: the fig. MF : the fig. NH. Hyp. :. the fig. MF : the fig. SR :: the fig. MF : the fig. NH, v. 1. .'. the fig. SR = the fig. NH. And since the figs. SR and NH are similar and similarly situated, .'. PR = GH*. Now AB : CD :: EF : PR: Constr. :. AB : CD :: EF : GH. Q. E. D. * Euclid here assumes that if two similar and similarly situated figures are equal, their homologous sides are equal. The proof is easy and may be left as an exercise for the student. DEFINITION*. When there are any number of magnitudes of the same kind, the first is said to have to the last the ratio compounded of the ratios of the first to the second, of the second to the third, and so on up to the ratio of the last but one to the last magnitude. [Book v. Def. 12.] BOOK vi. PROP. 23. 345 PROPOSITION 23. THEOREM. Parallelograms which are equiangular to one another have to one another the ratio which is compounded of the ratios of their sides. A, D H ^Ao B K L M E F Let the par m AC be equiangular to the par m CF, having the L. BCD equal to the _ EGG : then shall the par AC have to the par m CF the ratio com- pounded of the ratios BC : CG and DC : CE. Let the par" 18 be placed so that BC and CG are in a st. line; then DC and CE are also in a st. line. I. 14. Complete the par m DG. Take any st. line K, and to BC, CG, and K find a fourth proportional L; vi. 12. and to DC, CE, and L take a fourth proportional M ; then BC : CG :: K : L, and DC : CE :: L : M. But K : M is the ratio compounded of the ratios K : L and L : M, v. Def. 12. that is, K : M is the ratio compounded of the ratios BC : CG and DC : CE. Xo\v the par'" AC : the par m CH :: BC : CG vi. 1. :: K : L, Constr. and the par" 1 CH : the par" 1 CF :: DC : CE vi. 1. :: L : M, Constr. .'., ex (equali, the par" 1 AC : the par" 1 CF :: K : M. v. 14. But K : M is the ratio compounded of the ratios of the sides; .'. the par m AC has to the par" 1 CF the ratio compounded of the ratios of the sides. Q. E. D. EXERCISE. The areas of two triangles or parallelograms are to one another in the ratio compounded of the ratios of their bases and of their altitudes. 346 EUCLID'S ELEMENTS. PROPOSITION 24. THEOREM. Parallelograms about a diagonal of any parallelogram are similar to the whole parallelogram and to one another. Let ABCD be a par of which AC is a diagonal; and let EG, HK be par" 15 about AC : then shall the par" 13 EG, HK be similar to the par ABCD, and to one another. For, because DC is par 1 to GF, .'. the ^LADC = the ^AGF; 1.29. and because BC is par 1 to EF, .'. the L. ABC = the Z.AEF; I. 29. and each of the L. s BCD, EFG is equal to the opp. L BAD, .'. the L BCD = the /.EFG; [i. 34. .'. the par m ABCD is equiangular to the par AEFG. Again in the A s BAG, EAF, because the L ABC the ^_AEF, 1.29. and the L. BAG is common; .'. A s BAC, EAF are equiangular to one another; i. 32. .'. AB : BC :: AE : EF. VI. 4. But BC = AD, and EF = AG ; I. 34. /. AB : AD :: AE : AG; and DC : CB :: GF : FE, and CD : DA :: FG : GA, .'. the sides of the par 1 "* ABCD, AEFG about their equal angles are proportional; .'. the par" 1 ABCD is similar to the par m AEFG. vi. Def. 2. In the same way it may be proved that the par ABCD is similar to the par 1 " FHCK, .'. each of the par 1115 EG, HK is similar to the whole par ra ; .'. the par" 1 EG is similar to the par" 1 HK. vi. 21. Q. E. D. BOOK VI. PROP. 25. 347 PROPOSITION 25. PROBLEM. To describe a rectilineal figure which shall be equal to one and similar to another rectilineal figure. B\ D C F H K Let E and S be two rectilineal figures: it is required to describe a figure equal to the fig. E and similar to the fig. S. On AB a side of the fig. S describe a par m ABCD equal to S, and on BC describe a par m CBGF equal to the fig. E, and having the L CBG equal to the L DAB: I. 45. then AB and BG are in one st. line, and also DC and CF in one st. line. Between AB and BG find a mean proportional HK; vi. 13. and on H K describe the fig. P, similar and similarly situated to the fig. S: vi. 18. then P shall be the figure required. Because AB : HK :: HK : BG, Constr. .'. AB : BG :: the fig. S : the fig. P. vi. 20, Cor. But AB : BG :: the par m AC : the par m BF; .'. the fig. S : the fig. P :: the par 1 " AC : the par m BF; v. i. and the fig. S =the par ra AC; Constr. .'. the fig. P =the par m BF = the fig. E. Constr. And since, by construction, the fig. P is similar to the fig. S, .'. P is the rectil. figure required. Q. E. F. 348 EUCLID'S ELEMENTS. PROPOSITION 26. THEOREM. If two similar parallelograms have a common angle, and be similarly situated, they are about the same diagonal. B Let the par 1 ABCD, AEFG be similar and similarly situated, and have the common angle BAD: then shall these par ms be about the same diagonal. Join AC. Then if AC does not pass through F, let it cut FG, or FG produced, at H. Join AF; and through H draw HK par 1 to AD or BC. I. 31. Then the par" 18 BD and KG are similar, since they are about the same diagonal AHC; vi. 24. .'. DA : AB :: GA : AK. But because the par 1 " 5 BD and EG are similar; /////>. .'. DA : AB :: GA : AE; vi. Def. 2. .'. GA : AK :: GA : AE; .'. AK = AE, which is impossible; .'. AC must pass through F; that is, the par 1 " 8 BD, EG are about the same diagonal. Q.E. D. BOOK VI. PROP. 30. 349 Obs. Propositions 27, 28, 29 being cumbrous in form and of little value as geometrical results are now very generally omitted. DEFINITION. A straight line is said to be divided in extreme and mean ratio, when the whole is to the greater segment as the greater segment is to the less. [Book vi. Def. 4.] PROPOSITION 30. PROBLEM. To divide a given Straight line in extreme and mean ratio. Let AB be the given st. line: it is required to divide it in extreme and mean ratio. Divide AB in C so that the rect. AB, BC may be equal to the sq. on AC. n. 11. Then because the rect. AB, BC - the sq. on AC, .'. AB : AC :: AC : BC. vi. 17. Q. E. p. EXERCISES. 1. ABCDE is a regular pentagon; if tlie lines BE and AD inter- sect in O, shew that each of them is divided in extreme and mean ratio. 2. If the radius of a circle is cut in extreme and mean ratio, the greater segment is equal to the side of a regular decagon inscribed in the circle. 350 EUCLID'S ELEMENTS. PROPOSITION 31. THEOREM. In a right-angled triangle, any rectilineal figure described on the hypotenuse is equal to the sum of the two similar and similarly described figures on the sides containing the right angle. Let ABC be a right-angled triangle of which BC is the hypotenuse; and let P, Q, R be similar and similarly described figures on BC, CA, AB respectively: then shall the fig. P be equal to the sum of the tigs. Q and R. Draw AD perp. to BC. Then the A s CBA, ABD are similar; vi. 8. .'. CB : BA :: BA : BD; .'. CB : BD :: the fig. P : the fig. R,vi.20,CW. .'., inversely, BD : BC :: the fig. R : the fig. P. v. 2. In like manner DC : BC :: the fig. Q : the fig. P; .'. the sum of BD, DC : BC :: the sum of figs. R, Q : fig. P; v. 15. but BC = the sum of BD, DC; .'. the fig. P = the sum of the figs. R and Q. Q. K.D. NOTE. This proposition is a generalization of the 47th Prop, of Book i. It will be a useful exercise for the student to deduce the general theorem from the particular case with the aid of Prop. 20, Cor. 2. EXERCISES ON PROP. 31. 351 EXERCISES. 1. In a right-angled triangle if a perpendicular be drawn from the right angle to the opposite side, the segments of the hypotenuse are in the duplicate ratio of the sides containing the right angle. 2. If, in Proposition 31, the figure on the hypotenuse is equal to the given triangle, the figures on the other two sides are each equal to one of the parts into which the triangle is divided by the perpen- dicular from the right angle to the hypotenuse. 3. AX and BY are medians of the triangle ABC which meet in G : if XY be joined, compare the areas of the triangles AGB, XGY. 4. Shew that similar triangles are to one another in the duplicate ratio of (i) corresponding medians, (ii) the radii of their inscribed circles, (iii) the radii of their circumscribed circles. 5. DEF is the pedal triangle of the triangle ABC ; prove that the triangle ABC is to the triangle DBF in the duplicate ratio of AB to BD. Hence shew that the fig. AFDC : the A BFD :: AD 2 : BD 2 . 6. The base BC of a triangle ABC is produced to a point D such that BD : DC in the duplicate ratio of BA : AC. Shew that AD is a mean proportional between BD and DC. 7. Bisect a triangle by a line drawn parallel to one of its sides. 8. Shew how to draw a line parallel to the base of a triangle so as to form with the other two sides produced a triangle double of the given triangle. 9. If through any point within a triangle lines be drawn from the angles to cut the opposite sides, the segments of any one side will have to each other the ratio compounded of the ratios of the segments of the other sides. 