3^ IN MEMORIAM FLORIAN CAJORI EUCLID'S ELEMENTS OF GEOMETRY BOOKS I— VI. SonDon: C j. clay and SONS, CAMBKIDGE UNIVEKSITY PEESS WAREHOUSE, AVE MARIA LANE. CAMBRIDGE: DEIGHTON, BELL, AND CO. LEIPZIG: F. A. BROCKHAUS. NEW YORK : MACMILLAN AND CO. SPECIMEN COZY, ^itt iress iMatftematiral ^mt& EUCLID'S ELEMENTS OF GEOMETKY EDITED FOR THE SYNDICS OF THE PRESS H. M. TAYLOR, M.A. FELLOW AND FORMERLY TUTOR OF TRINITY COLLEGE, CAMBRIDGE. BOOKS I— VI. CAMBRIDGE AT THE UNIVEESITY PRESS 1893 [All Rifihts I'eserved.] Cambritifie : PBINTED BY C. J. CLAY, M.A. AND SONS, AT THE UNIVERSITY PRESS. \ r3 NOTE. The Special Board for Mathematics in the University of Cambridge in a Eeport on Geometrical Teaching dated May 10, 1887, state as follows: ' The majority of the Board are of opinion that the rigid adherence to Euclid's texts is prejudicial to the interests of education, and that greater freedom in the method of teaching Geometry is desirable. As it appears that this greater freedom cannot be attained while a knowledge of Euclid's text is insisted upon in the examinations of the University, they consider that such alterations should be made in the regulations of the examina- tions as to admit other proofs besides those of Euclid, while following however his general sequence of propositions, so that no proof of any proposition occurring in Euclid should be accepted in which a subsequent proposition in Euclid's order is assumed.' On March 8, 1888, Amended Regulations for the Previous Examination, which contained the following provision, were approved by the Senate : ' Euclid's definitions will be required, and no axioms or postulates except Euclid's may be assumed. The actual proofs of propo- sitions as given in Euclid will not be required, but no proof of any proposition occurring in Euclid will be admitted in which use is made of any proposition which in Euclid's order occurs subsequently.' And in the Regulations for the Local Examinations conducted by the University of Cambridge it is provided that : •Proofs other than Euclid's will be admitted, but Euclid's Axioms will be required, and no proof of any proposition will be accepted which assumes anything not proved in preceding propositions in Euclid.' PREFACE TO BOOKS I. AND II. IT was with extreme diffidence that I accepted an invi- tation from the Syndics of the Cambridge University Press to undertake for them a new edition of the Elements of Euclid. Though I was deeply sensible of the honour, which the invitation conferred, I could not but recognise the great responsibility, which the acceptance of it would entail. The invitation of the Syndics was in itself, to my mind, a sign of a widely felt conviction that the editions in common use were capable of improvement. Now improve- ment necessitates change, and every change made in a work, which has been a text book for centuries, must run the gauntlet of severe criticism, for while some will view every alteration with aversion, others will consider that every change demands an apology for the absence of more and greater changes. I will here give a short account of the chief points, in which this edition differs from the best known editions of the Elements of Euclid at present in use in England. While the texts of the editions of Potts and Todhunter are confessedly little more than reprints of Simson's English version of the Elements published in 175G, the text of the present edition does not profess to be a translation from the Greek. I began by retranslating the First Book : but there proved to be so many points, in which I thought it M3061G5 * vi PREFACE. desirable to depart from the original, tiiat it seemed best to give up all idea of simple translation and to retain merely the substance of the work, following closely Euclid's sequence of Propositions in Books I. and II. at all events. Some of the definitions of Euclid, for instance trapezium^ rhomboid, gnomon are omitted altogether as unnecessary. The word trapezium is defined in the Greek to mean " ant/ four sided figure other than those already defined,^' but in many modern works it is defined to be "« quadrilateral, which has one pair of parallel sides." The first of these definitions is obsolete, the second is not universally ac- cepted. On the other hand definitions are added of several words in general use, such as ^^eWme^er, parallelogram, diagonal, which do not occur in Euclid's list. The chief alteration in the definitions is in that of the word figure, which is in the Greek text defined to be " that ivhich is enclosed hy one or more boundaries." I have preferred to define a figure as "a combination of points, lines and surfaces." That Euclid's definition leads to diffi- culty is seen from the fact that, though Euclid defines a circle as "a figure contained by one line...", he demands in his postulate that "a circle may be described...". Now it is the circumference of a circle which is described and not the surface. Again, when two circles intersect, it is the circumferences which intersect and not the surfaces. I have rejected the ordinarily received definition of a square as " a quadrilateral, whose sides are equal, and whose angles are right angles." There is no doubt that, when we define any geometrical figure, we postulate the possibility of the figure ; but it is useless to embrace in the definition more properties than are requisite to determine the figure. The word axiom is used in many modern works as applicable both to simple geometrical propositions, such as " two straight lines cannot enclose a space," and to proposi- PREFACE. vii tions, other than geometrical, accepted without demonstra- tion and true universally, such as ^Hhe whole of a thing is greater than a part " These two classes of propositions are often distinguished by the terms "geometrical axioms" and "general axioms." I prefer to use the word axiom as appli- cable to the latter class only, that is, to simple propositions, true of magnitudes of all kinds (for instance 'Hhings which are equal to the same thing are equal to one anothefi'"'^), and to use the term postulate for a simple geometrical proposi- tion, whose truth we assume. When a child is told that A weighs exactly as much as B, and B weighs exactly as much as C, he without hesitation arrives at the conclusion that A weighs exactly as much as C. His conviction of the validity of his conclusion would not be strengthened, and possibly his confidence in his conclusion might be impaired, by his being directed to appeal to the authority of the general proposition " things which are equal to the same thing are equal to one another." I have therefore, as a rule, omitted in the text all reference to the general statements of axioms, and have only introduced such a statement occa- sionally, where its introduction seemed to me the shortest way of explaining the nature of the next step in the demonstration. If it be objected that all axioms used should be clearly stated, and that their number should not be unnecessarily extended, my reply is that neither the Greek text nor any edition of it, with Avhich I am acquainted, has attempted to make its list of axioms perfect in either of these respects. The lists err in excess, inasmuch as some of the axioms therein can be deduced from others : they err in defect, inasmuch as in the demonstrations of Propositions conclu- sions are often drawn, to support the validity of which no appeal can be made to any axiom in the lists. viii PREFACE. Under the term postulate I have included not only what may be called the postulates of geometrical operation, such as ^Ht is assumed that a straight line may he drawn from any point to any other point^^ but also geometrical theorems, the truth of which we assume, such as ''Hwo straight lines cannot have a common part.^^ The postulates of this edition are nine in number. Postulates 3, 4, 6 are the postulates of geometrical operation, which are common to all editions of the Elements of Euclid. Postulates 1, 5, 9 are the Axioms 10, 11, 12 of modern editions. Postulates 2, 7, 8 do not appear under the head either of axioms or of postulates in Euclid's text, but the substance of them is assumed in the demonstrations of his propositions. Postulate 9 has been postponed until page 51, as it seemed undesirable to trouble the student with an attempt to unravel its meaning, until he was prepared to accept it as the converse of a theorem, with the proof of which he had already been made acquainted. It may be mentioned that a proof of Postulate 5, ^^all right angles are equal" is given in the text (Proposition lOB), and that therefore the number of the Postulates might have been diminished by one : it was however thought necessary to retain this Postulate in the list, so that it might be used as a postulate by any person who might prefer to adhere closely to the original text of Euclid. One important feature in the present edition is the greater freedom in the direct use of "the method of super- position " in the proofs of the Propositions. The method is used directly by Euclid in his proof of Proposition 4 of Book I., and indirectly in his proofs of Proposition 5 and of every other Proposition, in which the theorem of Proposition 4 is quoted. It seems therefore but a slight alteration to adopt the direct use of this method in the PREFACE. ix proofs of any theorems, in the proofs of which, in Euclid's text, the theorem of Proposition 4 is quoted. It may of course be fairly objected that it would be more logical for a writer, who uses with freedom the method of superposition, to omit the first three Propo- sitions of Book I. To this objection my reply must be that it is considered undesirable to alter the numbering of the Propositions in Books I. and II. at all events. No doubt a work written merely for the teaching of geometry, without immediate reference to the requirements of candi- dates preparing for examination, might well omit the first three Propositions and assume as a postulate that "a circle may he described with any point as centre^ and with a length equal to any given straight line as radius^" instead of the postulate of Euclid's text (Postulate 6 of the pre- sent edition), " a circle may he described with any point as centre and with any straight line drawn from that j)oi7it as radius." The use of the words "each to each" has been aban- doned. The statement that two things are equal to two other things each to each, seems to imply, according to the natural meaning of the words, that all four things are equal to each other. Where we wish to state briefly that A has a certain relation to a, B has the same relation to 6, and C has the same relation to c, we prefer to say that A, B, C have this relation to «, h, c respectively. The enunciations of the Propositions in Books I. and II. have been, with some few slight exceptions, retained throughout, and the order of the Propositions remains unaltered, but different methods of proof have been adopted in many cases. The chief instances of alteration are to be found in Propositions 5 and 6 of Book I., and in Book II. The use of what may be called impossible figures, such as occurred in Euclid's text in the proofs of Propositions X PREFACE. 6 and 7 of Book I. has been avoided. It seems better to prove that a line cannot be drawn satisfying a certain condi- tion without making a pretence of doing what is impossible. Two Propositions (10 A and 10 B), have been introduced to shew that, if the method of superposition be used, we need not take as a postulate ^^ all rigid angles are equal to one another,'^ but that we may deduce this theorem from other postulates which have been already assumed. Another new Proposition introduced into the text is Proposition 26 A, '[if two triangles have two sides equal to two sides, and the angles opposite to one ^;azV of equal sides equal, the angles opposite to the other pair are either equal or supplementary/," which may be described, with reference to Euclid's text, as the missing case of the equality of two triangles. It is intimately connected with what is called in Trigonometry " the ambiguous case " in the solution of triangles. Another new Proposition (41 A) is the solution of the problem " to construct a triangle equal to a given rectilineal figure." It appears to be a more practical method of solving the general problem of Proposition 45 "^o construct aj)aral- lelogram equal to a given rectilineal figure, having a side equal to a given straight line, and having an angle equal to a given angle," to begin with the construction of a triangle equal to the given figure rather than to follow the exact sequence of Euclid's propositions. In the notes a few ''Additional Propositions" have been introduced containing important theorems, which did not occur in Euclid's text, but with which it is desirable that the student should become familiar as early as possible. Also outlines have been given of some of tho many different proofs which have been discovered of ■ Pythagoras's Theorem. They may be found interesting and useful as exercises for the student. PREFACE. xi Euclid's proofs of many of the Propositions of Book II. are unnecessarily long. His use of the diagonal of the square in his constructions in Propositions 4 to 8 can scarcely be considered elegant. It is curious to notice that Euclid after giving a demonstration of Proposition 1 makes no use whatever of the theorem. It seems more logical to deduce from Proposition 1 those of the subsequent Propositions whicli can be readily so deduced. In Book II. outlines of alternative proofs of several of the Propositions have been given, which may be developed more fully and used in examinations, in place of the proofs given in the text. Some of these proofs are not, so far as I know, to be found in English text books. The most interesting ones are those of Propositions 12 and 13. Some, which I thought at first were new, I have since found in foreign text books. The Propositions in the text have not been distin- guished by the words "Theorem" and "Problem." The student may be informed once for all that the word theorem is used of a geometrical truth which is to be demonstrated, and that the word problem is used of a geometrical construction which is to be performed. Although Euclid always sums up the result of a Propo- sition by the words oinp ISct Sct^ai or ottc/j t3et Trotryo-ai, there seems to be no utility in putting the letters q.e.d. or Q.E.F. at the end of a Proposition in an English text- book. The words " Quod erat demonstrandum " or " Quod erat faciendum " in a Latin text were not out of place. "When the book is opened, the reader will see as a rule on the left hand page a Pi-oposition, and on the opposite page notes or exercises. The notes are either appropriate to the Proposition they face or introductory to the one next succeeding. The exercises on the right hand page are. xii PREFACE, it is hoped, in all cases capable of being solved by means of the Proposition on the adjoining page and of preceding Propositions. They have been chosen with care and with the special view of inducing the student from the com- mencement of his reading to attempt for himself the solution of exercises. For many Propositions it has been difficult to find suitable exercises : consequently many of the exercises have been specially manufactured for the Propositions to which they are attached. Great pains have been taken to verify the exercises, but notwithstanding it can scarcely be hoped that all trace of error has been eliminated. It is with pleasure that I record here my deep sense of obligation to many friends, who have aided me by valuable hints and suggestions, and more especially to A. R. Forsyth, M.A., Fellow and Assistant Tutor of Trinity College, Charles Smith, M.A., Fellow and Tutor of Sidney Sussex College, R. T. Wright, M.A., formerly Fellow and Tutor of Christ's College, my brother-in-law the Reverend T. J. Sanderson, M.A., formerly Fellow of Clare College, and my brother- W. W. Taylor, M.A., formerly Scholar of Queen's College, Oxford, and after- wards Scholar of Trinity College, Cambridge. The time and trouble ungrudgingly spent by these gentlemen on this edition have saved it from many blemishes, which would otherwise have disfigured its pages. I shall be grateful for any corrections or criticisms, which may be forwarded to me in connection either with the exercises or with any other part of the work. H. M. TAYLOR. Trinity College, Cambridge, October 1, 1889. PREFACE TO BOOKS III AND IV. IN Book III. the chief deviation from Euclid's text will be found in the first twelve Propositions, where a good deal of rearrangement has been thought desirable. This rearrangement has led to some changes in the sequence of Propositions as well as in the Propositions themselves ; but, even with these changes, the first twelve Propositions will be found to include the substance of the whole of the first twelve of Euclid's text. The Propositions from 13 to 37 are, except in unim- portant details, unchanged in substance and in order. The enunciation of the theorem of Proposition 36 has been altered to make it more closely resemble that of the complementary theorem of Proposition 35. An additional Proposition has been introduced on page 186 involving the principle of the rotation of a plane figure about a point in its plane. It is a principle of which extensive use might with advantage be made in the proof of some of the simpler' properties of the circle. It has not however been thought desirable to do more in this edition than to introduce the student to this method and by a selection of exercises, which can readily be solved by its means, to indicate the importance of the method. PREFACE TO BOOKS V. AND VI. of the Greek text is printed in brackets at the lieacl of each Proposition. In Book VI a slight departure from Euclid's text is made in the treatment of similar figures. The definition of similar polygons which is adopted in this work brings into prominence the important property of the fixed ratio of their corresponding sides. Its use has the great merit of tending at once to simplicity and brevity in the proofs of many theorems. The numbering of the Propositions in Book VI remains unchanged : Propositions 27, 28, 29 are omitted as in many of the recent English editions of Euclid, and iti several cases a Proposition which consists of a theorem and its converse is divided into two Parts. Proposition 32 of Euclid's text, which is a very special case of no great interest, has been replaced by a simple but important theorem in the theory of similar and similarly situate figures. The chief difficulty with respect to the additions which have been made to Book VI was the immense number of known theorems from which a selection had to be made. I have attempted by means of two or three series of Propositions arranged in something like logical sequence to introduce the student to important general methods or well-known interesting results. One series gives a sketch of the theory of transversals, and the properties of harmonic and anharmonic ranges and pencils, and leads up to Pascal's Theorem. Another series deals with similar and similarly situate figures and leads up PREFACE TO BOOKS V. AND VI. to Gergoime's elegant solution of the problem to describe a circle to touch three given circles. These are followed by an introduction to the method of Inversion, an account of Casey's extension of Ptolemy's Theorem, some of the im- portant properties of coaxial circles, and Poncelet's Theorems relating to the porisms connected with a series of coaxial circles. No attempt has been made to represent the very large and still increasing collection of theorems connected with the "Modern Geometry of the Triangle." I hereby acknowledge the great help I have received in this portion of my work from friends, and especially from Dr Forsyth and from my brother Mr J. H. Taylor. To the latter I am indebted for the Index to Books I — VI, which I hope may prove of some assistance to persons using this edition. H. M. T. TuiNiTY College, Cambridge, March 16, 1893. THE ELEMENTS OF GEOMETRY. T. E. BOOK I. Definition 1. That 'which has ^wsition but not magni- tude is called a point. The word point is used in many different senses. We speak in ordinary language of the point of a pin, of a pen or of a pencil. Any mark made with such a point on paper is of some definite size and is in some definite position. A small mark is often called a spot or a dot. Suppose such a spot to become smaller and smaller ; the smaller it becomes the more nearly it resembles a geometrical point : but it is only when the spot has become so small that it is on the point of vanishing altogether, i.e. when in fact the spot still has position but has no magnitude, that it answers to the geometrical definition of a point. A point is generally denoted by a single letter of the alphabet: for instance we speak of the point A. Definition 2. That which has position and length but neither breadth nor thickness is called a line. The extremities of a line are points. The intersections of lines are points DEFINITIONS. 3 The word line also is used in many different senses in ordinary language, and in most of these senses the main idea suggested is that of length. For instance we speak of a line of railway as connecting two distant towns, or of a sounding line as reaching from the bottom of the sea to the surface, and in so speaking we seldom think of the breadth of the railway or of the thickness of the sounding line. When we speak of a geometrical line, we regard merely the length: we exclude the idea of breadth and thickness altogether : in fact we consider that the cross-section of the line is of no size, or in other words that the cross-section is a geometrical point. If a point move with a continuous motion from one position to another, the path which it describes during the motion is a line. Definition 3. That which has position, length and breadth hut not thickness is called a surface. The boundaries of a surface are lines. The intersections of surfaces are lines. The word surface in ordinary language conveys the idea of ex- tension in two directions : for instance we speak of the surface of the Earth, the surface of the sea, the surface of a sheet of paper. Although in some cases the idea of the thickness or the depth of the thing spoken of may be present in the speaker's mind, yet as a rule no stress is laid on depth or thickness. When we speak of a geome- trical surface we put aside the idea of depth and thickness altogether. We are told that it takes more than 300,000 sheets of gold leaf to make an inch of thickness ; but although the gold leaf is so thin, it must not be regarded as a geometrical surface. In fact each leaf however thin has always two bounding surfaces. The geometrical surface is to be regarded as absolutely devoid of thickness, and no number of surfaces put together would make any thickness whatever. Definition 4. That tohich has jjosition, length, breadth and thickness is called a solid. The boundaries of solids are surfaces. 1—2 4 BOOK I. Definition 5. Any combination of points, lines, and surfaces is called a figure. Definition 6. A line which lies evenly between points on it is called a straight line. This is Euclid's definition of a straight line. It cannot be turned to practical use by itself. We supplement the definition, as Euclid did, by making some assumptions the nature of which wUl be seen hereafter. Postulates. There are a few geometrical propositions so obvious that we take the truth of them for granted, and a few geometrical operations so simple that we assume we may perform them when we please without giving any explanation of the process. The claim we make to use any one of these propositions, or to perform any one of these operations, is called a postulate. Postulate 1. Two straight lines cannot enclose a space. This postulate is equivalent to Two straight lines cannot intersect in more than one point. Postulate 2. Two straight lines cannot have a common pa/rt. If two straight lines have two points ^ , jB in common, they must coincide between A and 5, since, if they did not, the two straight lines would enclose a space. Again, they must coincide beyond A and B, since, if they did not, the two straight lines would have a common part. Hence we conclude that Tioo straight lines, which have two poi^its in common, are coincident throughout their length. Thus two points on a straight line completely fix the position of the line. Hence we generally denote a straight line by mentioning two points on it, and when the straight line is of finite length, we generally denote it by mentioning the points which are its two extremities. DEFINITIONS. 5 For instance, if P and Q be two points on a straight line, the Hne is called the straight line PQ or the straight line QP, or sometimes more shortly PQ or QP: and the straight line which is terminated by two points P and Q is called in the same way PQ or QP. It may be remarked that, when merely the actual length of the straight line is under discussion, we use PQ or QP indifferently: but that, when we wish to consider the direction of the line, we must carefully distinguish between PQ and QP. Postulate 3. A straight line may he drawn from any point to any other point. Postulate 4. A finite straight line may be produced at either extremity to any length. The demands made in Postulates 3 and 4 are in practical geometry equivalent to saying that a ' straight edge ' may be used for drawing a straight line from one point to another and for producing a straight line to any length. We assume, as Euclid did, that it is possible to shift any geometrical figure from its initial position unchanged in shape and size into another position. Test of Equality of Geometrical Figures. The criterion of the equality of two geometrical figures, which we shall use in most cases, is the possibility of shifting one of the figures, unchanged in shape and size, so that it exactly fits the place which the other of the figures occupies. (See Def. 21.) This method of testing the equality of geometrical figures is generally known as the method of superposition. Test of equality of straight lines. Two straight lines AB, CD are said to be equal, when it is possible to shift either of them, say AB, so that it coincides with the other CD, the end A on and the end B on I), or the end A on 1) and the end B on C. 6 BOOK L Addition of Lines. Having defined the equality of straight lines, we proceed to explain what is meant by the addition of straight lines. D If in a straight line we take points ^, B, C, D in order, we say that the straight line ^C is the sum of the two straight lines AB^ BG (or of any two straight lines equal to them), and that the straight line AB is the difference of the two straight lines AC, BC (or of any two straight lines equal to them). In the same way we say that the straight line AD is the sum of the three straight lines AB, BG, GD. Again, if AB be equal to BG, we say th&t AG is double of AB or ofBG. Definition 7. A surface which lies evenly between straight lines on if is called a plane. This is Euclid's definition of a plane : there is the same difficulty in making use of it that there is in making use of his definition of a straight line. Consequently this definition has by many modern editors been replaced by the following, which perhaps merely expresses Euclid's meaning in other words : A surface such that the straight li7ie joining any two j)oints in the surface lies wholly in the surface is called a plane. Definition 8. A figure, which lies xoholly in one plane, is called a plane figure. All the geometrical propositions in the first six books of the Elements of Euclid relate to figures in one plane. This part of Geometry is called Plane Geometry. DEFINITIONS. 7 Definition 9. Two straight lines in the same plane, which do not meet however far tliey may he jn-oducecl both ways, are said to he parallel* to one another. D Definition 10. A plane angle is the inclination to one another of tico straight lines which meet hut are not in tlie same straight line. The idea of an angle is one which it is very difficult to convey by the words of a definition. We will content ourselves by explaining some few things connected with angles. If two straight lines AB, AG meet at A, the amount of their divergence from one another or their inclination to one another is called the angle which the lines make loith one another or the angle hetiueen the lines, or the angle contained by the lines. The angle formed by the straight lines AB, AG ia generally de- nominated BAG, or GAD, the middle letter always denoting the point where the lines meet, and the letters B and G denoting any two points in the straight lines AB, AC. It must be carefully noted that the magnitude of the angle is not affected by the length of the straight hnes^B, AG. The point A, where the two straight lines AB and AC, which form the angle BAG, meet, is called the vertex of the angle BAG. If there be only two straight lines meeting at a point A, the angle formed by the lines is sometimes denoted by the single letter A. * Derived from Trapd "by the side of" undaWriXas "one another' irapdWtjXoL ypafxfxai "lines side by side". 8 BOOK T. Test of Equality of Angles. Two angles are said to be equal, when it is possible to shift the straight lines forming one of the angles, unchanged in position relative to each other, so as to exactly coincide in direction with the straight lines forming the other angle. D A /B G For instance, the angles ABC, DEF will be equal, if it be pos- sible to shift AB, BG unchanged in position relative to each other, so that B coincides with E, and so that also either BA coincides in direction with ED and BG with EF, or BA coincides in direction with EF and BG with ED. If a straight line move in a plane, while one point in the line remains fixed, the line is said to turn or revolve about the fixed point. If the revolving line move from any one position to any other position, it generates an angle, and the amount of turning from one position to the other is the measure of the magnitude of the angle between the two positions of the line. For instance each hand of a watch, as long as the watch is going, is turning uniformly round its fixed extremity, and is generating an angle uniformly. This mode of regarding angles enables us to realize that angles are capable of growing to any size and need not be limited (as in most of the propositions in Euclid's Elements they are supposed to be) to magnitudes less than two right angles. (See Def. 11.) Addition of Angles. If three straight lines AB, AG, AD meet at the same point, we say that the angle BAD is the sum of the two angles BAG, GAD (or of any two angles equal to them). In the same way we say that the angle BAG is the difference of the two angles "^ "" BAD, GAD (or of any two angles equal to them). Two angles such as BAG, GAD, which have a common vertex and one common bounding line, are called adjacent angles. DEFINITIONS. 9 Definition 11. If two adjacent angles made hy two. straight lines at the pohit where they meet be equal, each of these angles is called a right angle, aiid the straight lines are said to be at right angles to each other. Either of two straight lines which Ci I> are at right angles to each other is said to be perpendicular to the other. b If a straight line AE be drawn from a point A at right angles to a given straight line CD, the part AE intercepted between the point and the straight line is commonly called tlie perpendicular from the point A on the straight line CD. Euclid uses as a postulate. Postulate 5. All right angles are equal to one another. It is not necessary to assume this proposition, since it can be proved by the method of superposition. A proof will be found on a sub- sequent page. (p. 37) Definition 12. An angle less than a right angle is called an acute angle. A71 angle greater tha^i a right angle and less than two right angles is called an obtuse angle. Definition 13. A line, ivhich is such that it can be described by a moving point starti7ig from any j)oint of the line and returning to it again, is called a closed line. A figure composed wholly of straight lines is called a rectilineal figure. llie straight lines, which form a closed rectilineal figure, are called the sides of the figure. The sum of the lengths of the sides of any figure is called the perimeter of the figure. The poifit, where two adjacent sides meet, is called a vertex or an angular point of the figure. The angle formed by two adjacent sides is called an angle of the figure. 10 BOOK I. A straight line joining any two vertices of a closed recti- lineal figure^ which are not extremities of the same side, is called a diagonal*. The surface contained within a closed figure is called the area of the figure. A closed rectilineal figure, which is such that the whole figure lies 07i one side of each of the sides of the figure, is called a convex figure. A closed rectilineal figure is in general denoted by naming the letters, which denote its vertices, in order: for instance the five-sided figure in the diagram is denoted by the letters A, B, G, D, E, in order: i.e. it might be called the figure ABODE, or the figure CBAED. A, B, G, D, E are its vertices. AB, BG, GD, BE, EA are its sides. ABG, BGD, GDE, BE A, EAB are its angles. AG, AD are two of its diagonals. It will be observed that a closed figure has the same number of angles as it has sides. If a closed figure have an even number of sides, we speak of a pair of sides as being opposite, and of a pair of angles as being opposite. If a closed figure have an odd number of sides, we speak of an angle as being opposite to a side and vice versa. For instance in the quadrilateral ABGD the side AD is said to be opposite to the side BG, and the angle BAD opposite to the angle BGD, but in the five-sided figure ABGDE the side GD is said to be opposite to the angle BAE, and the angle AED opposite to the side BG. Derived from 5td "through", and yosvla "an angle". DEFINITIONS. 11 Definition 14. A figure, all the sides of which are equals is called equilateral. A figure, all the angles of which are equal, is called equiangular. A figure, which is both equilateral and equiangular, is called regidar. Definition 15. A closed rectilineal figure, which has three sides*, is called a triangle. A closed rectilineal figure, which has four sides, is called a quadrilateraL A closed rectilineal figure, which has more than four sides, is called a polygon f- Definition 16. A triangle, which has two sides equal, is called isosceles J. A triangle, which has a right angle, is called right-angled. The side opposite to the right angle is called the hypo- tenuse §• * A figure, which has three sides, must also have three angles. It is for this reason called a triangle. t Derived from ttoXi^s "much" and -^wvla "an angle". X Derived from taos "equal" and ctk^Xos "a leg". § Derived from vir6 "under" and TeLveiv "to stretch", r/ viro- relvovcra ypa/x/j.'/i "the line subtending" or "stretching across" (the right angle). 12 BOOK I. A triangle^ which has an obtuse angle^ is called obtuse- angled. A tria7igle, which has three acute angles, is called acute- angled. Definition 17. A quadrilateral^ which has four sides equal, is called a rhombus. Definition 18. A quadri- lateral, whose opposite sides are parallel, is called a parallelo- gram. . Definition 19. ^ parallelogra7n, one of whose angles is a right angle, is called a rectangle. It will be proved later that each angle of a rectangle is a right angle. Definition 20. A rectangle, which has two adjacent sides equal, is called a square. It will be proved later that all the sides of a square are equal. DEFINITIONS. 13 Definition 21. Two figures are said to be ec[ual in all respects, when it is possible to shift one unchanged in shajye and size so as to coincide with the other. The figures ABODE, FGHKL are equal in all respects, if it be possible to shift ABODE so that the vertices A, B, O, D, E may coincide with the vertices J^, G, H, K, L respectively : in which case the sides of the two figures must be equal, AB, BO, OD, DE, EA to EG, GH, HK, KL, LF respectively, and the angles must be equal, ABO, BOD, ODE, DEA, EAB to FGH, GHK, HKL, RLE, LEG respectively. Definition 22. A plane closed lioie, which is such that all straight lines drawn to it from a fixed point are equal, is called a circle. This point is called the centre the circle. oj It will be proved hereafter that a circle has only one centre. A straight line drawn from the centre of the circle to the circle is called a radius. A straight line drawn through the centre and terminated both ways by the circle is called a, diameter. It will be proved hereafter that three points on a circle completely fix the position and magnitude of the circle: hence we generally denote a circle by mentioning three points on it; for instance the circle in the diagram might be called the circle BDE, or the circle DBO. The one assumption which we make with reference to describing a circle is contained in the following postulate : u BOOK I. Postulate 6. A circle inay he described with any point as centre and with any straight line drawn from that point as radius. Postulate 7. Any straight line drawn through a point within a closed Jigure Tnust, if produced far enough, intersect the figure in two points at least. In the diagram we have three specimens of closed figures each with a point A inside the figure. It is easily seen that any straight line through A must intersect the figure in two points at least: in the case of two of the figures a straight line cannot intersect the figure in more than two points : but in the third case, a straight line can be drawn to intersect the figure in four points. Postulate 8. Any line joining two points one within and the other without a closed figure must intersect the figure in one point at least. It follows that Any closed line drawn through two points one within and the other witJiout a closed Jigure must intersect the Jigure in ttco points at least. --^^..^B In the diagram we have three specimens of closed figures with two points A^B, one inside and the other outside the figure. DEFINITIONS. 15 It is easily seen that any line joining A and B must intersect the' figure in one point at least, and that any closed line drawn through A and B must intersect the figure in two points at least : in two of the cases in the diagram either of the paths represented by part of the dotted line joining A and B intersects the figure in one point only and the closed line drawn intersects the figure in two points only : but in the third case one of the paths from ^ to i? represented by part of the dotted line intersects the figure in one point only, while the path represented by the other part of the dotted line intersects the figure in three points, and the closed line drawn through A and B intersects the figure in four points. Axioms. There are a number of simple propositions generally admitted to be true universally, i.e. with reference to magnitudes of all kinds. Such propositions were called by Euclid koivoX evvoiat, "common notions": they are now usually denominated axioms'*', a^najxara, as being propositions claimed without demonstration. The following are examples of such axioms : Things which are equal to tlie same thing are equal to one another. If equals he added to equals, the wholes are equal. If equals he taken froin equals, the remaiiiders are equal. Doubles of equals are equal. Halves of equals are equal. The whole of a thing is greater than a part. If one thing he greater than a second and the second greater than a third, the first is greater than the third. Such propositions as the above we shall use freely in the following pages without further remark. * Dr Johnson in his English Dictionary defined an axiom as "a proposition evident at first sight, that cannot be made plainer by demonstration." 16 BOOK I. PROPOSITION 1. On a given finite straight liite to construct an equilateral triangle. Let ABhe the given finite straight line: it is required to construct an equilateral triangle on AB. Construction. With A as centre and AB as radius, describe the circle BCD. (Post. 6.) With B as centre and BA as radius, describe the circle ACE. These circles must intersect : (Post.- 8.) let them intersect in C. VH^^ \ ^ Draw the straight lines CA, CB : (Post. 3.) then ABC is a triangle constructed as required. Proof. Because A is the centre of the circle BCD^ AC \& equal to AB. (Def. 22.) And because B is the centre of the circle ACE, BC is equal to BA. Therefore CA, AB, BC are all equal. Wherefore, the triangle ABC is equilateral, and it has been constructed on the given finite straight line AB. I'ROPOSITION- 1. 17 It is assumed in this proposition that the two circles intersect. It is easily seen that they must intersect in two points. "We can take either of these points as the third angular point of an equilateral triangle on the given straight line ; there are thus two triangles which can be constructed satisfying the requirements of the proposition. We say therefore that the problem put before us in this proposition admits of two solutions. We shall often have occasion to notice that a geometrical problem admits of more than one solution, and it is a very useful exercise to consider the number of possible solutions of a particular problem. For the future we shall generally use the abbreviated expression "draw AB^' instead of "draw the straight line AB''' or "draw a straight line from the point A to the point £." EXEKCISES. 1. Produce a straight line so as to be (a) twice, {b) three times, (c) five times, its original length. 2. Construct on a given straight line an isosceles triangle, such that each of its equal sides shall be (a) twice, {h) three times, (c) six times, the length of the given line. 3. Prove that, if two circles, whose centres are A, B, and whose radii are equal, intersect in C, D, the figure ABCD is a rhombus. T, B. 18 BOOK L PROPOSITION 2. From a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line : it is required to draw from A a straight line equal to BG. Construction. Draw AB, the straight line from A to one of the extremities of BG ', (Post. 3.) on it construct an equilateral triangle ABB. (Prop. 1.) With B as centre and BG as radius, describe the circle GEF^ (Post. 6.) meeting BB (produced if necessary) at E. (Post. 7.) With B as centre and BE as radius, describe the circle EGH^ meeting DA (produced if necessary) at G : then AG i^ Si, straight line drawn as required. Proof. Because B is the centre of the circle GEF^ BG is equal to BE. (Def. 22.) Again, because D is the centre of the circle EGH^ ZXy is equal to BE; and because ABB is an equilateral triangle, BA is equal to BB ; (Def. 14.) therefore AG \^ equal to BE. And it has been proved that BG is equal to BE; therefore AG \^ equal to BG. Whorefore, /rom the given point A a straight line AG has been drawn equal to the given straight line BG. PROPOSITION 2. 19 It is assumed in this proposition that the straight line BB inter- sects the circle GEF. It is easily seen that it must intersect it in tioo points. It will be noticed that in the construction of this proposition there are several steps at which a choice of two alternatives is afforded: (1) we can draw either AB ov AC as the straight line on which to construct an equilateral triangle : (2) we can construct an equilateral triangle on either side of AB: (3) if DB cut the circle in E and J, we can choose either DE or DI as the radius of the circle which we describe with D as centre. There are therefore three steps in the construction, at each of which there is a choice of two alternatives: the total number of solutions of the problem is therefore 2 x 2 x 2 or eight. On the opposite page two diagrams are drawn, to represent two out of these eight possible solutions. It will be a useful exercise for the student to draw diagrams corresponding to some of the remaining six. EXEKCISES. 1. Draw a diagram for the case in which the given point is the middle point of the given straight line. 2. Draw a diagram for the case in which the given point is in the given straight line produced. 3. Draw from a given point a straight line (a) twice, (&) three times the length of a given straight line. 4. Draw from D in any one of the diagrams of Proposition 2 a straight line, so that the part of it intercepted between the two circles may be equal to the given straight line. Is a solution always possible ? 20 BOOK I. PROPOSITION 3. From the greater of two given straight lines to cut off a i^art equal to the less. Let AB and CD be the two given straight lines, of whicli AB i^ the greater : it is required to cut off from AB d^ part equal to CD. Construction. From A draw a straight line AE equal to CD; (Prop. 2.) with A as centre and ^jS' as radius, describe the circle EFG. (Post. 6.) The circle must intersect AB between A and B, for AB \^ greater than AE. Let F be the point of intersection : then AF is the part required. Proof. Because A is the centre of the circle EFG^ AE is equal to AF. (Del 22.) But AE was made equal to CD] (Construction.) therefore AF \& equal to CD. Wherefore, from AB the greater of two given straight lines a part A F has heeii cut off equal to CD the less. PROPOSITION 3. 21 The demand made in Postulate 6, that "a circle maybe described with any point as centre and with any straight line drawn from that point as radius," is equivalent, in practical geometry, to saying that a pair of compasses may be used in the following manner: the ex- tremity of one leg of a pair of compasses may be put down on any point A, the compasses may then be opened so that the extremity of the other leg comes to any other point and then a circle may be swept out by the extremity of the second leg of the compasses, the extremity of the first leg remaining throughout the motion on the point A. Compasses are also used practically for carrying a given length from any one position to any other : for instance, they would gene- rally be used to solve the problem of Proposition 3 by opening the compasses out till the extremities of the legs came to the points C, D : they would then be shifted, without any change in the opening of the legs, until the extremity of one leg was on A and the extremity of the other in the straight line AB. Euclid restricted himself much in the same way as a draughtsman would, if he allowed himself only the first mentioned use of the com- passes : the first three propositions shew how Euclid with this self- imposed restriction solved the problem, which without such a restric- tion could have been solved more readily. After the problems in the first three propositions have been solved, we may assume that we can draw a circle, as a practical draughts- man would, with any point as centre and with a length equal to any given straight line as radius. EXEECISES. 1. On a given straight line describe an isosceles triangle having each of the equal sides equal to a second given straight line. 2. Construct upon a given straight line an isosceles triangle having each of the equal sides double of a second given straight line. 3. Construct a rhombus having a given angle for one of its angles, and having its sides each equal to a given straight line. 22 BOOK L PROPOSITION 4. If two triangles have two sides of iJie one equal to two sides of the other, and also the angles contained hy those sides equal, the two triangles are equal m all resjjects. (See Def. 21.) Let ABC, DEF be two triangles, in which AB is equal to DE, and AC to DF, and the angle BAG is equal to the angle ^i>i^: it is required to prove that the triangles ABC, DEF are equal in all respects. Proof. Because the angles BAG, EDF are equal, it is possible to shift the triangle ABC so that A coincides with D, and AB coincides in direction with DE, and AC with DF. (Test of Equality, page 8.) If this be done, because ^^ is equal to DE, B must coincide with E \ and because AC is equal to DF, C must coincide with F. Again because B coincides with E and G with F, BG coincides with EF -, (Post. 2.) therefore the triangle ABC coincides with the triangle DEF, and is equal to it in all respects. Wherefore, if two triangles &c. PROPOSITION 4. 23 The proof of this proposition holds good not only for a pair of triangles such as ABC, DEF in the diagram: it holds good equally for a pair such as ABC, D'E'F', one of which must be reversed or turned over before the triangles can be made to coincide or fit exactly. In this proposition Euclid assumed Postulate 2, that two straight lines cannot have a common part. When the triangle ABC is shifted, so that ^ is on D and AB is on DE, there would be no justification for the conclusion that B must coincide with E, because ^^ is equal to DE, if it were possible for two straight lines to have a common part. In fact, two curved lines might be drawn from the point D starting in the same direction DE but leading to two totally distinct points E and F although the lines were of the same length. It is tacitly assumed, that if the lines be straight lines, this is im- possible. EXERCISES. 1. If the straight line joining the middle points of two opposite sides of a quadrilateral be at right angles to each of these sides, the other two sides are equal. 2. If in a quadrilateral ABCD the sides AB, CD be equal and the angles ABC, BCD be equal, the diagonals AC, BD are equal. 3. If in a quadrilateral two opposite sides be equal, and the angles which a third side makes with the equal sides be equal, the other angles are equal. 4. Prove by the method of superposition that, if in two quadri- laterals ABCD, A'B'C'D', the sides AB, BC, CD be equal to the sides A'B', B'C, CD' respectively, and the angles ABC, BCD equal to the angles A'B'C, B'C'D' respectively, the quadrilaterals are equal in all respects. 24 BOOK I. PROPOSITION 5. If two sides of a triangle he equal, the angles opjjosite to these sides are equal, and the angles made hy producing these sides beyond the third side are equal. Let ABC be a triangle, in which AB is equal to AC, and AB, AC are produced to D, E : it is required to prove that the angle ACB is equal to the angle ABC, and the angle BCE to the angle CBJ). Construction. Let the 6gure ABODE be turned over and shifted unchanged in shape and size to the position ahcde, A to a, B to b, C to c, D to d and E to e. A A Proof. Because the angles DAE, ead are ecjual, it is possible to shift the figure abode so that a coincides with A, and ae coincides in direction with AD, and ad with AE. (Test of Equality, page 8.) If this be done, because ac is equal to AB, c must coincide with B ; and because ab is equal to AC, b must coincide with C ; hence cb coincides with BC. (Post. 2.) Now because ace coincides in direction with A BD, and cb with BC, the angle acb coincides with the angle ABC, and the angle bee with the angle CBD ; therefore the angle acb is equal to the angle ABC, and the angle bee to the angle CBD. PROPOSITION 5. 25 But the angle ach is equal to the angle AGB^ and the angle hce to the angle BCE ; therefore the angle ACB is equal to the angle ABC^ and the angle BCE to the angle GBD. Wherefore, if two sides &c. Corollary 1. An equilateral triangle is also equi- angular. Corollary 2. If two angles of a triangle be unequal.^ the sides opposite to these angles are unequal. EXEKCISES. 1. The opposite angles of a rhombus are equal. 2. If a quadrilateral have two pairs of equal adjacent sides, it has one pair of opposite angles equal. 3. If in a quadrilateral ABGD, AB be equal to AD and EC to DC, the diagonal AG bisects each of the angles BAD, BCD. 4. If in a quadrilateral ABCD, AB be equal to ^D and jBC to DC, the diagonal BD is bisected at right angles by the diagonal AC. 5. Prove that the triangle, whose vertices are the middle points of the sides of an equilateral triangle, is equilateral. 6. Prove that the triangle, formed by joiuing the middle points of the sides of an isosceles triangle, is isosceles. 7. Prove by the method of superposition that, if in a convex quadrilateral ABCD, AB be equal to CD and the angle ABC to the angle BCD, AD is parallel to BC 26 BOOK I. PROPOSITION 6. If two angles of a triangle he equal, tlie sides opposite to these angles are equal. Let ABC be a triangle, in which tlie angle ABC is equal to the angle ACB : it is required to prove that AC is equal to AB. Construction. Let the triangle ABC be turned over and. shifted unchanged in shape and size to the position ahcpA to a, B to 6, and C to c. Pkoof. Because the sides BC, cb are equal, it is possible to shift the triangle acb so that cb coincides with BC, c with B, and b with C, (Test of Equality, page 5.) and so that the triangles acb, ABC are on the same side of BC. If this be done, because the angles ABC, acb are equal, ca must coincide in direction with BA ; and because the angles ACB, abc are equal, ba must coincide in direction with CA. And because two straight lines cannot intersect in more than one point, (Post. 1.) the point a, which is the intersection of ca and ba, must coincide with the point A, which is the intersection of BA and CA. Now because a coincides with A and c with B, ac coincides with AB and is equal to it. But ac is equal to AC ; therefore ^0 is equal to AB. Wherefore if two angles &c. Corollary. An equiangular triangle is also equilateral. PROPOSITION 6. 27 When in two propositions the hypothesis of each is the conclusion of the other, each proposition is said to be the converse of the other. The theorems in Propositions 5 and 6 are the converses of each other. It must not be assumed that the converse of a proposition is necessarily true. EXEECISES. 1. Shew that, if the angles ABC and ACB at the base of an isosceles triangle be bisected by the straight lines JSD and CD, DEC will be an isosceles triangle. 2. BAG is a triangle having the angle B double of the angle A. If BB bisect the angle B and meet ^C at D, BB is equal to AB. 3. Prove by the method of superposition that, if in two triangles ABC, A'B'C the angles ABC, BCA be equal to the angles A'B'C, B'C'A' respectively and the sides BC,B'C' he equal, the triangles are equal in all respects. 4. Prove by the method of superposition that, if in two quadri- laterals ABCB, A'B'C'B' the angles BAB, ABC,BCB be equal to the angles B'A'B', A'B'C, B'C'B' respectively, and the sides AB, BC be equal to the sides A'B', B'C respectively, the quadrilaterals are equal in all respects. 5. If in a quadrilateral ABCB, AB be equal to AB and the angle ABC to the angle ABC, then BC is equal to BC, and the diagonal AC bisects the quadrilateral and two of its angles. 28 BOOK I. PROPOSITION 7. If two points on the saine side of a straight line he equi- distant from one point in the line, they cannot he equidistant from any other point in the liiie. Let AB he> ^ given straight line, and C, D be two points on the same side of it equidistant from the point A : it is required to prove that C, D cannot be equidistant from any other point in the line. Construction. Take any other point B in the line, and draw BC, BD. Proof. Because C and D are two different points, either (1) the vertex of each of the triangles ABC, ABD must be outside the other triangle, (2) the vertex of one triangle must be inside the other, or (3) the vertex of one triangle must be on a side of the other. (1) First let the vertex of each triangle be without the other. Because AD is equal to AC, the angle ACD is equal to the angle ADC. (Prop. 5.) But the an^le ACD is greater than the angle BCD, and the angle BDC is greater than the angle ADC : therefore the angle BDC is greater than the angle BCD ; therefore BC, BD are unequal. (Prop. 5, Coroll. 2.) (2) Next let the vertex D of one triangle ABD be within the other triangle ABC : produce AC, AD to B, F. (Post. 4.) PROPOSITION 7. 29 Then because in the triangle ACD, AC is. equal to AD, the angles ECD, FDC made by producing the sides AC, AD are equal. (Prop. 5.) But the angle ECD is greater than the angle BCD, and the angle BDC is greater than the angle FDC ; therefore the angle BDC is greater than the angle BCD; therefore BC, BD are unequal, (Prop. 5, Coroll. 2.) (3) Next let the vertex D of one triangle lie on one of the sides BC of the other : then BC, BD are unequal. Wherefore, if two points 07i the same side &c. 30 BOOK I. PROPOSITION 8. If two triangles have three sides of the one equal to three sides of the other, the triangles are equal in all respects. Let ABC, DEF be two triangles, in which AB i^ equal to DE, AG to DF, and BC to EF\ it is required to prove that the triangles ABC, DEF are equal in all respects. Proof. Because the sides BC, EF are equal, it is possible to shift the triangle ABC, so that BC coincides with EF, B with E and C with F, (Test of Equality, page 5.) and so that the triangles ABC, DEF are on the same side of EF. If this 1)6 done, A must coincide with D : for there cannot be two points on the same side of the straight line EF equidistant from E and also equidistant from F, (Prop. 7.) Now because A coincides with i>, and B coincides with E, (Constr.) AB coincides with BE. (Post. 2.) Similarly it can be proved that AC coincides with DF. Therefore the triangle ABC coincides with the triangle DEF, and is equal to it in all respects. Wherefore, if two tricmgles is equal to BD. Wherefore, the given finite straight line AB is bisected at the point D. PROPOSITION 10. 35 EXEBCISES. 1. Prove that there is only one point of bisection of a given finite straight line. 2. If two circles intersect, then the straight line joining their centres bisects at right angles the straight line joining their points of intersection. 3. Draw from the vertex of a triangle to the opposite side a straight line, which shall exceed the smaller of the other sides as much as it is exceeded by the greater. 3—2 u BOOK I. PEOPOSITION 10 A. From the same point in a given straight line and on the same side of it, only one straight line can be drawn at right angles to the given straight line. From the point C in the straight line AB let the straight line CD be drawn at right angles to AB : it is required to prove that no other straight line can be drawn from G at right angles to AB on the same side of it. Construction. Draw from C within the angle ACB any straight line CE, D Proof. Because the angles DOB, DC A are equal, and the angle ECB is greater than the angle DCB, and the angle DC A is greater than the angle EC A ; therefore the angle ECB is greater than the angle EC A ; therefore CE is not at right angles to AB. (Del 11.) Similarly it can be proved that no straight line drawn from C within the angle DCB can be at right angles to^^. Therefore no straight line other than CD drawn from C can be at right angles to ^^ on the same side of it. Wherefore, from the same point kc. PROPOSITION ion. 37 PROPOSITION 10 B. All right angles are equal to one another. Let the straight lines AB^ CD meet at E and make the angles CEA, CEB right angles; (Def. 11.) and let the straight lines FG, HK meet at L and make the angles HLF^ II LG right angles : it is required to prove that the angle CEA is equal to the aniifle II LF. Proof. If the figure ABODE be shifted, so that E coincides with L, and the line ^^ in direction with FG^ and so that ECy LII are on the same side of EG : then EC must coincide with LH, for at the same point L in FG on the same side of it there cannot be two straight lines at right angles to EG; (Prop. 10 A.) therefore the angle A EC coincides with the angle FLH, and is equal to it. Wherefore, all right angles &c. 38 BOOK I. PROPOSITION 11. To draw a straight line at right angles to a given straight line from a given point in it. Let ABh^ the given straight line, and G the given point in it : it is required to draw from C a straight line at right angles to^^. Construction. Take any point D in. AG ^ and from GB cut off GE equal to GD. (Prop. 3.) On DE construct an equilateral triangle DFE^ (Prop. 1.) and draw GF : then GF is a straight line drawn as required. Z> Proof. Because in the triangles DGF, EGF, DG is equal to EG, GF to GF, and FD to FE, (Constr.) the triangles are equal in all respects ; (Prop. 8.) therefore the angle DGF is equal to the angle EOF, and they are adjacent angles; therefore each of these angles is a right angle ; and the straight lines are at right angles to each other. (Def. 11.) Wherefore, GF has been drawn at right angles to the given straight line AB,fro7n the given point G in it. PROPOSITION 11. 39 In the definition of a circle (Def. 22) we meet with the idea of a point, which moves subject to a given condition, the condition being that the point is always to be at a given distance from a given point, i.e. from the centre of the circle. The 'path of such a moving point, or the place {locus}, at some position on which the point must always be and at any position on which the point may be, is called the locus of the point. Hence we say in the case just mentioned that the locus of a point which is at a given distance from a given point is a circle. As a further illustration of this idea, let us consider the locus of a point, which moves subject to the condition that it is always to be equidistant from two given points. Let A, B he two given points, and let P be a point equidistant from A and B, i.e. let PA be equal to PB, Draw AB and take G the middle point otAB. Draw PC. Then because in the triangles PGA, PGB, PA is equal to PB, PC to PC, and ^C to BG, the two triangles are equal in all respects: (Prop. 8.) therefore the angle PGA is equal to the angle PCB, and since they are adjacent angles each is a right angle. It follows that, if P be equidistant from A and B, it must lie on the straight line GP which bisects AB a.t right angles. Every point on GP satisfies this condition. We may state the result of this proposition thus : The locus of a point equidistant from tico given points is the straight line which bisects at right angles the straight line joining the given points. EXEECISES. 1. The diagonals of a rhombus bisect each other at right angles. 2. Find in a given straight line a point equidistant from two given points. Is a solution always possible? 3. Find a point equidistant from three given points. 4. In the base JBC of a triangle ABG any point D is taken. Draw a straight line such that, if the triangle ABG be folded along this straight line, the point A shall fall upon the point D, 40 BOOK T. PROPOSITION 12. To draw a straight line at right angles to a given straight line from a given point without it. Let AB be the given straight line, and C the given point without it : it is required to draw from C a straight line at right angles to^^. Construction. Take any point D on the side of AB away from (7, draw C7i>, and with G as centre and CD as radius, describe the circle J^EF, _ (Post. 6.) meeting AB (produced if necessary) at E and F. Draw CE, OF, and bisect the angle EOF by the straight line CG meeting AB s^tG'. (Prop. 9.) then (76^ is a straight line drawn as required. Proof. Because in the triangles EGG, FGG, EC is equal to EG, and CG to CG, and the angle EGG is equal to the angle FGG, the triangles are equal in all respects ; (Prop. 4.) therefore the angle CGE is equal to the angle CGF, and they are adjacent angles. Therefore the straight lines CG, AB are at right angles to each other. (Def. 11.) Wherefore, CG has been drawn at right angles to the given straight line AB from the given point C without it. PROPOSITION 12. 41 In the construction for this proposition it is said that the point D is to be taken on the side of AB mcay from G. This restriction is introduced as a means of ensuring the intersection of the circle DEF with the straight line AB. EXEKCISES. 1. Through two given points on opposite sides of a given straight line draw two straight lines, which shall meet in the given line and include an angle bisected by that line. In what case can there be more than one solution ? 2. From two given points on the same side of a given straight line, draw two straight lines, which shall meet at a point in the given line and make equal angles with it. 3. Prove by the method of superposition that, if the perpendiculars on a given straight line from two points on the same side of it be equal, the straight line joining the points is parallel to the given line. 42 BOOK I. PROPOSITION 13. The sum of the angles, which one straight line makes with another straight line on one side of it, is equal to two right angles. Let the straight line AB make with the straight line CD, on one side of it, the angles ABC, ABD : it is required to prove that the sum of these angles is equal to two right angles. If the angle ABC be equal to the angle ABD, each of them is a right angle, and their sum is equal to two right angles. Construction. If the angles ABC, ABD be not equal, from the point B draw BE at right angles to CD ; (Prop. 11.) BE cannot coincide with BA ; let it lie within the angle ABD. D Proof. Now the angle CBE is the sum of the angles CBA, ABE; to each of these equals add the angle EBD ; then the sum of the angles CBE, EBD is equal to the sum of the angles CBA, ABE, EBD. Again, the angle DBA is equal to the sum of the angles DBE, EBA ; to each of these equals add the angle ABC ; then the sum of the angles DBA, ABC is equal to the sum of the angles DBE, EBA, ABC. And the sum of the angles CBE, EBD has been proved to be equal to the sum of the same three angles. PROPOSITION 13. 43 Therefore the sum of the angles CBE, EBD is equal to the sum of the angles DBA, ABC. But CBE, EBD are two right angles ; (Constr.) therefore the sum of the angles DBA, ABC is equal to two right angles. "Wherefore, the sum of the angles &c. Corollary. The sum of the four angles, which two intersecting straight lines make ivith one another, is equal to four right angles. EXEECISES. 1. Prove in the manner of Proposition 13 that, if A, B, C, D be four points in order on a straight line, the sum of AB, BD is equal to the sum of ^C, CD. 2. If one of the four angles, which two intersecting straight lines make with one another, be a right angle, all the others are right angles. 3. Prove by the method of superposition that only one perpen- dicular can be drawn to a given straight line from a given point without it. 4. Prove by the method of superposition that, if two right- angled triangles have their hypotenuses equal and two other angles equal, the triangles are equal in aU respects. 5. A given angle BAC is bisected; if CA be produced to G and the angle BA G bisected, the two bisecting lines are at right angles. 44 BOOK I. PROPOSITION 14. 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, these two straight lines are in the sa7ne straight line. At the point B in the straight line AB, let the two straight lines BC, BD, on opposite sides of AB, make the adjacent angles ABC, ABD together equal to two right angles : it is required to prove that BB is in the same straight line with CB. Construction. Produce GB to B. Proof. Because the straight line AB makes with the straight line GBE, on one side of it, the angles ABC, ABE, the sum of these angles is equal to two right angles. (Prop. 13.) But the sum of the angles ABC, ABD is equal to two right angles. Therefore the sum of the angles ABC, ABB is equal to the sum of the angles ABC, ABE. From each of these equals take away the angle ABC ; then the angle ABD is equal to the angle ABE ; therefore the line BD coincides in direction witli BE, and is in the same straight line with CB. Wherefore, if at a poirit ikc. PROPOSITION 14. 45 Definition. Two angles, which are together equal to two right angles, are called supplementary angles, and each angle is said to be the supplement of the other. Two angles, which are together equal to one right angle, are called complementary angles, a7id each angle is said to he the complement of the other. EXERCISES. 1. If E be the middle point of the diagonal AC of a quadri- lateral ABCD, whose opposite sides are equal, B, E, D lie on a straight line. 2. If OA, OB, OG, OD be four straight lines drawn in order from O, such that the angles BOG, BOA are equal and also the angles AOB, GOD, then the lines OA, OG are in the same straight line and also the Hues OB, OD. 3. If it be possible within a quadrilateral ABGD, whose opposite sides are equal, to find a point E such that EA, EC are equal, and EB, ED are equal, then AEG, BED are straight lines. 4. If it be possible within a quadrilateral ABGD, whose opposite sides are equal, to find a point E, such that EA, EB, EC, ED are equal, then the quadrilateral is equiangular. 46 BOOK I. PKOPOSITION 15. If two straight lines intersect, vertically opposite angles are equal. Let the two straight lines AB, CD intersect at E : it is required to prove that the angle AEG is equal to the angle DEB^ and the angle CEB to the angle AED. Proof. The sum of the angles CEA, AED, which AE makes with CD on one side of it, is equal to two right angles. (Prop. 13.) Again, the sum of the angles AED, DEB, which DE makes with ^i^ on one side of it, is equal to two right angles. Therefore the sum of the angles CEA, AED is equal to the sum of the angles AED, DEB. From each of these equals take away the common angle AED; then the angle CEA is equal to the angle DEB. Similarly it may be proved that the angle CEB is equal to the angle AED. Wherefore, if two straight lines &c. PROPOSITION 15. 47 EXEECISES. 1. If tlie diagonals of a quadrilateral bisect one another, oppo- site sides are equal. 2. In a given straight line find a point such that the straight lines, joining it to each of two given points on the same side of the line, make equal angles with it. 3. A, B are two given points; CD, DE two given straight lines : find points P, Q in CD, DE, such that AP, PQ are equally inclined to CD, and PQ, QB equally inclined to DE. 4. A straight line is drawn terminated by one of the sides of an isosceles triangle, and by the other side produced, and bisected by the base: prove that the straight lines thus intercepted between the vertex of the isosceles triangle, and this straight line, are together equal to the two equal sides of the triangle. 48 BOOK I. PROPOSITION 16. An exterior miyle of a triangle is greater than either of the interior opposite angles. Let ABC be a triangle, and let ACD be the exterior angle made by producing the side BG to D : it is required to prove that the angle ACB is greater than either of the interior opposite angles CBA^ BAG. Construction. Bisect AG -^.t E. (Prop. 10.) Draw BE and produce it to F^ making EF equal to EB^ (Prop. 3.) and draw FG. Proof. Because in the triangles AEB, GEF^ AE 'w, equal to GE^ and EB to EF, and the angle AEB is equal to the angle GEF, the triangles are equal in all respects ; (Prop. 4.) therefore the angle BAE (or BAG) is equal to the angle FGE. Now the angle EGD (or AGD) is greater than the angle EGF. Therefore the angle ACD is greater than the angle i^^C. Similarly it can be proved that the angle BGG^ which is made by producing AG and is equal to the angle AGD, is greater than the angle ABG Wherefore, an exterior angle is equal to the angle ADD. (Prop. 5.) Therefore the angle ABD is greater than the angle ACB. But the angle ABC is greater than the angle ABD; therefore the angle ABC is greater than the angle ACB. Wherefore, when two sides &c. PROPOSITION 18. 53 ADDITIONAL PEOPOSITION. The straight lines draicn at right angles to the sides of a triangle at their middle points meet in a point. Let ABC be a triangle and D, E, F the middle points of the sides jBC, CA, AB. Draw EO, FO* at right angles to CA, AB. Draw OA, OB, OC, OD. Because in the triangles AEO, GEO, AE is equal to CE, EO common, and the angle AEO is equal to the angle CEO, the triangles are equal in all respects; (Proi?. 4.) therefore ^0 is equal to CO. Similarly it can be proved that ylO is equal to BO; therefore jBO is equal to GO. Next because in the triangles BOD, COD, BO is equal to CO and BD to CD and OD is common, the triangles are equal in all respects ; (Prop. 8.) therefore the angle BDO is equal to the angle CDO, and 01) is at right angles to BC. Wherefore the straight line drawn at right angles to BC at its middle point D passes through 0, the intersection of the straight lines drawn at right angles to the other two sides at their middle points. EXEKCISES. 1. ABC is a triangle and the angle A i^ bisected by a straight line which meets BC at D; shew that BA is greater than BD, and CA greater than CD. 2. Prove that, if D be any point in the base BC between B and C of an isosceles triangle ABC, AD is less than AB. 3. Prove that, if AB, AG, AD be equal straight lines, and AG fall within the angle BAD, BD is greater than either BC or CD. 4. ABCD is a quadrilateral of which AD is the longest side and BC the shortest ; shew that the angle ABC is greater than the angle ADC, and that the angle BCD is greater than the angle BAD. 5. If the angle C of a triangle be equal to the sum of the angles A and B, the side AB is equal to twice the straight line joining G to the middle point of AB. * "We assume that the straight lines drawn at right angles to CA, AB at E and F meet. (See Exercise 2, page 51.) 54 BOOK L PROPOSITION 19. When two angles of a triangle are unequal, the greater angle has the greater side opposite to it. Let ABC be a triangle, of which the angle ABC is greater than the angle ACB : it is required to prove that the side ^C is greater than the side AB. Proof. AG must be either less than, equal to, or greater than AB. li AC were less than AB, the angle ABC would be less than the angle -4C7i; (Prop. 18.) but it is not ; therefore ^C is not less than AB. li AC were equal to AB, the angle ABC would be equal to the angle ACB ; (Prop. 5.) but it is not ; therefore ^C is not equal to AB. Therefore AC must be greater than AB. Wherefore, when two angles &c. PBOPOSITIOJV 19. 55 We leave it to the student to prove that, while a point P is moving along a straight line XY, the distance OP of the point P from a fixed point O outside the line is decreasing when P is moving towards H the foot of the perpendicular from on the line, and that OP is increasing when P is moving away from H. Assuming the ^J"' truth of this proposition, it follows that OH is less than each of the two straight lines OP^, OP^ where Pj, P.j are two positions of the point P close to H on either side of it. For" this reason we say that OH is a mlTilTmim value of OP. In the same way, if a geometrical quantity vary continuously, its magnitude in a position, where it is greater than in the positions close to it on either side, is called a maximum value. It will be seen that, if a quantity vary continuously, there must be between any two equal values of the quantity at least one maximum or minimum value. EXEKCISES. 