10. Draw a straight line parallel to the base of an isosceles tri- angle so as to cut off a triangle which has to the whole triangle the ratio of the base to a side. 11. Through a given point, between two straight lines containing a given angle, draw a line which shall cut off a triangle equal to a given rectilineal figure. Obs. The 32nd Proposition as given by Euclid is de- fective, and as it is never applied, we have omitted it. H. E. 23 352 EUCLID'S ELEMENTS. PROPOSITION 33. THEOREM. In equal circles, angles, whether at the centres or the cir- cumferences, have the same ratio as the arcs on which they stand: so also have the sectors. Let ABC and DEF be equal circles, and let BGC, EHF be angles at the centres, and BAG and EOF angles at the O ces ; then shall (i) the L BGC : the L. EHF :: the arc BC : the arc EF, (ii) the L. BAC : the L. EOF : : the arc BC : the arc EF, (iii) the sector BGC : the sector EHF :: the arc BC : the arc EF. Along the O 06 of the ABC take any number of arcs CK, KL each equal to BC; and along the O ce of the DEF take any number of arcs FM, MM, NR each equal to EF. Join GK, GL, HM, HN, HR. (i) Then the ^ s BGC, CGK, KGL are all equal, for they stand on the equal arcs BC, CK, KL: in. 27. .'. the L. BGL is the same multiple of the L. BGC that the arc BL is of the arc BC. Similarly the L. EHR is the same multiple of the /.EHF that the arc ER is of the arc EF. And if the arc BL = the arc ER, the /_BGL = the L. EHR; ill. 27. and if the arc BL is greater than the arc ER, the L. BGL is greater than the L. EHR; and if the arc BL is less than the arc ER, the L. BGL is less than the L. EHR. BOOK VI. PROP. 33. 353 Now since there are four magnitudes, namely the L S BGC, EHF and the arcs BC, EF; and of the antecedents any equimultiples have been taken, namely the L BGL and the arc BL; and of the consequents any equimultiples have been taken, namely the L. EHR and the arc ER: and it has been proved that the L. BGL is greater than, equal to, or less than the L. EHR according as BL is greater than, equal to, or less than ER; .'. the four magnitudes are proportionals; v. Def. 4. that is, the L. BGC : the L EHF : : the arc BC : the arc EF. (ii) And since the L BGC = twice the L BAC, Hi. 20. and the /_ EHF twice the L. EOF; .*. the _ BAC : the L. EOF : : the arc BC : the arc EF. v. 8. (iii) Join BC, CK; and in the arcs BC, CK take any points X, O. Join BX, XC, CO, OK. Then in the A s BGC, CGK, I BG=CG, Because -! GC = GK, [and the L. BGC = the L. CGK; m. 27. /. BC = CK; I. 4. and the A BG C = the A CG K. And because the arc BC = the arc CK, Constr. .'. the remaining arc BAC = the remaining arc CAK : .'. the z_BXC=the L COK; ill. 27. .'. the segment BXC is similar to the segment COK; HI. Def. and they stand on equal chords BC, CK; .'. the segment BXC = the segment COK. HI. 24. And the A BGC = the A CGK;' .'. the sector BGC - the sector CGK. 23-2 354 EUCLID'S ELEMENTS. B N M Similarly it may be shewn that the sectors BGC, CGK, KGL are all equal; and likewise the sectors EHF, FHM, MHN, NHR are all equal. .'. the sector BGL is the same multiple of the sector BGC that the arc BL is of the arc BC; and the sector EHR is the same multiple of the sector EHF that the arc ER is of the arc EF: And if the arc BL = the arc ER, the sector BGL = the sector EHR: Proved. and if the arc BL is greater than the arc ER, the sector BGL is greater than the sector EHR: and if the arc BL is less than the arc ER, the sector BGL is less than the sector EHR. Now since there are four magnitudes, namely, the sec- tors BGC, EHF and the arcs BC, EF; and of the antecedents any equimultiples have been taken, namely the sector BGL and the arc BL; and of the consequents any equimultiples have been taken, namely the sector EHR and the arc ER: and it has been shewn that the sector BGL is greater than, equal to, or less than the sector EHR according as the arc BL is greater than, equal to, or less than the arc ER; .'. the four magnitudes are proportionals; v. Def. 4. that is, the sector BGC : the sector EHF : : the arc BC : the arc EF. Q. E. D. BOOK VI. PROP. B. PROPOSITION B. THEOREM. If the vertical angle of a triangle be .bisected by a straight line which cuts the base, the rectangle contained by the sides of the triangle shall be equal to the rectangle contained by the segments of the base, together with the square on the straight, line which bisects the angle. Let ABC be a triangle having the L. BAG bisected by AD. then shall the rect. BA, AC -the rect. BD, DC, with the sq. on AD. Describe a circle about the A ABC, iv. 5. and produce AD to meet the O ce in E. Join EC. Then in the A s BAD, EAC, because the L. BAD = the t, EAC, Hyp. and the L ABD = the L. AEC in the same segment; HI. 21. .*. the A BAD is equiangular to the A EAC. i. 32. .'. BA : AD :: EA : AC; vi. 4. .*. the rect. BA, AC = the rect. EA, AD, vi. 16. = the rect. ED, DA, with the sq. on AD. ii. 3. But the rect. ED, DA = the rect. BD, DC; HI. 35. .'. the rect. BA, AC = the rect. BD, DC, with the sq. on AD. Q. E. D. EXERCISE. If the vertical angle BAG be externally bisected by a straight line which meets the base in D, shew that the rectangle contained by BA, AC together with the square on AD is equal to the rectangle contained by the segments of the base. 356 EUCLID'S ELEMENTS. PROPOSITION C. THEOREM. If from the vertical angle of a triangle a, straight line be drawn perpendicular to the base, the rectangle contained by the sides of the triangle shall be equal to the rectangle con- tained by the perpendicular and the diameter of the circle described about the triangle. Let ABC be a triangle, and let AD be the perp. from A to BC: then the rect. BA, AC shall be equal to the rect. contained by AD and the diameter of the circle circumscribed about the A ABC. Describe a circle about the A ABC; IV. 5. draw the diameter AE, and join EC. Then in the A s BAD, EAC, the rt. angle BDA = the rt. angle ACE, in the semicircle ACE, and the L ABD = the /_ A EC, in the same segment; ill. 21. .'. the A BAD is equiangular to the A EAC; I. 32. .'. BA : AD :: EA : AC; vi. 4. .*, the rect. BA, AC = the rect, EA, AD. vi. 16. Q.E.D. BOOK VI. PROP. D. 357 PROPOSITION D. THEOREM. TJie rectangle contained by the diagonals of a quadri- lateral inscribed in a circle is equal to the sum of the two rectangles contained by its opposite sides. Let ABCD be a quadrilateral inscribed in a circle, and let AC, BD be its diagonals: then the rect. AC, BD shall be equal to the sum of the rect- angles AB, CD and BC, AD. Make the L DAE equal to the L BAC; I. 23. to each add the L EAC, then the L DAC = the L BAE. Then in the A 8 EAB, DAC, the L EAB = the /. DAC, and the L ABE = the L. ACD in the same segment; nf. 21. .'. the triangles are equiangular to one another ; i. 32. .'. AB : BE :: AC : CD; VI. 4. .'. the rect. AB, CD the rect. AC, EB. vi 16. Again in the A s DAE, CAB, the L DAE = the /.CAB, Constr. and the L ADE = the L ACB, in the same segment, ill. 21. .'. the triangles are equiangular to one another ; \. 32. .'. AD : DE :: AC : CB; VI. 4. .'. the rect. BC, AD = the rect. AC, DE. VI. 16. But the rect. AB, CD = the rect. AC, EB. Proved. .'. the sum of the rects. BC, AD and AB, CD the sum of the rects. AC, DE and AC, EB ; that is, the sum of the rects. BC, AD and AB, CD - the rect. AC, BD. n. 1. Q. E. D. 358 EUCLID'S ELEMENTS. NOTE. Propositions B, C, and D do not occur in Euclid, but were added by Eobert Simson. Prop. D is usually known as Ptolemy's theorem, and it is the par- ticular case of the following more general theorem : The rectangle contained by the diagonals of a quadrilateral is less than the sum of tJie rectangles contained by its opposite sides, unless a circle can be circumscribed about the quadrilateral, in which case it is equal to that sum. EXERCISES. 1. ABC is an isosceles triangle, and on the base, or base pro- duced, any point X is taken : shew that the circumscribed circles of the triangles ABX, ACX are equal. 2. From the extremities B, C of the base of an isosceles triangle ABC, straight lines are drawn perpendicular to AB, AC respectively, and intersecting at D : shew that the rectangle BC, AD is double of ihe rectangle AB, DB. 3. If the diagonals of a quadrilateral inscribed in a circle are at right angles, the sum of the rectangles of the opposite sides is double the area of the figure. 4. ABCD is a quadrilateral inscribed in a circle, and the diagonal BD bisects AC : shew that the rectangle AD, A B is equal to the rect- angle DC, CB. 5. If the vertex A of a triangle ABC be joined to any point in the base, it will divide the triangle into two triangles such that their circumscribed circles have radii in the ratio of AB to AC. 6. Construct a triangle, having given the base, the vertical angle, and the rectangle contained by the sides. 7. Two triangles of equal area are inscribed in the same circle : shew that the rectangle contained by any two sides of the one is to the rectangle contained by any two sides of the other as the base of the second is to the base of the first. 