1. Prove that the hypotenuse of a right-angled triangle is greater than either of the other sides. 2. The base of a triangle is divided into two parts by the perpen- dicular from the opposite vertex : prove that each part of the base is less than the adjacent side of the triangle. 3. A straight line drawn from the vertex of an isosceles triangle to any point in the base produced is greater than either of the equal sides. 4. If D be any point in the side BG of a triangle ABC, then the greater of the sides AB, AG ia greater than AD. 5. The perpendicular is the shortest straight line which can be drawn from a given point to a given straight line ; and, of any two others, that which makes the smaDer angle with the perpendicular is the shorter. 6. The base of a triangle whose sides are unequal is divided into two parts by the straight line bisecting the vertical angle : prove that the greater part is adjacent to the greater side. 56 BOOK I. PROPOSITION 20. The sum of any two sides of a tria7igle is greater' than tJie third side. Let ABC be a triangle : it is required to prove that the sum of any two sides of it is greater than the third side; namely, the sum of CA, AB greater than BG ', the sum of AB, BC greater than CA ; the sum of BC, CA greater than AB. Construction. Produce any side BA to D, making AD equal to AC. (Prop, 3.) Draw DC. Proof. Because AC i^ equal to AD, the angle ADC is equal to the angle A CD. (Prop. 5.) But the angle BCD is greater than the angle ACD. Therefore the angle BCD is greater than the angle BDC. And because in the triangle BCD, the angle BCD is greater than the angle BDC ; BD is greater than BC. (Prop. 19.) Now because DA is equal to A C, BD, which is the sum of BA, AD, is equal to the sum of CA, AB. Therefore the sum of CA, AB is greater than BC. Similarly it can be proved that the sum of AB, BC is greater than CA ; and that the sum of BC, CA is greater than AB. Wherefore, the sura of any two sides &c. PROPOSITION 20. 57 The result of this proposition enables us to solve a great number of problems, of which the following is a specimen — To find in a given straight line XY a jioint P such that the sum of its distances PA, PB from two given poi7its A, B is a minimum. If the points ^, 7i be on opposite sides of AT, the straight line AB intersects AT in the point required. If ^, 5 be on the same side of AT, draw AH perpendicular to XY; produce AH to C, so that HC is equal to HA. Take any point P in AT. Draw BDC, DA, PA, PB, PC. Then it is easily proved that AP is equal to CP, and AD to CD. Therefore the sum of AP, PB is equal to the sum of CP, PB, and this is a minimum when P coincides with D. (Prop. 20.) Therefore D is the point required. From the diagram it is seen that the angle BDY is equal to the angle CDX, which is equal to the angle ADX. It appears therefore that when the sum of PA, PB is a minimum, the lines PA, PB make equal angles with XY. EXERCISES. 1. Prove that any three sides of any quadrilateral are greater than the fourth side. 2. If D be any point within a triangle ABC, the sum of DA, DB, DC is greater than half the perimeter of the triangle. 3. The sum of the four sides of any quadrilateral is greater than the sum of its two diagonals, 4. In a convex quadrilateral the sum of the diagonals is greater than the sum of either pair of opposite sides. 5. D is the rdiddle point oi BC the base of an isosceles triangle ABC, and E any point in AC. Prove that the difference of BD, DE is less than the difference of AB, AE. 6. The two sides of a triangle are together greater than twice the straight line drawn from the vertex to the middle point of the base. 7. Find in a given straight line a point such that the difference of its distances from two fixed points is a maximum. 58 BOOK I. PROPOSITION 21. If from the ends of the side of a triangle there he drawn two straight lines to a point within the triangle, the sum of these lines is less titan the sum of the other two sides of the triangle, but they contain a greater angle. Let ABC be a triangle; and from B, C, the ends of the side BC, let the two straight lines BJ), CD be drawn to a point D within the triangle : it is required to prove that the sum of BD, DC is less than the sum of BA, AC, but the angle BDC is greater than the angle BAC. Construction. Produce BD to meet AC n.t E. Proof. The sum of the two sides BA, AE oi the tri- angle BAE is greater than the third side BE. (Prop. 20.) To each of these unequals add EC ; then the sum of BA, ^C is greater than the sum of BE, EC. Again, the sum of the two sides CE, ED of the triangle CED is greater than the third side CD. To each of these unequals add DB; then the sum of CE, EB is greater than the sum of CD, DB. And it has been proved that the sum of BA, AC is greater than the sum of BE, EC ; therefore the sum of BA, AC is greater than the sum of BD, DC. Again, the exterior angle BDC of the triangle CDE is greater than the interior opposite angle CED. (Prop. 16.) And the exterior angle CEB of the triangle ABE is greater than the interior opposite angle BAE; therefore the angle BDC is greater than the angle BAC. Wherefore, if from the ends &c. P BO POSITION 21. 59 EXEKCISES. 1. If D be any point within a triangle ABC, the sum of I) A, DB^ DC is less than the perimeter of the triangle and greater than half the perimeter. 2. Prove that the perimeter of a triangle is less than the peri- meter of any triangle which is drawn completely surrounding it. 3. If two triangles have a common base and equal vertical angles, the vertex of each triangle lies outside the other triangle. 4. If from the angles of a triangle ABC, straight lines AOD, BOE, COF be drawn through a point within the triangle to meet the opposite sides, the perimeter of the triangle ABC is greater than two-thirds of the sum of AD, BE, CF. 5. ABD, ACD are two triangles on the same side of AD in which ^C is greater than AB. Prove that, if the angles ABD, ACD be both right angles or be equal obtuse angles, then BD is greater than DC. 60 BOOK I. PROPOSITION 22. 'fo coixstruct a triangle having its sides equal to three given straight Ihies. Let AB, CD, EF be the three given lines : it is required to construct a triangle whose sides are equal to AB, CD, EF, Construction. Produce one of the given lines CD both ways, and cut off CG equal to AB, (Prop. 3.) and DR to EF. With C as centre and CG as radius describe the circle GKL, and with D as centre and Dll as radius describe the circle HKM. Let these circles intersect in K : Drawee, DK: then CKD is a triangle drawn as required. Proof. Because G is the centre of the circle GKL, CK is equal to CG ; and CG is equal to AB. (Constr.) Therefore CK is equal to AB. Again, because D is the centre of the circle HKM, D K is equalto D H ; and DH is equal to EF. (Constr.) Therefore DK is equal to EF. Therefore the three lines KC, CD, DK are equal to the , three AB, CD, EF respectively. Wherefore, the triangle KCD has been constructed having its sides equal to the three given straight lines AB, CD, EF. PROPOSITION 22. 61 It may l?e observed that it is not possible to construct a triangle which shall have its sides equal to any three given straight lines. In Proposition 20 it has been proved that any two sides of a triangle are together greater than the third side. It follows therefore that it is impossible to construct a triangle having its sides equal to three given straight lines, except when the given straight lines are such that amj two of them are greater than the third or the greatest line is less tJian the sum of the other tioo. We see therefore that in this proposition we have to solve a pro- blem, which admits of solution only when the given lines satisfy a certain condition. We shall meet with many other problems in which the geometrical quantities given in the i)roblem (for that reason generally called the data), must satisfy some condition in order that the problem may admit of solution. It will be a useful exercise for the student to investigate such conditions when they exist. EXEECISES. 1. Prove that the two circles drawn in the construction of Pro- position 22 will always intersect, provided that the sum of any two of the given straight lines is greater than the third. 2. How many different shaped triangles could be made of 8 dif- ferent lines whose lengths are respectively 2, 2, 2, 3, 3, 4, 4, 5 inches? 3. Construct a right-angled triangle, having given the hypotenuse and one side. 4. Construct a quadrilateral equal in all respects to a given quadrilateral. 62 BOOK I. PROPOSITION 23. From a given point in a given straight line to draw a straight line making with the given straight line an angle equal to a given angle. Let ABC be the given straight line, B the given point in it, and DBF the given angle : it is required to draw from B a straight line making with ABC an angle equal to the angle DEF. Construction. In ED^ EF take any points G^ H^ and draw GH. From BC cut oiff BK equal to EH, (Prop. 3.) and construct the triangle LBK, having the side BK equal to EH, BL equal to EG, and KL equal to HG : (Prop. 22.) then BL is a straight line drawn as required. Proof. Because in the triangles BLK, EGH, KB is equal to HE, BL to EG, and LK to GH, the triangles are equal in all respects ; (Prop. 8. ) therefore the angle KBL (or CBL^ is equal to the angle HEG (or FED). "Wherefore, from the given point B in the given straight line ABC a straight line BL has been drawn 7naking with the straight line ABC an angle KBL equal to the given angle FED. PROPOSITION 23. 63 EXEKCISES. 1. Construct a triangle, having given the base and each of the angles at the base. 2. Make an angle double of a given angle. 3. If one angle of a triangle be equal to the sum of the other two, the triangle can be divided into two isosceles triangles. 4. Construct a triangle, having given the base, one of the angles at the base, and the sum of the sides. 64 BOOK I. PROPOSITION 24. If two sides of one triangle he equal to two sides of anotJier and the angle contained' hy the two sides of the one he greater than the angle contained hy the two sides of the other, the third side of the one is greater than the third side of the other. Let ABC, DEF be two triangles, in which AB is equal to DE and AC to DF, and the angle BAC is greater than the angle EDF : it is required to prove that the third side BC is greater than the third side EF. Construction. Of the two sides DE, DF let DF be one which is not less than the other. From the point D in the straight line DE, draw DC making with BE the angle EDG equal to the angle BAC, (Prop. 23.) and make DG equal to DF. (Prop. 3.) Draw EG meeting DF in //. Proof. Because DF is. not less than DE, and DG is equal to DF, DG is not less than DE. And because in the triangle DEG, DG is not less than DE, the angle DEG is not less than the angle DGE. (Props. 5 and 18.) Next, because DHG is the exterior angle of the triangle i>-E'^, it is greater than the interior opposite angle DEG. (Prop. 16.) Therefore the angle DUG is greater than the angle DGH. And because in the triangle DHG, the angle DHG is greater than the angle DGH, DG is greater than DH. (Prop. 19.) PROPOSITION 24. 65 But DG is equal to DF. Therefore DF is greater than DH, or the point F lies outside the triangle DEG. Next because the sum of DH, HG, two sides of the triangle DHG, is greater than the third side DG, and the sum of FII, HE, two sides of the triangle ElIF^ is greater than the third side EF ; the sum of DH, HG, FH, HE is greater than the sum of DG, EF; i.e. the sum of DF, EG is greater than the sum of DG, EF. Take away the equals DF, DG ; then EG is greater than EF. Now the triangles EDG, BAG are equal in all respects. (Prop. 4.) Therefore JBC, which is equal to EG, is greater than EF. Wherefore, if two sides (fee. EXEECISES. A point P moves along the circumference of a circle from one extremity A of a, diameter AB to the other extremity B ; prove that throughout the motion (a) AP is increasing and BP is decreasing ; (b) if be any point in AB nearer A than i>, OP is increasing ; (c) if be any point in BA produced, OP is increasing. T. E. BOOK I PROPOSITION 25. If two sides of one triangle be equal to two sides of another^ and the third side of the one he greater than the third side of the other, the angle opposite to the third side of the one is greater than the angle opposite to the third side of the other. Let ABC, DEF be two triangles, in which AB i's, equal to DE, and AG to DF, and BC is greater than EF : it is required to prove that the angle BA C is greater than the angle EDF. Proof. The angle BAG must be either greater than, equal to, or less than the angle EDF. If the angle BAG were equal to the angle EDF, BG would be equal to EF', (Prop. 4.) but it is not ; therefore the angle BAG is not equal to the angle EDF. Again, if the angle BAG were less than the angle EDF, BG would be less than EF; (Prop. 24.) but it is not ; therefore the angle BAG is not less than the angle EDF. Therefore the angle BAG is greater than the angle EDF. Wherefore, if two sides &c. PROPOSITION- 25. 67 EXEKCISES. 1. If D be the middle point of the side BC oi & triangle ABC, in which ^O is greater than AB, the angle ADC is an obtuse angle. 2. If in the sides AB, AC of a triangle ABC, in which AC is greater than AB, points D, E be taken such that BD, CE are equal, CD is greater than BE. 3. If in the sides AB, AC produced of a triangle ABC, in which AC IB greater than AB, points D, E be taken such that BD, GE are equal, BE is greater than CD. 4. If in the side AB and the side AC produced of a triangle ABC points D and E be taken, such that BD, CE are equal, BE is greater than CD. 5—2 68 BOOK I. PROPOSITION 26. Part 1. If two triangles have two angles of the one equal to tioo angles of the other ^ and the side adjacent to the angles in the one equal to the side adjacent to the angles in the other, the triangles are equal in all respects. Let ABC, DEF be two triangles, in which the angle ABC is equal to the angle DEF, and the angle BCA is equal to the angle EFD, and the side BG adjacent to the angles ABC, BCA is equal to the side EF adjacent to the angles DEF, EFD : it is required to prove that the triangles ABC, DEF are equal in all respects. Proof. Because the sides BC, EF are equal, it is possible to shift the triangle ABC, so that BC coincides with EF, B with E and C with F, (Test of Equality, page 5.) and the triangles are on the same side of EF. If this be done, because BC coincides with EF, and the angle ABC is equal to the angle DEF, BA must coincide in direction with ED. Similarly it may be proved that CA must coincide in direction with FD. Therefore the point A, which is the intersection of BA, CA, must coincide with D, which is the intersection of ED, FD. Next, because A coincides with D, and B with E, AB must coincide with DE. (Post. 2.) Similarly AG must coincide with DF. PROPOSITION 26. PART 1. 69 Therefore the triangle ABC coincides with the triangle DEF, and is equal to it in all respects. Wherefore, if two triangles &c. EXEKCISES. 1. If AD be the bisector of the angle BAG, and BDG be drawn at right angles io AB, AB is equal to AG. 2. AB, AG are any two straight lines meeting at A : through any point P draw a straight line meeting them at E and F, such that AE may be equal to AF. 3. If upon the same base AB two triangles BAG, ABB be con- structed, having the angle BAG equal to ABB, and ABG equal to BAD, then the triangles BDG, AGD are equal in all respects. 4. If the opposite sides of a quadrilateral be equal, the diagonals bisect each other. 5. If the straight line bisecting the vertical angle of a triangle be at right angles to the base, the triangle is isosceles. 70 BOOK I. PROPOSITION 26. Part 2. If two triangles have two angles of the one equal to two angles of the other ^ and the sides opposite to a pair of equal angles equal, the triangles are equal in all respects. Let ABC, DEF be two triangles, in which the angle ABC is equal to the angle DEF, and the angle BOA equal to the angle EFD, and BA the side opposite to the angle BCA is equal to ED the side opposite to the angle EFD : it is required to prove that the triangles ABC, DEF are equal in all respects. Proof. Because the sides AB, DE are equal, it is possible to shift the triangle DEF, so that DE coincides with AB, D with A and E with By and the triangles are on the same side of AB. If this be done, because ED coincides with BA, and the angle DEF is equal to the angle ABC, EF must coincide in direction with BC. Now F cannot coincide with any point G in BC, since the angle AGB the exterior angle of the triangle AGO is greater than the interior and opposite angle ACB, (Prop. 16.) which is equal to the angle DFE. Again F cannot coincide with any point H in BC produced, since the interior and opposite angle AHB of the triangle ACH is less than the exterior angle ACB, (Prop. 16.) which is equal to the angle DFE. PROPOSITION 26. PART 2. 71 Therefore F must coincide with 6', EF with BC, and DF with ^(7; therefore the triangle DEF coincides with the triangle ABC, and is equal to it in all respects. Wherefore, if two triangles kc. ADDITIONAL PEOPOSITION. The straight lines, which bisect the angles of a triangle, meet in a point. Let ABC be a triangle. Bisect the angles ABC, BCA by the straight lines BI, CI*. Draw IL, IM, IN perpendicular to the sides. Because in the triangles IBN, IBL the angle IBN is equal to the angle IBL, and the angle INB to the angle ILB, and BI is common, the triangles are equal in all respects : (Prop. 26, Part 2.) therefore IN is equal to IL. Similarly it can be proved that IM is equal to IL : therefore IN is equal to 131. Next because in the right-angled triangles IAN, lAM the hypotenuse lA is common, and IN is equal to IM, the triangles are equal in all respects : (Exercise 5, page 49.) therefore the angle IAN is equal to the angle I AM, and I A is the bisector of the angle BAG. Therefore the bisector of the angle BA C passes through the intersection of the bisectors of the angles ABC, BCA. EXERCISES. 1. The perpendiculars let fall on two sides of a triangle from any point in the straight line bisecting the angle between them are equal to each other. 2. In a given straight line find a point such that the perpendicu- lars drawn from it to two given straight lines which intersect are equal. 3. Through a given point draw a straight line such that the per- pendiculars on it from two given points may be on opposite sides of it and equal to each other. * It is assumed that these lines intersect. 72 BOOK I. PROPOSITION 26 A. If two triangles have two sides equal to two sides, and the angles opposite to 07ie p)ciir of equal sides equals the angles opposite to the other pair are either equal or supplementary. Let ABC, DEF be two triangles, in which AB is equal to DE, and BC to EF, and the angle BAG is equal to the angle EDF : it is required to prove that the angles ACB, DEE are either equal or supplementary. (Def. page 45.) Of the two sides AG, DE, let AG h^ not greater than BE. B Fig. 1. Proof. Because the sides AB, DE are equal, it is possible to shift the triangle ABC, so that AB coincides with DE, A with D and B with E, (Test of Equality, page 5.) and so that the two triangles ABC, DEF are on the same side of DE. If this be done, because AB coincides with DE, and the angle BAG is equal to the angle EDF, AG must coincide in direction with DE. Because AG is, not greater than DE, G must coincide either (1) with F or (2) with G some point iuDF. (Fig 1.) If G coincide with F, then BC coincides with EF, (Post. 2.) and the triangle ABC with the triangle DEF, and the two triangles are equal in all respects ; therefore the angle A GB is equal to the angle DFE. (Fig. 2.) Again, if G coincide with G, because BC is equal to EG, and EF is equal to BC, EG is equal to EF, PROPOSITION- 26 A. 73 And because in the triangle EFG, EG is equal to EF, the angle EFG is equal to the angle EGF. (Prop. 5.) Now the angles DGE, EGF are together equal to two right angles, i.e. are supplementary; (Prop. 13.) therefore the angles DGE^ EFG are supplementary ; and the angle DGE is equal to the angle AGB; therefore the angles ACB, DFE are supplementary. Wherefore, if two triangles &c. Corollary. When two triangles have two sides equal to two sides, and the angles opposite to one pair of equal sides equal to one another, they are equal in all respects, provided that of the angles opposite to the second pair of equal sides, (1) each be less than a right angle, (2) each be greater than a right angle, or (3) one of them be a right angle. EXEKCISES. 1. If the straight line bisecting the vertical angle of a triangle also bisect the base, the triangle is isosceles. 2. If two given straight lines intersect, and a point be taken equally distant from each of them, it lies on one or other of the two straight lines which bisect the angles between the given straight lines. 3. Prove that two right-angled triangles are equal in all respects, if the hypotenuse and a side of the one be respectively equal to the hypotenuse and a side of the other. 4. If two exterior angles of a triangle be bisected, and from the point of intersection of the bisecting lines a straight line be drawn to the third angle, it bisects that angle. 5. If two triangles have two sides equal to two sides, and the angles opposite to the greater sides equal, the triangles are equal in all respects. 6. Construct a triangle having given two sides and the angle opposite to one of them. Is this always possible ? 74 BOOK I. On Equal Triangles. It is in many cases convenient to denominate the sides BC, CA, AB oi a triangle ABC by the small letters a, 6, c respectively. Here a, b, c stand for the sides of the triangle opposite to the angles A, B, C respectively. Using this notation we may sum up the results of Propositions 4, 8, 26 Part 1, 26 Part 2, and 26 A as follows : Two triangles ABC^ A'B'C are equal in all respects, (I) if a = a\ h^h\ and C=C", (Prop. 4.) (II) if a = a', h = h', and c = c\ (Prop. 8.) (III) if ^ = ^', B = B', and c = c', (Prop. 26, Part 1.) (IY)if^=^', B = B\s,iid a = a\ (Prop. 26, Part 2.) (Y) a a = a'j b = b', and A = A'j and if in addition (1) -5 and B' be each less than a right angle, or (2) B and B' be each greater than a right angle, or (3) either B or B' be a right angle. (Prop. 26 A.) The six quantities, the angles A, B, C, and the sides a, b, c, are often denominated the parts of the triangle ABC. It will be observed that the equality of three pairs of parts is always required to ensure the equality in all respects of two triangles, but that the equality of three pairs of parts is not always sufficient. By the theorem of Proposition 32 it can be shewn that, if any two of the equations A = A', B = B\ C = C\ be true, the third is also true : from this we conclude that the set of equations A=A', B^B\ C =-C\ is insufficient to deter- mine the equality of the triangles, and that the two cases III. and IV. are virtually the same. 75 On the angles made by one straight line with two OTHERS. When a straight line ABGD intersects two other straight lines EBF, GO II, the angles ABE, ABF, DCG, BCH outside the two lines EFj Gil are called exterior angles ; the angles GBB, CBF, BCG, BCH inside the two lines FF, GH are called interior angles ; a pair of interior angles on opposite sides of ABGD are called alternate angles. There are two pairs of alternate angles in the diagram, EBC, BCH; CBF, BCG. A pair of angles, one at B and the other at C, one exterior and the other interior, on the same side of ABGD are sometimes called corresponding angles. There are four pairs of corresponding angles, ABF, BCH; ABE, BCG ; BCH, CBF; and DCG, CBE, the first angle in each pair being an exterior angle, and the second the interior. 76 BOOK 1. PROPOSITION 27. If a straight line, meeting two other straight lines in the same plane, make two alternate angles equal, the two straight lines are parallel. Let the straight line EF, meeting the two straight lines AB, CD in the same plane, make the alternate angles AEi\ EFD equal to one another : it is required to prove that AB, CD are parallel. Proof. AB, CD cannot meet when produced beyond B and D ; for if they did, the exterior angle AEF of the triangle formed by them and EF would be greater than the interior opposite angle EFD (Prop. 16.); but it is not. Similarly it can be proved that AB, CD cannot meet when produced beyond A and C. But those straight lines in the same plane which do not meet however far they may be produced both ways, are parallel. (Del 9.) Therefore AB, CD are parallel. Wherefore, if a straight lifie &c. PROPOSITION 27. 77 EXEKCISES. 1. No two straight lines drawn from two angles of a triangle and terminated by the opposite sides can bisect one another. 2. Two straight lines at right angles to the same straight line are parallel. 3. Prove Proposition 27 by the method of superposition. 78 BOOK L PROPOSITION 28. If a straight line intersecting two other straight lines, 'make an exterior angle equal to the interior and opposite angle on the same side of the line ; or if it make two interior angles on the same side together equal to two right angles, the two straight lines are parallel. Let the straight line EF, intersecting the two straight lines AB, CD, (1) make the exterior angle EGB equal to the interior and opposite angle on the same side GHD, or (2) make the interior angles on the same side BGH, GHD together equal to two right angles : it is required to prove that AB, CD are parallel. Proof. (1) Because the angle EGB is equal to the angle GHD, and the angle EGB is equal to the angle AGH, (Prop. 15.) the angle AGH is equal to the angle GHD ; and they are alternate angles ; therefore AB, CD are parallel. (Prop. 27.) (2) Because the angles BGH, GHD are together equal to two right angles, and the angles AGH, BGH are together equal to two right angles, (Prop. 13.) the angles AGH, BGH are together equal to the angles BGH, GHD. Take away the common angle BGH; then the angle AGH is equal to the angle GHD ; and they are alternate angles ; therefore AB, CD are parallel. (Prop. 27.) Wherefore, if a straight line &c. PROPOSITION 28. 79 EXEECISES. 1. If a straight line intersecting two other straight lines make two external angles on the same side of the line together equal to two right angles, the two straight lines are parallel. 2. If a straight line intersecting two other straight lines make two external angles on opposite sides of the line equal, the two straight lines are parallel. 3. If a straight line intersecting two other straight lines make two corresponding angles equal, the two straight lines are parallel. 80 BOOK I. PROPOSITION 29. If a straight line intersect two parallel straight lines, it makes alternate angles equal, it makes each exterior angle equal to the interior and opposite angle on the same side of the liiie, and it also makes interior angles on the same side together equal to two right angles. Let the straight line EF intersect the two parallel straight lines AB, CD : it is required to prove that (1) the alternate angles AGff, GHD are equal, (2) the exterior angle EGB is equal to the interior and opposite angle GHD on the same side of EF, and (3) the two interior angles BGH, GHD on the same side of EF are together equal to two right angles. Proof. (1) Because AGTT, BGII are the angles which EF makes with AB on one side of it, the sum of the angles AGH, BGII is equal to two right angles. (Prop. 13.) Therefore, if the angles AG II, GHD were unequal, the sum of the angles BGH, GHD would not be equal to two right angles ; and since these are the interior angles which the straight lines AB, CD make with EF on one side of it, AB, CD would not be parallel. (Post. 9, page 51.) But AB, CD are parallel ; therefore the angle A Gil is equal to the angle GHD. (2) But the angle AGH is equal to the angle EGB; (Prop. 15.) therefore the angle EGB is equal to the angle GHD. (3) Add to each of the equal angles EGB, GHD the angle BGII; PROPOSITION 29. 81 then the angles EGB^ BGH are together equal to the angles BGH, GHD. But the angles EGB, BGH are together equal to two right angles. Therefore the angles BGH^ GHD are together equal to two right angles. Wherefore, if a straight line (fee. Corollary. All the angles of a rectangle are right angles. (See Def. 19.) EXEKCISES. 1. Any straight line parallel to the base of an isosceles triangle makes equal angles with the sides. 2. If through any point equidistant from two parallel straight lines, two straight lines be drawn cutting the parallel straight lines, they will intercept equal portions of these parallel straight lines. 3. If the straight line bisecting an exterior angle of a triangle be parallel to a side, the triangle is isosceles. 4. If DE, DF drawn from D any point in the base ^C of an isosceles triangle ABC, to meet AB, AG in E, F he parallel to AG, AB, the perimeter of the parallelogram AEDF is constant. T. E. 82 BOOK I. PROPOSITION 30. Straight lines parallel to the same straight line are parallel to each other. Let each of the straight lines AB, CD be parallel to EF : it is required to prove that AB^ CD are parallel to one another. Construction. Draw a straight line GHK intersecting AB, CD, EFin G, H, K respectively. Proof. Because GHK intersects the parallels AB, EF^ the angle GKF is equal to the angle AGH. (Prop. 29.) Again, because GK intersects the parallels CD, EF, the angle GHD is equal to the angle GKF. (Prop. 29.) Therefore the angle A GH is equal to the angle GHD ; and they are alternate angles ; therefore AB is parallel to CD. (Prop. 27.) Wherefore, straight lines &c. PROPOSITION 30. 83 EXERCISES. 1. Two intersecting straight lines cannot both be parallel to the same straight line. 2. Only one straight line can be drawn through a given point parallel to a given straight line. » 3. If two straight lines, each of which is parallel to a third straight line, meet, the two lines are coincident throughout their length. 4. If a straight line intersect one of two parallel straight lines, it must intersect the other. G— 2 84 BOOK L PROPOSITION 31. To draw through a given point a straight line parallel to a given straight line. Let A be the given point, and BC the given straight line: it is required to draw through A a straight line parallel to BO, Construction. In BO take any point D, and draw AD ; from the point A in the straight line ^i> on the side of AD remote from draw A£J making the angle DAE equal to the angle ADO ; (Prop. 23.) and produce the straight line UA to F : then EF ig the straight line required. Proof. Because the straight line AD meets the two straight lines BO, FF, and makes the alternate angles FAD, ADO equal, FF is parallel to BO. (Prop. 27.) Wherefore, the straight line FAF has been drawn through the given point A, parallel to the given straight line BO. PROPOSITION 31. 85 EXERCISES. 1. Find a point 5 in a given straight line CD, such that, if AB be drawn to B from a given point Ay the angle ABC will be equal to a given angle. 2. Draw through a given point between two intersecting straight lines a straight line so that it is bisected at the point. 3. ABCD is a quadrilateral having BG parallel to AD ; shew that its area is the same as that of the parallelogram which can be formed by drawing through the middle point of DC a straight line parallel to AB. 4. AG, BG are two given straight lines: it is required to draw a straight line from a given point P to AC, so that it is bisected hyBG. 5. Construct a triangle having given two angles, and the length of the perpendicular from the third angle on the opposite side. 6. Construct a right-angled triangle, having given one side and the angle opposite. 86 BOOK I. PROPOSITION 32. An exterior angle of a triangle is equal to the sum of the two interior opposite angles ; and the sum of the three interior angles of a triangle is equal to two right angles. Let ABC be a triangle : it is required to prove that (1) the exterior angle ACD made by producing the side BG is equal to the sum of the two interior opposite angles CAB^ ABC, and (2) the sum of the three interior angles ABC^ BCA^ CAB is equal to two right angles. Construction. Through the point C draw CE parallel to^^. (Prop. 31.) Proof. (1) Because AC meets the parallels BA, CE, the alternate angles BAC, ACE are equal. (Prop. 29.) Again, because BD meets the parallels BA, CE, the exterior angle ECD is equal to the interior opposite angle ABC. (Prop. 29.) And the angle ACE was proved to be equal to the angle BAC] therefore the whole angle ACD is equal to the sum of the two angles CAB, ABC. (2) To each of these equals add the angle BCA ; then the sum of the angles ACE, ACB is equal to the sum of the three angles ABC, BCA, CAB. But the sum of the angles ACE, ACB is equal to two right angles; (Prop. 13.) therefore also the sum of the angles ABC, BCA, CAB is equal to two right angles. Wherefore, An exterior angle &lq. FEOPOSITIO.Y 32. 87 Corollary. 2%e sum of the interior angles of any con- vex rectilineal figure of n sides is less hy four right angles than 2n right angles. This may be proved in either of the following ways : In fig. (1), where straight lines are drawn from any point within the figure to the vertices, the angles of the n triangles so formed are equal to the angles of the figure together with the angles at 0, which are equal to four right angles. In fig. (2), where all the diagonals from one vertex E are drawn, the angles of the w - 2 triangles so formed are together equal to the angles of the figure. B (2) D EXEKCISES. 1. Straight lines AD, BE, CF are drawn within the triangle ABC making the angles DAB, EBC, FGA all equal to one another. If AD, BE, CF do not meet in a point, the angles of the triangle formed by them are equal to those of the triangle ABC. 2. Trisect a right angle. 3. Trisect a quarter of a right angle. 4. If A be the vertex of an isosceles triangle ABC, and BA be produced to D, so that AD is equal to BA, and DC be drawn: then BCD is a right angle. 5. A straight line drawn at right angles to BC the base of an isosceles triangle ABC cuts AB in D and CA produced in E: prove that AED is an isosceles triangle. 6. Construct a right-angled triangle having given the hypotenuse and the sum of the sides. 7. The line joining the right angle of a right-angled triangle to the middle point of the hypotenuse is equal to half the hypotenuse. 8. The locus of the vertices of all right-angled triangles which have a common hypotenuse is a circle. 88 BOOK I. PROPOSITION 33. If two sides of a convex quadrilateral he equal and parallel, the other sides are equal and parallel. Let ABDG be a quadrilateral, in which the sides AB, CD are equal and parallel : it is required to prove that the sides AC, BD are equal and parallel. Construction. Draw one of the diao^onals BC. Proof. Because ^^ is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal. (Prop. 29.) Because in the triangles ABC, DCB, AB is equal to DC, and BC to CB, and the angle ABC to the angle DCB, the triangles are equal in all respects; (Prop. 4.) therefore the angle ACB is equal to the angle DBC, and CA to BD. And because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, ACis parallel to BD. (Prop. 27.) And it was proved to be equal to it. Wherefore, if two sides &c. PROPOSITION 33. 89 EXEECISES. 1. Draw a straight line so that the part intercepted between two given straight lines is equal to one given straight line and parallel to another. 2. If a quadrilateral have two of its opposite sides parallel, and the two others equal but not parallel, any two of its opposite angles are together equal to two right angles. 3. If a straight line which joins the extremities of two equal straight lines, not parallel, make the angles on the same side of it equal to each other, the straight line which joins the other extremities will be parallel to the first. 4. If from D any point in the base BC of an isosceles triangle ABC, BE, DF be drawn perpendicular to the sides, then the sum of DE, DF is constant. 90 BOOK I. PROPOSITION 34. Opposite sides of a parallelogram are equal, and opposite angles are equal ; and a diagonal of a parallelogram bisects its area. Let ACDB be a parallelogram, of which BC is a diagonal : it is required to prove that (1) opposite sides are equal, AB to CD, and AC to BD ; (2) opposite angles are equal, B AC to BDC and A BD to ACD ; aiid (3) the diagonal BC bisects the area of the parallelogram. Proof. Because ^^ is parallel to CJJ, and BC meets them, the alternate angles ABC, BCD are equal. (Prop. 29.) And because ^C is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal. (Prop. 29.) Now because in the two triangles ABC, DCB, the angle ABC is equal to the angle DCB, and the angle BCA to the angle CBD, and the side BC adjacent to the equal angles in each is common to both, the triangles are equal in all respects. (Prop. 26, Part 1.) Therefore ^^ is equal to DC, AC equal to DB, and the angle BAC equal to the angle CDB. And because the angle ABC is equal to the angle DCB, and the angle CBD to the angle BCA, the whole angle ABD is equal to the whole angle DCA, And the angle BAC has been proved to be equal to the angle CDB. Therefore in the parallelogram AD (1) opposite sides are equal and (2) opposite angles are equal. PROPOSITION 34. 91 Again, it has been proved that the triangles ABC, DCB are equal in all respects: therefore (3) the diagonal BC bisects the area of the paral- lelogram AD. Wherefore, opposite sides tkc. Corollary 1. All the sides of a square are equal. Corollary 2. The angles made hy a pair of straight lines are equal to the angles made hy any pair of straight lines parallel to them. A parallelogram ABCD is often spoken of as the parallelogram AG, or the parallelogram BD, or more simply as -40 or BD, when there is no danger of confusion with the diagonal ^C or with the diagonal BD. EXERCISES. 1. Prove that, if the diagonals of a quadrilateral bisect one another, the quadrilateral is a parallelogram. Prove also the converse. 2. If two sides of a quadrilateral be parallel and the other two equal but not parallel, the diagonals are equal. 3. If in a quadrilateral the diagonals be equal and two sides be parallel, the other sides are equal. 4. Find in a side of a triangle the point from which straight lines drawn parallel to the other sides of the triangle and terminated by them are equal. 5. Prove that every straight line which bisects the area of a parallelogram must pass through the intersection of its diagonals. 6. Construct a triangle whose angles shall be equal to those of a given triangle, and whose area shall be four times the area of the given triangle. 7. ABCD is a parallelogram having the side ^D double oiAB: the side ^jB is produced both ways to E and F till each produced part equals AB, and straight lines are drawn from G and D to E and F so as to cross within the figure : shew that they will meet at right angles. 8. If be any point within a parallelogram ABGD, the sum of the triangles OAB, OCD is half the parallelogram. 9. Divide a given straight line into n equal parts, where n is a whole number. 92 BOOK I. PROPOSITIOISr 35. Two parallelograms^ which have one side common and the sides opposite to the common side in a straight line, are equal in area. Let A BCD, EBGF be two parallelograins, which have a common side BC, and the sides AD, EF in a straight' line : it is required to prove that ABGD, EBGF are equal in area. A B D F A D :E fa DE F Peoof. Because ABGD is a parallelogram, AB is equal to DG, (Prop. 34.) and because EBGF is a parallelogram, BE is equal to GF ; and because AB is parallel to DG, and BE to GF, the angle ABE is equal to the angle DGF. (Prop. 34, Coroll. 2.) And because in the triangles ABE, DGF, AB\^ equal to DG, and BE to GF, and the angle ABE to the angle DGF, the triangles are equal in all respects. (Prop. 4.) Take from the area ABGF, the equal areas FDG, EAB\ then the remainders are equal, that is, the parallelograms ABGD, EBGF are equal in area. Wherefore, two parallelograms &c. PROPOSITION 35. 93 The propositions in the remaining part of the First Book of Euclid and those in the Second Book relate chiefly to cases of equality of the areas of two figures. The test of equality to which we have hitherto always appealed has been that of the possibility of shifting one figure so that it exactly coincides with the other. In this case the figures are equal in all respects, but we say that two figures are equal in area also, when it is possible to shift all the parts of the area of one figure, so that they together exactly fit the area of the second figure. It will be observed that this is the test made use of in Pro- position 35. For the future we shall often, when there is no danger of ambiguity, speak of the equality of two figures when we mean only equality of area, and we shall often speak of a figure when we mean only the area of the figure. EXEKCISES. 1. Construct a rectangle equal to a given parallelogram. 2. Construct a rhombus equal to a given parallelogram. 3. Construct a parallelogram to be equal to a given parallelogram in area and to have its sides equal to two given straight lines. Is this always possible ? 94 BOOK I. PROPOSITION 36. Two j^cLt'cdl^logramSj which have two sides equal and in a straight lirie and also have the sides opposite to the equal sides in a straight line, are equal. Let A BCD, EFGH be two parallelograms, which have their sides BC, FG equal and in a straight line, and also their sides AD, EH in a straight line : it is required to prove that ABCD, EFGH are equal. Construction. Draw BE, GH. Proof. Because BC is equal to FG, and FG to EH, (Prop. 34.) BC is equal to EH; and they are parallel. Because the two sides BC, EH of the convex quadrilateral EBCH are equal and parallel, the other sides BE, CH are equal and parallel ; (Prop. 33.) therefore EBCH is a parallelogram. Now because EBCH and ABCD have the side BC common, and the sides AD, EH in a straight line, EBCH is equal to ABCD. (Prop. 35.) Similarly it can be proved that EBCH is equal to EFGH, Therefore the parallelograms ABCD, EFGH are equal. Wherefore, two parallelograms &c. PROPOSITION 36. 95 ADDITIONAL PROPOSITION. The straight lines, drawn from the vertices of a triangle per- pendicular to the opposite sides, meet in a point*. Let ABC be a triangle, and AL, BM, GN be drawn perpendicular to BG, GA, AB respectively. Draw the straight lines FAE, DBF, BCD parallel to BC, GA, AB respectively. Because BE is a parallelogram, AEis equal to BC ; (Prop. 34.) and because GF is a parallelogram, FA is equal to JSC; therefore FA is equal to AE. Again, because AL meets the parallels FAE, BLG, the angle FAL is equal to the angle ALG. (Prop. 29.) But the angle ^L(7 is a right angle; therefore the angle FAL is a right angle. Therefore AL is the straight line drawn at right angles to FE at its middle point. Similarly it can be proved that BM, GN are the straight lines drawn at right angles to FD, BE at their middle points. Now AL, BM, GN the straight lines drawn at right angles to the sides of the triangle DEF at their middle points meet in a point. (Add. Prop., page 53.) Therefore AL, BM, GN the straight lines drawn from the vertices of the triangle ABG perpendicular to the opposite sides meet in a point. EXERCISES. 1. Construct a parallelogram to be equal to a given parallelogram and to have one of its sides in a given straight line. 2. Construct a parallelogram to be equal to a given parallelogram and to have two of its sides in two given straight lines. * This point is often called the ortliocentre of the triangle. 96 BOOK I. PROPOSITION 37. Two triangles, wliich have one side common and the angular points opposite to the common side on a straight line parallel to it, are equal. Let ABC, DBC be two triangles, which have a common side BC, and their angular points A, D on a, straight line AD parallel to BG : it is required to prove that the triangles ABC, DBC are equal. Construction. Through B draw BE parallel to GA, and through C draw Ci^ parallel to BD, (Prop. 31.) meeting AD (produced if necessary) in B and F. Proof. Because the parallelograms EBGA, DBGF have a common side BG and the sides EA, DF in a straight line, EBGA is equal to DBGF. (Prop. 35.) And because the diagonal AB bisects the parallelogram EBGA, the triangle ABC is half of EBGA ; (Prop. 34.) and because the diagonal DC bisects the parallelogram DBGF, the triangle DBG is half of DBGF. Now the halves of equals are equal. Therefore the triangles ABC, DBG are equal. Wherefore, two triangles &c. PROPOSITION 37. 97 EXEKCISES. 1. If P be a point within a parallelogram ABCD, the difference of the triangles PAB, PAD is equal to the triangle PAC. 2. If P be a point outside a parallelogram ABCD, the sum of the triangles PAB, PAD is equal to the triangle PAC. 3. AB and ECD are two parallel straight lines: BF, BF are drawn parallel to AD, AE respectively: prove that the triangles ^BC, DBF are equal to one another. 4. ABC is a given triangle: construct a triangle of equal area, having AB for base and its vertex in a given straight line. 5. Points A, B, G are taken, one on each of three parallel straight lines: BG, GA, AB meet the lines through A, B, G respec- tively in a, b, c: prove that each of the triangles ABG, Abe, BcUy Gab, is equal to half the triangle abc. T. E. 98 BOOK I. PROPOSITION 38. Two triangles^ which have two sides equal and in a straight line and also have the angular points opposite to the equal sides on a straight line parallel to it, are equal. Let ABC, DEF be two triangles, which have their sides BC, EF equal and in a straight line, and their angular points A, D, on a straight line AD parallel to BF\ it is required to prove that the triangles ABC, DEF are equal. Construction. Through B draw BG parallel to CA, and through F draw FII parallel to ED, meeting AD (produced if necessary) in G and H. G A T) H Proof. Because the parallelograms GBCA, DEFH have their sides BG, EF equal and in a straight line, and also their sides GA, DH in a straight line, they are equal to one another. (Prop. 36.) Because the diagonal AB bisects the parallelogram GBCA, the triangle ABC is half of GBCA ] (Prop. 34.) and because the diagonal DF bisects the parallelogram DEFH, the triangle DEF is half of DEFH. Now the halves of equals are equal ; therefore the triangles ABC, DEF are equal. Wherefore, two triangles &c. Corollary. Two triangles, which have two sides equal and in a straight line and also have the angular points opposite to the equal sides coincident, are equal. PROPOSITION 38. 99 EXEECISES. 1. ABCD is a parallelogram; from any point P in the diagonal BD the straight lines PA, PC are drawn. Shew that the triangles PAB and PGB are equal in area. 2. The three sides of a triangle are bisected, and the points of bisection are joined; prove that the triangle is divided into four triangles, which are all equal to one another. 3. If the sides BG, CA, AB of a triangle ABC be produced to A', B', C respectively, so that CA' = BC, AB' = CA, AB = BC', prove that the area of the triangle A'B'C is seven times that of the triangle ABC. 4. Make a triangle such as to be equal to a given parallelogram, and to have one of its angles equal to a given angle. 5. If the sides AB, BC, CA of a triangle ABC be respectively bisected in c, a, h, and Aa, Cc intersect in P: then BPb is a straight line. 6. The sides AB, AC oi a triangle are bisected in D, E: CD, BE intersect in F. Prove that the triangle BFC is equal to the quadrilateral ADFE. 7. If AB, PQRS, CD be three parallel straight lines and^' P, Q, R, S be situate on AC, AD, BC, BD respectively, then PQ is equal to RS, and PR to QS. 8. A', B', C are the middle points of the sides of the triangle ABC, and through A, B, C are drawn three parallel straight lines meeting B'C, CA', A'B' va. a, h, c respectively; prove that the tri- angle ahc is half the triangle ABC and that he passes through A, ca through B, ah through G. 7—2 100 BOOK I. PROPOSITION 39. If two equal triangles have a common side and lie on the same side of it, the angular points opposite to the common side lie on a straight line parallel to it. Let ABC, BBC be two equal triangles, which have a common side BG, and lie on the same side of BC : it is required to prove that the angular points A, D oppo- site to the side BC lie on a straight line parallel to BC. Construction. Draw AD, and in BD or BD produced take any point E other than D, and draw AE, EC. Proof. Because the triangle DBC is not equal to the triangle EBC, and the triangle ABC is, equal to the triangle DBC, the triangle ABC is not equal to the triangle EBC. If AE were parallel to BC, the triangle A BC would be equal to the triangle EBC ; (Prop. 37.) but they are not equal ; therefore ^^ is not parallel to BC. But it is possible to draw a straight line through A parallel to BC; (Prop. 31.) therefore AD is parallel to BC. Wherefore, if two equal triangles &c. PROPOSITION 39. 101 Additional Proposition. Each side of a triangle is double of the straight line joining the middle points of the other sides and is parallel to it. Let ABC be a triangle, D, E, F the middle points of the sides BG, GA, AB. Draw BE, GF, EF, FD, DE. Because the two triangles BFG, AFG have their sides BF, AF equal and in a straight line, and the point G common, the triangles are equal ; (Prop. 38, Coroll.) therefore the triangle BFG is half of the triangle ABG, Similarly it can be proved that the triangle BEG is half of the triangle ABG. Therefore the triangle BFG is equal to the triangle BEG. Next because the triangles BFG, BEG are equal and have a common side BG, the straight line FE joining their vertices is parallel to BG. (Prop. 39.) Similarly it can be proved that DF is parallel to GA and ED to AB. Again because BFED is a parallelogram, BD is equal to FE. (Prop. 34.) And because DFEG is a parallelogram, DC is equal to FE: therefore BG is double of FE. EXERCISES. 1. The middle points of the sides of any quadrilateral are the angular points of a parallelogram. 2. Of equal triangles on the same base, the isosceles triangle has the least perimeter. 3. Two triangles of equal area stand on the same base and on opposite sides: shew that the straight line joining their vertices is bisected by the base or the base produced. 4. The triangle ABG is double of the triangle EBG: shew that, if AE, BG produced if necessary meet in D, then AE is equal to ED. 5. If the straight lines joining the middle points of two of the sides of a triangle to the opposite vertices be equal, the triangle is isosceles. 102 BOOK L PROPOSITION 40. If two equal triangles have two sides equal and in a straight line, and if the triangles lie on the same side of this line, the angular points opposite to the equal sides lie on a straight line parallel to the first straight line. Let ABC, DEF be two equal triangles, which have equal sides BC, EF in a straight line and lie on the same side of /ii^: it is required to prove that the angular points A, D oppo- site to BC, EF lie on a straight line parallel to BF. Construction. Draw AD, and in ED or ED produced take any point G other than D, and draw AG, GF. Proof. Because the triangle DEF is not equal to the triangle GEF, and the triangle ABC is equal to the triangle DEF, the triangle ABC is not equal to the triangle GEF. Ji AG were parallel to BF, the triangle ABC would be equal to the triangle GEF ; (Prop. 38.) but they are not equal ; therefore AG is not parallel to BF. But it is possible to draw a straight line through A parallel to BF; (Prop. 31.) therefore AD is parallel to BF. Wherefore, if two equal triangles &c. PROPOSITION 40. 103 Additional Proposition. The straight lines joining the vertices of a triangle to the middle -points of the opposite sides meet in a point* which is for each line the point of trisection further from the vertex. Let ABC be a triangle, and D, E, F he the middle points of the sides BC, CA, AB. Draw BE, GF and let them intersect in G. Bisect BG, CG in M, N, and draw F3I, MN, NE. In the triangle ABC, BG is double of FE and is parallel to it. (Add. Prop., page 101.) In the triangle GBC, BG is double of MN and is parallel to it. Therefore FE is equal and parallel to MN. {Prop. 30.) Therefore FMNE is a parallelogram. (Prop. 33.) Now the diagonals of a parallelogram bisect each other. (Exercise 1, page 91.) Therefore GE is equal to GM, which is equal to MB. Therefore BG is double of GE. Similarly GG is double of GF. Similarly it can be proved that AD passes through G, and that ^6^ is double of GD. EXEECISES. 1. A point P is taken within a quadrilateral ABCD: prove that, if the sum of the areas of the triangles PAB, PCD be independent of the position of P, ABCD is a parallelogram. 2. The locus of a point P such that the sum of the areas of the two triangles PAB, PBC is constant, is a straight line parallel to AC. 3. AB, CD are two given straight lines: the locus of a point P such that the sum of the two triangles PAB, PCD is constant, is a straight line. 4. Trisect a given straight line. * This point is often called the centre of gravity or the centroid of the triangle. 104 BOOK I. PROPOSITION 41. If a parallelogram and a triangle have a common side, and the angular point of the triangle opposite to the common side lie on the same straight line as the opposite side of the parallelogram^ the panrallelogram is double of the triangle. Let ABCD be a parallelogram and EBG be a triangle which have a common side BO, and let the angular point E of the triangle lie in the same straight line as the side AD of the parallelogram : it is required to prove that the parallelogram ABCD is double of the triangle EBC. Construction. Draw AC. ^ E Proof. Because the triangles ABC^ EBG^ have a common side BC^ and ^^ is parallel to BC^ the triangles ABC^ EBC are equal. (Prop. 37.) And because the diagonal AC bisects the parallelogram ABCD, the parallelogram ABCD is double of the triangle ABC. (Prop. 34.) Therefore the parallelogram ABCD is double of the triangle EBC. Wherefore, if a parallelogrann &c. PROPOSITION 41. 105 EXERCISES. 1. ABCD is a parallelogram; from D draw any straight line DFG meeting jBC at F and AB produced at G; draw AF and CG : shew that the triangles ABF, CFG are equal. 2. If P be a point in the side AB, and Q a point in the side DC of a parallelogram ABCD, then the triangles PCD, QAB are equal in area. 3. The area of any convex quadrilateral is double that of the parallelogram whose vertices are the middle points of the sides of the quadrilateral. 4. The sides BC, CA, AB of a triangle ABC are trisected in the points D, d; E, e; F, f respectively: prove that the area of the hexagon DdEeFf is two-thirds that of the triangle ABC. 106 BOOK I. PROPOSITION 41 A. To construct a triangle equal to a given rectilineal figure. Let ABGDEFG be the given rectilineal figure : it is required to construct a triangle equal to ABGDEFG, Construction. Draw one of the diagonals AG such that with two adjacent sides of the figure AB^ BC it forms a triangle ABG. Through the vertex B draw BF parallel to GA^ to meet GA produced in P. Draw PC. Proof. Because the triangles FAG^ BAG have a common side AG^ and their angular points F, B on a. straight line parallel to ^C: the two triangles FAG, BAG are equal. (Prop. 37.) Add to each the figure AGDEFG ; then the figure FGDEFG is equal to the figure ABGDEFG. Now the sides of the figure FGDEFG are fewer by one than the sides of the figure ABGDEFG ; therefore by con- tinued application of this process we can construct a series of figures all equal to the given figure, the sides of each figure being fewer by one than the sides of the figure last preceding. We shall thus ultimately obtain a triangle equal to the given rectilineal figure. PROPOSITION 41 A. 107 It will be seen that by the method adopted in Proposition 41 A a triangle can be constructed equal to a given rectilineal figure of 4 sides by using the process once, to a figure of 5 sides by using it twice, and to a figure of n sides by using it n — 3 times. EXEECISES. 1. On one side of a given triangle construct an isosceles triangle equal to the given triangle. 2. On one side of a given quadrilateral construct a rectangle equal to the quadrilateral. 3. Construct a triangle equal in area to a given convex five-sided figure ABODE : AB is to be one side of the triangle and AE the direction of one of the other sides. 4. Bisect a given (1) parallelogram, (2) triangle, (3) quadri- lateral by a straight line drawn through a given point in one side of the figure. 5. ABCD is a given quadrilateral: construct a quadrilateral of equal area, having AB for one side, and another side on a given straight line parallel to AB. 6. ABCD is a given quadrilateral : construct a triangle, whose base shall be in the same straight line as AB, its vertex at a given point P in CD, and its area equal to that of the given quadrilateral. 108 BOOK I. PROPOSITION 42. To construct a parallelogram equal to a given triangle, and having an angle equal to a given angle. Let ABC be the given triangle, and D the given angle : it is required to construct a parallelogram equal to ABC, and having an angle equal to D. Construction. Bisect BC at E : draw AB, and from the point B, in the straight line BC, draw BF making the angle CBF equal to the angle I> ; (Prop. 23.) through A draw AFG parallel to BC meeting BF in F, and through C draw CG parallel to BF meeting A FG in G : then FBCG is a parallelogram constructed as required. Proof. Because the opposite sides of the quadrilateral FBCG are parallel, FBCG is a parallelogram. Because the triangles ABB, ABC have the sides BB, BC equal and in a straight line, and the angular point A common, the triangle ABB is equal to the triangle ABC ; (Prop. 38, Coroll.) therefore the triangle ABC is double of the triangle ABC. Because the parallelogram FBCG and the triangle ABC have a common side BC and the point A lies on the same straight line as the side FG, (Prop. 41.) the parallelogram FBCG is double of the triangle ABC. Therefore the parallelogram FBCG is equal to the triangle ABC, and it has an angle C^^ equal to thfe given angle D. Wherefore a parallelogram FBCG has been constructed equal to the given triangle ABC, and having an angle CBF equal to the given angle J). PROPOSITION 42. 109 EXERCISES. 1. On one side of a given triangle construct a rectangle equal to the triangle. 2. On one side of a given triangle construct a rhombus equal to the triangle. Is this always possible ? 3. On one side of a given triangle as diagonal construct a rhom- bus equal to the triangle. 110 BOOK I. PROPOSITION 43. Complements of parallelograms about a diagonal of a parallelogram, are equal. Let ABCD be a parallelogram, of which AC is a dia- gonal ; and EH, GF are parallelograms about A C ; and KB, KD the complements : (See note on page 111.) it is required to prove that KB is equal to KD. Proof. Because BD is a parallelogram, and AC ^ diagonal, the triangle ABC is equal to the triangle ADC. (Prop. 34.) Now the triangle ABC is equal to the two triangles AEK, KGC and the parallelogram KB ; and the triangle ADC is equal to the two triangles AUK, KFC and the parallelogram KD. Therefore the two triangles AEK, KGC and the parallelo- gram KB are together equal to the two triangles AHK, KFC and the parallelogram KD. Again, because EH is a parallelogram and AK 2^, diagonal, the triangle AEK is equal to the triangle AHK; (Prop. 34.) and because GF is a parallelogram, and KC a diagonal, the triangle KGC is equal to the triangle KFC. (Prop. 34.) Therefore taking away equals from equals, the remainder, the complement KB, is equal to the remainder, the com- plement KD. Wherefore, complements of parallelograms &c. PROPOSITION 43. Ill If through a point ^ on a diagonal ^C of a parallelogram ABC I), straight lines HKG, EKF be drawn parallel to the sides AB, BG respectively to meet the sides AT), BG, AB, DG in H, G, E, F respec- tively ; then EH, OF are called parallelograms about the diagonal AG, and the parallelograms EG, FH are called complements of these parallelograms. EXERCISES. 1. Prove that in the diagram of Proposition 43, the following are pairs of equal triangles: ABK and ADK; AEG and AHG; AKG and AKF. 2. The diagonals of parallelograms about a diagonal of a paral- lelogram are parallel. 3. Parallelograms about a diagonal of a square are squares. 4. If ABGD, AEFG be two squares so placed that the angles at A coincide, then A, F, G lie on a straight line. 5. If through E a point within a parallelogram ABGD straight lines be drawn parallel to AB, BG, and the parallelograms AE, EG be equal, the point E lies in the diagonal BD. 112 BOOK I. PROPOSITION 44. To construct a 'parallelogram equal to a given 'parallelo- gram, having an wiigle equal to an angle of the given paral- lelogram, and having a side equal to a given straight line. Let ABCD be the given parallelogram, and EF the given straight line : it is required to construct a parallelogram equal to ABGB, having an angle equal to the angle BAD, and having a side equal to EF. Construction. Produce DA to G, and make AG equal to EF. (Prop. 3.) Through G draw HGK parallel to AB meeting CB pro- duced in K. (Prop. 31.) Draw KA and produce it to meet CD produced in L, and through L draw LMH parallel to DAG to meet BA produced in M and HGK in H : then MA GH is a parallelogram constructed as required. Proof. Because LGKH is a parallelogram, KL a dia- gonal, and MG, BD complements of parallelograms about the diagonal KLy MG is equal to BD. (Prop. 43.) Again, because the straight lines BAM, DAG intersect at A, the angle MAG is equal to the vertically opposite angle BAD. (Prop. 15.) And AGi^ equal to EF. (Constr.) Wherefore, a parallelogram MAG H has been constructed equal to the given parallelogram ABCD, having an angle MAG equal to the angle BAD and having a side AG equal to the given straight line EF. FliOFOSITION i4. 113 EXERCISES. 1. On a given straight line construct a rectangle equal to a given rectangle. 2. On a given straight line construct a rhombus equal to a given triangle. Is this always possible ? 3. Construct a rectangle equal to the sum of two given rect- angles. 4. Construct a rectangle equal to the difference of two given rectangles. T. E. 114 BOOK L PROPOSITION 45. To construct a 'parallelogrmn equal to a given 7'ectilineal figure, having a side equal to a given straight line, and having an angle equal to a given angle. Let A be tlie given rectilineal figure, B the given angle, and C the given straight line : it is required to construct a parallelogram equal to the ligure A, having an angle equal to the angle B, and liaving a side equal to C. Construction. Construct the triangle DEF equal to the figure A. (Prop. 41 A.) Construct the parallelogram GIIKL equal to the triangle DEF, having the angle GHK equal to the angle B. (Prop. 42.) Construct the parallelogram MNPQ equal to the parallelo- gram GIIKL, having an angle MNP equal to the angle GHK, and having the side Jl/iV equal to C. (Prop. 43.) ■E F H K N F Proof. Because the triangle DEF is equal to the figure A, (Constr.) and the parallelogram GK is equal to the triangle DEF, (Constr.) and the parallelogram MP is equal to the parallelogram GK, (Constr.) the parallelogram MP is equal to the figure A. Because the angle GHK is equal to the angle B, (Constr.) and the angle MNP is equal to the angle GHK, (Constr.) the angle MNP is equal to the angle B. And MN is equal to C (Constr.) Wherefore a parallelogram MNPQ has been constructed equal to the given rectilineal figure A, having the side MN equal to tlie given straight line G, and having the angle MNP equal to the given angle B. PROPOSITION 45. 115 EXEECISES. 1. Ou a given straight line as diagonal, construct a rhombus equal to a given triangle. 2. Construct a right-angled triangle, having given the hypotenuse and the perpendicular from the right angle on it. (See Exercise 8, page 87.) 3. Construct a rectangle equal to a given rectangle, and having a diagonal equal to a given straight line. 4. Construct a rectangle equal to the sum of two given triangles. 8—2 116 BOOK I. PROPOSITION 46. On a given straight line to construct a square. Let -4^ be the given straight line : it is required to construct a square on AB. Construction. From the point A draw ^C at right angles to ^jB; (Prop. 11.) and make AC equal to AB ] (Prop. 3.) through B draw BD parallel to AG, (Prop. 31.) and through C draw CD parallel to AB meeting BD in D : then ABDC is a square constructed as required. Proof. Because CD is parallel to AB, ajid BD to A Q the figure ABDC is a parallelogram. (Def. 18.) Again the angle CAB is a right angle j therefore the parallelogram ABDC is a rectangle. (Def. 19.) Again, the adjacent sides AC, AB are equal ; (Constr.) therefore the rectangle ABDC is a square. (Del 20.) Wherefore, ABDC is a square constructed on the given straight line AB. PROPOSITION 46. 117 EXEECISES. 1. If two squares be equal in area, their sides are equal. 2. The squares on two equal straight lines are equal in all respects. 3. If in the sides AB, BC, CD, DA of a square points E, F, G, H be taken so that AE, BF, CG, DH are equal : then EFGH is a square. 4. If the diagonals of a quadrilateral be equal and bisect each other at right angles, the quadrilateral is a square. 5. On the sides AC, BC of a triangle ABC, squares ACDE, BCFG are constructed : shew that the straight lines AF and BD are equal. 6. Construct a square so that one side shall be in a given straight line and two other sides shall pass through two given points. 7. Construct a square so that two opposite sides shall pass through two given points, and its diagonals intersect at a third given point. 8. Prove that the straight line, bisecting the right angle of a right-angled triangle, passes through the intersection of the diagonals of the square constructed on the outer side of the hypotenuse. 118 BOOK I. PROPOSITION 47. In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other sides. Let ABC be a right-angled triangle, having the right angle BA C : it is required to prove that the square on BC is equal to the sum of the squares on BA^ AC. Construction. On BC on the side away from A con- struct the square BDEC, (Prop. 46.) and similarly on BA, AC construct the squares BAGF, ACKH; through A draw A L parallel to BD meeting DE in L ; (Prop. 31.) and draw AD, EC. Proof. Because each of the angles BAC, BAG is a right angle, the two straight lines AC, AG, on opposite sides of AB, make with it at A the adjacent angles together equal to two right angles; therefore CA is in the same straight line with AG. (Prop. 14.) The angle DBC is equal to the angle EBA, for each of them is a right angle. (Prop. 10 A.) PROPOSITION 47. 119 Add to each of these equals the angle A EC ; then the angle DBA is equal to the angle FBC. And because in the triangles ABD, FBC, AB is equal to FB and BD to BC ■ and the angle ABD is equal to the angle FBC ] the triangles ABD, FBC are equal in all respects. (Prop. 4.) Because the parallelogram BL and the triangle ABD have a common side BD and A is in the same straight line as the side of BL opposite to BD, (Prop. 41.) the parallelogram BL is double of the triangle ABD. And because the square GB and the triangle FBC have a common side FB, and C is in the same straight line as the side of GB opposite to FB, (Prop. 41.) the square GB is double of the triangle FBC. Now the doubles of equals are equal. Therefore the parallelogram BL is equal to the square GB. Similarly it can be proved that the parallelogram CL is equal to the square HC. Therefore the whole square BDEC, which is the sum of the rectangles BL, CL, is equal to the sum of the two squares GB, HC. And the square BDEC is constructed on BC, and the squares GB,HConBA,AC. Therefore the square on the side BC is equal to the sum of the squares on the sides BA, AC. Wherefore, in a right-angled triangle kc. EXEKCISES. 1. Construct a square equal to the difference of two given squares. 2. The diagonals of a quadrilateral intersect at right angles. Prove that the sum of the squares on one pair of opposite sides is equal to the sum of the squares on the other pair. 3. If be the point of intersection of the perpendiculars drawn from the angles of a triangle upon the opposite sides, the squares on OA and BC are together equal to the squares on OB and GA^ and also to the squares on 0(7 and AB. 4. Divide a given straight line into two parts so that the sum of the squares on the parts may be equal to a given square. 5. Divide a given straight line so that the difference of the squares on the parts is equal to a given square. 120 BOOK L The proof of the theorem "the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other sides," which we have given in the text of the 47th proposition, is attributed to Euclid. Tradition however says that the first person to dis- cover a proof of the truth of the theorem was Pythagoras, a Greek philosopher who lived between 570 and 500 B.C. The theorem is in consequence often quoted as the Theorem of Pythagoras. "What was the nature of Pythagoras' proof is not known. The theorem is one of great importance and a large number of proofs of its truth have been discovered. It is advisable that the student should be made acquainted with some proofs besides the one given in the text. We have made a selection of live proofs of the theorem : in each case not attempting to give the complete proof, but merely giving hints of the line of argument to be used, and leaving the student to develope it more fully. Proof I. Take a right-angled triangle ABC, and on the side AB away from C construct the square ABDE, and on the hypotenuse ^C on the same side as B construct the square ACFG. From F draw FH perpendicular to BD, and FK perpendicular to ED pro- duced. It may be proved that (1) CBD is a straight line, (2) G lies in DJ^;, (3) the triangles ABC, AEG, CHF, GKF are all equal, (4) HK is a square and equal to the square on BC. PROPOSITIOK 47. 121 Proof II. Take a right-angled triangle ABC and on the sides AB, BC, GA away from C, A, B construct the squares BADE, CBFG, ACHK. Through L the intersection of the diagonals AE, BD of the square on the larger side AB, draw iHLTf. perpendicular to CA and OLP parallel to CA . Take Q, R, S, T the middle points of the sides AC, CH, HK, KA of the square on the hypotenuse. Through Q, S draw QUV, SWX parallel to BC, and through R, T draw RVW, TXU parallel to AB. It may be proved that (1) all the quadrilaterals LMEO, LOBN, LNAP, LPDM, AQUT, CRVQ, HSWR, KTXS are equal to one another, (2) the quadrilateral UVWX is a square, (3) the squares CF, UW B.re equal. 122 BOOK 7. Proof III. Take two equal squares ABCD, EFGH. Take any point I in AT), and measure off BK, CL, DM, EN, EO each equal to AI. Draw IK, KL, LM, MI ; through N draw NQP parallel to EF, and through draw OQR parallel to EH. Draw QF, QH. E N H M O Q F P G It may be proved that (1) the square ABCD is divided into one square IKLM and four equal right-angled triangles, (2) the square EFGH is divided into two squares EOQN, QPGR and four equal triangles, (3) all the triangles are equal to each other, (4) the square IL is equal to the sum of the squares ON, PR, (5) the three squares IL, ON, PR are squares on the hypotenuse and on the sides of one or other of the equal triangles. Proof IV. Take a right-angled triangle ABC and on the hypo- tenuse BC on the same side as A construct the square BCED, and on the sides CA, AB away from B, C construct the squares CAHK, ABFG. Through A draw ML AN perpendicular to BC, and produce EG, KH to meet MEAN. G N PROPOSITION- 47. 123 It may be proved that (1) D lies in FG, (2) E lies in KH produced, (3) the rectangle BL, and the square AF are each equal to the parallelogram AD^ (4) the rectangle GL and the square AK are each equal to the parallelogram AE. Proof V. Take a right-angled triangle ABG : on the sides AB, BC, CA away from C, A, B construct the squares BADE, GBFG, ACHK. On HK construct a triangle HLK equal in all respects to the triangle ABC having HL parallel to AB, and KL to CB. Draw FE, GB, BD, BL. It may be proved that (1) GBD is a straight line. (2) the triangles FBE, GBA are equal, (3) all the quadrilaterals GFEB, GCAD, BCHL, LKAB are equal to one another. 124 BOOK I. PROPOSITION 48. If the square on one side of a triangle he equal to the sum of the squares on the other sides, the angle contained hy these two sides is a right angle. Let the square on BC, one of the sides of the triangle ABC, be equal to the sum of the squares on the other sides BA, AC: it is required to prove that the angle BAC is a right angle. Construction. From the point A draw AD at right angles to ^ C ; (Prop .11.) and make AD equal to BA ; (Prop. 3.) and draw DC. Proof. Because DA is equal to BA, the square on DA is equal to the square on BA. To each of these equals add the square on AC; then the sum of the squares on DA, AC \^ equal to the sum of the squares on BA, AC. Now because the angle DAC is a right angle, the square on DC is equal to the sum of the squares on DA, AC. (Prop. 47.) And the square on BC is equal to the sum of the squares on BA, AC. (Hypothesis.) Therefore the square on DC is equal to the square on BC, and DC is equal to BC. And because in the triangles DAC, BAC, DA is equal to BA, AC to AC, and CD to CB, the triangles are equal in all respects ; (Prop. 8.) PROPOSITION 48. 125 therefore the angle DAG is equal to the angle BAG. Now DAG is a right angle ; (Constr.) therefore BAG is a right angle. Wherefore, if the square ttc. EXEKCISES. 1. If the difference of the squares on two sides of a triangle be equal to the square on the third side, the triangle is right angled. 2. The locus of a point, such that the difference of the squares on its distances from two given points is equal to a given square, is a straight line. 3. Prove by means of Proposition 48 that the straight lines, drawn from the vertices of a triangle perpendicular to the opposite sides, meet in a point. 4. If the sum of the squares on two opposite sides of a quadri- lateral be equal to the sum of the squares on the other two sides, the diagonals of the quadrilateral intersect at right angles. 5. ABCD is a quadrilateral such that, if any point P be joined to A, B, C, D, the sum of the squares on PA, PC is equal to the sum of the squares on PB, PD : prove that ABCD is a rectangle. 126 BOOK /. MISCELLANEOUS EXERCISES. 1. How many diagonals can be drawn through the same vertex in (1) a quadrilateral, (2) a hexagon, (3) a polygon of n sides? 