8. A circle is described round an equilateral triangle, and from any point in the circumference straight lines are drawn to the angular points of the triangle : shew that one of these straight lines is equal to the sum of the other two. 9. ABCD is a quadrilateral inscribed in a circle, and BD bisects the angle ABC : if the points A and C are fixed on the circumference of the circle and B is variable in position, shew that the sum of AB and BC has a constant ratio to BD. THEOREMS AND EXAMPLES ON BOOK VI. I. ON HARMONIC SECTION. 1. To divide a given straight line internally and externally so that its segments may be in a given ratio. H, y /*^ L M A PX/B Q i Let AB be the given st. line, and L, M two other st. lines which determine the given ratio: it is required to divide AB internally and externally in the ratio L : M. Through A and B draw any two par 1 st. lines AH, BK. From AH cut off Aa equal to L, and from BK cut off Bb and Bb' each equal to M, Bb' being taken in the same direction as Aa, and Bb in the opposite direction. Join ab, cutting AB in P ; join ab', and produce it to cut AB externally at Q. Then AB is divided internally at P and externally at Q, so that AP : PB = L : M, and AQ: QB = L: M. The proof follows at once from Euclid vi. 4. Obs. The solution is singular; that is, only one internal and one external point can be found that will divide the given straight line into segments which have the given ratio. 360 EUCLID'S ELEMENTS. DEFINITION. A finite straight line is said to be cut harmonically when it is divided internally and externally into segments which have the same ratio. A P B Thus AB is divided harmonically at P and Q, if AP : PB = AQ : QB. P and Q are said to be harmonic conjugates of A and B. If P and Q divide AB internally and externally in the same ratio, it is easy to shew that A and B divide PQ internally and externally in the same ratio : hence A and B are harmonic conjugates of P and Q. Example. The base of a triangle is divided harmonically by the internal and external bisectors of the vertical angle : for in each case the segments of the base are in the ratio of the other sides of the triangle. [Euclid vi. 3 and A.] Obs. We shall use the terms Arithmetic, Geometric, and Harmonic Means in their ordinary Algebraical sense. 1. If AB is divided internally at P and externally at Q in the same ratio, then AB is the harmonic mean between AQ and A P. For by hypothesis AQ .. , alternately, AQ that is, AQ QB=AP : PB; AP = QB : PB, AP=AQ-AB : AB-AP, which proves the proposition. 2. If AB is divided liarmonically at P and Q, and O is the middle point of AB; , then shall OP . OQ = OA 2 . A O P B Q For since AB is divided harmonically at P and Q, .-. AP : PB = AQ : QB; .-. AP-PB : AP+PB=AQ-QB : AQ + QB, or, 2OP : 2OA = 2OA : 2OQ; .-. OP.OQ=OA 3 . Conversely, it OP.OQ=OA ? , it may be shewn that AP : PB = AQ : QB; that is, that AB is divided harmonically at P and Q. THEOREMS AND EXAMPLES ON BOOK VI. 361 3. The Arithmetic, Geometric and Harmonic means of tico straight lines may be thus represented graphically. In the adjoining figure, tvro tan- gents AH, AK are drawn from any external point A to the circle PHQK ; HK is the chord of contact, and the st. line joining A to the centre O cuts the o ce at P and Q. Then (i) AO is the Arithmetic mean between AP and AQ : for clearly (ii) AH is the Geometric mean between AP and AQ: for AH 2 = AP . AQ. m. 36. (iii) AB is the Harmonic mean between AP and AQ: for OA.OB^OP 2 . Ex. 1, p. 233. .. AB is cut harmonically at P and Q. Ex. 1, p. 360. That is, AB is the Harmonic mean between AP and AQ. And from the similar triangles OAH, HAB, OA : AH = AH : AB, /. AO. AB = AH 2 ; vi. 17. /. the Geometric mean between two straight lines is the mean propor- tional between their Arithmetic and Harmonic means. 4. Given the base of a triangle and the ratio of the other sides, to find the locus of the vertex. Let BC be the given base, and let BAG be any triangle standing upon it, such that BA : AC the given ratio : it is required to find the locus of A. Bisect the L BAG internally and externally by AP, AQ. Then BC is divided internally at P, and externally at Q, so that BP : PC= BQ : QC = the given ratio; .'. P and Q are fixed points. And since AP, AQ are the internal and external bisectors of the /BAG, /. the / PAQ is a rt. angle; /. the locus of A is a circle described on PQ as diameter. EXEKCISE. Given three points B, P, C in a straight line: find the locus of points at winch BP ami PC subtend equal angles. 362 EUCLID'S ELEMENTS. DEFINITIONS. 1. A series of points in a straight line is called a range. If the range consists of four points, of which one pair are har- monic conjugates with respect to the other pair, it is said to be a harmonic range. 2. A series of straight lines drawn through a point is called a pencil. The point of concurrence is called the vertex of the pencil, and each of the straight lines is called a ray. A pencil of four rays drawn from any point to a harmonic range is said to be a harmonic pencil. 3. A straight line drawn to cut a system of lines is called a transversal. 4. A system of four straight lines, no three of which are concurrent, is called a complete quadrilateral. These straight lines will intersect two and two in six points, called the vertices of the quadrilateral ; the three straight lines which join opposite vertices are diagonals. THEOREMS ON HABMOXIC SECTION. 1. If a transversal is drawn parallel to one ray of a harmonic pencil, the other three rays intercept equal parts upon it: and con- versely. 2. Any transversal is cut harmonically by the rays of a harmonic pencil. 3. In a harmonic pencil, if one ray bisect the angle between the other pair of rays, it is perpendicular to its conjugate ray. Conversely if one pair of rays form a right angle, then they bisect internally and externally the angle between the other pair. 4. If A, P, B, Q. and a, p, b, q are harmonic ranges, one on each of two given straight lines, and if Aa, Pp, Bb, the straight lines which join three pairs of corresponding points, meet at S ; then will Qq also pass through S. 5. If two straight lines intersect at A, and if A, P, B, Q and A, p, b, q are two harmonic ranges one on each straight line (the points corresponding as indicated by the letters}, then Pp, Bb, Qq will be con- current : also Pq, Bb, Qp ivill be concurrent. 6. Use Theorem 5 to prove th-at in a complete quadrilateral in which the three diagonals are drawn, the straight line joining any pair of opposite vertices is cut harmonically by the other two diagonal's. THEOREMS AND EXAMPLES ON BOOK VI. 363 II. ON CENTRES OP SIMILARITY AND SIMILITUDE. 1. If any two unequal similar figures are placed so that their homologous sides are parallel, the lines joining corresponding points in the two figures meet in a point, whose distances from any two corre- sponding points are in the ratio of any pair of homologous sides. A' Let ABCD, A'B'C'D' be two similar figures, and let them be placed so that their homologous sides are parallel; namely, AB, BC, CD, DA parallel to A'B', B'C', C'D', D'A' respectively: then shall A A', BB', CC', DD' meet in a point, whose distances from any two corresponding points shall be in the ratio of any pair of homologous sides. Let A A' meet BB', produced if necessary, in S. Then because AB is par 1 to A'B'; Hyp. .'. the A 5 SAB, SA'B' are equiangular; .-. SA : SA' = AB : A'B'; vi. 4. .-. AA' divides BB', externally or internally, in the ratio of AB to A'B'. Similarly it may be shewn that CC' divides BB' in the ratio of BC to B'C'. But since the figures are similar, BC : B'C'=AB : A'B'; .-. AA' and CC' divide BB' in the same ratio; that is, AA', BB', CC' meet in the same point S. In like manner it may be proved that DD' meets CC' in the point S. .-. A A', BB', CC', DD' are concurrent, and each of these lines is divided at S in the ratio of a pair of homologous sides of the two figures. Q. E. D. COR. If any line is drawn through S meeting any pair of homolo- gous sides in K and K', the ratio SK : SK' is constant, and equal to the ratio of any pair of homologous sides. NOTE. It will be seen that the lines joining corresponding points are divided externally or internally at S according as the correspond- ing sides are drawn in the same or in opposite directions. In either case the point of concurrence S is called a centre of similarity of the two figures. 364 EUCLID'S ELEMENTS. 2. A common tangent STT' to two circles ivhose centres are C, C', meets the line of centres in S. If through S any straight line is drawn meeting these two circles in P, Q, and P', Q', respectively, then the radii CP, CO. shall be respectively parallel to C'P', C'Q'. Also the rectangles SQ . SP', SP . SQ' shall each be equal to the rectangle ST . ST'. Join CT, CP, CQ and C'T', C'P', C'Q'. Then since each of the / " CTS, C'T'S is a right angle, in. 18. .-. CT is par 1 to CT'; /. the A* SCT, SC'T' are equiangular; .-. SC : SC' = CT : C'T' = CP : C'P'; /. the A 8 SCP, SC'P' are similar; vi. 7. /. the /SCP = the L SC'P'; .-. CP is par 1 to C'P'. Similarly CQ is par 1 to C'Q'. Again, it easily follows that TP,' TQ are par 1 to T'P', T'Q' respectively; .-. the A 8 STP, ST'P' are similar. Now the rect. SP . SQ = the sq. on ST; in. 37. .-. SP : ST=ST : SQ, vi. 16. andSP : ST--.SP' : ST'; .