2. How many different diagonals can be drawn to (1) a quadri- lateral, (2) a hexagon, (3) a polygon of n sides? 3. If two straight lines bisect each other at right angles, any point in either of them is equidistant from the extremities of the other. 4. A straight line drawn bisecting the angle contained by two equal sides of a triangle bisects the third side at right angles. 5. Two isosceles triangles GAB^ DAB are on the same base AB: shew that the triangles ACD, BCD are equal in all respects. 6. Prove by the method of superposition that, if two isosceles triangles have the same vertical angle, their bases are parallel. 7. It ABC, DBC he two triangles equal in all respects on oppo- site sides of BC, then AD is perpendicular to BC and is bisected by it. 8. The angle BAG of a triangle ABG is bisected by a straight line which meets BC in D, and from AB on AB produced AE is cut off equal to ^C : prove that DE is equal to DG. 9. If two circles whose radii are equal cut in A, B and if the line joining their centres when produced meet the circumference in C, D, prove that AGBD is a rhombus. 10. Prove by the method of superposition that, if the opposite angles of a quadrilateral be equal and one pair of opposite sides be equal, the other sides are equal. 11. Prove by the method of superposition that, if a quadrilateral has two pairs of adjacent angles equal, it has one pair of opposite sides equal. 12. Two adjacent sides of a quadrilateral are equal, and the two angles which they form with the other sides are together equal to the angle between the other sides. Prove that one diagonal of the quadri- lateral is equal to a side. 13. In a triangle BAG is the greatest angle. Prove that, if a point D be taken in AB, and a point E in AG, DE is less than BG. 14. If AD be drawn perpendicular from the vertex A to the opposite side BG of a triangle ABG in which ^C is greater than AB, then DC is greater than BD, and the angle DAG is greater than the angle BAD. 15. How many different triangles can be formed by taking three lines out of six lines whose lengths are 2, 3, 4, 5, 6, 7 inches re- spectively ? MISCELLANEOUS EXERCISES. 127 16. Find the point the sum of whose distances from the four angular points of a convex quadrilateral is a minimum. 17. AB is a given finite straight line. From C the middle point of AB, CI) is drawn in any direction and of any length. Prove that AD, BIJ together are greater than twice CD. 18. From a given point draw a straight line making equal angles with two given intersecting straight lines. How many such lines can be drawn? 19. In a given straight line find a point equally distant from two given straight lines. In what case is a solution impossible ? 20. How many equalities must be given between the sides and the angles of (1) two quadrilaterals, (2) two hexagons, (3) two polygons of n sides, before the conclusion can be drawn that the figures are equal in all respects ? 21. In the triangle ABC the angles at B and C are equal ; m and I are points in AC produced and on AB resi^eetively, and Im is joined cutting BC in O. Prove that, if the sum ofAl and Am be double AB, then jBO is greater than CO. 22. The side BC of a triangle ABC is produced to a point D ; the angle ACB is bisected by the straight line CE which meets AB at E. A straight line is drawn through E parallel to BC, meeting AC at F, and the straight line bisecting the exterior angle ACD at G. Shew that EF is equal to FG. 23. A straight line drawn at right angles to BC the base of an isosceles triangle ABC cuts the side AB at D and CA produced atE : shew that AED is an isosceles triangle. 24. If in the base of a triangle ABC, there be taken any two points P and Q equidistant from the extremities of the base, and if through each of the points P, Q two straight lines be drawn parallel to AB, AC, so as to form two parallelograms having PA, QA for diagonals ; these two parallelograms are equal in area. 25. Find a point equidistant from each of three straight lines in a plane which do not coincide in direction with the sides of any triangle that can be drawn in the plane. Is the construction required in this last problem always possible ? 26. From a point P outside an angle BA C draw a straight line cutting the straight lines AB, ^C in points D and E such that PD may be equal to DE. 27. If one acute angle of a triangle be double of another, the triangle can be divided into two isosceles triangles. 28. If in a triangle ABC, ACB be a right angle, and the angle CAB be double the angle ABC, then AB is double AC. 128 BOOK I. 29. P is a point in the side CD of a square ABGD such that AP is equal to the sum of PC and CB and Q is the middle point of CD. Prove that the angle BAP is twice the angle QAD. 30. AOB, COD are two indefinite straight lines intersecting each other in the point O, and P is a given point in the plane of these lines. It is required to draw through the point P a straight line PXY cutting AB in X and CD in Y, in such a manner that OX may be equal to OY. Can this problem be solved in more than one way? 31. Construct an equilateral triangle one of whose angular points is given and the other two lie one on each of two given straight lines. 32. The sides AB, AC of a, triangle are bisected in D, E, and BE, CD are produced to F, G, so that EF is equal to BE, and DG to CD : prove that FAG is a straight line. 33. In a plane triangle an angle is acute, right or obtuse, according as the straight line joining the angle to the middle point of the opposite side is greater than, equal to, or less than half that side. 34. The difference of the angles at the base of a triangle is double the angle between the perpendicular to the base, and the bisector of the vertical angle. 35. AC is the longest side of the triangle ABC. Find in AC a. point D such that the angle ADB shall be equal to twice the angle ACB. 36. ABCD is a quadrilateral: the bisectors of the angles ABD, ACD meet at F: prove that the angle BFC is half the sum of the angles BAG, BDC. 37. Prove that, if points L, M, N be taken in the sides BC, CA, AB of a triangle ABC, such that the triangles ANM, NBL, MLG are equiangular to each other, they are equiangular also to the triangles ABC and LMN. 38. If one angle at the base of a triangle be double the other, the less side is equal to the sum or the difference of the segments of the base made by the perpendicular from the vertex, according as the first angle is greater or less than a right angle. 39. Any convex pentagon ABCDE has each of its angular points joined to the non-contiguous points and a star-shaped figure ACEBD is formed: prove that the sum of the angles of the figure ACEBD is two right angles. 40. In the figure of Proposition 1, if AB produced both ways meet the circles again in D and E and CD, CE be drawn, CDE will be an isosceles triangle having one angle four times each of the other angles. 41. From the extremities of the base of an isosceles triangle straight lines are drawn perpendicular to the sides; shew that the angles made by them with the base are each equal to half the vertical angle. MISCELLANEOUS EXERCISES. 129 42. The sides AB, AG of & given triangle ABC are bisected at the points E, F; & perpendicular is drawn from A to the opposite side, meeting it at D. Shew that the angle FDE is equal to the angle BAG. 43. AB, AG are two given straight lines, and P is a given point in the former : it is required to draw through P a straight line to meet ^C at Q, so that the angle APQ may be three times the angle AQP. 44. Construct a right-angled triangle having given the hypotenuse and the difference of the sides. 45. From a given point it is required to draw to two parallel straight lines, two equal straight lines at right angles to each other. 46. Construct a triangle of given perimeter, having its angles equal to those of a given triangle. 47. Given one angle, and the opposite side, and the sum of the other sides, construct the triangle. 48. If two triangles on the same side of a common base have their sides terminated in opposite extremities of the base equal, the line joining the vertices will be parallel to the common base. 49. If an exterior angle of a triangle be bisected, and also one of the interior and opposite angles, the angle contained by the bisecting lines is equal to half the other interior and opposite angle. 50. If the perpendicular drawn from one of the equal angles of an isosceles triangle upon the opposite side divide the angle into two parts, one of which is double of the other, the vertical angle of the triangle is either one half or four-fifths of a right angle. 51. The sides AB, AG oi a, triangle ABG are produced to E, JF, and the angles GBE, BGF are bisected by the straight Hues BD, GD: prove that the angle BDG is half the sum of the angles ABC, AGB. 52. In a triangle ABG points D, E are taken in ^C7 such that AD is equal to AB, and that GE is equal to GB, and a point F is taken in ^jB such that BF is equal to i^C: prove that the angle EBD is equal to the angle BGF. 53. ABG is a triangle, and from a point D in AB the straight line DEF is drawn meeting BG in E and AG produced in F; shew that the angles between the straight lines bisecting the angles ABE, ADE are equal to the angles between the straight lines bisecting the angles AGE, AFE. 54. Find a point in a given straight line such that its distance from a given point is double of its distance from a given straight line through the given point. 55. Construct an equilateral triangle, having given its altitude. T. E. 9 130 BOOK I. 56. The straight line bisecting the exterior angle at the vertex A of a triangle ABC will meet the base BG produced beyond B, or beyond 0, according as ^C is greater or less than AB. 57. If from any point within an isosceles triangle perpendiculars be let fall on the base and the sides, the sum of these perpendiculars is less than the altitude of the triangle, if the vertical angle be less than the angle of an equilateral triangle. 58. ABC is a triangle right-angled at A, and on AB, AG are described two equilateral triangles ABD, AGE (both on the outside or both on the inside), and DB and EG or those produced meet in F; shew that A is the ortho-centre of the triangle DEF. 59. If the angle between two adjacent sides of a parallelogram be increased, while their lengths do not alter, the diagonal through their point of intersection will diminish. 60. If straight lines be drawn from the angles of any parallelo- gram perpendicular to any straight line which is outside the parallelo- gram, the sum of those from one pair of opposite angles is equal to the sum of those from the other pair of opposite angles. 61. If a six-sided plane rectilineal figure have its opposite sides equal and parallel, the three straight lines joining the opposite angles will meet at a point. 62. The vertical angle GAB of a triangle ABG is bisected by ADE, and BE, GF are drawn perpendicular to ADE : prove that the middle point of BG is equidistant from E and F. 63. Find in a side of a triangle a point such that the sum of two straight lines drawn from the point parallel to the other sides and terminated by them is equal to a given straight line. 64. If the angular points of one parallelogram lie on the sides of another parallelogram, the diagonals of both parallelograms pass through the same point. 65. If the straight line joining two opposite angles of a parallelo- gram bisect the angles, the parallelogram is a rhombus. 66. Draw a straight line through a given point such that the part of it intercepted between two given parallel straight lines may be of given length. 67. Bisect a parallelogram by a straight line drawn through a given point within it. 68. Shew that the four triangles into which a parallelogram is divided by its diagonals are equal in area. 69. Straight lines bisecting two adjacent angles of a parallelogram intersect at right angles. 70. Straight lines bisecting two opposite angles of a parallelogram are either parallel or coincident. MISCELLANEOUS EXERCISES. 131 71. Find a point such that the perpendiculars let fall from it on two given straight lines shall be respectively equal to two given straight lines. How many such points are there ? 72. ABCD is a quadrilateral having BG parallel to AD, E is the middle point of DC ; shew that the triangle AEB is half the quadri- lateral. 73. If the sides of a triangle be trisected and straight lines be drawn through the points of section adjacent to each angle so as to form another triangle, this is equal to the original triangle in all respects. 74. If two opposite sides of a parallelogram be bisected and two straight lines be drawn from the points of bisection to two opposite angles, the two lines trisect the diagonal which passes through the other two angular points. 75. BAG is a right-angled triangle, A being the right angle. AGDE, BGFG are squares on ^O and BG. AG produced meets DF in K. Prove that DF is bisected in K, and that AB is double of GK. 76. In a triangle ABG on AG, BG on the sides of them away from B, A, squares AGDE, BGFG are constructed ; prove that, if AG produced bisect DF, BAG is a right angle. 77. A straight line PQ drawn parallel to the diagonal ^C of a parallelogram ABGD meets AB in P and BG in Q; shew that the other diagonal BD bisects the quadrilateral BPDQ. 78. If the opposite angles of a quadrilateral be equal, the oppo- site sides are equal. 79. If in two parallelograms ABGD, EFGH, AB be equal to EF, BG to FO, and the angle ABG to the angle EFG, then the parallelo- grams are equal in all respects. 80. If the straight line AB be bisected in G, and AD and BE be drawn at right angles to AB, and AD be taken equal to AG, and DE be drawn at right angles to DG, DE is double of DG. 81. ABG is a given triangle : construct a triangle of equal area, having its vertex at a given point in BG and its base in the same straight line as AB. 82. In the right-angled triangle ABG, the sides AB, AG which contain the right angle are equal. A second right-angled triangle is constructed having the sides containing the right angle together equal to AB, AG, but not equal to one another. Prove that this triangle is less than the triangle ABG. 83. If two triangles have two sides of the one equal to two sides of the other, and the sum of the two angles contained by these sides equal to two right angles, the triangles are equal in area. 9-2 132 BOOK I. 84. If a triangle be described having two of its sides equal to the diagonals of any quadrilateral, and the included angle equal to either of the angles between these diagonals, then the area of the triangle is equal to the area of the quadrilateral. 85. ABC is a triangle ; find the locus of a point such that the sum of the areas OAB, OBG, OCA is constant and greater than the area of ^£(7. 86. Any point P is joined to the middle point of AD. On AP, DP squares AFQQ', DP RE' are described on the sides remote from D, A respectively. Prove that QR is double of and perpendicular to OP. 87. ABGD is a parallelogram; E the point of bisection of AB\ prove that AG, DE being joined will each pass through a point of trisection of the other. 88. In every quadrilateral the intersection of the straight lines which join the middle points of opposite sides is the middle point of the straight line which joins the middle points of the diagonals. 89. The line joining the middle points of the diagonals of the quadrilateral ABGD meets AD and BG in. E and F. Shew that the triangles EBG, FAD are each half the quadrilateral. 90. PQR is a straight line parallel and equal to the base BG of a triangle meeting the sides in P and Q. Shew that the triangles BPQ, AQR are equal. 91. Two straight lines AB and GD intersect at E, and the tri- angle AEG is equal to the triangle BED; shew that BG ia parallel to^D. 92. Construct the smallest triangle, which has a given vertical angle, and whose base passes through a given point. 93. In the base JO of a triangle take any point D ; bisect AD, DG, AB, BG at the points E, F, G, H respectively: shew that EG is equal and parallel to FH. 94. Given the middle points of the sides of a triangle, construct the triangle. 95. ABG ia & triangle. The side GA is bisected in D and E is the point of trisection of the side BG which is nearer to B. Shew that the line joining A to E bisects the line joining B to D. 96. Shew how by means of Prop. 40 to produce a finite straight line beyond an obstacle which cannot be passed through directly. 97. BAG is a fixed angle of a triangle, and (i) the sum, (ii) the difference of the sides AB, AG is given; shew that in either case the locus of the middle point of UC is a straight line. MISCELLANEOUS EXERCISES. 133 98. Through a fixed point two straight lines OPQ, OP'Q' are drawn meeting two fixed parallel straight lines. Prove that, if PQ' and P'Q meet in R, the locus of 22 is a straight line, and that OR bisects PP' and QQ'. 99. ABC is a triangle and on the side BG & parallelogram BDEC is described, and the parallelogram whose adjacent sides are AI), AB is completed and also that whose adjacent sides are AE^ AG; shew that the sum or the difference of the two latter parallelograms is equal to the first. 100. ABG is a triangle; ADF, GFE are the perpendiculars let fall from A and G, one on the internal bisector of the angle B and the other on the external bisector: the area of the rectangle BDFE is equal to that of the triangle. 101. If on the sides AB, BG, GA of any triangle, squares ABEF, BGGH, GAEL be constructed, and EH, GL, KF be drawn, then all the triangles ABG, BEH, GGL, AKF are equal to one another. 102. Construct a square, which shall have two adjacent sides passing through two given points, and the intersection of diagonals at a third given point. 103. ABG is a triangle right-angled at A and K is the corner of the square on AB opposite to A and H the corner of the square on AG opposite to A. If AB be produced to D, so that AD is equal to GK and ^ C be produced to E, so that AE is equal to BH, then GD is equal to BE. 104. Upon BG, GA, AB, sides of the triangle ABG, perpen- diculars are drawn from a point D, meeting the sides, or the sides produced, in E, F, G respectively. Prove that the sum of the squares on AG, BE, GF is equal to the sum of the squares on BG, GE, AF. 105. Divide a given straight line into two parts such that the square on one of them may be double the square on the other. BOOK 11. DEFINITIONS. 1)efinition 1. A rectangle is said to be contained "by the two sides which contain any one of its angles. The expression, that a rectangle is contained hy two straight lines AB, BG is of course a faulty one, as the area of the rectangle is contained by the four sides of the figure. The expression must be considered merely as an abbreviated form of the statement that the rectangle has for two of its adjacent sides the two straight lines AB, BG. It can easily be proved that, if two adjacent sides of one rectangle be equal to two adjacent sides of another rectangle, the two rectangles are equal in all respects. (See Exercise 79, page 131.) A rectangle, two of whose adjacent sides are equal to two straight lines AB, CD, is therefore often said to be the rectangle contained by AB, CD ; such a rectangle is often denominated simply the rectangle AB, CD. It is clear that the rectangle AB^ CD is the same as the rectangle CD, AB. Also it may be noticed that, ii AB h^ equal to CD, the rectangle AB, CD is equal to the square on AB, or to the square on CD. 135 In Arithmetic or in Algebra, if we wish to represent a given length, we take a definite length, for instance an inch, as a unit of length and we express the given length by the number of units of length contained in it. In the same way, if we wish to represent a given area, we take a definite area, for instance a square inch, as a unit of area, and we express the given area by the number of units of area contained in it. It is easily seen that, if a rectangle have 2 inches in one side and 3 inches in an adjacent side, its area consists of 2 x 3 or 6 squares, each square having one inch as its side, and similarly that, if a rectangle have m units of length in one side and n units of length in an adjacent side, its area consists of wm squares, each square having a unit of length as its side. Thus in Arithmetic or in Algebra the area of a rectangle is repre- sented by the product of the numbers, which represent the lengths of two adjacent sides. If the rectangle be a square, its area is represented by the square of the number, which represents the length of a side. In consequence of this connection between Algebra and Geometry, there is a certain correspondence between the theorems and problems of the Second Book of Euclid, and theorems and problems in Algebra. A short statement of a corresponding proposition in Algebra is given as a note to each Proposition, in which statement each straight line is represented by a corresponding letter, and each area by a corresponding product. 136 BOOK II. PROPOSITION 1. If there he two straight lines, one of which is divided into any two parts, the rectangle contained hy the two straight lines is equal to the sum of the rectangles contained hy the undivided line, and each of the parts of the divided line*. Let AB and CD be two straight lines ; and let CD be divided into any two parts at the point E : it is required to prove that the rectangle contained by AB, CD is equal to the sum of the rectangles contained by AB, CE, and by AB, ED. Construction. From the point C draw CF at right angles to CD; (I. Prop. 11.) and make CG equal to AB ; (I. Prop. 3.) through G draw GHK parallel to CD ; (I. Prop. 31.) and through D, E draw DK, EH parallel to CG meeting GHK in K II. E D Proof. Each of the figures CK, CH, EK is a parallel- ogram, (Constr.) and each of the figures has one angle a right angle ; (I. Prop. 29, Coroll.) therefore each of the figures is a rectangle. Now the rectangle CK is equal to the sum of the rect- angles CH, EK. * The algebraical equivalent of this theorem is the equation a{h-\-c)=.ah-\-ac. PROPOSITION 1. 137 But CK is contained by AB, CD^ for it is contained by CG^ CD, and CG is equal to AB. And CH is contained liy yl^, 0^, for it is contained by CG, CE, and CG is equal to AB. And ^A'^ is contained by AB, ED, for it is contained by EH, ED, and EH is equal to CG, which is equal to AB. Therefore the rectangle contained by AB, CD is equal to the sum of the rectangles contained by AB, CE, and by AB, ED. Wherefore, iftJiere he two straight lines o AD. It may be proved that (1) AL, BK, FN, EM are each equal to the rect- angle AB, AC, and (2) MP is equal to the square on CB, and hence that the sum of the square on ^C and four times the rectangle AB, GB is equal to the square on AD. EXERCISES. 1. Prove that the square on a straight Hue is nine times the square on one third of the line. 2. If a straight line be bisected and produced to any point, the square on the whole line thus produced is equal to the square on the part produced together with twice the rectangle contained by the line and the line made up of the half and the part produced. 162 BOOK IL PROPOSITION 9. If a straight line he divided into two equals and also into two unequal parts, the sum of the squares on the two unequal parts is double of the sum of the squares on half the line and on the line between the points of section'^. Let the straight line AB he divided into two equal parts at the point C, and into two unequal parts at the point D : it is required to prove that the sum of the squares on AD^ DB is double of the sum of the squares on AC, CD. A C D B 1 1 Proof. Because AD is divided at C, the square on A JD is equal to the sum of the squares on AC, CD and twice the rectangle AC, CD. (Prop. 4.) And because CB is divided at D, the sum of the square on DB and twice the rectangle CB, CD is equal to the sum of the squares on CB, CD. (Prop. 7.) Add these pairs of equals ; then the sum of the squares on AD, DB and twice the rectangle CB, CD is equal to the sum of the squares on AC, CD, CB, CD and twice the rectangle AC, CD. Take away from these equals twice the rectangle CB, CD, and twice the rectangle AC, CD, which are equal ; then the sum of the squares on AD, DB is equal to the sum of the squares on AC, CD, CB, CD, that is, is equal to twice the sum of the squares on AC, CD. Wherefore, if a straight line &c. * The algebraical equivalent of this theorem is the equation PROPOSITION 9. 153 The theorems of Propositions 9 and 10 may both be included in one enunciation : thus, The sum of the squares on the sum and on the difference of two given straight lines is equal to twice the sum of the squares on the lines. The straight lines in both propositions are A C, CD : the only difference being that in Prop. 9. ^C is the greater, and in Prop. 10. CD is the greater. For an outline of an alternative proof of Propositions 9 and 10, see page 155. EXEKCISES. 1. A straight line is divided into two parts, such that the diagonal of the square on one of these parts is equal to the whole line. If a square be constructed whose side is the difference between the aforesaid part and half the given line, its diagonal is equal to the other of the two parts into which the line is divided. 2. Deduce a proof of ii. 9 from the result of ii. 5. 3. If a straight line be divided into two equal and also into two unequal parts, the squares on the two unequal parts are equal to twice the rectangle contained by the two unequal parts together with four times the square on the line between the points of section. 154 BOOK II. PROPOSITION 10. If a st7'aight line be bisected and produced to any point, the sum of the squares on the whole line thus produced and on the part produced is double of the sum of the squares on half the line and on the line m^ade up of the half and the part produced*. Let the straight line ^^ be bisected at C, and produced toD: it is required to prove that the sum of the squares on AD, BD is double of the sum of the squares on AC, CD. A C B D 1 1 Proof. Because AD is divided at C, the square on AD is equal to the sum of the squares on AC, CD and twice the rectangle AC, CD; (Prop. 4.) and because CD is divided at B, the sum of the square on BD and twice the rectangle CB, CD is equal to the sum of the squares on CB, CD. (Prop. 7). Add these equals ; then the sum of the squares on AD, BD and twice the rectangle CB, CD is equal to the sum of the squares on AC, CD, CB, CD and twice the rectangle AC, CD. Take away from these equals twice the rectangle CB, CD, and twice the rectangle A C, CD, which are equal ; then the sum of the squares on AD, BD is equal to the sum of the squares on AC, CD, CB, CD, that is, is equal to twice the sum of the squares on AC, CD. Wherefore, if a straight line &c. * The algebraical equivalent of this theorem is the equation PROPOSITION 10. 155 Outline of AlterTiative Proof of Propositions 9 a7ul 10. In a straight line A BCD, take AB equal to CD. Through A, B, C, D draw AJS, BI\ CG, DII at right angles to ABCD. Take AK equal to AB, KL to BC and LE to AB. Draw EFGH, LQRS, KMNP parallel to ABCD. E F G H Q R M N A B It may be proved that C D (1) EQ, ND are each equal to the square on AB, (2) Z(7, FP are each equal to the square on A C, (3) QN is equal to the square on BC, (4) ED is the square on AD^ and (5) the sum of ED and ^iV is equal to the sum of EQ, ND, LC, and FP, and hence that the sum of the squares on AD, BC (which are tlie sum and the difference of ^C and AB) is equal to twice the sum of the squares on AC, AB. EXEECISES. 1. In AB the diameter of a circle take two points C and D equally distant from the centre, and from any point E in the circum- ference draw EG, ED: shew that the squares on EG and ED are together equal to the squares on ^C and AD. 2. It in BG the base of a triangle a point D be taken such that the squares on AB and BD are together equal to the squares on AG and GD, then the middle point of AD will be equally distant from £ and G. 3. A square BDEG is described on the hypotenuse BG of a right- angled triangle ABG : shew that the squares on DA and AG are together equal to the squares on EA and AB. 4. AB is divided into any two parts in G, and AG, BG are bisected in D, E : prove that the square on AE and three times the square on BE are equal to the square on BD and three times the square on ^D. 156 BOOK II. PROPOSITION 11. To divide a given straight line into two parts, so thai tJie rectangle contained hy the whole and one part may he equal to the square on the other part^. Let ABhe the given straight line : it is required to divide it into two parts in a point H, so that the rectangle contained by the whole line AB and a part HB may be equal to the square on the other part AH. CoNSTKUCTiON. On AB construct the square ABDC ; (I. Prop. 46.) bisect AC at E -, (I. Prop. 10.) draw BE ; produce CA to F, and make j^i^" equal to EB ; (I. Prop. 3.) and on AF construct the square AFGII \ (I. Prop. 46.) then AB \s, divided at H so that the rectangle AB, HB is equal to the square on AH. Produce GH to meet CD at K. G H J^' A E C Proof. Because ACis bisected at E, and produced to F, the sum of the rectangle FC, FA, and the square on AE is equal to the square on FE. (Prop. 6.) But FE is equal to EB. (Constr.) Therefore the sum of the rectangle FC, FA, and the square on AE is equal to the square on EB. But, because the angle EAB is a right angle, the square on EB is equal to the sum of the squares on AE, AB. (I. Prop. 47.) * The algebraical equivalent of this problem is to find the smaller root of the quadratic equation ax=:{a~x)^, or the positive root of the quadratic equation a{a-x) — x^. PROPOSITION 11. 157 Therefore the sum of the rectangle FC, FA, and the square on AE, is equal to the sum of the squares on AE, AB. Take away from each of these equals the square on AE ] then the rectangle FC, FA is equal to the square on AB. But the figure FK is the rectangle contained by FC, FA, for FG is equal to FA ; (Constr.) and AD \s> the square on AB ; therefore FK is equal to AD. Take away from these equals the common part AK; then FH is equal to ED. But HD is the rectangle contained by AB, HB, for -4^ is equal to BD ; (Constr.) and FH is the square on AH ; therefore the rectangle AB, HB is equal to the square on AH. Wherefore the straight line AB is divided at II, so that the rectaiigle AB, II B is equal to the square on AH. EXEKCISES. 1. Shew that in a straight line divided as in II. 11 the rectangle contained by the sum and the difference of the parts is equal to the rectangle contained by the parts. 2. If the greater segment of the line divided in this proposition be divided in the same manner, the greater segment of the greater segment is equal to the smaller segment of the original line. 3. Prove that when a straight line is divided as in this pro- position the square on the line made up of the given line and the smaller part is equal to five times the square on the larger part. 4. Prove that in the figure of this proposition the squares on AB, HB are together equal to three times the square on AH, and that the difference of the squares on AB, AH ia equal to the rect- angle AH, AB. 5. Produce a given straight line, so that the rectangle contained by the whole line thus produced and the given line may be equal to the square on the part produced. 158 BOOK II. PROPOSITION 12. In an obtuse-angled triangle, if a perpendicular be drawn from either of the acute angles to the opposite side jyroduced, the square on the side opposite the obtuse angle is greater than the sum of the squares on tJte other sides, by twice the rect- angle contained by the side on which, when produced, the per- pendicular falls, and the part of the produced side inter- cepted between the perpendicular and the obtuse angle. Let ABC be an obtuse-angled triangle, and let the angle ACB be the obtuse angle; from the point A let AD be drawn perpendicular to BC produced : it is required to prove that the square on AB is greater than the sum of the squares on AC, CB, by twice the rectangle BC, CD. Proof. Because BD is divided into two parts at C, the square on BD is equal to the sum of the squares on BC, CD, and twice the rectangle BC, CD. (Prop. 4.) To each of these equals add the square on DA ; then the sum of the squares on BD, DA is equal to the sum of the squares on BC, CD, DA, and twice the rectangle BC, CD. But, because the angle at Z) is a right angle, the square on BA is equal to the sum of the squares on BD, DA, and the square on CA is equal to the sum of the squares on CD, DA. (I. Prop. 47.) Therefore the square on BA is equal to the sum of the squares on BC, CA, and twice the rectangle BC, CD ; that is, the square on BA is greater than the sum of the squares on BC, CA by twice the rectangle BC, CD. Wherefore in an obtuse-angled triangle &,c. PROPOSITION 12. 159 Outline of Alternative Proof. On the sides BC, CA, AB of an obtuse-angled triangle ABC^ in which tlie angle BAG is obtuse, constioict the squares BBEC, CFG A, AHKB. Draw AL^ BAI, CN perpendicular to 7iC, CA, AB and produce them to meet the opposite sides (produced if neces- sary) of the squares in P, Q, R. Draw AD, CK. D P It may be proved that (1) the triangle ABD is equal to the triangle KBC in all respects, and (2) the rectangle BP is equal to the rectangle BR, and similarly that CQ is equal to CP, and ^i? to ^^, and hence that the square on BC is greater than the sum of the squares on CA, AB by twice the rectangle AR or AQ. EXEECISES. 1. The sides of a triangle are 10, 12, 15 inches: prove that it is acute-angled. 2. On the side BG of any triangle ABC, and on the side of BG remote from A, a square BDEG is constructed. Prove that the difference of the squares on AB and ^C is equal to the difference of the squares on AD and AE. 3. G is the obtuse angle of a triangle ABG, and D, E the feet of the perpendiculars drawn from A , B respectively to the opposite sides produced: prove that the square on AB is equal to the sum of the rectangles contained by BC, BD and AC, AE. 4. ABG is a triangle having the sides AB and AG equal ; AB is produced beyond the base to D so that BD is equal to AB; shew that the square on CD is equal to the square on AB, together with twice the square on BG. 160 BOOK II. PROPOSITION 13. I71 any triangle^ the square on a side subtending an acute angle, is less tha7i the sum of the squares on the other sides, by twice the rectangle contained by either of these sides, and the part of the side intercepted between the perpendicular let fall on it from the opposite angle, and the acute angle. Let ABC be a triangle, and let the angle ABC be an acute angle; let AD hQ drawn perpendicular to BG and meet it (produced if necessary) in D : it is required to prove that the square on AC is less than the sum of the squares on AB, BC by twice the rectangle BC, BD. Either (1) D lies in BC, or (2) i> coincides with G, or (3) D lies in BC produced. (1) (2) (3) Proof. Because fig. (1) BC is divided in D, fig. (2) B is the same point as C, or fig. (3) BD divided in C, the sum of the squares on BC, BD is equal to the sum of the square on CD, and twice the rectangle BC, BD. (I. Prop. 47 and II. Prop. 7.) To each of these equals add the square on DA ; then the sum of the squares on BC, BD, DA is equal to the sum of the squares on CD, DA and twice the rectangle BC, BD. But because the angle BDA is a right angle, the square on AB is equal to the sum of the squares on BD, DA, and the square on AC is equal to the sum of the squares on CD, DA. (I. Prop. 47.) Therefore the sum of the squares on BG, AB is equal to the sum of the square on AC and twice the rectangle BG, BD : that is, the square on AC is less than the sum of the squares on AB, BG by twice the rectangle BG, BD. Wherefore, in any triangle &c. PROPOSITION 13. 161 Outline of Alternative Proof. On the sides BC, CA, AB of an acute-angled triangle ABC construct the squares BDEC, CFG A, AHKB. Draw ALy BM, CN perpendicular to BCy CA, AB, and produce tliera to meet the opposite sides of the squares in P, Q, R. Draw AD, CK, It may be proved that (1) the triangle ABD is equal to the triangle KBC in all respects, and (2) the rectangle BP is equal to the rectangle BR, and similarly that CQ is equal to CP, and AR to AQ, and hence that the square on BG is less than the sum of the squares on CA, ABhy twice the rectangle AQ or AR. EXERCISES. 1. In any triangle the sum of the squares on the sides is equal to twice the square on half the base together with twice the square on the straight line drawn from the vertex to the middle point of the base. 2. The base of a triangle is given : find the locus of the vertex, when the sum of the squares on the two sides is given. 3. The sum of the squares on the sides of a parallelogram is equal to the sum of the squares on the diagonals. 4. In any quadrilateral the squares on the diagonals are together equal to twice the sum of the squares on the straight lines joining the middle points of opposite sides. 5. The squares on the sides of a quadrilateral are together greater than the squares on its diagonals by four times the square on the straight line joining the middle points of its diagonals. T. E. 11 162 BOOK II. PROPOSITION 14. To jind the side of a square equal to a given rectangle* Let ABCD be the given rectangle : it is required to find the side of a square equal to ABCD. Construction. If two adjacent sides BA, AD he equal, the rectangle is a square, and BA or AD is the line required. But if they be not equal, produce one of them BA to E, and make AE equal to ^Z>; (I. Prop. 3.) bisect BE at F; (I. Prop. 10.) and with i^^as centre and FB as radius, describe the circle BGE, and produce DA to meet the circle in G ; then AG is the line required. Draw FG. Proof. Because BE is divided into two equal parts at F, and into two unequal parts at A, the sum of the rectangle BA, AE and the square on FA is equal to the square on FE. (Prop. 5.) But FE is equal to FG. Therefore the sum of the rectangle BA, AE and the square on FA is equal to the square on FG. * The algebraical equivalent of this problem is to find the positive root of the quadratic equation 9? — ah. PROPOSITION 14. 163 But because the angles at A are right angles, the square on FG is equal to the sum of the squares on FA, AG ; (I. Prop. 47.) therefore the sum of the rectangle BA, AE and the square on FA, is equal to the sum of the squares on FA, AG. Take away from each of these equals the square on FA ; then the rectangle BA, AE is equal to the square on AG. But A BCD is the rectangle contained by BA, AE, since it is contained by BA, AD, and AE \& equal to AD. Therefore A BCD is equal to the square on AG. Wherefore the straight line AG has been found, which is the side of a square equal to the given rectangle ABCD. Corollary 1. To find the side of a square equal to a given rectilineal figure. Construct a rectangle, i.e. a parallelogram, which has an angle equal to a right angle, equal to the given figure, (I. Prop. 45), and then use Proposition 14. Corollary 2. The square on the perpendicular from any point of a circle on any diameter is equal to the rectangle contained by the parts of that diameter intercepted between its extremities and the perpendicular. EXEKCISES. 1. Construct a rectangle equal to a given square, and having (1) the sum (2) the difference of two of its adjacent sides equal to a given straight line. 2. The largest rectangle, the sum of whose sides is given, is a square. 3. Construct a rectangle equal to a given square such that the difference of two adjacent sides shall be equal to a given straight line. 4. Construct a right-angled triangle equal to a given rectilineal figure and such that one of the sides containing the right angle is double of the other. 5. Produce a given straight line AB both ways to C and D so that the rectangles CA, AD and CB, BD may be equal respectively to two given squares. 164 BOOK 11. MISCELLANEOUS EXERCISES. 1. In a triangle whose vertical angle is a right angle a straight line is drawn from the vertex perpendicular to the base: shew that the square on either of the sides adjacent to the right angle is equal to the rectangle contained by the base and the segment of it adjacent to that side. 2. If ABC be a triangle whose angle ^ is a right angle, and BE, CF be drawn bisecting the opposite sides, four times the sum of the squares on BE and CF is equal to five times the square on BG. 3. The hypotenuse AB of a right-angled triangle ABC 19 tri- sected in the points D, E; shew that, if CD, CE be joined, the sum of the squares on the three sides of the triangle CDE is equal to two thirds of the square on AB. 4. ABCD is a rectangle, and points E, F are taken in BC, CD respectively. Prove that twice the area of the triangle AEF together with the rectangle BE, DF is equal to the rectangle AB, BC. 5. On the outside of the hypotenuse BC, and the sides CA, AB of a right-angled triangle ABC, squares BDEC, AF, and AG are described: shew that the squares on DG and EF are together equal to five times the square on BC. 6. On the outside of the hypotenuse BC of a right-angled tri- angle ABC and on the sides GA, AB squares BDEC, AF, AG are constructed: prove that the difference of the squares on DG and EF is three times of the difference of the squares onAB and AG. 7. In the hypotenuse AB of a right-angled triangle ACB, points D and E are taken such that AD is equal to AC and BE to BC; prove that the square on DE is equal to twice the rectangle BD, AE. 8. One diagonal ^C of a rhombus ABCD is divided into any two parts at the point P ; shew that the rectangle AP, PC is equal to the difference between the squares on ^i^ &nd PB. 9. In a given straight line find a point such that the difference of the squares upon the straight lines joining it to two given points may be equal to a given rectangle. In what cases is this problem impossible? 10. If a straight line be drawn through one of the angles of an equilateral triangle to meet the opposite side produced, so that the rectangle contained by the whole straight line thus produced and the part of it produced is equal to the square on the side of the triangle, the square on the straight line so drawn will be double the square on a side of the triangle. 11. The square on any straight line drawn from the vertex of an isosceles triangle to any point in the base is less than the square on a side of the triangle by the rectangle contained by the segments of the base. MISCELLANEOUS EXERCISES. 165 12. In the figure of II. 11, BE and CH meet at ; shew that .40 is at right angles to CH. 13. Divide a given straight line so that the square on one part is equal to the rectangle contained by the other part and another given straight line. 14. Divide a given straight line so that the rectangle contained by the parts shall be equal to the rectangle contained by the whole line and the difference of the parts. 15. How many triangles can be formed by choosing three lines out of six whose lengths are 3, 4, 5, 11, 12 and 13 inches? How many of these triangles are (1) obtuse-angled, (2) right-angled, (3) acute- angled ? 16. Prove that the sum of the squares on the straight lines drawn from any point to the middle points of the sides of a triangle is less than the sum of the squares on the straight lines drawn from the same point to the angular points of the triangle by one quarter the sum of the squares on the sides of the triangle. 17. ABBG is a parallelogram whose diagonals intersect in 0. OL, OM are drawn at right angles to AB, AG meeting them in L, M respectively; prove that the sum of the rectangles AB, AL and AC, AM is double the square on AO. 18. In a triangle ABC the angles B and C are acute: if E, F be the points where perpendiculars from the opposite angles meet the sides AC, AB, the square on BC is equal to the rectangle AB, BF, together with the rectangle A C, CE. What change is made in the theorem, if B be an obtuse angle? 19. Prove that the locus of a point, whose distance from one of two fixed points is double that from the other, is a circle. 20. At B in AB two equal straight lines BC, BD are drawn making equal angles with AB and AB produced. Shew that the difference of the squares on AC and AD is equal to twice the rectangle AB, CD. 21. If the squares on the sides of a quadrilateral be equal to the squares on the diagonals, it must be a parallelogram. 22. ABC is a. triangle having the angle C greater than the angle B, and D a point in the base BC, such that the sum of the squares on AB, ^C is twice the sum of the squares on AD, DB. Shew that either D is the middle point of the base, or that its distance from the foot of the perpendicular from A on BC is one half of BC. 23. If a figure be composed of a rhombus and the square described outwards on one of its sides, the side of the equivalent square is equal to half the sum of the diagonals of the rhombus. 24. ABC is a triangle in which C is a right angle, and DE is drawn from a point D in ^C perpendicular to AB; shew that the rectangle AB, AE is equal to the rectangle AC, AD. 166 BOOK 11. 25. ABC is an acute-angled triangle ; perpendiculars AFD, BPE are drawn on EG, GA from the opposite angles. Prove that the rect- angle BD, DG is equal to the rectangle AD, PD. 26. A, B, G, D are the angular points of a parallelogram, taken in order. If there be a point P such that the sum of the squares on PA and PG be equal to the sum of those on PB and PD, ABGD is a rectangle. 27. A, B, G,D are fixed points, and P is a point such that the sum of the squares on PA, PB, PG, PD is constant; prove that P lies on a circle, the centre of which is at the point where the straight line joining the middle points of AB, GD cuts the straight line joining the middle points of AD, BG. 28. A, B, G, D are fixed points, and P is a point such that the sum of the squares on PA , PB is equal to the sum of the squares on PG, PD. Prove that the locus of P is a straight line at right angles to the line joining the middle points of AB, GD, and passes through the intersection of the lines drawn perpendicular to either of the pairs of lines AG, BD or AD, BG at their middle points. 29. In the plane of a triangle AEG find a point P such that the sum of the squares on AP, BP and GP may be a minimum. 30. Produce a given straight line so that the rectangle contained by the whole line thus produced and the produced part may be equal to the square on another given straight line. CAMBRIDGE : PRINTBU BY C. J. CLAT, M.A. AND SONS, AT THE UNIVERSITY PRESS. BOOK III. DEFINITIONS. Definition 1. Any part of a circle is called an arc. The line which has been defined (i. Def. 22) as a circle is often spoken of as the circumference of the circle. The reason of this is that a circle is defined in many books as the part of the plane contained by the line, which is then called the circumference. Half of a circle is called a semicircle. It will be proved hereafter that a diameter bisects a circle, i.e. divides it into two equal arcs. (See page 175.) Definition 2. A straight line joining two points on a circle is called a chord of the circle. The straight line joining the extremities of an arc is called the chord of the arc. The figure formed of an arc and the chord of the arc is called a segment of the circle. In the diagram the straight lines AB, BC, CA are chords of the circle ABC \ AFB, BDC, CEA are arcs. The straight line AB is the chord of the arc AFB, and it is also the cliord of the arc ACB. The figure formed' of the arc BFEC and the chord BG is called the segment BFEC or BFAG, or more often BFC or BEG or BAG; and the figure formed of the arc BDG and the chord BC is called the segment BDC. 168 BOOK III. Definition 3. Tke angle contained by two chords joining a point in an arc of a circle to the extremities of the arc is called an angle in the arc, and the arc is said to contain the angle. An angle in an arc is often spoken of as an angle in the segment formed by the arc and the chord of the arc, and the segment is said to contain the angle. The angle BAG is said to be an angle in the arc (or in the segment) BFC and the angle ACB an angle in the arc (or in the segment) ADB ; and the arc (or the seg- ment) BFC is said to contain the angle BA C, and the arc (or the segment) ADB to contain the angle ACB. An angle in an arc is said to stand on the arc which forms the remainder of the circle. The angle BAC is said to stand on the arc BDC, and the angle ABC on the arc AEC. Definition 4, Arcs, which contain equal angles, are said to be similar; and likewise segments, which contain equal angles, are said to be similar. In the diagram the arcs ABC, DEF are said to be similar, when the contained angles ABC, DEF are equal; and also the segments ABC, DEF are said to be similar when the contained angles ABC, DEF are equal. DEFINITIONS. 169 Definition 5. A jjointj whose distance from the centre of a circle is less than the radius of the circle, is said to he within the circle ; and a jjoint, whose distance from the centre of a circle is greater tJian the radius of the circle, is said to he without the circle. In the diagram the point P is within the circle, its distance GP from the centre C being less than the radius GA , and the point Q, is without the circle, its distance CQ from the centre G being greater than the radius GB. It is clear that any line drawn from a point P within a circle to a point Q without the circle must intersect the circle once at least (I. Postulate 7, page 14). In the diagram the straight line PQ meets the circle in the points R and S, and the straight line and the circle inter- sect at each of those points. Definition 6. A sti'aight line and a circle, which pass through a poi7it hut do not intersect there, are said to touch one another at the point. The sti'aight line is called a tan- gent to the circle. In the diagram the circle ABG and the straight line DCE pass through the point C, but do not intersect there : they touch at the point C, and DE is a tangent to the circle at the point G. T. E. 12 170 BOOK III. In the diagram the circles PRS, QRS raeet at the points R and S, and the circles intersect at each of those points : for instance, points on the circle PES near R on one side of R lie within the circle QRS and on the other side of R without it. Definition 7. Two circles^ which jmss through a 2^oint hut do not intersect there, are said to touch one another at the point. In each of the figures in the diagram the circles ABC, BDE pass through the point B, but do not intersect there: all points on the circle ABC near B lie without the circle BDE: and all points on the circle BDE near B in figure (1) lie without the circle ABC, and in figure (2) lie within the circle ABC. (See remarks on the contact of circles on pages 199 and 201.) Definition 8. Circles which have the same point for a centre are said to be concentric. DEFINITIONS. 171 Deb'INITION 9. The ferpeftidicular clrmvn to a straight line from a point is called the distance of the straight line from the point. If the distances of two straight lines from a point he equals the straight lines are said to he equidistant from the point. In the diagram, if the straight lines OM, ON be drawn from the point perpendicular to the two straight lines AB, CD, then OM, ON are called the distances of the two straight lines AB, CD from the point 0. If OM, ON be equal, the straight Unes AB, CD are said to be equidistant from the point 0. If the distances of two straight lines from a point he unequal, the line the distance of which is the longer is said to he farther from the point, and the line the distance of which is the shorter is said to he nearer to the point. In the diagram, if the straight lines OM, ON be drawn from the point O perpendicular to the two straight lines AB, CD, and if OM be less than ON, then CD is said to be farther from the point O than AB, and AB is said to be nearer to the point than CD. 12—2 172 BOOK III. Definition 10, If all the angular points of a rectilineal figure lie on a circle, the figure is said to be inscribed in the circle, and the circle is said to he described about the figure. In the diagram, if the angular points of the triangle ABC lie on the circle ABC, the triangle ABC is said to be inscribed in the circle ABC, and the circle ABC is said to be described about the tri- angle ABC. Similarly, if the angular points D, E, F, G of the quadrilateral DEFG lie on the circle DEFG, the quadrilateral D£2^(r is said to be inscribed in the circle DEFG, and the circle DEFG is said to be described about the quadrilateral DEFG. If all the sides of a rectilineal figure touch a circle, the figure is said to he described about the circle, and the circle is said to he inscribed in the figure. In the diagram, if the sides BC, CA, AB of the triangle ABC touch the circle DEF at the points D, E, F respectively, the triangle ABC is DEFINITIONS. 173 said to be described about the circle DEF, and the circle DEF is said to be inscribed in the triangle ^UC Similarly, if the sides LG, GH, HK, KL of the quadrilateral GHKL touch the circle MNPQ at the points M, N, P, Q respectively, the quadrilateral GHKL is said to be described about the circle MNPQ, and the circle MNPQ is said to be inscribed in the quadrilateral GHKL. 174 BOOK III. PROPOSITION 1. A circle cannot have more than 07ie ce^itre. Let A be a centre of the given circle BCD : it is required to prove that no other point can be a centre of the circle BCD, Construction. Take any point E within the circle and draw AE, and produce AE both ways to meet the circle, beyond A in C, and beyond E in D. Proof. Because ^ is a centre of tlie circle, AC is equal to AD. (I. Def. 22.) Now EC is greater than AC, which is only a part of it, and J^jD, which is only a part of AD, is less than AD; therefore EC is greater than ED. But a centre of a circle is a point from which all straight lines drawn to the circle are equal ; (I. Def. 22.) therefore E cannot be a centre of the circle. Similarly it can be proved that no point other than A can be a centre. Wlierefore, a circle cannot have &c. The definition of a circle implies that the figure has a centre (I. Def. 22): it is here proved that it cannot have more tlian one centre: we shall therefore for the future speak of the centre of a circle. FROPOSITIOX 1. 175 PKOPOSITION lA. A diameter bisects a circle. Let A BCD be a circle, the centre and AOC a diameter: it is required to prove that the arcs ABC, ADC are equal. Construction. Draw any two radii OP, OQ making equal angles POC, QOC with DC. (I. Prop. 23.) Proof. It is possible to shift the figure AOCQD by turning it over so that AOC is not shifted, and so that the arcs ADC, ABC lie on the same side of AC. If this be done, because the angle QOC is equal to the angle POC, OQ must coincide in direction with OP'. and because OQ i^ equal to OP, Q must coincide with P. Similarly it can be proved that every point on the arc ADC must coincide with some point on the arc ABC, and every point on the arc ABC with some point on the arc ADC. Therefore the arc ADC coincides with the arc ABC, and is equal to it. Wherefore, a dicuneter bisects &,c. EXEECISES. 1. Prove by superposition that circles, which have equal radii, are equal. 2. Prove by superposition that equal circles have equal radii. 3. Two circles, which have a common centre, but whose radii are not equal, cannot meet. 4. Prove by superposition that two diameters at right angles divide a circle into four equal arcs. 176 BOOK III. PROPOSITION 2. Jf a straight line bisect a chord of a circle at right angles, the line passes through the centre. Let ^^ be a chord of the circle ABC, and let CDE be the straight line which bisects AB at right angles: it is required to prove that CDIJ passes through the centre of the circle ABC. Construction. Take any point G not in CJS and on the same side of CJS as B. Draw AG cutting CB in JI, and draw GB, HB. C Proof. Because in the triangles ADH, BDH, AD is equal to BJ), and DH to BE, and the angle ADH is equal to the angle BDlf, the triangles are equal in all respects; (I. Prop. 4.) therefore HA is equal to HB. Therefore GA, which is the sum of GH, HA, is equal to the sum of GH, HB. And the sum of GH, HB is greater than GB; (I. Prop. 20.) therefore GA is greater than GB. But all straight lines drawn from the centre to a circle are equal. (I. Def. 22.) Hence the point G cannot be the centre of the circle. Similarly it can be proved that no point on the same side of GE as A can be the centre; therefore the centre must be in CE. Wherefore, if a straight line &c. PROPOSITION 2. Corollary 1. Only one chord drawn through a point toithin a circle which is 7iot the centre can he bisected at the point. Corollary 2. If two chords of a circle bisect each other ^ their point of intersection is the centre. EXEECISES. 1. The straight line, which joins the middle points of two parallel chords of a circle, passes through the centre. 2. The locus of the middle points of parallel chords in a circle is a straight line. 3. Two equal parallel chords of a circle are equidistant from the centre. 4. Every parallelogram inscribed in a circle is a rectangle. 5. The diagonals of a rectangle inscribed in a circle are diameters of the circle. 6. If PQ, RS be two parallel chords of a circle and if PR, QS in- tersect in U and if PS, QR intersect in F, then UV passes through the centre. 178 BOOK 11 L PROPOSITION 3. If a straight liyie he drawn from the centre of a circle to the 7nid(Ue point of a chord, which is not a diameter, it is at right angles to the chord. Let ABC be a circle and F its centre, and let D be the middle point oi AB a chord, Avhich is not a diameter : it is required to prove that FD is at right angles to AB. Construction. Draw BA, EB. Proof. Because in the triangles ADE, BDE, AD \^ equal to BD, DE to DE, and EA to EB, the triangles are equal in all respects; (I. Prop. 8.) therefore the angle ADE is equal to the angle BDE, and they are adjacent angles. Therefore the straight lines EI), AB are at right angles to each other. (I. Def. 11.) Wherefore, if a straight line &c. PROPOSITION 3. 179 EXEKCISES. 1. Why are the words "which is not a diameter" inserted in enunciation of Proposition 3 ? 2. The straight Une, which joins the middle points of two parallel chords of a circle, is at right angles to the chords. 3. Circles are described on the sides of a quadrilateral as diame- ters: shew that the common chord of the circles described on two adjacent sides is parallel to the common chord of the other two circles. 4. A straight line is drawn intersecting two concentric circles : prove that the portions of the straight line which are intercepted between the circles are equal. 5. A straight line cuts two concentric circles in P, E, and Q, S: prove that the rectangle PQ, QR and the rectangle FQ, PS are con- stant for all positions of the line. 180 BOOK III. PROPOSITION 4. If a straight liiie he drawn from the centre of a circle at right angles to a chord, it bisects the chord. Let A BC be a circle, and D its centre, and let the straight line DE he drawn at right angles to AB a chord, which is not a diameter*: it is required to prove that DE bisects AB, that is, that J^ is equal to BE. Construction. Draw DA, BB. Proof. Because DB is equal to DA, each being a radius of the circle, the angle DAB is equal to the angle DBA. (I. Prop. 5.) And the angle DEA is equal to the angle DEB, each being a right angle. Then because in the triangles EAD, EBD, the angle EAD is equal to the angle EBD, and the angle DEA to the angle DEB, and ED, a side opposite to a pair of equal angles, is common, the triangles are equal in all respects ; (I. Prop. 26, Part 2.) therefore ^^ is equal to BE. Wherefore, if a straight line &c. The case when the chord is a diameter requires no proof. PROPOSITION 4. 181 In Propositions 2, 3, 4 we have to deal with tliree properties of a straight line : (a) the passing through the centre of a circle, (6) the being at right angles to a given chord, (c) the bisecting the given chord. It is proved in these propositions that, if a straight line have any two of these three properties, it necessarily has the third property. Proposition 2 deduces {a) from (6) and (c) ; Proposition 3 deduces ih) from (c) and (a); Proposition 4 deduces (c) from (a) and (Z>). EXEKCISES. 1. Two chords are drawn through a point on a circle equally inclined to the radius drawn to the point : prove that the chords are equal. 2. If ABPQ, ABBS be two circles and BB, QS be any two parallel straight lines drawn through the points of section, then PB, QS are equal. 3. If A and B be two fixed points and P move so that the per- pendicular from A on BP bisects BP, the locus of P is a circle. 4. Draw through a point of intersection of two circles a straight line to make equal chords in the two circles. 5. The locus of the middle points of all chords drawn through a fixed point on a circle is a circle. 0. If two circles PAB, QAB intersect each other at A, the locua of the middle point of a straight line PQ drawn through J is a circle. 182 BOOK III. PROPOSITION 5. To find the centre of a given circle. Let ABC be a given circle: it is required to find its centre. Construction. Draw any two chords which are not parallel and which cut the circle in A^ B, and in C, D. Bisect AB and CD at E and F; (I. Prop. 10.) and draw EG^ EG at right angles to AB, CD respectively meeting* at G : (I. Prop. 11.) then G is the centre of the circle ABC. Proof. Because the straight line EG bisects the chord AB at right angles, EG passes through the centre; (Prop. 2.) and because the straight line EG bisects the chord CD at right angles, EG passes through the centre. (Prop. 2.) Therefore the point G, where the two lines EG, EG meet, is the centre. " Wherefore, the centre G of the given circle ABC has been found. * The lines must meet. See Ex. 2, p. 51. PROPOSITION 5. 183 Outline of Alternative Construction. Draw any cliord AB, of the circle ABC. Bisect AB \n D, and draw EDF at right angles to AB, meeting the circle in E and F. Bisect EF in G. It may be proved that (1) the centre of the circle ABC is in EF; (2) no other point but G can be the centre. EXEKCISES. 1. Draw all the lines, which are wanted to find the centre of a given circle, (a) in the method given in the text ; (6) when the two chords in this method meet on the circle; (c) in the alternative method above. Which method requires the fewest lines ? 2. Draw through a given point within a circle a chord such that it is bisected at the point. 3. Describe a circle with a given centre to cut a given circle at the extremities of a diameter. 184 BOOK II L PROPOSITION 6. Every chord of a circle lies within the circle. Let .4^ be the chord joining any two points A, B on tlie circle ABC: it is required to prove that any point on the chord AB between A and B is within the circle. Construction. Pind the centre D of the circle; (Prop. 5.) take any point E on AB between A and B and draw DA, BE, DB, Proof. Because in the triangle DAB, DB is equal to DA, the angle DAB is equal to the angle DBA; (I. Prop. 5.) but the exterior angle DEB of the tri- angle DAE is greater than the inte- rior opposite angle DAE; (I. Prop. 16.) therefore the angle DEB is greater than the angle DBE. And because in the triangle DEB, the angle DEB is greater than the angle DBE, the side DB is greater than the side DE ; (I. Prop. 19.) that is, DE is less than DB which is a radius of the circle. Therefore the point E is within the circle. (Def. 5.) But E is any point on the chord AB between A and B, and ^^ is the chord joining any two points on the circle. Wherefore, every chord of a circle &c. EXEKCISES. 1. If a chord of a circle be produced, the produced parts lie without the circle. 2. Describe a circle which shall pass through two given points, and which shall have its radius equal to a given straight line greater in length than half the distance between the points. How many such circles are possible? 3. Draw a straight line to cut two equal circles in P, Q and B, S so that the straight lines PQ, QR, RS may be equal. What condition is necessary that such a straight line can be drawn ? PROPOSITIONS 6 AND 7. 185 PROPOSITION 7. If two circles have a cojiimon pointy tliey cannot have the same centre. Let the two circles ABC, ADE meet one another at the point A: it is required to prove that they cannot have the same centre. Construction. Find F the centre of one of the circles ABC. (Prop. 5.) Draw any straight line FCE meeting the circles at two distinct points C and E, and draw FA. Proof. Because F is the centre of the circle ABC\ FC is equal to FA. (I. Def. 22.) But FE is not equal to FC ; therefore FE is not equal to FA ; that is, two straight lines FE^ FA drawn to the circle ADE from the point F are not equal. Tlierefore F is not the centre of the circle ADE. (I. Def. 22.) Wherefore, if two circles tfec. Corollary. 2^wo concentric circles cannot have a coimrhon point, T. E. 13 186 BOOK III. Definition. A point is said to rotate about anotJier point, when the first point moves along a circle, of which the second point is the centre. A finite straight line is said to rotate about a point, when each of its extremities moves along a circle, of which the point is the centre, while the line remains of constant length. A plane figure is said to rotate about a point, when each oftvjo points fixed in the figure moves along a circle, ofwhicJb the point is the centre, while the figure remains unchanged in shape and size. ADDITIONAL PKOPOSITION. Any finite straight line may he shifted from any one position in a plane to any other by rotation about some point in the plane. Let AB, A'B' be any two positions of a finite straight line in a plane : it is required to prove that the line can be shifted from the position AB to the position A'B' hy rotation about some point in the plane. Draw AA'^ BB' and bisect them in M, N, and draw MO, NO at right angles to AA',BB' meeting in 0. Draw OA, OB, OA', OB'. Because in the triangles AOM, A'OM, AM is equal to A'M, and OM to OM, and the angle OMA to the angle OMA', the triangles are equal in all respects; (I. Prop. 4.) therefore OA is equal to OA'. '^ B Similarly it can be proved that OB is equal to OB'. ROTATION. 187 Again, because in the triangles OAB, OA'B', OA is equal to OA', OB to OB', and AB to A'B', the triangles are equal in all respects ; (I. Prop. 8.) therefore the angle AOB is equal to the angle A'OB' : add to each the angle BOA' ; then the angle AOA' is equal to the angle BOB'. It appears therefore that the triangle OAB can be shifted into the position OA'B' by being turned in its own plane round the point O through an angle AOA' or BOB' ; therefore AB can be shifted to A'B' by rotation round the point O. EXERCISES. 1. About what point must AB one side of a parallelogram ABCD rotate in order to take (1) the position CD, (2) the position DC? ^ 2. Prove that, when a straight line rotates about a point, every point in the line rotates about the point through the same angle. 3. Any triangle can be shifted from any one position to any other position, which it can occupy in the same plane without being turned over, by rotation about some point in the plane. 4. Prove that, when a plane figure rotates about a point, every point in the figure rotates about the point through the same angle. 5. Describe an equilateral triangle of which one angular point is given and the others lie on two given straight lines. How many solutions are there ? 6. Construct an equilateral triangle, one of whose angular points is given and the other two lie one on each of two given circles. Find the limits of the position of the given point which admit of a possible solution. 7. Construct a square to have one vertex at a fixed point and two opposite vertices on two given straight lines. 13—2 188 BOOK HI. PROPOSITION 8. Part 1. Of all straiyht lines drawn to a circle from a 2>oint o7i the circle^ the line ivhich is a diameter is the greatest; and of any two others, the one which subtends the greater angle at tlie centre is the greater. Let CD^ be a given circle, A its centre, and J5 any point on the circle; let BAU be a diameter, and let BC, BI) be any other two straight lines drawn from B to the circle, and of the angles BAG, BAD subtended by BC, BD at A let the angle BAD be the greater: it is required to prove that BE is greater than BD, and BD greater than BC. Proof. Because AE is equal to AD, therefore BE, which is the sum of BA, AE, is equal to the sum of BA, AD: but the sum of BA, AD is greater than BD ; (I. Prop. 20.) therefore BE is greater than BD. 'Next, because in the triangles BAD, BAC, AD i^ equal to AC, and BA to BA, and the angle BAD is greater than the angle BAC, therefore BD is greater than BC. (I. Prop. 24.) Wherefore, of all straight lines &c. Corollary. A diameter is the greatest chord of a circle. PROPOSITION 8. PART 1. 189 EXERCISES. 1. If two chords of a circle subtend equal angles at the centre, they are equal. 2. If two chords of a circle be equal, they subtend equal angles at the centre. 3. Of any two chords in a circle the one which subtends the greater angle at the centre is the greater. 190 BOOK III. PROPOSITION 8. Part 2. Of all straight lines drawn to a circle from an internal imint not the centre^ the one which passes through the centre is tJie greatest^ and the one which when produced passes through the centre is the least ; and of any two others^ the one which subtends the greater angle at the centre is the greater. Let ODE be a given circle, A its centre and B any other internal point ; let BA produced beyond A cut the circle in E^ and produced beyond B in F^ and let BC^ BD be any other two straight lines drawn from B to the circle, and of the angles BAG, BAD subtended by BG, BD at A let the angle BAD be the greater: it is required to prove that BE is greater than BD^ BD greater than BG^ and BG greater than BF. Proof. Because AE is equal to AD, therefore BE^ which is the sum of BA, AE, is equal to the sum of BA, AD; but the sum of BA, AD is greater than BD; (I. Prop. 20.) therefore BE is greater than BD. Next, because in the triangles BAD, BAG, AD is equal to AG, BA to BA, and the angle BAD is greater than the angle BAG, therefore BD is greater than BG. (I. Prop. 24.) Again, because the sum of BG, BA is greater than AG, (I. Prop. 20.) and AG is, equal to AF, which is the sum of BF, BA, therefore the sum of BG, BA is greater than the sum of BF, BA. Therefore BG is greater than BF. Wherefore, of all straight lines (fee, PROPOSITION 8. PART 2. 191 EXEKCISES. 1. If two straight lines drawn to a circle from an internal point not the centre be equal, they subtend equal angles at the centre. 2. If two straight lines drawn to a circle from an internal point not the centre subtend equal angles at the centre, they are equal. 3. If each of two equal straight lines have one extremity on one of two concentric circles and the other extremity on the other, the lines subtend equal angles at the common centre. .19:2 BOOK III. PROPOSITION 8. Part 3. Of all straight lines drawn to a circle from a?i external 'point, the one which passes through the centre is the gi'eatest, and the one which when produced passes through the centre is the least ; and of any two others, the one which subtends the greater angle at the centre is the greater. Let ODE be a given circle, A its centre and B a given external point; let BA cut the circle in F and let BA produced cut the circle in E, and let BD, BG be any other two straight lines drawn from B to the circle, and of the angles 7i^C, BAD subtended by ^(7, ^i> at^ let the angle BAD be the greater : it is required to prove that BE is greater than BD, BD greater than BG, and BG greater than BF. Proof. Because AE is equal to AD, therefore BE, which is the sum of BA, AE, is equal to the sum of BA, AD : but the sum of BA, AD is greater than BD ; (I. Prop. 20.) therefore BE is greater than BD. Next, because in the triangles BAD, BAG, AD is equal to AG, and BA to BA, and the angle BAD is greater than the angle BAG, therefore BD is greater than BG. (I. Prop. 24.) Again, because the sum of BG, GA is greater than BA, (I. Prop. 20.) which is the sum of BF, FA , and because CA is equal to FA, therefore BG is greater than BF. Wherefore, of all straight lines &c. PROPOSITION 8. PART 3. 193 We conclude from the results of the several Parts of Proposition 8 that, if be a fixed point on the diameter 4(7 of a circle ABCD nearer to A than to G and P be a point which is capable of motion along the circum- ference in the direction represented by the arrow, while P is moving along the arc ABC from A to C the distance OP increases continuously from OA to OG, and while P is moving along the arc GDA from G to A, the distance OP decreases continuously from OG to OA. We say therefore that OG is a maximum value of OP, and OA is a minimum value. (See remarks on page 55.) It may be observed here that, if P travel round the circle any number of times, it passes G and A alternately. It appears therefore that here maximum and minimum values occur alternately. The occurrence of maximum and minimum values alternately is true generally in the case of quantities which vary continuously, i.e. quantities whose magnitude changes without suffering any abrupt changes. EXERCISES. 1. Find the shortest distance between two points one on each of two circles which do not meet. 2. A and B are two fixed points; it is required to find a point P on a given circle, so that the sum of the squares on AP and BP may be the least possible. Under what conditions is the solution indeterminate? 194 BOOK III. PROPOSITION 9. From a point not tlie centre not Tnore than two equal straight lines can he drawn to a circle, one on each side of the straight line drawn from that point to the centre. Let J. be a given point, and BCD a given circle, and let AB, AD be two equal straight lines drawn from A to the circle : it is required to prove that no other straight line equal to ^^ or AD can be drawn from A to the circle. Construction. Find E the centre of the circle; (Prop. 5.) draw EA, EB, ED. J> D Proof. Take C any point of the circle on the same side of AE as AB. Because B and C are equidistant from E, they cannot be equidistant from A. (I. Prop. 7.) Therefore there cannot be another straight line equal to AB drawn from A to the circle on the same side of AE as AB. Similarly it can be proved that there cannot be another straight line equal to AD drawn from A to the circle on the same side of AE as AD. Wherefore, from a point equi- distant from the centre : it is required to prove that ^^ is equal to CD. Construction. Find E the centre of the circle ABCD; (Prop. 5.) and from E draw EF, EG at right angles to A B, GB. (T. Prop. 12.) Draw EA, EG. Proof. Because the straight line EF is drawn through the centre of the circle at right angles to the chord AB^ it bisects it ; (Prop. 4.) that is, ^^ is double of AF. Similarly it may be proved that GB is double of CG. Next, because the angles AFE^ GGE are right angles, the square ot\. AE is equal to the sum of the squares on AF, FE, and the square on GE is equal to the sum of the squares on GG, GE, (I. Prop. 47.) And because ^^ is equal to GE, the square on AE is equal to the square on GE. Therefore the sum of the squares on A F, FE is equal to the sum of the squares on GG, GE. Because EF is equal to EG, the square on EF is equal to the square on EG; therefore the square on ^i^ is equal to the square on GG ; therefore AF \^ equal to GG. PROPOSITION U. PART 2. 