-. ST : SQ = SP': ST'; .-. the rect. ST . ST'= SQ . SP'. In the same way it may be proved that the rect. SP . SQ'=the rect. ST . ST'. Q. E. I). COB. 1. It has been proved that SC : SC' = CP : C'P'; thus the external common tangents to the two circles meet at a point S which divides the line of centres externally in the ratio of the radii. Similarly it may be shewn that the transverse common tangents meet at a point S' which divides the line of centres internally in the ratio of the radii. COB. 2. CC'is divided harmonically at S and S'. DEFINITION. The points S and S' which divide externally and internally the line of centres of two circles in the ratio of their radii are called the external and internal centres of similitude respectively. THEOREMS AND EXAMPLES ON BOOK VI. 365 EXAMPLES. 1. Inscribe a square in a given triangle. 2. In a given triangle inscribe a triangle similar and similarly situated to a given triangle. 3. Inscribe a square in a given sector of circle, so that two angular points shall be on the arc of the sector and the other two on the bounding radii. 4. In the figure on page 278, if Dl meets the inscribed circle in X, shew that A, X, D l are collinear. Al. A is H 3. To prove that the locus of the intersection of tangents drawn to a circle at the extremities of all chords which pass through a given point is the polar of that point, Let A be the given point within the circle, of which O is the centre. Let H K be any chord passing through A; and let the tangents at H and K intersect at P : it is required to prove that the locus of P is the polar of the point A. I. To shew that P lies on the polar of A. Join OP cutting HK in Q. Join OA : and in OA produced take the point B, so that OA . OB = (radius) 2 , n. 14. Then since A is fixed, B is also fixed. Join PB. 24 368 EUCLID'S ELEMENTS. Then since H K is the chord of contact of tangents from P, /. OP . OQ= (radius) 2 . Ex. i. p. 233. But O A . O B = (radius) 2 ; Constr. :. OP.OQ=OA.OB : /. the four points A, B, P, Q are coneyclic. /. the L " at O. and B together = two rt. angles. in. 22. But the L at Q is a rt. angle; Constr. :. the L at B is a rt. angle. And since the point B is the inverse of A ; Constr. .-. PB is the polar of A; that is, the point P lies on the polar of A. II. To shew that any point on the polar of A satisfies the given conditions. Let BC he the polar of A, and let P be any point on it. Draw tangents PH, PK, and let HK be the chord of contact. Now from Ex. 1, p. 366, we know that the chord of contact HK is the polar of P, and we also know that the polar of P must pass through A ; for P is on BC, the polar of A : Ex. 2, p. 367. that is, HK passes through A. /. P is the point of intersection of tangents drawn at the ex- tremities of a chord passing through A. From I. and II. we conclude that the required locus is the polar of A. NOTE. If A is without the circle, the theorem demonstrated in Part I. of the above proof still holds good ; but the converse theorem in Part n. is not true for all points in BC. For if A is without the circle, the polar BC will intersect it; and no point on that part of the polar which is within the circle can be the point of intersection of tangents. We now see that (i) The Polar of an external point with respect to a circle is the chord of contact of tangents drawn from it. (ii) The Polar of an internal point is the locus of the intersections of tangents drawn at the extremities of all chords which pass through it. (iii) The Polar of a point on the circumference is the tangent at that point. THEOREMS AND EXAMPLES ON BOOK VI. 369 The following theorem is known as the Harmonic Property of Pole and Polar. 4. Any straight line drawn through a point is cut harmonically by the point, its polar, and the circumference of the circle. Let AHB be a circle, P the given point and HK its polar; let Paqb be any straight line drawn through P meeting the polar at q and the o ce of the circle at a and b : then shall P, a, q, b be a harmonic range. In the case here considered, P is an external point. Join P to the centre O, and let PO cut the o ce at A and B : let the polar of P cut the O M at H and K, and PO at Q. Then PH is a tangent to the AHB. Ex. 1, p. 366. From the similar triangles OPH, HPQ, OP : PH = PH : PQ, .-. PQ.PO=PH 2 = Pa . Pb. :. the points O, Q, a, b are concyclic: /. the /aQA = the / nbO = the L Oab i. 5. = the / OQ.b, in the same segment. And since QH is perp. to AB, .-. the /aQH = the L &QH. .'. Qg and QP are the internal and external bisectors of the / aQb : .'. P, a, q, b is & harmonic range. Ex. 1, p. 360. The student should investigate for himself the case when P is an internal point. Conversely, it may be shewn that if through a fixed point P any secant is drawn cutting the circumference of a given circle at a and b, and if q is the harmonic conjugate at P with respect to a, b; then the locus o/q is the polar of P with respect to the given circle. [For Examples on Pole and Polar, see p. 370.] DEFINITION. A triangle so related to a circle that each side is the polar of the opposite vertex is said to be self-conjugate with respect to the circle. 24-2 370 EUCLID'S ELEMENTS. EXAMPLES ON POLE AND POLAR. 1. The straight line which joins any two points is the polar with respect to a given circle of the point of intersection of their polars. 2. The point of intersection of any two straight lines is the pole of the straight line which joins their poles. 3. Find the locus of the poles of all straight lines which pass through a given point. 4. Find the locus of the poles, with respect to a given circle, of tan- gents drawn to a concentric circle. 5. If two circles cut one another orthogonally and PQ be any diameter of one of them; shew that the polar of P with regard to the other circle passes through QL. 6. If two circles cut one another orthogonally, the centre of each circle is the pole of their common chord with respect to the oilier circle. I. Any two points subtend at the centre of a circle an angle equal to one of the angles formed by the polars of the given points. 8. O is the centre of a given circle, and AB a fixed straight line. P is any point in AB; find the locus of tlie point inverse to P with respect to the circle. 9. Given a circle, and a fixed point O on its circumference: P is any point on the circle: find the locus of the point inverse to P with respect to any circle whose centre is O. 10. Given two points A and B, and a circle whose centre is O; shew that the rectangle contained by OA and the perpendicular from B on tJie polar of A is equal to the rectangle contained by OB and the perpendicular from A on the polar of B. II. Four points A, B, C, D are taken in order on the circumference of a circle; DA, CB intersect at P, AC, BD at Q. and BA, CD in R : shew that the triangle PQR is self-conjugate with respect to the circle. 12. Give a linear construction for finding the polar of a given point with respect to a given circle. Hence find a linear construction for drawing a tangent to a circle from an external point. 13. If a triangle is self-conjugate with respect to a circle, tJie centre of the circle is at the orthocentre of the triangle. 14. The polars, with respect to a given circle, of the four points of a harmonic range form a harmonic pencil : and conversely. THEOREMS AND EXAMPLES OX BOOK VI. 371 IV. ON THE RADICAL AXIS. 1. To find the locus of points from which the tangents drawn to two given circles are equal. Fig. 1. Fig. 2. Let A and B be the centres of the given circles, whose radii are a and b ; and let P be any point such that the tangent PQ drawn to the circle (A) is equal to the tangent PR drawn to the circle (B) : it is required to find the locus of P. Join PA, PB, AQ, BR, AB; and from P draw PS perp. to AB. Then because PQ=PR, .-. PQ 2 =PR 2 . But PQ 2 =PA 2 -AQ 2 ; and PR 2 = PB 2 -BR 2 : i. 47. .-. PA 2 -AQ 2 =PB 2 -BR- ; that is, PS 2 + AS 2 -a 2 =PS 2 + SB 2 -& 2 ; 1.47. or, AS 2 -a 2 =SB 2 -6 2 . Hence AB is divided at S, so that AS 2 - SB 2 = a 2 - fc 2 : .'. S is a, fixed point. Hence all points from which equal tangents can be drawn to the two circles lie on the straight line which cuts AB at rt. angles, so that the difference of the squares on the segments of AB is equal to the difference of the squares on the radii. Again, by simply retracing these steps, it may be shewn that in Fig. 1 every point in SP, and in Fig. 2 every point in SP exterior to the circles, is such that tangents drawn from it to the two circles are equal. Hence we conclude that in Fig. 1 the whole line SP is the required locus, and in Fig. 2 that part of SP which is without the circles. In either case SP is said to be the Radical Axis of the two circles. 372 EUCLID'S ELEMENTS. COBOLLABY. If the circles cut one another as in Fig. 2, it is clear that the Radical Axis is identical with the straight line which passes through the points of intersection of the circles; for it follows readily from in. 36 that tangents drawn to two intersecting circles from any point in the common chord produced are equal. 2. The Radical Axes of three circles taken in pairs are concurrent. Z, Let there be three circles whose centres are A, B, C. Let OZ be the radical axis of the s (A) and (B) ; and OY the Radical Axis of the & a (A) and (C), O being the point of their intersection : then shall the radical axis of the 0" (B) and (C) pass through O. It will be found that the point O is either without or within all the circles. I. When O is without the circles. From O draw OP, OQ, OR tangents to the s (A), (B), (C). Then because O is a point on the radical axis of (A) and (B) ; Hyp. .: OP = OQ. And because O is a point on the radical axis of (A) and (C), Hyp. .: OP = OR, .-. OQ=OR; .-. O is a point on the radical aids of (B) and (C), i.e. the radical axis of (B) and (C) passes through O. II. If the circles intersect ia such a way that O is within them all ; the radical axes are then the common chords of the three circles taken t\vo and two ; and it is required to prove that these common chords are concurrent. This may be shewn indirectly by m. 35. DEFINITION. The point of intersection of the radical axes of three circles taken in pairs is called the radical centre. THEOREMS AND EXAMPLES ON BOOK VI. 373 A. To draw the radical axis of two given circles. Let A and B be the centres of the given circles : it is required to draw their radical axis. If the given circles intersect, then the st. line drawn through their points of intersection will be the radical axis. [Ex. 1, Cor. p. 372.] But if the given circles do not intersect, describe any circle so as to cut them in E, F and G, H : Join EF and HG, and produce them to meet in P. Join AB; and from P draw PS perp. to AB. Then PS shall be the radical axis of the " (A), (B). DEFINITION. If each pair of circles in a given system have the same radical axis, the circles are said to be co-axal. EXAMPLES. 1. Shew that the radical axis of two circles bisects any one of their common tangents. 2. If tangents are drawn to two circles from any point on their radical axis; shew that a circle described ivith this point as centre and any one of the tangents as radius, cuts both the given circles ortho- gonally. 3. O is the radical centre of three circles, and from O a tangent OT is drawn to any one. of them: shew that a circle whose centre is O and radius OT cuts all the given circles orthogonally. 4. If three circles touch one another, taken two and two, shew that their common tangents at the points of contact are concurrent. 374 EUCLID'S ELEMENTS. 5. If circles are described on the three sides of a triangle ax diameter, their radical centre is the orthocentre of the triangle. 6. All circles which pass through a fixed point and cut a given circle orthogonally, pass through a second fixed point. 7. Find the locus of the centres of all circles ivhich pass through a given point and cut a given circle orthogonally. 8. Describe a circle to pass through two given points and cut a given circle orthogonally. 9. Find the locus of the centres of all circles which cut tico given circles orthogonally. 10. Describe a circle to pass through a given point and cut two given circles orthogonally. 11. The difference of the squares on the tangents drawn from any point to two circles is equal to twice the rectangle contained by the straight line joining their centres and the perpendicular from the given point on their radical axis. 12. In a system of co-axal circles which do not intersect, any point is taken on the radical axis ; shew that a circle described from this point as centre with radius equal to the tangent drawn from it to any one of the circles, will meet the line of centres in two fixed points. [These fixed points are called the Limiting Points of the system.] 13. In a system of co-axal circles the tico limiting points and the points in which any one circle of the system cuts the line of centres form a harmonic range. 14. In a system of co-axal circles a limiting point has the same polar with regard to all the circles of the system. 15. If two circles are orthogonal any diameter of one is cut harmonically by the other. Obs. In the two following theorems we are to suppose that the segments of straight lines are expressed numerically in terms of some common unit ; and the ratio of one such segment to another will be denoted by the fraction of which the first is the numerator and the second the denominator. THEOREMS AND EXAMPLES OX BOOK VI. 375 V. ON TRANSVERSALS. DEFINITION. A straight line drawn to cut a given system of lines is called a transversal. 1. If three concurrent straight lines are drawn from the angular po'nts of a triangle to meet the opposite sides, then the product of three alternate segments taken in -order is equal to the product of the other three segments. B B Let AD, BE, CF be drawn from the vertices of the A ABC to intersect at O, and cut the opposite sides at D, E, F: then shall BD . CE . AF = DC . EA . FB. By similar triangles it may be shewn that BD : DC = the alt. of AAOB : the alt. of A AOC; BD_ AAOB A DC~ A AOC' similarly, and CE EA AF FB A BOG ABOA ; ACOA or, ACOB* Multiplying these ratios, we have BD CE AF DC ' EA ' FB : BD.CE.AF=DC. EA . FB. C A ' CO ' Q. . D. The converse of this theorem, which may be proved indirectly, is very important : it may be enunciated thus : // f ,hree straight lines draion from the vertices of a triangle cut the opposite sides so that the product of three alternate segments taken in order is equal to the product of the other three, then the three straight lines are concurrent. That is, if BD . CE . AF= DC . EA . FB, then AD, BE, CF are concurrent. 376 EUCLID'S ELEMENTS. 2. If a transversal is drawn to cut the sides, or the sides produced, of a triangle, the product of three alternate segments taken in order is equal to the product of the other three segments. B C D\ B C D\ Let ABC be a triangle, and let a transversal meet the sides BC, CA, AB, or these sides produced, at D, E, F: then shall BD . CE . AF = DC . EA . FB. Draw AH par 1 to BC, meeting the transversal at H. Then from the similar A S DBF, HAF, BD_HA FB ~ AK : and from the similar A S DCE, HAE, CE_EA DC~HA : BD CE EA .-. , by multiplication, ^ . = ; thai BD.CE.AF DCTEATFB- 1 ' or, BD.CE.AF=DC.EA.FB. Q.E. D. NOTE. In this theorem the transversal must either meet two sides and the third side produced, as in Fig. 1 ; or all three sides pro- duced, as in Fig. 2. The converse of this Theorem may be proved indirectly: If three points are taken in two sides of a triangle and the third side produced, or in all three sides produced, so that the product of three alternate segments taken in order is equal to the product of tlie other three segments, the three points are collinear. The propositions given on pages 103 106 relating to the concur- rence of straight lines in a triangle, may be proved by the method of transversals, and in addition to these the following important theorems may be established. MISCELLANEOUS EXAMPLES ON BOOK VI. 377 DEFINITIONS. (i) If two triangles are such that three straight lines joining corresponding vertices are concurrent, they are said to be co- polar. (ii) If two triangles are such that the points of intersection of corresponding sides are collinear, they are said to be co-axiaL THEOBEMS TO BE PEOVED BY TRANRVF.BBAT.B. 1. The straight lines which join the vertices of a triangle to the points of contact of the inscribed circle (or any of the three escribed circles) are concurrent. 2. The middle points of the diagonals of a complete quadrilateral are collinear. 3. Co-polar triangles are also co-axial; and conversely co-axial triangles are also co-polar. 4. The six centres of similitude of three circles lie three ly three on four straight lines. MISCELLANEOUS EXAMPLES ON BOOK VI. 1. Through D, any point in the base of a triangle ABC, straight lines DE, DF are drawn parallel to the sides AB, AC, and meeting the sides at E, F: shew that the triangle AEF is a mean proportional between the triangles FBD, EDC. 2. If two triangles have one angle of the one equal to one angle of the other, and a second angle of the one supplementary to a second angle of the other, then the sides about the third angles are proportional. 3. AE bisects the vertical angle of the triangle ABC and meets the base in E ; shew that if circles are described about the triangles ABE, ACE, the diameters of these circles are to each other in the same ratio as the segments of the base. 4. Through a fixed point O draw a straight line so that the parts intercepted between O and the perpendiculars drawn to the straight line from two other fixed points may have a given ratio. 378 EUCLID'S ELEMENTS. 5. The angle A of a triangle ABC is bisected by AD meeting BC in D, and AX is the median bisecting BC: shew that XD has the same ratio to X B as the difference of the sides has to their sum. 6. AD and AE bisect the vertical angle of a triangle internally and externally, meeting the base in D and E; shew that if O is the middle point of BC, then OB is a mean proportional between OD and OE. 7. P and Gl are fixed points; AB and CD are fixed parallel straight lines; any straight line is drawn from P to meet AB at M, and a straight line is drawn from Gt parallel to PM meeting CD at N : shew that the ratio of PM to GIN is constant, and thence shew that the straight line through M and N passes through a fixed point. 