207 And it has been proved that AB \& double of AF^ and CD oiCG, Therefore AB \& equal to CD. Wherefore, chords of a circle &c. Parts 1 and 2 of Proposition 14 are the converses of each other. EXEECISES. 1. lu a circle chords, which are equidistant from the centre, subtend equal angles at the centre. 2. In a circle chords, which subtend equal angles at the centre, are equal. 208 BOOK III. PROPOSITION 15. Part 1. Of any two chords of a circle the otie which is the greater is the nearer to the centre. Let AB^ CD be two chords of the circle ABC I), of which -4 ^ is greater than CD : it is required to prove that ^^ is nearer to the centre than CD. Construction. Find £J the centre of tlie circle ABCD; (Prop. 5.) and from E draw UF, EG at right angles to AB, CD. (I. Prop. 12.) Draw EA. EG. Proof. Because the straight line EF is drawn through the centre at right angles to the chord AB, AF is equal to FB, (Prop. 4.) and AB is double of AF. Similarly it can be proved that CD is double of CG. But ^^ is greater than CD ; therefore AF is greater than CG. Next, because the angles AFE, CGE are right angles, the square on AE is equal to the sum of the squares on AF, FE, and the square on CE is equal to the sum of the squares on CG, GE. (I. Prop. 47.) And because AE is equal to CE, the square on ^4^ is equal to the square on CE. Therefore the sum of the squares on AF, FE is equal to the sum of the squares on CG, GE ; PROPOSITION 15. PART 1. 209 Because AF is greater than CG, the square on AF \^ greater than the square on CG\ therefore the square on FE is less than the square on GE, Therefore FE is less than GE, that is, ^^ is nearer to tlie centre than CD. Wherefore, of any two chords lie on AB, one of the lines which contain the angle BAG. Construction. Draw DG. - Proof. Because DA is equal to DG, the angle DGA is equal to the angle DAG; (I. Prop. 5.) therefore the sum of the angles DAG, DGA is double of the angle DAG. But the angle BDG is equal to the sum of the angles DAG, DGA; (I. Prop. 32.) therefore the angle BDG is double of the angle DAG. Next, let the centre D lie within (fig. 2) or without (fig. 3) the angle BAG. Construction. Draw AD and produce it to meet the circle in B. Proof. It follows from the first case, that the angle UDG is double of the angle BAG, and that the angle HDB is double of the angle UAB ; PROPOSITION 20. 221 therefore in (fig. 2) the sum of the angles EDO, EBB is double of the sum of the angles EAG, EAB, and in (fig. 3) the difference of the angles EDO, EDB is double of the difference of the angles EAG^ EAB ; therefore in all cases the angle BDG is double of the angle BAG. Wherefore, the angle vjhich an arc &c. In the diagram of Proposition 20 in each of the figures the angle BDC is double of the angle BAC. Now it is easily seen that although in figures (1) and (3) the angle BAC is restricted to values less than a right angle, and the angle BDG in consequence to values less than two right angles, in figure (2) the angle BAG is restricted only to values less than two right angles and the angle BDG in consequence only to values less than four right angles. It appears therefore that, if we wish not to destroy the generality of the theorem of Proposi- tion 20, we must allow our definition of an angle to include angles w;hich^are equal to two or greater than two right angles; there is nothing inconsistent with a strict adherence to Euclid's methods in doing so. ^"~~ ■ EXERCISES. 1. Two circles, whose centres are A and D, touch externally at E : a third circle, whose centre is B, touches them internally at C and F: prove that the angle ADB is double of the angle EGF. 2. If AB be a fixed diameter and DE an arc of constant length in a fixed circle, and the straight lines AE, BD intersect at P, the angle APB is constant. 3. If ^BC be a triangle inscribed in a circle and the angle BAG be bisected hy AD, which meets the circle in D, then the diameter through D will bisect BG &t right angles. 4. AB is a diameter and PQ any chord of a circle cutting AB within the circle, and AL is drawn perpendicular to PQ. Prove that the angle LAB is equal to the sum of the angles PAB, QAB. 222 BOOK III. PROPOSITION 21. Angles in the same arc of a circle are equal. Let ABCD be a circle, and BAC^ BBC be two angles in the same arc BADC: it is required to prove that the angles BAG, BBC are equal. Construction. Find the centre E; (Prop. 5.) and draw EB, EG. Proof. Because the angle which the arc BFG of the circle subtends at the centre is double of the angle which it subtends at the circumference, the angle BEG is double of the angle BAG., and also the angle BEG is double of the angle BBC ; * (Prop. 20.) therefore the angle BAG is equal to the angle BBG. Wherefore, angles in the same arc &c. Corollary. If a straight line joining two points subtend equal angles at two other points on the same side of the linCj the four points lie on a circle. ' Let the straight line BG subtend equal angles at the two points ^, J9 on the same side of BG. If a circle be described about the triangle BAG'^', the * That it is possible to describe a circle through the three vertices of a triangle appears in the Additional Proposition on page 53. PROPOSITION 21. 223 circle must cut BD again at some point not on the same side of ^C as B. (Prop. 6.) Now take H any point but D in BD or BD produced and draw HD. Then the angle BHC cannot be equal to the angle BDC, (I. Prop. 16.) and therefore cannot be equal to the angle BAC. But angles in the same arc of a circle are equal. (Prop. 21.) Therefore the circle BAC cannot meet BD in H ; that is, it must meet it in D. In some books in the proof of Proposition 21, the result of Pro- position 20 is quoted as if it were true only in the case of arcs greater than a semicircle: that is, as if the angle BEC, which the arc BFC subtends at the centre, were restricted to magnitudes less than two right angles. The general truth of the theorem is then deduced as a consequence. "We leave this deduction to the student as an exercise. EXEECISES. 1. The locus of a point at which a given straight line subtends a constant angle is an arc of a circle. 2. If of three concurrent straight lines inclined at given angles to one another two pass through two fixed points, the third also passes through a third fixed point. 3. If two sides of a triangle of constant shape and size pass through two fixed points, the third ahvays touches a fixed circle. 4. If two sides of a triangle of constant shape and size always touch two fixed circles, the third side always touches a fixed circle. 5. If ABC be an equilateral triangle described in a circle whoso centre is 0, and if -40 produced meet the circle in D, then OD,BC bisect each other. 6. Two circles ADB, ACB intersect in points^ andU. Through A any chord BAG is drawn, and BC, BD are joined, and the angle DBG is internally bisected by a line BE which meets BG in E. Shew that E lies on a fixed circle. 7. If ABG be an isosceles triangle on the base BG, inscribed in a circle, and P, Q be points on the arcs AG,AB respectively of the circle such that AQ i^ parallel to BP, then GQ is parallel to AP. 8. If the diagonals of a quadrilateral inscribed in a circle be at right angles, the perpendicular from their intersection on any side bisects the opposite side. 224 BOOK III. PROPOSITION 22. The sum of two opposite angles of a convex quadrilateral inscribed in a circle is equal to two rigid angles. Let A BCD be a quadrilateral inscribed in the circle ABCD: it is required to prove that the sum of the angles ABG^ ADC is equal to two right angles, and that the sum of the angles BAD^ BCD is equal to two right angles. Construction. Draw AC, BD. Proof. Because the angles BCA, BDA are in the same arc BCDA^ the angle BCA is equal to the angle ^Z>^ j (Prop. 21.) and because the angles CAB, CDB are in the same arc CDAB, the angle CAB is equal to the angle CDB. (Prop. 21.) Therefore the sum of the angles BCA, CAB is equal to the sum of the angles BDA, CDB, that is, to the angle ADC. To each of these equals add the angle J ^C : then the sum of the angles ABC, BCA, CAB is equal to the sum of the angles ABC, ADC. But because the angles ABC, BCA, CAB are the angles of a triangle, their sum is equal to two right angles. (I. Prop. 32.) Therefore the sum of the angles ABC, ADC is equal to two right angles. Similarly it can be proved that the sum of the angles BAD, BCD is equal to two right angles. Wherefore, the sum of two opposite angles &c. PROPOSITION 22. 225 Corollary. If the sum of two opposite angles of a convex quadrilateral he equal to two right angles., the vertices of the quadrilateral lie on a circle. Let ABCD be a convex quadrilateral in which the sum of the angles BAD^ BCD is equal to two right angles. If a circle be described about the triangle BAD^ the circle must cut AC again in some point not on the same side of BD as A. (Prop. 6.) Now take H any point but C in AC ov AC produced and draw HB, HD. Then the angle BUD cannot be equal to the angle BCD. (I. Prop. 21.) Therefore the sum of the angles BAD., BHD cannot be equal to two right angles. But the sum of two opposite angles of a convex quadri- lateral inscribed in a circle is equal to two right angles. (Prop. 22.) Therefore the circle BAD cannot meet AC in H ', that is, it must meet it in C. EXERCISES. 1. If the sides AB, DC of a quadrilateral ABCD inscribed in a circle be produced to meet at E, the triangles AEC, BED are equi- angular to one another. 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. If PQRS, pqrs be two circles, and PprR, QqsS be chords such that P, I), q, Q lie on a circle, then R, r, s, S lie on a circle. 4. If any two consecutive sides of a convex hexagon inscribed in a circle be respectively parallel to their opposite sides, the remaining sides are parallel to each other. 5. If any arc of a circle described on the side JB(7 of a triangle ABC cut BA, CA produced if necessary in P and Q, PQ is always parallel to a fixed straight line. 6. £ is a point on one of the diagonals ^C of a parallelogram ABCD. Circles are described about DEA and BEG. Shew that BD passes through the other point of intersection of the circles. 226 BOOK III. PROPOSITION 23. Two arcs of circles, y)hich have a common chord and are on the same side of it, cannot he similar unless they are coincident. Let ABC, ADC be two arcs of circles, which have a common chord A (7, and are on the same side of it : it is required to prove that ABC, ADC cannot be similar arcs, unless they are coincident. Construction. Draw through A, one of the extremities of the chord AB, any straight line ABD to meet the arcs in B, D; and draw CB, CD. . G Proof. If the points B, D do not coincide, one of the angles ABC, ADC is an exterior angle and the other an interior angle of the triangle BCD ; therefore the angle ABC is not equal to the angle ADC, (I. Prop. 16.) and therefore the arc ABC is not similar to the arc ADC. (Def. 4.) It has now been proved that, if any straight line ABD meet the arcs in two points B and D which are not coincident, the arcs cannot be similar. Therefore, if the arcs be similar, every straight line drawn through A must meet the arcs in two coincident points, that is, the arcs ABC, ADC must coincide. Wherefore, two arcs of circles &c. PROPOSITION 23. 227 EXERCISES. 1. If on opposite sides of the same straight line there be two arcs of circles, which contain supplementary angles, the arcs are parts of the same circle. 2. Prove that, if two circles have three points in common, the circles are coincident. 228 BOOK III. PROPOSITION 24. Similar arcs of circles^ which have equal chords, are equal. Let ABGy DEF be two similar arcs of circles, which have equal chords AC, DF : it is required to prove that the arcs ABC, DEF are equal. Proof. Because the chords AC, BF are equal, it is possible to shift the figure ABC, so that AC coincides with DF, A with D, and C with F, (i. Test of Equality, page 5) and so that the arcs ABC, DEF are on the same side of DF. If this be done, the arc ABC must coincide with the arc DEF, for the arcs ABC, DEF are similar, and two arcs of circles, which have a common chord, and are on the same side of it, cannot be similar unless they coincide. (Prop. 23.) Therefore the arcs ABC, DEF are equal. Wherefore, similar arcs of circles &c. PROPOSITION 24. 229 EXEBCISES. 1. If D be a point in the side BG of a triangle ABG whose sides AB, AG are equal, the circles described about the triangles ABD, AGD are equal. 2. Find a point P within an equilateral triangle ABG, such that the circles described about the triangles PBG, PGA, PAB may be all equal. 3. Find a point P in the plane of a triangle ABG such that the circles described about the triangles PBG, PGA, PAB maybe equal. 230 BOOK III. PROPOSITION 25. TofiTid the centre of the circhy of which a given arc is a part. Let ABC be a given arc : it is required to find the centre of the circle, of which the arc ABC is a part. Construction. Draw AC, and bisect it at J). (I. Prop. 10.) At I> draw DB at right angles to ^0 cutting the arc at B. (I. Prop. 11.) Draw ABj bisect it at B, and at B draw BF at right angles to AB meeting BD or BD produced at F*; then F is the centre required. (1) B (V B (s) B /e/ ^ \ ^\ D \ {/ \ J A ^ G D Proof. Because DB bisects the chord il(7 at right angles, DB passes through the centre; (Prop. 2.) and because EF bisects the chord J.^ at right angles, EF passes through the centre. (Prop. 2.) Now two straight lines cannot intersect in more than one point. (I. Post. 1.) Therefore F, the point of intersection of BD and EF, is the centre. Wherefore, the centre of the circle, of which the given arc ABC is a part, has been found. * The lines must meet, see Ex. 2, p. 51. In figure (2) F coincides with D, PROPOSITION 25. 231 PROPOSITION 25 A. Equal circles have equal radii. If two circles be equal, it is possible to shift one of them so as to coincide with the other. (I. Def. 21, page 13.) Let this be done. Then, because a circle cannot have more than one centre, (Prop. 1.) the centres of the two coincident circles must be coincident : and therefore all the radii of both circles are equal. Wherefore, equal circles (kc. EXERCISES. 1. Having given two arcs of circles, shew how to find whether they are parts of the same circle. 2. Having given two arcs of circles, find whether they are parts of concentric circles. 3. Having given two arcs of circles, find whether one circle lies wholly within the other. 232 BOOK III. PROPOSITION 26. In equal circles the arcs^ on which equal angles at the centres standi are equal ; and the arcs, on which equal angles at the circumferences stand, are equal. Let ABCD, EFGE be two equal circles, and let AKC, ELG be two angles at the centres standing on the arcs ABC, EHG, and let ABC, EFG be two angles at the cir- cumferences standing on the same arcs : and let (1) the angles A KG, ELG be equal, (2) the angles ABC, EFG be equal : it is required in either case to prove that the arcs ADC, EHG are equal. Construction. Draw AC, EG, Proof. Because the angle AKC is double of the angle ABC, (Prop. 20.) and the angle ELG is double of the single EFG, (Prop. 20.) in case (1) because the angles AKC, ELG are equal, the angles ABC, EFG are equal, and in case (2) because the angles ABC, EFG are equal, the angles AKC, ELG are equal. Now because the circles are equal, their radii AK, KC, EL, LG are equal. (Prop. 25 A.) Therefore in both cases (1) and (2), because in the triangles AKC, ELG, AK \& equal to EL, KC to LG, and the angle AKC to the angle ELG, the triangles are equal in all respects ; (I. Prop. 4.) therefore AC i^ equal to EG. PROPOSITION^ 26. 233 And because the arcs ABC, EFG^ which contain equal angles, have equal chords AC, EG, the arcs ABC, EFG are equal : (Prop. 24.) but the circles ABCD, EFGH are equal ; therefore the remaining arcs ADC, EHG are equal. Wherefore, in equal circles kc. Corollary. In the same circle the arcs, on which equal angles at the centres stand, are equal; and the arcs, on which equal angles at tlie circumferences stand, are equal. EXERCISES. 1. If PQ, MS he a pair of parallel chords in a circle, then the arcs PS, QR are equal, and the arcs PR, QS are equal. 2. A quadrilateral is inscribed in a circle, and two opposite angles are bisected by straight lines meeting the circumference in P and Q ; prove that PQ is a diameter. 3. If through P any point on one of two circles, which intersect in A and B, the straight lines PA, PB be drawn and produced if necessary to cut the other circle in Q and R, the arc QR is x>i constant length. 4. The internal bisectors of the vertical angles of all triangles, on the same base and on the same side of it, which have equal vertical angles, pass through one fixed point and the external bisectors through another fixed point. 5. If through one of the points of intersection of two equal circles a straight line be drawn terminated by the circles, the straight lines joining its extremities with the other point of intersection are equal. T. K. 16 234 BOOK III. PROPOSITION 27. In equal circles, anyles coidaiited hij a7'cs, which are of equal length, are equal. Let A BCD, EFGH be equal circles, and let ABC, EFG be arcs of equal lengtli : it is required to prove that the arcs ABC, EFG contain equal angles. Construction. Find the centres K, L of the circles ABGD, EFGH, (Prop. 5.) and draw AK, KG, EL, LG. Proof. Because the circles ABCD, EFGH are equal, their radii AK, KG, EL, LG are equal. (Prop. 25 A.) And because AK is equal to EL, it is possible to shift the fi^Mve ABCD K so that A coincides with E, and K with L, and so that the parts of the arcs ABC, EFG near E are on the same side of EL. If this be done, because the radii of the circles are equal and their centres coincide, the circles must coincide ; and because the circles coincide, and the parts of the arcs ABC, EFG near E are on the same side of EL, those parts of the arcs coincide ; and because the arcs are of equal length and have one extremity common, therefore the other extremity must be common, that is, the point C must coincide with the point G, PROPOSITION 27. 235 Therefore the arc ABC coincides with the arc EFG ; and angles in the two arcs are then angles in the same arc and therefore equal. (Prop. 21.) Wherefore, in equal circles &c. Corollary. In the same circle^ angles contained by arcs, which are of equal length, are equal. EXERCISES. 1. The straight lines joining the extremities of two equal arcs of a circle are parallel or are equal. Can they be both parallel and equal ? 2. The straight lines bisecting any angle of a quadrilateral in- scribed in a circle and the opposite exterior angle, meet on the circle. 3. If from any point on a circle a chord and a tangent be drawn, the perpendiculars on them from the middle point of either of the arcs subtended by the chord are equal to one another. 16—2 236 BOOK III. PROPOSITION 28. In equal circles^ arcs cut off by chords, tvhich are equal to one another, are of equal length, tlie greater equal to the greater and the less equal to the less. Let ABCD, EFGH be equal circles, and let AC\ EG be equal chords which cut off the two greater arcs ABC, EFG, and the two less arcs ADC, EHG : it is required to prove that the arcs ABC, EFG are of equal length, and that the arcs ADC, EHG are of equal length. Construction. Find K, L, the centres of the circles ABCD, EFGH, (Prop. 5.) and draw AK, KC, EL, LG. Proof. Because the circles are equal, their radii AK, KC, EL, LG are equal; (Prop. 25 A.) and because in the triangles AKC, ELG, KA is equal to LE, KC to LG, and ^C to EG, the triangles are equal in all respects; (I. Prop. 8.) therefore the angle AKC is equal to the angle ELG. But in equal circles the arcs, on which equal angles at the centres stand, are equal; (Prop. 26.) therefore the arc ADC is equal to the arc EHG. But the circle ABCD is equal to the circle EFGH\ therefore the arc ABC is equal to the arc EFG. Wherefore, in equal circles &c. PROPOSITION 28. 237 Corollary. In the scmie circle, arcs cut off by chords, which are equal to one another, are of equal length, the greater equal to the greater a7id the less equal to the less. EXERCISES. 1. A triangle is turned about its vertex till one of the sides passing through the vertex is in the same straight line as the other previously was. Prove that the line joining the vertex with the in- tersection of the two positions of the base, produced if necessary, bisects the angle between these two positions. 2. Find a point on one of two given equal circles, such that, if from it two tangents be drawn to the other circle, the chord joining the points of contact is equal to the chord of the first circle formed by joining its points of intersection with the two tangents produced. Determine the conditions of the possibility of a solution of the problem. 238 BOOK III. PROPOSITION 29. In equal circles, chords, hy which arcs of equal length are subtended, are equal. Let ABCD, EFGII be equal circles, and let AC, EG be chords by wliicli ADC, EIIG, arcs of equal length are sub- tended : it is required to prove that the chords AC, EG are equal. Construction. Find K, L, the centres of the circles ABCD, EFGII, (Prop. 5.) and draw AK, EL. Proof. Because the circles ABCD, EFGH are equal, their radii AK, EL are equal. (Prop. 25 A.) And because AK \% equal to EL, it is possible to shift the figure ABCDK so that A coincides with E, and K with L, and so that the parts of the arcs ABC, EFG near E are on the same side of EL. If this be done, because the radii of the circles are equal and their centres coincide, the circles must coincide; and because the circles coincide, and the parts of the arcs ABC, EFG near E are on the same side of EL, those parts of the arcs coincide ; and because the arcs are of equal length and have one extremity common, therefore the other extremity must be common, that is, the point C must coincide with the point G. PROPOSITION 29. 239 Therefore the chord AC coincides with the cliord EG and is equal to it. Wlierefore, w equal circles ; (I. Prop. 10.) at ]) draw DB at right angles to AC meeting the arc at B: (I. Prop. 11.) the arc ABC is bisected as required at the point B. Draw AB, BC. Proof. Because in the triangles ADB, CDB^ A J) is equal to CD, and DB to DB, and the angle ADB to the angle CDB, the triangles are equal in all respects; (I. Prop. 4.) therefore ^^ is equal to CB. But arcs cut off by equal chords are equal, the greater equal to the greater, and the less equal to the less; (Prop. 28, Coroll.) and because BD, if produced, is a diameter, (Prop. 2.) each of the arcs AB, CB is less than a semicircle, and therefore the arc AB is, the smaller of the two arcs cut oif by the chord AB, and the arc CB the smaller of those cut off by the chord CB ; therefore the arc AB is equal to the arc CB. Wherefore, the given arc ABC is bisected at B. PROPOSITION 30. 241 EXERCISES. 1. Find the triangle of maximum area which can be inscribed in a given circle having a given chord for one side. 2. Prove that the triangle of maximum area inscribed in a circle is equilateral. 3. The greatest quadrilateral which can be inscribed in a circle is a square. 4. Having given a regular polygon of any number of sides in- scribed in a circle, inscribe a regular polygon of double the number of sides. 5. li ABC an arc of a circle less than a semicircle be bisected in B, and AB produced meet CI) which is drawn at right angles io BG in J), and the tangents at A and C meet in A', then B, C, D, E lie on a circle. 242 BOOK III. PROPOSITION 31. An angle in a semicircle is a right angle ; an angle in an arc, which is greater than a semicircle, is less than a right angle ; and an angle in an arc^ which is less than a semi- circle, is greater than a right angle. Let ABODE be a circle, of which ABD is a semicircle, ADC an arc greater than a semicircle, and ABC an arc less than a semicircle : it is required to prove that the angle in the semicircle ABD is a right angle; that the angle in the arc ADC is less than a right angle, and that the angle in the arc ABC is greater than a right angle. Construction. Take any point B in the arc ABC and any point E in the semicircle AED and draw AB, BC, CD^ DE, EA, AC, AD. Proof. Because ACDE is a quadrilateral inscribed in a circle, the sum of the angles ACD, AED is equal to two right angles. (Prop. 22.) But because each of the angles ACD, AED is contained by a semicircle, the angle ACD is equal to the angle AED ; (Prop. 27, Coroll.) therefore each of them is a right angle. Next, because the angle ACD of the triangle ACD is a right angle, the angle ADC is less than a right angle, (I. Prop. 17.) and it is an angle in the arc ADC. PROPOSITION 31. 243 Again, because ABCD is a quadrilateral inscribed in a circle, the sum of the angles ABC, ADC is equal to two right angles. (Prop. 22.) And the angle ADC has been proved to be less than a riglit angle ; therefore the angle ABC is greater than a right angle, and it is an angle in the arc ABC. Wherefore, an angle in a semicircle is an arc described as required. Proof. Because in the triangles FEA, FEB, EA is equal to EB, and FE to FE, and the angle FEA to the angle FEB, the triangles are equal in all respects ; (I. Prop. 4.) therefore FA is equal to FB. * The lines must meet. See Ex. 2, p. 51. PROPOSITION 33. 247 Therefore the circle AGH passes through B. Again, because AD is drawn from A at right angles to the radius AF, AD touches the circle; (Prop. 16.) and Ijecause the chord AB is drawn from the point of contact of the tangent AD, the angle in the alternate arc AGB is equal to the angle DAB, (Prop. 32.) that is, to the angle C. Wherefore, on the given straight line AB the arc AGB /ms been described containing an angle equal to the given angle C. EXEECISES. 1. Find a point at which each of two given finite straight lines subtends a given angle. 2. Construct a triangle, having given the base, the vertical angle, and the foot of the perpendicular from the vertex on the base. 3. Having given the base and the vertical angle of a triangle, construct the triangle which will have the maximum area. 4. Find a point within a given triangle ABC, so that, if ^O, BO, CO be joined, the angles OAB, OBC, OCA shall be all equal. 5. Construct a triangle, having given the base, the vertical angle, and the altitude. 6. Find the locus of a point at which two given equal straight lines AB, BG subtend equal angles. 248 BOOK III. PROPOSITION 34. To cut off from a given circle an arc containi^ig 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 circle ABC an arc con- taining an angle equal to the angle D. Construction. Take any point A on the circle, and through A draw the straight line EAF to touch the circle at A. (Prop. 17.) From A draw AB making the angle BAE equal to the angle jD, and cutting the circle again at B : (I. Prop. 23.) then the arc ACB, on the side oi AB away from E^ is an arc cut off as required. Proof. Because the chord ^^ is drawn from the point of contact A of the tangent EAF, the angle EAB is equal to the angle in the alternate arc ACB. (Prop. 32.) And the angle EAB is equal to the angle D : therefore the angle in the arc ACB is equal to the angle D. "Wherefore, the arc ACB has been cut off from the given circle ABC containing an angle equal to the given angle D. PROPOSITION 34. 249 Through a given point two chords can be drawn which will cut off arcs containing an angle equal to a given angle. Outline of Alternative Construction. Through A draw any chord AP, and from P draw PR making the angle RPA equal to the given angle D, and cutting the circle again &tQ. It may be proved that the arc ABQ, measured from A on the side of ^P opposite to that on which PR is drawn, is an arc cut off as required. EXERCISES. 1. In a given circle inscribe an equiangular triangle. 2. Inscribe in a given circle a triangle, so that one angle may be a half of a second angle and a third of the third angle. 3. Inscribe in a given circle a right-angled triangle, so that one of its acute angles may be three times the other, T. K. 17 250 BOOK III, PROPOSITIOISr 35. If two chords of a circle intersect at a point within the circle^ the rectangle contained by the segments of one chord, is equal to the rectangle contained by the segynents of the other chord. Let ABGD be a circle and AG^ BD two chords inter- secting at the point E within the circle : it is required to prove that the rectangle contained by AEj EC is equal to the rectangle contained by BE, ED. Construction. If the point E be the centre, it is clear that AE, EC, BE, ED are equal, each being a radius, and that the rectangles AE, EG and BE, ED, each of which is equal to the square on a radius, are equal. If E be not the centre, find the centre ; (Prop. 5.) draw OF &t right angles to AG, (I. Prop. 12.) and draw OE, OG. Proof. Because OF is drawn from the centre at right angles to the chord AG, therefore AF is equal to FG. (Prop. 4.) And because EC is the sum of FG, EF, and ^^ is the difference of AF, EF, that is, of FG, EF, therefore the rectangle AE, EG is equal to the difference of the squares on FG, EF. (II. Prop. 5.) But because the angles at F are right angles, the sum of the squares on OF, FG is equal to the square on OG, and the sum of the squares on OF, EF is equal to the square on OE; (I. Prop. 47.) PROPOSITION 35. 251 therefore the difference of the squares on FG^ EF is equal to the difference of the squares on 0C\ OE. Therefore the rectangle AE, EC is equal to the difference of the squares on 0(7, OE. Similarly it can be proved that the rectangle BE, ED is equal to the difference of the squares on OB, OE, that is, is equal to the difference of the squares on OC, OE, since OC is equal to OB. Therefore the rectangle AE, EC is equal to the rectangle BE, ED. Wherefore, if two cJwrds of a circle kc. There are two special cases which should be noticed by the student, one case, when the points E, F coincide, i.e. when one chord bisects the other ; the other case, when the points 0, F coincide, i. e. when one chord is a diameter. In Proposition 35 the distances between the ends of a chord and a point in the chord are spoken of as segments of the chord. In Pro- position 36 it will be noticed that the expression segments of a chord has been used of the distances between the ends of the chord and a point taken in the chord produced. In the first case the chord is equal to the sum of the segments, in the second to the difference of the segments. EXEBCISES. 1. Prove the converse of Proposition 35, i.e. that, if -4C, BD be two straight lines intersecting at E such that the rectangles AE, EG, and BE, ED are equal, then A, B, 0, D lie on a circle. 2. If through any point in the common chord of two intersecting circles there be drawn any two other chords, one in each circle, their four extremities all lie on a circle. 3. Draw through a given point within a circle a chord, one of whose segments shall be four times as long as the other. When is this possible? 4. Divide a given straight line into two parts, so that the rect- angle contained by the parts may be equal to a given rectangle. 5. A, B, G are three points on a circle, D is the middle point of BG and AD produced meets the circle in E: prove that the sum of the squares on AB, AG is double of the rectangle AD, AE. 17—2 252 BOOK III. PROPOSITION 36. If two chords of a circle when produced intersect at a point without the circle, the rectangle contained by the segments of one chord is equal to the rectangle contained by the segments of the other chord. Let ABDG be a circle and let GA^ DB be two chords, which intersect, when produced beyond A and B^ at the point E without the circle : it is required to prove that the rectangle contained by EA, EC is equal to the rectangle contained by EB, ED. Construction. Find the centre; (Prop. 5.) drav/ OF at right angles to ACy (I. Prop. 12.) and draw OE, 00. Proof. Because OF is drawn from the centre at right angles to the chord AG, therefore AF is equal to FG. (Prop. 4.) And because EG is the sum of EF, FG, and EA is the difference of EF, AF, that is, of EF, FG, therefore the rectangle EA, EG is equal to the difference of the squares on EF, FG. (II. Prop. 6.) But because the angles at F are right angles, the sum of the squares on OF, FE is equal to the square on OF, and the sum of the squares on OF, FG is equal to the square on OG ; (I. Prop. 47.) therefore the difference of the squares on EF, FG is equal to the difference of the squares on OE, OG. PROPOSITION 36. 253 Therefore the rectangle EA, EG is equal to the difference of the squares on OE, 00. Similarly it can be proved that the rectangle EB^ ED is equal to the difference of the squares on OE, OD, that is, is equal to the difference of the squares on OE^ OC^ since OD is equal to OC. Therefore the rectangle EA^ EC is equal to the rectangle EB, ED. Wherefore, if two chords- of a circle &c. There are two special cases which should be noticed by the student, one case, when the points O, F coincide, i.e. when one chord is a diameter; the other case, when the points JB, D coincide, i.e. when one chord is a tangent. The statement of the theorem in the latter case appears in the Corollary. Corollary. If a chord of a circle be produced to any point, the rectangle contained hy the segments of the chord is equal to the square on, the tangerit drawn to tJie circle from the point. This result is seen at once on considering the tangent as the limiting position of the secant. EXEECISES. 1. Prove the converse of Proposition 36, i.e. that, if EAC, EBD be two straight lines intersecting at E such that the rectangles EA,EG and EB, ED are equal, then A, B, C, D lie on a circle. 2. If two circles intersect each other, their common chord bisects their common tangents. 3. From a given point as centre describe a circle cutting a given straight line in two points, so that the rectangle contained by their distances from a given point in the straight line may be equal to a given square. 4. If yi BO be a triangle and D a point in ^C such that the angle ABD is equal to the angle ACB, then the rectangle AC, AJD ia equal to the square on AB. 5. If from each of two given points, a pair of tangents be drawn to a given circle, the middle points of the chords joining the points of contact of each pair of tangents lie on the circumference of a circle passing through the two given points. 254 BOOK III. PROPOSITION 37. If from an extei'nal point there he drawn to a circle two straight lines, one- of which cuts the circle in two points and the other meets it, and if the recta^iyle contained hy the seg- ments of the chord on the line which cuts the circle he equal to the square on the line which meets the circle^ the line which 7neets the circle is a tangent to it. Let ABCD be a circle and E an external point; and let EAC be a straight line cutting the circle ait A, C and EB a straight line meeting it at B, such that the rect- angle contained by EA, EC is equal to the square on EB: it is required to prove that EB touches the circle. Construction. Pind the centre ; (Prop. 5.) from E draw ED to touch the circle at Z> ; (Prop. 17.) and draw OB, OD, OE. Proof. Because ED is a tangent, and OD is the radius, the angle EDO is a right angle. (Prop. 18.) Because EAC cuts the circle and ED touches it, the rectangle EA, EC is equal to the square on ED; (Prop. 36, Coroll.) and the rectangle EA, EC is equal to the square on EB; therefore the square on EB is equal to the square on ED; therefore EB is equal to ED. PROPOSITION 37. 255 Again, because in the triangles EBO, EDO^ EB is equal to ED, and OB to OD, and OE to OE, the triangles are equal in all respects; (I. Prop. 8.) therefore the angle EBO is equal to the angle EDO. But EDO is a right angle ; therefore the angle EBO is a right angle. And because BE is at right angles to the radius OB, BE touches the circle. (Prop. 16.) Wlierefore, if from an external lyoint &c. EXEKCISES. 1. If three circles meet two and two, the common chords of each pair meet in a point. 2. If three circles touch two and two, the tangents at the points of contact meet at a point. 3. If the tangents drawn to two intersecting circles from a point be equal, the common chord of the circles passes through the point. 4. Describe a circle which shall touch a given straight line at a given point, and shall cut off from another given straight line a chord of a given length. 5. On OP, the straight line drawn from a given point to P a point on a given straight Hne, a point Q is taken such that the rectangle OP, OQ is constant: prove that the locus of Q is a circle. 6. On OP, a chord of a given circle drawn from a given point O, a point Q is taken such that the rectangle OP, OQ is constant: prove that the locus of Q is a straight line. 256 BOOK III. PROPOSITION 37 A. If two triangles he equiangular to 07ie another, the rectangle contained hy any side of tlie 07ie and any side of tlie oilier is equal to the rectangle contained by the cor- 7'esj)ondi7ig sides *. Let ABC, DEF be two triangles which are equiangular to one another, having the angles at A, B, C equal to the angles at D, E, F respectively : it is required to prove that the i-cctangle AB, EF is equal to the rectangle BC, DE. Construction. In AB, CB produced beyond B, take points G, II such that BG is equal to EF, and BH to ED; (I. Prop. 3.) and draw Gil. Proof. Because the angle GBH is equal to the angle CBA, (I. Prop. 15.) and the angle FED is equal to the angle CBA, the angle GBH is equal to the angle FED. Because in the triangles BGII, EFD, BG is equal to EF and BH to ED, and the angle GBH is equal to the angle FED, the triangles are equal in all respects: (I. Prop. 4.) therefore the angle BGII is equal to the angle EFD ; but the angle EFD is equal to the angle BCA, therefore the angle AGH is equal to the angle ACH; therefore the points A, C, G, II He on a circle. (Prop. 21, CoroU.) Therefore the rectangle AB, BG is equal to the rectangle CB, BH; (Prop. 35.) that is, the rectangle -4^, EF is equal to the rectangle BC, DE. Wherefore, if two triangles &c. * In two triangles which are equiangular to one another, two sides are said to correspond when they arc opposite to equal angles. PTOLEMY'S THEOREM. 257 PROPOSITION 37 B. The rectangle contained hy the diagonals of a convex quadrilateral inscribed in a circle is equal to the sum of the rectangles contained hy pairs of opposite sides*. Let ABCD be a quadrilateral inscribed in a circle and AC, BD be its diagonals : it is required to prove that the rectangle AC, BD is equal to the sum of the rectangles AB, CD and BC, AD. Construction. From B in BA, on the same side of BA as CD, draw BE making the angle ABE equal to the angle CBD, and meeting AC in E. (I. Prop. 23.) Proof. Because the angle BAC is equal to the angle -Si) C, (Prop. 21.) and the angle ABE is equal to the angle DBC, (Constr.) therefore the triangles ABE, DBC are equiangular to one another ; (I. Prop. 32.) therefore the rectangle AB, CD is equal to the rectangle AE, BD. (Prop. 37 A.) Again, because the angle ABE is equal to the angle DBC, the angle ABD is equal to the angle EBC ; and the angle BDA is equal to the angle BCA (i. e. BCE ), (Prop. 21.) therefore the triangles ABD, EBC are equiangular to one another; (I. Prop. 32.) therefore the rectangle AD, BC is equal to the rectangle EC, BD; but it has been proved that the rectangle AB, CD is equal to the rectangle AE, BD. Therefore the sum of the rectangles AB, CD and A D, BC is equal to the sum of the rectangles AE, BD and EC, BD, that is, to the rectangle AC, BD. (II. Prop. 1.) Wherefore, the rectangle contained &c. * Thi.<5 theorem is attributed to Ptolemy, a Greek geometer of Alexandria, who died about a.d. 160. 258 BOOK III. ADDITIONAL PEOPOSITION. If a straight line be drawn through a given point to cut a given circle, the intersection of the tangents at the two points of section always lies on a fixed straight line*. Let FRS be any straight line drawn through a given point P to cut a given circle, whose centre is O, in E and S. Let QR, QS be the tangents at R, S. Draw OQ intersecting RS in M\ and draw OP, and draw QH per- pendicular to OP or OP produced. Because the angle at 31 is a right angle (Ex. 3, page 217), and the angle at If is a right angle, the points P, H, M, Q lie on a circle ; fig. 1 (Prop. 21, Coroll.) and fig. 2 (Prop. 22, CoroU.) therefore the rectangle OH, OP is equal to the rectangle OM, OQ. (Prop. 36.) But because QS is a tangent at S, the angle OSQ is a right angle, (Prop. 18.) and the angle at 31 is a right angle, therefore the rectangle 031, OQ is equal to the square on OS; (I. Prop. 47.) Therefore the rectangle OP, OH is equal to the square on OS. But OP and OS are both constants, therefore OH" is a constant, and the point Q always lies on a fixed straight line, i.e. the line drawn through the fixed point H at right angles to OP. * This line is called the polar of the given point, and the point is called the pole of the line with respect to the circle. POLES AND POLARS. 259 It has now been proved that, if a point and a straight line be such that the straight line joining the centre of a circle to the point is at right angles to the line, and the rectangle contained by the distances of the point and the line from the centre is equal to the square on the radius of the circle, the point is the pole of the line, and the line the polar of the point with respect to the circle. In the diagram is the centre of the circle : if is a point in OP such that the rectangle OP, OH is equal to the square on the radius, and HQ is at right angles to OP. P is the pole of HQ, and HQ is the polar of P. It will be observed that, if P be without the circle (fig. 1), the polar cuts the circle : if P be on the circle (fig. 2), the polar is the tangent to the circle, and if P be within the circle (fig. 3), the polar does not cut the circle. 260 BOOK III. ADDITIONAL PKOPOSITION. If a quadrilateral he inscribed in a circle, the square on the straight line joining the points of intersection of opposite sides is less than the sum of the squares on the straight lines joining those points to the centre of the circle by tiolce the square on the radius of the circle. Let A BCD be a quadrilateral inscribed in a circle whose centre is O; and let the sides AB, CD meet in Q and the sides AD, BC in JR. Draw QR and draw AS making the angle MAS equal to the angle RQD and meeting RQ in S. (I. Prop. 23.) (1) Because in the triangles RAS, RQD, the angle RAS is equal to the angle RQD, the angle RSA is equal to the angle RDQ; (I. Prop. 32.) therefore the points S, A, D, Q lie on & circle. (Prop. 21 or 20, Coroll.) Therefore the rectangle RS, RQ is equal to the rectangle RA, RD. (Prop. 36.) Also it can be proved that the points R, S, A, B lie on a circle. Therefore the rectangle QS, QR is equal to the rectangle QA, QB. (Prop. 35 or 36.) Therefore the square on QR, in figure (1), being the sum of the rectangles RS, RQ and QS, QR, is equal to the sum of the rectangles RA, RD and QA, QB; and in figure (2), being the difference of the rectangles RS, RQ, and QS, QR, is equal to the difference of the rectangles RA, RD and QA, QB. Since the rectangle RA, RD is equal to the difference of the squares on RO and the radius, (Prop. 36.) POLES AND FOLARS. 261 and the rectangle QA, QB in figure (1) is equal to the difference of the squares on QO and the radius, (Prop. 36.) and in figure (2) is equal to the difference of the squares on the radius and QO; (Prop. 35.) it follows that in both cases the square on QR is less than the sum of the squares on QO, RO by twice the square on the radius. ADDITIONAL PEOPOSITION. If one pair of opposite sides of a quadrilateral inscribed in a circle intersect at a fixed point, the other pair of opposite sides intersect on a fixed straight line *. Let ABGD be a quadrilateral inscribed in a circle, whose centre is ; and let the sides AB, CD meet in Q and AD, BG in R. Because Q and R are the intersections of opposite sides of a quadri- lateral inscribed in the circle, the square on QR is less than the squares on OQ, OR by twice the square on the radius; (Add. Prop, page 260.) therefore the difference of the squares on QR, OQ is equal to the dif- ference of the square on OR and twice the square on the radius, which is a constant, if the point R be fixed. Therefore the locus of the point Q is a straight line. (Ex. 2, page 125.) * We leave to the student as an exercise the proof that this line is the polar of the fixed point. 262 BOOK III. ADDITIONAL PKOPOSITION. If one point He on the polar of another point, the second point lies on the polar of the first point. Let P, Q be two points such that Q lies on the polar of P, i. e. if QH be drawn perpendicular to OP, the rectangle OH, OF is equal to the square on the radius. Construction. Draw P/r perpendicular to 0^. (I. Prop. 12.) Proof. Because the angles at H and K are right angles, Q, K, P, H lie on a circle; (Prop. 22, Coroll.) therefore the rectangle OQ, OK is equal to the rectangle OH, OP, (Prop. 36.) and therefore to the square on the radius ; and KP is at right angles to 0K\ therefore KP is the polar of Q, or, in other words, P lies on the polar of Q. POLES AND POLARS. 263 EXERCISES. 1. Prove that the polar of a point without a circle is the straight line joining the points of contact of tangents drawn from the point to the circle. 2. If be the centre of a circle, and the polar of a point P cut PO in H, and any straight line through P cut the circle in B and S, then the polar bisects the angle RHS. 3. If a straight line PQR cut a circle in Q and R and cut the polar of P in K, and if 31 be the middle point of QR, then the rect- angles PQ, PR and PK, PM are equal. 4. If P, Q, R, S he the points of contact of the sides AB, BC, CD, DA of a quadrilateral -4 BCZ) with an inscribed circle, the straight lines AC, BD, PR, QS are concurrent. 5. Shew how to draw two tangents to a given circle from a given external point by means of straight lines only. 6. Shew how to draw a tangent to a given circle at a given point on it by means of straight lines only. 264 BOOK III. ADDITIONAL PKOPOSITION. The locus of a point from which tangents drawn to two given circles are equal is a straight line*. Let P be a point such that PQ, PR tangents drawn to two given circles are equal. Find the centres A, B of the circles; (Prop. 5.) draw AB, AP, AQ,BP, BR, and draw PH perpendicular to AB. (I. Prop. 12.) Because PQ, PR are tangents the angles at Q and R are right angles. Therefore the sum of the squares on PQ, AQ is equal to the square on AP, (I. Prop. 47.) and the sum of the squares on PR, BR is equal to the square on BP; therefore the difference of the squares on AQ, BR is equal to the difference of the squares on AP, BP, But because the angles at H are right angles, the difference of the squares on AP, BP is equal to the difference of the squares on AH, HB. Therefore the difference of the squares on AH, HB is equal to the difference of the squares on A Q, BR, which is a constant ; therefore If is a fixed point, and the straight line HP on which P lies is drawn through H at right angles to AB the line of the centres, and is therefore a fixed straight line. * This line is called the Radical Axis of the two circles. This name was given to the line by L. Gaultier de Tours, a French geo- meter. See Journal de Vecole Polytechnique, torn. ix. p. 139 (1813). RADICAL AXIS. 265 EXERCISES. 1. Prove that the radical axis of two intersecting circles passes through their points of intersection. 2. What is the radical axis of two circles which touch each other? 3. Prove that the middle points of the four common tangents of two circles external to each other lie on a straight line. 4. Prove that the radical axes of three circles taken two and two together meet in a point*. 5. Shew how to draw the radical axis of two circles which do not meet. 6. Draw a circle passing through a given point and cutting two given circles so that its chords of intersection with the two circles may each pass through given points. 7. is a fixed point outside a given circle: find a straight line such that each of the tangents drawn from any point P in that line to the circle shall be equal to PO. 8. Draw a straight line in a given direction so that chords cut from it by two given circles may be equal. 9. Prove that the difference of the squares of the tangents from any point to two circles is equal to twice the rectangle under the dis- tance between their centres and the distance of the point from their radical axis. 10. Through two given points draw a circle to cut a given circle in such a way that the angle contained in the segment cut off the given circle may be equal to a given angle. * This point is called the Radical Centre of the three circles. T. E. 18 266 BOOK III. Definition. Two circles or other curves, ivhich meet at a jmint, are said to meet at the angle at which their tangents at the point meet. Two circles or other curves are said to he orthogonal or to cut orthogonally at a point, when they intersect at right angles at the point. ADDITIONAL PEOPOSITION. If the square on the distance hetioeen the centres of tico circles he equal to the sum of the squares on the radii, the circles are orthogonal. liQi A, B be the centres of two circles GPD, EPF, which intersect at P, and are such that the square on ^B is equal to the sura of the squares on AP, BP. Because the square on AB is equal to the sum of the squares on AP, BP, . the angle .^PJB is a right angle. (I. Prop. 48.) And BP is a radius of the circle EPF ; therefore AP touches the circle EPF. Similarly it can be proved that BP touches the circle CPD ; therefore the circles GPD, EPF are orthogonal. CoBOLLABY. The radius of each of two orthogonal circles drawn to a point of intersection is a tangent to the other circle. ORTHOGONAL CIRCLES. 267 EXEBCISES. 1. A circle, which passes through a given point and cuts a given circle orthogonally, passes through a second fixed point. 2. Describe a circle to cut a given circle orthogonally at two given points. 3. Describe a circle through two given points to cut a given circle orthogonally. 4. Two chords AB, BC of a circle ACBB, of which AB is a diameter, intersect at E: a. circle described round CDE will cut the circle ACBB at right angles. 5. Two circles cut each other at right angles in A, B; P is any point on one of the circles, and the lines PA, PB cut the other circle in Q, 22: shew that QR is a diameter. 6. The internal and external bisectors of the vertical angle A of the triangle ABC meet the base in D and E respectively. Prove that the circles described about the triangles ABD and ABE cut at right angles, as also do those described about the triangles ACD and ACE. 18—2 268 BOOK III. ADDITIONAL PKOPOSITION. Every circle^ which cuts two given circles orthogonally^ has its centre on the radical axis of the given circles, and if it cut the straight line joining their centres, it cuts it in two fixed points. Let ^ , jB be the centres of two given circles and let P be the centre of a circle, which cuts the given circles orthogonally at Q and E. Draw AB, PQ, PR, and draw PH perpendicular to AB. Because the circles cut ortho- gonally at Q, PQ is a tangent at Q. (Add. Prop, page 266, Coroll.) Similarly it can be proved that PR is a tangent at R. ButPQis equal to Pi?; therefore P is a point on the radical axis of the given circles, and therefore H is a fixed point. (Add. Prop, page 264.) Next, let the circle whose centre is P cut the line AB in M, N. Because the circles cut orthogonally at Q, AQ ia a. tangent to the circle QMNR at Q, and therefore the square on ^Q is equal to the rectangle AM, AN; but because PH is at right angles to MN^ MH is equal to NH; (Prop. 4.) and the rectangle AM, AN is equal to the difference of the squares on AH, MH; (II. Prop. 6.) therefore the square on ^Q is equal to the difference of the squares on AH, MH. Now the lines AQ, AH are of constant length ; therefore MH (or NH) is of constant length. Therefore the points M and N are at a constant distance from H, which is a fixed point ; therefore the points M and N are fixed points. ORTHOGONAL CIRCLES. 269 We leave it to the student as an exercise to prove that : if the given circles be external to each other, the points M and N are real, one within each of the given circles ; if the circles touch externally, M and N coincide with the point of contact ; if the circles intersect, the circle, whose centre is P, does not intersect the line ^J5 in real points ; if one circle touch the other internally, the points are again real, and they coincide with the point of contact ; if one circle lie wholly within the other, the points M and N are both real, one within both circles and the other without both circles. EXERCISES. 1. Draw a circle to cut three given circles orthogonally. 2. Prove that every pair of circles, which cut two given circles orthogonally, has the same radical axis. 3. Of four given circles three have their centres in the same straight line, and the fourth cuts the other three orthogonally; prove that the radical axis of each pair of the three circles is the same. 4. ABCD is a quadrilateral inscribed in a circle; the opposite sides AB and DC are produced to meet at F', and the opposite sides BC and AD at E: shew that the circle described on EF as diameter cuts the circle ABCD a,t right angles. 5. Find a point such that its polar with respect to each of two given circles is the same. 270 BOOK III. ADDITIONAL PKOPOSITION. The middle points of the sides of a triangle and the feet of the perpendiculars from the angular points on the opposite sides lie on a circle. Let D, jy, F be the middle points of the sides BC, CA, AB of a triangle ABC, and L, 31, N the feet of the perpendiculars on them from A, B, C. y. Draw FL, LB, FD, DE. . _ ^ '0( %. Then because ALB is a right-angled triangle, and F is the middle point oi AB, FL is equal to FA ; (Ex. 7, page 87.) therefore the angle FLA is equal to the angle FAL. (I. Prop. 5.) Similarly it can be proved that the angle A LE is equal to the angle LAE ; therefore the angle FLE is equal to the angle BAG. Again because FD is parallel to AG and DE to BA, (Add. Prop, page 101.) FAED is a parallelogram, and the angle FDE is equal to the angle BA C; (I. Prop. 34.) therefore the angle FLE is equal to the angle FDE. Therefore L, D, E, F lie on a circle. (Prop. 21, Coroll.) Similarly it can be proved that M, D, E, F lie on a circle, and that N, D, E, F lie on a circle. But only one circle can be described through the three points D, E, F; therefore these three circles are coincident. Therefore the six points L, M, N, D, E, F lie on a circle. THE NINE POINT CIRCLE. 271 ADDITIONAL PEOPOSITION. The circle through the middle points of the sides of a triangle passes through the middle points of the straight lines joining the angular points of the triangle to the orthocentre. Let D, E, F be the middle points of the sides BG, CA, AB of a triangle ABC. Draw AL, BM, GN perpendicular to BG, GA, AB, intersecting at 0. Bisect AO, BO, GO at a, b, c. (Add. Prop, page 95.) A ^i nJ- Z^ \ f!/ //A\ ^\^ /k In the triangle OBG, D, c, h are the middle points of the sides, and L, M, N are the feet of the perpendiculars from the vertices on the opposite sides ; therefore L, D, c, 31, N, h lie on a circle. (Add. Prop, page 270.) Similarly it can be proved that L, c, E, 21, a, N lie on a circle, and that L, M, a, N, F, b lie on a circle. But only one circle can be described through the three points L,M,N; therefore these circles are coincident. Therefore the nine points L, D, c, E, M, a, N, F, b lie on a circle*. * This circle is called the Nine Point Circle of the triangle. 272 BOOK III. ADDITIONAL PKOPOSITION. The feet of the perpendiculars drawn from any point on a circle to the three sides of a triangle inscribed in the circle lie on a straight line*. Let ABC be a triangle and PL, PM, PN be the perpendiculars from a point P on the circle ABC to BG, CA, AB. Draw LN, NM. Because PLB, PNB are right angles, a circle can be described about PNLB ; (III. Prop. 21, Coroll.) therefore the angles PNL, PBL are supple- mentary. (III. Prop. 22.) But the angles PAC, PBG are supplementary; (III. Prop. 22.) therefore the angle PNL is equal to the angle PAC. Again because PMA, PNA are right angles, a circle can be described about PNAM; (III. Prop. 22, Coroll.) therefore the angle PN3I is equal to the angle PAM. Therefore the sum of the angles PNL, PNM is equal to the sum of the angles PAC, PAM, that is, to two right angles. (I. Prop. 13.) Therefore LN, NM are in the same straight line. (I. Prop. 14.) EXERCISES. 1. If PL, PM, PN be the perpendiculars drawn from P a point on the circle ABC to the sides BC, CA, AB of an inscribed triangle, and straight lines PI, Pm, Pn be drawn to meet the sides in I, in, n such that the angles LPl, MPni, NPn are equal and measured in the same sense, then I, m, n are collinear. 2. P is a point on the circle circumscribing the triangle ABC. The pedal line of P cuts AG and BC in 31 and L. Y is the foot of the perpendicular from P on the pedal line. Prove that the rectangles PY, PC, and PL, PM are equal. * This line is called the Pedal Line. Its discovery is attributed to Dr Robert Simson, and it is in consequence also called Simson's Line. MISCELLANEOUS EXERCISES. 1. If any point P on a fixed circle be joined to a fixed point J, and AQ be drawn equal to AP at a constant inclination PAQ to AP, the locus of Q is an equal circle. 2. Draw a straight line from one circle to another, to be equal and parallel to a given straight line. 3. Construct a parallelogram, two of whose angular points are given and the other two lie on two given circles. Discuss the number of possible solutions of the problem in the different cases which may occur. 4. Find the locus of the centre of a circle whose circumference passes through two given points. 5. Shew that the straight lines drawn at right angles to the sides of a rectilineal figure inscribed in a circle from their middle points intersect at a fixed point. 6. Determine the centre of a given circle by means of a ruler with parallel edges whose breadth is less than the diameter of the circle. 7. A circle is described on the radius of another circle as diameter. Prove that any chord of the greatest circle drawn from the point of contact is bisected by the lesser circle. 8. Two circles DFG, GEO whose centres are A and B intersect at C\ through G, DGE and FGG are drawn equally inclined to AB: shew that DE and FG are equal. 9. AB and GD are two chords of a circle cutting at a point E within the circle ; AB is produced to H so that BH is equal to BE. The circles AEG and AGH cut BG in K and L; prove that B is the middle point of KL. 10. Through either of the points of intersection of two given circles draw the greatest possible straight line terminated both ways by the two circumferences. 11. Through two points A, B on the same diameter of a circle and equidistant from its centre two parallel straight lines AP, BQ are drawn towards the same parts, meeting the circle in P and Q : shew that PQ is perpendicular to AP and BQ. 274 BOOK HI. 12. A is a fixed point in the circumference of a circle and^BC an inscribed triangle such that the sum of the squares on AB,AG is constant ; shew that the locus of the middle point of BG is a. straight line. 13. From one of the points of intersection of two circles straight lines are drawn equally inclined to the common chord of the circles : prove that the portions of these lines, intercepted between the other points in which they meet the circumference's of the circles, are equal. 14. Describe a circle of given radius to touch two given circles. Discuss the number of possible solutions. 15. Describe three circles to have their centres at three given points and to touch each other in pairs. How many solutions are there ? 16. Describe a circle to touch two given concentric circles and to pass through a given point. Discuss the number of solutions. Is a solution always possible? 17. li AB, CD be two equal chords of a circle, one of the pairs of lines AD, BG and AG, BD are equal, and the other pair parallel. 18. If two equal chords AB, GD of a circle intersect at E, AE is equal to GE or DE. 19. A quadrilateral is described about a circle : shew that two of its sides are together equal to the other two sides. 20. Shew that every parallelogram described about a circle is a rhombus. 21. If the tangents at two points where a straight line meets two circles, one point being on each circle, be parallel, the other pair also are parallel. 22. Two straight lines ABD, ACE touch a circle at B and C; if DE be joined, DE is equal to BD and GE together: shew that DE touches the circle. 23. When an equilateral polygon of an even number of sides is described about a circle, the alternate angles are equal. 24. If a quadrilateral be described about a circle, the angles sub- tended at the centre of the circle by any two opposite sides of the figure are together equal to two right angles. 25. Two radii of a circle at right angles to each other when pro- duced are cut by a straight line which touches the circle : shew that the tangents drawn from the points of section are parallel to each other. 26. If the straight line joining the centres of two circles, which are external to one another, cut them in the points A, B, G, D, the squares on the common tangents to the two circles are equal to the rectangles BD, AG, and BG, AD. What is the corresponding theorem for two intersecting circles ? MISCELLANEOUS EXERCISES. 275 27. APB is an arc of a circle less than a semicircle ; tangents are drawn at A and B and at any intermediate point P; shew that the sum of the sides of the triangle formed by the three tangents is invariable for all positions of P. 28. Given a circle, and a straight line in its plane ; draw a tangent to the circle, which shall make a given angle with the given straight line. 29. Find the point in the circumference of a given circle, the sum of whose distances from two given straight lines at right angles to each other, which do not cut the circle, is the greatest or least possible. 30. A straight line is drawn touching two circles : shew that the chords are parallel which join the points of contact and the points where the straight line through the centres meets the circumferences. 31. From the centre C of a circle, CA is drawn perpendicular to a given straight line AB, which does not meet the circle, and in ^C a point P is taken, such that AF is equal to the length of the tangent li'om A : prove that, if Q be any point in AB, QF is equal to the length of the tangent from Q. 32. If a quadrilateral, having two of its sides parallel, be described about a circle, a straight line drawn through the centre of the circle, parallel to either of the two parallel sides, and terminated by the other two sides, is equal to a fourth part of the perimeter of the figure. 83. With a given point as centre describe a circle to cut a given circle at right angles. How must the point be situated that this may be possible ? 34. If two circles touch and a chord be drawn through the point of contact, the tangents at the other points where the chord meets the circles are parallel. 35. From a given point A without a circle whose centre is O draw a straight line cutting the circle at the points B and C, so that the area BOG may be the greatest possible. 36. Two circles FAB, QAB cut one another at A : it is required to draw through A a straight line FQ so that PQ may be equal to a given straight line. 37. Two given circles touch one another externally in the point P, and are touched by the line AB in the points A and B respectively. Shew that the circle described on AB as diameter passes through the point P, and touches the line joining the centres of the two given circles. 38. From a point without a circle draw a line such that the part of it included within the circle may be of a given length less than the diameter of the circle. 276 BOOK III. 39. When an equilateral polygon is described about a circle it must necessarily be equiangular if the number of sides be odd, but not otherwise. 40. One circle touches another internally at the point A : it is' required to draw from A a straight line such that the part of it between the circles may be equal to a given straight line, which is not greater than the difference between the diameters of the circles. 41. If a hexagon circumscribe a circle, the sums of its alternate sides are equal. 42. If a polygon of an even number of sides circumscribe a circle, the sum of the alternate sides is equal to half the perimeter of the polygon. 43. Under what condition is it possible to describe a circle to touch the two diagonals and two opposite sides of a quadrilateral ? 44. If a parallelogram be circumscribed about a circle with centre 0, and a line be drawn touching the circle and cutting the sides of the parallelogram, or the sides produced in A, B, C, D, prove that the angles AOB, BOG, and COD are, each of them, equal to a half of one or other of the angles of the parallelogram. 45. If two equal circles be placed at such a distance apart that the tangent drawn to either of them from the centre of the other is equal to a diameter, they will have a common tangent equal to the radius. 46. Draw a straight line to touch one given circle so that the part of it contained by another given circle shall be equal to a given straight line not greater than the diameter of the latter circle. 47. AB is the diameter and G the centre of a semicircle : shew that the centre of any circle inscribed in the semicircle is equidis- tant from G and from the tangent to the semicircle parallel to AB. 48. A circle is drawn to touch a given circle and a given straight line. Shew that the points of contact are always in the same straight line with one or other of two fixed points in the circumference of the given circle. 49. Describe a circle to touch a given circle and a given straight line. How many solutions are there ? 50. A series of circles is described passing through one angular point of a parallelogram and a fixed point on the diagonal through that angular point. Shew that the sum of the squares on the tan- gents from the extremities of the other diagonal is the same for each circle of the series. 51. A quadrilateral is bounded by a diameter of a circle, the tangents at its extremities, and a third tangent : shew that its area is equal to half that of the rectangle contained by the diameter and the side opposite to it. MISCELLANEOUS EXERCISES 277 52. If the centres of two circles which touch each other externally be fixed, the external common tangents of those circles will touch another circle of which the straight line joining the fixed centres is the diameter. 53. TP^ TQ are tangents from T to two circles which meet in A and PQ meets the circles in P', Q', and the angles PAP', QAQ' are equal ; find the locus of T. 54. Describe a circle of given radius passing through a given point and touching a given straight line. How many solutions may there be? 55. Describe a circle of given radius touching a given straight line and a given circle. How many solutions may there be? 56. Given one angle of a triangle, the side opposite it, and the sum of the other two sides, construct the triangle. 57. AOB, COD are two diameters of a circle whose centre is 0, and they are mutually perpendicular. If P be any point on the cir- cumference, shew that CP and DP are the internal and external bisectors of the angle APB. 68. If AB and CD be two perpendicular diameters of a circle and P any point on the arc A GB, shew that D is equally distant from PA, PB. 59. If two circles intersect each other, prove that each common tangent subtends, at the two common points, angles which are sup- plementary to each other. 60. From one of the points of intersection of two equal circles, each of which passes through the centre of the other, a straight line is drawn to intersect the circles in two other points : prove that these points form an equilateral triangle with the other point of intersec- tion of the two circles. 61. A series of circles touch a fixjed straight line at a fixed point: shew that the tangents at the points where they cut a parallel fixed straight line all touch a fixed circle- 62. Two points P, Q are taken in two arcs described on the same straight line AB, and on the same side of it; the angles PAQ, PBQ are bisected by the straight lines AR, BR meeting at R: prove that the angle ARB is constant. 63. APB is a fixed chord passing through P a point of intersec- tion of two circles AQP, PBR; and QPR is any other chord of the circles passing through P: shew that AQ and RB when produced meet at a constant angle. 64. A,B,C,D are four points on a circle. Prove that the four f)oints where the perpendiculars from any point O to the straight ines AB and CD meet AC and BD lie on a circle. 278 BOOK III. 65. Any number of triangles are described on the same base BG, and on the same side of it having their vertical angles equal, and perpendiculars, intersecting at D, are drawn from B and G on the opposite sides; find the locus of D. 66. Let and G be any fixed points on the circumference of a circle, and OA any chord; then, if AG be joined and produced to B, so that OB is equal to OA, the locus of B is an equal circle. 67. If ABG be a triangle, AD and BE the perpendiculars from A and B upon BG and AC, DF and EG the perpendiculars from D and E upon AG and BG, then FG is parallel to AB. 68. The four circles which pass through the middle points of the sides of the four triangles formed by two sides of a quadrilateral and one of its two diagonals intersect in a point. 69. In a circle two chords of given length are drawn so as not to intersect, and one of them is fixed in position; the opposite extremities of the chords are joined by straight lines intersecting within the circle: shew that the locus of the point of intersection will be a portion of the circumference of a circle, passing through the extremities of the fixed chord. 70. The centre C of a circle BPQ lies on another circle APQ of which PBA is a diameter. Prove that PG is parallel to BQ. 71. Through one of the points of intersection of two circles, centres A and B, a chord is drawn meeting the circles at P and Q respectively. The lines PA, QB intersect in G. Find the locus of C. 72. At each extremity of the base of a triangle a straight line is drawn making with the base an angle equal to half the sum of the two base angles; prove that these lines will meet on the external bisector of the vertical angle. 73. In the figure of Euclid i. 47 the feet of the perpendiculars drawn from the centre of the largest square upon the three sides of the given right angled triangle are collinear. 74. Three points A, B, G are taken on a circle. The tangents at B and G meet in T. If from T a straight line be drawn parallel to AB, it meets ^0 in the diameter perpendicular to AB. 75. If ABG he a triangle inscribed in a circle, PQ a diameter, and perpendiculars be let fall from P on the two sides meeting in A, and from Q on those meeting in B, the lines joining the feet of the two sets of perpendiculars will intersect at right angles. 76. Through any point on the side BG of an equilateral tri- angle ABG, OK, OL are drawn parallel to AB, AG respectively to meet AG, AB respectively in K and L: the circle through 0, K and L cuts AB, AG again in P and Q. Prove that OPQ is an equilateral triangle. MISCELLANEOUS EXERCISES. 279 77. Two chords AB, CD of a circle intersect in E. From EB, ED, produced if necessary, parts EF, EG are cut off respectively equal to ED, EB. Prove that FG is parallel to CA. 78. If ABC be an equilateral triangle and D be any point on the circumference of the circle ABC, then one of the three distances DA^ DB, DC is equal to the sum of the other two. 79. If through E a point of intersection of two circles AGE, BDE two straight lines AB and GD be drawn terminated by the circles, then A G and BD cut one another at a constant angle. 80. If AD, BE, GF be the perpendiculars from the angles A,B,G of a triangle on the opposite sides, these lines bisect the angles of the triangle DEF. 81. AB is a common chord of the segments AGB, ADEB of two circles, and through G any point on AGB are drawn the straight lines AGE, BGD: prove that the arc DE is of invariable length. 82. Divide a, circle into two parts so that the angle contained in one arc shall be equal to twice the angle contained in the other. 83. A quadrilateral is inscribed in a circle : shew that the sum of the angles in the four arcs of the circle exterior to the quadrilateral is equal to six right angles. 84. A, B, G, D are four points taken in order on the circum- ference of a circle ; the straight lines AB, GD produced intersect at P, and AD, BG at Q: shew that the straight lines which bisect the angles APC, AQC are perpendicular to each other. 85. A quadrilateral can have one circle inscribed in it and another circumscribed about it: shew that the straight lines joining the opposite points of contact of the inscribed circle are perpendicular to each other. 86. If D, E, F be three points on the sides BC, GA, AB of a triangle, the perimeter of the triangle DEF is least when D, E, F are the feet of the perpendiculars from A, B, G on BG, GA, AB. 87. Shew that the four straight lines bisecting the angles of any quadrilateral form a quadrilateral which can be inscribed in a circle. 88. If a polygon of six sides be inscribed in a circle, the sum of three alternate angles is equal to four right angles. 89. AB is a. diameter of a circle, GD a chord perpendicular to it. A straight line through A cuts the circle in E, and GD produced in F: prove that the angles AEO, DEF are equal. 90. If be a point within a triangle A BG, and OD, OE, OF be drawn perpendicular to BG, GA, AB, respectively, the angle BOG is equal to the sum of the angles 5yi C, £D^. .. - 280 BOOK III. 91. AOB, GOB are two chords of a circle which intersect within the circle at 0. Through the point A a straight line ^^ is drawn to meet the tangent to the circle at the point C, so that the angle AFC is equal to the angle BOC. Prove that OF is parallel to EG. 92. The sums of the alternate angles of any polygon of an even number of sides inscribed in a circle are equal. 93. Through a point C in the circumference of a circle two straight lines AOB^ DGE are drawn cutting the circle at B and E: shew that the straight line which bisects the angles ACE, DGB meets the circle at a point equidistant from B and E. 94. AOB, COB are two diameters of the circle ABGB, at right angles to each other. Equal lengths OE, OF, are measured off along OA, OB respectively. Shew that BF produced cuts BE at right angles, and that these two straight lines, when produced, intercept between them one fourth of the circumference of the circle. 95. If the extremities of the chord of a circle slide upon two straight lines, which intersect on the circumference, every point in the circumference will trace out a straight line. 96. Any point P is taken on a given arc of a circle described on a line AB, and perpendiculars AG and BH are dropped on BP and AP respectively ; prove that GH touches a fixed circle. 97. OA and OB are two fixed radii of a given circle at right angles to each other and POQ is a variable diameter; prove that the locus of the intersection of PA and QB is a circle equal to the given one. 98. Describe a circle touching a given straight line at a given point, such that the tangents drawn to it from two given points in the straight line may be parallel. 99. Describe a circle with a given radius touching a given straight line, such that the tangents drawn to it from two given points in the straight line may be parallel. 100. Two circles intersect at the points A and B, from which are drawn chords to a point G in one of the circumferences, and these chords, produced if necessary, cut the other circumference at B and E : shew that the straight line BE cuts at right angles that diameter of the circle ABC which passes through G. 101. If squares be described externally on the sides and the hypo- tenuse of a right-angled triangle, the straight line joining the inter- section of the diagonals of the latter square with the right angle is perpendicular to the straight line joining the intersections of the diagonals of the two former. 102. G is the centre of a given circle, GA a straight line less than the radius ; find the point of the circumference at which GA subtends the greatest angle. MISCELLANEOUS EXERGLSES. 281 103. If two chords of a circle meet at a right angle within or with- out a circle, the squares on their segments are together equal to the square on the diameter. 104. On a given base BC a triangle ABC is described such that AG is equal to the perpendicular from B upon AG. Find the locus of the vertex A. 105. Draw from a given point in the circumference of a circle, a chord which shall be bisected by its point of intersection with a given chord of the circle. 106. Through any fixed point of a chord of a circle other chords are drawn; shew that the straight lines from the middle point of the first chord to the middle points of the others will meet them all at the same angle. 107. The two angles at the base of a triangle are bisected by two straight lines on which perpendiculars are drawn from the vertex: shew that the straight line which passes through the feet of these perpendiculars will be parallel to the base and will bisect the sides. 108. A point P is taken on a circle of which AB is a fixed chord; a parallelogram is described of which AB and AP are adjacent sides: find the locus of the middle points of the diagonals of the paral- lelogram. Find the maximum length of the diagonal drawn through A. 109. ^ is a fixed point on a circle whose centre is and BOB is a diameter. The tangents at A and D meet in L. Shew that the locus of the intersection of LB with the perpendicular from A on OB is a circle. 110. ABG is a triangle inscribed in a circle ; AB, BE, perpendicu- lars io BG and AG, meet in 0: AK \b 2, diameter of the circle: prove that GK is equal to BO, 111. A straight line touches a circle at the point P and QR is a chord of a second circle parallel to this tangent ; PQ , PR cut the first circle in S, T, and the second circle in U, F; prove that ST, UV are parallel to each other. 112. A, B, G are three points on a circle, the bisectors of the angle BAG and the angle between BA produced and AG meet BG and BG produced in E and F respectively ; shew that the tangent at A bisects EF. 113. A circle is described passing through the right angle of a right-angled triangle, and touching the hypotenuse at its middle point : prove that the arc of this circle, cut off by either side of the triangle, is divided at the middle point of the hypotenuse into two parts one of which is double of the other. 114. If through the point of contact P of two circles two straight lines PQq, PRr be drawn to meet the circles in Q, R, and q, r respec- tively ; then the triangles PQR, Pqr are equiangular. T. E. 19 282 BOOK III. 115. The angle between two chords of a circle, which intersect within the circle, is equal to half the sum of the angles subtended at the centre by the intercepted arcs. 116. ABCD is a parallelogram and any straight line is drawn cutting the sides AB, CD in P, Q respectively. The circle passing through A, P, and Q cuts AD in S, and ^C in T. Shew that the circles circumscribing the triangle DSQ, CTQ touch one another at the point Q. 117. If one circle touch another internally, any chord of the greater circle which touches the less is divided at the point of its con- tact into segments which subtend equal angles at the point of contact of the two circles. 118. Two circles are drawn touching a circle, whose centre is C, in P and Q respectively and intersecting in PQ produced, and again in B. Prove that the angles CBP, CRQ are equal. 119. The angle between two chords of a circle, which intersect when produced without the circle, is equal to half the difference of the angles subtended at the centre by the intercepted arcs. 120. Construct a triangle, having given the base, the vertical angle, and the length of the straight line drawn from the vertex to the middle point of the base. 121. ^ is a given point: it is required to draw from A two straight lines which shall contain a given angle and intercept on a given straight line a part of given length. 122. A and B are two points, XY a given straight line of un- limited length, not passing through A or B ; find the point (or points) in XY at which the straight line AB subtends an angle equal to a given angle. Also state when this problem is impossible. 123. The chords of two intersecting circles which are bisected at any point of the common chord are equal. 124. Find a point in the tangent to a circle at one end of a given diameter, such that when a straight line is drawn from this point to the other extremity of the diameter, the rectangle contained by the part of it without the circle and the part within the circle may be equal to a given square not greater than that on the diameter. 125. If perpendiculars be drawn from the extremities of the diameter of a circle upon any chord or any chord produced, the rect- angle under the perpendiculars is equal to that under the segments between the feet of the perpendiculars and either extremity of the chord. 126. Two circles intersect and any straight line ACBD cuts them in .4 , JB and (7, D respectively. If JB be a point on the line such that the rectangles contained by AC, BE and J5D, CE are equal, the locus of jE is a straight line. MISCELLANEOUS EXERCISES. 283 127. ABB is an isosceles triangle having the side AB equal to the side BD ; ^C is drawn at right angles to ^i? to meet BD produced in <7, and the bisector of the angle B meets AG in 0. Shew that the square on AB is equal to the difference of the rectangles CB, BD and CA, AO. 128. Through a given point without a circle draw, when possible, a straight line cutting the circle so that the part within the circle may be equal to the part without the circle. What condition is necessary in order that a real solution may be possible? 129. Two equal circles have their centres at A and B: is a fixed point outside those circles. A is the centre of a third circle whose radius is equal to OB : prove that a right-angled triangle can be con- structed having its sides equal to the tangents from to the three circles. 130. Prove that, if a straight line drawn parallel to the side BC of a triangle ABC cut the sides AB, AC in D, E respectively, the rect- angle contained by AB^ AE is equal to the rectangle contained by AG, AD. 131. Construct a triangle, having given the base, the vertical angle, and the length of the straight line drawn from the vertex to the base bisecting the vertical angle. 132. Through two given points on the circumference of a given circle draw the two parallel chords of the circle which shall contain the greatest rectangle. 133. A straight line PAQ is drawn through A one of the points of intersection of two given circles to meet the circles again in P and Q: find the line which makes the rectangle AP, AQ a maximum. 134. Produce a given straight line so that the rectangle contained by the whole line thus produced, and the part produced, shall be equal to the square on another given line. 135. Two circles intersect in 0, and through a straight line ORS is drawn cutting the circles again in J^ and S. SO is produced to P, so that the rectangle OP, RS is constant. Shew that the locus of P is a straight line. 136. If from the foot of the perpendicular from each of the angular points of a triangle on the opposite side a perpendicular be drawn to each of the other sides, the feet of the six perpendiculars so drawn lie on a circle. 137. AB, GD are chords of a circle intersecting at 0, and AG, DB meet at P. If circles be described about the triangles AOG, BOD, the angles between their tangents at will be equal to the angles at P, and their other common point will lie on OP. 19—2 284 BOOK III. 138. If ABC be a triangle, D, E,F the feet of the perpendiculars from A, B, G on the opposite sides, their point of intersection, the rectangles DO, DA, and DE, DF are equal. 139. A straight line AD is drawn bisecting the angle ^1 of a triangle ABC and meeting the side BG in D. Find a point E in BQ produced either way such that the square on ED may be equal to the rectangle contained by EB, EG. 140. If D, D'; E, E' ; F, F' be pairs of points on the sides BG, GA, AB of a triangle respectively, such that D, D', E, E' are con- cyclic, E, E', F, F' are concyclic, F, F', D, D' are concyclic, then D, D', E, E'y F, F' are concyclic. 141. In the straight line PQ a point R is taken, and circles are described on PR, RQ as diameters ; shew how to draw a line through P such that the chords intercepted by the two circles may be equal. 142. If M be the middle point of PQ, where P and Q are points without a given circle, the sum of the squares on the tangents to the circle from P and Q is equal to twice the sum of the square on the tangent from 31 and the square on PM. 143. A circle FDG touches another circle BDE in D and a chord ^P of the latter in F: prove that the rectangle FA, FB is equal to the rectangle contained by GE and the diameter of FDG, where GE is drawn perpendicular to AB at its middle point G and on the same side of it as the circle FDG. 144. Given the base and the vertical angle of a triangle, prove that the locus of the centre of the nine-point circle is a circle. 145. If a circle be circumscribed to a triangle, the middle point of the base is equally distant from the orthocentre and the point diametrically opposite the vertex. Also these three points are in the same straight line. BOOK IV. DEFINITIONS. Definition 1. A figure of five sides is called a pentagon, one of six sides is called a hexagon, one of eight sides is called an octagon, one of ten sides is called a decagon, one of twelve sides is called a dodecagon*. Definition 2. When each of the angular points of one rectilineal figure lies on one of the sides of a second recti- lineal figure^ and each of the sides of the second figure passes through one of the angular points of the first figure, the first figure is said to he inscribed in the second figure, and the second figure is said to he described about the first figure. * Derived from irhre "five," ?^ "six," oKTiii "eight," Mkol "ten, ddodcKa "twelve," respectively, and ywuia "an angle." 286 BOOK IV PROPOSITION 1. To draw a chord of a given circle equal to a given straight line. Let ABC be the given circle, and D the given straight line : it is required to draw a chord of the circle ABG equal toi>. Construction. Take any point A on the circle ABC, and from A draw AE equal to D. (I. Prop. 2.) If E lie on the circle ABC, what is required is done, for in the circle ABG the chord AE is drawn equal to B. But if E do not lie on the circle ABG^ with A as centre and ^^ as radius describe the circle ECF cutting the circle ABG at F. Draw AF. ^i^ is a chord drawn as required. Proof. Because A is the centre of the circle EGF, AFi^ equal to AE. But AE is equal to D. (Constr.) Therefore ^i^is equal to i>, and it is a chord of the circle ABG. Wherefore, a chord A F of the given circle ABG has been drawn equal to the given straight line D. PROPOSITION 1. 287 It is clear that it is not possible to draw a chord of a given circle to be equal to a given straight line, if the given line be greater than the diameter of the circle (III. Prop. 8, Part 1) ; and further that, if a solution be possible, in general two chords can be drawn from a given point equal to the given line. In the diagram, if the two circles intersect in C, the chord AG also is equal to the given line. EXEKCISES. 1. In a given circle draw a chord parallel to one given straight line and equal to another. .2. On a given circle find a point such that, if chords be drawn to it from the extremities of a given chord, their sum shall be equal to a given straight line. How many solutions are there in the different cases which may occur ? 288 BOOK IV. PROPOSITION 2. To inscribe in a given circle a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle : it is required to inscribe in the circle ABC a triangle equi- angular to the triangle DEF. Construction. Take any point A on the circle, and through A draw the chord ^^ to cut off the arc AGB containing an angle equal to the angle DFE, and through A draw the chord ^C to cut off the arc ABC containing an angle equal to the angle DEF*. (III. Prop. 34.) Draw BC : the triangle ABC is inscribed as required. Proof. Because the arc ACB contains an angle equal to the angle DFE, (Const r.) and the angle ACB is contained by the arc ACB, the angle ACB is equal to the angle DFE. Similarly it can be proved that the angle ABC is equal to the angle DEF. And because the sum of three angles of a triangle is equal to two right angles, (I. Prop. 32.) and the angles ACB, ABC are equal to the angles DFE^ DEF respectively, * It must be noticed that the arcs AGB, ABC are measured in opposite directions along the circumference from the point A. PROPOSITION 2. 289 the remaining angle BAG of the triangle ABC is equal to the remaining angle EDF of the triangle DEF ; therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, a triangle ABC equiangular to the triangle DEF has been inscribed in the given circle ABC. Since the arc ABC may be measured in either direction along the circumference from A, we see that two triangles equiangular to a given triangle can be inscribed in a given circle so as to have a vertical angle equal to a given angle of the triangle at a given point on the circle, and that six triangles equiangular to a given triangle can be inscribed in a given circle, so as to have one of its vertical angles at the given point on the circle. EXEKCISES. 1. Prove that all triangles inscribed in the same circle equi- angular to each other are equal in all respects. 2. The altitude of an equilateral triangle is equal to a side of an equilateral triangle inscribed in a circle described on one of the sides of the original triangle as diameter. 3. ABC, A'B'C are two triangles equiangular to each other in- scribed in a circle AA'BB'GC. The pairs of sides BC, B'C ; GA, C'A'; AB, A'B' intersect in a, b, c respectively. Prove that the triangle abc is equiangular to the triangle ABC. 290 BOOK IV. PROPOSITION 3. To describe about a given circle a triangle equiangular to a given triangle. Let ABC be the given circle and DEF the given triangle : it is required to describe about the circle ABC a triangle equiangular to the triangle DEF. Construction. Find the centre G of the circle ABCj (III. Prop. 5.) and draw any diameter HGA meeting the circle in A. At G in GH draw the straight lines GB^ GC on opposite sides of GH making the angles BGH^ CGH equal to the angles EFD, DEF, (I. Prop. 23.) meeting the circle in B, C. Through A, B, C draw MAN, NBL, LCM at right angles to GA, GB, GC respectively: (I. Prop. 11.) the triangle LMN is a triangle described as required. Proof. Because the sum of the angles of the quadri- lateral GBNA is equal to four right angles, (I. Prop. 32, Coroll.) and two of the angles GAN, GBN are right angles, the sum of the angles AGB, ANB is equal to two right angles. But the sum of the angles AGB, IIGB is equal to two right angles; (I. Prop. 13.) therefore the sum of the angles AGB, ANB is equal to the sum of the angles AGB, IIGB) therefore the angle ANB is equal to the angle HGB, that is, to the angle EFD. PROPOSITION 3. 291 Similarly it can be proved that the angle LMN i^ equal to the angle DEF; therefore the remaining angle NLM of the triangle LMN is equal to the remaining angle EDF of the triangle DEF. (I. Prop. 32.) Therefore the triangle LMN is equiangular to the triangle DEF. Again, because MN is drawn through A a point on the circle ABC at right angles to the radius AG^ MN touches the circle. (III. Prop. 16.) Similarly it can be proved that NL^ LM touch the circle. Therefore the triangle LMN is described about the circle ABC. Wherefore, a tricmgle LMN equiangular to the given triangle DEF has been described about the given circle ABC. EXERCISES. 1. Prove that all triangles described about the same circle equi- angular to each other are equal in all respects. 2. Describe a triangle about a given circle to have its sides parallel to the sides of a given triangle. How many solutions are there? 3. The angles of the triangle formed by joining the points of con- tact of the inscribed circle of a triangle with the sides are equal to the halves of the supplements of the corresponding angles of the original triangle. 4. If ABC, A'B'C be two equal triangles described about a circle in the same sense and the pairs of sides J5C,J5'C"; GA, G'A' ; AB, A'B' meet in a,h,c respectively, a, &, c are equidistant from the centre of the circle. 292 BOOK IV. PROPOSITION 4. To inscribe a circle in a given triangle. Let A BG be the given triangle : it is required to inscribe a circle in tlie triangle ABC. Construction. Bisect two of the angles ABC, BCA of the triangle ABC by BD, CD meeting at D, (I. Prop. 9.) and from D draw DE^ Di\ DG perpendicular to BC, CA, AB respectively. (I. Prop. 12.) With B as centre and DE^ DF, or DG as radius describe a circle : it will be a circle described as required. Proof. Because in the triangles DEB^ DGB., the angle DBE is equal to the angle DBG, (Constr.) and the angle DEB is equal to the angle DGB, (I. Prop. 10 B.) and the side BD is common, the triangles are equal in all respects; (I. Prop. 26, Part 2.) therefore DE is equal to DG. Similarly it can be proved that DE is equal to DF. Therefore the three straight lines DE, DF^ DG are equal to one another, and the circle described with D as centre, and DE, DF, or DG as radius passes through the ex- tremities of the other two ; and touches the straight lines BC, CA, AB, because the angles at the points E, F, G are right angles, and the PROPOSITION 4. 293 straight line drawn through a point on a circle at right angles to the radius touches the circle. (III. Prop. 16.) Therefore the straight lines AB, B'C, CA do each of them touch the circle, and therefore the circle is inscribed in the triangle ABC. Wherefore, the circle EFG has been imcrihed in the given triangle ABC. EXERCISES. 1. The base of a triauglc is fixed, and the vertex describes a circle passing through the extremities of the base: find the locus of the centre of the inscribed circle. 2. If a polygon be described about a circle, the bisectors of all its angles meet in a common point. 3. Describe a circle to touch a given circle and two given tan- gents to the circle. 4. Construct a triangle, having given the base, the vertical angle and the radius of the inscribed circle. 5. Find the centre of a circle cutting off three equal chords from the sides of a triangle. 6. The triangle whose vertices are the three points of contact of the inscribed circle with the sides of a triangle, is always acute- angled. 294 BOOK IV. It can be proved in the same manner as in Proposition 4 that, if the angles at B and C of the triangle be bisected externally by BJ)^ , CD^ , meeting at i>, , and perpendiculars, i)j^j, D^F^, Dfi^ be drawn, the circle described with D^ as centre and either of the three lines i^,^,, -^i^j, ^fii ^^ radius will touch the three sides of the triangle. Such a circle satisfies the definition (III. Def. 10) of an inscribed circle. The circles are however generally distinguished thus, the circle IJFG, which lies wholly within the triangle ABC, is called the inscribed circle, whereas the circle E^Ffi^ is called an escribed circle, and is said to be escribed beyond the side BO, to distinguish it from the two other circles which can, in a similar manner, be escribed beyond CA and beyond AB respectively. PROPOSITION L 295 EXERCISES. 1. Prove that the radius of the inscribed circle of a triangle is less than the radius of any one of the escribed circles. 2. Prove that the greatest of the escribed circles of a triangle is that which is escribed beyond the greatest side, and the least, beyond the least side. 3. If the centres of the escribed circles of a triangle be joined, and the points of contact of the inscribed circle be joined, the two tri- angles so formed are equiangular to each other. 4. A circle touches the side BC of a triangle ABC and the other two sides produced : shew that the distance between the points of con- tact of the side BC with this circle and with the inscribed circle, is equal to the difference between the sides AB and AG. 5. Construct a triangle, having given its base, one of the angles at the base, and the distance between the centre of the inscribed circle and the centre of the circle touching the base and the sides produced. 6. Prove that, ii A,B he two fixed points on a circle and P a variable point, the locus of the centre of each of the escribed circles of the triangle APB is a circle. 7. The centre of the inscribed circle of a triangle is the ortho- centre of the triangle formed by the centres of the escribed circles. 296 BOOK IV. PROPOSITION 5. 2^0 describe a circle about a given triangle. Let ABC be the given triangle : it is required to describe a circle about the triangle ABG. Construction. Bisect two of the sides AB^ AG oi the triangle ABG, at B, E, (I. Prop. 10.) and draw DF^ EF at right angles to AB, AG meeting at F. (I. Prop. 12.) Draw FA J and with F as centre and FA as radius describe a circle: this is a circle described as required. F must lie either in BG (fig. 2) or not in BG (figs. 1 and 3). If F do not lie in BG, draw FB, FG. Proof. Because in the triangles FDA, FDB, AD \^ equal to BD, (Constr.) and DF is equal to DF, and the angle ADF is equal to the angle BDF, (Constr.) the triangles are equal in all respects; (I. Prop. 4.) therefore FA is equal to FB. Similarly it can be proved that FA is equal to FG. Therefore the circle described with F as centre and FA as radius passes through the points B and G, and is de- scribed about the triangle ABG. "Wherefore, a circle ABG has been described about the given triangle ABG* PROPOSITION 5. 297 The construction of Proposition 5 shews that only one circle can be described about a given triangle, a theorem which has already been established otherwise. (III. Prop. 9, CorolL 2.) The circle ABC is often spoken of as the circumscribed circle of the triangle ABC. Propositions 4 and 5 solve problems of the same nature ; each shews how to describe a circle to satisfy three given conditions. The problem of Proposition 4 to describe a circle to touch three given straight lines, admits of 4 solutions; Proposition 5, to describe a circle to pass through three given points, admits of but a single solution. A circle can be described to satisfy three (and not more than three) independent conditions, but it will be found that the solution is not always unique : if the problem be one which can be solved by geometrical methods, the number of solutions will be found to be 1 or 2or4 = 2x2or8 = 2x2x2or some higher power of 2. The number 2 occurs in one of its powers from the fact that at each step of the solution where choice is possible, the choice lies between the two intersections of a circle and a straight line or the two intersections of two circles. If it be required to describe a circle to touch four or more given straight lines, or to pass through four or more given points, relations of some kind must exist between the positions of the lines or of the points in order that a solution may be possible. EXEECISES. 1. Inscribe in an equilateral triangle another equilateral triangle having each side equal to a given straight line. 2. Shew how to cut off the corners of an equilateral triangle, so as to leave a regular hexagon. 3. The sides AB, AG oi a. triangle are produced and the exterior angles are bisected by straight Hues meeting in 0: if a circle be described about the triangle BOG, its centre will be on the circle described about the triangle ABG. T. E. 20 298 BOOK IV. PROPOSITION 6. To inscribe a square in a given circle. Let ABCD be the given circle : it is required to inscribe a square in the circle ABCD. CpNSTRUCTiON. Find the centre E of the circle ABCD, (III. Prop. 5.) and draw two diameters AEC, BED at right angles to one another. (I. Prop. 11.) Draw ^5, BG, CD, DA: the quadrilateral ABCD is a square inscribed as required. Proof. Because the angle BEC is double of the angle BAG, and the angle AED is double of the angle ACD, (III. Prop. 20.) and the angle BEC is equal to the angle AED, (I. Prop. 15.) therefore the angle BAG is equal to the angle AGD. And because AC meeting AB, CD makes the alternate angles BA C, A CD equal, AB, CD are parallel. (I. Prop. 27.) Similarly it can be proved that AD, BG are parallel. Therefore the quadrilateral ABCD is a parallelogram. Again, because ABC is an angle in a semicircle ABC, the angle ABC is a right angle. (III. Prop. 31.) Therefore the parallelogram ABCD is a rectangle. (I. Def. 19.) PROPOSITION 6. 299 Again, because in the triangles AEB, CEB^ AE is equal to CE, BE to BE, and the angle AEB to the angle GEB, the triangles are equal in all respects; (I. Prop. 4.) therefore BA is equal to BC. Therefore the rectangle ABCD is a square. (I. Def. 20.) Wherefore, a square ABCD has been inscribed iJi the given circle ABCD. EXERCISES. 1. Inscribe a regular octagon in a given circle. 2. Shew how to cut off the comers of a square so as to leave a regular octagon. 3. Inscribe in a given square, a square to have its sides equal to a given straight line. 20—2 300 BOOK IV. PROPOSITION 7. To describe a square about a given circle. Let ABCD be the given circle : it is required to describe a square about it. Construction. Find E the centre of the circle ABCD^ (III. Prop. 5.) and draw two diameters AEC, BED at right angles to one another. (I. Prop. 11.) Draw GAF, HGK parallel to BD, and GBH, FDK parallel to AG: the quadrilateral PGHK is a square described as required. Q A F B r V H K Proof. Because GF, HK are each parallel to BD, GF, HK are parallel to each other. (I. Prop. 30.) Similarly it can be proved that Gil J FK are parallel to each other ; therefore FGHK is a parallelogram. Again, because GAEB is a parallelogram, (Constr.) the angle AGB is equal to the angle AEB, (I. Prop. 34.) which is a right angle; (Constr.) therefore the parallelogram FGHK is a rectangle. (I. Def. 19.) Again, because GBDF is a parallelogram, GF is equal to BD, a diameter of the circle. Similarly it can be proved that GH is equal to AG, a diameter of the circle; therefore GF is equal to GH. Therefore the rectangle FGHK is a square. (I. Def. 20.) PROPOSITION 7. 301 Again, because A£) intersects the parallel lines GF, BD, the angle GAB is equal to the alternate angle AED^ (I. Prop. 29.) which is a right angle ; (Constr.) therefore 6^^^ touches the circle. (III. Prop. 16.) Similarly it can be proved that GBH^ HCK, KDF touch tlie circle. Wherefore, a square FGHK has been described about the given circle A BCD. EXEBCISES. 1. Describe a regular octagon about a given circle. 2. Prove that the area of a circumscribed square of a circle is double that of an inscribed square. 3. If two circles be such that the same square can be inscribed in one and described about the other, the circles must be concentric. Is any other condition necessary ? 4. If a parallelogram admit of a circle being inscribed in it and another circle being described about it, the parallelogram must be a square. 302 BOOK IV. PROPOSITION 8. To iyiscrihe a circle in a given square. Let A BCD be the given square : it is required to inscribe a circle in it. Construction. Draw AC, BD intersecting in E. From E draw EF, EG perpendicular to J) A, AB two of the sides of the square, (I. Prop. 12.) and with E as centre and EF or EG as radius describe a circle : it is a circle inscribed as required. Proof. Because CD is equal to AD, (1. Prop. 34, Coroll. 1.) the angle CAD is equal to the angle ^ C/>. (I. Prop. 5.) And because AG meets the parallels AB, DC, the angle BA C is equal to the alternate angle A CD ; (I. Prop. 29.) therefore the angle BAG is equal to the angle DAG. And because in the triangles GAE, FAE, the angle GAE is equal to the angle FAE, and the angle AGE to the angle AFE, and AE equal to AE, the triangles are equal in all respects ; (I. Prop. 26, Part 2.) therefore EG is equal to EF, and therefore the circle described with E as centre and EF or EG as radius passes through the extremity of the other, and touches the two sides DAy A B. (III. Prop. 16.) PROPOSITION 8. 303 Similarly it can be proved that this circle touches each of the sides ^(7, CX): it is therefore inscribed in the square ABCD. Wherefore, a circle FGH has been inscribed in the given square ABCD. EXEECISES. 1. Prove that a circle can be inscribed in any rhombus. 2. Two opposite sides of a convex quadrilateral are together equal to the other two. Shew that a circle can be inscribed in the quadrilateral. 3. AD, BE are common tangents to two circles ABC, DEC, that touch each other ; shew that a circle may be inscribed in the quadrilateral ABED, and a circle may be described about it. 304 BOOK IV. PROPOSITION 9. To describe a circle about a given square. Let ABCD be the given square: it is required to describe a circle about it. Construction. Draw AC, BD intersecting at E; and with E as centre and EA^ EB^ EC or ED as radius describe a circle : it is a circle described as required. Proof. Because in the triangles BAC, DAC, BA is equal to EA, (I. Prop. 34, Coroll. 1.) and BC to EC, and AC to AC, the triangles are equal in all respects; (I. Prop. 8.) therefore the angle BAC is equal to the angle BAC, or the angle BAC is half of the angle BAD. Similarly it can be proved that the angle ABD is half of the angle ABC. But the angle BAD is equal to the angle ABC', (I. Prop. 29, Coroll. and I. Prop. 10 B.) therefore the angle BAE is equal to the angle ABE, Therefore BE is equal to AE. (I. Prop. 6.) Similarly it can be proved that CE and DE are each of them equal to AE or BE. Therefore the circle described with E as centre and one of four lines EA, EB, EC, or ED as radius passes through the extremities of the other three, and is described about the square ABCD. Wherefore, a circle ABCD has been described about the given square ABCD. PROPOSITION 9. 305 EXEECISES. 1. A point is taken without a square such that the angles sub- tended at it by three sides of the square are equal : shew that the locus of the point is the circumference of the circle circumscribing the square. 2. Find the locus of a point at which two given sides of a square subtend equal angles. 3. If a quadrilateral be capable of having a quadrilateral of minimum perimeter inscribed in it, it must admit of a circle being inscribed in it. 4. ABCD is a quadrilateral inscribed in a circle, and its diagonals, AC, BD intersect at right angles in E; K, L, M, N are the feet of the perpendiculars from E on the sides of the quadrilateral. Shew that KLMN can have circles inscribed in it and described about it. 306 BOOK IV. PROPOSITION 10. To construct a triangle having each of two angles double of the third angle. Construction'. Take any straight line AB, and divide it at (7, so that the rectangle AB, BC may be equal to the square on AC. (II. Prop. 11.) With centre ^ and radius AB describe the circle BDE, and draw a chord BD equal to AG, (Prop. 1.) and draw DA : the triangle ABD is a triangle constructed as required. Draw DC, and about the triangle ACD describe the circle ACD. (Prop. 5.) Peoof. Because the rectangle AB, BC is equal to the square on AC, and AC is equal to BD, the rectangle A B, BC is equal to the square on BD. And because from the point B without the circle ACD, BCA is drawn cutting it in C and A, and BD is drawn " meeting it in D, and the rectangle AB, BC is equal to the square on BD, therefore^/) touches the circle ACD. (III. Prop. 37.) And because BD touches the circle ACD, and DC is, drawn from the point of contact D, the angle BDC is equal to the angle DAC. (III. Prop. 32.) To each of these add the angle CDA ; then the whole angle BDA is equal to the sum of the angles CDA, DAC. But the angle BCD is equal to the sum of the angles CDA, DAG; (I. Prop. 32.) therefore the angle BDA is equal to the angle BCD. PROPOSITIOF 10. 307 But because AD is equal to AB, the angle ABD is equal to the angle ADB\ (I. Prop. 5.) therefore the angle DBA is equal to the angle BCD. And because the angle DBO is equal to the angle BCD, DC is equal to DB. (I. Prop. 6.) But DB is equal to CA', (Constr.) therefore CA is equal to CD; therefore the angle CD A is equal to the angle CAD. (I. Prop. 5.) Therefore the sum of the angles CA D, CD A is double of the angle CAD. But the angle BCD is equal to the sum of the angles CAD, CDA', (I. Prop. 32.) therefore the angle BCD is double of the angle CAD. And the angle BCD lias been proved equal to each of the angles BDA, DBA ; therefore each of the angles BDA, DBA is double of the angle BAD. Wherefore, a triangle ABD has been constructed havirtg each of two angles ABD, ADB double of the third angle BAD. It will be observed that the smaller angle of the triangle constructed in this proposition is equal to a fifth of two right angles. EXERCISES. 1. Describe an isosceles triangle having each of the angles at the base one third of the vertical angle. 2. Divide a right angle into five equal parts. 3. In the figure of Proposition 10, if the two circles cut again at E, then BE is equal to DC. 4. In the figure of Proposition 10, the circle AC J) is equal to the circle described about the triangle ABD. 5. In the figure of Proposition 10, if AF be the diameter of the smaller circle, DF is equal to a radius of the circle which circum- scribes the triangle BCD. 6. If in the figure of Proposition 10, the circles meet again in E, then GE is parallel to BD. 308 BOOK IV. PROPOSITION 11. To inscribe a regular pentagon in a given circle. Let ABODE be the given circle : it is required to inscribe a regular pentagon in the circle ABODE. Construction. Construct a triangle FGH having each of the angles at G^ H double of the angle at F; (Prop. 10.) in the circle ABODE, inscribe the triangle AOD^ equi- angular to the triangle FGH, so that the angles OAD, ADC, DO A may be equal to the angles GFH, FHG, EOF re- spectively. (Prop. 2.) Bisect the angles AOD, ADO by the straight lines OE, DB, (I. Prop. 9.) and draw OB, BA, AE, ED : then ABODE is a pentagon inscribed as required. Proof. Because each of the angles AOD, ADO is double of the angle OAD, (Constr.) and they are bisected by OE, DB, the five angles ADB, BDO, OAD, DOE, EOA are equal. But equal angles at the circumference of a circle stand on equal arcs; (III. Prop. 26, Coroll.) therefore the five arcs AB, BO, OD, DE, EA are equal. And the chords, by which equal arcs are subtended, are equal ; (III. Prop. 29, Coroll.) therefore the five straight lines AB, BO, OD, DE, EA are equal ; therefore the pentagon ABODE is equilateral. Again, the arc ED is equal to the arc BA ; to each of these add the arc AE; PROPOSITION 11. 309 then the whole arc AED is equal to the whole arc BAE. And the angle AED is contained by the arc AED, and the angle BAE by the arc BAE ; therefore the angle AED is equal to the angle BAE. (III. Prop. 27, CoroU.) Similarly it can be proved that each of the angles ABC, BCD, CDE is equal to the angle AED or BAE; therefore the pentagon ABODE is equiangular. Therefore ABODE is a regular pentagon. "Wherefore, a regular pentagon ABODE has been inscribed in the given circle ABODE. The following is a complete Geometrical construction for inscribing a regular decagon or pentagon in a given circle. Find the centre. Draw two diameters AOC, BOB at right angles to one another. Bisect OB in E. Draw AE and cut off EF equal to OE. Place round the circle ten chords equal to AF. These chords will be the sides of a regular decagon. Draw the chords joining five alternate vertices of the decagon ; they will be the sides of a regular pentagon. We leave the proof of this as an exercise for the student. EXERCISES. 1. A regular pentagon is inscribed in a circle, and alternate angular points are joined by straight lines. Prove that these lines will form by their intersections a regular pentagon. 2. If ABCBE be a regular pentagon, AC EBB is a regular star shaped pentagon, each of whose angles is equal to two-fifths of a right angle. 3. ABCBE is a regular pentagon; draw ^C7 and BB, and let BB meet AG sX F\ shew that AC is equal to the sum of AB and BF. 4. li AB, BC, and CB be sides of a regular pentagon, the circle which touches AB and CB at B and C passes through the centre of the circle inscribed in the pentagon. 310 BOOK IV. PROPOSITION 12. To describe a regular pentagon about a given circle. Let ABCDE be the given circle : it is required to describe a regular pentagon about the circle ABCDE. Construction. Let A^ B,C, D, E be the angular points of a regular pentagon inscribed in the circle ; (Prop. 11.) so that AB, BC, CD, DE, EA are equal arcs. Find the centre F ; (III. Prop. 5.) draw FA, FB, EC, FD, FE, and draw GAH, HBK, KCL, LDM, MEG at right angles to FA, FB, EC, ED, EE respectively; (I. Prop. 11.) then GHKLM is a pentagon described as required. Draw FK, EL, Proof. Because in the triangles FBK, ECK, FB is equal to EC, and FK to FK, and the angle FBK equal to the angle ECK, each being a right angle, (I. Prop. 10 B.) and each of the angles FKB, EEC therefore is less than a right angle, (T. Prop. 17.) the triangles are equal in all respects ; (I. Prop. 26 A, CoroU.) therefore KB is equal to KG, and the angle BEK equal to the angle CEK, and the angle BKE to the angle CKF; therefore KF bisects each of the angles BEG, BKC. Similarly it can be proved that LF bisects each of the angles CED, CLD. PROPOSITION 12. 311 Again, because the arc BG is equal to the arc CD^ the angle BFC is equal to the angle CFD. (III. Prop. 27, Coroll.) And because the angle KFC is half of the angle BFC, and the angle LFC is half of the angle CFD; therefore the angle KFC is equal to the angle LFC. Now because in the triangles KFC, LFC, the angle KFC is equal to the angle LFC, and the angle KCF to the angle LCF, and FC to FC, the triangles are equal in all respects; (T. Prop. 26, Part 1.) therefore KC is equal to LC, and the angle FKC equal to the angle FLC. Now it has been proved that KB is equal to KC, and that KL is double of KC ; and it can similarly be proved that KII is double of KB ; therefore UK is equal to KL. Similarly it can be proved that any two consecutive sides of GHKLM are equal ; therefore the pentagon GHKLM is equilateral. And because it has been proved that the angles FKC, FLC are equal, and that the angle BKC is double of the angle FKC, and the angle CLD double of the angle FLC, therefore the angle BKC is equal to the angle CLD. Similarly it can be proved that any two consecutive angles of GHKLM are equal. Therefore the pentagon GHKLM is equiangular. The pentagon is therefore regular. And because each side is drawn at right angles to a radius of the circle at its extremity, it touches the circle; (III. Prop. 16.) therefore the pentagon is described about the circle ABODE. Wherefore, a regular pentagon GHKLM has been de- scribed about the given circle ABODE. 312 BOOK IV. PROPOSITION 13. To irtscribe a circle in a given regular pentagon. Let ABODE be the given regular pentagon : it is required to inscribe a circle in ABODE. Construction. Bisect any two consecutive angles of the pentagon ABC, BCD by BE, OE (I. Prop. 9.) meeting at E; draw EG, EH, EK, EL, EM perpendicular to AB, BO, CD, DE, EA respectively. (I. Prop. 12.) With E as centre and EG, EH, EK, EL or EM as radius describe a circle: it is a circle inscribed as required. Draw AE. G K D Proof. Because in the triangles ABE, OBE, AB is equal to CB^ (Hypothesis.) and BE to BE, and the angle ABE to the angle OBE, (Constr.) the triangles are equal in all respects ; (I. Prop. 4.) therefore FA is equal to EC, and the angle BAE is equal to the angle BOE. Again, because the angle BAE is equal to the angle BOD, (Hypothesis.) and the angle BAE has been proved equal to the angle BOE, and the angle BOE is half of the angle BOD, (Constr.) therefore the angle BAE is half of the angle BAE, ov AE bisects the angle BAE. PROPOSITION 13. 313 Similarly it can Ije pi'oved that EF, DF bisect the angles AED, EDO; therefore the bisectors of all the angles of the pentagon meet in a point. Again, because in the triangles FGH, FCK^ the angle FUG is equal to the angle FKC^ and the angle FOE to the angle FGK^ (Constr.) and FG to FG, the triangles are equal in all respects; (I. Prop. 26, Part 2.) therefore FH is equal to FK. Similarly it can be proved that the perpendiculars on any two consecutive sides are equal to one another : therefore FG^ FH^ FK, FL, FM are equal, and the circle described with F as centre and one of the five lines FG, FH, FK, FL or FM as radius passes through the ex- tremities of the other four ; and because the angles at G, II, K, L, M are right angles, it touches AB, EG, CD, BE, EA. (III. Prop. 16.) Wherefore, a circle GHKLM has been inscribed hi the given regular pentagon ABODE. EXERCISES. 1. How many conditions are necessary in order that a given pentagon may admit of a circle being inscribed in it ? 2. Prove that the bisectors of all the angles of any regular polygon meet in a point. T. 12. 21 314 BOOK IV. PROPOSITION 14. To describe a circle about a given regular pentagon. Let ABODE be tbe given regular pentagon : it is required to describe a circle about ABODE. Construction. Bisect any two consecutive angles of the pentagon ABO, BOD, by BF, OF (I. Prop. 9.) meeting at F, and draw FA, FE, FD ; with F as centre and FA, FB, FO, FD or FE as radius describe a circle : it will be a circle described as required. Proof. Because the angle ABO is equal to the angle BOD, (Hypothesis.) and the angle FBO is half of the angle ABO, and the angle FOB is half of the angle BOD, therefore the angle FBO is equal to the angle FOB; therefore FG is equal to FB. (I. Prop. 6.) Again, because in the triangles ABF, OBF, AB is equal to OB, (Hypothesis.) BE to BF, and the angle ABF to the angle OBF, (Gonstr.) the triangles are equal in all respects ; (I. Prop. 4.) therefore FA is equal to FG, and the angle FAB to the angle FOB. And because the angle FAB is equal to the angle FOB, and the angle BAE to the angle BOD, (Hypothesis.) PROPOSITION U. 315 and the angle FOB is half of the angle BCD^ (Constr.) therefore the angle FAB is half of the angle BAE; and FA bisects the angle BAF. Similarly it can be proved that FD, FE bisect the angles CDE, DEA respectively, and that FD and FE are each of them equal to FA or FC', therefore FA, FB, FC, FD, FE are all equal, and the circle described with F as centre and one of the five lines FA, FB, FC, FD, FE as radius passes through the extremities of the other four, and is described about the pentagon ABCDE. Wherefore, a circle ABCDE has been described about the given regular pentagon ABCDE. EXERCISES. 1. Describe a regular decagon to have five of its vertices coin- cident with those of a given regular pentagon. 2. How many conditions are necessary in order that a given pentagon may admit of a circle being described about it ? State the conditions. 3. Shew how to cut off the corners of a regular pentagon so as to leave a regular decagon. 21— -2 S16 BOOK IV. PROPOSITION 15. 2^0 iyiscribe a regular hexagon in a given circle. Let ABC DBF be the given circle : it is required to inscribe a regular hexagon in the circle ABGDEF. Construction. Find G the centre of the circle ; (III. Prop. 6.) draw any diameter AGD^ and with A as centre and AG as radius describe the circle 6?^i^ intersecting the circle ABC DBF in B and F. Draw BG, FG and produce them to meet the circle again in E and C, and draw AB, BC, CD, DE, EF, FA : then ABODE F is a hexagon inscribed as required. Proof. Because G is the centre of the circle ABCDEF, GB is equal to GA. And because A is the centre of the circle BGF, AB i& equal to AG. Therefore AB, BG, GA are all equal. Therefore the angles AGB, BAG^ GBA are all equal. (I. Prop. 5, CoroU. 1.) But the sum of these three angles is equal to two right angles; (I. Prop. 32.) therefore the angle AGB is equal to one-third of two right angles. Similarly it can be proved that the angle EGA is equal to one-third of two right angles. PROPOSITION 15. 317 But the sum of the three angles BGA^ AGF, FOE is equal to two right angles;. (I, Prop. 13.) therefore the angle FGE is one-third of two right angles, and therefore the angles EGA, AGF, FGE are equal. And because opposite vertical angles are equal, (I. Prop. 15.) the angles opposite to these are equal ; therefore all the angles AGE, FGE, EGD, DGC, CGB, EGA are equal. Therefore the arcs AF, FE, ED, DC, CB, BA are equal. (III. Prop. 26, Coroll.) Therefore the chords AF, FE, ED, DC, CB, BA are equal. (HI. Prop. 29, Coroll.) Therefore the liexagon ABC DEE is equilateral. Again, because the arcs BAF, AFE, FED, EDC, DCB, CBA are equal, the angles BAF, AFE, FED, EDC, DCB, CBA in those arcs are equal. (III. Prop. 27, Coroll.) Therefore the hexagon ABC DEE is equiangular : it is therefore regular, and it is inscribed in the circle ABCDEF. Wherefore, a regular liexagon ABCDEF has been in- scribed in the given circle ABCDEF. EXEECISES. 1. Inscribe a regular dodecagon in a given circle. 2. If ABCDEF be a regular hexagon, and AC, BD, CE, DF, EA, FB be drawn, they will form another regular hexagon of one third the area. 3. The perimeter of the inscribed equilateral triangle of a circle is three quarters the perimeter of the circumscribed regular hexagon. 4. Six equal circles can be described each touching a given circle and two of the others. 318 BOOK TV. PROPOSITION 16. To inscribe a regular polygon of fifteen sides in a given circle. Let A BCD be the given circle : it is required to inscribe a regular polygon of fifteen sides in the circle A BCD. Construction. Let ^, (7 be two angular points of an equilateral triangle inscribed in the circle, (Prop. 2.) and let A^B,DhQ three angular points of a regular pentagon inscribed in the circle. (Prop. 1 1 . Draw CD, and place round the circle fifteen chords AL, LM... each equal to CD. The figure ALM ... is a polygon inscribed as required. Proof. If the whole circle contain fifteen equal parts, the arc ABC, which is a third of the circle, contains five such parts, and the arc ABD, which is made up of the arcs AB, BD, each of which is a fifth of the circle, contains six such parts ; therefore the arc CD^ which is the difference of the arcs ABD, ABC, consists of one such part; therefore the arc CD is one-fifteenth of the circle. And because the arcs AL, LM, .,., which are the shorter arcs cut off* by equal chords AL, LM, ..., are equal, (III. Prop. 28, Coroll.) each of the arcs AL, LM, ..., is one fifteenth of the circle. PROPOSITION 16. 319 Therefore the extremity of the last chord coincides with the point A, and the extremities of the chords which have been placed round the circle exactly divide the circle into fifteen equal parts. The figure ALM... therefore is equilateral. And as each of the angles is contained by an arc made up of two of the fifteen equal parts, all the angles are equal ; (III. Prop. 27, Coroll.) therefore the figure ALM... is equiangular. Wherefore, a regular polygon of fifteen sides han been in- scribed in the given circle AJiCD. EXEECISES. 1. Prove that ia the figure of Proposition 16 a vertex of the inscribed regular polygon of fifteen sides coincides with each of the vertices of the regular figures used in the construction. 2. Prove that, if all but one of the bisectors of the angles of a polygon meet in a point, they all do so, and a circle can be inscribed in the polygon. 3. Prove that, if all but one of the rectangular bisectors of the sides of a polygon meet in a point, they all do so, and a circle can be described about the polygon. 320 BOOK IV. When a regular polygon of any number of sides is given, we can inscribe a regular polygon of the same number of sides in a given circle, and we can also describe a regular polygon of the same number of sides about a given circle. Moreover we can always inscribe a circle in a given regular polygon and describe a circle about it. Methods have now been given for the construction of regular figures of 3, 4, 5, 6, and 15 sides. When any regular polygon is given we can construct a regular polygon of double the number of sides by describing a circle about the polygon and bisecting the smaller arcs subtended by the sides of the given polygon, and so on in succession for each duplication of the number of sides. Thus we see that we can by EucUd's methods construct regular polygons of 3 x 2"', 4x2"', 5x2" and 15 x 2'^ sides, where « is any positive integer, in- cluding zero. It was proved by Gauss* in the year 1801 that by purely geometrical methods those regular polygons can be con- structed, the number of whose sides is a prime f number of the form 2'* + !. This general law relating to the number of sides includes the case of the triangle {n = l) and the pentagon (n=2); the next two cases are those of the polygons which have 17 sides {n = 4) and 257 sides {n = 7) respectively. * Disquisitiones Arithmeticce (sectio septima). t A prime number is an integer which is not divisible without remainder by any integer except itself and unity. MISCELLANEOUS EXERCISES. 1. To inscribe a square in a given parallelogram. 2. On a given circle find a point such that, if chords be drawn to it from the extremities of a given chord of the circle, their difference shall be equal to a given straight line less than the given chord. 3. AB is a fixed chord of a circle whose centre is O, and CD is any other chord equal to AB. The extremities of these chords are joined by straight lines. Prove that, if the joining lines meet each other, their point of intersection lies on the circle which circumscribes the triangle AOB. 4. Through each angular point of a triangle two straight lines are drawn parallel to the straight lines joining the centre of the cir- cumscribed circle to the other angular points of the triangle. Prove that these six straight lines form an equilateral hexagon, which has three pairs of equal angles.. 5. If two equilateral triangles be described about the same circle, they will form an equilateral hexagon whose alternate angles are equal. 6. Describe about a given circle a quadrilateral equiangular to a given quadrilateral. 7. If the diameter of one of the escribed circles of a triangle be equal to the perimeter of the triangle, the triangle is right-angled. 8. The diameter of the inscribed circle of a right-angled triangle is equal to the difference between the sum of the two smaller sides and the hypotenuse. 9. Construct a triangle having given the centres of its inscribed circle and of two of its escribed circles. 10. The side AB of a triangle ABC touches the escribed circles at (xj, Gg, Gg: prove that G^G^ is equal to BC, and G^G^ to CA. 11. If a circle be inscribed in a right-angled isosceles triangle, the distance from the centre of the circle to the right angle will be equal to the difference between the hypotenuse and a side. 322 BOOK IV. 12. Describe a circle to touch each of two given straight lines and to have its centre at a given distance from a third given straight line. 13. Three circles are described, each of which touches one side of a triangle ABC, and the other two sides produced. If D be the point of contact of the side BC, E that of AC, and F that of AB, then AE is equal to BD, BF to CE, and CB to AF. 14. If the diagonals of the quadrilateral ABCD intersect at right angles at 0, the sum of the radii of the inscribed circles of the tri- angles A OB, BOC, COD, DO A is equal to the difference between the sum of the diagonals and the semi-perimeter of the quadrilateral. 15. Having given the hypotenuse of a right-angled triangle and the radius of the inscribed circle, construct the triangle. 16. If the inscribed circle of a triangle ABC touch the sides AB, AC qX the points D, E, and a straight line be drawn from A to the centre of the circle meeting the circumference at G, the point G is the centre of the inscribed circle of the triangle ADE. 17. Two sides of a triangle whose perimeter is constant are given in position : shew that the third side always touches a fixed chcle. 18. The points of contact of the inscribed circle of a triangle are joined ; and from the angular points of the triangle so formed perpen- diculars are drawn to the opposite sides : shew that the triangle of which the feet of these perpendiculars are the angular points has its sides parallel to the sides of the original triangle. 19. Four triangles are formed by three out of four given points on a given circle : shew that a circle may be described so as to pass through the centres of the inscribed circles of the four triangles. 20. The rectangle of the segments of the hypotenuse of a right- angled triangle made by the point of contact of the inscribed circle is equal to the area of the triangle. 21. If on the sides of any triangle three equilateral triangles be constructed, the centres of the inscribed circles of these triangles are the vertices of an equilateral triangle. 22. Describe three equal circles to touch each other and a given circle. 23. Construct an isosceles triangle, having its base equal to the greater, and the diameter of its inscribed circle equal to the less of two given straight lines. 24. In a given right-angled triangle, the lengths of the sides con- taining the right angle are 6 and 8 feet respectively. Find the lengths of the segments into which the hypotenuse is divided by the circle inscribed in the triangle. MISCELLANEOUS EXAMPLES. 323 25. Two triangles ABC, DEF are inscribed in the same circle so that AB, BE, GF meet in one point ; prove that, if O be the centre of the inscribed circle of one of the triangles, it will be the ortho- centre of the other. 26. A circle B passes through the centre of another circle A ; a triangle is described circumscribing A and having two angular points on B : prove that the third angular point is on the line of centres. 27. A circle is escribed to the side i3(7 of a triangle ABC touching the other sides in F and (r. A tangent DE is drawn parallel to UC meeting the sides in D, E. DE is found to be three times BC in length. Shew that DE is twice AF. 28. The sum of the diameters of the inscribed and the circum- scribed circles of a right-angled triangle is equal to the sum of the sides containing the right angle. 29. Prove that two circles can be described with the middle point of the hypotenuse of a right-angled triangle as centre to touch the two circles described on the two sides as diameters. 30. The perpendicular from A on the opposite side i>C of a triangle ABC, meets the circumference of the circumscribed circle in G. If P be the point in which the perpendiculars from the angles upon the opposite sides intersect, then PG is bisected by BC. 31. Through G, the middle point of the arc ACB of a circle, any chord GP is drawn, cutting the straight line AB in Q. Shew that the locus of the centre of the circle circumscribing the triangle BQP is a straight line. 32. The distance of the orthocentre of a triangle from any vertex is double of the distance of the centre of the circumscribed circle from the opposite side. 33. Two equilateral triangles ABC, DEF are inscribed in a circle whose centre is 0. AC, DF intersect in P, and AB, DE in Q. Prove that either POQ is a straight line, or a circle can be described about APOQD. 34. Given the base, the difference of the angles at the base and the radius of the circumscribing circle of a triangle, shew how to con- struct the triangle. 35. From the vertices of a triangle draw straight lines which shall form an equilateral hexagon whose area is double that of the triangle. 36. If on each side of an acute-angled triangle as base, an isosceles triangle be constructed the sides of each being equal to the radius of the circumscribed circle, the vertices of these triangles form the vertices of a triangle equal in all respects to the original triangle. 37. If O be the orthocentre of the triangle ABC, the triangle formed by the centres of the circles OBG, OCA, GAB is equal to the triangle ABC in all respects. 324 BOOK IV. 38. The ends of a straight line AB move along two fixed straight lines in a plane ; prove that there is a point, in rigid connection with AB, which describes a circle. 39. The circle through B, C and the centre of the circle inscribed in the triangle ABC meets the sides AB, AC again in E, F; prove that EF touches the inscribed circle. 40. Let ABC be a triangle, the centre of the inscribed circle, and 0', 0", 0'", the centres of the escribed circles situated in the angles A, B, C, respectively; prove (1) that the circumscribed circle passes through the middle points of the lines 00', 00", 00'"; (2) that the four points 0, B, C, 0' lie on a circle which has its centre on the circumscribed circle; (3) that the points 0", B, C, 0'" lie on a circle whose centre is on the circumscribed circle. 41. The triangle of least perimeter which can be inscribed in a given acute-angled triangle is the triangle formed by joining the feet of the perpendiculars from the angular points on the opposite sides. 42. Describe a circle to touch a given straight line, and pass through two given points. 43. Describe a circle to pass through two given points and cut off from a given straight line a chord of given length. 44. Describe a circle to pass through two given points, so that the tangent drawn to it from another given point may be of a given length. 45. If I, be the centres of the inscribed and the circumscribed circles of a triangle ABC, and if AI be produced to meet the circum- scribed circle in F, then OF bisects BC. 46. If I be the centre of the inscribed circle of the triangle ABC, and AI produced meet the circumscribed circle at F, FB, FI, and FG are all equal. 47. Construct a triangle having given one angular point and the centres of the inscribed and the circumscribed circles. 48. is the centre of the circumscribed circle of a triangle ABC ; D, E, F are the feet of the perpendiculars from A, B, C, on the op- posite sides: shew that OA, OB, OC are respectively perpendicular to EF, FD, BE. 49. The four circles each of which passes through the centres of three of the four circles touching the sides of a triangle are equal to one another. 50. Construct a triangle having given an angle and the radii of the inscribed and the circumscribed circles. 51. From the vertex of a triangle draw a straight line to the base so that the square on the straight line may be equal to the rectangle contained by the segments of the base. MISCELLANEOUS EXAMPLES. 325 52. Four triangles are formed by three out of four given straight lines ; shew that the circumscribed circles of these triangles all pass through a common point. 53. The straight line joining the middle points of the arcs of a circle cut off by two sides of an inscribed equilateral triangle is trisected by those sides. 54. The perpendicular from an angle of an equilateral triangle on the opposite side is equal to three quarters of the diameter of the circumscribed circle. 55. If the inscribed and the circumscribed circles of a triangle be concentric, the triangle is equilateral. 56. The angle C of the triangle ABC is a right angle. P is the intersection of the diagonals of a square on AC, and Q of those of a square on BC. Prove that the circumscribed circle of the triangle ABC passes through the intersection of PQ with a perpendicular to AB drawn through the middle point of AB. 57. Describe three circles to touch in pairs at three given points. 68. A rhombus is described about a given rectangle. Prove that its centre coincides with that of the rectangle and that each of its angular points lies either on a fixed straight line or on a fixed circle. 59. Describe an isosceles triangle such that thi'ee times the verti- cal angle shall be four times either of the other angles. 60. li ABCDE be a regular pentagon, and AC, BD intersect at O, then ^0 is equal to DO, and the rectangle AC, CO is equal to the square on BC. 61. Prove that the difference of the squares on a diagonal and on a side of a regular pentagon is equal to the rectangle contained by them. 62. If with one of the angular points of a regular pentagon as centre and one of its diagonals as radius a circle be described, a side of the pentagon will be equal to a side of the regular decagon inscribed in the circle. 63. If ABCDE be a pentagon described about a circle, and F be the point of contact of AB, then twice AF is equal to the difference of the sum of AB, AE, CD and the sum of BC, DE. 64. AOA', BOB' are two diameters of a circle at right angles to one another. £0 is bisected at C and AC cuts the circle on BO as diameter in D and E. Circles having A as centre and AD, AE as radii are described cutting the original circle in jP and F^, G and G' : prove that A'GFF'G' is a regular pentagon. 65. A regular hexagon ABCDEF is inscribed in a circle. A second circle is described through A and B to cut the first circle at right angles : and a third circle is described through A and C to cut the first circle at right angles. Prove that the diameter of the third circle is three times that of the second. 326 BOOK IV. 66. If the alternate angles of an equilateral hexagon be equal to one another, a drcle can be inscribed in the hexagon. 67. Construct a regular polygon of 2n sides of equal perimeter with a given regular polygon of n sides. 68. Any equilateral figure which is inscribed in a circle is also equiangular. 69. Prove that a polygon which is described about one and inscribed in another of two concentric circles must be regular. 70. Shew that it is always possible to describe about a circle a polygon equiangular to any given polygon. Will the two polygons be necessarily similar? 71. Find the locus of the centre of the circumscribing circle of a triangle, when the vertical angle and the sum of the sides containing it are given. 72. Given the circumscribed circle, an escribed circle and the centre of the inscribed circle, construct the triangle. 73. Given an angular point, the circumscribing circle and the orthocentre, construct the triangle. 74. Describe a square about a given quadrilateral. How many solutions are there? 75. Construct a square so that each side shall touch one of four given circles. How many solutions are there ? CAMBBinaE: PBIITTBD BY C. J. CLAY, M.A. &SON8, AT THE UNIVEBSITY PEB8S. BOOK V. DEFINITIONS. Definition 1. If one magnitude he equal to another magnitude of the same kind repeated twice, thrice or any number of times, the first is said to be a multiple of the second, and the second is said to be an aliquot part or a measure of the first. If one magnitude A be equal to m times another magnitude B of the same kind {m being an integer, i.e. a whole number), A is said to be the m^'' multiple of B, and B the m* part of A. If A be any multiple of B and if C be the same multiple of D, then A and C are said to be equimultiples of B and D. The magnitudes treated of in Book V. are not necessarily Geome- trical magnitudes : but they are assumed to be such that any magnitude can be supposed to be repeated as often as desired, in other words, that any multiple we please of a magnitude can be taken. They are assumed also to be such that any one taken twice is greater than it is alone ; such quantities as those which are called in Algebra either negative or imaginary are excluded from consideration. The capital letters A, B, C,D &c. will be used to denote magnitudes, and the small letters m, n, p, q &c. to denote whole numbers. When a magnitude A is spoken of, the letter A is supposed to repre- sent the magnitude itself. Definition 2. The relatioyi of one magnitude to another of the same kind with respect to the multiples of the seco7id or of aliquot parts of the second, ivhich the first is greater than, equal to, or less than, is called the ratio of the first to the second. It is difficult to convey a precise idea of "ratio" by a definition. The student will gradually acquire a firmer grasp of the meaning of the term as he proceeds. It is important to bear in mind that the difference between two magnitudes is not their ratio. 328 BOOK V. Whenever the ratio of one magnitude to another is spoken of, it is necessarily implied, although it may not always be express?7 stated, that the two magnitudes are of the same kind. In the ratio of one magnitude to another, the first is called the antecedent and the second the consequent of the ratio. In the ratio of A to B, A is the antecedent and B the consequent. The ratio of a magnitude to an equal magnitude is called a ratio of equality, and is said to be equal to unity; the ratio of a magnitude to a less magnitude is called & ratio of greater inequality, and is said to be greater than unity; the ratio of a magnitude to a greater magnitude is called a ratio of less inequality, and is said to be less than unity. The ratio of one diameter to another diameter of the same circle is a ratio of equality : the ratio of a diagonal to a side of a square is a ratio of greater inequality : the ratio of the area of a circle to the area of a square described about the circle is a ratio of less inequality. Definition 3. If one magnitude repeated any number of times be greater than, equal to, or less than a second 7nagni- tude repeated any other number of times, the ratio of the first magnitude to the second magnitude is said to be greater than, equal to, or less than the ratio of the second number to the first number. li A, B he two magnitudes such that m times A is equal to n times B, the ratio of ^ to £ is equal to the ratio of n to m. Similarly, if m times A be greater than n times B, the ratio of ^ to iJ is greater than that of n to m ; and if m times A be less than n times J5, the ratio of ^ to B is less than that of n to m. Definition 4. When two magnitudes of the same kind are such that some measure of the first is equal to some measure of the second, the two magnitudes are said to be commensurable. Two magnitudes of the same kind, which are not com- mensurable, are said to be incommensurable. DEFINITIONS. 329 li AyByChe three magnitudes of the same kind such that C is the w*^ part of A and G is the n*'^ part of B, or, in other words, such that A is equal to m times C and B is equal to 7i times C, then ^ and B have a common measure C, and therefore are commensurable. If the ratio of ^ to 5 be equal to the ratio of one integer to another, say that of n to w, and the ratio of C to D be also equal to that of n to m, the ratio of ^ to B is equal to that of C to D : and similarly, if the ratio of ^ to 5 be equal to that of n to m, and the ratio of C to D be greater or less than that of n to m, the ratio of il to £ is less or greater respectively than that of C to D. A complete method is thus afforded of testing the equality or the inequality of the ratios of pairs of commensurable magnitudes: but the same method is not applicable to incommensurable magnitudes. Now it will be manifest from what has been said that, if we have four magnitudes A, B, C, D, of which A and B are incommen- surable, the ratio of G to D cannot be equal to that of A to B, unless G and D be also incommensurable. It is possible to find two magnitudes of the same kind that are not commensurable. It can be proved that a diagonal and a side of the same square are such a pair of magnitudes, and also that the circumference and a diameter of the same circle are another such pair of magnitudes. The question arises how the ratios of two pairs of such incommensurable magnitudes are to be compared. It is easy to prove that a diagonal of a square is greater than once and less than twice a side ; these inequalities give a very rough comparison of the lengths of the two lines. It can be proved that 10 times a diagonal is greater than 14 times and less than 15 times a side : these inequalities give a less rough comparison of the lengths. Again, it can be proved that 100 times a diagonal is greater than 141 times and less than 142 times a side: these inequalities give a still less rough comparison of the lengths of the two lines. These facts are represented by saying that the ratio of a diagonal to a side is greater than the ratio of 1 to 1 and less than that of 2 to 1 : greater than the ratio of 14 to 10 and less than that of 15 to 10 : greater than the ratio of 141 to 100 and less than that of 142 to 100. Pairs of ratios of greater and greater numbers might be quoted, between which the ratio of a diagonal to a side always lies: but no two numbers can be found such that the ratio in question is equal to the ratio of the numbers. T. E. 22 330 BOOK V. In the following definition of the equality of two ratios, the case of incommensurable magnitudes is included as well as that of com- mensurable magnitudes. Definition 5. If four rtuignitudes he such that, when any equimultiples whatever of the first and the third are taken, and also any equimultiples whatever of the second and the fourth, the multiples of the first and the third are simul- taneously either both greater than, or both equal to, or both less than the multiples of the second and the fourth respec- tively, the ratio of the first magnitude to the second is said to he equal to the ratio of the third to the fourth. When the ratio of the first of four magnitudes to the second is equal to that of the third to the fourth, the mag- nitudes are said to be proportionals or in proportion. When four magnitudes are proportionals, the first is said to he to the second as the third to the fourth. Let A, B, G, D he four magnitudes, of which A and B are of the same kind, and C and D are of the same kind, and let any equimul- tiples whatever of A and C, say m times A and m times C, be taken, and any equimultiples whatever of B and Z), say n times B and n times D ; then, if m times A be greater than n times B and also m times G greater than n times D, or else m times A be equal to n times B and also w times G equal to n times D, or else m times A be less than n times B and also m times G less than n times D, for every possible pair of whole numbers m and n, the ratio of ^ to J5 is equal to the ratio of G to D, and A, B, G, D are proportionals. The fact that four magnitudes A, B, G, D are in proportion is denoted by saying that A has to B the same ratio that G has to D, or that the ratio of .4 to B is equal to that of G to D, or that A is to i? as C to D : it is expressed still more concisely by the notation A:B = G:D. Note. It will be observed that when four magnitudes A , B, G, D DEFINITIONS. 331 are defined in order as proportionals, i.e. ^1 is to JB as C to D, they are at the same time defined as proportionals also in the three several orders 5, A, D, C\ C, D, A, B ; and D, C, B, A; that is to say, it follows from the definition that, if any one of the four proportions A : B = C : D, B : A=D : C, C : D = A : B, D : C = B : A exist, the other three exist also. It follows at once from Definition 5 that if, of A, B, C, D, four magnitudes, A be equal to C and B be equal to D, the ratio of J^ to i? is equal to the ratio of C to D ; and further that if, of ^4, 7?, C, three magnitudes, A be equal to B, the ratio of ^ to C is equal to the ratio of B to C, and also the ratio of C to J is equal to the ratio of C to B. Definition 6. When /our magnitudes are proportionals, the first and the third, the antecedents, are said to he homologous to one another, and the second and the fourth, the consequents, are also said to he homologous to one an- other. In the proportion 4 is to £ as O to D, the antecedents A and G are homologous to one another and the consequents B and D are homologous to one another. Definition 7. If it he possible to take equimultiples of the first and the third of four magnitudes and equimultiples of the second and the fourth, such that the multiple of the first is greater than that of the second and the multiple of the third not greater than that of the fourth, the ratio of the first to the second is said to he greater than that of the third to the fourth ; and the ratio of the third to the fourth is said to he less than the ratio of the first to the second. As an example the ratios of the numbers 2 to 3 and 5 to 8 may be taken. If the 5^^ multiples of the first and the third be taken and the S"^ multiples of the second and the fourth, the multiples in order are 10, 9, 25, 24 : here 10 is greater than 9 and 25 greater than 24 : but equality between the ratios is not thereby established. 