8. C is the middle point of an arc of a circle whose chord is AB ; D is any point in the conjugate arc : shew that AD + DB : DC :: AB : AC. 9. In the triangle ABC the side AC is double of BC. If CD, CE bisect the angle ACB internally and externally meeting AB in D and E, shew that the areas of the triangles CBD, ACD, ABC, CDE are as 1, 2, 3, 4. 10. AB, AC are two chords of a circle; a line parallel to the tangent at A cuts AB, AC in D and E respectively: shew that the rectangle AB, AD is equal to the rectangle AC, AE. 11. If from any point on the hypotenuse of a right-angled triangle perpendiculars are drawn to the two sides, the rectangle contained by the segments of the hypotenuse will be equal to the sum of the rectangles contained by the segments of the sides. 12. D is a point in the side AC of the triangle ABC, and E is a point in AB. If BD, CE divide each other into parts in the ratio -i : 1, then D, E divide CA, BA in the ratio 3 : 1. 13. If the perpendiculars from two fixed points on a straight line passing between them be in a given ratio, the straight line must pass through a third fixed point. 1-4. PA, PB are two tangents to a circle; PCD any chord through P : shew that the rectangle contained by one pair of opposite sides of the quadrilateral ACBD is equal to the rectangle contained by the other pair. 15. A, B, C are any three points on a circle, and the tangent at A meets BC produced in D : shew that the diameters of the circles circumscribed about ABD, ACD are as AD to CD. MISCELLANEOUS EXAMPLES ON BOOK VI. 379 16. AB, CD are two diameters of the circle ADBC at right angles to each other, and EF is any chord; CE, CF are drawn meeting AB produced in G and H : prove that the rect. CE, HG = the rect. EF, CH. 17. From the vertex A of any triangle ABC draw a line meeting BC produced in D so that AD may be a mean proportional between the segments of the base. 18. Two circles touch internally at O; AB a chord of the larger circle touches the smaller in C which is cut by the lines OA, OB in the points P, Q: shew that OP : OQ : : AC : CB. 19. AB is any chord of a circle; AC, BC are drawn to any point C in the circumference and meet the diameter perpendicular to AB at D, E: if O be the centre, shew that the rect. OD, OE is equal to the square on the radius. 20. YD is a tangent to a circle drawn from a point Y in the diameter AB produced; from D a perpendicular DX is drawn to the d'.ameter: shew that the points X, Y divide AB internally and ex- ternally in the same ratio. 21. Determine a point in the circumference of a circle, from which lines drawn to two other given points shall have a given ratio. 22. O is the centre and OA a radius of a given circle, and V is a fixed point in OA ; P and Q are two points on the circum- ference on opposite sides of A and equidistant from it; QV is pro- duced to meet the circle in L : shew that, whatever be the length of the arc PQ, the chord LP will always meet OA produced in a fixed point. 23. EA, EA' are diameters of two circles touching each other externally at E ; a chord AB of the former circle, when produced, touches the latter at C', while a chord A'B' of the latter touches the former at C : prove that the rectangle, contained by AB and A'B', is four times as great as that contained by BC' and B'C. 24. If a circle be described touching externally two given circles, the straight line passing through the points of contact will intersect the line of centres of the given circles at a fixed point. 25. Two circles touch externally in C ; if any point D be taken without them so that the radii AC, BC subtend equal angles at D, and DE, DF be tangents to the circles, shew that DC is a mean proportional between DE and DF. 380 EUCLID'S ELEMENTS. 26. If through the middle point of the base of a triangle any line be drawn intersecting one side of the triangle, the other produced, and the line drawn parallel to the base from the vertex, it will be divided harmonically. 27. If from either base angle of a triangle a line be drawn intersecting the median from the vertex, the opposite side, and the line drawn parallel to the base from the vertex, it will be divided harmonically. 28. Any straight line drawn to cut the arms of an angle and its internal and external bisectors is cut harmonically. 29. P, Q. are harmonic conjugates of A and B, and C is an external point : if the angle PCQ is a right angle, shew that CP, CQ are the internal and external bisectors of the angle ACB. 30. From C, one of the base angles of a triangle, draw a straight line meeting AB in G, and a straight line through A parallel to the base in E, so that CE may be to EG in a given ratio. 31. P is a given point outside the angle formed by two given lines AB, AC: shew how to draw a straight line from P such that the parts of it intercepted between P and the lines AB, AC may have a given ratio. 32. Through a given point within a given circle, draw a straight line such that the parts of it intercepted between that point and the circumference may have a given ratio. How many solutions does the problem admit of? 33. If a common tangent be drawn to any number of circles which touch each other internally, and from any point of this tangent as a centre a circle be described, cutting the other circles ; and if from this centre lines be drawn through the intersections of the circles, the segments of the lines within each circle shall be equal. 34. APB is a quadrant of a circle, SPT a line touching it at P; C is the centre, and PM is perpendicular to CA: prove that the A SCT : the A ACB : : the A ACB : the A CMP. 35. ABC is a triangle inscribed in a circle, AD, AE are lines drawn to the base BC parallel to the tangents at B, C respectively shew that AD = AE, and BD : CE : : AB 2 : AC 2 . 36. AB is the diameter of a circle, E the middle point of the radius OB ; on AE, EB as diameters circles are described ; FQL is a common tangent touching the circles at P and Q, and AB produced at L : shew that BL is equal to the radius of the smaller circle. MISCELLANEOUS EXAMPLES ON BOOK VI. 381 37. The vertical angle C of a triangle is bisected by a straight line which meets the base at D, and is produced to a point E, such that the rectangle contained by CD and CE is equal to the rectangle contained by AC and CB: shew that if the base and vertical angle be given, the position of E is invariable. 38. ABC is an isosceles triangle having the base angles at B and C each double of the vertical angle: if BE and CD bisect the base angles and meet the opposite sides in E and D, shew that DE divides the triangle into figures whose ratio is equal to that of AB to BC. 39. If AB, the diameter of a semicircle, be bisected in C and on AC and CB circles be described, and in the space between the three circumferences a circle be inscribed, shew that its diameter will be to that of the equal circles in the ratio of two to three. 40. O is the centre of a circle inscribed in a quadrilateral A BCD ; a line EOF is drawn and making equal angles with AD and BC, and meeting them in E and F respectively : shew that the triangles AEO, BOF are similar, and that AE: ED = CF : FB. 41. From the last exercise deduce the following: The inscribed circle of a triangle ABC touches AB in F; XOY is drawn through the centre making equal angles with AB and AC, and meeting them in X and Y respectively: shew that BX : XF = AY : YC. 42. Inscribe a square in a given semicircle. 43. Inscribe a square in a given segment of a circle. 44. Describe an equilateral triangle equal to a given isosceles triangle. 45. Describe a square having given the difference between a diagonal and a side. 46. Given the vertical angle, the ratio of the sides containing it, and the diameter of the circumscribing circle, construct the triangle. 47. Given the vertical angle, the line bisecting the base, and the angle the bisector makes with the base, construct the triangle. 48. In a given circle inscribe a triangle so that two sides may pass through two given points and the third side be parallel to a given straight line. 49. In a given circle inscribe a triangle so that the sides may pass through three given points. 382 EUCLID'S ELEMENTS. 50. A, B, X, Y are four points in a straight line, and O is such a point in it that the rectangle OA, OY is equal to the rectangle OB, OX: if a circle be described with centre O and radius equal to a mean proportional between OA and OY, shew that at every point on this circle AB and XY will subtend equal angles. 