22—2 332 BOOK V. If the ll**^ multiples of the first and the third be taken, and the 7*** multiples of the second and the fourth, the multiples in order are 22, 21, 56, 56 : here 22 is greater than 21 and 55 not greater than 56 : and the fact is established that the ratio of 2 to 3 is greater than that of 5 to 8. Definition 8. The ratio of the first of three magnitudes of the same kind to the third is said to be compounded of the ratio of the first to the second ayid the ratio of the second to the third. The ratio of the first of three magnitudes of the same kind to the third is also said to he the ratio of the ratio of the first magnitude to the second to the ratio of the third magnitude to the second. If A, B, C be three magnitudes of the same kind, the ratio of Aio G \B compounded of the ratio ^ to J5 and the ratio B to C. Further, the ratio of A to C is said to be the ratio of the ratio A to B to the ratio C to B. Definition 9. If the first of a number of magnitudes of the same kind be to the second as the second to the third and as the third to the fourth and so on, the magnitudes are said to be in continued proportion. . ^ i If a number of magnitudes be in continued proportion, the ratio of the first to the third is said to be duplicate of the ratio of the first to the second, and the ratio of the first to the fourth is said to he triplicate of the ratio of the first to the second. If four magnitudes be in proportion, the first and the fourth are called the extremes and the second and the fourth the means of the proportion. If three magnitudes he iii continued proportion, the first and the third are called the extremes and the second the mean of the proportion ; also the second is called a mean proportional between the first and the third, and the third is called a third proportional to the first and tJie second. PROPOSITION 1. 333 PROPOSITION 1. [Euclid Elem. Prop. 4.] If four magnitudes in order he proportionals and any equimultiples of the antecedents he taken, and any equi- multiples of the consequents, the four multiples are pro- portionals in the same order as the magnitudes* * Let the magnitudes A, B, C, D be proportionals, and let m, n be two given numbers : it is required to prove that m times A, n times B, tn times (7, n times D are proportionals. Construction. Let p, q be any two numbers, and let m times A be E, ?* times B be F, m times C he G and n -^ times i> be /^ 1' > A .^ 13 ^^G- "^ J '>> f+ Proof. Because E^ G are equimultiples of A, (7, "^ p times E, p times 6^ are equimultiples of ^, (7, no matter what number p may be ; and because F, H are equimultiples of B, D, q times F, q times H are equimultiples of B, D, no matter what number q may be ; and because A is to B a,s C to i>, p times j^ and ^ times G are both greater than, both equal to or both less than q times F and q times if respectively, for all values of p and q; (Def. 5.) therefore ^ is to i^ as 6^ to if ; (Def. 5.) that is, m times A is to n times B as m times C to ?i times i>. Wherefore, if four magnitudes, &c. * Algebraically. If a : &=c : d, then wia -.nb-mc: nd. 334 BOOK V. PROPOSITION 2. [Euclid Elem. Prop. 8.] The greater of two magnitudes has to a third magnUude a greater ratio than the less has; and a third magnitude has to the less of two other magnitudes a greater ratio than it has to the greater*. Let A, B, C be three magnitudes of the same kind, of which A is greater than B : it is required to prove that A has to (7 a greater ratio than B has to 0, and that C has to ^ a greater ratio tlian G has to A. Construction. Let the excess of A over ^ be i>; take the m^^ equimultiples of B, D, such that each is greater than (7, and of the multiples of C let the p^"^ multiple be the first which is greater than m times B^ and let n be the number next less than ^9. Proof. Because m times B is not less than n times G, and 7n times D is greater than C ', (Constr.) therefore the sum of m times B and m times D is greater than the sum of n times G and G, that is, m times A is greater than p times G ; and m times B is less than p times G; (Constr.) therefore A has to C a greater ratio than B has to G. (Del 7.) Next, because p times G is less than m times Aj and p times C is greater than in times B, therefore G has to -ff a greater ratio than C has to A. (Def. 7.) "Wherefore, the greater, &c. * Algebraically. If a > 6, then a i o-b : c and c : b>c : a. FUOFOSITIONS 2, 3. 335 PROPOSITION 3. [Euclid Elem. Prop. 9.] If the ratio of the first of three magnitudes to the third he equal to the ratio of the second to the thirds the first magni- tude is equal to the second** Let yl, ^, (7 be three magnitudes of the same kind such that ^ is to C as i? is to C : it is required to prove that A is equal to B. Proof. Because, any magnitude greater than B has to C a greater ratio than B has to C, (Prop. 2.) and A has to C the same ratio as B to C, A cannot be greater than B. Again, because any magnitude less than B has to C a less ratio than B to C, (Prop. 2.) and A has to C the same ratio as B to C, A cannot be less than B. Therefore A must be equal to B. Wherefore, if the ratio of the first ^ &c. * Algebraically, li a : c = h : c, then a = 6. 336 BOOK F. PROPOSITION 4. Part 1. [Euclid Elem. Prop. 10.] If the ratio of the first of three magnitudes to the third he greater than the ratio of the second to the third, the fi^'st magnitude is greater than the second*. Let A, B, G be three magnitudes of the same kind, such that A has to C a greater ratio than B to C : it is required to prove that A is greater than B. Proof. Because the ratio of ^ to C is greater than that of B to (7, there are some equimultiples, say the m*'^ multiples, of A, B and some multiple, say the n^^ multiple, of (7, such that m times A is greater than n times C, and m times B not greater than n times (7; (Def. 7.) therefore there is some number m such that m times A is greater than m times B', therefore A is greater than B. Wherefore, if the ratio of the first ^ &c. PROPOSITION L Part 2. [Euclid Elem. Prop. 10.] If the ratio of the first of three m>agnitudes to the second he greater than the ratio of the first to the third, the second magnitude is less than the thirdj. Let A, B, (7 be three magnitudes of the same kind, such that A has to B a, greater ratio than ^ to C: it is required to prove that B is less than G. Proof. Because the ratio of ^ to J5 is greater than that of A to (7, there is some multiple, say the m^^ multiple, of A, and there are some equimultiples, say the n^^ multiples, of B, G, such that m times A is greater than n times B, and m times A not greater than n times G; (Def. 7.) therefore there is some number n such that n times G is greater than n times B ; therefore G is greater than B. Wherefore, if the ratio of the first, &c. * Algebraically. If a : ob : c, then a>b. + Algebraically. If a : 6>a : c, then 6 : it is required to prove that ^ is to ^ as E to F. Proof. Because ^ is to ^ as (7 to D, in times A and rti times G are both greater than, both equal to or both less than n times B and n times D respectively, for all values of m and n\ (Def. 5.) and because jE^ is to ^ as C to D, m times E and 7?i times C are both greater than, both equal to, or both less than 71 times F and n times D respectively, for all values of m and n', therefore m times A and Tti times E are both greater than, both equal to or both less than n times B and n times F respectively, for all values of 7n and n; therefore ^ is to i? as ^ to F. (Def. 5.) Wherefore, ratios, which are equal, &c. * Algebraically, li a :h=^c -. dy and e •.f=c\d, then a : h = e :/. 338 BOOK V, PROPOSITION 6. [Euclid Elem. Prop. 12.] If any riumber of ratios be eqical, each ratio is equal to the ratio of the sum of the antecedents to the sum of the con- sequents^* Let A^ -S, C, i), E^ F be any number of magnitudes of the same kind, such that the ratios of A to B, C to J), E to F are equal : it is required to prove that -4 is to ^ as the sum of A, C, E to the sum of B, I), F. Construction. Take any equimultiples, say the m^^ multiples, oi A, C, E^ and any equimultiples, say the n^^ multiples, of B, i>, F. Proof. Because ^ is to ^ as C to D, and also as E to F^ therefore m times A, tn times C and m times E are simultaneously all greater than, all equal to or all less than n times B^ n times D and n times F respectively, for all values of in and n\ (Def. 5.) therefore m times A and m times the sum of A, C, E are simultaneously all greater than, all equal to or all less than n times B and n times the sum of B, D, F respectively, for all values of m and n. Therefore A is to B as the sum of A, (7, E to the sum of B, I), F. (Def. 5.) Wherefore, if any number of ratios, &c. Corollary. The ratio of two mojgnitudes is equal to tlie ratio of any two equimultiples ofthem^. * Algebraically, li a:h = c : d=e :f, then a : h — aJrC-\-e : & + flf+/. t Algebraically, a :h=ma imh. PROPOSITIONS 6, 7. 339 PROPOSITION 7. [Euclid Elem. Prop. 13.] If the first of three ratios be equal to the second and the second greater than tJie third, the first is greater than the third*- Let A, Bj Cy />, E, F be six magnitudes, such that the ratio of ^ to ^ is equal to that of C to Z>, and the ratio of (7 to Z> is greater than that of US' to F: it is required to prove that the ratio of J to -5 is greater than that of E to F. Proof. Because the ratio of C to i> is greater than that of E to F, it is possible to find some equimultiples, say the m^^ multiples, of C and E, and some equimultiples, say the ri*'^ multiples, of D and F, such that m times C is greater than n times D and m times E not greater than n times F. (Def. 7.) Again, because ^ is to -6 as C to i>, m times A and m times C are simultaneously both greater than, both equal to, or both less than n times B and n times D respectively, for all values of m and n. (Def. 5.) Therefore for some values of m and n m times A is greater than n times B and m times E not greater than n times F, Therefore the ratio of ^ to i? is greater than that of E to F. (Def. 7.) Wherefore, if the first of three ratios, tfec. * Algebraically. If a\h = C'.d, and c'.d>e:f, then a : h>e :/. 340 BOOK V. PROPOSITION 8. [Euclid Elem. Prop. 14.] If the first of four 'proportionals of the same kind he greater than the third, the second is greater than the fourth; if the first be equal to tlie second, the third is equal to the fourth; if the first he less than the second, the third is less than the fourth*. Let A, B, C, D be four magnitudes of the same kind such that A is to B as C to D: it is required to prove that, if A be greater than C, B is greater than i>, and, if A be equal to C, B is equal to D, and, if A be less than G, B is less than D. Proof. First. Let A be greater than C. Because A, B, C are three magnitudes and A is greater than C, the ratio of ^ to ^ is greater than that of C to ^ ; (Prop. 2.) but the ratio of ^ to ^ is equal to that oi C to D; therefore the ratio of C to Z> is greater than that of C to ^; (Prop. 7.) therefore B is greater than D. (Prop. 4, Part 2.) Next. Let A be equal to C. Because A is to B a^s G to JJ, Bisto A a.s Bto G; (Del 5. Note.) and A is equal to G; therefore ^ is to C as i) to (7; therefore B is equal to D. (Prop. 3.) Lastly. Let A be less than G. Because A is to B as G to D, G is to D as A to B; (Del 5. Note.) therefore, by the first case, if G be greater than A, D is greater than B, that is, if A be less than G, B is less than D. Wherefore, if the first, &c. * Algebraically. If a i b = c : d, then a> = <:c according as b> = <.d. PROPOSITIONS 8,9. 341 PROPOSITION 9. [Euclid Elem. Prop. 16.] If the Jirst of four magnitudes of the same kind he to the second as the third to the fourth^ then also the first is to the third as the second to the fourth*. Let A, B, C, D be four magnitudes of the same kind such that A is to ^ as (7 to /) : it is required to prove that ^ is to C as J5 to D. Construction. Take any equimultiples, say the m^^ multiples, of A, B, and any equimultiples, say the n^^ multiples, of C, D. Proof. Because m times ^ is to m times -6 as ^1 to B^ and because n times C is to n times i> as C to ^, (Prop. 6. CoroU.) and A is to B as C to B: (Hypothesis.) therefore m times ^ is to m times B as n times C to n times B. (Prop. 5.) Therefore m times A and m times B are both greater than, both equal to or both less than n times C and n times D respectively, for all values of ??^ and n ; (Prop. 8.) therefore ^ is to C as ^ to B. (Dei. 5.) Wherefore, if the first, &c. * Algebraically. If a : b = c : d, then a : c = b i d. 342 BOOK V. PROPOSITION 10. [Euclid Elem. Prop. 17.] If the suin of the first and the second of four magnitudes he to the second as the sum of the third and the fourth to the fourth^ the first is to the second as the third to the fourth*. Let A, B, 0, D be four magnitudes, A and B being of the same kind and C and D of the same kind, such that the sum of A and B is to B as the sum of C and D to D: it is required to prove that ^ is to -5 as C to D. th Construction. Take any equimultiples, say the m* multiples, oi A, B^ C, D and any equimultiples, say the n^ multiples, of ^, i>; then the sums of m times B and n times B and of m times D and n times D are equimultiples of B and D respectively. Proof. Because the sum of A and ^ is to ^ as the sum of C and D to Z>, (Hypothesis.) therefore m times the sum of A and B and m times the sum of C and D are simultaneously both greater than, both equal to or both less than the sum of m and n times B and the sum of m and n times D respectively, for all values of m and n: (Def. 5.) therefore m times A and m times G are simultaneously both greater than, both equal to or both less than n times B and n times D respectively, for all values of m and n ; therefore J^ is to ^ as C to D. (Del 5.) Wherefore, if the su7n, &c. * Algebraically. If a + b : b=c + d : d, then a : b = c : d. PROPOSITIONS 10, 11. 343 PROPOSITION 11. [Euclid Elem. Prop. 18.] If the first of four magnitudes he to the second as the third to tJie fourth, then the sum of the first and the second is to t/ie second as the sum of the third and tlie fourth to the fourth*. Let -4, j8, C, i> be four magnitudes, A and B being of the same kind, and C and D of the same kind, such that .4 is to ^ as (7 to Z> : it is required to prove that the sum of A and ^ is to J5 as the sum of C and D to D. Construction. Take any equimultiples, say the mP^ multiples, of A^ G and any equimultiples, say the n^^ multiples, of B, D. Proof. Because ^ is to -6 as C to Z), (Hypothesis.) therefore m times A and m times G are simultaneously both greater than, both equal to or both less than n times B and n times B respectively, for all values of m and n\ (Def. 5.) therefore m times the sum of A and B and m times the sum of G and D are simultaneously both greater than, both equal to or both less than the sum of m and n times B and the sum of m and n times D respectively, for all values of m and n. And it is manifest that m times the sum of A and B and m times the sum of G and D are simultaneously both greater than ]) times B and 'p times D respectively, for all values of m and />, where m is not less than p. Therefore m times the sum of A and B and m times the sum of G and D are simultaneously both greater than, both equal to or both less than p times B and p times I) respectively, for all values of m and jo; therefore the sum of A and jB is to ^ as the sum of G and D to D. (Def. 5.) Wherefore, if tJie first, kc. * Algebraically. If a : & = c : d, then a + 6 : &=c + d : d. 344 BOOK V. PROPOSITION 12. [Euclid Elem. Prop. 19.] If the sum of the first and the second of four magnitudes he to the sum of the third and the fourth as the second to the fourth^ the first is to the second as the third to the fourth'*' . Let A^ B, C, D he four magnitudes of the same kind, such that the sum of A and B is to the sum of C and D a.s B to D : it is required to prove that A is to ^ as (7 to Z>. Proof. Because the sum of A and B is to the sum of C and i) as i5 to J), the sum of A and -5 is to ^ as the sum of C and D to B. (Prop. 9.) Therefore ^ is to ^ as C is to B. (Prop. 10.) Wherefore, if the sum, &c. Algebraically. If a + b : c + d = b : d, then a ; h = c : d. PROPOSITIONS 12, 13. 345 PROPOSITION 13. [Euclid Elem. Prop. 20.] If the first of six magnitudes be to the second as the fourth to the fifths and the second he to the third as the fifth to the sixth, then the first and the fourth are hoth greater than, both equal to, or both less tJian the third and the sixth respectively"^. Let A, B, G, D, E, i^be six magnitudes, A, B, C being of the same kind and D, E, F of the same kind, such that A is to ^ as i) to E, and ^ is to (7 as ^ to i^ : it is required to prove that A and D are both greater than, both equal to or both less than C and F respectively. Proof. First, let A be greater than C. Because ^ is to ^ as i> to E, (Hypothesis.) and the ratio of ^ to ^ is greater than that of G to B, (Prop. 2.) the ratio oi Dto E is greater than that of G to B; (Prop. 7.) and because C is to ^ as 2^ to E; (Dei. 5, Note.) the ratio oi D to E is greater than that of i^to E; (Prop. 7.) therefore D is greater than F. (Prop. 4.) Secondly, because the magnitudes are proportionals when taken in the orders A, B, D, E ; B, G, E, F, they are also proportionals when taken in the orders E, E, A, B ; E, F,B,G; (Def. 5, Note.) therefore by the first case, if D be greater than F, A is greater than G. Lastly. The magnitudes are also proportionals when taken in the orders G, B, F,E; B, A, E, D ; (Def. 5, Note.) therefore by the first and second cases, if G be greater than A, F \^ greater than D, and if F be greater than D, G is greater than A ; therefore A and D are both greater than, both equal to or both less than G and F respectively. Wherefore, if the first, tfec. * AlgebraicaUy. If a : h = d : e and b : c = e :/, ^^^^ ^ - ^ then a> = = to F. Construction. Take any equimultiples, say the m^^ multiples, of ^, D, and any equimultiples, say the n^^^ multiples, of B, E, and any equimultiples, say the p^^ multiples, of C, F. Proof. Because ^ is to jB as /> to E, and ^ is to C as ^ to F; therefore m times A is to n times B&sm times Dion times E, and n times -5 is to p times C as n times E to p times F. (Prop. 1.) Therefore m times A and m times D are both greater than, both equal to or both less than ]} times C and p times F respectively, for all values of m and jo. (Prop. 13.) Therefore A is to C as J) to F. (Del 5.) Wherefore, if the first, &g. Corollary. Ratios which are duplicate of equal ratios are equal f. * Algebraically. It a : b — d : e and b : c = e : f, then a : c = d :f. t Algebraically. If a : b = b : c and d : e = e :/ and a : b==d : e, then a : c = d : /. PROPOSITION U, 15. 347 PROPOSITION 15. If the first of six magnitudes have to the second a greater ratio than the fourth to the fifth, and the second have to the third a greater ratio than the fifth to the sixth, then the first lias to the third a greater ratio than the fourth to the sixth'''. Let A, B, C, D, E, F be six magnitudes, A, B, C being of the same kind, and D, E, F of the same kind, such that A has to j5 a greater ratio than D to E, and B has to 6' a greater ratio than E to F: it is required to prove that A has to (7 a greater ratio than D to F. Construction. Because the ratio of ^ to -5 is greater than that of D to E, it is possible to find some equi- multiples, say the m*^ multiples, of A and i>, and some equi- multiples, say the n^^ multiples, of B and E, such that m times A is greater than n times B, and 7Ai times D not greater than n times E: (Def. 7.) and because the ratio oi B to C is greater than that of E to F, it is possible to find some equimultiples, say the p^^ multiples, of B and E, and some equimultiples, say the q^^ multiples, of C and F, such that 2) times B is greater than q times (7, and p times E not greater than q times F. (Def. 7.) Let p times m be r and n times q be s, and let n times B he H and p times B be K. Proof. Because m times A is greater than n times B, and p times m is r, and n times B is ^, therefore r times A is greater than p times ZT ; and because p times ^ is greater than q times C, and p times B is K and n times g' is s, therefore n times ^ is greater than s times C ; and because n times B is ^, and /> times B is ^, jo times H is equal to n times ^; therefore r times A is greater than s times (7. Similarly it can be proved that r times D is not greater than s times /'^ ; therefore A has to C a greater ratio than Z> to F. (Def. 7.) Wherefore, if the first, &c. * Algebraically. If a : bxl : e and b : oe : f, then a : c> d : /. 23—2 348 BOOK PROPOSITION 16. Ratios, of which equal ratios are duplicate, are equal*. Let A, B, C, D, E, F be six magnitudes, A, B, C being of the same kind, and D, E, F of the same kind, such that ^ is to i? as ^ to C, and D is to E as E to F, and also ^ is to C as i) to F\ it is required to prove that ^ is to ^ as /> to E. Proof. If the ratio of ^ to 5 were greater than that of D to E, then also, since ^ is to j5 as ^ to C, and Z> is to ^ as J^" to F, the ratio oi B to C would be greater than that of E to F; (Prop. 7.) therefore the ratio of ^ to C would be greater than that of DtoF. (Prop. 15.) Similarly, if the ratio of ^ to ^ were less than D to E, then the ratio of A to C would be less than that of D to F; therefore ^ is to ^ as i> to E. Wherefore, ratios, of which &c. * Algebraically. U a : c = d :f and a : b = b : c and d : e = e : /, then a : b = d : e, negative quantities being excluded. BOOK YL DEFINITIONS. It is often convenient to speak of closed rectilineal figures as a class. The wording of definition 15 of Book i. (page 11) implies that the term polygon does not include a triangle or a quadrilateral. This restriction for the future will not be maintained, and any closed rectilineal figure, no matter what the number of its sides may be, will be included under the term polygon. Definition 1. When the angles of one polygon taken in order are equal to the angles of another taken in order, the two polygons are said to be equiangular to one another. F A The polygons ABODE, FGHKL are equiangular to one another, if the angles at ^, 5, C, D, E be equal to the angles at F, G, H, K, L respectively. Pairs of vertices A, F; B, G; &c., at which the angles are equal, are corresponding vertices: and pairs of sides AB, FG; BC, GH; &c. joining corresponding pairs of vertices are corresponding sides. In this definition there is one more condition of equality than is necessary. If n - 1 of the angles of a polygon of n sides be equal to w - 1 of the angles of another polygon of n sides, the remaining angles must be equal. (See I. Prop. 32, Coroll.) Definition 2. When the ratio of a side of one of two polygo7is, which are equiangular to one another, to the cor- responding side of the other is the same for all pairs of cor- responding sides, the polygons are said to he similar to one another. 350 BOOK VL The polygons ABODE, FGHKL are similar to one another, if the angles &t A, B, C, D, E be equal to the angles at F, G, H, K, L respectively, and if also all the ratios of AB to FG, BG to GH, CD to HK, DE to KL, EA to LF be equal to one another. It will be seen hereafter that there are in this definition three more conditions of equality than are necessary. One unnecessary condition is contained in the statement that the polygons are equi- angular to one another. Other two unnecessary conditions are con- tained in the statement that all the ratios are equal. For it can be proved that in general, if in two equiangular polygons all but two of the ratios of corresponding sides be equal, all the ratios are equal. Definition 3. If in each of two given finite straight lines a point he taken such that the segments of the first line are in the same ratio as the segments of the second line^ the two lines are said to he cut proportionally hy the points. ^'^ P ^^' B .B A ^ P ■i D D In the diagram (figure 1) the points P, Q cut the straight lines AB, CD proportionally, if AP be to PB as CQ to QD. This definition is extended also to the case, where the points P and Q are in the lines AB, CD produced (figure 2). It must however be noticed that both points P, Q must be in the lines themselves, or both points in the lines produced : otherwise the lines are not said to be cut proportionally. In figure 1 the points P, Q are said to cut the lines AB, CD intemaUy; in figure 2 the points P, Q are said to cut the lines AB, CD extemaUy. DEFINITIONS. 351 Definition 4. In some cases, where one side of a tri- angle is specially distinguished from the other tivo sides, that side is called the base of the triangle aoid the perpendicular* upon that side from the ojyposite vertex is called the altitude of the triangle. Similarly, one side of a parallelogram is sometimes called the base and tlie perpendicular distance between it and the opposite side the altitude of the parallelogram. Definition 5. When a straight line is divided into tivo parts, so that the whole is to one part as that part to the other part, the line is said to be divided in extreme and mean ratio. Definition 6. The figure formed of an arc of a circle and the radii drawn to its extremities is called a sector of the circle. The angle between the radii, which is subtended by the arc, is called the angle of the sector. B (i) ^ G A If be the centre of the circle, of which the arc ABC is a part, and OA, OC be radii, the figure OABC is a sector. In figure 1 the angle of the sector is less than two right angles, in figure 2 the angle of the sector is greater than two right angles. Definition 7. Points lying on a straight line are said to be collinear. A set of such p)oints is called a range. Straight lines passing through a point are said to be concurrent. A set of such lines is called a pencil. T]ie lines are called the rays of the pencil and the ])oint is called the vertex of the pencil. A set of points ABCD... lying on a straight line is called the range ABCD.... * (Seel. Def. 11, p. 9.) 352 BOOK VL A set of straight lines drawn from a point to a series of points A, B, C, D... is called the pencil (ABCD...). Definition 8. Four points on a straight line^ such that one pair divide the straight line joining the other pair internally and externally in the same ratio, are called a harmonic range. In the diagram, if AP be to PB as AQ to QB, then A, P, B, Q is a harmonic range. P O B Because AP is to PB b,9 AQ to QB, therefore QB is to BP as QA to AP (V. Def. 5, Note, and V. Prop. 9), that is, the points B, A divide the distance QP internally and externally in the same ratio. The two points which form either pair of a harmonic range are said to be conjugate to one another. The points A, B and P, Q are two pairs of conjugate points. Definition 9. A pencil of four rays passing through the four points of a harmonic range is called a harmonic pencil. Two rays, which pass through a pair of conjugate points of a harmonic range, are called conjugate rays of the pencil. Definition 10. Four points on a straight line, such that one pair divide the straight line joining the other pair internally and externally in different ratios, are called an anharmonic range. The ratio of the ratio of internal division to the ratio of external division is called the ratio of the anharmonic range. If two anharmonic ranges have equal ratios, they are called like anharmonic ranges. Definition 11. A pencil of four rays passing through the four points of an anharmonic range is called an anhar- monic pencil. DEFINITIONS. 353 The ratio of the range is also said to he the ratio of the pencil. If two pencils have equal ratios, they are called like an- harmonic pencils. Quadrilaterals are often divided into three classes (1) convex, (2) re-entrant, (3) cross, the natures of which appear from the adjoin- ing diagrams. If the sides of a quadrilateral be produced both ways, the character of the complete figure which is so formed is independent of the class to which the quadrilateral belongs, as is evident from the adjoining diagram. Such a figure is called a complete quadrilateral, of which the following is a definition. Definition 12. The figure formed hy four infinite straight lines is called a complete ciuadrilateral. The straight line joining the intersection of one pair of lines to the intersection of the other pair is called a diagonal. There are three such diagonals. In the last diagram PP\ QQ', RR' are diagonals of the complete quadrilateral there represented. 354 BOOK VL PROPOSITION 1. The ratio of two triangles of the same altitude is equal to the ratio of tlieir bases. Let the triangles ABC, ADE be two triangles of the same altitude, that is, having a common vertex A and their bases BC, BE in a straight line : it is required to prove that the triangle ABC is to the triangle ADE as BC to DE. Construction. In CB produced, take any number of straight lines BE, EG each equal to BC, and in DE pro- duced, any number EU, HK, KL each equal to DE\ and draw AE, AG, AH, AK, AL. H K Proof. Because BC, EB, GE are equal, the triangles ABC, AEB, A GEare equal. (I. Prop. 38, Coroll.) Therefore the triangle AGC is the same multiple of the triangle ABC, that GC is of BC. Similarly it can be proved that the triangle AD Lis the same multiple of the triangle ADE^.. that DL is of DE. Again, if GC be equal to DL, the triangle AGC is equal to the triangle ADL, (I. Prop. 38, Coroll.) and if GC be greater or less than DL, the triangle AGC is greater or less respectively than the triangle ADL. Therefore since of the four magnitudes the triangles ABC, ADE and the lines BC, DE, the triangle AGC and the line GC are any equimultiples wliatever of the first and PROPOSITION 1. 355 the third, and the triangle ADL and the liiie DL are any equimultiples whatever of the second and tlie fourth, and it has been proved that the triangle AGC and GC are both greater than, both equal to or both less than the triangle ADL and DL respectively ; therefore the triangle ABC is to the triangle ADE as BG to DE. (V. Def. 5.) Wherefore, the ratio of two triangles, &c. Corollary 1. TJie ratio of two triangles of equal alti- tudes is equal to the ratio of their hoses. Corollary 2. The ratio of two triangles of equal bases is equal to the ratio of their altitudes. Corollary 3. The ratio of two parallelograms of equal altitudes is equal to the ratio of their ba^es. Each parallelogram is double of the triangle on the same base and of the same altitude. The ratio of the parallelograms therefore is equal to the ratio of the triangles. (V. Prop. 6, Coroll.) EXEECISES. 1. The diagonals of a convex quadrilateral, two of whose sides are parallel and one of them double of the other, cut one another at a point of trisection. 2. The sum of the perpendiculars on the two sides of an isosceles triangle from any point of the base is constant. 3. If straight hues AO, BO, CO be drawn from the vertices of a triangle ABC, and AO produced cut BC in D, the triangles AOB, AOC have the same ratio a.a BD, DC. 4. If in the sides BC, CA of a triangle points D, E be taken, such that BD is twice DC, and CE twice EA, and the straight lines AD, BE intersect in O, then the areas of the triangles EOA, AOB, BODy ABC are in the ratios of the numbers 1, 6, 8, 21. 356 BOOK VI, PROPOSITION 2. Part 1. If a straight line he parallel to one side of a triangle^ it cuts the other sides proportionally. Let the straight line DE be parallel to the side BO of the triangle ABC, and cut the sides AB, AC or these sides produced in i), E respectively : it is required to prove that BD is to DA as CE to EA. Construction. Draw BE, CD. Proof. Because the two triangles BDE, CDE have the side DE common, and BC is parallel to DE, the triangles BDE, CDE are equal. (I. Prop. 37.) Therefore the triangle BDE is to the triangle ADE as the triangle CDE to the triangle ADE. (V. Def. 5, Note.) But the triangle BDE is to the triangle ADE as BD to DA; (Prop. 1.) and the triangle CDE is to the triangle ADE as CE to EA ; (Prop. 1.) therefore BD is to DA as CE to EA. (V. Prop. 5.) Wherefore, if a straight line &c. Corollary. Because BC is parallel to DE a side of the triangle ADE, it follows that DB is to BA as EC to CA ; therefore also A B ifi to AD r^ AC to AE. (V. Props. 10, and 11.) PROPOSITION 2. PART 1. 357 EXEECISES. 1. A straight line drawn parallel to BC, one of the sides of a triangle ABC, meets AB at D and AC at E; if BE and CD meet at F, then the triangle ADF is equal to the triangle AEF. 2. If in a triangle ABC a straight line parallel to BC meet AB at D and ^C at E, and if i?E and CD meet at jP: then AF produced if necessary will bisect BC and DE. 3. Through D, any point in the base of a triangle ABC, straight lines DE, DF are drawn parallel to the sides AB, AC, and meet the sides at E, F: shew that the triangle AEF is a mean proportional between the triangles FBD, EDC. 4. If two sides of a quadrilateral be parallel, any straight line drawn parallel to them will cut the other sides proportionally. 5. If a straight line EF, drawn parallel to the diagonal ^ C of a parallelogram ABGD, meet AD, DC, or those sides produced, in E and F respectively, then the triangle ABE is equal in area to the triangle BCF. 6. ABC is a triangle, and through D, a point in AB, DE is drawn parallel to BC meeting ^C in ^. Through C a line CF is drawn parallel to BE, meeting AB produced in F. Prove that AB is a mean proportional between AD and AF. 7. Through a given point within a given angle draw a straight line such that the segments intercepted between the point and the lines which form the angle may have to one another a given ratio. 8. Find a point D in the side ^B of a triangle ABC such that the square on CD is in a given ratio to the rectangle AD, DB. 358 J300K VI. PROPOSITION 2. Part 2. If a straight line cut two sides of a triangle proportion- all't/j it is parallel to the third side. Let the straiglit line DE cut the sides AB^ AC oi the triangle ABC, or these sides produced, proportionally in i), E respectively, so that BB is to DA as CE to EA : it is required to prove that DE is parallel to BC. Draw BE, CD. Proof. Because BD is to DA as CE to EA, (Hypothesis.) and as BD to DA so is the triangle BDE to the triangle ADE, (Prop. 1.) and as CE to EA so is the triangle CDE to the triangle ADE', (Prop. 1.) therefore the triangle BDE is to the triangle A DE as the triangle CDE to the triangle ADE; (V. Prop. 5.) i.e. the triangles BDE, CDE have the same ratio to the triangle ADE ; therefore the triangles BDE, CDE are equal. (V. Prop. 3.) But these triangles have a common side DE and lie on the same side of it ; therefore BC is parallel to DE. (I. Prop. 39.) Wherefore, if a straight line &c. PROPOSITION 2. PART 2. 359 Corollary. If AB be to AD as AC to AE^ it follows that BD is to DA as C^ to EA (V. Props. 10 and 11.) and therefore BG is parallel to AD. EXEECISES. 1. Prove that there is only one point which divides a given straight line internally in a given ratio, and only one point which divides a given straight line externally in a given ratio. 2. If DBF be a triangle inscribed in a triangle ABC and have its sides parallel to those of ABC, then D, E, F must be the middle points of £C, CA, AB. 3. From a point Fj in the common base of two triangles ACB, ADB, straight lines are drawn parallel to AC, AD, meeting BC, BD &t F, G: shew that FG is parallel to CD. 4. If two given distances PQ, RS be measured off on two fixed parallel straight lines, then the locus of the intersection of each of the pairs PS, QR and PR, QS is a parallel straight line. 5. On three straight lines OAP, OBQ, OCR the points are chosen so that AB, PQ are parallel and BC, QR are parallel ; prove that AC, PR also are parallel. 6. If two opposite sides AB, DC of a quadrilateral A BCD be parallel, any straight line PQ which cuts AD, BC proportionally must be parallel to AB and DC. 7. Take D, E, the middle points of the sides CA, CB of a triangle ; join D and E, and draw AE, BD, intersecting in ; then the areas of the triangles DOE, EOB, BOA are in continued proportion. 360 BOOK VI PROPOSITION 3. Part 1. If an angle of a triangle he bisected internally or externally hy a straight line which cuts the opposite side or that side produced, the ratio of the segmeiits of that side is equal to the ratio of the other sides of the triangle. Let the angle BAG of the triangle ABC be bisected internally or externally by the straight line AD which cuts in D the opposite side BC (fig. 1) or BC produced (fig. 2): it is required to prove that BD is to DC as BA to AC. Construction. Through C draw CE parallel to DA to meet BA produced or BA in E; and take in BA or BA produced a point F on the side of A away from E, Proof. Because AC intersects the parallels AD, EC, the angle DAC is equal to the angle ACE; (T. Prop. 29.) and because FAE intersects the parallels AD, EG, the angle FAD is equal to the angle AEG. (I. Prop. 29.) And the angle DAG is equal to the angle FAD; (Hypothesis.) therefore the angle AEG is equal to the angle AGE; therefore ^C is equal to AE. (I. Prop. 6.) Now because AD is parallel to EC, one of the sides of the triangle BEG, BD is to DC as BA to AE ; (Prop. 2.) and AG has been proved equal to AE; therefore BD is to DC as BA to AC. Wherefore, if an angle &c. PROPOSITION 3. PART I. 361 EXEECISES. 1. ^ BO is a triangle which has its base BG bisected in D. BE, DF bisect the angles ADC, ADB meeting AG, AB in E, F. Prove that EF is parallel to BC. 2. If AD bisect the angle BAG, and meet BG in D, and DE, DF bisect the angles ADG, ADB and meet AG, AB in E, F respectively, then the triangle BEF is to the triangle GEF as BA is to A G. 3. An internal point is joined to the vertices of a triangle ABG. The bisectors of the angles BOG, GOA, AOB meet BC, GA, AB respectively in D, E, F: prove that the ratio compounded of the ratios AE to EG, CD to DB, and BF to FA is unity. 4. One circle touches another internally at O. A straight line touches the inner circle at G, and meets the outer one in A, B: prove that OA is to OjB as ^0 to GB. 5. The angle ^ of a triangle ABG is bisected by AD which cuts the base at D, and is the middle point of BC : shew that CD has the same ratio to OB that the difference of the sides has to their sum. 6. AD and AE bisect the interior and the exterior angles at A of a triangle ABG, and meet the base at D and E ; and O is the middle point of BG: shew that OB is a mean proportional between OD and OE. 7. If A, B, G be three points in a straight line, and D a point at which AB and BG subtend equal angles, then the locus of D is a circle. T. B. 24 362 . BOOK Vl. PROPOSITION 3. Part 2. If a straight line drawn through a vertex of a triangle cut the opposite side or that side produced, so that the ratio of the segments of that side is equal to the ratio of the other sides of the triangle, tJie straight line bisects the vertical angle internally or externally. Let the straight line AD drawn through A one of the vertices of the triangle ^i5C cut the opposite side BC or BG produced in D, so that BD is to DC as BA to AG: it is required to prove that AD bisects the angle at A internally or externally. Construction. Through G draw GE parallel to DA to meet BA produced (fig. 1) or BA (fig. 2) in E; and take in BA or BA produced a point F on the side of A away from E. Proof. Because DA is parallel to GE one of the sides of the triangle BEG, BD is to DG as BA to AE. (Prop. 2.) And BD is to DG as BA to AG', (Hypothesis.) therefore BA i^ to AG as BA to AE', (Y. Prop. 5.) therefore ^^ is equal to AG ', (Y. Prop. 3.) therefore the angle AGE is equal to the angle AEG. (I. Prop. 5.) Again, because AG intersects the parallels AD, EG, the angle DAG is equal to the angle AGE; (I. Prop. 29.) and because FAE intersects the parallels AD, EC, the angle FAD is equal to the angle AEG', (I. Prop. 29.) therefore the angle DAG is equal to the angle FAD. Wherefore, if a straight line &c. PROPOSITION 3. PART 2. 363 EXERCISES. 1. The bisector of the angle BAG of a triangle ABC meets BG in D; a straight line EGF parallel to BG meets AB, AD, AG in E, G, F respectively; prove that EG is to GF as BD to DG. 2. The sides AB, AG oi a. given triangle ABG are produced to any points D, E, so that DE is parallel to BG. The straight line DE is divided at F so that DF is to jPE as BD is to C£ : shew that the locus of i^ is a straight line. 3. If a chord of a circle AB be divided at G so that ^ C is to GB as AP to PB, where P is a point on the circle : then a circle can be de- scribed to touch AB at G and the given circle at P. 4. ABGD is a quadrilateral : if the bisectors of the angles at A and G meet in jBD, then the bisectors of the angles at B and D meet in^C. 5. If ^, B, C, D be four points in order on a straight line, such that AB is to BG as AD to DG, and P be any point on the circle described on BD as diameter, then PB, PD are the bisectors of the angle APG. 24—2 364 BOOK VI. PROPOSITION 4. If two triangles be equiangular to one another, they are similar. Let tlie triangles ABC, DEF be equiangular to one an- other : it is required to prove that the ratios AB to DE, BG to EF and CA to FD are equal. Construction. Of the two lines BA, ED let BA be the greater*. In BA take BG equal to ED, and in BG take BII equal to EF; and draw GH. Proof. Because in the triangles GBH, DEF, BG is equal to ED, and BH to EF, and the angle GBU to the angle DEF, the triangles are equal in all respects; (I. Prop. 4.) therefore the angle BGH is equal to the angle EDF] and the angle EDF is equal to the angle BAG; (Hypothesis.) Hence the angle BGH is equal to the angle BAG, and AC is parallel to GH ; (I. Prop. 28.) therefore BA is to BG as BG to BH -, (Prop, 2, Part 1, Coroll.) and GB is equal to DE, and BH to EF-, therefore AB is to DE as BG to ^i^. Similarly it can be proved that either of these ratios is equal to the ratio GA to FD. Wherefore, if two triangles &c. * The case when BA is equal to ED has already been dealt with. (I. Prop. 26.) PROPOSITION 4. 365 EXEKCISES. 1. A common tangent to two circles cuts the straight line joining the centres externally or internally in the ratio of the radii. 2. If AB, CD, two parallel straight lines, be divided proportionally by P, g, so that AP is to PB as GQ to QD, then AG, PQ, BD meet in a point. 3. ABGD is a parallelogram ; P and Q are points in a straight line parallel to AB; PA and QB meet at R, and PD and QG meet at S ; shew that RS is parallel to AD. 4. The tangents at the points P and Q of a circle intersect in T: if from any other point R of the circle the perpendiculars RM, RN be drawn to the tangents TP and TQ, and the perpendicular RL be drawn to the chord PQ, then RL is a mean proportional between RM and RN. 5. A straight line, parallel to the side BG of a triangle ABG, meets the sides A,B, AG (or those sides produced) at D and E. On DE is constructed a parallelogram DEFG, and the straight lines BG, GF (produced if necessary) meet each other at S. Prove that AS is parallel to DG or EF. 6. Inscribe an equilateral triangle in a given triangle, so as to have one side parallel to a side of the given triangle. 7. If two triangles have their bases equal and in the same straight line, and also have their vertices on a parallel straight line, any straight line parallel to their bases will cut off equal areas from the two triangles. 8. In a given triangle ABG draw a straight line PQ parallel to AB meeting AG, BG in P, Q, so that PQ may be a mean proportional between BQ, QG. 9. Two circles intersect at A, and a straight line is drawn bisect- ing the angle between the tangents at A. Prove that the segments of the line cut off by the circles are proportional to the radii. 10. If AGB, BGD be equal angles, and DB be perpendicular to BG and BA to AG, then the triangle DBG is to the triangle ABG as DG is to GA . 366 BOOK VI. PROPOSITION 5. If the ratios of the three sides of one triangle to the three sides of another triangle he equals the triangles are equi- angular to one another. Let the given triangles ABC, DEF be such that the ratios AB to DE, BG to EF and CA to FD are equal: it is required to prove that the triangles ABC, DEF are equiangular to one another. Construction. On the side of EF away from D draw EG J FG making the angles FEG, EFG equal to the angles GBA, BOA respectively. (I. Prop. 23.) Proof. Because in the triangles ABC, GEF, the angles ABC, BOA are equal to the angles GEF, EFG, the triangles ABC, GEF are equiangular to one another, (I. Prop. 32.) and therefore AB is to GE as BC to EF. (Prop. 4.) And AB is to DE as BC to EF; (Hypothesis.) therefore AB is to GE as AB to DE; (V. Prop. 5.) therefore GE is equal to DE. (Y. Prop. 3.) Similarly it can be proved that GF is equal to DF. Then because in the triangles DEF, GEF, DE, EF, FD are equal to GE, EF, FG respectively, the triangles are equal in all respects; (I. Prop. 8.) therefore the triangles DEF, GEF are equiangular to one another ; and the triangle GEF was constructed so as to be equi- angular to the triangle ABC ; therefore the triangles ABC, DEF are equiangular to one another. Wherefore, if the ratios &c. PROPOSITION 5. 367 PROPOSITION 5. A. If one pair of angles of two triangles he equal and another pair of angles he supplementary, tJie ratios of the sides opposite to these pairs of angles are eqvxil. Let ABC, DBF be two triangles in which the angles ABC, DBF are equal, and the angles ACB, DFE are sup- plementary : it is required to prove that AB is to DE as ^ C to DF. Construction. Of the two angles ACB, DFE, let ACB be the less. With A as centre and AC as radius describe a circle cutting BC in G ; and draw AG. Proof. Because ^C is equal to AG, the angle AGC is equal to the angle ACG ; (I. Prop. 5.) and the angle AGB is the supplement of the angle AGC, and the angle DFE is the supplement of the angle ACB ; (Hypothesis.) therefore the angle AGB is equal to the angle DFE; and the angle ABG is equal to the angle DEF ; (Hypothesis.) therefore the triangles ABG, DEF are equiangular to one another; (I. Prop. 32.) therefore AB is to DE as AG to DF; (Prop. 4.) and ^C is equal to AG ; (Constr.) therefore AB is to DE as AC to DF. Wherefore, if one pair of angles, kc. 368 BOOK VI. PEOPOSITION 6. Jf the ratios of two sides of one triangle to two sides of another triangle he equal, and also the angles contahied hy those sides he equal, the triangles are equiangular to one another. Let ABC, DEF be two triangles in which ^i? is to DE as BG to EF, and the angle ABC is equal to the angle DEF: it is required to prove that the triangles ABC, DEF are equiangular to one another. Construction. Of the two lines BA, ED let BA be the greater*. In BA take BG equal to ED, and in BG take BU equ9,l to EF; and draw GH. Proof. Because BA is to ED as BG to EF^ and BG is equal to ED, and BH to EF, therefore BA is to BG as BG to BH ; therefore GH i& parallel to AC, .(Prop. 2, Part 2, Coroll.) and the angle BGH is equal to the angle BAG. (I. Prop. 29.) Again, because in the triangles DEF, GBH, ED is equal to BG and EF to BH, and the angle DEF to the angle GBH, the triangles are equal in all respects ; (I. Prop. 4.) * The case when BA is equal to ED has already been dealt with, (I. Prop. 4.) PROPOSITION 6. 369 therefore the angle EDF is equal to the angle BGH, and therefore to the angle BAC. And the angles at B and E are equal ; (Hypothesis.) therefore the triangles ABC, DEF are equiangular to one another. (I. Prop. 32.) Wherefore, if the ratios kc. EXERCISES. 1. Shew that the locus of the middle points of straight lines parallel to the base of a triangle and terminated by its sides is a straight line. 2. CAB, CEB are two triangles having the angle B common and the sides CA, CE equal ; if BAE be produced to D and ED be taken a third proportional to BA,AC, then the triangle BDG is similar to the triangle BAG. 3. From a point E in the common base of the triangles ACB, ADB, straight lines are drawn parallel to AC, AD, meeting BC, BD in F and G ; shew that FG, CD are parallel. 4. C is a point in a given straight line AB, and AB is produced to 0, so that CO is a mean proportional between AO and BO. If P be any point on a circle described with centre and radius OC, then the angles APG, BPC are equal. 6. If a point be taken within a parallelogram ABCD, such that the angles OBA, ODA are equal, then the angles OAD, OCD are equal. 6. If two points P, Q, be such that when four perpendiculars PM, Pm, QN, Qn are dropped upon the straight lines AMN, Amn, PM is to Pm as QN to Qn, then P and Q lie on a straight line through A. 7. If on the three sides of any triangle, equilateral triangles be described either all externally or all internally, the centres of the circles inscribed in these triangles are the vertices of an equilateral triangle. 8. The straight line OP joining a fixed point O to a variable point P on a fixed circle is divided in Q in a constant ratio ; prove that the locus of Q is a circle. 9. Given the base and the vertical angle of a triangle, find the locus of the intersection of bisectors of sides. 370 BOOK VI. PROPOSITION 7. If the ratios of two sides of one triangle to two sides of another triangle he equal, and also the angles opposite to one pair of these sides he equal, the angles opposite to the other pair of sides are equal or supplementary. Let ABG, DEF be two triangles, in which AB is to DE as BC to EF, and the angle BAG is equal to the angle EDF: it is required to prove that the angles ACB, DFE are either equal or supplementary. Construction. On the side of EF 2,yf2iy from D, draw EG making the angle FEG equal to the angle GBA, and draw FG making the angle EFG equal to the angle EGA. (I. Prop. 23.) Proof. Because the triangles ABG, GEF are equi- angular to one another, (I. Prop. 32.) AB is to GE as AG to GF; (Prop. 4.) and AB is to DE as BG to EF-, (Hypothesis.) therefore AB is to GE as AB to DE, (Y. Prop. 5.) and GE is equal to DE. (V. Prop. 3.) Now because in the triangles GEF, DEF, GE is equal to DE and EF to EF, and the angle EGF to the angle EDF', (for each is equal to the angle at A) PROPOSITION 7. 371 therefore the angles GFU, DFE are either equal or supple- mentary; (I. Prop. 26, A.) and the angle GFE is equal to the angle ACB ; (Constr.) therefore the angles ACB^ DFE are either equal or sup- plementary. Wherefore, if the ratios ttc. Corollary. When two of the ratios of a side of one triangle to the corresponding side of another triangle W^e equal, and also the angles opposite to one pair of these sides equal, the triangles are equiangular to one another, provided that of the angles opposite to the second pair of sides, (1) each be less than a right angle, (2) each be greater than a right angle, or (3) one of them be a right angle. (I. Prop. 2Q A, Coroll.) EXEKCISE. Prove that, if ABCD, EFGH be two quadrilaterals, such that the angles ABC, ABC are equal to the angles EFG, ERG respectively, and the ratios AB to EF, BC to EG, CD to GH are equal, and if the angles BAD, FEH be both acute angles, then the quadrilaterals are similar. 372 BOOK VI. PROPOSITION 8. In a right-angled triangle, if a perpendicular be drawn froTYi the opposite vertex to the hypotenuse, the perpendicular is a mean proportional between the segments of the hypo- tenuse, and each of the sides of the triangle is a mean pro- portional between the hypotenuse and the segment of it adjacent to that side. * Let ABC be a right-angled triangle, and let AD be drawn perpendicular to the hypotenuse BC : it is required to prove that BD is to DA as AD to DC, that BC is to BA as BA to BD, and that BC is to CA as CA to CD. Proof. Because in the triangles ABC, DBA, the right angle BAC is equal to the right angle BDA, and the angle ABC is equal to the angle DBA, therefore the triangles ABC, DBA are equiangular to one another. (I. Prop. 32.) Similarly it can be proved that the triangles DAC, ABC are equiangular to one another. Therefore the triangles DBA, DAC are equiangular to one another. Now, because the triangles DBA, DAC are equiangular to one another, BD is to DA as AD to DC ] (Prop. 4.) and because the triangles ABC, DBA are equiangular to one another, BC is to BA as BA to BD', (Prop. 4.) and because the triangles ABC, DAC are equiangular to one another, BC is to CA^^AC to CD. (Prop. 4.) Wherefore, in a right-angled triangle &c. PEOFOSITION 8. 373 EXEECISES. 1. If the perpendicular drawn from the vertex of a triangle to the base be a mean proportional between the segments of the base, the triangle is right-angled. 2. If a triangle whose sides are unequal can be divided into two similar triangles by a straight line joining the vertex to a point in the base, the vertical angle must be a right angle. 3. If CD, CE, the internal and the external bisectors of the angle at C in a triangle ABC having a right angle at A, cut BA in D and jK respectively, then AC is a mean proportional between AD, AE. 4. A perpendicular AD is drawn to the hypotenuse BC of a right- angled triangle from the opposite vertex A : and perpendiculars DE, DF are drawn from D to the sides AB, AC ; prove that a circle will pass through the four points B, E, F, G. 5. On the tangent to a circle at A two points C and B are taken such that AC IB equal to CB : the straight lines joining B, C to F, the opposite extremity of the diameter through A, cut the circle in D, E respectively ; prove that AE is to ED as FA to FD. 6. A chord CD is drawn parallel to a diameter AB of a circle, and AC, AD are produced to cut the tangent a.t B in E, F respectively ; prove that the sum of the rectangles AC, CE and AD, DF is equal to the square on AB. 7. If ^ be a point outside a circle and B be the middle point of the chord of contact of tangents drawn from A, and P, Q be any two points on the circle, then PA is to QA as PB to QB. 8. Two circles intersect in A, B; from B perpendiculars BE, BF are drawn to their diameters AC, AD; prove that C, E, F, D lie on a circle, which is cut at right angles by the circle whose centre is A and radius AB. 9. The circumference of one circle passes through the centre of another circle. If from any point of the former circle two straight lines be drawn to touch the latter circle, the straight line joining the points of contact is bisected by the common chord of the two circles. 374 BOOK VI. PROPOSITION 9. From a given finite straight line to cut off any aliqtwt part required. Let AB be the given finite straight line : it is required to cut off from AB a given aliquot part, say the ?^*^ part. Construction. From A draw any straight line AC making an angle with AB, and in it take any point D, and cut off ^^ the same multiple oi AB that ^^ is of the part to be cut off, i. e. take AU equal to n times AD. Draw BB, and draw DF parallel to it meeting AB inF: then AF is the part required. Proof. Because FD is parallel to BF, one of the sides of the triangle ABE, AB is to AF as AE to AD ; (Prop. 2, Part 1, Coroll.) and AE is equal to n times AD; therefore -4^ is equal to n times AF. Therefore AF is the n*^ part of AB. Wherefore, yVom the given straight line AB, AF the part required has been cut off. PROPOSITION 10. 375 PROPOSITION 10. To divide a given finite straight line similarly to a given divided straight line. Let ^^ be a given straight line and CD another given straight line divided in E : it is required to divide AB similarly to CD. Construction. Draw AF making an angle with AB\ cut off A G, GH equal to CE^ ED respectively. Draw HB, and draw GK parallel to HB meeting ABinK: then ^^ is divided at K similarly to CD at E. Proof. Because GK is parallel to HB one of the sides of the triangle AHB^ AK is to KB as AG to GH; (Prop. 2.) and AG is equal to CE, and GH to ED. (Constr.) Therefore ^Z is to KB as CE to ED. Wherefore, the straight line AB has been divided at K similarly to the straight line CD at E. EXERCISES. 1. If three straight lines passing through a point cut two parallel straight lines ABC, PQR in A, P; B, Q; G, R, then the lines A 0, PR are similarly divided in B, Q. 2. Draw a straight line through a given point A, so that the perpendiculars upon it from two other given points B and G may be in a given ratio. 3. Draw through two given points on a circle two parallel chords which shall have a given ratio to one another. 376 BOOK VI. PROPOSITION 11. lines To find a third proportional to two given finite straight Let AB, CD be two given straight lines : it is required to find a third proportional to AB^ CD. Construction. Draw from any point P a pair of straight lines PE, PF making an angle with one another, and from PE cut off PG, GH equal to AB^ CD respectively and from PF cut off PK equal to CD. Draw GK and draw HL parallel to GK meeting PF in L : then KL is a third proportional to AB^ CD. Proof. Because GK is parallel to HL one of the sides of the triangle PHL, PG is to GH as PK to KL; (Prop. 2.) and PG is equal to AB and GH and PK are each equal to CD; therefore AB is to CD as CD to KL. Wherefore to the two given straight lines AB^ CD a third proportional KL has been found. PROPOSITION 12. 377 PKOPOSITION 12. To find a fourth proportional to three given straight lines. Let AB, CD, UFhe three given straight lines : it is required to find a fourth proportional to AB, CD, EF. Construction. Draw from any point P a pair of straight lines PG, PH making an angle with one another; and from PG cut off PK, KL equal to AB, CD respectively, and from PH cut off PM equal to EF. Draw KM, and draw LN parallel to KM meeting PH in N: then MN is a fourth proportional to AB^ CD, EF. Proof. Because KM is parallel to LN one of the sides of the triangle PLN, PK is to KL as PM to MN -, (Prop. 2.) and PK is equal to AB, KL to CD, and PM to EF-, therefore ^^ is to CD as EF to MN. Wherefore to the three given straight lines AB, CD, EF, a fourth proportional MN has been found. EXEKCISE. 1. is a point on a straight line AB\ find a point B in AB produced, such that BA is to BB as CA to GB. T. E. 25 378 BOOK VI. PROPOSITION 13. To find a mean proportional between two given straight lines. Let AB, CD be two given straight lines: it is required to find a mean proportional between J^ and CD. Construction. Draw any straight line and from it cut oQ EF, FG equal to AB^ CD respectively. Describe a circle on EG as diameter and draw FII at right angles to EG meeting the circle in H: then FH is a mean proportional between AB and CD. Draw EH, HG. Proof. Because EHG is a semicircle, the angle EHG is a right angle; (III. Prop. 31.) and because HF is the perpendicular from H on the hypo- tenuse of the right-angled triangle EHG, EF is to FH as FH to FG', (Prop. 8.) and EF is equal to ^^ and FG to CD; therefore AB is to FH as FH to CD, Wherefore, between the two given straight lines AB, CD a mean proportional FH has been found. PROPOSITION 13. 379 EXEKCISES. 1. Find a mean proportional between two given straight lines by the use of the theorem of Proposition 37 of Book iii. 2. Divide a given finite straight line into two parts, so that their mean proportional may be of given length. 3. Construct an isosceles triangle equal to a given triangle and having the vertical angle equal to one of the angles of the given triangle. 4. Find a third proportional to two given straight lines by a method similar to that of Proposition 13. 25—2 380 BOOK VI. Definition. If the ratio of a side of one polygon to a side of another 'polygon he equal to the ratio of an adjacent side of the second to an adjacent side of the Jtrst^ those sides are said to he reciprocally proportional. PROPOSITION 14. Part 1. If two parallelograms, which have a pair of equal angles, he equal in area, their sides ahout the equal angles are reci- procally proportional. Let ABGD, EFGH be two parallelograms, which have the angles at B and H equal, and which are equal in area : it is required to prove that AB is to HG as EH to BC. Construction. From AB, GB produced cut oSBN, BL equal to HG, HE, and complete the parallelograms AL, LN. L M Proof. Because in the parallelograms LN, EG, LB is equal to EH, and BN to HG, and the angle LBN to the angle EHG, therefore the parallelograms LN, EG are equal in area. (I. Props. 4 and 34.) And the area of EG is equal to the area of ^C; therefore the area oi AL is to the area of LN as the area oi AL to the area of AC. And AB is to BN as the area of AL to the area of LN, and LB is to BC as the area of ^Z to the area oi AC ; (Prop. 1, Coroll. 3.) therefore AB is to BN as LB to BC, (V. Prop. 5.) that is, AB is to HG as EH to BC. Wherefore, if two parallelograms &c. PROPOSITION U. 381 PROPOSITION 14. Part 2. If two parallelograms^ which have a pair of equal angles, have their sides about the equal angles reciprocally propor- tional, the parallelograms are equal in area. Let ABCD, EFGH be two parallelograms, which have the angles at B, H equal and in which ^^ is to EG as EH to PC: it is required to prove that the parallelograms A PCD, EFGH are equal in area. Construction. From AP, CB produced cut off PN, PL equal to HG, HE and complete the parallelograms AL, LN. Proof. Because in the parallelograms LN, EG, LP is equal to EH, and PN to HG, and the angle LPN to the angle EHG, therefore the parallelograms LN, EG are equal in area. (I. Props. 4 and 34.) Because AP is to HG as EH to PC, and PN is equal to HG and LP to EH, therefore ^^ is to PN as LP to PC. And ^i5 is to BN as the area of ^Z to the area of LN, and LB is to BC as the area of AL to the area oi AC; (Prop. 1, Coroll. 3.) therefore the area of AL is to the area of LN as the area of ^Z to the area of AG. (V. Prop. 5.) Therefore the area of LN is equal to the area of AG. And the parallelograms LN, EG are equal in area ; therefore the parallelograms AC, EG are equal in area. Wherefore, if two parallelograms ^ AG to PR. (V. Prop. 5.) Again, because in the triangles DAG, SPR, the ratios of -4i> to PS and AG to PR are equal, and the angles DAG, SPR are equal; (Constr.) therefore the triangles DAG, SPR are equiangular to one another, and GD is to RS as AG to PR; (Prop. 6.) therefore GD is to RS as AB to PQ. (V. Prop. 5.) PROPOSITION 18. 389 Again because the triangles ABC^ PQR are equiangular to one another, (Const r.) the angle ACB is equal to the angle PRQ ; and because the triangles DAC, SPR have been proved equiangular to one another, the angle ACD is equal to the angle PRS ; therefore the angle BCD is equal to the angle QRS, Similarly it can be proved that the ratio of any other corresponding pair of sides of the polygons ABODE, PQRST is equal to that of AB to PQ, and that any corresponding pair of angles are equal. Wherefore, on the given straight line AB, the polygon ABODE has been constructed similar to the given polygon PQRST, so that AB, PQ are corresponding sides. EXEECISES. 1. Given the length of the line joining the middle point of a side of a square with an end of the opposite side; determine, by any method, the length of a diagonal of the square. 2. Inscribe in a given triangle a second triangle so that its sides may be parallel to three given straight lines. In how many ways can this be done ? 3. In a triangle ABC inscribe a square so that two of its vertices may be on BC and the other two on AB, AC. 4. In a semicircle inscribe a square, so that two corners may lie in the diameter and two on the circumference. 5. In a given sector of a circle inscribe a square so that two corners may lie on the arc and one on each of the bounding radii. 6. In a given sector inscribe a square so that two corners may be on one of the bounding radii, one on the other bounding radius and one on the arc. 390 BOOK VI. PROPOSITION 19. Similar triangles are to one another in the ratio duplicate of the ratio of two corresponding sides. Let ABC, BEF be similar triangles and BG, EF be corresponding sides: it is required to prove that the triangle ABC is to the triangle DEF, in the ratio duplicate of the ratio of BC to EF. Construction. Find a third proportional to BC, EF and from BC cut off BG equal to it. Draw A G. Proof. Because the triangles ABC, BEF are similar, AB is to BE as BC to EF ; (Prop. 4.) and BG is to EF as EF to BG; (Constr.) therefore AB is to BE as EF to BG\ (V. Prop. 5.) and the angle ABG is equal to the angle BEF \ therefore the triangles ABG, BEF are equal in area. (Prop. 15, Part 2.) And the triangle ^^C is to the triangle ^^6r as BC to BG ; therefore the triangle ABC is to the triangle BEF as BC to BG; And because BC is to EF as EF to BG, BC has to BG the ratio duplicate of the ratio of BC to EF. (V. Def. 9.) Therefore the triangle ABG has to the triangle BEF the ratio duplicate of the ratio of BC to EF. Wherefore, similar triangles &c. Corollary. If three straight lines be proportionals, the first is to the third as any triangle on the first to a similar triangle on the second. PROPOSITION 19. 391 EXEKCISES. 1. Through a point within a triangle three straight lines are drawn parallel to the sides, dividing the triangle into three triangles and three parallelograms: if the three triangles be equal to each other in area, each is one-ninth of the original triangle. 2. An isosceles triangle is described having each of the angles at the base double of the third angle: if the angles at the base be bisected, and the points where the lines bisecting them meet the opposite sides be joined, the triangle will be divided into two parts having the same ratio as the base to the side of the triangle. 3. ABC is a triangle, the angle A being greater than the angle B : a point D is taken in BG, such that the angle CAD is equal to B. Prove that CD is to CB in the ratio duplicate of the ratio of AD to AB. 4. The sides BC, CA, AB of an equilateral triangle ABC are divided in the points D, E, F so that the ratios BD to DC, CE to EA and ^i^ to FB are each equal to 2 to 1. Find the ratio of the triangle DEF to the triangle ABC. 6. If a straight line AB be produced to a point G so that AB is a mean proportional between AG and GB, then the square on AB is to the square on BG as AB to the excess of AB over BG. 6. Find a mean proportional between the areas of two similar right-angled triangles which have one of the sides containing the right angle common. 7. Bisect a given triangle by a line parallel to its base. 8. Bisect a given triangle by a line drawn perpendicular to its 9. Divide a given triangle into two parts, having a given ratio to one another, by a straight line parallel to one of its sides. 10. ABG is a triangle; AB is produced to E: AD is a straight line meeting BG in D : BF is parallel to ED and meets AD in F: construct a triangle similar to ABG and equal to AEF, 392 BOOK VI. PROPOSITION 20. A pair of similar polygons may he divided into pairs of similar triangles^ each pair having the same ratio as the polygons. Let ABC BE, PQRST be a pair of similar polygons: it is required to prove that the polygons can be divided into pairs of similar triangles. Construction. Take any point L within the polygon ABODE, and draw LA, LB, LC, LB, LE. Within the polygon PQRST, draw PX, QX making the angles QPX, PQX. equal to the angles BAI-j, ABL re- spectively, and draw XR, XS, XT. Proof. Because the triangles LAB, XPQ are equi- angular to one another, (Constr.) LB is to XQ as AB to PQ ; (Prop. 4.) and because the polygons ABGBE, PQRST are similar, AB is to PQ as BO to QR ; (Def. 2.) therefore LB is to XQ as BC to QR. (V. Prop. 5.) Again because the polygons ABCBE, PQRST are similar, the angle ABC is equal to the angle PQR ; and the angle ABL is equal to the angle PQX ; (Constr.) therefore the angle LBC is equal to the angle XQR. Therefore the triangles LBC, XQR are equiangular to one another, (Prop. 6.) and therefore similar. (Prop. 4.) Similarly it can be proved that the triangles LOB, LBE, LEA are similar to the triangles XRS, XST, XTP respectively. PROPOSITIOJS' 20. 393 Again, because ^5 is to PQ as BG to QR; and because the triangle LAB is to the triangle XPQ in the ratio duplicate of the ratio of ^j5 to PQ, and the triangle LBC is to the triangle XQR in the ratio duplicate of the ratio of BG to QR, (Prop. 19.) therefore the triangle LAB is to the triangle XPQ as the triangle LBG to the triangle XQR. (V. Prop. 14. Coroll.) Similarly it can be proved that each of the ratios of the triangles LGD, LDE, LEA to the triangles XRS, XST, XTP respectively is equal to the ratio of the triangle LAB to the triangle XPQ. Therefore the polygon ABGDE is to the polygon PQRST as the triangle LAB to the triangle XPQ. (V. Prop. 6.) Wherefore, a pair of similar polygons olygon8 are to one anotlter in the ratio duplicate oftlie ratio of two corresponding sides. EXERCISES. 1. If ABC be a right-angled triangle and CD be drawn perpen- dicular to the hypotenuse, then AD is to DB as the square on ^C to the square on CB. 2. If a straight line be drawn from each corner of a square to the nearer point of trisection of the next side of the square in order, so as to form a square, this square will be two-fifths of the original square. What will be the area of the new square, if the lines be drawn to the further point of trisection ? T. E. 26 394 BOOK VI PROPOSITION 21. Polygons which are similar to the same polygon are similar to 07ie another. Let each of the polygons ABC..., FGH...^ be similar to the polygon PQR... : it is required to prove that ABG...^ FGH... are similar to one another. Proof. Because the polygons ABC..., PQR... are similar, the angle ABC is equal to the angle PQR, and AB is to PQ as BC to QR', (Def. 2.) and because the polygons FGH...^ PQR... are similar, the angle FGH is equal to the angle PQR, and PQ is to FG as QR to GH. Therefore the angle ABC is equal to the angle FGH, and AB is to FG as BC to GH. (V. Prop. 14.) Similarly it can be proved that every pair of corre- sponding angles of the polygons ABC..., FGH... are equal and that the ratios of all pairs of corresponding sides are equal. Therefore the polygons ABC..., FGH... are similar. "Wherefore, polygons which are similar as EF to GH, and AKLB has to CMND the ratio duplicate of the ratio of AB to CD, (Prop. 20, Coroll.) and EPQRF has to GSTUH the ratio duplicate of the ratio oi EF to GH, therefore AKLB is to CMND as EPQRF to GSTUH. (V. Prop. 14, Coroll.) Wherefore, if four straight lilies &c. EXERCISE. 1. Perpendiculars are let fall from two opposite angles of a rect- angle on a diagonal : shew that they will divide the diagonal into equal parts, if the square on one side of the rectangle be double that on the other. PEOPOSITION 22. 397 PROPOSITION 22. Part 2. If the ratio of two similar polygons similarly described on the first and the second of four straight lines he equal to the ratio of two similar polygons similarly described on the third and the fourth^ the four straight lines are propor- tionals. Let AB, CD, EF, GUhe four given straight lines, and let AKLB, CMND be two similar polygons similarly de- scribed on AB, CD, and EPQRF, GSTUH be two similar polygons similarly described on EF, GH, and let AKLB, CMND, EPQRF, GSTUH be proportionals : it is required to prove that -4^ is to CD as EF to GH. Proof. Because AKLB has to CMND the ratio dupli- cate of the ratio of AB to CD, (Prop. 20, Coroll.) and EPQRF has to GSTUH the ratio duplicate of the ratio of EF to GH, and AKLB is to CMND as EPQRF to GSTUH, therefore AB is to CD as EF to GH. (V. Prop. 1 6.) Wherefore, if the ratio . Similarly it can be proved that the ratios of all pairs of corresponding sides of the parallelograms AEFG, ABCD are equal. Therefore the parallelograms are similar. Wherefore, a parallelogram, &c. PROPOSITION 24. 401 EXERCISES. 1. Prove that, in the figure of VI. 24, EG and BD are parallel. 2. Prove that, if in the figure of VI. 24, EF, OF produced cut CD, CB in fl, K, then HG, GA, KE meet in a point. 3. Prove that, if two similar quadrilaterals ABCD, AEFG be so placed that ABE, ADG are straight lines, then the points A, F, C lie on a straight hne. 4. In a given triangle inscribe a rhombus which shall have one of its angular points at a given point in the base, and a side on that base. 5. Construct a parallelogram similar to a given parallelogram, so that two of its vertices are on one side of a given triangle and the other vertices on the other two sides. 402 BOOK VI, PROPOSITION 25. To construct a polygon shnila?' to a given polygon and equal to another given polygon. Let ABODE be one given polygon, and FGffX smother: it is required to construct a polygon similar to ABODE and equal to FGHK. Construction. Construct on AB a rectangle AL equal to ABODE, and on BL construct a rectangle LM equal to FGHK. (I. Prop. 45.) Find FQ a mean proportional between AB and BM, (Prop. 13.) and on FQ construct a polygon PQRST similar to ABODE, so that FQ, AB are corresponding sides: (Prop. 18.) then FQRST is a polygon constructed as required. D K B L M P Q Proof. Because ABODE is to FQRST in the ratio duplicate of the ratio of ^^ to FQ, and J ^ is to BM in the ratio duplicate of the ratio of ^^ toP^; therefore ABODE is to FQRST as AB to BM; and ^^ is to BAI as the rectangle AL to the rectangle LM, that is, as ABODE to FGHK Therefore FQRST is equal to FGHK; and it was constructed similar to ABODE. "Wherefore, apolygoji FQRST has been constructed similar to the polygon ABODE and equal to the polygon FGHK. EXEECISES. 1. Construct a sqimre equal to a given equilateral triangle. 2. Construct an equilateral triangle equal to a given rectangle. PMOPOSITION 25. 403 In Proposition 4 it was proved that, if two tHangles be equiangular to one another, they are similar. Hence the condition of the equality of the ratios of corresponding sides, which appears in Definition 2, is unnecessary in the case of two triangles, which are equiangular to one another. If we take the case of two polygons ABODE, PQRST of more than three sides, which are equiangular to one another, and which are such that all but two of the ratios of pairs of corresponding sides are equal, say AB to PQ, BC to QR, CD to RS, where two adjacent sides are omitted, we can prove that the polygons are similar. Take any point L within ABCD, and draw LA, LB, LC, LD, DA. Within the polygon PQRS draw PX, QX, making the angles QPX, PQX equal to the angles BAL, ABL respectively, and draw XR, XS, SP. It can be proved, as in Proposition 20, that the triangles ALB, BLC, CLD are similar to the triangles PXQ, QXR, RXS respectively ; therefore the angles ALB, BLC, CLD are equal to the angles PXQ, QXR, RXS respectively, and therefore the angle ALD is equal to the angle PXS; also each of the ratios LA to XP^ LB to XQ, LC to XR and LD to XS is equal to the ratio of AB to PQ, and therefore ^L is to PX as LD to XS. Therefore tTie triangles ALD, PXS are similar, (Prop. 6.) and AD is to PS as LA to XP. Hence the two triangles AED, PTS are equiangular to one another ; therefore they are similar, and each of the ratios DE to ST, EA to TP is equal to the ratio of AD to PS, (Prop. 4) which is equal to the ratio of LA to XP, and therefore to the ratio AB to PQ. In this case therefore the two polygons are similar. This method reduces the case, where the two sides whose ratios are omitted are adjacent, to the similar case of quadrilaterals (Ex. 1, page 395). A similar method will reduce the case, where the two sides whose ratios are omitted are not adjacent, to the similar case of quadrilaterals (Ex. 2, page 395). The two cases together justify the remark on Definition 2, page 850. 404 BOOK VI. PROPOSITION 26. If two sirnilar parallelograms have a comynon angle and be similarly jjlaced, one is about the diagonal of the other. Let the parallelograms ABCD, AEFG be similar and similarly placed and have a common angle at A : it is required to prove that the points A, F, C lie on a straight line. Construction. Draw AF and AG. A E Proof. Because the parallelograms AEFG, ABGD are similar, AG is to AB as GF to DO; (Def. 2.) and the angle AGF is equal to the angle ADC; therefore the triangles AGF, ADC are equiangular to one another. (Prop. 6.) Therefore the angle GAF is equal to the angle DAC, i.e. the three points A, F, G lie on a straight line. Wherefore, if two similar parallelograms &c. EXEEGISE. 1. Inscribe in a given triangle a parallelogram similar to a given parallelogram so as to have two corners on one side and one on each of the other sides of the triangle. PEOPOSITION 30. 405 PROPOSITION 30. To divide a given straight line in extreme and mean ratio. Let AB be the given straight line: it is required to divide it in extreme and mean ratio. Construction. Divide AB at the point C into two parts so that the rectangle AB^ BC may be equal to the square on AC. (II. Prop. 11.) B Proof. Because the rectangle AB, BG is equal to the square on AC, AB is to AC as AC to BC. (Prop. 17.) Wherefore, the given straight line AB has been divided at C in extreine and mean ratio. EXERCISES. 1. Two diagonals of a regular pentagon which meet within the figure divide each other in extreme and mean ratio. 2. Divide a given straight line into two parts so that any triangle described on the first part may have to a similar and similarly de- scribed triangle on the second part the ratio which the whole has to the second part. 406 BOOK Vl. PKOPOSITION 31. A polygon on the hypotenuse of a right-angled triangle is equal to the sum of the polygons siTriilarly described on the other sides. Let ABC be a right-angled triangle having the right angle BAG: and let BDC, GEA^ AFB be similar polygons similarly- described on BG, GA, AB respectively : it is required to prove that the polygon BBG is equal to the sum of the polygons CUA^ AFB. Proof. Because BBG has to GBA the ratio duplicate of the ratio of BG to GA, and the square on BG has to the square on GA the ratio duplicate of the ratio of BG to GA; (Prop. 20, Coroll.) therefore BBG is to GBA as the square on BG to the square on GA; therefore BBG is to the square on BG as GBA to tlie square on GA. (V. Prop. 9.) PROPOSITIOir 31. 4or Similarly it can be proved that BDC is to the square on BC as AFB to the square on AB. Therefore BDC is to the square on BC as the sum of CEA^ AFB to the sum of the squares on CA, AB; (V. Prop. 6.) and the square on BC is equal to the sum of the squares on CA, AB; (I. Prop. 47.) therefore BDC is equal to the sum of CEA, AFB. Wherefore, a polygon tfec. EXERCISES. 1. Divide a given finite straight line into two parts so that the squares on them shall be to one another in a given ratio. 2. Construct an equilateral triangle equal to the sum of two given equilateral triangles. 3. On two given lines similar triangles are described ; construct a similar triangle equal to the difference of the given triangles. 4. Construct a triangle equal to the sum of three given similar triangles and similar to them. 6. Construct a polygon equal to the sum of any number of similar polygons and similar to them. 408 BOOK VI. PROPOSITION 32. If two triangles have sides parallel in pairs, the straight lines joining the corresponding vertices meet in a point. Let ABC, DEF be two triangles such that the sides BC, GA, AB are parallel to the sides EF, FD, DE respec- tively : it is required to prove that the straight lines joining the pairs of points A, Dy B, E; C, F meet in a point. Construction. Draw AD, BE and let them meet in G; and draw GC, GF. Proof. Because AB is parallel to DE, the angles GAB, GBA are equal to the angles GDE, GED respectively ; (I. Prop. 29.) therefore the triangles GAB, GDE are equiangular to one another; (I. Prop. 32.) therefore GB is to GE as BA to ED; (Prop. 4.) and because the triangles ABC, DEF are equiangular to one another, (I. Prop. 34, Coroll. 2.) BA is to ED as BC to EF; (Prop. 4.) therefore GB is to GE as BC to EF; (V. Prop. 5.) and the angle GBC is equal to the angle GEF \ (I. Prop. 29.) therefore the triangles GBC, GEE are equiangular to one another; (Prop. 6.) therefore the angles BGC, EGF are equal, that is, the points C, F, G lie on a straight line, or, in other words, AD, BE, GF meet in a point. Wherefore, if two triangles &c. PROPOSITION 32. 409 It will be seen at once that, if in the diagram of Proposition 32 AB be equal to BE, then the straight lines AD, BE do not meet at any point at a finite distance, in other words, they are parallel. Also, because the triangles ABC, DEF are similar, if AB be equal to DE, then also BG is equal to EF, and therefore BE and CF are parallel. Hence we must consider the case when the two triangles are similarly placed and equal as a special case in which the point of intersection of the lines joining the corresponding vertices is at an infinite distance. EXEKCISES. 1. If two similar triangles be similarly placed on two parallel straight lines, the lines joining corresponding vertices meet in a point. 2. If any two similar polygons have three pairs of corresponding sides parallel, the straight lines joining the corresponding vertices meet in a point. 3. AB is a fixed diameter of a circle ABC: PQ is a straight line parallel to AB and of constant length, which moves so that its middle point traces out the circle ABC; find the locus of the intersection of AP, BQ and of AQ, BP. 4. Prove that, if the corresponding sides of ABCD, EFGH two squares be parallel, the straight lines AE, BF, CG, DH pass through a point, and AG, BH, CE, DF pass through another point. T. E. 27 410 BOOK VI. PROPOSITION 33. Part 1. In equal circles angles at the centres have the same ratio as the arcs on which they stand. Let BCD, MNO be two given equal circles, and let BA 6', MLN be two angles at their centres : it is required to prove that the angle BAG is to the angle MLN as the arc BC to the arc MN. Construction. From A draw any number of radii AD, AE, AF making the angles CAD, DAE, EAF each equal to the angle BAC; and from L draw any number of radii LO, LP, LQ, LR making the angles NLO, OLP, PLQ, QLR each equal to the angle MLN. Proof. Because the angles BAC, CAD, DAE, EAF are all equal, the arcs BG, CD, DE, EF are all equal; (III. Prop. 26.) therefore the angle BAF and the arc BF are equimultiples of the angle BAC and the arc BC. Similarly it can be proved that the angle MLR and the arc MR are equimultiples of the angle MLN and the arc MN. And, because the circles are equal, if the angle BAF be equal to the angle MLR, the arc ^i^is equal to the arc MR, (III. Prop. 26.) and if the angle BAF he greater or less than the angle MLR, the arc BF is greater or less respectively than the arc MR. Therefore the angle BAC is to the angle MLN as the arc BC to the arc MN. (V. Def. 5.) Wherefore, in equal circles &c. PROPOSITION 33. PART 1. 411 Corollary. In equal circles angles at the circumferences have the same ratio as the arcs on which they stand. The angles at the centres are double of the angles at the circumferences, and therefore have the same ratio. (V. Prop. 6, Coroll.) In the construction of Proposition 33 there is nothing to limit the magnitude of the multiple angles BAF, 31 LR ; they may be greater than two right angles, greater than four right angles, or greater than any multiple of four right angles, and at the same time the multiple arcs BF, MR will be greater than half the circumference of the circle, greater than the circumference, or greater than any multiple of the circumference. In the Third Book (page 221) we had occasion to remark that the admittance of angles equal to or greater than two right angles was not inconsistent with Euclid's methods. We may now go further and say that the admittance of angles without any restric- tion whatever on their magnitude is essential to his method. The validity of the proof of this Proposition depends on the possibility of choosing any multiples we please of the angles BAG, MLN, that is, of taking the multiple angles BAF^ MLR as large as we please. 27—2 412 BOOK VI. PROPOSITION 33. Part 2. In equal circles^ the areas of sectors have the same ratio as their angles. Let BCD, MNO be two given equal circles, and let BAC, MLN be two angles at their centres : it is required to prove that the angle BAG is to the angle MLN as the sector BAG to the sector MLN. Construction. From A draw any number of radii AD, AE, J/' making the angles CAD, DAE, EAF ea>c\\ equal to the angle BAC ; and from L draw any number of radii LO, LP, LQ, LR making the angles NLO, OLP, PLQ, QLR each equal to the angle MLN. Proof. Because the angle CAD is equal to the angle BAC, it is possible to shift the figure CAD so that AC will be on AB, and AD on AC ] if this be done, then the point C will coincide with B and D with C, and therefore the arc CD with the arc BC. (III. Prop. 23.) Therefore the sector CAD coincides with the sector BAC and is equal to it in all respects. Similarly it can be proved that the sectors DAE, EAF are each equal to the sector BAC. Therefore the angle BAF and the sector BAF are equi- multiples of the angle BAC and the sector BAC. Similarly it can be proved that the angle MLR and the sector MLR are equimultiples of the angle MLN and the sector MLN. FBOPOSITIOK 33. PART 2. 413 And it can be proved as before that, if the angle BAF be equal to the angle MLR, the sector BAF in equal to the sector MLR; and, if the angle BAF he greater or less than the angle MLR, the sector is greater or less respectively than the sector MLR; therefore the angle BAG is to the angle MLN as the sector BAC to the sector MLN. (V. Def. 5.) Wherefore, in equal circles tkc. Corollary. In equal circles the areas of sectors have the same ratio as the arcs on which they stand. 414 BOOK VI. PROPOSITION 34. If an angle of a triangle he bisected by a straight line which cuts the opposite side, the rectangle contained by the segments of that side is less than the rectangle contained by the other sides by the square on the line. Let the angle BAC of the triangle ABC be bisected by the straight line AD, which cuts BC at D : it is required to prove that the rectangle BD, BC is less than the rectangle BA, AC hy the square on AD. Construction. Describe the circle ABC; (IV. Prop. 5.) produce AD to meet the circle at E, and draw EG. Proof. Because in the triangles BAD, BAC, the angle BAD is equal to the angle EAC, (Hypothesis.) and the angle ABD to the angle AEC, (III. Prop. 21.) therefore the triangles are equiangular to one another ; therefore BA is to EA ^^ AD to AC \ (Prop. 4.) therefore the rectangle BA, AC \s, equal to the rectangle EA, AD, (Prop. 16.) that is, to the rectangle ED, DA together with the square on AD. (II. Prop. 3.) And the rectangle ED, DA is equal to the rectangle BD, DC', (III. Prop. 35.) therefore the rectangle BD, DC is less than the rectangle BA, AC hy the square on AD. Wherefore, if an angle &c. PROPOSITION 34. 415 EXERCISES. 1. If an angle of a triangle be bisected externally by a straight line which cuts the opposite side produced, the rectangle contained by the segments of that side is greater than the rectangle contained by the other sides by the square on the line. 2. Prove that, if the internal and the external bisectors of the vertical angle of a triangle ABC cut BC in D and E, then the square on BE is equal to the difference of the rectangles EB, EC and BE, BG. 3. If I be the centre of the inscribed circle of a triangle ABC and AI produced cut the circumscribed circle ABC in E, then the rectangle contained by AI, IE is equal to twice the rectangle con- tained by the radii of the circumscribed and the inscribed circles, (See Ex. 46, page 324.) 4. If Ij be the centre of the circle of the triangle ABC escribed beyond BC and AI^ cut the circumscribed circle ABC in E, then the rectangle contained by AI-^, I^E is equal to twice the rectangle con- tained by the radii of the circumscribed and the escribed circles. 5. ^^ is the base of a triangle ABC whose sides are segments of a line divided in extreme and mean ratio. CP the bisector of the angle C, and CQ the perpendicular from C on AB meet AB in P and Q. Prove that the square on CP is equal to twice the rectangle contained by PQ and AB. 416 BOOK VI. PROPOSITION 35. If a perpendicular he drawn from a vertex of a triangle to tJie opposite side, the rectangle contained hy the oilier sides of the triangle is equal to the rectangle contained hy the perpendicular and the diameter of the circle described about the triangle. Let AD be the perpendicular drawn from the vertex A of the given triangle ABC to the opposite side BG \ it is required to prove that the rectangle contained by AB, AG is equal to the rectangle contained by AD and the diameter of the circle described about ABG. Construction. Describe the circle ABC; (IV. Prop. 5.) draw the diameter AB and draw jEC. Proof. Because in the triangles BAD, EAC, the angle ABD is equal to the angle AEG, (III. Prop. 21.) and the angle ADB to the angle AGE; (III. Prop. 31.) therefore the triangles are equiangular to one another, (I. Prop. 32.) and BA is to EA as AD to AG; (Prop. 4.) therefore the rectangle BA, AG is equal to the rectangle EA, AD. (Prop. 16.) Wherefore, if a perpendicular (fee. PROPOSITION 35 A. 417 PROPOSITION 35 A. The ratio of twice the area of a triangle to the rectangle contained hy two of the sides is equal to the ratio of the thir-d side to the diameter of the circumscribed circle of tlie triangle. Let ABC be a triangle : it is required to prove that twice the area of ABC is to the rectangle contained by AC, BC as AB to the diameter of the circle described about ABC. Construction. Describe the circle ABC; draw the diameter AJE, draw AD perpendicular to BC and draw JSC. Proof. Because in the triangles BAD, EAC, the angle ABD is equal to the angle A EC, (III. Prop. 21.) and the angle ADB to the angle ACE ; (III. Prop. 31.) therefore the triangles are equiangular to one another, (I. Prop. 32.) and yli) is to ^(7 as ^^ to AE. (Prop. 4.) Therefore the rectangle AD, BC is to the rectangle A C, BC as AB to AE; (Prop. 1.) and the rectangle AD, BC is equal to twice the area of the triangle ABC ; therefore twice the area of the triangle ABC is to the rectangle AC, BCas AB to the diameter of the circle ABC. Wherefore, the ratio &,c. 418 BOOK VI. ADDITIONAL PKOPOSITION 1. If a straight line cut the three sides of a triangle produced if necessary, the ratio compounded of the ratios of the segments of the sides taken in order is equal to unity.* Let the sides BC, CA, AB ot the triangle ABC he cut by the straight line LMN in L, M, N respectively. Through C draw CZ parallel to LMN to meet ABN in Z. Because ZG, NML are parallel, AM is to MC as AN to NZ, and CL is to LB as ZN to NB ; (Prop. 2.) therefore the ratio compounded of the ratios AM to MC and CL to LB is equal to the ratio compounded of the ratios AN to NZ and ZN to NB, i.e. the ratio AN to NB ; (V. Def. 8.) therefore the ratio compounded of the ratios AM to MC, CL to LB and BN to NA is equal the ratio compounded of the ratios AN to NB and NB to AN, that is, to the ratio AN to AN, i.e. to unity. (V. Def. 2.) * This theorem is attribute 1 to Menelaus, a Greek Geometer, who lived in the latter part of the first century a.d. THEOREM OF MENELAUS. 419 Definition. A straight line drawn to cut a series of lines is often called a transversal. The straight line LMN in Additional Proposition 1 is a transversal of the triangle ABC. EXEECISES. 1. Points E, F are taken in the sides AC, AB of a triangle such that AE is twice EC and BF is twice FA ; FE produced cuts BG in D ; find the ratio BD to DC. 2. If the bisectors of the angles B, C of a triangle ABC meet the opposite sides in D and E, and if the straight line DE produced meet BC produced in F, then the external angle at A is bisected by^F. 3. BB is the perpendicular let fall from one end of the base upon the straight line bisecting the vertical angle BAC of a tri- angle. liBA be three times as long as AC, AD will be bisected at the point E, where it cuts the base. 4. If a side J?C of a triangle ABC be bisected by a straight line which meets the sides AB, AC, produced if necessary, in D and E respectively, then AE is io EC b.b AD to DB. 5. If one side of a given triangle be produced and the other shortened by equal quantities, the line joining the points of section will be divided by the base in the inverse ratio of the sides. 6. In the sides AB, AC oi o, triangle ABC two points D, E are taken such that /)D is equal to CE ; DE, BC are produced to meet at F: shew that AB ib to AC as EF to DF. 420 BOOK VI. The converse of the theorem on page 418 may be stated as follows : — ADDITIONAL PROPOSITION 2. If three points be taken on the sides of a triangle (either one on a side produced and the other two on sides, or else all three on sides produced), such that the ratio compounded of the ratios of the segments of the sides taken in order is equal to unity, the three points lie on a straight line. Let three points X, M, N be taken on the sides BG, CA, AB of a triangle ABC, either all on sides produced (fig. 2) or one L on a side produced, and the others M, N on sides (fig. 1) such that the ratio compounded of the ratios AM to MG, GL to LB and BN to NA is equal to unity. Draw LM and let it produced cut AB in P. Then the ratio compounded of the ratios AM to MG, GL to LB and BP to PA is equal to unity; (Add. Prop. 1.) and the ratio compounded of the ratios AM to MG, GL to LB and BN to NA is equal to unity; (Hypothesis.) therefore the ratio BP to PA is equal to the ratio BN to NA ; therefore BP is to BA as BN to BA ; (V. Prop. 10 or 11) therefore BP is equal to BN; (V. Prop. 3.) that is, P coincides with N, or, in other words, L, M, N are in a straight line. THEOREM OF MENELAUS, 421 EXEKCISES. 1. The inscribed circle of a triangle ABC touches the sides BC, CA, AB at D, E, F; EF, FD, DE produced meet BC, CA, AB in L, M, N : prove that L, M, N are coUinear. 2. An escribed circle of a triangle ABC touches the side BC at D and the sides AC, AB produced &t E, F; ED, FD produced cut AB, AC in K, H respectively; prove that FE, BC, KH meet in a point. 3. If AB, CD, EF be three parallel straight Unes, and AC, BD meet in N, CE, DF meet in L, and EA, FB meet in M, then L, M, N lie on a straight line. 4. If D, E, F be the points of contact with BC, CA, AB of the inscribed circle, or of any one of the escribed circles of the triangle ABC, the lines AD, BE, CF pass through a point. 5. If D be the point of contact of the inscribed circle of a triangle ABC with BC, and E, F the points of contact of escribed circles with CA produced and BA produced respectively, then AD, BE, CF meet in a point. 6. If one escribed circle of a triangle ABC touch AC in F &nd BA produced in G and another escribed circle touch AB in H and CA produced in K, then FH, KG produced cut BC produced in points equidistant from the middle point of BC. 422 BOOK VI. ADDITIONAL PROPOSITION 3. If three straight lines be drawn from the vertices of a triangle meet- ing in a point and cutting the opposite sides or the sides produced, the ratio compounded of the ratio of the segments of the sides taken in order is equal to unity* Let the straight lines A 0, £0, CO be drawn from the vertices of the triangle ABC meeting in 0, and cutting BC, CA, AB in D, E, F respectively. Through C draw HCG parallel to AB to meet BO, AO produced in H, G. Then because the triangles A OF, GOG are equiangular to one another, AF is to GC as FO to CO ; (Prop. 4.) and because the triangles FOB, COH are equiangular to one another, FB is to CH as FO to CO ; therefore AF is to GC as FB to CH, (V. Prop. 5.) and therefore AF is to FB as GC to CH. (V. Prop. 9.) And because the triangles CEH, AEB are equiangular to one another, CEi^ioAE as CH io AB-, and because the triangles BDA, CDG are equiangular to one another, BD i^ to CD BLBBAioCG', \ therefore the ratio compounded of the ratios AF to FB, CE to EA, and BD to DC is equal to the ratio compounded of the ratios GC to CH, CH to AB, and AB to CG, which is equal to unity. * This theorem was first published in the year 1678 by Giovanni Ceva, an Italian. CEVA'S THEOREM. ^23 In Additional Proposition 3 it has been proved that, if through the vertices A, B, C of a triangle three concurrent straight lines ADy BE, CF be drawn meeting the sides BC, CA, AB in D, E, F, the ratio of BD to DC is equal to the ratio compounded of the ratios J^i^'to FA and AE to EC. In Additional Proposition 1 it has been proved that, if the straight line FE be drawn and produced to meet BC produced in L, the ratio BL to LC is equal to the ratio compounded of the ratios BE to FA and AE to EC. Therefore BD is to DC as BL to LC, or, in other words, BDCL is a harmonic range. It is a remarkable fact that, although the theorem on page 418 had been known as early as the 1st century, the theorem on page 422, which seems a very natural complement to the other, should not have been discovered until the 17th century. EXERCISES. 1. Prove Ceva's Theorem for a triangle ABC and a point O, (1) when lies between AB produced and AC produced, (2) when O lies between BA produced and CA produced. 2. Prove Ceva's Theorem by using the result of Ex. 3, page 355. 3. Prove Ceva's Theorem by the use of Menelaus' Theorem, considering in the figure of Add. Prop. 3 COF a transversal of the triangle ABD and BOE a transversal of the triangle ADC. 4. D, E, F are the points in which the bisectors of the angles A,B, C of & triangle cut the opposite sides ; prove that, if BC be equal to half the sum of the sides AB, AC, then EF bisects AD. 424 BOOK VL The converse of the theorem on page 422 may be stated as follows ; — ADDITIONAL PKOPOSITION 4. If three straight lines he drawn through the vertices of a triangle cutting the opposite sides {either all three sides, or else one side and the other two sides produced) so that the ratio compounded of the ratios of the segments of the sides taken in order is equal to unity^ the three straight lines meet in a point. Let three straight lines AD, BE, CF be drawn from the vertices A, B, C of a triangle ABC to cut the opposite sides in D, E, F respectively, so that the ratio compounded of the ratios AF to FB, BD to DC and CE to EA is equal to unity. Let AD, CF meet in : draw BO and produce it to meet CA in P. Then because the ratio compounded of the ratios AF to FB, BD to DC and CP to PA is equal to unity, (Add. Prop. 3.) and also the ratio compounded of the ratios AF to FB, BD to DC and CE to EA is equal to unity; (Hypothesis.) therefore the ratio CP to PA is equal to the ratio CE to EA ; therefore CP is to CA as CE to CA ; (V. Prop. 10 or 11.) therefore CP is equal to CE ; (V. Prop. 3.) therefore P coincides with E, or, in other words, AD, BE, CF meet in a point. CENTROID OF WEIGHTS. 425 If in the sides BG, GA, AB of a triangle, points D, E, F be taken such that BD is to D (7 as w to m, GE is to EA as Z to n, and AF is to FB as m to I, where Z, m, n are any three integers, the straight lines AD, BE, GF meet in a point, say 0. (Add. Prop. 4.) It is proved by Add. Prop. 1 that the ratio of AO to OD is equal to the ratio compounded of the ratios AE to EG and GB to BD, that is, of the ratios n to I and m + 7i to n\ therefore ^0 is to OD as m-\-n to I. Similarly it appears that BO is to OE as n + l to ?« and that GO is to Oi^ as Z + m to n. It follows that, if we divide BG in D so that J5D is to DG as n to w, and then divide D^ in so that DO is to OA as Z to m + n, we arrive at the same point, as if we divide GA in E so that GE is to EA as Z to w, and then divide EB in so that jBO is to OS as m to w + Z, or as if we divide AB in F so that ^jP is to FB as m to I, and then divide EG in so that FO is to OC as n to Z + m. This point is called the centroid of weights I, in, n bX A, B, G respectively. It appears that the position of the centroid of three weights is independent of the order in which the weights are taken, or, in other words, the centroid of three weights is a unique point. It is not difl&cult to see that this proposition can be extended to any number of weights, so that we may state the proposition in the general form, the centroid of a number of given weights is a unique point. EXERCISE. 1. From the vertex A oi a. triangle ABG a straight line is drawn cutting BG in D, and the angles BDA, GDA are bisected by straight lines cutting AB, AG in F, E respectively: prove that AD, BE, GF intersect in a point. T. E. 28 426 BOOK VI. ADDITIONAL PBOPOSITION 5. The locus of a point, the ratio of whose distances from two given points is constant, is a circle*. Let A, B he two given points and P a point such that the ratio of AP to BP is equal to the given ratio. Draw PA, PB ; and draw PC, PD the internal and the external bisectors of the angle APB meeting AB in C and AB produced in D. G B Because PC, PB are the bisectors of the angle APB, therefore the ratios of ^C to GB and AD to BB are equal to the ratio of AP to PB ; and the ratio of AP to PB is constant ; therefore G and D are two fixed points. (Ex. 1, page 359.) And because PG, PD are the bisectors of the angle APB, the angle GPD is a right angle. (Ex.5, page 43. ) Therefore every point on the locus of P must lie on the circle upon the fixed line GD as diameter. Next we will prove that every point of the circle belongs to the locus. Let P be any point on the circle described on GD as diameter. Draw PA, PG ; and draw PE making the angle GPE equal to the angle GPA and meeting GD at E ; then AP is to PE aq AG to GE. (Prop. 3, Part 1.) Again, because GPD is a right angle, PD is the external bisector of the angle APE ; therefore AP is to PE as AD to DE. * This theorem is attributed to Apollonius of Perga, a Greek geometer, who lived in the latter part of the third century b.c. CIRCLE OF APOLLONIUS. 427 Therefore ^C is to CE as AB to BE ; (V. Prop. 5.) and ^C is to GB as AB to BE; therefore CE is to EB as CB to BB, and iJ coincides with B^ which is a fixed point. (Ex. 1, page 359.) Therefore AP is to PB in the fixed ratio AG io CB for every point P on the circle. We may state the result of this proposition thus: — If a circle he described upon the straight line joining two conjugate points of a harmonic range as a diameter, the ratio of the distances of a point on the circle from the other pair of conjugate points is constant. EXEKCISES. 1. Prove that, if A, B he two fixed points and P be a point such that PA is equal to m times PB, (1) if m vanish, the locus reduces to the point A ; (2) if m be equal to unity, the locus is the straight line which bisects AB &t right angles ; (3) if m be infinitely great, the locus reduces to the point B; (4) if m be greater than unity, the locus is a circle excluding A and including B ; (5) if m be less than unity, the locus is a circle including A and excluding B ; (6) if m be greater than unity, the greater the value of m, the less the circle; (7) if m be less than unity, the less the value of m, the less the circle ; (8) the loci for two different values of m do not intersect. 2. A and B are the centres of two circles. A straight line PQ parallel to AB meets the circles in P and Q : find the locus of the point of intersection of AP and BQ. 3. Find a point such that its distances from three given points may be in given ratios. 4. Prove that, if a map be laid flat on another map of the same district on a larger scale, there is one place in the district which is represented in the two maps by points which are superposed one on the other. 28—2 428 BOOK VI. ADDITIONAL PKOPOSITION 6. If the line hetioeen one pair of conjugates of a harmonic range be bisected^ the square on half the line is equal to the rectangle contained by the segments of the line hetioeen the other pair of conjugates made by the point of bisection. Let AGBB be a harmonic range, such that ^0 is to CB as AD to BB, so that A, B are one pair of conjugates and C, D the other. First, let be the middle point of CD, the line between one pair of conjugates. Describe the circle on CD as diameter. Take any point P on the circle, and draw PA, PC, PB, PO. Because the angle OPG is equal to the angle OGP, the sum of the angles OPB, BPG is equal to the sum of the angles GAP, GPA ; (I. Prop. 32.) and because AP is to PB as ^C to CB, (see page 427.) the angle BPG is equal to the angle GPA ; therefore the angle OPB is equal to the angle GAP; therefore OP touches the circle described about APB; (Converse of III. Prop. 32.) therefore the square on OP, which is equal to the square on OG, is equal to the rectangle OA, OB. (III. Prop. 36, Coroll.) HARMONIC RANGE. 429 Secondly, let C/ be the middle point of AB, the line between the second pair of conjugates. The rectangle OA, OB is equal to the difference between the squares on 00' and O'B, (II. Prop. 10.) and the rectangle OA, OB is equal to the square on 0C\ therefore the square on OG is equal to the difference between the squares on 00' and O'B. Therefore the square on O'B is equal to the difference between the squares on 00' and 00, which is equal to the rectangle O'G, O'D. (H. Prop. 10.) Note. All the circles, which are the loci of the point P for different values of the ratio AP to BP, have their centres in the line AB, and, since the rectangle O'G, O'D is equal to the square on the tangent from 0' to the circle GPD, the straight line which bisects AB at right angles is the radical axis of every pair of such circles. Such a series of circles is called Coaxial. The points A, B are called the limiting points of the series of circles. EXEECISES. 1. If two circles be described upon the straight lines joining the two pairs of conjugate points of a harmonic range as diameters, the circles cut orthogonally. 2. A common tangent to two given circles is divided harmonically by any circle which is coaxial with the given circles. 430 BOOK VI. ADDITIONAL PEOPOSITION 7. A chord of a circle is divided harmonically by any point on it and the polar of the point. Let PQ be a chord of the given circle CQPD : take A any point on PQ produced and draw the diameter ACD. Let B be the point such that A GBD is a harmonic range. Draw PB, BQ and draw BR at right angles to AB meeting PQ Because A CBD is a harmonic range and is the middle point of CD, the rectangle OA, OB is equal to the square on OC; (Add. Prop. 6.) therefore BR is the polar of A. (see page 259.) Because AP is to PB as AG to CB, and AQ is to QB as AG to GB ; (Add. Prop. 5.) therefore AP is to PB as AQ to QB ; and PB is to PA as QB to AQ ; (V. Def. 5 note.) therefore PB is to BQ a.s PA to AQ; (V. Prop. 9.) therefore BA is the external bisector of the angle PBQ ; (Prop. 3, Part 2.) therefore BR is the internal bisector, and PB is to RQ as PB to BQ, (Prop. 3, Part 1.) and therefore as PA to AQ; therefore AQRP is a harmonic range. HARMONIC PROPERTY OF POLAR. 431 It may be remarked that in the diagram on page 430 the point A is taken outside the circle. If the point were inside the circle, say R, its polar would intersect PQB in A. (Add. Prop, on page 262.) Hence the theorem is established generally. EXEKCISES. 1. Prove that, if AGBD be a harmonic range, and if be the middle point of CD, then ^C is to Ci? as ^0 to OC. 2. Establish the theorem of page 426 by proving that in the figure of page 430 the triangles ABQ, APO are similar and also that the triangles ABP, AQO are similar. 3. If any straight line PQR be drawn touching one given circle at Q and cutting another at P, R, the segments PQ, QR subtend equal or supplementary angles at either of the limiting points of the coaxial system to which the given circles belong. 4. If any straight line PQES be drawn cutting two given circles of a coaxial system in P, S and Q, R, the segments PQ, RS subtend equal or supplementary angles at either of the limiting points. 432 BOOK VI. ADDITIONAL PEOPOSITION 8. If a pencil he drawn from a point to the four points of a harmonic range and if a straight line he drawn through one of the points parallel to the ray which passes through the conjugate point, the part of the line intercepted hetween the rays through the other pair of points is hisected at the point. Let ABGD be a harmonic range, and be any point not in the straight line AT). Let OA, OB, OD be drawn*, and let FCH be drawn parallel to ^0, meeting OB, OD, produced if necessary, in F, H. Because the triangles OAB, FOB are equiangular to one another, OA is to FC as AB to BG; (Prop. 4.) and because the triangles OAD, HCD are equiangular to one another, OA is to HC as AD to DC. And because ABCD is a harmonic range, AB ia to BC &s AD to DC; therefore OA is to FC as OA to HG; (V. Prop. 5.) therefore FG is equal to HG. (V. Prop. 3.) Note. The converse of this proposition is true, viz. if the line FH be bisected at C, then ABGD is a harmonic range. EXEECISE. 1. Give a construction to find the fourth point of a harmonic range when three points are given. * The ray OG of the pencil {ABGD) is omitted in the figure, as it is not wanted in the proof. Similar omissions will be met with elsewhere. HARMONIC RANGES AND PENCILS. 433 ADDITIONAL PEOPOSITION 9. The points^ in which a harmonic pencil form a harmonic range. cut by any straight line^ Let {A BCD) be a pencil drawn through the points of the har- monic range ABCD : let abed be any other straight line cutting the rays OA, OB, OC, OD in a, b, c, d respectively. Through C, c draw GCF, gcf parallel to OA cutting the rays OB, OD, produced if necessary, in F, G and /, g. Because ABCD is a harmonic range, and GCF is parallel to OA, therefore FC is equal to CG. (Add. Prop. 8.) And because /c^r is parallel to FCG, therefore fc is equal to eg. (Ex. 1, page 369.) And because fcg is parallel to Oa and fc is equal to eg, abed is a harmonic range. (Note, p. 432.) EXEKCISES. 1. The pencil formed by joining the four angular points of a square to any point on the circumscribing circle of the square is a harmonic pencil. 2. Give a construction for drawing the fourth ray of a harmonic pencil, when three rays are given. 3. Draw a harmonic pencil of which the rays pass through the angular points of a rectangle, and one of which is given in direction. 4. CA, CB are two tangents to a circle; E is the foot of the perpendicular from B on AD the diameter through A ; prove that CD bisects BE. 434 BOOK VI. ADDITIONAL PEOPOSITION 10. If two harmonic ranges have two corresponding points, one in each range, coincident, the straight lines joining the other pairs of corresponding points pass through a point. Let AGED, Achd be two harmonic ranges, of which the point A is a common point. Draw Gc, Bb and let them, produced if necessary, meet in 0, and draw OD. Because O {AGBD) is a harmonic pencil, if Acb cut OD in d', then Acbd' is a harmonic range ; (Add. Prop. 9.) therefore Ac is to cb as Ad' to d'b ; but Acbd is a harmonic range ; therefore Ac is to cb as Ad to db; therefore Ad' is to d'b as Ad to db, (V. Prop. 5.) and d' coincides with d; (Ex. 1, page 359.) i.e. Gc, Bb, Dd meet in a point. EXEECISE. 1. Prove that the intersections of the pairs of straight lines (76, Be; Bd, Db; Dc, Gd in the above figure lie on a straight line which passes through A. HARMONIC RANGES AND PENCILS. 435 ADDITIONAL PKOPOSITION 11. If two harmonic pencils have two corresponding rays, one of each pencil, coincident, the intersections of the other three pairs of corresponding rays lie on a straight line. Let (ABCD), 0' (abed) be two harmonic pencils, of which OAaO' is a common ray. Let OB, O'b meet in Q, and OC, O'c in R ; draw QR and let it meet 00' in P, OD in S, and O'd in s. Then because O (ABCD) is a harmonic pencil, PQRS is a harmonic range; (Add. Prop. 9.) therefore PQ is to QR as PS to SR ; and because 0' (abed) is a harmonic pencil, PQRs is a harmonic range ; therefore PQ is to QR as Ps to sR ; therefore Ps is to sR as PS to SR, (V. Prop. 5.) and the points S, s coincide; (Ex. 1, page 359.) i.e. the intersections of OB, O'bi OC, O'c, and OD, O'd are collinear. EXEECISE. 1. Prove that the straight lines joining the intersections of the pairs of straight lines OB, O'c and OC, O'b; OC, O'd and OD, O'c; OD, O'b and OjB, O'd intersect in a point which lies on the line 00'. 436 BOOK VI. ADDITIONAL PKOPOSITION 12. If a pencil be drawn from a point to the four points of an anharmonic range and if a straight line be drawn through one of the points parallel to the ray which passes through a second point, the part of it intercepted between the rays through the other pair of points will be divided in a constant ratio at the first point. Let ABCD be an anharmonic range and be any point not in the straight line AB. Let OA, OB, OD be drawn and FCH be drawn parallel to ^O meeting OB, OD, produced if necessary, in F, H, Because the triangles OAB, FOB are equiangular to one another, OA is to FC a& AB to BG ; (Prop. 4.) and because the triangles OAD, HCD are equiangular to one another, OA is to HC as AD to DC ; therefore the ratio of the ratio OA to FC to the ratio OA to HC is equal to the ratio of the ratio AB to BC to the ratio AD to DC, which is constant ; i. e. the ratio HC to FC is constant and is equal to the ratio of the range ABCD. (Def. 10.) EXEKCISES. 1. If ABCD, ABCE be two like anharmonic ranges, then the points D, E coincide. 2. If the ratio of the range ABCD be equal to the ratio of the range ADGB, the range ABCD is harmonic. ANHARMONIC RANGES AND PENCILS. 437 ADDITIONAL PKOPOSITION 13. The points in which an anharmonic pencil is cut by any straight line form an anharmonic range of constant ratio. Let {ABCD) be a pencil drawn through the points of the an- harmonic range ABCD. Let ahcd be any other straight line cutting the rays OA, OB, OC, OD in a, b, c, d respectively. Through C, c draw GCH, gch parallel to OA cutting the rays OD, OB in G, H and g, h. Because ABCD is an anharmonic range, and GCH is parallel to OA, therefore the ratio of GC to CH is the ratio of the range ABCD ; (Add. Prop. 12.) and because abed is an anharmonic range and gch is parallel to Oa, therefore the ratio of ^c to ch is the ratio of the range abed ; (Add. Prop. 12.) and because gch is parallel to GCH, gc is to ch&B GC to CH; (Ex. 2, page 365.) therefore abed is an anharmonic range of ratio equal to that of the range ABCD. EXEKCISES. 1. Find a point on a given straight line such that lines drawn from it to three given points shall intercept on any parallel to the given line lengths having a given ratio. 2. Three points F, G, H are taken on the side BC of a triangle ABC : through G any line is drawn cutting AB and AC in L and 31 respectively; FL and HM intersect in K; prove that K lies on a fixed straight line passing through A. 438 BOOK VI. ADDITIONAL PEOPOSITION 14. If two like anharmonic ranges have two corresponding points^ one in each range, coincident, the straight lines joining the other pairs of corresponding points pass through a point. Let ACBD, Acbd, be two like anharmonic ranges, of which the point ^ is a common point. Draw Cc, Bb, and let them, produced if necessary, meet in ; and draw OD. Because 0{ACBD) is an anharmonic pencil, if Acb cut OD in d', then Acbd' is an anharmonic range of ratio equal to that of the pencil; (Add. Prop. 13.) and Acbd is a like anharmonic range; therefore the ratio of the ratio Ac to cb to the ratio Ad' to d'b is equal to the ratio of the ratio Ac to cb to the ratio Ad to db ; therefore the ratio Ad' to d'b is equal to the ratio Ad to db, and d' coincides with d, (Ex. 1, page 359.) i.e. Cc, Bb, Dd meet in a point. EXEECISE. 1. Prove that the intersections of the pairs of straight lines Cb, Be ; Bd, Db ; Dc, Cd in the above figure lie on a straight line through A, ANHARMONIC RANGES AND PENCILS. 439 ADDITIONAL PKOPOSITION 15. If two like anharmonic pencils have two corresponding rays, one in each pencil, coincident, the intersections of the other three pairs of corresponding rays lie on a straight line. Let (ABCD), 0' (abed) be two like anharmonic pencils, of which OAaO' is a common ray. Let OB, O'b meet in Q, and OC, O'c in R; draw QR and let it meet 00' in P, OD in S and O'd in s. Because PQRS is a transversal of the anharmonic pencil {ABCD), PQRS is a range of ratio equal to that of the pencil; (Add. Prop. 13) and because PQRs is a transversal of the anharmonic pencil 0' {abed), PQRs is a range of ratio equal to that of the pencil ; and because O {ABCD), 0' {abed) are like anharmonic pencils, (Hypothesis) therefore PQRS, PQRs are two like anharmonic ranges ; therefore the points S, s coincide; (Ex. 1, page 436) i.e. the intersections of OB, O'b; OC, O'c; and OD, O'd are coUinear. EXEKCISE. 1. Prove that the straight lines joining the intersections of the pairs of straight lines OB, O'c and OC, O'b; OC, O'd and OD, O'c; OD, O'b and OB, O'd intersect in a point which lies on the straight line 00'. 440 BOOK VI, ADDITIONAL PKOPOSITION 16. The anharmonic ratio of the pencil formed by joining four given points on a circle to any fifth point on the same circle is constant. Let A, B, G, B he four given points on a circle, and let be any fifth point on the circle, and let the pencil O (ABCD) be drawn. Take 0' any other point in the same arc AD as ; then the angles AO'B, BO'C, CO'D are equal to the angles AOB, BOG, COD respectively; and the pencil 0' {ABGD) is equal* to the pencil O (ABGD). Next take 0' any point in the arc AB. Then the angles BO'G, GO'D are equal to the angles BOG, GOD respectively, and the angle between O'B and AO' produced, say O'A^, is equal to the angle AOB, and the pencil O' (A-^BGD) is equal to the pencil (ABGD). Similarly it can be proved that the pencil is the same for all positions of O' on the circle. EXEECISE. 1. The locus of the vertex of a harmonic pencil, whose rays pass through the angular points of a square, is the circumscribed circle of the square. * In the sense that one pencil can be shifted so that its rays coincide with the rays of the other pencil. (I. Def. 21.) ANHARMONIG PROPERTIES OF A CIRCLE. 441 ADDITIONAL PEOPOSITION 17. The anhai-nwnic ratio of the range formed by the intersections of four given tangents to a circle by any fifth tangent to the same circle is constant. Let Aa, Bh, Cc, Dd be the tangents at four given points, A, B, C, D on a circle, and let them cut the tangent at any fifth point T on the circle in a, b, c, d. Find P the centre, and take any point on the circle. Draw PA, PB, PT, Pa, Pb. Because the angle APT is equal to twice the angle AOT, (in. Prop. 20) and also equal to twice the angle aPT, the angle ^OT is equal to the angle aPT. Similarly it can be proved that the angle BOT is equal to the angle bPT ; therefore the angle AOB is equal to the angle aPb. Similarly it can be proved that the angles BOG, COD are equal to the angles bPc, cPd ; therefore the pencil (ABCD) is equal to the pencil P (abed), and the pencil {ABCD) has a constant ratio; (Add. Prop. 16) therefore the pencil P (abed) has a constant ratio ; and therefore the range abed has a constant ratio. (Add. Prop. 13) EXEECISE. 1. If a straight line cut the four sides of a square in a harmonic range, it touches the inscribed circle of the square. T. E. 29 442 BOOK VI. ADDITIONAL PKOPOSITION 18. The anharmonic ratio of the pencil formed by joining four points on a circle to any fifth point on the circle is the same as the ratio of the rectangles contained by the chords that join the points. Let A, B, G, Dhe four given points on a circle. Draw AB, AC, AD, BC, CD. In AD take Ab equal to AB; draw Bb and let it be produced to meet the circle in 0; draw OA, OC, OD and let OG cut AD in c; draw be parallel to CD to meet OG in e, and draw Ae. Because be is parallel to CD, the angle bee is equal to the angle OGD, which is equal to the angle OAc ; (III. Prop. 21) therefore A, b, e, lie on a circle. Therefore the angle Aeb is equal to the angle AOB, which is equal to the angle A CB ; and the angle Abe is equal to the supplement of the angle AOe, which supplement is equal to the angle ABC ; therefore the triangles Abe, ABC are equiangular to one another; and ^6 is equal to AB ; (Constr.) therefore be is equal to BC. Now the anharmonic ratio of the pencil {ABCD) is equal to that of the range AbcD, (Add. Prop. 13) which is equal to the ratio of the ratio Ab to be to the ratio AD to Dc, that is, to the ratio compounded of the ratios Ab to be and Do to AD, which is the ratio of the rectangle Ab, cD to the rectangle AD, be. (See page 399.) THE SIX AN HARMONIC RATIOS. 443 And because he is parallel to CD, the triangles DcG, bee are equiangular to one another ; therefore cD is to be as CD to be ; (Prop. 4) therefore the ratio of the rectangle Ab, eD to the rectangle AD, be is equal to the ratio of the rectangle AB, CD to the rectangle AD, be. Therefore the anharmonic ratio of the pencil O (ABCD) is equal to the ratio of the rectangle AB, CD to the rectangle AD, BC. The anharmonic ratio of a range ABCD is defined to be (Definition 10) the ratio of the ratio AB to BC to the ratio AD to DC, which, by Definition 8 of Book V., is equal to the ratio compounded of the ratios AB to BC and DC to AD ; and this last ratio has, on page 399, been shewn to be equal to the ratio of the rectangle AB, CD to the rectangle AD, BC. Now it can be proved (Ex. 1, page 137) that, if ABCD be a range, the sum of the rectangles AB, CD and AD, BC ia equal to the rectangle AC, BD. If therefore any two of these three rectangles be given, the third is at once found. There are six ratios, of which one of these rectangles is the antecedent and another the consequent; if any one of these ratios be given, the other five ratios are at once found. Now in the definition the ratio of the rectangle AB, CD to the rectangle AD, BC is defined as the anharmonic ratio of the range ABCD. There is no reason against adopting any other of the six ratios as the ratio of the range, but it is important strictly to adhere throughout an investigation to one and the same ratio. EXEKCISES. 1. The anharmonic ratio of the range formed by the intersections of four given tangents to a circle with any fifth tangent is equal to the ratio of the rectangles contained by the chords which join the points of contact of the given tangents. 2. Two fixed points D, E are taken on the diameter AB of a circle, and P any point on the circumference; perpendiculars AM, BN are let fall on PD, PE ; prove that the ratio of the rectangle PM, PN to the rectangle AM, BN is constant. 29—2 444 BOOK VI. ADDITIONAL PEOPOSITION 19. If two tnangles he such tJiat the straight lines joining their vertices in pairs pass through a point, the intersections of pairs of correspond- ing sides lie on a straight line. Let ABC, abc be two given triangles such that the straight lines Aa, Bb, Cc meet in a point 0. Let the pairs of sides BG, be; CA, ca; AB, ah, produced if necessary, meet in D, E, F respectively, and let OBh cut AC, ac in H,h. Because EAHC, EaJic cut the same pencil 0{EAHG), EAHC, Eahc are like anharmonic ranges ; (Add. Prop. 13), therefore the pencils B (EAHC), h (Eahc) are like anharmonic pencils; and they have a common ray BHhb ; therefore the intersections of the pairs of rays BC, he ; BE, bE ; BA, ba lie on a straight line; (Add. Prop. 15); that is, D, E, F lie on a straight line. Note. The point is often called the pole of the triangles ABC, abc, and the straight Hne DEE the axis of the triangles. This theorem may then be enunciated thus, compolar triangles are coaxial. Compolar triangles are often said to be in perspective. TRIANGLES IN PERSPECTIVE. 445 ADDITIONAL PEOPOSITION 20. If two triangles he such that the intersections of their sides taken in pairs lie on a straight line, the straight lines joining pairs of corre- sponding vertices meet in a point.* Let ABC, abc be two given triangles such that the pairs of sides BC, be; CA, ca; AB, ab intersect in three points D, E, F lying on a straight line. Let the straight lines Aa, Bb, Cc be drawn and let Bb cut AC, ac, DEF in H, h, K respectively. Because the pencils B {EFKD), b (EFKD) have a common trans- versal EFKD, they are like anharmonic pencils ; and because EAHC is a transversal of the pencil B (EFKD), and Eahc is a transversal of the pencil b {EFKD), EAHC, Eahc are like anharmonic ranges ; (Add. Prop. 13) and they have a common point E ; therefore the lines Aa, Hh, Cc meet in a point, (Add. Prop. 14) that is, Aa, Bb, Cc meet in a point. Note. The above theorem may be enunciated thus, coaxial tri- angles are compolar. * The theorems of this and the preceding pages are attributed to Gerard Desargues (born at Lyons 1593, died 1662)x 446 BOOK VL ADDITIONAL PEOPOSITION 21. If a hexagon he inscribed in a circle, the intersections of -pairs of opposite sides lie on a straight line. * Let ABCDEF be a hexagon inscribed in a given circle. Let the pairs of sides AB, DE ; BG, EF -, CD, FA meet in L, iH, N respectively, and let AB, CD meet in P, and BC, DE in Q. ^ L M Because A, B, G,D, E, F are points on a circle, A (BCDF), E (BCDF) are like anharmonic pencils; (Add. Prop. 16) and because PCDN is a transversal of the pencil A (BCDF), and BCQM is a transversal of the pencil E (BCDF), PCDN, BCQM are like anharmonic ranges ; (Add. Prop. 13) and they have a common point C ; therefore the lines PB, DQ, NM meet in a point, (Add. Prop. 14) that is, AB, DE, NM meet in a point, or, in other words, the points L, M, N lie on a straight line. EXEECISES. 1. If the tangents to the circumscribed circle of a triangle ABG at A, B, C meet the sides BC, CA, AD, in L, M, N, then L, M, N lie on a straight line. 2. If ACE, BDF be two ranges, then the intersections of the pairs. of straight lines AB, DE ; BC, EF ; CD, FA are coUinear. * This theorem is true for any conic. It was discovered at the age of sixteen by Blaise Pascal (born at Clermont 1623, died at Paris, 1662). PASCAL'S AND BBIANCHON'S THEOREMS. 447 ADDITIONAL PEOPOSITION 22. If a hexagon be described about a circle, the straight lines joining opposite vertices pass through a point.* Let ABCDEF be a hexagon described about a given circle. Let AB meet CD, DE in L, M and EF meet BC, CD in N, P. Because the six sides of the hexagon are tangents to a circle, the points where the sides AB, EF are cut by the remaining sides of the hexagon form like anharmonic ranges, (Add. Prop. 17) that is, ABLM, FNPE are like anharmonic ranges ; therefore D {ABLM), C{FNPE) are like anharmonic pencils; and they have a common ray LGDP-, therefore the pairs of rays DA, GF; DB, CN; DM, CE intersect on a straight line, (Add. Prop. 15) that is, DA , CF meet in a point on the line BE, or, in other words, AD, BE, GF meet in a point. EXERCISE. 1. If the inscribed circle of a triangle ABG touch the sides BG, GA, AB in D, E, F, then AD, BE, GF meet in a point. * This theorem also is true for any conic. It was discovered by Charles Julien Brianchon (born at Sevres, 1785). 448 BOOK VI. SIMILAE FIGUKES. If two similar figures be placed so that their corresponding sides are parallel in pairs, the figures are then said to be similar and similarly situate. It appears from Proposition 32 that, if two triangles be similar and similarly situate, then the lines joining pairs of corresponding vertices meet in a point. It is easily proved that, if two polygons of any number of sides be similar and similarly situate, then the lines joining pairs of corresponding vertices meet in a point, and further that the ratio of the distances of this point from any pair of cor- responding points of the two figures is constant, and is equal to the ratio of a pair of corresponding sides of the two figures. Such a point is called in consequence a centre of similitude of the two figures. -- It is also easily seen that, if a point be a centre of similitude of two figures, it is also a centre of similitude of any two figures similarly described with reference to the first pair of similar figures ; for instance, if a point be a centre of similitude of two triangles, it is also a centre of similitude of the circumscribed circles of the triangles, and also a centre of similitude of the inscribed circles and so on. If a pair of corresponding points lie on the same side of the centre of similitude, it is called a centre of direct similitude: if a pair of corresponding points lie on opposite sides of the centre of similitude, it is called a centre of inverse similitude. It must be noticed that the centre of direct similitude is at an infinite distance in the case when the two similar figures are equal. As an example we will prove an important theorem with reference to the centres of similitude of the circumscribed circle and the Nine Point circle of a triangle : The orthocentre and the centroid of a triangle are the centres of direct and inverse similitude of the circumscribed circle and the Nine Point circle of the triangle. CENTRES OF SIMILITUDE. 449 If ABC be a triangle, and D, E, Fhe the middle points of its sides, the triangles ABC, DEF are equiangular to one another, and the ratio of AB to DE is equal to the ratio 2 to 1. (Add Prop., page 101.) Also the radii of the circles ABC, DEF are in the ratio of 2 to 1. Because the lines AD, BE, CF meet in a point G, and the ratios ^G to GD, BG to GE, CG to GF are each equal to 2 to 1, (Add Prop., page 103.) A G is the centre of inverse similitude of the two circles ABC, DEF. Again, if P be the orthocentre of the triangle ABC, and a, 6, c be the middle points of PA, PB, PC, the ratios PA to Pa, PB to Pb, PC to Pc are each equal to 2 to 1. Therefore P is the centre of direct similitude of the triangles ABC, abc, and therefore of the circles ABC, abc. Now the circles abc, DEF are identical. (Add. Prop., page 271.) Therefore P is the centre of direct simihtude of the circles ABC, DEF. Now, if 0, 0' be the centres of the circles ABC, DEF, since the radii of the circles are in the ratio 2 to 1, their centres of similitude lie in the straight line 00', and divide the distance internally and externally in the ratio 2 to 1. That is, P, 0', G, lie on ar straight line, and PO is equal to twice PO', and GO is equal to twice GO'. 450 BOOK VI. ADDITIONAL PEOPOSITION 23. Every straight line which passes through the extremities of two parallel radii of tivo fixed circles passes through a centre of similitude of the circles. Let AP, BQ be two parallel radii of two given circles, whose centres are A, B. Draw ABy PQ and let them, produced if necessary, meet at S. (2) Because the triangles APS, BQS are equiangular to one another, AS is to BS as AP to BQ ; (Prop. 4) that is, S is a point, which divides the distance AB externally (fig. 1) or internally (fig. 2) in the ratio of the radii; therefore ^S is a fixed point. (Ex. 1, page 359.) Again, because the triangles APS, BQS are equiangular to one another, SP is to SQ as AP to BQ ; (Prop. 4) that is, the ratio of the distances SP to SQ is a constant ratio ; therefore S ib & centre of similitude of the circles. In figure 1, where the two radii AP, BQ are drawn in the same sense, Sis in the line of centres AB produced, and the two distances SP, SQ are drawn in the same direction. Sis a. centre of direct similitude. In figure 2, where AP, BQ are drawn in opposite senses, S is in the line of centres AB, and the two distances SP, SQ are drawn in opposite directions. Sis a. centre of inverse similitude. CENTRES OF SIMILITUDE. 451 ADDITIONAL PROPOSITION 24. If a circle he drawn to touch two given circles, the straight line which passes through the points of contact passes through one of the centres of similitude of the given circles. Let a circle be drawn to touch two given circles, whose centres are A, B, at P, Q. Draw -4P, BQ and let them, produced if necessary, meet at *. Draw PQ and let it, produced if necessary, meet the circles again atP', Q': draw^Q'. Because the circles touch at P, the centre of the circle which is described to touch the given circles must lie in ^P ; (III. Prop. 10, Coroll.) similarly it must lie in PQ ; therefore is the centre ; therefore the angle OPQ is equal to the angle OQP, which is equal to the angle BQQ' and therefore to the angle BQ'Q\ therefore APO, BQ' are parallel ; (I. Prop. 28) therefore PQ passes through a centre of simiUtude. (Add. Prop. 23.) * If ^P, PQ be parallel, so that the point is infinitely distant, and the radius of the circle which touches the given circles at P, Q is infinitely large, PQ becomes one of the common tangents of the given circles, which common tangents pass through S. 452 BOOK VL ADDITIONAL PKOPOSITION 25. If two straight lines he drawn through a centre of similitude of two given circles to cut the circles, a pair of chords of the two circles join- ing pairs of inverse* points intersect on the radical axis of the given circles, and a pair of chords of the two circles joining pairs of cor- responding points are parallel. Let >S be a centre of similitude of two given circles PQqp, P'Q'q'p' and let SPQP'Q', Spqp'q' be any two straight lines drawn through S cutting the circles ; P, Q' ; Q, P' ; p, q' ; q, p' being pairs of inverse points and P, P'\ Q, Q' ;p, p'\ q, q' being pairs of correspond- ing points. Draw Pq, Q'p', and let them, produced if necessary, meet at R. Because S ia a. centre of similitude, Sp is to 8p' as SQ to SQ'; therefore Qp, Q'p' are parallel ; (Prop. 2, Part 2) therefore the angles PQp, PQ'p' are equal j (I. Prop. 29) and the angles PQp, Pqp are equal; (III. Prop. 21) therefore the angles PQ'p'-, Pqp are equal. Therefore the four points P, q, p', Q' lie on a circle, (III. Prop. 22, Coroll.) and the rectangle RP, Rq is equal to the rectangle RQ', Rp'; (III. Prop. 36) therefore the squares on the tangents drawn from R to the two circles are equal, and consequently R lies on the radical axis of the given circles. (Add. Prop, page 264.) * For this name see page 460. CENTRES OF SIMILITUDE. 453 ADDITIONAL PEOPOSITION 26. IJ a straight line he drawn through a centre of similitude of two given circles, the rectangle contained by the distances of two inverse points is constant. Let /S be a centre of similitude of two given circles, whose centres are A, B. Draw MNS a common tangent through S (see note on page 451), and let SQ'QP'P be any straight line through S cutting the circles in P, P' and Q, Q'. Draw AP, BQ, AM, BN. Because the rectangle SP, SP' is equal to the square on SM, SP is to SM as SM to SP'. (Prop. 17, Part 2.) And because S ia a, centre of similitude, SP is to SQ as SM to SN; and therefore SP is to SM a.a SQ to SN; (V. Prop. 9) therefore SQ is to SN as SM to SP'; (V. Prop. 5) therefore the rectangle SQ, SP' is equal to the rectangle SM, SN. (Prop. 16, Part 1.) Similarly it can be proved that the rectangle SP, SQ' is equal to the rectangle SM, SN. EXEBCISE. 1. Prove that in the above figure the rectangle PQ, P'Q' is equal to the square on MN, 454 BOOK VI. ADDITIONAL PEOPOSITION 27. The six centres of similitude of three given circles taken in pairs lie three by three on four straight lines *. Let A, B, C he the centres of three given circles; let a, b, c be the radii of the A, B^ G circles, and let D, D' be the centres of direct and of inverse similitude of the^, G circles, E, E' those of the G, A circles, and F, F' those of the ^, J5 circles. Because BD is to DO as b to c, and GE is to EA as e to a, and AF is to FB as a to b, (Add. Prop. 23) therefore the ratio compounded of the ratios BB to DG, GE io EA and AF to FB, is equal to the ratio compounded of the ratios 6 to c, c to a and a to b, that is to unity ; therefore DEF is a straight line. (Add. Prop. 2.) Similarly it can be proved that DE'F', D'EF', D'E'F are straight lines, and also that the lines of each of the sets AD', BE', GF'; AD', BE, GF ; AD, BE', GF; AD, BE, GF' meet in a point. (Add. Prop. 4.) * These lines are called the axes of similitude of the three circles. CIRCLES TOUCHING TWO CIRCLES. 455 ADDITIONAL PEOPOSITION 28. If a point on the radical axis of two given circles he joined to the points of contact of a circle, which touches both the given circles, by straight lines which cut the circles again, another circle can be de- scribed to touch the given circles at the points of section. Let be a point on the radical axis of the given circles A, B; and let a circle touch them at P, Q. Draw OP, OQ and let them meet the A, B circles in p, q. Draw PT, QS, pt, qs the tangents to the circles at P, Q, jp, q and draw 8 Because is on the radical axis of the circles A, B, the rectangle OP, Op is equal to the rectangle OQ, Oq; therefore P, Q, q, p lie on a circle. (Ex. 1, page 253.) The difference of the angles tpO, sqO is equal to the difference of the angles TPO, SQO, which is equal to the difference of the angles PQO, QPO, which is equal to the difference of the angles qpO, pqO ; (in. Prop. 22) therefore the angle tpq is equal to the angle sqp. It is therefore possible to describe a circle touching the circles A, B atp, q. CoROLLABY. If the radical centre of three given circles he joined to the points of contact of a circle, which touches all the three given circles, by straight lines which cut the circles again, another circle can be described to touch the given circles at the points of section. (See Ex. 140, page 284.) 456 BOOK VL ADDITIONAL PEOPOSITION 29. If two circles he drawn to touch three given circles, so that the straight line joining the tioo points of contact on each of the given circles passes through the radical centre of the given circles, the radical axis of the pair of circles is one of the axes of similitude of the three given circles. Let PQR, pqr be a pair of circles touching three given circles A, By G &t P, p; Q, q; R, r, so that Pp, Qq, Rr pass through the radical centre of the circles A,B,C. (Add. Prop. 28, Coroll.)' Draw PQi pq and let them, produced if necessary, meet in F. Because the circles PQR, pqr touch the circles A, B, the lines PQ, pq pass through one of the centres of similitude of the circles A and B ; (Add. Prop. 24) therefore the rectangle FP, FQ is equal to the rectangle Fp, Fq ; (Add. Prop. 26.) therefore i^ is a point on the radical axis of the circles PQR, pqr. Similarly it can be proved that a centre of similitude of each of the pairs of circles B, C and C, A lies on the radical axis of PQR, pqr. Therefore the radical axis of the circles PQR, pqr is an axis of similitude of the circles A, B, G. CIRCLES TOUCHING THREE CIRCLES. 457 ADDITIONAL PKOPOSITION 30. If a pair of chords of two given circles intersect in the radical axis, and if the chords intersect the polars of one of the centres of similitude in two points collinear loith the centre of similitude, then the points of intersection of the chords ivith the circles lie two by two on two straight lines through the centre of similitude. Let S be one of the centres of similitude of two given circles A, J5; let L, M be two points collinear with S, on the polars of S with respect to the circles A, B\ and let be a point on the radical axis. Draw OL, and let it intersect the circle A in P, p. Draw SP, Sp, and let them meet the circle A in P', p', and let the corresponding points to P, P', p, p\ where the lines meet the circle B be Q', Q, q\ q. Because L is on the polar of S, the line P^ passes through L. (Add. Prop, page 261.) Because S is the centre of similitude, and the intersections of Pp, P'p\ and of Q'q', Qq are corresponding points, and Pp, P'p' inter- sect at L, and L, 31 are corresponding points, therefore Qq, Q'q' intersect at M; and because S is the centre of similitude, Pp, Qq intersect on the radical axis, (Add. Prop. 25) that is, Qq must pass through ; and it also passes through M. Therefore the points Q, q obtained by the above construction are the points where the straight line OM cuts the circle B, that is, if OL cut the circle ^ in P, ^ and OM cut the circle B in Q,q, then PQ, pq pass through S. T. E. 30 458 BOOK VI. ADDITIONAL PBOPOSITION 31. To describe a circle to touch three given circles.* Let A, B, G be three given circles. Find the radical centre of the circles A, B, C, and draw DEF one of their axes of similitude. Find the poles L, M, N of the line DEF with respect to the circles A,B,C respectively. Draw OL, OM, ON and let them, produced if necessary, cut the circles A, B, C respectively in P, ^; Q, q; R, r. Because L, M are the poles of DEF with respect to the circles A, J5, therefore L, 31 are on the polars of F with respect to the circles A,Bi and because F is a centre of similitude of the two circles, therefore LMF is a straight line. Therefore the line PQ passes through F. (Add. Prop. 30.) * This solution of the problem is due to Joseph Diez Gergonne (born at Nancy 1771, died at Montpellier 1859). CIRCLES TOUCHING THREE CIRCLES. 459 Therefore a circle can be described through P, Q to touch the given circles at P, Q. (Add. Prop. 24.) Similarly it can be proved that circles can be described through Q, P, and through P, P to touch at each pair of points. Therefore these three circles are identical (see Ex. 140, page 284), that is, the circle PQR touches the given circles at P, Q, P. Similarly the circle pqr touches the given circles at p, q, r. Since there are four axes of similitude, and since two circles can be obtained by the foregoing construction from each axis of similitude, there are 2 x 4 or 8 circles which can be described to touch three given circles. It is readily seen that, if three circles touch a fourth circle, there are two distinct possible types of configuration of the three circles relatively to the circle which they touch : (1) the three circles may lie on the same side of the fourth circle, (2) two of the three circles may lie on one side and the third on the other side of the fourth circle. Since any one of the three given circles may be the one which lies by itself on the one side of the fourth circle, the type (2) may be sub- divided into three different groups. It can be proved without much difficulty that of the eight circles which can be described to touch three given circles A, B, C, there are four pairs of circles, such that ^, P, lie on the same side of each circle of one pair ; A lies on one side, and P, C on the other, of each circle of a second pair ; P lies on one side, and C, A on the other, of each circle of a third pair ; and C lies on one side, and -4, P on the other, of each circle of a fourth pair. Each of these four pairs of circles is obtained by the above con- struction from one of the axes of similitude of the given circles. 30—2 400 BOOK VI. INVEKSION. 1. We will now proceed to give an account of a geometrical Method called Inversion, and we will do so without stating the theorems which we are about to establish in the formal way in which theorems have been stated heretofore. We will further depart from our former method by making use of the arithmetical method of representing geometrical magnitudes (see page 135), so that we shall not be debarred from using fractions to represent the ratios of geometrical quantities, and if necessary we shall use the signs ordinarily used in Algebra to signify addition, subtraction, and the other elementary operations. 2. Definition. If be a fixed point and P any other point, and if on the straight line OP (produced if necessary) we take a point P' such that OP .OP' = a\ where a is a constant, then each of the points P, P' is called the inverse of the other with respect to the circle whose centre is and radius a. The straight line OP is often called the radius vector of the point P. The point is called the pole of inversion and a the radius of inversion. If the point P trace out a curve, the curve which is the locus of P' is called the inverse of the curve which is the locus of P. INVERSION. 461 3. If we consider the points P', P", which are the inverses of P with respect to the same pole and different radii of inversion a, b, since OP. OP'=a? and OP . OP" = h^, therefore OP' : OP" = a? : 62. It follows that since the points P', P" lie on the same radius vector, and OP' : OP" is a constant ratio, the loci of P', P" are two similar curves of which is a centre of similitude. 4. If P, P', and Q, Q! be two pairs of inverse points, then since OP . OP'=a^ and OQ . OQ' = a\ therefore OP . 0F= OQ . OQ' ; it follows that P, Q, Q', P' lie on a circle, and the angle OPQ is equal to the angle OQ'P'. Again, if the line OQQ' approach nearer and nearer to the posi- tion of OPP', the lines QP, Q'F in the limit are the tangents to the curves which are the loci of P and P' at P and P' respectively. (See page 217.) We may state this result in the form : Two inverse curves at two inverse points cut the radium vector through the pole of inversion at the same angle on opposite sides. It follows that any two curves cut at the same angle as their inverse curves at the inverse point. (See Definition on page 266.) 5. From the definition of inversion it follows at once that a straight line through the pole inverts into itself. Three points A, B, C in the same radius vector, whose distances OA, OB, 00 are such that 0A + 0C=20B, invert into three points A', B'y C\ whose distances OA', OB', OC are such that 1 _1___2^ OA''^ 0C~ OB'' 462 BOOK VL Three magnitudes such as OA, OB, OG are said to be in Arith- metical Progression, and three magnitudes such as 0A\ 0B\ OC are said to be in Harmonical Progression. Because _L+^^^A^, therefore OB' . 00' +0 A' . OB' = 20 A' : OC ; whence OA' .B'C' = OC' . A'B', andOA' :A'B' = OC' : G'B'; therefore OA'B'C is a harmonic range. Also three points A, B, C in the same radius vector, whose distances OA, OB, OC are such that OA . OC=OB^, invert into three points A', B', C, whose distances are such that OA' . OC'=OB'^. Three magnitudes such as OA, OB, OG are said to be in Geometrical Progression. It may be noticed that three magnitudes a, b, c are in Arithmetical, Geometrical, or Harmonical Progression according as a — b : b — c = a : a, a-b : b-c = a : b, or a-b :b-c = a: c respectively. 6. Let P be a point on a fixed straight line ; draw OA perpendicular to the line, and on OA, produced if necessary, take the point A' such that OA.OA' = a'^. Let P' be the inverse of P, so that OP . OP' = a^ : then OP.OP'=.OA.OA'; therefore P, A, A', P' lie on a circle, and the angle 0P'^'= the angle O^P = a right angle; therefore the locus of P' is a circle whose diameter is OA'. Hence the theorem : A straight line which does not pass through the pole of inversion inverts into a circle which passes through the pole and touches there a parallel to the given line. INVERSION. 463 7. Again let P be any point on a circle, of which OA is a diameter. Take in OA, produced if necessary, the point A' such that OA.OA'=a?. Let F be the inverse of P, so that OP. OP'=a^; then OP.OP'=OA.OA'; therefore P, A, A\ P', lie on a circle, and the angle P'A'0=the angle OPA=a, right angle, i.e. the locus of P' is a straight line through A' the inverse of A drawn at right angles to OA'. Hence the theorem : A circle which passes through the pole of inversion inverts into a straight line parallel to the tangent at the pole. 8. Again, let P be any point on a circle, which does not pass through the pole, and P' be the inverse point so that OP . OP' = a^ : let OP, pro- duced if necessary, cut the circle in Q ; find A the centre and draw QA , and let OA, produced if necessary, cut P'B drawn parallel to QA in B. Because OP . OQ=t% where t represents the length of the tangent from to the given circle, and OP . OP' = a\ therefore OP' : OQ = a^: t'^=OB : OA=BP' : AQ ; therefore the locus of P' is a circle such that B is its centre and is one of the centres of similitude of it and the given circle. Hence the theorem : A circle, which does not pass through the pole of inversion, inverts into a circle which does not pass through the pole, and is sv^h that the pole is a centre of similitude of the two circles. 464 BOOK VI. 9. Again, if P, P' and Q, Q' be two pairs of inverse points, OP. OP'=OQ.OQ' = a\ the triangles OPQ, OQ'P' are similar. P'Q'^OQ: ^ OQ'.OQ _ a^ PQ~OP~OP. OQ ~ OP.OQ' Therefore P'Q:=a OP.OQ 'OP.OQ' Thus the distance between two points in a figure is expressed in terms of the distance between their inverse points and the distances of the inverse points from the pole. 10. Now let us take the ordinary definition of a circle, i.e. the locus of a point P such that its distance PC from a fixed point G is constant. Let F be the inverse point of P ; take C the inverse point of C. If we please we may in this investigation choose the radius of inversion (see page 461) equal to the tangent drawn from to the original circle ; then P' lies on the same circle as P, and OG.OO'=OT^--^OP. OP'. Therefore the triangles OP'G', OOP are similar, and OP':P'C'=OG:CP. Hence the theorem (which has been already proved otherwise. Add. Prop. 5) : The locus of a point the distances of which from two fixed points are in a constant ratio is a circle. It will be seen that the straight line TT\ which is the polar of O, and the circle on 00 as diameter are inverses of each other. INVERSION. 465 11. Again, it is known from Book III. that all straight lines which cut a circle at right angles pass through the centre, and also that all straight lines through the centre cut the circle at right angles. If therefore we invert a figure consisting of a series of straight lines cutting a series of concentric circles (centre C) at right angles, we shall obtain a series of coaxial circles passing through the pole of inversion and through C the inverse point of G, and cutting each of a second series of circles at right angles. (Add. Prop, page 268.) Hence the theorem : A system of concentric circles inverts into a system of coaxial circles. In the above diagram is the pole of inversion, and the tangent from to the circle PQR (or P'Q'R') is taken as the radius of inver- sion so that the circle inverts into itself. The points 0, A, B in the left-hand figure are supposed to coincide with 0, B', ^i' respectively in the right-hand figure : the figures are drawn apart merely for the sake of clearness. Further, if we invert the last system with respect to any point, we shall get a system of exactly the same nature ; viz. a system of circles passing through two fixed points and cutting each of another system of circles at right angles. Hence the theorem : A system of coaxial circles inverts into another system of coaxial circles, and the limiting points into the limiting points. 