51. O is a fixed point, and OP is any line drawn to meet a fixed straight line in P; if on OP a point Q is taken so that OQ to OP is a constant ratio, find the locus of Q. 52. O is a fixed point, and OP is any line drawn to meet the circumference of a fixed circle in P; if on OP a point Q is taken so that OQto OP is a constant ratio, find the locus of Q. 53. If from a given point two straight lines are drawn including a given angle, and having a fixed ratio, find the locus of the extremity of one of them when the extremity of the other lies on a fixed straight line. 54. On a straight line PAB, two points A and B are marked and the line PAB is made to revolve round the fixed extremity P. C is a fixed point in the plane in which PAB revolves; prove that if CA and CB be joined and the parallelogram CADB be completed, the locus of D will be a circle. 55. Find the locus of a point whose distances from two fixed points are in a given ratio. 56. Find the locus of a point from which two given circles sub- tend the same angle. 57. Find the locus of a point such that its distances from two intersecting straight lines are in a given ratio. 58. In the figure on page 364, shew that QT, P'T' meet on the radical axis of the two circles. 59. ABC is any triangle, and on its sides equilateral triangles are described externally : if X, Y, Z are the centres of their inscribed circles, shew that the triangle XYZ is equilateral. 60. If S, I are the centres, and R, r the radii of the circumscribed and inscribed circles of a triangle, and if N is the centre of its nine- points circle, prove that (i) SI 2 = R 2 -2Rr, (ii) NI=R-r. Establish corresponding properties for the escribed circles, and hence prove that the nine-points circle touches the inscribed and escribed circles of a triangle. SOLID GEOMETRY. EUCLID. BOOK XI. DEFINITIONS. FROM the Definitions of Book I. it will be remembered that (i) A line is that which has length, without breadth or thickness. (ii) A surface is that which has length and breadth, without thickness. To these definitions we have now to add : (iii) Space is that which has length, breadth, and thickness. Thus a line is said to be of one dimension ; a surface is said to be of two dimensions ; and space is said to be of three dimensions. The Propositions of Euclid's Eleventh Book here given establish the first principles of the geometry of space, or solid geometry. They deal with the properties of straight lines which are not all in the same plane, the relations which straight lines bear to planes which do not contain those lines, and the relations which two or more planes bear to one another. Unless the contrary is stated the straight lines are supposed to be of indefinite length, and the planes of infinite extent. Solid geometry then proceeds to discuss the properties of solid figures, of surfaces which are not planes, and of lines which can not be drawn on a plane surface. H. E. 25 384 EUCLID'S ELEMENTS. LINES AND PLANES. 1. A straight line is perpendicular to a plane when it is perpendicular to every straight line which meets it in that plane. NOTE. It will be proved in Proposition 4 that if a straight line is perpendicular to two straight lines which meet it in a plane, it is also perpendicular to every straight line which meets it in that plane. A straight line drawn perpendicular to a plane is said to be a normal to that plane. 2. The foot of the perpendicular let fall from a given point on a plane is called the projection of that point on the plane. 3. The projection of aline on a plane is the locus of the feet of perpendiculars drawn from all points in the given line to the plane. i. ^Xl Thus in the above figure the line ab is the projection of the line AB on the plane PQ. It will be proved hereafter (see page 420) that the projection of a straight line on a plane is also a straight line. DEFINITIONS. 385 4. The inclination of a straight line to a plane is the acute angle contained by that line and another drawn from the point at which the first line meets the. plane to the point at which a perpendicular to the plane let fall from any point of the first line meets the plane. Thus in the above figure, if from any point X in the given straight line AB, which intersects the plane PQ at A, a perpen- dicular Xx is let fall on the plane, and the straight line Ax6 is drawn from A through x, then the inclination of the straight line AB to the plane PQ is measured by the acute angle BA&. In other words : The inclination of a straight line to a plane is the acute angle contained by the given straight line and Us* projection on the plane. AXIOM. If two surfaces intersect one another, they meet in a line or lines. 5. The common section of two intersecting surfaces is the line (or lines) in which they meet. NOTE. It is proved in Proposition 3 that the common section of two planes is a straight line. Thus AB, the common section of the two planes PQ, XY is proved to be a straight line. 25-2 386 EUCLID'S ELEMENTS. 6. One plane is perpendicular to another plane when any straight line drawn in one of the planes perpendicular to the common section is also perpendicular to the other plane. Thus in the adjoining figure, the plane EB is perpendicular to the plane CD, if any straight line PQ, drawn in the plane EB at right angles to the common section AB, is also at right angles to the plane CD. 7. The inclination of a plane to a plane is the acute angle contained by two straight lines drawn from any point in the common section at right angles to it, one in one plane and one in the other. Thus in the adjoining figure, the straight line AB is the com- mon section of the two inter- secting planes BC, AD; and from Q, any point in AB, two straight lines QP, QR are drawn perpendicular to AB, one in each plane: then the inclination of the two planes is measured by the acute angle PQR. NOTE. This definition assumes that the angle PQR is of constant magnitude whatever point Q is taken in AB : the truth of which assumption is proved in Proposition 10. The angle formed by the intersection of two planes is called a dihedral angle. It may be proved that two planes are perpendicular to one another when the dihedral angle formed by them is a right angle. DEFINITIONS. 387 8. Parallel planes are such as do not meet when pro- duced. 9. A straight line is parallel to a plane if it does not meet the plane when produced. 10. The angle between two straight lines which do not meet is the angle contained by two intersecting straight lines respectively parallel to the two non-intersecting lines. Thus if AB and CD are two straight lines which do not meet, and ab, be are two intersecting lines parallel respectively to AB and CD ; then the angle between AB and CD is measured by the angle abc. 11. A solid angle is that which is made by three or more plane angles which have a common vertex, but are not in the same plane. A solid angle made by three plane angles is said to be trihedral ; if made by more than three, it is said to be polyhedral. A solid angle is sometimes called a corner. 12. A solid figure is any portion of space bounded by one or more surfaces, plane or curved. These surfaces are called the faces of the solid, and the inter- sections of adjacent faces are called edges. 388 EUCLID'S ELEMENTS. POLYHEDRA. 13. A polyhedron is a solid figure bounded by plane faces. Obs. A. plane rectilineal figure must at least have three sides ; or four, if two of the sides are parallel. A polyhedron must at least have four faces ; or, if two faces are parallel, it must at least have five faces. 14. A prism is a solid figure bounded by plane faces, of which two that are opposite are similar and equal polygons in parallel planes, and the other faces are paralle- lograms. The polygons are called the ends of the prism. A prism is said to be right if the edges formed by each pair of adjacent parallel- ograms are perpendicular to the two ends ; if otherwise the prism is oblique. 15. A parallelepiped is a solid figure bounded by three pairs of parallel plane faces. Fig. 1. Fig. 2. A parallelepiped may be rectangular as in fi<: fig. 2. 1, or oblique as in DEFINITIONS. 389 16. A pyramid is a solid figure bounded by plane faces, of which one is a polygon, and the rest are triangles having as bases the sides of the polygon, and as a com- mon vertex some point not in the plane of the polygon. The polygon is called the base of the pyramid. A pyramid having for its base a regular polygon is said to be right when the vertex lies in tb.3 straight line drawn perpendicular to the base from its central point (the centre of its inscribed or cir- cumscribed circle). 17. A tetrahedron is a pyramid on a triangular base : it is thus con- tained by four triangular faces. 18. Polyhedra are classified according to the number of their faces ; thus a hexahedron has six faces ; an octahedron, has eight faces ; a dodecahedron has twelve faces. 19. Similar polyhedra are such as have all their solid angles equal, each to each, and are bounded by the same number of similar faces. 20. A Polyhedron is regular when its faces are similar and equal regular polygons. 390 EUCLID'S ELEMENTS. 