466 BOOK VI. 12. A circle can be inverted into itself with respect to any point as pole of inversion, if the tangent to the circle from the point be taken as the radius of inversion. Any two circles can be inverted into themselves with respect to any point on their radical axis as the pole of inversion. Any three circles can be inverted into themselves with respect to their radical centre as the pole of inversion. Hence we conclude that when one circle is drawn to touch two given circles at P and Q, if be any point on the radical axis of the given circles, the lines OP, OQ will cut the circles again in two points P', Q\ such that another circle can be described to touch the given circles at P', Q'. We also conclude that when one circle is drawn to touch three given circles at P, Q, B, if be the radical centre, the lines OP, OQ, OB cut the circles again in three points P', Q\ B' such that a circle described through them wiU touch the circles at P', Q\ B'. These two theorems have been proved before (page 455). 13. If two circles PQO, pqO be described to touch two given circles A, B &t P, p; Q, q, and to touch each other at 0, then the centre of similitude F, which is the point of intersection of the straight lines PQ, pq, must lie on the common tangent to the circles at 0. (Add. Prop. 29.) Hence, if be taken as the pole of inversion, the two circles OPQ, Opq will invert into two parallel common tangents to the inverse circles (page 423), and therefore the given circles will invert into a pair of equal circles. And since FO^=FP .FQ=FM . FN, if FMN be a common tangent to the given circles, (Add. Prop. 26) we see that the pole may be chosen anywhere on the circle, whose centre is F and whose radius is a mean proportional between the tangents from the point F to the circles A, B. Hence with any point on two definite circles as pole of inversion we can invert two given circles into two equal circles. It follows that with any one of certain definite points as pole of inversion, three given circles can he inverted into three equal circles. INVERSION. 467 14. If we take A\ B', C three points in order on a straight line, the relation A'B' + B'G'=A'C' exists between the segments. If we invert with respect to any pole 0, the three inverse points A, B, C will lie on a circle through O, and the chords will satisfy the relation AB BC AG , .^.^ oaTob + obTog = oaTog ' ^''' P^^" ^^^)' or AB.OG + BG . OA = AG . OB, which is Ptolemy's Theorem. (III. Prop. 37 B.) Again if we take A\ B\ G', D', four points in order on a straight line, the relation A'B' . G'D' + A'D' . B'G'=A'G' . B'D' exists between the segments. (Ex. 1, page 137.) If we invert with respect to any pole 0, the four inverse points A, B, G, D will he on a circle through 0, and the chords will satisfy the relation, AB GD AD BG _ AG BD OA . OB ' OG . OD'^ OA . OD ' OB . 0G~ OA . OG ' OB . OD * or AB.GD + AD.BG=AC.BD, which also is a form of Ptolemy's Theorem. im BOOK VI. PEAUCELLIEE'S CELL. 15. Before we leave the subject of inversion, it will be as well to explain the nature of a simple piece of mechanism, by the use of which the inverse of any given curve can be drawn. The figure represents such an instrument ; OA, OB are two equal rods and AP, PB, BP\ P'A, four other equal rods, all six being freely hinged together at the points 0, A, B, P, P\ This instrument is generally called a Peaucellier's Cell*. Because APBP' is a rhombus, its diagonals bisect each other at right angles at M; (Ex. 1, page 39)- and because OAB is an isosceles triangle, and M is the middle point of AB, OM is at right angles to AB. Therefore OPP' is a straight Hne, and the rectangle OP . OP'=OAP-MP^ (II. Prop. 6) = OA^-AP^ = a constant. If therefore the point be fixed, and P be made to trace out any curve, P' will trace out the inverse curve. * This mechanical invention is due to A. Peaucellier, Capitaine du G^nie (a Nice), who proposed the design of such an instrument as a question for solution in the Nouvelles Annales 1864 (p. 414). His solution was published in the Nouvelles Annales 1873 (pp. 71 — 8). CASETS THEOREM. 469 We will now proceed to prove a theorem which establishes a relation between six of the common tangents of paii-s of four circles which touch a fifth given circle, of a kind similar to that which Ptolemy's Theorem establishes between the chords joining the points of contact. Let a circle whose centre is touch two given circles whose centres are ^, B at P, Q ; let PQ, produced if necessary, cut the circles A, B again in P', Q' and pass through their centre of similitude F ; and let pqF be one of their common tangents through F. (Add. Prop. 24 note.) Because P is a centre of similitude of the circles, pP' is parallel to qQ, and pP to qQ' ', (Add. Prop. 25) therefore ^2 : FQ =Fq : FQ, &ndLpq :PQ' = Fq : FQ'; therefore pq^ : P'Q . PQ' = Fq'^ : FQ . FQ' ; and PgS^PQ . FQ' ; therefore pq^= P'Q . PQ'. (Ex. 1, page 453.) Again because P is a centre of similitude, OPA is parallel to Q'B, and OQB to P'A ; therefore OA : OP = P'Q : PQ, and OB : OQ=PQ' : PQ; therefore OA . OB : OP . OQ = PQ' . P'Q : PQ^, or OA .OB : OP^=pq^ : PQ^; therefore, if OA . OB=OL\ then OL : OP=pq : PQ. (V. Prop. 16.) 470 BOOK VI. Similarly, if two other circles, whose centres are C, D, touch the same circle, centre 0, at R, S, and rs be their common tangent, OC . OD : OP^~ = rs^: RS^ ; therefore, if OC . OD = OM^, then OM :OP=rs :BS; (V. Prop. 16) therefore OL . OM : OP^=pq .rs: PQ. RS, If we denote the common tangent to the two circles which touch the fifth circle at P, Q by {PQ), and so on for the other pairs of circles, we may write this proportion, OL . OM: OP^={PQ) . (RS) : PQ . RS. It can be proved in a similar manner that, if OL"^=OA . OC and OM'^=OB . OD, then OL' . OM' : OP^={PR) . (QS) : PR . QS. Now 0L2 : OL'^=OA . OB : OA . OC=OB : OC, and OM'^ : OM^=OB . OD : OC . OD = OB : OC ■ therefore OL^ : OL'^=OM'^ : 0M% and 0L'.0L'=0M' '.OM; (V. Prop. 16) therefore OL . 0M=0L' . OM'. (Prop. 16.) Hence we have (PQ) . (RS) : PQ . RS={PR) . {QS) : PR . QS and similarly = {PS) . {QR) : PS . QR. And because, if PQRS be a convex quadrilateral, PR . QS=PQ . RS + PS . QR (Ptolemy's theorem), therefore (PR) . {QS) = {PQ) . {RS) + {PS) . {QR}. This theorem is generally known as Casey's Theorem*. It may be observed that the common tangent of each pair of circles, which appears in the equation, is that tangent which passes through the same centre of similitude of the circles as the chord joining the points of contact of the circles with the fifth circle. * This theorem was discovered by John Casey (born at Kilbenny, County Cork, 1820, died at Dublin 1890). CASETS THEOREM. 471 It is readily seen that, if four circles touch a fifth circle, there are three distinct possible types of configuration of the four circles relatively to the circle which they touch ; (1) the four circles may lie on the same side of the fifth circle, (2) three of the four circles may lie on one side and the fourth circle on the other side, (3) two of the four circles may lie on one side and the other two on the other side. The converse of Casey's Theorem may be stated in the following manner : if an equation of the form of the equation in Casey's Theorem exist between the common tangents of four circles taken in pairs, the common tangents being chosen in accordance with one of the three possible types of configuration, then the four circles touch a fifth circle. The truth of this converse theorem which is often assumed without any attempt at proof can be proved, but the proof of it is thought to be beyond the scope of this work. 472 BOOK ri. CONTINUITY. Let us consider a variable point P on a given straight line, on which A, B are two fixed points. It is seen at once that, (1) if P be outside AB beyond A , then the excess of PB over PA is equal to ^7? ; (2) if P be in AB, then the sum of AP and PB is equal to AB, and (3) if P be outside AB beyond B, then the excess of ^P overPP is equal to AB. (1) (2) (3) We can write these results in the forms (1) PA+AB = PB, (2) AB=AP + PB, (3) AB + BP=AP. Here we observe that, while P changes from one side of A to the other, the distance PA, which vanishes when P coincides with A, changes sides in the equation, which otherwise remains unchanged; and again that while P changes from one side of B to the other, the distance PB similarly changes sides in the equation. A geometrical theorem consists, in many cases, of a proof that a certain equation exists between a number of geometrical magnitudes, such equation remaining unchanged in form for variations in the geometrical magnitudes involved, consistent with the conditions to which they are subject. CONTINUITY. 473 It is found in many of such theorems, as in the illustration which we have just given, that, if subject to continuous variation of some chosen geometrical magnitude some other magnitude con- tinuously diminish and vanish, then in the equation which applies to the configuration determined by the next succeeding values of the chosen variable magnitude, the magnitude which has vanished appears on the opposite side of the equation. This fact is due to the absence of any sudden changes in the magnitudes under considera- tion. The general law that no sudden change occurs is often spoken of as THE PRINCIPLE OF CONTINUITY. Let as consider a variable point P on a given straight line, on which ^ is a fixed point ; and let us consider any equation between variable geometrical magnitudes, one of which is PA the distance between P and A. The principle of continuity leads us to expect that, if P in the variation of its position pass from one side to the other of A, the sign of PA in the equation will change. In other words, we may consider the equation to remain unchanged in form, if we resolve to represent by the expression PA not only the distance between P and A, but also the fact that the distance is measured from P towards A. This result is at once obtained by resolving that -PA shall represent a distance equal to PA and measured in the opposite direction; in other words, that AP= -PA. Let us return to the consideration of a variable point P on a given straight line, on which A, 6 are two fixed points. It appears that the equation which exists between the distances between the points takes different forms according as P is (1) in BA produced, (2) in AB, or (3) in AB produced. If we allow the use of the minus sign, we may write these equations, (1) AB + PA-PB = 0, (2) AB-AP-PB = 0, (3) AB-AP + BP=0, where each symbol such &s AB represents merely the length of a line measured in the same direction as AB. It is at once seen that, if we adopt the convention that PQ = -QP, all these equations are the same ; each may be written AB = AP+PB, or AB + BP + PA=:0. T. E. 31 474 BOOK VL The first form expresses that the operation of passing from ^ to £ is the same as passing from ^ to P and then from P to J5 ; the second form expresses that the aggregate result of the operations of passing from Aio B^ and then from 5 to P and then from P to A is to arrive at the point A of starting ; both of which facts are true for all combinations of three points A,B^ P ons. straight line. The results of the theorems contained in Propositions 5 and 6 of Book II. become the same, if we take into account the fact that the distance BD is measured in opposite directions in the two figures : and similarly the results of Propositions 12 and 13 of Book II. become the same, if we take into account the sign of CD. As a further illustration of the Principle of Continuity we will take Ptolemy's Theorem. Let us consider a variable point P on a circle, on which A, B, G are three fixed points. It is proved in III. Prop. 37 B, that (1) if P be in the arc ^P, AB.PC=BC.PA + CA.PB; (2) if P be in the arc BC, BG.PA = CA,PB + AB.PC; and (3) if P be in the arc CA, GA . PB=AB . PG + BG . PA. These equations may be written (1) AB.PG-BG.PA-GA,PB = 0, (2) AB.PG-BG.PA + GA.PB = 0, (3) AB. PG + BG. PA -GA.PB = 0. Hence, while P passes along the arc from one side of B to the other, the sign of PB, which vanishes when P coincides with P, changes sign in the equation, which otherwise remains unchanged, and so on for passage through C or ^. PORISMS. 475 POEISMATIC PEOBLEMS. In some cases, when a Geometrical problem is submitted for solution, it is found that some relation between the geometrical magnitudes or figures which are given is necessary in order that a solution may be possible, and that, if one solution be possible, and therefore the relation exist, the number of possible solutions is infinite. The solution is then said to be indeterminate. Such a problem is called a porism. "We might take as an illustration of such a problem the problem, to construct a triangle which shall be inscribed in one and described about another of two given concentric circles. It is easily seen that, if one such triangle exist, the radius of the first circle must be equal to twice the radius of the second, and that then every equilateral triangle which is inscribed in the first circle is also described about the second circle. Again we might take the problem, to construct a triangle so that each vertex shall lie on one of three given concentric circles, and that each side shall touch a fourth given concentric circle. It is easily proved by means of the method of rotation (page 186) that, if such a triangle be possible, an infinite number of such triangles are possible. We will now proceed to find the relation which must exist between two circles which are not concentric, in order that it may be possible to construct a triangle which shall be inscribed in one and described about the other, and we shall prove that, if it be possible to construct one such triangle, it is possible to construct an infinite number. 31—2 476 BOOK VI. The square on the straight line joining the centres of the circum- scribed circle and the inscribed circle of a triangle is less than the square on the radius of the circumscribed circle by twice the rectangle contained by the radii of the two circles. Let ABC be the circumscribed circle of the triangle ABC and DEF the inscribed circle touching AB at F. Find 0, 1 the centres of the circles ABC, DEF. Draw AI, IB, BO, IF and let AI, BO produced meet the circle ABC at H, K. Draw BH, HK. Because the angle FAI (or BAH) is equal to the angle HKB, and the angle AFI is equal to the angle KHB, the triangles FAI, HKB are equiangular to one another; therefore the rectangle AI, BH is equal to the rectangle KB, IF. (III. Prop. 37 A.) Because the angle BIH is equal to the sum of the angles BAI, ABI, which is equal to the sum of the angles HBC, CBI, that is, to the angle IBH ; therefore BH is equal to IH ; therefore the rectangle AI, BH is equal to the rectangle AI, IH, which is equal to the difference of the squares on OB, 01; (III. Prop. 35) therefore the difference of the squares on OB, 01 is equal to the rect- angle KB, IF, that is, to twice the rectangle contained by the radii. PORISM OF TWO CIRCLES. ill If two circles he such that one triangle can he constructed that is inscribed in one circle and circumscrihed ahout the other circle, an infinite number of such triangles can he constructed. Let ABC be a triangle: and let its circumscribed and inscribed circles be described, and let I be the centre of the inscribed circle. Take any other point A' on the cir- cumscribed circle ABC; draw AI, A' I and produce them to cut the circle ABC oX H,H'. With centre H' and radius H'l draw a circle cutting the circle ABC in B\ C. Draw A'B\ B'C, C'A'. Because H'B' is equal to H'l, the angle H'lB' is equal to the angle H'B' I; therefore the sum of the angles IB' A', lA'B' is equal to the sum of the angles H'B'C, C'B'I. (I. Prop. 32.) And because the chords H'B', H'C are equal, the angle lA'B' is equal to the angle H'B'C; therefore the angles IB' A', C'B'I are equal, that is, B'l is the bisector of the angle A' B'C. And A'l is the bisector of the angle B'A'C ; therefore I is the centre of the inscribed circle of the triangle A'B'C. Because AIH, A'lH' are two chords of the circle ABC, the rectangle AI, III is equal to the rectangle A'l, IH'\ and the rectangle AI, IH is equal to twice the rectangle contained by the radius of the circle ABC and the radius of the inscribed circle of the triangle ABC, (page 476) and the rectangle A'l, IH' is equal to twice the rectangle contained by the radius of the circle ABC and the radius of the inscribed circle of the triangle A'B'C. (page 476.) Therefore the radii of the inscribed circles of the triangles ABC, A'B'C are equal: and the circles have a common centre I. Therefore the triangle A'B'C has the same circumscribed circle and the same inscribed circle as the triangle ABC. 478 BOOK VL We now proceed to prove a theorem which might fairly have been included at an earlier stage. Let PAB be any triangle and ^ be a point in AB, such that mAE=nEB. Here E is the centroid of weights m and ns^i A and B. (See page 425. ) Draw PM perpendicular to -4ij and draw PE. Because PA^=PE^ + EA^ - 2ME . AE ; (IL Prop. 13) and PB^ = PE^ + EB^ + 2ME . EB, (II. Prop. 12) and because mAE = nEBy and therefore mME . AE = nME .EB; therefore mPA^ + nPB^ ={m + n) PE^ + mEA 2 + nEB^. Hence the theorem : The sum of any multiples of the squares on the distances of any point from two given points is equal to the sum of the same multiples of the squares on the distances of the given points from the centroid of weights at the given points proportional to those multiples, together with the sum of the multiples of the square on the distance of the point from the centroid. Because mAE=nEB, therefore mAE^ = nAE . EB, and mAE^ + nEB^ =nAE .EB + nEB^ = nAB.EB. We may therefore write the result of this theorem in the form mPA^ + nPJ52 = (m + n) PE^ + nBE . BA or nPB^ - nBE . BA = (m + n) PE^ - mPA^ COAXIAL CIRCLES. 479 Next, let us assume P a point such that the tangents P'l\ PS drawn from it to two given circles, whose centres are A, B, satisfy the equation mPT=nPS. Take two points 0, 0' in AB, such that AO . AO'=AT^ and BO . BO' = BS^'. (Add. Prop, page 268.) Because viPT=nPS, mWr- = n''PS^; therefore m" {PA^-A T^) = n^ {PB"^ - BS^) , or m^{PA'^-AO.AO') = n'^{PB^-BO.BO'). Now by the last theorem, if qAO=pOO' and rOO' = qO'B, q{PB^-BO.BO') = {r + q)PO'^-rPO^, and q {PA^ - AO . AO') = {p + q) PO"^ -pPO'\ Therefore n2 {q + r) PO'^ - n^rPO^=m^ {p + q) PO^ - m^pPO'^, or {m2 {p + q) + nV} P0^= {n^ {q + r)+ m'^p} PO'^. Therefore PO is to PO' in a constant ratio, and therefore the locus of P is one of the system of coaxial circles, of which 0, 0' are the limiting points. (See Note on page 429.) In consequence of the very algebraical character of the proof which has just been given, we will give another proof of the same theorem depending in a great measure on the theory of similitude. 480 BOOK VI. Let us again assume P is a point such that the tangents PT^ PS drawn to two given circles, whose centres are A, B, satisfy the equation mPT=nPS. There must be some point Q in AB, such that, if QH, QK be the tangents to the circles, 'mQH = nQK\ therefore m^PT^ = 'nPPS^, and m^QH^=n^QK^; therefore m^{PT^- QH 2) = n^ (P^^ - QK^) , or m^ (P^2 -AQ^)= r? {PB^ - BQ^). Now draw two circles, with centres A and B, and radii AQ, BQ, and let PQ cut these circles in L, M respectively; then PA^-AQ^ = PQ.PL, and PP2 - QB^ = PQ . PM; (III. Prop. 36) therefore rn^PQ . PL = n^PQ . PM, or mPPL = n^PM. Therefore m^ {PQ + QL) = n^ {PQ - QM), or {n^-vi^)PQ = m^QL + n^QM. Therefore {n^ - m^) PQ : QM= m^QL + n^QM : QM = mPQA+n^QB : QB, since QL : QM= QA : QB. From this it follows that the ratio QP : Q3I is constant ; therefore the locus of P is a circle which passes through Q, and has its centre in the line AB. (See page 450.) Hence the theorem : the locus of a point, such that the tangents drawn from it to two given circles are in a constant ratio, is a circle. COAXIAL CIRCLES. 481 Next, let P be a point on the locus such that FT, Ft, the tangents drawn to two given circles, are in a given ratio. Draw Ss a common tangent to the two given circles. Let Ss cut the radical axis of the given circles in M, and the circle which is the locus of P in ^ and R. Because both Q and R are points on the locus of P, QS is to Qs as RS to Rs ; therefore SQsR is a harmonic range. And because M is the middle point of Ss, MS^ = Ms^ = MQ.MR, and MQ . MR is equal to the square on the tangent from M to the locus of P. It follows therefore that M is a point on the radical axis of each of the pairs of circles; and since their centres are collinear, the circles have a common radical axis. Hence the theorem : The circle which is the locus of a point, such that the tangents draion from it to tioo given circles are in a constant ratio, belongs to the same coaxial system as the given circles. 482 BOOK VL We will now prove the theorem : If two opposite sides of a quadrilateral, which is inscribed in a circle, touch another circle, the other sides of the quadrilateral touch a third circle coaxial with the other two. Let ABCB be a quadrilateral inscribed in a given circle ; and let AB, CD touch another given circle at P, Q. Draw PQ and let it be produced to meet AD, BCin It, S. Because the angles BPS, DQR are equal and the angles PBS, QDR are equal, (III. Prop. 21) therefore the triangles BPS, DQR are equiangular to one another; therefore BP is to BS as DQ to DR. (Prop. 4.) Similarly it can be proved that the triangles APR, GQS are equi- angular to one another and that AP is to AR as CQ to CS. Again, because the angles BSP, DRQ are equal, a circle can be described to touch BG, DA at S, R. Next, because in the two triangles ARP, BSP, the angles APR, BPS are equal, and the angles ARP, BSP are supplementary; therefore AP is to AR as BP to BS. (Prop. 5 A.) Therefore the ratios of the tangents drawn from the four points A, B, G, D to the two circles PQ, RS are equal. Therefore the four points A, B, G, D aM lie on a circle coaxial with the two circles PQ, RS ; (page 481) that is, the circle ABGD is coaxial with the circles PQ, RS, or, in other words, the circle RS is coaxial with the circles ABGD and PQ. FONCELET'S THEOREMS. 483 Next, let ABC, A'B'C be two triangles inscribed in a circle, such that AB, A'B' touch a second circle and J5C, B'C a third circle belonging to the same coaxial system. Because ABB' A' is a quadrilateral inscribed in the circle ABC, and AB, B'A' touch another circle, therefore AA', BB' touch a circle of the same coaxial system ; and because BCG'B' is a quadrilateral inscribed in the circle ABC, and BG, C'B' touch another circle of the system ; therefore BB\ CC touch a circle of the system. Therefore AA', BB', CC touch the same circle of the system. And because AA'C'C is a quadrilateral inscribed in the circle ABC, and AA', CC touch another circle of the system, therefore AC, A'C touch a circle of the system; that is, A'C always touches that circle of the system which retouches, or, in other words, A'C touches a definite circle of the system. Hence the theorem : If two sides of a triangle inscribed in a given circle touch given circles of the same coaxial system as the first circle, the third side touches a fourth fixed circle of the system.* Corollary. If all the sides but one of a polygon inscribed in a given circle touch given circles of the same coaxial system as the first circle, the reviaining side touches another fixed circle of the system.* * These theorems are due to Jean Victor Poncelet (born at Metz 1788, died at Paris 18G7). 484 BOOK VI. MISCELLANEOUS EXERCISES. 1. Two triangles ABC, BCD have the side BC common, the angles at B equal, and the angles ACB, BDC right angles. Shew that the triangle ABC is to the triangle BCD as AB to BD. 2. The rectangle contained by two straight lines is a mean pro- portional between the squares described upon them. 3. Any polygons whatsoever described about a circle are to one another as their perimeters. 4. The sum of the perpendiculars drawn from any point within an equilateral triangle on the three sides is invariable. 5. In a parallelogram E, F, G, H are the middle points of the Bides AB, BC, CD, DA ; if AF, AG, CE, CH be drawn, the parallelo- gram formed by them is one-third of the parallelogram ABCD. 6. ABC is a triangle, D any point in AB produced ; E a point in BC, such that CE is to EB as AD to BD. Prove that DE produced bisects A C. 7. ABC, ABD are triangles on the same base, and CD meets the base in E ; then CE is to DE as the triangle ABC to the triangle ABD. 8. Triangles of unequal altitudes are to each other in the ratio compounded of the ratios of their altitudes and their bases. 9. If triangles ABC, AEF have a common angle A, the triangle ABC is to the triangle AEF as the rectangle AB, AC to the rectangle AE, AF. 10. O is the centre of the circle inscribed in a triangle ABC, and BO, CO meet the opposite sides in D, E respectively. Prove that the triangles BOE, COD are to one another in the ratio of the rectangles AE,AB; AD, AC. 11. If in the sides of a triangle BC, CA, AB, points D, U, F be taken such that BD is twice DC, CE twice EA, and AF twice FB, and AD, BE, CF intersect in pairs in P, Q, R, then the areas of the triangles PQR, ABC are in the ratio of 1 to 7. 12. A point and a straight line being given, to draw a line parallel to the given line such that all the lines drawn through the point may be cut by the parallels in a given ratio. 13. If from a point in the base BC of a triangle OM and ON he drawn parallel to the sides AB and AC respectively, then the area of the triangle AMN is a mean proportional between the areas of BNO and CMO. 14. If P, Q be two points within a parallelogram ABCD, and if PA, QB meet in R, and PD, QC meet in S, and if PQ be parallel to AB^ then RS is parallel to AD. MISCELLANEOUS EXERCISES. 485 15. A point P is taken on the bisector of the angle BAG of the triangle ABC between A and the base; prove that, if ^C be greater than AB, the ratio PC to PB is greater than the ratio ^C to CB. 16. On a circle of which AB is a diameter take any point P. Draw PC, PD on opposite sides of AP, and equally inclined to it, meeting AB at C and D : prove that AC is to BC as AD is to BD. 17. Apply VI. 3 to solve the problem of the trisection of a finite straight line. 18. AB is a diameter of a circle, CD is a chord at right angles to it, and E is any point in CD ; AE and BE are drawn and produced to cut the circle at F and G : shew that the quadrilateral GFDG has any two of its adjacent sides in the same ratio as the remaining two. 19. ABCD is a quadrilateral having the opposite sides AD, BG parallel; ^ is a point in AB, and the straight lines EC, ED are drawn; AF is drawn parallel to EC meeting CD in F; shew that BF is parallel to ED. 20. If a straight line AD be drawn bisecting the angle BAG of the triangle BAG, and meeting BC in D, and FDC be drawn perpen- dicular to AD, to meet AB and AG produced if necessary, in F and E respectively, and EG he drawn parallel to BC, meeting AB in G, then BG is equal to BF. 21. The side AB oi the triangle ABC is produced to G , and the angle CBG bisected by BE meeting AC in E: the angle EBG is bisected by BL and the exterior angle of the triangle EBG got by producing EB is bisected by BF. Shew that, if BL be parallel to ^ C and BF meet AG in F, CE is a mean proportional between GA and GF. 22. Each acute angle of a right-angled triangle and its cor- responding exterior angle are bisected by straight lines meeting the opposite sides ; prove that the rectangle contained by the portions of those sides intercepted between the bisecting lines is four times the square on the hypotenuse. 23. A and B are fixed points, and AP, BQ are parallel chords of a variable circle, which passes through the fixed points. If the ratio of AP to BQ be constant, then the loci of P and Q are each a pair of circles, and the sum of the radii of two of these circles, and the difference of the radii of the other two, are independent of the magni- tude of the ratio. 24. If two chords AB, AC, drawn from a point A in the circum- ference of the circle ABC, be produced to meet the tangent at the other extremity of the diameter through A in D, E respectively, then the triangle AED is similar to the triangle ABC. 25. AB is the diameter of a circle, E the middle point of the radius OB ; on AE, EB as diameters circles are described ; PQL is a common tangent meeting the circles at P and Q, and AB produced at L : shew that BL is equal to the radius of the smaller circle. 4.SQ BOOK VL 26. From B the right angle of a right-angled triangle ABC, Bp is let fall perpendicular to AC ; from p, pq is let fall perpendicular to BA ; from q, qr is let fall perpendicular to AC, and so on; prove that Bp+pq + &c. : AB = AB + AC : BC. 27. An isosceles triangle is described on a side of a square and the vertex joined with the opposite angles : the middle segment of the side has to either of the outside segments double of the ratio of the altitude of the triangle to its base. 28. A straight line BE is drawn parallel to the side BC oi a, triangle ABC. Q is a point in BC such that the rectangle BC, CQ is equal to the square on DE, and CR is taken equal to DE in BC pro- duced. Prove that AR is parallel to DQ. 29. ABC, DEF are triangles, having the angle A equal to the angle D ; and AB is equal to DF : shew that the areas of the triangles are as ^C to DE. 30. CA, CB are diameters of two circles which touch each other externally at C; a chord AD of the former circle, when produced, touches the latter at E, while a chord BF of the latter, when pro- duced, touches the former at G: shew that the rectangle contained by AD and BF is four times that contained by DE and FG. 31. If AA' B'B, BB' C'C, GC A' A be three circles, and the straight lines AA', BB', CC cut the circle A'B'C again in a, /3, y re- spectively, the triangle a^-y will be similar to the triangle ABC. 32. On the two sides of a right-angled triangle squares are described : shew that the straight lines joining the acute angles of the triangle and the opposite angles of the squares cut off equal segments from the sides, and that each of these equal segments is a mean proportional between the remaining segments. 33. ABA'B' is a rectangle inscribed in a circle and AB is twice A'B ; ^C is a chord equal to AB and meeting B'A' in F and BA' in E ; prove that AF:AE = CF : CA. 34. If BD, CD are perpendicular to the sides AB, AC ot a triangle ABC and CE is drawn perpendicular to AD, meeting AB in E, then the triangles ABC, ACE are similar. 35. Describe a circle touching the side BC of the triangle ABC and the other two sides produced; and shew that the distance between the points of contact oi BC with this circle and the inscribed circle is equal to the difference between AB and AC. 36. A straight line AB is divided into any two parts at C, and on the whole straight line and on the two parts of it equilateral triangles ADB, ACE, BCF are described, the two latter being on the same side of the straight line, and the former on the opposite side; G, H, K are the centres of the circles inscribed in these triangles : shew that the angles AGH, BGK are respectively equal to the angles ADC, BDCt and that GHK is an equilateral triangle. MISCELLANEOUS EXERCISES 487 37. Two circles, centres A and B, touch one another at G. A straight line is drawn cutting one circle in P and Q and the other circle in R and S. Prove that the ratio of the rectangle PR, PS to the square on CP is constant. 38. AB is the diameter of a circle, E the middle point of the radius OB ; on AE, EB as diameters circles are described. PQL is a common tangent, meeting the circles in P and Q, and AB produced in L. Shew that BL equals the radius of the lesser circle. 39. If through the vertex, and the extremities of the base of a triangle, two circles be described, intersecting one another in the base, or the base produced, their diameters are proportional to the sides of the triangle. 40. D is the middle point of the base BC of an isosceles triangle, OF perpendicular to AB, DE perpendicular to CF, EG parallel to the base meets AD in G; prove that EG is to GA in the triplicate ratio ofBD to DA. 41. Two straight lines and a point between them are given in position; draw two straight lines from the given point to terminate in the given straight lines, so that they shall contain a given angle and have a given ratio. 42. Four lines, AB, CD, EF, GH, drawn in any directions, inter- sect in the same point P ; then if from any point m in one of these lines, another be drawn parallel to the next in order, cutting the re- maining two in p and q ; the ratio mp :pq is the same in whichever line the point m is taken. 43. If P, Q be the points of intersection of a variable circle drawn through two given points A, B with a fixed circle, prove that the ratio AP . AQ : BP . BQ ia constant. 44. A quadrilateral is divided into four triangles by its diagonals; shew that if two of these triangles are equal, the remaining two are either equal or similar. 45. ABCD is a quadrilateral inscribed in a circle, E, F, G are the points of intersection of AB and CD, AC and BD, AD and BC re- spectively. K is the foot of the perpendicular let fall from F on EG. Prove that KA : KB=zFA : FB. 46. Divide a given finite straight line similarly to a given divided straight line parallel to the first line. 47. If two parallel straight lines AB, CD be divided proportionally at P, Q, so that AP is to PB as CQ to QD, then the straight lines AC, PQ, BD meet in a point. 48. BA C, DAE are similar equal triangles, BAD and CAE being straight lines; and the parallelograms of which BC and DE are diagonals are completed. Prove that the lines drawn to complete the parallelograms themselves form a parallelogram whose diagonal passes through A. 488 BOOK VI. 49. If, in similar triangles, any two equal angles be joined to the opposite sides by straight lines making equal angles with homologous sides; these lines shall have the same ratio as the sides on which they fall, and shall divide those sides proportionally. 50. APB, CQD are parallel straight lines, and ^P is to PB as DQ to QC, prove that the straight lines PQ, AD, BG meet in a point. 51. Describe a circle which shall pass through a given point and touch two given straight lines. 52. AD the bisector of the base of the triangle ABC is bisected in E, BE cuts ^C in i^, prove that AF :FC '.:1:2. 53. A straight line drawn through the middle point of a side of a triangle divides the other sides, one internally, the other externally in the same ratio. 54. In the triangle ABC there are drawn AD bisecting BC, and EF parallel to J5C and cutting AB, AC in E, F. Shew that BF and CE intersect in AD. 55. In the triangle ACB, having C a right angle, AD bisecting the angle A meets CB in D, prove that the square on ^C is to the square on AD as £C to 2BD. 56. 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. 57. A, B, C are three fixed points in a straight line ; any straight line is drawn through C; shew that the perpendiculars on it from A and B are in a constant ratio. 58. 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. 59. Through a given point draw a straight line, so that the parts of it intercepted between that point and perpendiculars drawn to the straight line from two other given points may have a given ratio. 60. A tangent to a circle at the point A intersects two parallel tangents at B, C, the points of contact of which with the circle are D, E respectively ; and BE, CD intersect at F : shew that AF is parallel to the tangents BD, CE. 61. P and Q 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 Q parallel to P3I meeting CD at N : shew that the ratio of PM to QN is constant, and thence shew that the straight hne through 31 and N passes through a fixed point. 62. If two circles touch each other, and also touch a given straight line, the part of the straight line between the points of con- tact is a mean proportional between the diameters of the circles. MISCELLANEOUS EXERCISES. 489 63. If at a given point two circles intersect, and their centres lie on two fixed straight lines which pass through that point, shew that whatever be the magnitude of the circles their common tangents will always meet in one of two fixed straight lines which pass through the given point. 64. From the angular points of a parallelogram ABCD perpen- diculars are drawn on the diagonals meeting them at E, F, G, H respectively: shew that EFGH is a parallelogram similar to ABCD. 65. ABODE is a regular pentagon, and AD, BE intersect at O : shew that a side of the pentagon is a mean proportional between AO and AD. 66. ACB is a triangle, and the side ^C is produced to D so that CD is equal to AC, and BD is joined: if any straight line drawn parallel io AB cuts the sides AC, CB, and from the points of section straight lines be drawn parallel to DB, shew that these straight lines will meet AB at points equidistant from its extremities. 67. If a circle be described touching externally two given circles, the straight line passing through the points of contact will intersect the straight line passing through the centres of the given circles at a fixed point. 68. A and B are two points on the circumference of a circle of which C is the centre ; draw tangents at A and B meeting at T ; and from A draw AN perpendicular to CB : shew that BT is to BC as BN is to NA. 69. Find a point the perpendiculars from which on the sides of a given triangle shall be in a given ratio. 70. A quadrilateral ABCD is inscribed in a circle and its dia- gonals AC, BD meet at 0. Points P, Q are taken in AB, CD such that AP is to PB &a AO to OB, and CQ is to QD as CO to OD ; prove that a circle can be described to touch AB, CD at P, Q. 71. Prove that the diagonals of the complements of parallelo- grams about a diagonal of a parallelogram meet in the diagonal of the parallelogram. 72. Through a point G of the side CD of a parallelogram ABCD are drawn AG and BG meeting the sides in E and F; and GH is drawn parallel to EF, meeting AF in H ; prove that FH is equal to^D. 73. Any regular polygon inscribed in a circle is a mean propor- tional between the inscribed and circumscribed regular polygons of half the number of sides. 74. If two sides of a parallelogram inscribed in a quadrilateral be parallel to one of the diagonals of the quadrilateral, then the other sides of the parallelogram are parallel to the other diagonal. 75. 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 other two together. T. E. 32 490 BOOK VI. 76. ABC is a triangle. At ^ a straight line JD is drawn making the angle CAD equal to CBA, and at C the straight line CD is drawn making the angle ACD equal to BAC. Shew that AD is a fourth proportional to AB, BG and CA. 77. If bd be drawn cutting the sides AB, AD and the diagonal AG of the parallelogram ABCD in b, d, and c respectively, so that Ab is equal to Ad, then the sum of AB, AD is to the sum of Ab, Ad as AG to twice Ac. 78. Having given the base of a triangle and the opposite angle, find that triangle for which the rectangle contained by the perpen- diculars from the ends of the base on the opposite sides is greater than for any other. 79. Through each angular point of a quadrilateral a straight line is drawn perpendicular to the diagonal which does not pass through that point, shew that the parallelogram thus formed is similar to the parallelogram formed by joining the middle points of the sides of the given quadrilateral. 80. ABCD is a quadrilateral inscribed in a circle; BA, CD pro- duced meet in P, and AD, BG produced in Q. Prove that PC is to PB as QA to QB. 81. Through D, any point in the base of a given triangle ABC, straight lines DE, DF are drawn parallel to the sides AB, AC and meeting the sides in E, F and EF is drawn ; shew that the triangle AEF is a mean proportional between the triangles FBD, EDG. 82. If through the vertex and the extremities of the base of a triangle two circles be described intersecting each other in the base or the base produced, their diameters are proportional to the sides of the triangle. 83. Draw a straight line such that the perpendiculars let fall from any point in it on two given straight lines may be in a given ratio. 84. In any right-angled triangle, one side is to the other, as the excess of the hypotenuse above the second, to the line cut off from the first between the right angle and the line bisecting the opposite angle. 85. AB is a fixed straight line, C a moving point, and CD a line parallel to AB ; a variable line PQR is drawn cutting AC in P, BG in Q and CDin R', prove that if the ratios AP to PC, and BQ to QC he constant, GR is of constant length. 86. If I, Ii be the centres of the inscribed circle of a triangle ABC and of the circle escribed beyond BG, the rectangle AI, AI^ is equal to the rectangle AB, AG. 87. If I, Ij be the centres of the inscribed circle of a triangle ABC and of the circle escribed beyond BG, and Z>, E be the points of contact of those circles with AB, then ID is to DB as EB to EI^. MISCELLANEOUS EXERCISES. 491 88. If Ij, I2 ^6 *h® centres of the circles of a triangle ABC escribed beyond BG, CA respectively and E, F be their points of contact with AB, then I^E is to EB as BF to I^F. 89. is a fixed point in a given straight line OA, and a circle of given radius moves so as always to be touched by OA ; a tangent OP is drawn from O to the circle, and in OP produced PQ is taken a third proportional to OP and the radius : shew that as the circle moves along OA, the point Q will move in a straight line. 90. On AB, AC, two adjacent sides of a rectangle, two similar triangles are constructed, and perpendiculars are drawn to AB, AC from the angles which they subtend, intersecting at the point P. If AB, AC he homologous sides, shew that P is in all cases in one of the diagonals of the rectangle. 91. If at any two points A, B ; AC, BD be drawn at right angles to AB on the same side of it, so that AB is & mean proportional between AG and BD; the circles described on AC, BD as diameters will touch each other. 92. One circle touches another internally at A, and from two points in the line joining their centres, perpendiculars are drawn intersecting the outer circle in the points B, C, and the inner in D, E; shew that AB : AG = AD : AE. 93. Find a straight line such that the perpendiculars on it from three given points shall be in given ratios to each other. 94. Divide a given arc of a circle into two parts, so that the chords of these parts shall be to each other in a given ratio. 95. CAB, GEB are two triangles having a common angle GBA, and the sides opposite to it CA, GE equal. If BA be produced to D, and ED be taken a third proportional to BA, AC, then the triangle BDG is similar to the triangle BAG. 96. One side of a triangle is given, and also its points of inter- section with the bisector of the opposite angle and the perpendicular from the opposite vertex ; construct the triangle. 97. The diameter of a circle is a mean proportional between the sides of an equilateral triangle and a regular hexagon which are described about the circle. 98. If two regular polygons of the same number of sides be con- structed, one inscribed in and the other described about a given circle, and a third of double the number of sides be inscribed in the circle, this last is a mean proportional between the other two. 99. A triangle DEF is inscribed in a triangle ABC so that DE, DF are parallel to BA, CA respectively; prove that the triangle DEF is to the triangle ABC as the rectangle BD, DC to the square on BC. 32—2 492 BOOK VI. 100. The vertical angle of a triangle is bisected by a straight line which meets the base at D, and is produced to meet the circle ABC at E ; prove that the rectangle contained by CD and CE is equal to the rectangle contained by ^C and CB. 101. A straight line is divided in two given points, determine a third point, such that its distances from the two given points may be proportional to its distances from the ends of the line. 102. AB'iBQ. diameter of a circle, PQ a chord perpendicular to AB, O any point on the circle; OP, OQ meet AB in R and S; prove that the rectangle AR . BS is equal to the rectangle A S . BR. 103. A, B are two fixed points and P a variable point on a circle, AA', BB' are drawn parallel to a fixed line to meet the circle in A', B' : the fixed line meets AB' in D, A'B in D\ AP in E, BP in E' \ prove that BE . B'E' is constant. 104. Prove that, if ABCD be a quadrilateral not inscriptible in a circle, a point E exists such that the rectangle AB, CD is equal to the rectangle AE, BD and the rectangle AD, BC is equal to the rectangle CE, BD. Hence prove the converse of Ptolemy's Theorem. 105. BE and CF are perpendiculars upon AD the bisector of the angle ^ of a triangle ABC. The area of the triangle is equal to either of the rectangles AE, CF or AF, BE. 106. If the exterior angle CAE of a triangle be bisected by the straight line AD which likewise cuts the base produced in D ; then BA . AC, the rectangle of the sides, is less than the rectangle BD . DC by the square on AD. 107. ABC is an isosceles triangle, the side AB being equal to ^C; F is the middle point of BC; on any straight line through A perpen- diculars EG and CE are drawn: shew that the rectangle AC, EF is equal to the sum of the rectangles EC, EG and FA, EG. 108. Describe a circle which shall pass through a given point and touch a given straight line and a given circle. 109. Divide a triangle into two equal parts by a straight line at right angles to one of the sides. 110. If a straight line drawn from the vertex of an isosceles triangle to the base, be produced to meet the circumference of a circle de- scribed about the triangle, the rectangle contained by the whole line so produced and the part of it between the vertex and the base shall be equal to the square on either of the equal sides of the triangle. 111. Two straight lines are drawn from a point A to touch a circle of which the centre is E ; the points of contact are joined by a straight line which cuts EA at H; and on HA as diameter a circle is described : shew that the straight lines drawn through E to touch this circle will meet it on the circumference of the given circle. MISCELLANEOUS EXERCLSES 493 112. Two triangles BAD, BAG have the side BA and the angle A common: moreover the angle ABB is equal to the angle ACB: shew that the rectangle contained by AC, BD is equal to that contained by AB, BC. 113. ABCD is a quadrilateral in a circle; the straight lines CE, DE which bisect the angles ACB, ABB, cut BD and AC d.t F and G respectively.- shew that EF is to EG as ED is to EC. 114. A Bqn&reDEFG is inscribed in a right-angled triangle ABC, so that D, G are in the hypotenuse AB oi the triangle E in AC, and F in CB : prove that the area of the square is equal to the rectangle AD, BG. 115. 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. 116. AB is a diameter, and P any point in the circumference of a circle; AP and BP are joined and produced if necessary; from any point G in AB a straight line is drawn at right angles to AB meeting AP at D and BP at E, and the circumference of the circle at F: shew that CD is a third proportional to CE and CF. 117. If F be a point in the side CB of a right-angled triangle and CD, FE be perpendiculars on the hypotenuse AB, then the sum of the rectangles AD, AE and CD, EF is equal to the square on AC. 118. In the figure of II. 11 shew that four other straight lines besides the given straight line are divided in the required manner. 119. A straight line CD bisects the vertical angle C of a triangle ABC, and cuts the base in D, on AB produced a point E is taken equidistant from C and D : prove that the rectangle AE . BE is equal to the square on DE. 120. If the perpendicular in a right-angled triangle divide the hypotenuse in extreme and mean ratio, the less side is equal to the alternate segment. 121. ABC is a right-angled triangle, CD a perpendicular from the right angle upon AB; shew that if ^C is double of BC, BD is one- fifth of AB. 122. Through a given point draw a chord in a given circle so that it shall be divided at the point in a given ratio. Find the limiting value of the ratio. 123. ABCD is a parallelogram; from B a straight line is drawn cutting the diagonal ^C at F, the side DC at G, and the side AD pro- duced at E : shew that the rectangle EF, EG is equal to the square on BF. 124. Find a point in a side of a triangle from which two straight lines drawn, one to the opposite angle, and the other parallel to the base, shall cut off towards the vertex and towards the base, equal triangles. 494 BOOK VI. 125. On a chord AB of a circle any point P is taken : on AP, PB any two similar and similarly situated triangles APE, PBF are con- structed, and the straight line £i^ joining the vertices of these triangles is produced to meet AB produced in Q. If any circle be described touching AB at P the common chord of these two circles passes through Q. 126. With a point A in the circumference of a circle ABC as centre, a circle PBG is described cutting the former circle at the points B and C; any chord AD of the former meets the common chord BC at E, and the circumference of the other circle at : shew that the angles EPO and DPO are equal for all positions of P. 127. It is required to cut off from one given line a part such that it may be a mean proportional between the remainder and another given line. 128. Construct a square so that its vertices shall lie on four of the sides of a regular pentagon. 129. Shew how to divide a given triangle into any number of equal parts by lines parallel to the base. 130. Divide a given triangle by a straight line drawn in a given direction into two parts whose areas shall be in a given ratio. 131. If E be the intersection of the diagonals of a quadrilateral ABCD, which has the sides AB and CD parallel, then (i) the straight line joining the middle points oi AB and CD passes through E ; (ii) if P be any point in DB produced, the rectangles PB, EC and PD, EA are together equal to the rectangle PE, AC. 132. Two quadrilaterals ABCD, ABEF in which BC, CD, DA are equal to BE, EF, FA respectively, are on the same side of AB. Prove that if the rectangles OA, OD and OB, OC be equal, where O is the point of intersection of the bisectors of the angles, DAF, CBE, then the quadrilaterals are equal in area. 133. Prove that the area of a quadrilateral, whose sides are all of given lengths, is a maximum when two opposite angles of the quadri- lateral are supplementary. 134. Having four given finite straight lines, construct the quadri- lateral of maximum area which can be formed with them taken in a given order for sides. 135. Either of the complements is a mean proportional between the parallelograms about the diameter of a parallelogram. 136. Shew that one of the triangles in the figure of iv. 10 is a mean proportional between the other two. 137. The sides AB and ^C of a triangle ABC are produced to D and E respectively, and DE is joined. A point F is taken in J5C such that BF : FC = AB.AE : AC. AD, prove that, if AF be joined and produced, it will pass through the middle point of DE. MISCELLANEOUS EXERCISES. 495 138. In any triangle ABC, if BB be taken equal to one-fourth of BG, and CE one-fourth of ^C, the straight line drawn from C through the intersection of BE and AD will divide AB into two parts, which are in the ratio of nine to one. 139. Lines drawn from the extremities of the base of a triangle intersecting on the line joining the vertex with the middle point of the base, cut the sides proportionally ; and conversely. 140. D is the middle point of the side BC of a triangle ABC, and P is any point in AD\ through P the straight lines BPE, CPF are drawn meeting the other sides at E, F: shew that EF is parallel to BC\ 141. ABC is a triangle and D, E, F points in the sides BC, GA, AB respectively such that AD, BE, GF meet in ; prove that the ratio ^0 to OD is equal to the sum of the ratios AF to FB and AE to EC. 142. Through any point O within triangle ABC straight lines AO, BO, CO are drawn cutting the opposite sides in D, E, F. EF, FD, DE are produced to meet the sides again inG, H, K. Prove that circles on DG, EH, FK as diameters pass through the same two points. 143. Find a point without two circles such that the tangents drawn therefrom to the circles shall contain equal angles. 144. Prove that the locus of a point, at which two given parts of the same straight line subtend equal angles is two circles. 145. Find on a given straight line AB two points P, Q such that APQB is a harmonic range, and the ratio AP to PQ is equal to a given ratio. 146. P is a point on the same straight line as the harmonic range ABCD ; prove that o^_PB PD AG~ BC'^ DC' 147. A chord AB and a diameter CD of a circle cut at right angles. If P be any other point on the circle, P {A CBD) is a harmonic pencil. 148. If two circles touch one another externally and from the centre of one tangents be drawn to the other, the chord joining the points in which the first circle is cut by the tangents, will be an har- monic mean between the radii. 149. A, B, G, A', B', C are two sets of three points lying on two parallel straight lines ; prove that the intersections of the three pairs of hnes AA', B'C; BB', C'A ; CC, A'B lie on a straight line. 150. P and Q are any two points in AD, BC two opposite sides of a parallelogram ; X and Y are the respective intersections of AQ, BP, and DQ, GP', prove that XY, produced, bisects the parallelogram. 496 BOOK VI. 151. If the sides of a quadrilateral circumscribing a circle touch at the angular points of an inscribed quadrilateral, all the diagonals meet in a point. 152. The square inscribed in a circle is to the square inscribed in the semicircle as 5 to 2. 153. Describe a rectangle which shall be equal to a given square and have its sides in a given ratio. 154. In a given rectangle inscribe a rectangle whose sides shall have to one another a given ratio. 155. Describe a triangle similar to a given triangle, one angular point being on the bounding diameter of a given semicircle, and one of the sides perpendicular to this diameter, and the other two angular points on the arc of the semicircle. 156. If M, N be the points at which the inscribed and an escribed circle touch the side ^C of a triangle ABC and if BM be produced to cut the escribed circle again at P, then NP is a diameter. 157. Shew that in general two circles can be described to cut two lines AB, AC &\, given angles and to pass through a fixed point P. If T, T' be the centres of these circles, then PA bisects the exterior angle Tcr. 158. From a given point outside two given circles which do not meet, draw a straight line such that the portions of it intercepted by the circles shall be in the same ratio as their radii. 159. A\ B', C are the middle points of the sides of the triangle ABC, and P is the orthocentre and PA', PB', PC produced meet the circumscribing circle in A", B'\ C"; prove that the triangle A"B"G" is equal in all respects to the triangle ABC. 160. If through any point in the arc of a quadrant, whose radius is R, two circles be drawn touching the bounding radii of the quadrant, and r, r' be the radii of these circles, then ri-'—BP'. 161. Let two circles touch one another, and a common tangent be drawn to them touching them in P, Q. If a pair of parallel tangents be drawn to the two circles meeting PQ in A, B, and if the line joining their points of contact meet PQ in G, then the ratio of AP to BQ is either equal to or duplicate of the ratio of PC to QC, according as one or another pair of parallel tangents is taken. 162. If three circles touch a straight line one of the circles which touches the three circles passes through their radical centre. 163. Two circles cut in the points A, B; any chord through A cuts the circles again at P and Q ; shew that the locus of the point dividing PQ in a constant ratio is a circle passing through A and B. MISCELLANEOUS EXERCISES 497 164. AB and AC are two fixed straight lines, and is a fixed point. Two circles are drawn through 0, one of which touches AB and ^C at D and E respectively, and the other touches them at F and G respectively. Prove that the circles passing round OBF and OEG touch one another at 0. 165. Prove that the locus of the middle points of the sides of all triangles which have a given orthocentre and are inscribed in a given circle is another circle. 166. From any point P on the circle ABC a pair of tangents PQ, PR are drawn to the circle DEF, and the chord QR is bisected in S. Shew that the locus of /S is a circle; except when the circle ABC passes through the centre of the circle DEF, when the locus of S is a straight line. 167. It is required to describe a circle through two given points A, B and to touch a given circle which touches AB at D. Find C the centre of the circle : draw CA, CB, and draw AO, BO at right angles to CA, CB respectively. Prove that CD produced will cut the circle in a point P, such that the circle APB will touch the given circle at P. 168. O is the radical centre of three circles. Points A, B, C are taken on the radical axes and AB, BC, CA are drawn. Prove that the six points in which these meet the three given circles lie on a circle. If radii vectores are drawn from O to these six points, they meet the three given circles in six points on a circle and its common chords with the three circles meet in pairs on OA, OB, OC. 169. If from a given point S, a perpendicular SY he drawn to the tangent PF at any point P of a circle, centre C, and in the line MP drawn perpendicular to CS, or in MP produced, a point Q be taken so that MQ = SY, Q will lie on one of two fixed straight lines. 170. The diagonals AC, BD, of a quadrilateral inscribed in a circle cut each other in E. Shew that the rectangle AB, BC is to the rectangle AD, DC as BE to ED. 171. The square on the straight line joining the centres of the circumscribed circle and an escribed circle of a triangle is greater than the square on the radius of the circumscribed circle by twice the rectangle contained by the radii of the circles. (See p. 476.) 172. If one triangle can be constructed such that one of two given circles is the circumscribed circle and the other of the given circles is one of its escribed circles, an infinite number of such triangles can be constructed. (See p. 477.) 173. A, B, C are three circles: prove that, if the common tangent of A and C be equal to the sum of the common tangents of A and B and of B and C, the three circles touch a straight line. 174. Three circles A, B, C touch a fourth circle : prove that the ratio of the common tangent of A and B to the common tangent of A and C is independent of the radius of A. 498 BOOK VI. 175. If ABGD be a quadrilateral inscribed in a circle and a second circle touch this at A , and the tangents to it from B, C, D be Bb, Gc, Dd, then the rectangle BD, Cc is equal to the sum of the rectangles BC, Dd and CI), Bb. 176. ABCD is a quadrilateral inscribed in a circle, F the inter- section of the diagonals: shew that the rectangle AF, FD is to the rectangle BF, FG as the square on ^D to the square on BG. 177. The diagonals of a quadrilateral, inscribed in a circle, are to one another in the same ratio as the sums of the rectangles con- tained by the sides which meet their extremities. 178. The sides of a quadrilateral ABGD produced meet in P and Q. Prove that the rectangles PD, DQ ; AD, DG ; PB, BQ ; AB, BG are proportionals. 179. Prove that the locus of a point such that the square on its distance from a fixed point varies as its distance from a fixed straight line is a circle. 180. A quadrilateral circumscribes a circle. Shew that the rectangles contained by perpendiculars from opposite angles upon any tangent are to one another in a constant ratio. INDEX. Acute-angled triangle, 12. Algebra and Geometry connected, 134. Aliquot part, definition of, 327. Alternate angles, definition of, 75. — arc, definition of, 244. Altitude, 351. Ambiguous case, preface, x. — extension of, 367. Angle, acute, 9. — in an arc, definition of, 168. — in a segment, 168. — obtuse, 9. — of intersection of curves, 266. — not limited to 2 rt. z s (ex- tension of application of the term), 221. — plane, 7. — right, 9. — of a sector, 351. Angles, addition of, 8. — of any magnitude, essential to Euclid's method, 411. — all right angles equal, 37. — test of equality of, 8. Anharmonic pencils, having com- mon ray, 439. transversals of, 487. — property of four tangents to circle, 441. — property of four points on circle, 440. — ratio of four points on circle, 442. — ranges, having common point, 438. property of, 436. Antecedent, definition of, 328. Apollonius, circle of, 426. Arc, definition of, 167. Arcs, direction of measurement of, 289. Area, definition of, 10. Arithmetical Progression, 462. Away from, why used, 33, 41. Axiom, use of term, preface, vi. Axioms, definition of, 15. — examples of, 15. — the list of, imperfect, preface, vii. Axis of similitude defined, 454. — of triangles, 444. — radical, 264. of pair of circles touching three circles, 456. Base of a parallelogram, 351. — of a triangle, 351. Brianchon's theorem, 447. Casey's theorem, 469 — 471. Centre, definition of, 13. Cf. 174. — of direct similitude, when infinitely distant, 454. — radical, 265. — of similitude, (a) direct, {h) inverse, 455. condition that chords in- tersecting on radical axis should pass through inverse points with respect to, 457. rectangle under distances to inverse points on two circles from, 453. Centroid, centre of gravity, de- finition of, 103. — of weights, 425. Ceva, theorem of, 422. Ceva's theorem, converse of, 424. Chord, definition of, 167. — of an arc, 167. a circle, 167. contact of circle touching two circles, 451. Chords of two circles intersecting on radical axis, 452. Circle, definition of, 13. — escribed, definition of, 294. — inscribed, definition of, 172, 294. — of Apollonius, 426. — Nine-point, 271. 500 BOOK VI. Circles, coaxial, definition of, 429. — construction for circles to touch three, 458. — equal, have equal radii, 231. — four pairs ofcircles touch three, 459. — orthogonal, 266. — pair of, touching two circles, 455. touching three circles, 455. — touching three circles, radical axis of pair of, 456. Circumference, definition of, 167. Circumscribed circle, definition of, 297. Cf. 172. Coaxial circles, properties of, 268, 465, 479, 481, 482, 483. — triangles, definition of, 444. are compolar, 445. CoUinear points, 351. Commensurable, definition of, 328. Complete quadrilateral, 353. Complement. Complementary angles, definitions of, 45. Complements of parallelograms, &c.. 111. Compolar triangles, definition of, 444. are coaxial, 444. Concentric, definition of, 170. Concurrent lines, definition of, 351. examples of, 53, 71, 95, 103, 422, 424. Configuration of three circles touching a fourth circle, 459. — of four circles touching a fifth circle, 471. Conjugate points, 352. — rays, 352. Consequent, definition of, 328. Contact, external, defined, 199. — internal, definition of, 201. Contain. Arcs contain angles, 168. Contained, a rectangle is, 134. Continuity, principle of, 472, 473, 474. Converse propositions, 27. not necessarily true, 27. Convex figure, definition of, 10. Corollary, =a further inference from facts proved. Correspondence of sides of AS, 256. Corresponding angles, definition of, 75. — vertices, 349. Curve, inverse of a, 460. Curves, angle between, 266. Data, definition of, 61. — often conditioned, 61. Decagon, definition of, 285. Desargues, theorems due to, 444, 445. Diagonals, of a complete quadri- lateral, 353. — definition of, 10. Diameter, definition of, 13. — bisects circle, 175. Direction of measurement of arcs, 289. Distance from a point, definition of, 171. — between centres of inscribed and circumscribed circles of triangle, 476. Distances of point from two points, sum of multiples of squares on, 478. Dodecagon, definition of, 285. DupHcate ratio, definition of, 332. Each to each, why discarded, pi-e- face, ix. Enunciation, joint for Props. 5 and 6, Book II., 145. — joint for Props. 9 and 10, Book II., 153. Equal in all respects, definition of, 13. EquaUty of ratios, definition of, 330. remarks on, 329. — ratio of, 328. Equiangular triangles, property of sides of pair of, 256. Equilateral figure, 11. INDEX. 501 Equimultiples, definition of, 327. Euclid, by his Postulates restrict- ed himself in his use of instru- ments, 21. — in Prop. 4, Book I., assumes that two lines cannot have a common part, 23. Example of the principle of con- tinuity, 474. Exterior angles, definition of, 75. Extreme and mean ratio, 351. Extremes, definition of, 832. Figure, definition of, preface, vi. — circumscribed, 172, 285. — convex, definition of, 10. — equiangular, 11. — equilateral, 11. — inscribed, 172, 285. — plane, 6. — rectilinear, 9. — regular, 11. Figures, impossible, preface, ix. Gauss, what regular polygons can be inscribed in a circle, 320. Geometrical Progression, 462. Gergonne's construction for cir- cles touching three circles, 458. Gnomon, why discarded, preface, vi. Harmonic pencils, having com- mon ray, 435. transversals of, 433. — property of polar, 430. — range, defined, 352. property of, 428, 432. — ranges, having common point, 434. Harmonical Progression, 462. Hexagon, definition of, 285. Homologous, definition of, 331. Hypotenuse, definition of, 11. Impossible figures, preface, ix. Incommensurable, definition of, 328. Inequality, ratio of greater in- equality, 328. — ratio of less inequality, 328. Interior angles, definition of, 75. Inverse curves, definition of, 460. mechanical mode of draw- ing, 468. — of circle through pole, 463. — of circle not through pole, 463. — of straight hne, 462. Inversion, definition of, 460. — angle between two curves not altered by, 461. — distance between two points in terms of inverse points, 464. — coaxial circles invert into co- axial circles, 465. — concentric circles invert into coaxial circles, 465. — limiting points invert into limiting points, 465. — locus of point P, where mP.4 = ?iP^, 464. — one circle may be its own in- verse, 466. — pole of, 460. — Ptolemy's theorem proved by, 467. — radius of, definition of, 460. — two circles may be their own inverses, 466. — two circles, how inverted into equal circles, 466. — two points and their inverses lie on a circle, 461. — three circles may be their own inverses, 466. — three circles, how inverted into equal circles, 466, Isosceles, definition of, 11. Letters, used to represent magni- tudes, 327. Like anharmonic pencils, 353. — anharmonic ranges, 352. Limit, explained, 217. Limiting points of a series of co- axial circles, 429. Line, definition of, 2. — pedal, definition of, 272. — Simson's, 272. — straight, definition of, 4. 502 BOOK 71, Lines, addition of, 6. — cut proportionally (1) extern- ally, (2) internally, 350. Locus, definition of, and example, 39. — of points, distances from which to two fixed points are equal, 39. distances from which to two fixed points are in a fixed ratio, 426, 464. tangents from which to two circles are equal, 264. tangents from which to two circles are in a fixed ratio, 479, 480, 481. Magnitude of angles not limited by Euclid's method, 411. Magnitudes represented by letters, 327. Maximum value, definition of, 55. example of, 190, 192, 193, 195. Maximum and minimum, illus- tration of, 55. — and minimum values occur alternately, 193. Mean proportional, definition of, 332. Means, definition of, 332. Measure, definition of, 327. Menelaus, theorem of, 418. converse of, 420. Method of Superposition, 5. Minimum value, definition of, 55. example of, 55, 57, 193, 195. Multiple, definition of, 327. — definition of m'^^, 327. Nine-point circle, definition of, 271. properties of, 270, 271, 448, 449. Notation of proportion, 330. Obtuse-angled triangle, definition of, 12. Octagon, definition of, 285. Orthocentre, definition of, 95. Orthogonal circles, 266—268. Orthogonally, system of circles cutting two circles, 268. Parallel, definition of, 7. Parallelogram, definition of, 12. Parallelograms about the dia- gonal of a parallelogram. 111, Part, definition of w^^^ 327. Parts of a triangle, definition of, 74. Pascal's theorem, 446. Peaucellier's cell, 468. Pedal line, 272. Pencil, anharmonic, 352, — definition of, 351. — how denoted, 351. — harmonic, 352. — like anharmonic, 353. Pencils, equality of, test of, 440. Pentagon, definition of, 285. — regular, construction for, 309. Perimeter, definition of, 9. Perpendicular, definition of, 9. Perspective, triangles in, 444. Plane, definition of, 6. Point, angular, 9. — definition of, 2. Points, conjugate, defined, 352. — limiting, of a series of coaxial circles, 429. Polar, definition of, 258. Pole, definition of, 258. — of inversion, 460. — of triangles, 444, Polygon, definition of, 11. — the term extended to include triangles and quadrilaterals, 349. Polygons, equiangular, 349. — number of conditions of equi- angularity of, 349. — number of necessary conditions of similarity of, 350. — regular, which can be inscribed in a circle, 320. — similar, 349. Poncelet's theorems, 483. Porism, definition of, 475. — examples of, 475 — 477. INDEX. 503 Porism of inscribed and circum- scribed circles, 477. — condition for, 476. — of set of coaxial circles, 483. Postulate, definition of, 'preface, viii, 4. — I. Two straight lines cannot enclose a space, 4. — II. Two straight lines cannot have a common part, 4, — • III. A straight line may be drawn from any point to any other point, 5. — IV. A finite straight line may be produced to any length, 5. — V. All right angles equal, preface, viii, 9. — "VI. A circle may be described with any centre, and with any radius, preface, ix, 14. — VII. Any straight line drawn through a point within a closed figure must, if produced far enough, intersect the figure in two points at least, 14. — VIII. Any line joining two points one within and the other without a closed figure must intersect the figure in one point at least, 14. — IX. If the sum of the two in- terior angles, which two straight lines make with a given straight line on the same side of it, be not equal to two right angles, the two straight lines are not parallel, 51. Principle of continuity, definition of, 472. examples of, 474. Problem, definition oi, preface, xi. Problems often admit of several solutions, 17, 19, 249, 287. Progression, Arithmetical, Geo- metrical, Harmonical, 462. Proportion, proportionals, defini- tion of, 330. — continued, definition of, 332. — notation of, 330. Proportional, reciprocally, defini- tion of, 380. Proportionally cut, externally, 350. internally, 350. Ptolemy's Theorem, property of chords joining four points on circle, 257. Casey's extension of, 469» 470, 471. Pythagoras, Theorem of, 120. Quadrilateral, complete, 353. harmonic properties of, 423. — convex, 353. — cross, 353. — definition of, 11. — re-entrant, 353. Kadical axis, 264. — centre, 265. Kadius, definition of, 13. — of inversion, 460. — vector, definition of, 460. Eange, anharmonic, 352. — definition of, 351. — how denoted, 351. Eanges, like anharmonic, 352. Eatio of anharmonic range, 352, 443. — compounded, definition of, 332. independent of order of composition, 399. — definition of, 327. — of equality, 328. — of greater inequality, 328. — of less inequality, 328. — of a pencil, 353. — of ratios, 332. Eatios compounded, 332. — equality of, defined, 330. Eays, conjugate, 352. — of a pencil defined, 351. Eeciprocally proportional, 379 Eectangle, definition of, 12, 134. — how denominated, 134. Ee-entrant quadrilaterals, 353. Eelations between a line and a circle, 213. 504 BOOK VI. Belations between two circles, 203. BespectfuUy, how used, preface, ix. Ehomboid, why discarded, pre- face, vi. Ehombus, definition of, 12. Bight angles are equal, 37. Cf. preface, viii. Eight-angled triangle, definition of, 11. Eotation, point about point, 186. — line about point, 186. — plane figure about point, 186. Secant, definition of, 213. Sector, angle of a, 351. — of a circle, 351. Segment, definition of, 167. Segments of a chord, 251. Semicircle, definition of, 167. Sides, corresponding, 256, 349. definition of, 9. Similar arcs, definition of, 168. — figures, similar and similarly situate, 448. — segments, definition of, 168. Similarity of figures, condition of, 403. Similitude, axis of, defined, 454. — centre of, defined, 448. direct, definition of, 448. at infinite distance, 409, 451. inverse, definition of, 448. — centres of, of two circles, 450. of three circles, 454. of circle of nine points and circumscribed circle, 449. Simson's Hue, 272. Solid, definition of, 3. Square, definition of, 12. — ordinary definition, why dis- carded, preface, vi. Straight lines, test of equality of, 6. Superposition, method of, 5. note on, preface, viii. Supplement, supplementary an- gles, definitions of, 45. Surface, definition of, 3, Tangent, definition of, 169. — common to two circles, 219. — limit of, a secant, 217, 245, 253. examples of, 245, 253. Test of equality of angles, 8. geometrical figures, 5. straight lines, 5. Third Proportional, definition of, 332. Theorem of Apollonius, 426. — Brianchon's, 447. — Casey's, 469, 470, 471. — Ceva's, 422. — of Desargues, 444. — of Menelaus, 418. — Pascal's, 446. — Poncelet's, 482, 483. — Ptolemy's, 257. — of Pythagoras, 120. Touch, meaning of, 169. Transversal, definition of, 419. Trapezium, why discarded, pre- face, vi. Triangles, coaxial, 444. — compolar, 444. ♦ — definition of, 11. — equal, note on, 74. — inscribed in circle, 416. — missing case of equality of, preface, x. Triplicate ratio, definition of, 332. Units of length and of area, 135. Unique point, centroid of weights is a, 422. Vertex, 7. — definition of, 9. — of a pencil defined, 351. Weights, centroid of, 425. Within a circle, 169. Without a circle, 169. CAMBRIDGE: PRINTED BY C. J. CLAY, M.A. AND SONS, AT THE UNIVERSITY PRESS. CAMBRIDGE UNIVERSITY PRESS. THE PITT PRESS SERIES. *^* Many of the books in this list can be had in two volumes^ Text and Notes separately. I. GREEK. Aristophanes. Aves— Plutus— Ranae. By W. C. 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