21. It will be proved (see page 425) that there can only be five regular polyhedra. They are defined as follows. (i) A regular tetrahedron is a solid figure bounded by four plane faces, which are equal and equi- lateral triangles. (ii) A cube is a solid figure bounded by six plane faces, which are equal squares. (iii) A regular octahedron is a solid figure bounded by eight plane faces, which are equal and equilateral triangles. (iv) A regular dodecahedron is a solid figure bounded by twelve plane faces, which are equal and regular pentagons. DEFINITIONS. 391 (v) A regular icosahedron is a solid figure bounded by twenty plane faces, which are equal and equilateral triangles. SOLIDS OF REVOLUTION. 22. A sphere is a solid figure described by the revo- lution of a semicircle about its diameter, which remains fixed. The axis of the sphere is the fixed straight line about which the semicircle revolves. The centre of the sphere is the same as the centre of the semi- circle. A diameter of a sphere is any straight line which passes through the centre, and is terminated both ways by the surface of the sphere. 23. A right cylinder is a solid figure described by the revolution of a rectangle about one of its sides which remains fixed. The axis of the cylinder is the fixed straight line about which the rectangle revolves. The bases, or ends of the cylinder are the circular faces described by the two revolving opposite sides of the rectangle. 392 EUCLID'S ELEMENTS. 24. A right cone is a solid figure described by the revolution of a right- angled triangle about one of the sides containing the right angle which re- mains fixed. The axis of the cone is the fixed straight line about which the triangle revolves. The base of the cone is the circular face described by that side which revolves. The hypotenuse of the right-angled triangle in any one of its positions is called a generating line of the cone. 25. Similar cones and cylinders are those which have their axes and the diameters of their bases proportionals. BOOK XI. PROP. 1. 393 PROPOSITIOX 1. THEOREM. One part of a straight line cannot be in a plane and an- other part outside it. Q If possible;, let AB, part of the st. line ABC, be in the plane PQ, and the part BC without it. Then since the st. line AB is in the plane PQ, .*. it can be produced in that plane, i. Pest. 2. Produce AB to D; and let any other plane which passes through AD be turned about AD until it passes also through C. Then because the points B and C are in this plane, .'. the st. line BC is in it: I. Def. 5. .'. APC and ABD are in the same plane and are both st. lines ; which is impossible. i . Def. 3. .'. the st. line ABC has not one part AB in the plane PQ, and another part BC outside it. Q. E. D. NOTE. This proposition scarcely needs proof, for the truth of it follows almost immediately from the definitions of a straight line, and a plane. It should be observed that the method of proof used in this and the next proposition rests upon the following axiom. If a plane of unlimited extent turns about a fixed strait/lit line as an axis, it can be made to pass through any point in space. 394 EUCLID'S ELEMENTS. PROPOSITION 2. THEOREM. Any two straight lines which cut one another are in one plane: and any three straight lines, of which each pair inter- sect one another, are in one plane. B Let the two st. lines AB and CD intersect at E; and let the st. line BC be drawn cutting AB and CD at B and C: then (i) AB and CD shall lie in one plane, (ii) AB, BC, CD shall lie in one plane. (i) Let any plane pass through AB ; and let this plane be turned about AB until it passes through C. Then, since C and E are points in this plane, .'. the whole st. line CED is in it. I. Def. 5 and xi. 1. That is, AB and CD lie in one plane. (ii) And since B and C are points in the plane which contains AB and CD, .'. also the st. line BC lies in this plane. Q. E. D. COROLLARY. One, and only one, plane can be made to j)ass through two given intersecting straight lines. Hence the position of a plane is fixed, (i) if it passes through a given straight line and a given point outside it; Ax. p. 393_ (ii) if it passes through two intersecting straight lines ; xi. 2. (iii) if it passes through three points not collinear ; xi. 2. 'iv) if it passes through two parallel straight lines. i. Def. 25. BOOK XI. PROP. 3. 395 PROPOSITION 3. THEOREM. If two planes cut one another their common section is a line. Let the two planes XA, CY cut one another, and let BD be their common section: then shall BD be a st. line. For if not, from B to D in the plane XA draw the st. line BED; and in the plane CY draw the st. line BFD. Then the st. lines BED, BFD have the same extremities; .'. they include a space; but this is impossible. .'. the common section BD cannot be otherwise than a st. line. Q. E. D. Or, more briefly thus Let the planes XA, CY cut one another, and let B and D be two points in their common section. Then because B and D are two points in the plane XA, .'. the st. line joining B, D lies in that plane. I. Def. 5. And because B and D are two points in the plane CY, .'. the st. line joining B, D lies in that plane. Hence the st. line BD lies in both planes, and is therefore their common section. That is, the common section of the two planes is a straight line. Q. E. D. 396 EUCLID'S ELEMENTS. PROPOSITION 4. THEOREM. [Alternative Proof.] If a straight line is perpendicular to each of two straight lines at their point of intersection, it shall also be perpen- dicular to the plane in which they lie. and Let the straight line AD be perp. to each of the st. lines AB, AC at A their point of intersection: then shall AD be perp. to the plane in which AB AC lie. Produce DA to F, making AF equal to DA. Draw any st. line BC in the plane of AB, AC, to cut AB, AC at B and C; and in the same plane draw through A any st. line AE to cut BC at E. It is required to prove that AD is perp. to AE. Join DB, DE, DC; and FB, FE, FC. Then in the A s BAD, BAF, because DA = FA, and the common side AB is perp. to DA, FA .'. BD= BF. Similarly CD = CF. Now if the A BFC be turned about its base BC until the vertex F comes into the plane of the A BDC, then F will coincide with D, since the conterminous sides of the triangles are equal. I. 7. .'. EF will coincide with ED, that is, EF = ED. Const 1 ) I. 4. BOOK XI. PROP. 4. 397 Hence in the A S DAE, FAE, since DA, AE, ED = FA, AE, EF respectively. .*. the L DAE = the /_ FAE. I. 8. That is, DA is perp. to AE. Similarly it may be shewn that DA is perp, to every st. line which meets it in the plane of AB, AC : .'. DA is perp. to this plane. Q.E.D. PROPOSITION 4. THEOREM. [Euclid's Proof.] If a straight line is perpendicular to each of two straight lines at their point of intersection, it shall also be perpen- dicular to the plane in ivhich they lie. Let the st. line EF be perp. to each of the st. lines AB, DC at E their point of intersection : then shall EF be also perp. to the plane XY, in which AB and DC lie. Make EA, EC, EB, ED all equal, and join AD, BC. Through E in the plane XY draw any st. line cutting AD and BC in G and H. Take any pt. F in EF, and join FA, FG, FD, FB, FH, FC. Then in the A s A ED, EEC, because AE, ED = BE, EC respectively, Constr. and the z.AED = the L BEC ; 1.15. /. AD=BC, and the L DAE = the L CBE. 1.4. 398 EUCLID'S ELEMENTS. F In the A s AEG, BEH, because the L GAE = the L. HBE, Proved. and the _ AEG = the L. BEH, i. 15. and EA = EB; Constr. :. EG = EH, and AG = BH. i. 26. Again in the A s FEA, FEB, because EA = EB, and the common side FE is perp. to EA, EB: Hyp. .'. FA=FB. i. 4. Similarly FC = FD. Again in the A 3 DAF, CBF, because DA, AF, FD = CB, BF, FC, respectively, .'. the L DAF - the /.CBF. i. S. And in the A s FAG, FBH, because FA, AG, = FB, BH, respectively, and the L. FAG =the L. FBH, Proved. .'. FG = FH. i. 4. Lastly in the A S FEG, FEH, because FE, EG, GF=FE, EH, H F, respectively, .'. the L. FEG = the : ' #i#2 i g P ar ' to AB. And Af/j, B//., must intersect one another, since they are both in the plane of the "A AXB : let them intersect at the point G. Then by similar A S , AG : G'/j = AB //jj/., = AX Xft = 3 1. /. Bg 2 cuts Ar/j at a point G whose distance fromy cutting a prism by parallel planes are equal. 24. Three straight lines in space OA, OB, OC, are mutually at right angles, and their lengths are a, b, c : express the area of the triangle ABC in its simplest form. 25. Find the diagonal of a regular octahedron in terms of one of its edges. 26. Shew how to cut a cube by a plane so that the lines of section may form a regular hexagon. 27. Shew that every section of sphere by a plane is a circle. 28. Find in terms of the length of an edge the radius of a sphere inscribed in a regular tetrahedron. 29. Find the locus of points in a given plane at which a straight line of fixed length and position subtends a right angle. 30. A fixed point O is joined to any point P in a given plane which does not contain O; on OP a point Q is taken such that the rectangle OP, OQ is constant: shew that Q lies on a fixed sphere. JUL 78 2 1 1973 WAY 101W OCT 17,98? ANNEX University of California SOUTHERN REGIONAL LIBRARY FACILITY 305 De Neve Drive - Parking Lot 17 Box 951388 LOS ANGELES, CALIFORNIA 90095-1388 Return this material to the library from which it was borrowed. MAY b 1 SEL/EMS